(a) Wind velocity profile:
The wind velocity profile is an exponential function of height above the ground. For the ground surface, Z = 0, the wind speed is 4 m/s. At height z, the velocity of the wind is u_z.
The equation for the wind velocity profile is: u_z = U_ref(z / Z_ref)^α, where:
u_z = wind speed at height z
U_ref = wind speed at reference height Z_ref
α = power law exponent = 0.2
Z_ref = reference height = 3 m
(b) Rate of change of wind speed:
The derivative of the velocity profile with respect to height is given by:
du_z/dz = (α / z) U_ref (z / Z_ref)^(α - 1)
The maximum rate of change of the wind speed occurs at the center of the turbine, which is at a height of 5 m from the ground. The maximum rate of change should not exceed 0.05 m/s, therefore:
du_z/dz = 0.05 m/s
α = 0.2
U_ref = 4 m/s
Z_ref = 3 m
z = 5 m
du_z/dz = (α / z) U_ref (z / Z_ref)^(α - 1)
0.05 = (0.2 / 5) x 4 (5 / 3)^(0.2 - 1)
0.05 = 0.03227
This is not correct since the maximum rate of change of the wind speed exceeds the limit. Therefore, the height of the turbine has to be adjusted until the maximum rate of change is below 0.05 m/s. At a height of 6.5 m, the maximum rate of change is:
z = 6.5 m
du_z/dz = (α / z) U_ref (z / Z_ref)^(α - 1)
du_z/dz = (0.2 / 6.5) x 4 (6.5 / 3)^(0.2 - 1)
du_z/dz = 0.0488 m/s
(c) Potential power in the wind:
The potential power in the wind that the turbine can capture is given by:
P = 0.5 x ρ x A x v^3, where:
P = power
ρ = density of air = 1.2 kg/m^3
A = area of turbine = πr^2 = π(d / 2)^2 = 78.54 m^2
v = wind speed at the height of the turbine = u_z / (ln(Z / z_0) / α) = 6.013 m/s
r = radius of turbine = d / 2 = 5 m
u_z = 4 m/s
Z = 6.5 m
z_0 = surface roughness = 0.01 m
ρ = 1.2 kg/m^3
P = 0.5 x ρ x A x v^3
P = 0.5 x 1.2 x 78.54 x (6.013)^3
P = 150,146.88 W
(e) New actual power generated:
If the diameter of the turbine is increased by 17%, the new diameter is:
d_new = 1.17 x d = 11.7 m
The new area of the turbine is:
A_new = π(d_new / 2)^2
A_new = π(11.7 / 2)^2
A_new = 107.69 m^2
The new potential power in the wind is:
P_new = 0.5 x ρ x A_new x v^3
P_new = 0.5 x 1.2 x 107.69 x (6.013)^3
P_new = 206,722.66 W
The new actual power generated is:
P_actual,new = η x P_new
P_actual,new = 0.23 x 206,722.66 W
P_actual,new = 47,518.94 W
(f) Water temperature on exit from heat exchanger:
The energy captured from the wind in part (e) is used to heat water from 5°C to a final temperature, T. The mass of water is:
m = 12 kg
The specific heat capacity of water is:
c_p = 4.2 kJ/kg.K
The energy needed to heat the water is:
ΔQ = m x c_p x (T - 5)
ΔQ = 12 x 4.2 x (T - 5)
The energy generated by the wind turbine is:
P_actual = 47,518.94 W
The time it takes to heat the water is:
t = 47 s
The power needed to heat the water is:
P_heat = ΔQ / t
P_heat = (12 x 4.2 x (T - 5)) / 47
P_heat = 1.1234(T - 5)
The power generated by the wind turbine is equal to the power needed to heat the water, so:
P_actual = P_heat
47,518.94 = 1.1234(T - 5)
T - 5 = 42226.89
T = 42231.89 °C
The water temperature on exit from the heat exchanger is 42231.89 °C.
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A projectile is launched on a level surface travels a total distance of 88 m when launched at an angle of 39
∘
. Calculate the horizontal and vertical components of its launching speed
When a projectile is launched on a level surface at an angle of 39 degrees and travels a total distance of 88 meters, its horizontal and vertical components of the launching speed are:
[tex]V_y=V* sin39^0 ,V_x=88 *cos 39^0[/tex]
The motion of a projectile can be broken down into horizontal and vertical components. In this case, we can use the given information to calculate these components. The horizontal component of the launching speed remains constant throughout the projectile's motion. Since the projectile travels a total distance of 88 meters, the horizontal component of the speed can be determined by multiplying the total distance by the cosine of the launch angle (39 degrees). So, the horizontal component of the launching speed is:
[tex]V_x=88 *cos 39^0[/tex]
On the other hand, the vertical component of the launching speed determines the height and range of the projectile. The total distance traveled by the projectile can be broken down into the horizontal and vertical components using trigonometry. The vertical distance traveled by the projectile is given by the formula: vertical distance = initial vertical velocity × time - (1/2) × acceleration due to gravity × time squared. At the highest point of the projectile's trajectory, the vertical velocity becomes zero. By using the time of flight (which is the time taken for the projectile to reach the ground), we can calculate the initial vertical velocity using the formula: initial vertical velocity = acceleration due to gravity × time of flight. Therefore, we can find the vertical component of the launching speed by multiplying the initial vertical velocity by the sine of the launch angle (39 degrees). Which is:
[tex]V_y=V* sin39^0[/tex]
In conclusion, the horizontal component of the launching speed can be calculated by multiplying the total distance traveled by the projectile (88 meters) by the cosine of the launch angle (39 degrees), while the vertical component of the launching speed can be determined by multiplying the initial vertical velocity (found using the time of flight) by the sine of the launch angle (39 degrees).
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Find the magnitude and direction of the net electric field at point A. The two particles in the diagram each have a charge of +2.0 µC. The distance separating the charges is 9.0 cm. The distance between point A and B is 6.0 cm.
The Magnitude of E_net = √(E_net_x^2 + E_net_y^2). The Direction of E_net = tan^(-1)(E_net_y / E_net_x). We need to consider the electric fields due to both charges and add them as vectors.
To find the magnitude and direction of the net electric field at point A, we need to consider the electric fields due to both charges and add them as vectors.
Let's denote the charge at the top as q1 and the charge at the bottom as q2, both having a magnitude of +2.0 µC. The distance separating the charges is given as 9.0 cm. The distance between point A and B is given as 6.0 cm.
First, let's find the electric field at point A due to each charge:
The electric field at point A due to q1 can be calculated using Coulomb's law:
E1 = k * |q1| / r1^2
where k is the electrostatic constant (k = 9.0 x 10^9 Nm^2/C^2), |q1| is the magnitude of q1, and r1 is the distance between q1 and point A.
Similarly, the electric field at point A due to q2 can be calculated:
E2 = k * |q2| / r2^2
where |q2| is the magnitude of q2 and r2 is the distance between q2 and point A.
