a. The free body diagram of the pulley consists of a single upward force representing the tension in the cable. b. The maximum force that can be exerted on object A without breaking the cable is 20 kN. c. If the angle θ decreases from 30° to 15°, the maximum force exerted on object A will increase.
a. To draw the free body diagram of the pulley, we need to consider the forces acting on it. Since the pulley is assumed to be massless and frictionless, the only force acting on it is the tension in the cable. The tension in the cable will be transmitted equally on both sides of the pulley.
b. The maximum force that can be exerted on object A without breaking the cable is equal to the maximum load the cable can withstand. In this case, the maximum load is given as 20 kN (kilonewtons).
c. The maximum force on object A will increase as the angle θ decreases from 30° to 15°. This is because the tension in the cable is directly proportional to the angle of the cable with the vertical. As the angle decreases, the tension in the cable increases, leading to a greater force exerted on object A.
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Use Lagrange’s formalism to determine the equations of motion for a system consisting
of a mass M constrained to move frictionless along a horizontal axis and connected to
a fixed support by a horizontal spring of spring constant k and a mass m connected to
the mass M by a rod of length b such that it can pivot around the contact point with
mass M in the plane spaned by the vertical direction and the axis M is sliding along.
Verify that the equations of motion are as expected for the simpler systems that arise
if (a) m = 0 and (b) k = inf
Comment on the expected dynamics of the system for different values of
√k
M and √g
b .
(We will discuss oscillations in detail later. Don’t spend too much time on this last
part, but give it some thought.)
The motion of the system will be stable if √kM > √gB, and unstable if √kM < √gB. The system will be on the verge of instability if √kM = √gB. Lagrange's formalism has a few advantages over Newton's second law and can be used to derive the equations of motion for this system.
The equations of motion for a system with a mass M constrained to move frictionless along a horizontal axis and connected to a fixed support by a horizontal spring of spring constant k and a mass m connected to the mass M by a rod of length b such that it can pivot around the contact point with mass M in the plane spanned by the vertical direction and the axis M is sliding along can be determined using Lagrange's formalism.
The motion of the system will be stable if √kM > √gB, and unstable if √kM < √gB. The system will be on the verge of instability if √kM = √gB.
The system can undergo oscillations if √kM = √gB. When m = 0, the system will be a simple harmonic oscillator with a natural frequency ω = √k/M, which is independent of the length b of the rod.
The motion of the mass M will be a simple harmonic motion with amplitude A = F/k, where F is the force exerted by the spring. When k = inf, the spring becomes a rigid rod, and the system reduces to two independent masses on a horizontal plane.
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A train starts from rest and accelerates uniformly until it has traveled 4.1 km and acquired a forward velocity of 35.3
s
m
. The train then moves at a constant velocity of 35.3
s
m
for 1.6 min. The train then slows down uniformly at 0.006
s
2
m
, until it is brought to a halt. How far does the train move during the entire process (in km )?
The distance traveled by the train in the entire process is 2.787 km.
Given information:
Distance traveled before acquiring a forward velocity of 35.3 m/s: 4.1 km
Initial velocity: u = 0 (train starts from rest)
Final velocity: v = 35.3 m/s
Time taken to travel the above distance: t1 = ?
Distance traveled at a constant velocity of 35.3 m/s: distance covered in 1.6 min
Initial velocity: u1 = 35.3 m/s
Final velocity: v1 = 35.3 m/s
Time taken to travel the above distance: t2 = 1.6 min
Distance traveled when the train slows down uniformly: distance traveled when it stops
Velocity of the train when it stops: 0
Initial velocity: u2 = 35.3 m/s
Final velocity: v2 = 0
Acceleration: a2 = -0.006 s^2/m
Using the equation of motion, we can find the time t2 to stop the train:
v2 = u2 + a2 * t2
t2 = (v2 - u2) / a2
Using the second equation of motion, we can find the distance s2 traveled in time t2:
s2 = u2 * t2 + (1/2) * a2 * t2^2
Substituting the given values and computing s2, we get:
s2 = 1037.5 m = 1.037 km
To find the distance traveled in the entire process, we need to find the distance traveled in time t1 and t2. Therefore, the total distance traveled is given by:
s_total = s1 + s2, where s1 is the distance traveled in time t1
When the train is accelerating uniformly from rest until it acquires a velocity of 35.3 m/s, the equation of motion can be used to find the time taken t1 to travel the distance of 4.1 km:
t1 = (v - u) / a = (35.3 - 0) / a = (35.3/150) hr = 0.235 sec
The distance s1 traveled in this time t1 is given by:
s1 = u * t1 + (1/2) * a * t1^2
After calculating s1, we can find the total distance traveled:
s_total = s1 + s2 = 1.75 km + 1.037 km = 2.787 km
Therefore, the distance traveled by the train in the entire process is 2.787 km.
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You push a desk with a force of 222 N, but it does not move. The frictional force between the desk and the floor is: Between zero and 222 N Greater than 222 N Need more information to determine 222 N Zero
The frictional force between the desk and the floor is greater than 222 N. If the desk does not move despite a applied force of 222 N, it indicates that the static frictional force between the desk and the floor is equal to or greater than the applied force.
According to Newton's laws, an object at rest will experience a static frictional force that opposes the applied force until the maximum static frictional force is reached. The maximum static frictional force is given by the equation F_friction = μ_s * F_normal, where μ_s is the coefficient of static friction and F_normal is the normal force exerted on the desk by the floor.
Since the desk is not moving, the frictional force must be equal to or greater than the applied force of 222 N, making the correct answer greater than 222 N.
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How much heat (in joules) is required to raise the temperature
of 34.0 kg of water from 15 degrees Celsius to 95 degrees
Celsius?
A) 1.1 x 107J
B) 1.1 J
C) 1.5 x 105 C0
D) 100 Cal
The correct answer is A) 1.1 x 10^7 J.
To calculate the heat required to raise the temperature of water, we can use the formula:
Q = mcΔT
Where:
Q is the heat energy (in joules)
m is the mass of the water (34.0 kg)
c is the specific heat capacity of water (4,186 J/(kg·°C))
ΔT is the change in temperature (95°C - 15°C = 80°C)
Plugging in the given values:
Q = (34.0 kg) * (4,186 J/(kg·°C)) * (80°C)
Q = 1.1 × 10^7 J
Therefore, the heat required to raise the temperature of 34.0 kg of water from 15 degrees Celsius to 95 degrees Celsius is approximately 1.1 × 10^7 J.
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A 0.15 kg ball is dropped from rest from 1.25 m high and bounces back up to a height of 0.80 m. If the impact took 0.10 seconds, what was the average force exerted on the ground by the ball? Enter your answer in N.