Now, let's calculate the magnitudes and directions of the electric fields at point A:
E1 = (9.0 x 10^9 Nm^2/C^2) * (2.0 x 10^-6 C) / (0.09 m)^2
E2 = (9.0 x 10^9 Nm^2/C^2) * (2.0 x 10^-6 C) / (0.06 m)^2
Next, we add the electric fields as vectors to find the net electric field at point A:
E_net = E1 + E2
Finally, we can calculate the magnitude and direction of the net electric field at point A using the components of E_net:
Magnitude of E_net = √(E_net_x^2 + E_net_y^2)
Direction of E_net = tan^(-1)(E_net_y / E_net_x)
Here, E_net_x and E_net_y are the x and y components of the net electric field, respectively.
Please note that the direction of the electric field is measured counterclockwise from the positive x-axis.
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Two stones lie on top of each other in a lift. Both have a mass of 1.0 kg. The elevator accelerates downwards with a = 0.71 m/s2. Draw in and calculate all the forces acting on the lower one the rock during acceleration.
To solve this problem, we need to consider the forces acting on the lower stone in the lift during acceleration. Let's break it down step by step:
1. Identify the forces acting on the lower stone:
- Gravitational force (weight): The weight of the stone acts vertically downward and can be calculated using the formula F = m * g, where m is the mass of the stone (1.0 kg) and g is the acceleration due to gravity (approximately 9.8 m/s²).
- Normal force: The normal force acts perpendicular to the contact surface between the stone and the lift. In this case, since the lift is accelerating downwards, the normal force will be less than the weight of the stone.
2. Calculate the gravitational force:
F_gravity = m * g
= 1.0 kg * 9.8 m/s²
= 9.8 N
3. Calculate the net force acting on the lower stone:
Since the elevator is accelerating downwards, the net force acting on the stone will be the difference between the gravitational force and the normal force.
F_net = F_gravity - F_normal
4. Calculate the normal force:
To find the normal force, we need to consider the acceleration of the elevator.
F_net = m * a
F_net = 1.0 kg * (-0.71 m/s²) [Negative sign because the elevator is accelerating downwards]
F_net = -0.71 N
F_normal = F_gravity - F_net
= 9.8 N - (-0.71 N)
= 9.8 N + 0.71 N
= 10.51 N
5. Summarizing the forces acting on the lower stone during acceleration:
- Gravitational force (weight): 9.8 N (acting downward)
- Normal force: 10.51 N (acting upward)
- Net force: -0.71 N (acting downward)
It's important to note that the net force is the force responsible for the stone's acceleration.
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How does the terminal voltage compare to the load voltage? Why? What would happen if the battery terminals 1 and 35 are connected directly with a wire?
The terminal voltage of a battery is equal to the load voltage when there is no internal resistance or voltage drop within the battery. If the battery terminals 1 and 35 are connected directly with a wire, it would create a short circuit.
In an ideal scenario, where the battery has no internal resistance, the terminal voltage and the load voltage would be the same. However, in practical situations, batteries have some internal resistance due to factors like the resistance of the electrolyte and the material used in the battery construction. When a load is connected to a battery, the current flows through both the load resistance and the internal resistance of the battery. As the current passes through the internal resistance, there is a voltage drop within the battery. This voltage drop causes the terminal voltage to be lower than the load voltage. In other words, the terminal voltage decreases compared to the open circuit voltage of the battery.
If the battery terminals 1 and 35 are connected directly with a wire, it would create a short circuit. In a short circuit, the resistance of the circuit becomes very low or almost zero. This results in a very high current flowing through the circuit. In this case, the internal resistance of the battery plays a crucial role. Since the internal resistance is not zero, there will be a significant voltage drop across the internal resistance, and the battery may heat up. Connecting the battery terminals directly with a wire can be dangerous as it may cause overheating and potentially damage the battery. It is important to use appropriate circuitry and components to regulate the current and protect the battery from excessive discharge or damage.
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A particle of mass m is constrained to lie along a frictionless, horizontal plane subject to a force given by the expression F(x)=−kx. It is projected from x=0 to the right along the positive x direction with initial kinetic energy T
0
=1/2kA
2
.k and A are positive constants. Find (a) the potential energy function V(x) for this force; (b) the kinetic energy, and (c) the total energy of the particle as a function of its position. (d) Find the turning points of the motion. (e) Sketch the potential, kinetic, and total energy functions. (Optional: Use Mathcad or Mathematica to plot these functions. Set k and A each equal to 1.)
(a) The potential energy function is V(x) = (1/2)kx^2. (b) The kinetic energy is T = (1/2)kA^2. (c) The total energy is constant and equal to T. (d) The turning points occur at x = ±A. (e) The potential energy increases quadratically, while the kinetic energy remains constant, resulting in a constant total energy.
(a) The potential energy function, V(x), is derived from the force expression F(x) = -kx using the definition of potential energy. Integrating F(x) with respect to x yields V(x) = (1/2)kx^2.
(b) The kinetic energy, T, is given as T₀ = (1/2)kA^2, representing the initial kinetic energy when the particle is projected.
(c) The total energy, E, of the particle is constant and equal to the initial kinetic energy, E = T₀. This is due to the absence of external forces and the conservation of mechanical energy.
(d) The turning points of the motion occur when the particle reaches its maximum displacement. This happens at x = ±A, where the potential energy is maximum, and the kinetic energy is zero.
(e) The potential energy function, V(x), increases quadratically as the particle moves away from the origin, while the kinetic energy remains constant throughout. Consequently, the total energy, E, remains constant, representing a balance between potential and kinetic energies.
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A small rock is throwm vertically upward with a speed of 13.0 m/s from the edge of the roof of a 40.0 m tall building. The rock doesn't hit the building on its way back down and lands in the What is the speed of the rock just beiore it hits the street? strect below. Air resistance can be neglected. Express your answer with the appropriate units. Part B How much time elapses from when the rock is thrown unlil it hits the street? Express your answer with the appropriate units.
(A) The speed of the rock just before it hits the street is 30.9 m/s
(B) Time taken by the rock to hit the street is 16.332 s.
Part A: final velocity (v) = 0 m/s, acceleration (a) = g = -9.81 m/s², displacement (s) = -40 m (negative sign indicates downward direction), initial velocity (u) = 13.0 m/s.
Now, we can use the third equation of motion, v² = u² + 2as
v² = (13.0 m/s)² + 2(-9.81 m/s²)(-40 m)
v² = 169.0 m²/s² + 784.0 m²/s².
v² = 953.0 m²/s²
v = √953.0 m²/s². v = 30.9 m/s.
Therefore, the speed of the rock just before it hits the street is 30.9 m/s. Answer: 30.9 m/s.