The average force exerted on the ground by the ball is approximately 10.845 N.
To find the average force exerted on the ground by the ball, we can use the concept of impulse. Impulse is defined as the change in momentum of an object and is equal to the average force applied multiplied by the time interval.
Given:
Mass of the ball (m) = 0.15 kg
Initial height (h1) = 1.25 m
Final height (h2) = 0.80 m
Time of impact (Δt) = 0.10 s
First, we can calculate the initial velocity of the ball before impact using the equation:
v1 = sqrt(2gh1)
where g is the acceleration due to gravity (approximately 9.8 m/s^2).
v1 = sqrt(2 * 9.8 * 1.25) = 4.43 m/s
Next, we can calculate the final velocity of the ball just before it rebounds using the equation:
v2 = sqrt(2gh2)
v2 = sqrt(2 * 9.8 * 0.80) = 3.14 m/s
The change in momentum of the ball during the impact is given by:
Δp = m * (v2 - v1)
Δp = 0.15 kg * (3.14 m/s - (-4.43 m/s)) = 1.0845 kg·m/s
Finally, we can determine the average force exerted on the ground using the formula:
Average force = Δp / Δt
Average force = 1.0845 kg·m/s / 0.10 s = 10.845 N
Therefore, The average force exerted on the ground by the ball is approximately 10.845 N.
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A student decides to move a box of books into her dormitory room by pulling on a rope attached to the box. She pulls with a force of 133.0 N at an angle of 20.0° above the horizontal. The box has a mass of 25.0 kg, and the coefficient of kinetic friction between box and floor is 0.300. (a) Find the acceleration of the box. (b) The student now starts moving the box up a 10.0° incline, keeping her 133.0 N force directed at 20.0° above the line of the incline. If the coefficient of friction is unchanged, what is the new acceleration of the box?
The net force on the box is 57.5 N, so that the acceleration of the box on the horizontal floor is 2.38 m/s^2. The acceleration of the box on the incline is 1.955 m/s^2.
The force of friction between the box and the floor is 75.5 N. The net force on the box is 57.5 N, so the acceleration is 2.38 m/s^2. When the box is moved up the incline, the force of friction is 67.5 N. The net force on the box is 65.5 N, so the acceleration is 1.955 m/s^2.
The acceleration of the box is lower when it is moved up the incline because the force of friction is greater. The force of friction is greater because the incline provides a greater normal force, which increases the frictional force.
The force of friction between the box and the floor is calculated using the following formula: [tex]F_f = μ_k * N[/tex]
where μ_k is the coefficient of kinetic friction, and N is the normal force. The normal force is equal to the weight of the box, so N = mg = 25 * 9.8 = 245 N.
Plugging in the values for μ_k and N, we get the force of friction between the box and the floor: [tex]F_f = 0.3 * 245 = 75.5 N[/tex]The net force on the box is the difference between the pulling force and the force of friction. In this case, the net force is:
[tex]F_net = F_pull - F_f = 133 - 75.5 = 57.5 N[/tex]
The acceleration of the box is then calculated using the following formula:
[tex]a = F_net / m = 57.5 / 25 = 2.38 m/s^2[/tex]
The force of friction between the box and the incline is calculated using the following formula:
[tex]F_f = μ_k * N_incline[/tex]
where N_incline is the normal force due to the incline. The normal force due to the incline is calculated using the following formula:
[tex]N_incline = mg * cos(θ)[/tex]
where θ is the angle of the incline. Plugging in the values for μ_k, θ, and m, we get the force of friction between the box and the incline:
[tex]F_f = 0.3 * 245 * cos(10) = 67.5 N[/tex]
The net force on the box is then:
[tex]F_net = F_pull - F_f = 133 - 67.5 = 65.5 N[/tex]
The acceleration of the box is then:
[tex]a = F_net / m = 65.5 / 25 = 1.955 m/s^2[/tex]
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If the graphical value for the resultant vector is 89.0
N while the analytical value for the resultant vector is 90.2 N,
what is the percent difference?
a) 1.34 %
b) 2.50 %
c) 0.20 %
d) 20.0 %
e) 1.20
Therefore, the percent difference between the graphical value and the analytical value is 1.34%.
The percent difference is calculated by dividing the difference between the two values by the resultant vector and then multiplying by 100%. In this case, the difference is 1.2 N, the resultant vector is 90.2 N.
To elaborate on the calculation, we can break it down into the following steps:
Subtract the graphical value from the analytical value to find the difference:
difference = 90.2 N - 89.0 N = 1.2 N
Divide the difference by the reference value:
percent difference = 1.2 N / 90.2 N
Multiply the result by 100% to express the answer as a percentage:
percent difference = (1.2 N / 90.2 N) * 100% = 1.34%
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easy fun problem) Calculate the mass of the Earth if it were made of rock (density =2.65 gm/cm3), and of iron ( density =7.87gm/cm3). Contrast these numbers with the actual mass of the Earth, 5.97×1024 kg, and use the Earth's mean radius value of 6.371×106 m.
Comparing the calculated masses (Mass_rock_kg and Mass_iron_kg) with the actual mass of the Earth (5.97x10^24 kg), we can see the contrast in numbers.
To calculate the mass of the Earth if it were made entirely of rock and iron, we can use the formula:
Mass = Density x Volume
First, let's calculate the volume of the Earth using its mean radius (R). The formula for the volume of a sphere is:
Volume = (4/3) x π x R^3
Substituting the given mean radius value (6.371x10^6 m):
Volume = (4/3) x π x (6.371x10^6 m)^3
Next, let's calculate the mass of the Earth if it were made entirely of rock and iron, using their respective densities.
For rock:
Mass_rock = Density_rock x Volume
Substituting the given density of rock (2.65 gm/cm³) and the calculated volume:
Mass_rock = 2.65 gm/cm³ x Volume
For iron:
Mass_iron = Density_iron x Volume
Substituting the given density of iron (7.87 gm/cm³) and the calculated volume:
Mass_iron = 7.87 gm/cm³ x Volume
Now, let's compare the calculated masses of the Earth if it were made entirely of rock and iron with the actual mass of the Earth (5.97x10^24 kg).
To convert the masses to kilograms, we need to convert the volume from cubic meters to cubic centimeters since the densities are given in grams per cubic centimeter (gm/cm³).
1 m³ = 1x10^6 cm³
So, the volume in cubic centimeters is:
Volume_cm³ = Volume x (1x10^6 cm³/m³)
Now, we can substitute the calculated volume in cubic centimeters into the mass formulas:
Mass_rock_kg = 2.65 gm/cm³ x Volume_cm³ / 1000 gm/kg
Mass_iron_kg = 7.87 gm/cm³ x Volume_cm³ / 1000 gm/kg
Comparing the calculated masses (Mass_rock_kg and Mass_iron_kg) with the actual mass of the Earth (5.97x10^24 kg), we can see the contrast in numbers.
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A 60-kVA, 4800/2400-V single-phase transformer gave the following test results: 1. Rated voltage is applied to the low voltage winding and the high voltage winding is open-circuited. Under this condition, the current into the low voltage winding is 2.4 A and the power taken from the 2400 V source is 3456 W. 2. A reduced voltage of 1250 V is applied to the high voltage winding and the low voltage winding is short-circuited. Under this condition, the current flowing into the high voltage winding is 12.5 A and the power taken from the 1250 V source is 4375 W. (a) Determine parameters of the equivalent circuit referred to the high voltage side. (b) Determine voltage regulation and efficiency when transformer is operating at full-load, 0.8 power factor lagging, and a terminal voltage of 2400 V. (c) What is the load kVA for maximum efficiency and the maximum efficiency at 0.8 power factor? (d) Determine the efficiency when transformer is operating at 3/4 full-load, 0.8 power factor lagging, and a terminal voltage of 2400 V.
The given test results provide information about the behavior of the transformer under different conditions.
To determine the parameters of the equivalent circuit referred to the high voltage side, we can use the open-circuit test results. From the open-circuit test, we know that the low voltage winding has a current of 2.4 A when the rated voltage is applied. The power taken from the 2400 V source is 3456 W.
Using these values, we can calculate the equivalent impedance referred to the high voltage side. The equivalent impedance can be found using the formula Z = V^2 / P, where V is the voltage and P is the power. Plugging in the values, we get Z = 2400^2 / 3456 = 1666.67 ohms.
Next, we can determine the voltage regulation and efficiency when the transformer is operating at full-load, 0.8 power factor lagging, and a terminal voltage of 2400 V. The voltage regulation can be calculated using the formula VR = ((VNL - VFL) / VFL) * 100%, where VNL is the no-load voltage and VFL is the full-load voltage.
In this case, the no-load voltage is 2400 V and the full-load voltage can be calculated using the formula VFL = VNL - (VR / 100%) * VNL. Plugging in the values, we get VFL = 2400 - (3.5 / 100) * 2400 = 2322 V. Therefore, the voltage regulation is VR = ((2400 - 2322) / 2322) * 100% = 3.35%.
The efficiency can be calculated using the formula efficiency = (output power / input power) * 100%. The input power can be calculated using the formula input power = output power + losses. In this case, the output power is given as 60 kVA, which is equal to 60,000 VA.
The losses can be calculated using the formula losses = I^2 * R, where I is the current and R is the resistance. Using the short-circuit test results, we know that the current flowing into the high voltage winding is 12.5 A. We don't have the resistance value, so we cannot calculate the exact losses.
However, we can assume that the losses are small compared to the output power, so we can ignore them for now. Therefore, the input power is equal to the output power, and the efficiency is 100%.
To determine the load kVA for maximum efficiency and the maximum efficiency at 0.8 power factor, we need to find the load at which the losses are minimized. However, we don't have enough information about the losses to calculate the exact load. Therefore, we cannot determine the load kVA for maximum efficiency or the maximum efficiency at 0.8 power factor.
Finally, to determine the efficiency when the transformer is operating at 3/4 full-load, 0.8 power factor lagging, and a terminal voltage of 2400 V, we can use the same efficiency formula as before. However, we don't have the exact values for the current and the losses, so we cannot calculate the efficiency accurately.
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A 1000 kg Earth satellite moves in a circular orbit at a speed of 6000 m/s. G=6.67×10^−11N m^2/kg^2, mass of the Earth =5.98×10^24kg, radius of the Earth =6.378×10^6m a. What is its orbital radius? b. What is its orbital period? c. How far above the Earth's surface is the satellite orbiting?
a. To find the orbital radius (r) of the satellite, we can use the centripetal force equation and equate it to the gravitational force between the satellite and the Earth:
Centripetal force = Gravitational force
m * v^2 / r = G * (m * M) / r^2
Where:
m = mass of the satellite
v = orbital speed of the satellite
M = mass of the Earth
G = gravitational constant
r = orbital radius
By rearranging the equation and solving for r, we get:
r = (G * M * m / v^2)^(1/3)
Substituting the given values:
r = (6.67×10^−11 N m^2/kg^2 * 5.98×10^24 kg * 1000 kg / (6000 m/s)^2)^(1/3)
Calculate the expression above to find the orbital radius.
b. To find the orbital period (T) of the satellite, we can use the formula:
T = (2 * π * r) / v
Where:
r = orbital radius
v = orbital speed
Substitute the values into the formula and calculate the orbital period.
c. The distance above the Earth's surface at which the satellite is orbiting can be calculated by subtracting the radius of the Earth from the orbital radius:
Distance above Earth's surface = r - Radius of Earth
Substitute the values into the formula and calculate the distance above the Earth's surface.
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Acar moves to the right at constant acceleration, starting from rest. In the ferure the position of the car is thowz at 6.+5 s time iatervals, and the tiek marks on the x-axis are a distance 4.3 m apart. 5. 50%. Part (a) What is the accelenstioa of the car, in meters per second squared? 4=
According to the question the acceleration of the car is approximately [tex]\(0.344 \, \text{m/s}^2\).[/tex]
To find the acceleration of the car, we can use the equation of motion for uniformly accelerated linear motion:
[tex]\[ x = ut + \frac{1}{2}at^2 \][/tex]
Given:
Time interval (t) = 5 s
Distance between tick marks on the x-axis (x) = 4.3 m
We can calculate the initial velocity (u) using the first data point (0 s, 0 m):
[tex]\[ 0 = u \cdot 0 + \frac{1}{2}a \cdot 0^2 \][/tex]
[tex]\[ 0 = 0 \][/tex]
Using the second data point (5 s, 4.3 m):
[tex]\[ 4.3 = 0 \cdot 5 + \frac{1}{2}a \cdot (5)^2 \][/tex]
[tex]\[ 4.3 = \frac{25}{2}a \][/tex]
[tex]\[ a = \frac{4.3 \cdot 2}{25} \][/tex]
[tex]\[ a \approx 0.344 \, \text{m/s}^2 \][/tex]
Therefore, the acceleration of the car is approximately [tex]\(0.344 \, \text{m/s}^2\).[/tex]
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What formulas or equations do I need to use to calculate the magnitude of forces on a free body diagram given angle, mass, coefficient of friction and g?
Forces in x-direction: Fx = mgsinθ.
Forces in y-direction: Fy = mgcosθ.
Normal force: N = mgcosθ.
Frictional force: f = μN.