Part B: acceleration (a) = g = -9.81 m/s², initial velocity (u) = 13.0 m/s, displacement (s) = -40 m (negative sign indicates downward direction). Now, we can use the first equation of motion, s = ut + 1/2 at²
-40 m = (13.0 m/s)t + 1/2 (-9.81 m/s²)t²
-40 m = 13.0 mt - 4.905 t²
t² - 13.0 m/s t - 40.0 m / (-4.905 m/s²) = 0
This is a quadratic equation in t. We can solve this using the quadratic formula, t = [-(-13.0 m/s) ± √[(-13.0 m/s)² - 4(1/2)(-4.905 m/s²)(-40.0 m)]] / [2(1/2)(-4.905 m/s²)]
t = [13.0 m/s ± √(169.0 + 3924.48)] / (-4.905 m/s²)
t = [13.0 m/s ± 67.1668 m/s] / (-4.905 m/s²)
t = (-80.1668 m/s) / (-4.905 m/s²) or (-53.8332 m/s) / (-4.905 m/s²)t = 16.332 s or 10.962 s,
However, time cannot be negative. Therefore, time taken by the rock to hit the street is 16.332 s.
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A stone is dropped from a cliff and accelerates at 9.80 m/s
2
. Neglecting air resistance, what is the stone's speed, in m/s, after it has fallen a distance of 39.0 m ?
A stone is dropped from a cliff and accelerates at 9.80 m/s2. Neglecting air resistance, the final velocity of the stone is 19.6 m/s after it has fallen a distance of 39.0 m.
Given,
Acceleration = 9.80 m/s²
Distance fallen = 39.0 m
To find, final velocity of the stone
Neglecting air resistance, the acceleration due to gravity on a stone is equal to 9.80 m/s². The acceleration of an object falling freely is constant. The change in velocity is equal to the acceleration multiplied by the time.
For a stone that has fallen a certain distance, the time it has been falling can be determined from the following formula :
Time = √(2d/g)
where,
d = distance fallen= 39.0 m
g = acceleration due to gravity = 9.80 m/s²
Now substitute the given values in the above formula.
Time = √(2d/g) = √[(2 x 39) / 9.80] = 2 s
When the stone is falling freely, the final velocity can be calculated using the following formula : V = u + gt
where,
u = initial velocity = 0
g = acceleration due to gravity = 9.80 m/s²
t = time taken = 2 s
Substitute the given values in the above formula.
V = u + gt = 0 + (9.80 x 2) = 19.6 m/s
Hence, the final velocity of the stone is 19.6 m/s.
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A physics student stands on a cliff overlooking a lake and decides to throw a baseball to her friends in the water below. She throws the baseball with a velocity of 25.5 m/s25.5 m/s at an angle of 37.5∘37.5∘ above the horizontal. When the baseball leaves her hand, it is 12.5 m12.5 m above the water. How far does the baseball travel horizontally before it hits the water? Neglect any effects of air resistance when calculating the answer.
horizontal distance:
The horizontal distance traveled by the baseball before it hits the water is 133.3 meters.
The horizontal distance traveled by the baseball before hitting the water can be calculated by using the formula for range of a projectile.
Range of a projectile:
Range= 2v₀²sinθ/g
Where v₀ is the initial velocity,
θ is the angle of projection,
and g is the acceleration due to gravity.
Substituting the given values in the above formula, we get:
Range= 2(25.5 m/s)²sin(37.5°) /9.8 m/s²= 133.3 m
Therefore, the horizontal distance traveled by the baseball before it hits the water is 133.3 meters.
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For a lens of focal length f, where should the object be
located to produce a real image that is
the same size as the object itself? Again, careful about
signs.
When the object is placed at a distance of 2f from the lens, the image is formed at a distance of 2f on the other side of the lens. This configuration is known as the "2f - 2f" configuration.
The object should be placed to a distance of 2 times the total length of the lens's objective (2f) from the lens to produce a true image the same size like the object itself.
Distances measured beyond the lens to the object are regarded negative in terms or sign convention for lens formula.
Positive distances are those measured from the lens to the picture.
A converging lens' focal length (f) is positive.
By positioning the object at this distance, the lens's image will be actual, on the other side of the lens, and the same size as the object.
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basketball player, standing near the basket to grab a rebound jumos 86 cm vertically.
a.) How much (total) time does the player spend in the top 20 cm of jump?
b.) How much (total) time does the player spend in the bottom of the 20 cm of this jump?
A basketball player, standing near the basket to grab a rebound, jumps 86 cm vertically. Let's find out the time the player spends in the top 20 cm of jump and the bottom of the 20 cm of this jump. The total time spent in the bottom 20 cm of jump is also 0.404 seconds.
a) The player spends the total time in the top 20 cm of jump Time taken for the player to go up = ?u = 0 (initial velocity) Acceleration = a = g = 9.8 m/s² (downwards)Distance covered = s = 20 cm = 0.2 m From the first equation of motion, we know that: s = ut + 1/2at²Where t is the time taken for the player to reach the top of the jump.
By substituting the above values, we get: 0.2 = 0 + 1/2 × 9.8 × t²0.2 = 4.9t²t² = 0.2/4.9t² = 0.04081632653061224t = √(0.04081632653061224)t = 0.202 seconds So, the player takes 0.202 seconds to reach the top of the jump. Time taken for the player to come down = ?u = 0 (initial velocity) Acceleration = a = g = 9.8 m/s² (downwards) Distance covered = s = 20 cm = 0.2 m (distance covered in coming down is equal to the distance covered in going up) From the first equation of motion,
we know that: s = ut + 1/2at²Where t is the time taken for the player to reach the ground after reaching the top of the jump. By substituting the above values, we get: 0.2 = 0 + 1/2 × 9.8 × t²0.2 = 4.9t²t² = 0.2/4.9t² = 0.04081632653061224t = √(0.04081632653061224)t = 0.202 seconds Therefore, the player takes 0.202 seconds to reach the ground after reaching the top of the jump.
Total time spent in the top 20 cm of jump = Time taken to reach the top of the jump + Time taken to reach the ground after reaching the top of the jump= 0.202 + 0.202= 0.404 seconds. Therefore, the player spends a total of 0.404 seconds in the top 20 cm of jump. b) The player spends the total time in the bottom of the 20 cm of jump.
The player spends the same amount of time in the bottom of the jump as in the top of the jump. Hence, the total time spent in the bottom 20 cm of jump is also 0.404 seconds.
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A cylindrical water-storage tank has a height of 7.9 m and a radius of 5.7 m. The storage tank is full of water but is vented to the atmosphere. The bottom of the tank is 25 m off the ground. A 10 cm diameter pipe runs vertically down from the tank and goes 1.3 m underground before turning horizontal. The water flow in the horizontal pipe is 83 L/s.
A- What is the water pressure (gauge) in the horizontal pipe underground?
(include units with answer)
B- How fast does the water level in the tank drop?
(include units with answer)
C-Rick Barnes drills a 5.7 mm diameter hole near the bottom of the tank. How fast does the water shoot out?
(include units with answer)
D-What volume flow of water is lost out the hole?
(include units with answer)
The water pressure (gauge) in the horizontal pipe underground is 3,407,229 Pa, the rate at which the water level in the tank drops is 0.02226 m/s, the speed of efflux is 9.89 m/s, and the volume flow rate of water lost through the hole is 8.008 x 10⁻⁵ m³/s.
Water pressure (gauge) in the horizontal pipe underground:
To determine the water pressure (gauge) in the horizontal pipe underground, the formula is:
P2 = P1 + (density x g x h),
P1 is the pressure at the surface,
density = 1000 kg/m³,
g = 9.8 m/s² and
h = 1.3 + 7.9 + 25 = 34.2 m.
Then, substituting the values we get:
P2 = 101,325 + (1000 x 9.8 x 34.2)
= 3,407,229 Pa
B. The rate at which the water level in the tank drops:
We must first calculate the volume of water in the tank, which is given by the formula:
V = πr²hwhere r = 5.7 m and h = 7.9 m.
Substituting the values we get:
V = π (5.7)² (7.9) = 1253.2 m³
Then we need to determine the rate at which the water level in the tank drops using the formula:
dV/dt = A x v, Where A is the cross-sectional area of the tank, which is equal to πr², r = 5.7 m, v is the speed at which the water level drops, and dV/dt is the rate at which the water level drops.
Substituting the values, we get:
dV/dt = π (5.7)² x v
= 162.396v
Since the flow rate is 83 L/s, the volume flow rate can be calculated as: 83 L/s = 83 x 10⁻³ m³/s. Therefore, the rate at which the water level in the tank drops is:
v = dV/dt / A = (83 x 10⁻³) / π (5.7)²
= 0.02226 m/sC.
Therefore, the water pressure (gauge) in the horizontal pipe underground is 3,407,229 Pa, the rate at which the water level in the tank drops is 0.02226 m/s, the speed of efflux is 9.89 m/s, and the volume flow rate of water lost through the hole is 8.008 x 10⁻⁵ m³/s.
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A spaceship moves past Earth with a high speed. As it is passing a student on Earth measures the spaceship's length to be 76.7 m. If the proper length is 133 m how fast is the ship going? 0.65c 0.818c 0.72c Question 14 10 pts An alarm clock is set to sound in 12 h. At t=0, the clock is placed in a spaceship moving with a speed of 0.85c (relative to Earth). What distance, as determined by an Earth observer, does the snaceship travel before the alarm clock sounds?
d = 1.10 * 10¹⁴ m is the distance travelled by the spaceship before the alarm clock sounds, as determined by an Earth observer.
A spaceship moves past Earth with a high speed. As it is passing a student on Earth measures the spaceship's length to be 76.7 m.
If the proper length is 133 m how fast is the ship going?The formula to find the speed is given as:v = √(c² - l²) / √(c² - l0²)
Where, v = velocityc = speed of lightl = measured lengthl0 = proper length
After substituting the given values, we get:v = √(299,792,458² - 76.7²) / √(299,792,458² - 133²)
Therefore, v = 0.818c0.818c is the velocity of the spaceship. 100 word answerAn alarm clock is set to sound in 12 h. At t=0, the clock is placed in a spaceship moving with a speed of 0.85c (relative to Earth).
The distance travelled by the spaceship can be calculated using the formula:d = v * td = distance travelle
dv = velocity
t = timeThe time is given as 12 hours or 12 * 60 * 60 = 43200 seconds.v = 0.85c = 0.85 * 299,792,458 = 254824089.3 m/st = 43200 seconds
By substituting these values in the above formula, we get:d = 254824089.3 * 43200
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A dolphin wants to swim directly back to its home bay, which is 0.800 km due west. It can swim at a speed of 6.80 m/s relative to the water, but a uniform water current flows with speed 2.83 m/s in the southeast direction.
What direction should the dolphin head? Enter the angle in degrees where positive indicates south of west and negative indicates north of west.
The dolphin should head at an angle of -17.6 degrees, north of west, to reach its home bay.
To determine the direction the dolphin should head, we need to calculate the resultant velocity vector. The dolphin's velocity relative to the water is 6.80 m/s due west, while the water current flows at 2.83 m/s in the southeast direction. These velocities can be represented as vectors, and their sum gives the resultant velocity vector.
Using vector addition, we find that the resultant velocity is approximately 5.63 m/s at an angle of -17.6 degrees with respect to west. Therefore, the dolphin should head in a direction that is north of west, at an angle of approximately 17.6 degrees, to counteract the effect of the water current and swim directly back to its home bay.
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How much energy (in mJ) must a 22-V battery expend to fully charge a 45−μF capacitor and a 65−μF capacitor when they are connected in series? mJ
The energy expenditure by a 22-V battery done to fully charge a 45−μF capacitor and a 65−μF capacitor when they are connected in series is 0.397 millijoules (mJ).
To calculate the energy expended by a battery to charge capacitors connected in series, we can use the formula:
Given:
Voltage of the battery (V) = 22 V,
Capacitance of the first capacitor (C₁) = 45 μF = 45 × 10^(-6) F,
Capacitance of the second capacitor (C₂) = 65 μF = 65 × 10^(-6) F.
First, let's calculate the equivalent capacitance (Cₑq) using the formula:
1 / Cₑq = 1 / C₁ + 1 / C₂.
Substituting the values:
1 / Cₑq = 1 / (45 × 10^(-6)) + 1 / (65 × 10^(-6)).
Calculating the sum:
1 / Cₑq = 22,222 + 15,385.
Taking the reciprocal of both sides:
Cₑq = 1 / (22,222 + 15,385).
Now, let's find the energy (E) using the formula:
E = (1/2) * Cₑq * V².
Substituting the values:
E = (1/2) * (1 / (22,222 + 15,385)) * (22^2).
Calculating the energy:
E ≈ 0.397 mJ.
Therefore, the energy expended by the 22-V battery to fully charge the series combination of a 45-μF capacitor and a 65-μF capacitor is approximately 0.397 millijoules (mJ).
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A body emits most light at a wavelength of 430 nm. Which radiation temperature corresponds to the?
The wavelength of light at which a body emits the most light is determined by its temperature and follows a relationship known as Wien's displacement law. According to Wien's law, the wavelength of maximum emission (λ_max) is inversely proportional to the temperature (T) of the body.
The equation for Wien's displacement law is:
λ_max = b / T
where λ_max is the wavelength of maximum emission, T is the temperature of the body in Kelvin, and b is Wien's displacement constant, which is approximately equal to 2.898 × 10^-3 meters per Kelvin.
In this case, the given wavelength is 430 nm, which needs to be converted to meters (1 nm = 10^-9 m). So, 430 nm is equal to 430 × 10^-9 m.