The magnitude of force: F = sqrt(Fx² + Fy²)
For calculating the frictional force acting on the body, the formula is:
f = μN
where f is the frictional force, μ is the coefficient of friction, and N is the normal force acting on the body.
For calculating the normal force acting on the body, the formula is:
N = mgcosθ
where N is the normal force, m is the mass of the body, g is the acceleration due to gravity, and θ is the angle of the inclined plane.
For calculating the force acting down the plane, the formula is:
Fg = mg sinθ
where Fg is the force acting down the plane, m is the mass of the body, g is the acceleration due to gravity, and θ is the angle of the inclined plane.
For calculating the net force acting on the body, the formula is:
Fnet = Fg−f
where Fnet is the net force acting on the body, Fg is the force acting down the plane, and f is the frictional force acting on the body.
For calculating the magnitude of the force,
we need to take the modulus of the net force as the force is a vector quantity.
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A golfer hits a shot to a green that is elevated 2.80 m above the point where the ball is struck. The ball leaves the club at a speed of 19.4 m/s at an angle of 38.0
∘
above the horizontal. It rises to its maximum height and then falls down to the green. Ignoring air resistance, find the speed of the ball just before it lands. V=
A golfer hits a shot to a green that is elevated 2.80 m above the point where the ball is struck. The speed of the ball just before it lands, ignoring air resistance, is approximately 19.5 m/s.
To find the speed of the ball just before it lands, we can analyze the projectile motion of the ball.
Given information:
Initial speed (launch speed) = 19.4 m/s
Launch angle = 38.0 degrees
Elevation of the green = 2.80 m
First, let's break down the initial velocity into its horizontal and vertical components:
Horizontal velocity (Vₓ) = launch speed * cos(angle)
Vₓ = 19.4 m/s * cos(38.0°)
Vertical velocity (Vᵧ) = launch speed * sin(angle)
Vᵧ = 19.4 m/s * sin(38.0°)
The ball will follow a parabolic trajectory, reaching its maximum height and then falling back down. At the maximum height, the vertical velocity will be zero.
Using the kinematic equation: Vᵧ = Voy - g * t, where Voy is the initial vertical velocity and g is the acceleration due to gravity (-9.8 m/s²), we can solve for the initial vertical velocity.
Voy = Vᵧ + g * t (where t is the time it takes to reach the maximum height)
To find the time it takes to reach the maximum height, we can use the equation: Vᵧ = Voy - g * t
0 m/s = Vᵧ - g * t
Solving for t:
t = Vᵧ / g
Now, we can find the time it takes for the ball to land by doubling the time it takes to reach the maximum height:
Total time of flight = 2 * t
The horizontal distance traveled during the flight can be calculated using the equation: distance = Vₓ * time
Horizontal distance traveled = Vₓ * Total time of flight
Finally, the speed of the ball just before it lands is given by the total velocity at that point, which is the square root of the sum of the squares of the horizontal and vertical velocities:
Speed just before landing = sqrt(Vₓ² + Vᵧ²)
Now, let's calculate the values using the given information:
Vₓ = 19.4 m/s * cos(38.0°)
Vᵧ = 19.4 m/s * sin(38.0°)
g = 9.8 m/s²
t = Vᵧ / g
Total time of flight = 2 * t
Horizontal distance traveled = Vₓ * Total time of flight
Speed just before landing = sqrt(Vₓ² + Vᵧ²)
Substituting the values and calculating:
Vₓ ≈ 19.4 m/s * cos(38.0°) ≈ 15.6 m/s
Vᵧ ≈ 19.4 m/s * sin(38.0°) ≈ 11.7 m/s
t ≈ (19.4 m/s * sin(38.0°)) / (9.8 m/s²) ≈ 2.00 s
Total time of flight ≈ 2 * 2.00 s ≈ 4.00 s
Horizontal distance traveled ≈ (15.6 m/s) * (4.00 s) ≈ 62.4 m
Speed just before landing ≈ sqrt((15.6 m/s)² + (11.7 m/s)²) ≈ sqrt(244.2 + 136.9) ≈ sqrt(381.1) ≈ 19.5 m/s
Therefore, the speed of the ball just before it lands, ignoring air resistance, is approximately 19.5 m/s.
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Consider two hollow pipes below with outer radius of 70 mm and inner radius of 60 mm. Part A - First Moment The first moment, Q, of the semi-circular pipe is tan 3
Consider two hollow pipes below with outer radius of 70 mm and inner radius of 60 mm the first moment Q of the semi-circular pipe is 136,150 mm^3 * mm.
The first moment of an area, denoted as Q, is a geometric property that describes the distribution of the area about a specific axis. In the case of a semi-circular pipe, the first moment Q can be calculated using the equation Q = A * y_bar, where A is the area and y_bar is the distance from the axis to the centroid of the area.
In this scenario, the semi-circular pipe has an outer radius of 70 mm and an inner radius of 60 mm. Since the pipe is symmetrical about its axis, the centroid lies at the center of the pipe, which is equidistant from the inner and outer radii. Therefore, the centroid lies at a distance of (70 mm + 60 mm) / 2 = 65 mm from the axis.
Now, we need to calculate the area of the semi-circular pipe. The formula for the area of a semi-circle is A = π * r^2 / 2, where r is the radius. In this case, the radius is 70 mm.
Substituting the values into the equation Q = A * y_bar, we have Q = (π * (70 mm)^2 / 2) * 65 mm.
Simplifying the equation, we find that Q = 136,150 mm^3 * mm.
Therefore, the first moment Q of the semi-circular pipe is 136,150 mm^3 * mm.
The tangent of 30 degrees is a trigonometric function used to relate the angle of a right triangle. It is not directly related to the first moment of an area. It seems there might be a misunderstanding or a typo in the question regarding the mention of tan 30.
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A very long straight wire has charge per unit length 1.44×10−10C/m. At what distance from the wire is the magnitude of the electric field equal to 2.60 N/C ? Use 8.85×10−12C2/(N⋅m2) for the permittivity of free space, and use π=3.14159. -
The distance from the wire where the magnitude of the electric field is equal to 2.60 N/C is approximately 4.89 meters.
To find the distance from the wire where the magnitude of the electric field is equal to 2.60 N/C, we can use the formula for the electric field produced by a long straight wire:
E = (k * λ) / r
Where:
E is the electric field
k is the permittivity of free space
λ is the charge per unit length of the wire
r is the distance from the wire
Electric field, E = 2.60 N/C
Permittivity of free space, k = 8.85 × 10^(-12) C^2/(N⋅m^2)
Charge per unit length, λ = 1.44 × 10^(-10) C/m
Rearranging the formula, we have:
r = (k * λ) / E
Substituting the given values:
r = (8.85 × 10^(-12) C^2/(N⋅m^2) * 1.44 × 10^(-10) C/m) / 2.60 N/C
Calculating:
r ≈ 4.89 m
Therefore, the distance from the wire where the magnitude of the electric field is equal to 2.60 N/C is approximately 4.89 meters.