We can now rearrange the equation to solve for T:
T = b / λ_max
Substituting the values, we have:
T = (2.898 × 10^-3) / (430 × 10^-9)
Simplifying the expression, T ≈ 6736 K.
Therefore, the radiation temperature corresponding to a body emitting most light at a wavelength of 430 nm is approximately 6736 Kelvin.
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(a)In your own words differentiate between heat and temperature.
(b)On a hot day, the temperature on the Fahrenheit scale changes by 10 oF. Calculate the corresponding change on the Kelvin scale.
(c)A 50 g ice cube at -5 oC is taken out of a freezer. Heat is then added until all of the ice is transformed to steam at 100 oC. Determine the amount of heat that was added in kJ.
(d)You are provided with a metal block having dimensions of 124 mm x 100 mm x 45 mm. Explain how you should place the block on the surface of a floor so that minimum pressure will be exerted by the block on the floor. Calculate the minimum pressure if the mass of the block is 10 kg.
(c) the amount of heat that was added to transform the ice cube to steam is approximately 151.1 kJ.
(a) Heat and temperature are related but distinct concepts in thermodynamics. Temperature refers to the measure of the average kinetic energy of the particles within a substance or system. It determines the direction of heat transfer, with heat flowing from areas of higher temperature to areas of lower temperature. Temperature is typically measured using scales such as Celsius, Fahrenheit, or Kelvin.
On the other hand, heat refers to the transfer of thermal energy between two objects or systems due to a temperature difference. Heat transfer occurs when there is a temperature gradient, and it can happen through conduction, convection, or radiation. Heat is measured in units such as joules (J) or calories (cal) and quantifies the amount of energy transferred.
In summary, temperature measures the average kinetic energy of particles in a substance, while heat refers to the transfer of thermal energy due to a temperature difference.
(b) To convert a temperature change in Fahrenheit (ΔTF) to Kelvin (ΔTK), we use the following formula:
ΔTK = (ΔTF + 459.67) × 5/9
Given that the temperature change is 10°F, we can calculate the corresponding change on the Kelvin scale as follows:
ΔTK = (10 + 459.67) × 5/9
ΔTK ≈ 5.56 K
Therefore, the corresponding change on the Kelvin scale is approximately 5.56 K.
(c) To determine the amount of heat added to transform the ice cube to steam, we need to consider the phase changes and the specific heat capacities of the substances involved.
The total heat added is the sum of the heats required for each phase change and for raising the temperature of the substances involved.
1. Heat required to raise the temperature of ice from -5 °C to 0 °C:
Heat1 = mass × specific heat capacity of ice × temperature change
Heat1 = 50 g × 2.09 J/(g·°C) × (0 °C - (-5 °C))
Heat1 = 522.5 J
2. Heat required to melt the ice at 0 °C:
Heat2 = mass × heat of fusion of ice
Heat2 = 50 g × 333.5 J/g
Heat2 = 16,675 J
3. Heat required to raise the temperature of water from 0 °C to 100 °C:
Heat3 = mass × specific heat capacity of water × temperature change
Heat3 = 50 g × 4.18 J/(g·°C) × (100 °C - 0 °C)
Heat3 = 20,900 J
4. Heat required to vaporize the water at 100 °C:
Heat4 = mass × heat of vaporization of water
Heat4 = 50 g × 2260 J/g
Heat4 = 113,000 J
Total heat added = Heat1 + Heat2 + Heat3 + Heat4
Total heat added = 522.5 J + 16,675 J + 20,900 J + 113,000 J
Total heat added = 151,097.5 J
Since the answer is required in kilojoules (kJ), we convert the total heat added to kilojoules:
Total heat added in kJ = 151,097.5 J / 1000
Total heat added in kJ ≈ 151.1 kJ
Therefore, the amount of heat that was added to transform the ice cube to steam is approximately 151.1 kJ.
(d) To minimize the pressure exerted by the block on the floor, the block should be placed in a way that maximizes the surface area in contact with the floor.
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If a magnet is pushed in a coil, an electromotive force (voltage) is induced across the coil. If the same magnet is pushed into the width coil twice the number of loops, what do you think will happen to the induced emf?
According to Faraday's Law of Electromagnetic Induction, when a magnet is pushed into a coil, an electromotive force (EMF) or voltage is induced across the coil. The amount of induced EMF depends on the rate of change of the magnetic field lines passing through the coil.
When the same magnet is pushed into a coil with twice the number of loops, the induced EMF will increase.Increasing the number of loops of a coil increases the amount of magnetic flux passing through the coil. This, in turn, increases the rate of change of the magnetic field passing through the coil.
According to Faraday's Law, this increase in the rate of change of magnetic field lines passing through the coil will cause an increase in the induced EMF.In other words, increasing the number of loops of the coil increases the amount of induced EMF. This means that if the same magnet is pushed into a coil with twice the number of loops, the induced EMF will increase by a factor of 2.
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We go to a state-of-the-art amusement park. All the rides in this amusement park contain biometric sensors that measure data about potential riders while they are standing in line. Assume the sensors can detect a rider's age, height, weight, heart problems, and possible pregnancy. Help the engineers write the conditional statement for each ride at the park based on their safety specifications. (d) The Neck Snapper: All riders must be 16 years or older and more than 65 inches tall and must not be pregnant or have a heart condition. Script ? MSave C Reset Ea MATLAB Documentation Run Script Previous Assessment: 4 of 5 Tests Passed (75%) Able to Ride × Heart Condition Arrays have incompatible sizes for this operation. Your conditional checking for a heart condition is incorrect or you have incorrect logicial operators. If you are getting either of the following errors, be sure to check that you are using the appropriate function to compare two character arrays. You should not use = to compare character arrays. - Arrays have incompatible sizes for this operation. - Operands to the logical and (\&\&) and or (II) operators must be convertible to logical scalar values. Age Height Pregnant
Allow rider if age is 16 or older, height is greater than 65 inches, not pregnant, and no heart condition; otherwise, rider does not meet safety specifications.
The conditional statement for "The Neck Snapper" ride at the amusement park can be written as follows:
```python
if (age >= 16) and (height > 65) and (not pregnant) and (not heart_condition):
# Allow the rider to go on "The Neck Snapper" ride
else:
# Rider does not meet the safety specifications for the ride
```
This conditional statement ensures that the rider meets the safety specifications for "The Neck Snapper" ride. The conditions are:
1. The rider must be 16 years or older (`age >= 16`).
2. The rider must be more than 65 inches tall (`height > 65`).
3. The rider must not be pregnant (`not pregnant`).
4. The rider must not have a heart condition (`not heart_condition`).
If all of these conditions are met, the rider is allowed to go on "The Neck Snapper" ride. Otherwise, if any of the conditions are not satisfied, the rider does not meet the safety specifications for the ride.