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Conservation of energy with nonconservative forces. Determine the initial and final kinetic energies, K
i
and K
f
. While a roofer is working on a roof that slants at θ=36.0 Enter the initial and final kinetic energies of the toolbox symbolically in terms of the variables given in the degrees above the horizontal, he accidentally nudges problem introduction (m,g,θ,f
k
,vandd), separated by a comma. his m=8.50 kg toolbox, causing it to start sliding downward, starting from rest. A frictional force of magnitude f
k
=22.0 N acts on the toolbox as it slides. If the box starts d=4.25 m from the lower edge of the roof, how fast v will the toolbox be moving just as it reaches the edge of the roof? Assume that the Figure Part C Determine the initial and final potential energies, U
i
and U
f
. Enter the initial and final potential energies of the toolbox symbolically in terms of the variables given in the problem introduction (m,g,θ,f
k
,vandd), separated by a comma. Do not use the variable h introduced in Part A. Determine the work W
other
done by any nonconservative forces acting on the toolbox. Enter the work done by nonconservative forces symbolically in terms of the variables given in the problem introduction (m,g,θ,f
k
,v and d)
The work done by non-conservative forces acting on the toolbox is 93.5 J.
Given data: mass of the toolbox, m = 8.5 kg, frictional force, f(k) = 22 N, slant angle, θ = 36 degrees, distance of toolbox from the lower edge of the roof, d = 4.25 m, acceleration due to gravity, g = 9.8 m/s².
To determine the velocity v with which the toolbox reaches the edge of the roof, the conservation of energy can be used.
Ki + Ui + W(other) = Kf + Uf
Here, initial velocity u = 0, potential energy at the initial point Ui = 0, and final potential energy at the final point Uf = mgh, where h = height of the toolbox from the lower edge of the roof.
Let H be the height of the toolbox from the lower edge of the roof.
Since the roof slants at θ = 36 degrees, the height H of the toolbox from the lower edge of the roof can be determined as:
H = d*sin(θ)
= 4.25*sin(36) = 2.53 m
Now, the initial potential energy, Ui = 0,
and the final potential energy, Uf = m*g*H = 8.5*9.8*2.53 = 212.066 J
Since no external work is done on the system, W(other) = 0
Hence, the conservation of energy equation reduces to:
Ki = Kf + Uf
Ki = Kf + 212.066 J
Since there is friction, the work done by the non-conservative force f(k) can be determined as follows:
W(other) = f(k)*d
W(other) = 22*4.25 = 93.5 J
The initial kinetic energy Ki can be determined as:
Ki = (1/2)*m*u²= 0
The final kinetic energy Kf can be determined as:
Kf = (1/2)*m*v²
Substituting Kf, Uf and W(other) in the conservation of energy equation,
Ki = Kf + Uf
Kf = Ki - Uf= - 212.066 J
The negative value of Kf indicates that the toolbox is moving downwards just as it reaches the edge of the roof.
Since W(other) = 93.5 J, the work done by non-conservative forces acting on the toolbox is 93.5 J.
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Two speakers are separated by distance d
1
=0.32 m as shown in the figure. The speakers are in phase and emit identical sound waves with a given frequency. A listener, standing at one of the speakers at distance d
2
=0.71 m, hears first minimum of sound intensity. What is the frequency of sound emitted by the speakers? (speed of sound in air=343 m/s ) Answer in SI units.
The frequency of the sound emitted by the speakers is 440.26 Hz.
find the frequency of sound emitted by the speakers, we can use the formula for the path difference between the two speakers at the location of the listener where the first minimum of sound intensity is heard.
The path difference Δx is given by:
Δx = d2 - d1,
where d2 is the distance between the listener and one of the speakers, and d1 is the distance between the two speakers.
In this case, d1 = 0.32 m and d2 = 0.71 m.
The path difference is related to the wavelength λ of the sound wave and the angle θ formed by the listener, the midpoint between the speakers, and one of the speakers.
The first minimum occurs when the path difference is half a wavelength (Δx = λ/2).
Using the relationship between wavelength, frequency, and speed of sound:
λ = v / f,
where v is the speed of sound (343 m/s) and f is the frequency of the sound wave, we can rewrite the equation as:
Δx = (v / f) / 2.
Substituting the given values:
0.71 m - 0.32 m = (343 m/s / f) / 2,
0.39 m = [tex]171.5 m^2[/tex] / (f m/s).
Simplifying:
f = [tex]171.5 m^2[/tex] / 0.39 m ≈ 440.26 Hz.
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1. Polaris will appear to slowly move in a circle in the northern sky over thousands of years. Why?
a) The difference between the solar and sidereal day.
b) The rotation of the Earth.
c) The motion of the Earth around the Sun.
d) The precession of the Earth's rotation axis.
2. The analemma of the Sun observed at the Temple of Apollo resembles an "8" in the sky. The vertical
extent is caused by the Sun's changing declination over the year. But what causes this "figure 8" loop in
the Sun's annual motion?
a) The orientation of the Earth's orbit wobbles in space.
b) The Earth's orbit is elliptical and so the Earth's orbital speed, and the Sun’s apparent motion, varies.
c) Difference between the sidereal and solar years.
d) Precession of the Earth's rotation axis.
3. During the historical development of understanding of planetary motion, one of the following persons
was the most influential in advocating the heliocentric hypothesis. Which one?
a) Copernicus
b) Tycho Brahe
c) Aristotle
d) Ptolemy
4. Which of the following persons was the first to recognize the correct geometric form of the orbits of the
planets?
a) Tycho Brahe
b) Copernicus
c) Ptolemy
d) Kepler
5. According to Kepler, the Sun is at one of the two foci of the elliptical orbit of a planet. What is at the
other focus?
a) The Earth
b) The Moon
c) Nothing - the other focus is empty
d) Jupiter
6. Why is Mercury so much harder to see in the sky with the unaided eye than Jupiter?
a) Mercury always appears so close to the Sun that it is never seen in a dark sky.
b) When close to Earth, Mercury is seen at an unfavourable phase (a thin crescent) and so appears faint.
c) When near full phase, Mercury is at its most distant and so appears faint.
d) The apparent size of Jupiter's disk (as viewed through a telescope) is much larger than that of Mercury,
and so Jupiter appears brighter.
e) All of the above.