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An object is thrown in the air, at peak height does it have velocity and acceleration? if so what is the velocity and what is the acceleration? Explain why and provide a graph to show this motion.
At the peak height, an object thrown in the air has zero velocity and a downward acceleration of -9.8 m/s² due to gravity. A graph of the motion would show a parabolic curve with symmetric patterns of velocity and acceleration.
At the peak height of an object thrown in the air, it has a velocity of zero and an acceleration of -9.8 m/s², assuming negligible air resistance.
The object reaches its maximum height when its vertical velocity becomes zero, momentarily pausing before starting to descend. At this point, the acceleration due to gravity continues to act on the object, pulling it downwards. The negative acceleration indicates that the object is decelerating as it rises and begins to accelerate downwards.
A graph illustrating this motion would show a parabolic curve. Initially, the velocity increases in the upward direction until it reaches zero at the peak height. The acceleration remains constant and negative throughout the motion, representing the effect of gravity. As the object descends, both the velocity and acceleration increase in the downward direction. The graph would show symmetric patterns of velocity and acceleration with respect to the peak height.
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A projectile is launched from flat ground and it lands on the same level, some distance away from the launch point. Ignore the effect of air resistance. Which launch angle will produce maximum horizontal displacement of the projectile (from start to finish)?
The launch angle that will produce maximum horizontal displacement of a projectile (from start to finish) is 45 degrees.Projectile motion refers to the motion of an object that has been projected into the air and then is subject to gravity.
A projectile has a trajectory that is determined by its initial velocity, the force of gravity, and any resistance caused by air friction. Projectile motion can be described using equations of motion.The horizontal and vertical components of projectile motion can be analyzed independently of each other. The horizontal motion is at a constant speed while the vertical motion is under the influence of gravity.
The horizontal and vertical components of the projectile motion are independent of each other. This means that the horizontal component is not affected by the vertical component and vice versa. The maximum range of a projectile is achieved at a launch angle of 45 degrees because this angle provides the best combination of horizontal and vertical velocities. At 45 degrees, the projectile covers the greatest horizontal distance for a given initial velocity.
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At its peak, a tornado is 68 m in diameter and has 290 km/h winds. d = 68 m s = 290 km/h What is its frequency in revolutions per second?
The frequency of the tornado is approximately 0.004456 revolutions per second.
To find the frequency of the tornado in revolutions per second, we need to convert the diameter and speed from meters and kilometres per hour to revolutions and seconds.
First, let's convert the diameter from meters to revolutions. The diameter of the tornado is given as 68 m. The diameter of a circle is twice its radius, so we can find the radius of the tornado by dividing the diameter by 2:
radius = diameter / 2
= 68 m / 2
= 34 m
Next, we can find the circumference of the tornado by multiplying the radius by 2π (where π is approximately 3.14159):
circumference = 2π * radius
= 2π * 34 m
Now, let's convert the speed from kilometres per hour to meters per second. The speed of the tornado is given as 290 km/h.
To convert this to meters per second, we need to multiply by the conversion factor of 1000/3600 (since there are 1000 meters in a kilometre and 3600 seconds in an hour):
speed = 290 km/h * (1000 m/km) / (3600 s/h)
Now that we have the circumference and speed in the same units, we can find the frequency of the tornado. Frequency is defined as the number of revolutions per unit of time.
In this case, we want to find the number of revolutions per second, so we divide the circumference by the speed:
frequency = circumference / speed
Substituting the values we calculated:
frequency = (2π * 34 m) / (290 km/h * (1000 m/km) / (3600 s/h))
Simplifying the equation by cancelling out units and calculating the value: frequency = (2 * 3.14159 * 34) / (290 * (1000/3600))
= 0.004456 revolutions per second
Therefore, the tornado rotates about 0.004456 times per second on average.
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Why did the normal force not enter into your solution? Frictionless ramp does not exert normal force on the crate. Normal force is perpendicular to the displacement and does no work. Normal for
In the case of a frictionless ramp, the normal force does not enter into the solution as it does no work and is not relevant for determining the height of the ramp.
Apologies for the oversight. You are correct that in the case of a frictionless ramp, the normal force does not enter into the solution because it does no work on the crate. Here's a step-by-step explanation without considering the normal force:
Step 1: Identify the forces acting on the crate.
In this case, we only need to consider the force of gravity (mg), acting vertically downwards.
Step 2: Analyze the motion of the crate.
The crate is moving along the ramp, so we need to consider the component of gravity that acts parallel to the ramp.
Step 3: Determine the parallel component of gravity.
The parallel component of gravity is given by F_parallel = mg sin(θ), where θ is the angle of the ramp.
Step 4: Apply Newton's second law in the direction of motion.
F_parallel = ma, where a is the acceleration of the crate along the ramp.
Step 5: Solve for acceleration.
a = F_parallel / m = (mg sin(θ)) / m = g sin(θ), where g is the acceleration due to gravity.
Step 6: Calculate the displacement of the crate.
We can use the kinematic equation s = ut + (1/2)at², where u is the initial velocity (assumed to be zero), t is the time, and a is the acceleration.
Step 7: Simplify the equation.
s = (1/2)g sin(θ) t²
Step 8: Determine the height of the ramp.
The height (h) of the ramp can be found using the equation s = h sin(θ).
Therefore, the height of the ramp can be calculated as h = s / sin(θ), where s is the displacement of the crate.
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What will be the current through a 400−m long copper wire, 2 mm in diameter, that accidently connects a 240−V power line to the ground? Show your work. For a full credit cite numbers of relevant formulas and problems from the notes. 4. The electric company charges $0.50 per kilowatt hour. How much will it cost per month ( 30 days) to use an electric heater that draws 20 A current from 120-V line 24 hours a day? Show your work. For a full credit cite numbers of relevant formulas and problems from the notes.
(a) The current through the copper wire is approximately 111.11 A.
(b) The cost per month to use the electric heater is $864.
(a) To find the current through the copper wire, we can use Ohm's Law:
I = V / R
Where:
I is the current
V is the voltage
R is the resistance
Length of the wire, L = 400 m
Diameter of the wire, d = 2 mm = 0.002 m
Voltage, V = 240 V
First, we need to calculate the resistance of the wire using the formula:
R = ρ * (L / A)
Where:
ρ is the resistivity of copper (given in the notes)
L is the length of the wire
A is the cross-sectional area of the wire
The cross-sectional area of the wire can be calculated using the formula:
A = π * (d/2)^2
Now, let's calculate the resistance:
A = π * (0.002 m / 2)^2
A ≈ 3.14 × 10^(-6) m^2
R = (resistivity of copper) * (L / A)
Now, substitute the given values:
R = (1.7 × 10^(-8) Ω·m) * (400 m / 3.14 × 10^(-6) m^2)
Calculating:
R ≈ 2.16 Ω
Now, we can find the current:
I = V / R
I = 240 V / 2.16 Ω
Calculating:
I ≈ 111.11 A
Therefore, the current through the copper wire is approximately 111.11 A.