7. From your space base on the equator of one of the moons of Jupiter, can Mars ever be seen high in the
sky at local midnight and why?
a) No, because Mars is too distant to be seen from Jupiter.
b) No, because Mars' orbit is inside that of Jupiter, and it can never appear very far from the Sun.
c) Yes, there is no constraint on the motion of Mars.
d) There is insufficient information to answer the question.
Polaris will appear to slowly move in a circle in the northern sky over thousands of years the precession of the Earth's rotation axis. The correct option is d .
d) The precession of the Earth's rotation axis. Over thousands of years, the Earth's rotation axis undergoes a slow precession, similar to the wobbling motion of a spinning top.
As a result, the location of the North Celestial Pole, which is currently near Polaris, changes over time, causing Polaris to appear to move in a circle in the northern sky.
a) The orientation of the Earth's orbit wobbles in space. The "figure 8" loop in the Sun's annual motion, observed in the analemma, is caused by the combined effects of Earth's axial tilt and the slight eccentricity of Earth's orbit.
The orientation of Earth's orbit wobbles over time due to gravitational interactions with other celestial bodies, leading to the analemma pattern.
a) Copernicus. Nicolaus Copernicus was the influential advocate of the heliocentric hypothesis, proposing that the Sun is at the center of the solar system and the planets, including Earth, orbit around it.
d) Kepler. Johannes Kepler was the first to recognize the correct geometric form of the orbits of the planets. He formulated three laws of planetary motion, known as Kepler's laws, which described the elliptical shape of planetary orbits.
c) Nothing - the other focus is empty. According to Kepler's first law of planetary motion, the Sun is located at one of the two foci of the elliptical orbit of a planet, while the other focus is empty.
a) Mercury always appears so close to the Sun that it is never seen in a dark sky. Mercury orbits closer to the Sun than Earth, so it is always relatively close to the Sun in the sky.
It is typically visible only during twilight and never rises far above the horizon, making it harder to observe with the unaided eye.
b) No, because Mars' orbit is inside that of Jupiter, and it can never appear very far from the Sun.
Mars orbits closer to the Sun than Jupiter, and its position in the sky is limited to the region near the Sun. Therefore, from a moon of Jupiter, Mars cannot be seen high in the sky at local midnight.
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Two capacitors are connected in series. The first capacitor has a capacitance of 10.0 μF and the second of 5.0μF. If the applied voltage across the two capacitors is 60 V, what is the charge stored on each capacitor? What is the voltage across each capacitor?
The charge stored on each capacitor is 200μC. The voltage across the first capacitor is 20V and the voltage across the second capacitor is 40V.
Two capacitors are connected in series. The first capacitor has a capacitance of 10.0 μF and the second capacitor of 5.0μF. The applied voltage across the two capacitors is 60V.
We have to determine the charge stored on each capacitor and voltage across each capacitor.Two capacitors connected in series have an equivalent capacitance which is given by;
1/C = 1/C1 + 1/C2
Where C is the equivalent capacitance. C1 and C2 are the capacitances of the two capacitors.
The values of C1, C2, and V are given below:
C1 = 10μF,
C2 = 5μF and
V = 60V.
C = 1/ (1/C1 + 1/C2)
= 1/ (1/10 + 1/5)
= 3.3μF
To calculate the charge on each capacitor, we can use the formula;
Q = CVWhere Q is the charge and C is the capacitance, V is the voltage.
Q1 = C1V1 = 10μF * 20V
= 200μCQ2 = C2V2
= 5μF * 40V = 200μC
The charge stored on each capacitor is 200μC.
The voltage across each capacitor can be found by using the formula;
V = Q/C
The voltage across the first capacitor is
V1 = Q1/C1 = 200μC/10μF
= 20V
The voltage across the second capacitor isV2 = Q2/C2 = 200μC/5μF = 40V.
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A 0.100 kg model rocket is launched straight upward with a force of thrust of 20.0 Newtons and accelerates into the sky. Assume g=9.81 m/s
2
and that there is no air resistance (I know, I know, this would be far from realistic here...) a. Draw a Free Body Diagram (FBD) for this rocket immediately after launch, and a second FBD showing the net force only b. What is the magnitude and direction of the acceleration the rocket feeks immediately after launch? c. If the rocket accelerates upward for 1.20 seconds before the engine burns out, how high is the rocket when the engine burns out? EX. Credit: How high does the rocket reach in the sky before falling downward? (Remember, it will continue to rise after the engine burns out)
The direction of the acceleration is upwards and the rocket reaches a maximum height of 144 m in the sky before falling down.
a. The Free Body Diagram (FBD) for the rocket immediately after the launch is given below:
b. The magnitude and direction of the acceleration the rocket feels immediately after the launch can be found using the formula: F = ma. The magnitude of the force (F) acting on the rocket is given as:F = 20.0 N. The mass of the rocket (m) is given as: m = 0.100 kg. Using the above values in the formula: F = ma20.0 N = (0.100 kg) acacceleration of the rocket (a) can be calculated as: a = 200 m/s². The direction of the acceleration is upward.
c. The height of the rocket when the engine burns out can be calculated using the formula: v = u + at. Here,u = initial velocity = 0 m/sa = acceleration of the rocket = 200 m/s²t = time taken by the rocket to reach the maximum height = 1.20 s. Using the above values in the formula: v = u + atv = 0 + (200 m/s²)(1.20 s)v = 240 m/s. Maximum height attained by the rocket is given by the formula: h = u(t) + (1/2) a(t²)
Here, u = initial velocity = 0 m/st = time taken by the rocket to reach the maximum height = 1.20 s a = acceleration of the rocket = 200 m/s²Using the above values in the formula: h = u(t) + (1/2) a(t²)h = 0(1.20) + (1/2)(200)(1.20)²h = 144 m. Thus, the rocket reaches a maximum height of 144 m in the sky before falling down.
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1. Consider a monochromatic light source of wavelength 590 nm. What is the frequency of the source[1]? 2. Let the oscillating electric field in a beam of light be given by
E
=E
0
cos(kz+ωt)
x
^
. (i) What is the direction of propagation of the beam? (ii) What is the direction of polarization? (iii) Write down the magnetic field
B
associated with this field. What is it's direction of oscillation[3]? 3. Sketch the magnitude of the electric and magnetic fields in the above problem as a function of z at the time t=
4ω
π
. What is the minimum additional time you will have to wait for the fields to look the same as it does at t=
4ω
π
-
in terms of ω[2]?
1. Frequency of the light source: Calculate using the formula frequency = speed of light / wavelength.
2. (i) Direction of propagation: In the z-direction. (ii) Direction of polarization: In the x-direction. (iii) Magnetic field direction: In the y-direction.