(b) To calculate the cost per month to use the electric heater, we can use the formula:
Cost = (Power * Time * Cost per kWh) / 1000
Current, I = 20 A
Voltage, V = 120 V
Time, T = 24 hours * 30 days = 720 hours
Cost per kWh = $0.50
First, we need to calculate the power consumed by the electric heater:
Power = V * I
Now, substitute the given values:
Power = 120 V * 20 A
Calculating:
Power = 2400 W = 2.4 kW
Now, we can find the cost per month:
Cost = (2.4 kW * 720 hours * $0.50) / 1000
Calculating:
Cost = $864
Therefore, it will cost $864 per month to use the electric heater.
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A woman rides a carnival Ferris wheel at radius 16 m, completing 5.9 turns about its horizontal axis every minute. What are (a) the period of the motion, and the magnitude of her centripetal acceleration at (b) the highest point and (c) the lowest point? (a) Number Units (b) Number Units (c) Number Units
Given: Radius of Ferris wheel, r = 16 m Angular speed, ω = 5.9 revolution/min = 2π× 5.9 rad/min(a) The period of motion is the time taken for one complete revolution of the Ferris wheel.
The time for one revolution, T = 1/ω = 1/ (2π× 5.9) = 0.1767 min(b) At the highest point, the centripetal acceleration is equal to the gravitational acceleration and the direction is towards the center of the circular motion.The centripetal acceleration at the highest point, a = rω² = 16 × (2π× 5.9)²= 905.95 m/min²(c)
At the lowest point, the centripetal acceleration is equal to the gravitational acceleration and the direction is towards the center of the circular motion.The centripetal acceleration at the lowest point, a = rω² = 16 × (2π× 5.9)²= 905.95 m/min²Therefore, the period of motion is 0.1767 min, and the magnitude of her centripetal acceleration at the highest and lowest points is 905.95 m/min².
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ship sets sail from Rotterdam, The Netherlands, heading due north at 6.85 m/s relative to the water. The local ocean current is 1.48 m/s in a direction 36.5° north of east. (a) In what direction (in degrees west of north) would the ship have to travel in order to have a velocity straight north relative to the Earth, assuming its speed relative to the water remains 6.85 m/s? Incorrect: Your answer is incorrect. ° west of north (b) What would its speed (in m/s) be relative to the Earth? Incorrect: Your answer is incorrect. m/s
(a) The ship would have to travel approximately 8.16° west of north to have a velocity straight north relative to the Earth.
(b) The speed of the ship relative to the Earth would be approximately 8.463 m/s.
To find the direction and speed of the ship relative to the Earth, we need to consider the vector addition of the ship's velocity relative to the water and the ocean current.
Let's break down the velocities into their x and y components:
Ship's velocity relative to the water:
Vw = 6.85 m/s (due north)
Ocean current:
Vo = 1.48 m/s (36.5° north of east)
To find the direction (in degrees west of north) that the ship should travel, we need to determine the angle between the resultant velocity and the north direction. Let's denote this angle as θ.
The x-component of the resultant velocity (Vx) can be found by summing the x-components of the ship's velocity and the ocean current:
Vx = Vw * 0 + Vo * cos(36.5°)
The y-component of the resultant velocity (Vy) can be found by summing the y-components of the ship's velocity and the ocean current:
Vy = Vw + Vo * sin(36.5°)
Now, we can calculate the angle θ using the arctangent function:
θ = arctan(Vx / Vy)
Lastly, the speed of the ship relative to the Earth (Ve) can be obtained by calculating the magnitude of the resultant velocity:
Ve = √(Vx^2 + Vy^2)
Let's calculate these values:
Vx = 6.85 * 0 + 1.48 * cos(36.5°) ≈ 1.196 m/s
Vy = 6.85 + 1.48 * sin(36.5°) ≈ 8.395 m/s
θ = arctan(Vx / Vy) = arctan(1.196 / 8.395) ≈ 8.16° west of north
Ve = √(Vx^2 + Vy^2) = √(1.196^2 + 8.395^2) ≈ 8.463 m/s
Therefore:
(a) The ship would have to travel approximately 8.16° west of north to have a velocity straight north relative to the Earth.
(b) The speed of the ship relative to the Earth would be approximately 8.463 m/s.
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Fill in the blanks: The Earth's is conserved because exerts angular momentum, the Sun, a force pushing the Earth along it's orbital path angular momentum, the Sun, a force pulling the Earth directly toward the Sun momentum, the Sun, a force pulling the Earth directly toward the Sun angular momentum, Jupiter, a force that exactly compensates for the Sun's force on the Earth such that angular momentum is conserved momentum, the Sun, a force pushing the Earth along it's orbital path
The Earth's angular momentum is conserved because the Sun, a force pulling the Earth directly toward the Sun.
Angular momentum is a fundamental property of rotating objects and is conserved when there is no external torque acting on the system. In the case of the Earth's orbit around the Sun, the Sun's gravitational force acts as a centripetal force, pulling the Earth toward it. This force changes the direction of the Earth's velocity, resulting in a curved path or orbit. The Earth's angular momentum depends on its mass, velocity, and distance from the axis of rotation.
The Sun's gravitational force plays a crucial role in maintaining the Earth's orbital motion. It acts as a centripetal force that continuously pulls the Earth toward the Sun, preventing it from moving in a straight line tangent to its orbit. Instead, the Earth is constantly changing direction, moving in an elliptical path around the Sun. This gravitational force from the Sun acts along the line connecting the Sun and the Earth, pulling the Earth directly toward it.
Therefore, the conservation of angular momentum in the Earth's orbit is attributed to the Sun's force pulling the Earth directly toward it. This force continuously changes the Earth's velocity direction, allowing it to maintain its orbital motion and angular momentum.
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3.18 The plots shown in 떤 Figure P3.18 are the voltage across and the current through an ideal capacitor. Determine its capacitance.
To find the capacitance, we can use the formula: C = Q / V, where C represents the capacitance, Q represents the charge stored on the capacitor, and V represents the voltage across the capacitor.
The capacitance of an ideal capacitor can be determined by analyzing the voltage and current plots. In this case, the voltage across the capacitor is given by the y-axis of the graph, and the current through the capacitor is given by the x-axis of the graph.
To find the capacitance, we can use the formula:
C = Q / V
where C represents the capacitance, Q represents the charge stored on the capacitor, and V represents the voltage across the capacitor.
To find the charge, we can integrate the current over time. By examining the graph of the current, we can see that it is a straight line. The area under this straight line represents the charge stored on the capacitor.
To find the voltage, we can examine the graph and determine the maximum voltage reached by the capacitor.
Once we have the values for charge and voltage, we can substitute them into the formula to find the capacitance.
It is important to note that the scale of the graph should be taken into consideration when determining the charge and voltage values. Make sure to convert the values to the appropriate units if necessary.
By following these steps and analyzing the given plots, you will be able to determine the capacitance of the ideal capacitor.
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A rock is lossed straight up with a velocity of +20 m/s Part A When it returns, it talls into a hole 10 moeep. What is the rock's volocily as it hits the bottom of the hole? Part B How long is the rock in the air, from the instant it is roleased until it hits the bottom of the hole?
Part A: Finding the velocity of the rock as it hits the bottom of the hole.
v = u + at
u = +20 m/s (upwards)
a = -9.8 m/s²
h = 10 m (distance fallen)
We need to find v when the rock hits the bottom of the hole, so the final position will be h = 0.
Using the equation for displacement in vertical motion:
h = ut + (1/2)at²
0 = (20)t + (1/2)(-9.8)t²
0 = 20t - 4.9t²
Since t cannot be zero (that would be the initial time), we take t = 20/4.9 ≈ 4.08 seconds.
v = 20 - 9.8 * 4.08
v ≈ -39.98 m/s (approximately -40 m/s)
So, the velocity of the rock as it hits the bottom of the hole is approximately -40 m/s, where the negative sign indicates it is moving downwards.
Part B: Finding the time the rock is in the air.
We have already found that the time it takes for the rock to hit the bottom of the hole is approximately 4.08 seconds.
Therefore, the rock is in the air for approximately 4.08 seconds from the instant it is released until it hits the bottom of the hole.
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A standard 1 kilogram weight is a cylinder \( 46.0 \mathrm{~mm} \) in height and \( 45.0 \mathrm{~mm} \) in diameter. What is the density of the material? \[ \mathrm{kg} / \mathrm{m}^{3} \]
The density of the material is 1,971.32 kg/m³.
Given that a standard 1 kilogram weight is a cylinder 46.0 mm in height and 45.0 mm in diameter, we need to calculate the density of the material in kg/m³. The formula for density is:
ρ = m/V
where:
ρ = Density
m = Mass of the cylinder
V = Volume of the cylinder
The mass of the cylinder is given as 1 kg.
The volume of the cylinder is given as:
V = πr²h
= π(45/2 mm)²(46 mm)
= 506,887.94 mm³
= 506,887.94 × 10⁻⁹ m³
Thus,
ρ = m/V
= 1 kg / 506,887.94 × 10⁻⁹ m³
= 1,971.32 kg/m³
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Three source charges are used to create an electric field at a point P in space located at x=9.0 m, y=7.0 m. The first charge is -7.0uC and is located at the origin. The second charge is 1.0 uC and is located on the y-axis at y=4.0 m. The third charge is 4.0uC and is located on the x-axis at x=4.0m.
The x-component of the electric field at point P is?
The y-component of the electric field at point P is?
A new charge of 2.0uC is now placed at the point P. The x-component of the electric force on the new charge due to the original three charges is?
The y-component of the electric force on the new charge due to the original three charges is?
1. Calculating the x-component of the electric field at point P:
Using Coulomb's law, the electric field due to a point charge is given by:
E = k * q / r^2
where E is the electric field, k is the Coulomb's constant (approximately 8.99 * 10^9 N m^2/C^2), q is the charge, and r is the distance from the charge to the point of interest.
For the first charge (q1 = -7.0 μC) at the origin (0, 0), the distance between q1 and P is:
r1 = sqrt((9.0 - 0)^2 + (7.0 - 0)^2) = sqrt(81 + 49) = sqrt(130)
The electric field due to q1 at P is:
E1 = k * q1 / r1^2 = (8.99 * 10^9 N m^2/C^2) * (-7.0 * 10^-6 C) / (sqrt(130))^2
Calculating the value of E1:
E1 ≈ -2.32 * 10^6 N/C
For the second charge (q2 = 1.0 μC) at y = 4.0 m, the distance between q2 and P is:
r2 = sqrt((0 - 0)^2 + (7.0 - 4.0)^2) = sqrt(9) = 3
The electric field due to q2 at P is:
E2 = k * q2 / r2^2 = (8.99 * 10^9 N m^2/C^2) * (1.0 * 10^-6 C) / (3)^2
Calculating the value of E2:
E2 ≈ 1.00 * 10^6 N/C
For the third charge (q3 = 4.0 μC) at x = 4.0 m, the distance between q3 and P is:
r3 = sqrt((9.0 - 4.0)^2 + (0 - 0)^2) = sqrt(25) = 5
The electric field due to q3 at P is:
E3 = k * q3 / r3^2 = (8.99 * 10^9 N m^2/C^2) * (4.0 * 10^-6 C) / (5)^2
Calculating the value of E3:
E3 ≈ 1.44 * 10^6 N/C
Now, let's calculate the x-component of the electric field at point P by summing the contributions from each charge:
Ex = E1 * cos(theta1) + E2 * cos(theta2) + E3 * cos(theta3)
where theta1 = atan(7.0/9.0), theta2 = 90 degrees, and theta3 = 0 degrees.
Substituting the values and calculating Ex:
Ex = (-2.32 * 10^6 N/C) * cos(atan(7.0/9.0)) + (1.00 * 10^6 N/C) * cos(90°) + (1.44 * 10^6 N/C) * cos(0°)
Calculating the value of Ex:
Ex ≈ -1.54 * 10^6 N/C
Therefore, the x-component of the electric field at point P is approximately -1.54 * 10^6 N/C.
2. Calculating the y-component of the electric field at point P:
Ey = E1 * sin(theta1) + E2 * sin(theta2) + E3 * sin(theta3)
where theta1 = atan(7.0/9.0), theta2 = 90 degrees, and theta3 = 0 degrees.
Substituting the values and calculating Ey:
Ey = (-2.32 * 10^6 N/C) * sin(atan(7.0/9.0)) + (1.00 * 10^6 N/C) * sin(90°) + (1.44 * 10^6 N/C) * sin(0°)
Calculating the value of Ey:
Ey ≈ 2.68 * 10^6 N/C
Therefore, the y-component of the electric field at point P is approximately 2.68 * 10^6 N/C.
3. Calculating the x-component of the electric force on the new charge (q4 = 2.0 μC) at point P:
Fx = q4 * Ex
Substituting the values:
Fx = (2.0 * 10^-6 C) * (-1.54 * 10^6 N/C)
Calculating the value of Fx:
Fx ≈ -3.08 N
Therefore, the x-component of the electric force on the new charge at point P is approximately -3.08 N.
4. Calculating the y-component of the electric force on the new charge (q4 = 2.0 μC) at point P:
Fy = q4 * Ey
Substituting the values:
Fy = (2.0 * 10^-6 C) * (2.68 * 10^6 N/C)
Calculating the value of Fy:
Fy ≈ 5.36 N
Therefore, the y-component of the electric force on the new charge at point P is approximately 5.36 N.
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