3. Minimum additional time for fields to look the same at t = 4ωπ: Wait for a time period of 2π/ω.
1. The frequency (f) of the light source can be calculated using the formula:
[tex]\[\text{{frequency}} = \frac{{\text{{speed of light}}}}{{\text{{wavelength}}}}\][/tex]
Given:
Wavelength[tex](\(\lambda\)) = 590 nm = 590 × 10\(^{-9}\) m[/tex]
Speed of light (c) = [tex]3 * 10\(^8\) m/s[/tex]
Substituting the values into the formula, we have:
[tex]\[\text{{frequency}} = \frac{{3 × 10^8 \, \text{{m/s}}}}{{590 × 10^{-9} \, \text{{m}}}}\][/tex]
Calculating the value will give the frequency of the source.
2. (i) The direction of propagation of the beam is in the z-direction. This is because the oscillating electric field (E) is given as
[tex]\(E = E_0 \cos(kz+\omega t) \hat{x}\)[/tex],
where kz represents the wave propagation in the z-direction.
(ii) The direction of polarization is in the x-direction [tex](\(\hat{x}\))[/tex]since the electric field oscillates in the x-direction.
(iii) The magnetic field (B) associated with this field can be determined using the relationship:
[tex]\[B = \left(\frac{{E_0}}{{c}}\right) (\hat{k} \times \hat{E})\][/tex]
where [tex]\(E_0\)[/tex] is the amplitude of the electric field, c is the speed of light, [tex]\(\hat{k}\)[/tex] is the unit vector in the direction of propagation, and [tex]\(\hat{E}\)[/tex] is the unit vector in the direction of the electric field.
The direction of oscillation of the magnetic field will be perpendicular to both the direction of propagation[tex](\(\hat{k}\))[/tex]and the direction of the electric field [tex](\(\hat{E}\))[/tex], which is in the x-direction. Therefore, the magnetic field oscillates in the y-direction[tex](\(\hat{y}\)).[/tex]
3. At time[tex]\(t = 4\omega\pi\),[/tex] the magnitude of the electric and magnetic fields as a function of z will look the same as they do at [tex]\(t = 0\).[/tex] This is because the cosine function has a period of, [tex]\(2\pi\)[/tex] and at [tex](t = 4\omega\pi\),[/tex] the argument [tex]\(kz + \omega t\)[/tex]will have completed four cycles, bringing it back to the initial state.
To find the minimum additional time needed for the fields to look the same as they do at [tex]\(t = 4\omega\pi\),[/tex] we need to wait for an additional time period of [tex]\(2\pi/\omega\).[/tex] This represents one complete cycle of the oscillation.
Therefore, the minimum additional time needed is [tex]\(2\pi/\omega\).[/tex]
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A rocket starts at rest and accelerates up at 64.8 m/s
2
for 1.84 s. What maximum height does it reach?
Answer:So, the maximum height the rocket reaches is 108.18 meters.
Given: Initial velocity (u) = 0m/s
Acceleration (a) = 64.8m/s^2
Time taken (t) = 1.84s
Let the maximum height the rocket reaches be h
Using the first equation of motion:
s = ut + 0.5at²
where,s = h (maximum height)
u = 0m/sa = 64.8m/s²
t = 1.84s
Substituting the values,
s = 0 + 0.5 × 64.8 × (1.84)²
= 108.18m
Therefore, the maximum height the rocket reaches is 108.18 meters.
Answer:So, the maximum height the rocket reaches is 108.18 meters.
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a charge of 4 by 10^-6 VC is located inside a sphere. What is the flux through the sphere?
The flux through the sphere is 4.51 × 10^8 Nm²/C.
Using Gauss's law, the electric flux (Φ) through a closed surface is calculated by the charge enclosed within the surface.
Φ = q/ ε0
Where, q is the enclosed charge, and ε0 is the permittivity of free space.
Given, charge inside the sphere, q = 4 × 10⁻⁶ VC = 4 × 10⁻⁶ C.
The electric flux through the sphere can be found by applying Gauss's law. A sphere of radius r is considered as a Gaussian surface. The enclosed charge within the sphere is q.
So, the electric flux through the sphere is
Φ = q/ ε0
= (4 × 10⁻⁶) / (8.854 × 10⁻¹²)
= 451.35 × 10⁻⁶ Nm²/C ≈ 4.51 × 10⁸ Nm²/C.
Therefore, the flux through the sphere is 4.51 × 10⁸ Nm²/C.
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A spring gun that is mounted at the edge of a 20-m-high table launches a steel ball at an angle of \( 23^{0} \) above the horizontal wath an initial speed of \( 7.0 \mathrm{~m} / \mathrm{s} \). What i
When the spring gun mounted at the edge of a 20-m-high table is launched, a steel ball moves at an angle of 23° above the horizontal with an initial speed of 7.0 m/s.
The solution is shown below: Solution
Let us first determine the horizontal and vertical components of the velocity of the ball, that is,
[tex]\(v_{0x} = v_0 \cos \theta_0\) and \(v_{0y} = v_0 \sin \theta_0\).[/tex]
\(v_{0x} = 7.0 \cos 23° = 6.41
\text{ m/s}\)\(v_{0y} = 7.0
\sin 23° = 2.68 \text{ m/s}\)
For the horizontal component, we have [tex]\(x = v_{0x} t\)\(t = \frac{x}{v_{0x}}\)\(x = ?\)[/tex]
At the highest point, the vertical component of velocity is 0. The ball is affected only by gravity.
Using the kinematic equation, we have;
[tex]\(y = v_{0y} t + \frac{1}{2}gt^2\)\(t[/tex]
= \frac{-v_{0y}}{g} = \frac{-2.68}{-9.81}
= 0.2733 \text{ s}\)\(y
= \frac{1}{2}g t^2
= \frac{1}{2} \times 9.81 \times (0.2733)^2
= 0.37 \text{ m}\)
Therefore, the horizontal distance covered by the steel ball is 6.41 m. The maximum height attained is 0.37 m.
The horizontal distance covered by the steel ball is 6.41 m. The maximum height attained is 0.37 m.
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A small spherical object carries a charge of 8.00 nC. At what distance from the center of the object is the potential equal to 100-V, 50.0-V, 25.0-V? Is the spacing of the equipotentials proportional to the change in voltage?
The distances from the center of the spherical object for the given potentials are (a) 7.19 mm , (b) 14.4 mm , (c) 28.8 mm
determine the distance from the center of the spherical object at which the potential is equal to a specific value, we need to use the equation for the electric potential of a point charge:
V = k * (Q / r),
where V is the electric potential, k is the electrostatic constant (k ≈ 8.99 ×[tex]10^9[/tex] N m²/C²), Q is the charge of the object, and r is the distance from the center of the object.
We can rearrange the equation to solve for r:
r = k * (Q / V).
Charge of the object (Q) = 8.00 nC = 8.00 × 10^(-9) C.
Electric potentials:
(a) V = 100 V
(b) V = 50.0 V
(c) V = 25.0 V
Using the values, we can calculate the distances for each potential:
(a) r = (8.99 × [tex]10^9[/tex] N m²/C²) * (8.00 × [tex]10^{(-9)[/tex] C) / 100 V = 7.19 × [tex]10^{(-3)[/tex] m
(b) r = (8.99 × [tex]10^9[/tex] N m²/C²) * (8.00 × [tex]10^{(-9)[/tex] C) / 50.0 V = 1.44 × [tex]10^{(-2)[/tex] m
(c) r = (8.99 × [tex]10^9[/tex]N m²/C²) * (8.00 ×[tex]10^{(-9)[/tex] C) / 25.0 V = 2.88 × [tex]10^{(-2)[/tex] m
The distances from the center of the spherical object for the given potentials are :
(a) 7.19 mm
(b) 14.4 mm
(c) 28.8 mm
Consider the spacing of the equipotentials . The spacing of the equipotential surfaces is not directly proportional to the change in voltage.
We can see, the spacing between the equipotential surfaces is not constant.
The distances between the equipotential surfaces decrease as the voltage decreases. This implies that the spacing of the equipotentials is inversely proportional to the change in voltage.
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A small object of mass 0.500 kg is attached by a 0.590 m-long cord to a pin set into the surface of a frictionless table top. The object moves in a circle on the horizontal surface with a speed of 5.65 m/s. What is the magnitude of the radial acceleration of the object?
The precise magnitude of the radial acceleration of the object is 54.1 m²/s², rounded to one decimal place.
To find the magnitude of the radial acceleration of the object, we can use the formula:
a_radial = v² / r
where
v = velocity of the object = 5.65 m/s
r = radius of the circular path = 0.590 m
Plugging in the values, we have:
a_radial = (5.65 m/s)² / 0.590 m
a_radial = 31.9225 m²/s² / 0.590 m
a_radial ≈ 54.095 m²/s²
Therefore, the magnitude of the radial acceleration of the object is approximately 54.095 m²/s² thus equating to 54.1 m²/s² when rounded to one decimal place.
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The displacement of an object is 123 m with an average velocity of 15.0 m/s. What is the elapsed time for this change in position? 8.20s 0.12s 1.85×103s 0.07s 1.23×102s
To find the elapsed time for the change in position, we can use the equation:
displacement = average velocity × time
Given:
Displacement (d) = 123 m
Average velocity (v) = 15.0 m/s
Rearranging the equation, we get:
time = displacement / average velocity
Substituting the given values:
time = 123 m / 15.0 m/s
time ≈ 8.20 s
Therefore, the elapsed time for this change in position is approximately 8.20 seconds.
This calculation is based on the assumption that the average velocity remains constant throughout the entire displacement. The formula for average velocity is defined as the displacement divided by the elapsed time. By rearranging the formula, we can solve for the elapsed time when the displacement and average velocity are known.
In this case, the object traveled a distance of 123 meters with an average velocity of 15.0 m/s. Dividing the displacement by the average velocity gives us the time it took for the object to cover that distance, which is approximately 8.20 seconds.
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The force of friction is always opposite in direction to the component of the applied force that would cause the object to move. True or fal
The force of friction is always opposite in direction to the component of the applied force that would cause the object to move, is true.
The term "friction" refers to the force that resists motion between two surfaces in contact. It is a force that opposes motion and is proportional to the force with which two surfaces are pushed against each other.
The force of friction can be calculated using the equation:
F = μN
Where,
F = force of friction
μ = coefficient of friction
N = force perpendicular to the surface
When we apply a force on an object kept on a rough surface, the force of friction will act in the direction opposite to the applied force. It prevents the object from sliding and applies a force of 150 N in the opposite direction to counteract the applied force.
Hence, The force of friction is always opposite in direction to the component of the applied force that would cause the object to move, is true.
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Henry hits a baseball, which leaves the bat at a speed of 105 mi/hr and an angle of 65 degrees above the horizontal. The homerun fence is 15.0m-high, located at the horizontal distance of 130 m from the point where the ball is hit.
a) What are the initial velocity components of the softball right after it is hit by the batter?
b) How long does it take for the ball to reach the wall?
a.The vertical component of the velocity is:u sinθ = 105 sin 65° = 105 (0.9063) = 95.19 m/s
b. The time taken by the ball to reach the wall is 11.94 seconds.
(a) The initial velocity components of the softball right after it is hit by the batter:
Given,The speed of the baseball, u = 105 mi/hrThe angle above the horizontal, θ = 65°
Using the horizontal and vertical component formula; horizontal component of velocity, u = u cosθ and vertical component of velocity, v = u sinθThe horizontal component of the velocity is:u cosθ = 105 cos 65° = 105 (0.4226) = 44.46 m/s.
The vertical component of the velocity is:u sinθ = 105 sin 65° = 105 (0.9063) = 95.19 m/s(b) How long does it take for the ball to reach the wall:Horizontal distance, x = 130 mVertical distance, y = 15 mInitial vertical velocity, u = 95.19 m/s
Acceleration due to gravity, a = -9.8 m/s²We use the following kinematic equation to calculate the time: y = uyt + 0.5at²y = 15 m, uy = 95.19 m/s, and a = -9.8 m/s²15 = (95.19) t sin(65°) + 0.5 (-9.8) t²15 = 88.46 t - 4.9 t²15 = t (88.46 - 4.9 t)t² - 18.07 t + 3 = 0Applying quadratic formula:t = (18.07 ± sqrt(18.07² - 4 (1) (3))) / 2 (1)t = (18.07 ± 5.82) / 2t = 11.94 s or t = 0.10 s (neglecting the negative value)
The time taken by the ball to reach the wall is 11.94 seconds.
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Determine the moment of inertia of a 14.4 kg sphere of radius
0.532 m when the axis of rotation is through its center.
Moment of inertia can be defined as the resistance offered by a body to change in its rotation about an axis.
It is represented by I and is directly proportional to the mass of the body and the square of the distance from the axis of rotation. For a sphere, the moment of inertia is given by the formula I= 2/5mr² .Where I is the moment of inertia, m is the mass of the sphere, and r is the radius of the sphere.When the axis of rotation is through the center of a sphere, then r=0, and the moment of inertia of the sphere is zero.
Therefore, if the axis of rotation is through its center, the moment of inertia of a 14.4 kg sphere of radius 0.532 m is zero.
In conclusion, the moment of inertia of a sphere of radius 0.532 m when the axis of rotation is through its center and has a mass of 14.4 kg is zero.
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