A. it takes 93.3 seconds to accelerate from 22.2 m/s to 27.8 m/s.
B. the distance required to accelerate from 22.2 m/s to 27.8 m/s is 1873.67 meters.
Initial velocity, u = 22.2 m/s
Final velocity, v = 27.8 m/s
Acceleration, a = 0.06t (where t is the time taken to accelerate)
We need to find:
a)
Using the formula:
v = u + at
We can write this as:
t = (v - u) / a = (27.8 - 22.2) / 0.06
t = 93.3 seconds
Therefore, it takes 93.3 seconds to accelerate from 22.2 m/s to 27.8 m/s.
b)
The formula we can use is:
v² - u² = 2as
where s is the distance required to accelerate.
Using the values of v, u, and a from the given data:
v² - u² = 2as
(27.8)² - (22.2)² = 2(0.06)s
224.84 = 0.12s
s = 224.84 / 0.12
s = 1873.67 meters
Therefore, the distance required to accelerate from 22.2 m/s to 27.8 m/s is 1873.67 meters.
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A spark plug in an automobile engine consists of two metal conductors that are separated by a distance of 0.63 mm. When an electric spark jumps between them, the magnitude of the electric field is 5.3×10
7
V/m. What is the magnitude of the potential difference ΔV between the conductors? Number Units
The potential difference ΔV between the conductors is 3.339 × 10^4 volts.
To determine the magnitude of the potential difference ΔV between the conductors, you can use the formula:
ΔV = Ed
Where:
ΔV is the potential difference,
E is the magnitude of the electric field, and
d is the separation distance between the conductors.
Plugging in the given values:
E = 5.3 × 10^7 V/m
d = 0.63 mm = 0.63 × 10^(-3) m
Calculating ΔV:
ΔV = (5.3 × 10^7 V/m) × (0.63 × 10^(-3) m)
= 3.339 × 10^4 V
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When a fast car on a racetrack arrives to a banked curve which at 18 degrees, it slows down to speed of 80 m/s while moving along a track banked at 18 degrees. If the friction coefficient is 0.5, what minimum radius of curve will allow the car to take the turn without sliding down the road?
The minimum radius of curve that will allow the car to take the turn without sliding down the road is approximately 692.6 meters.
To find the minimum radius of the curve that will allow the car to take the turn without sliding down the road, we need to consider the forces acting on the car.
First, let's identify the forces involved:
1. The force of gravity, acting vertically downward.
2. The normal force, acting perpendicular to the surface of the track.
3. The friction force, acting parallel to the surface of the track.
Since the car is not sliding down the road, the friction force must be equal to or greater than the force that tends to make the car slide down the road, which is the component of the gravitational force acting along the track's surface.
The component of the gravitational force along the surface of the track is given by:
[tex]Fg_{parallel}[/tex] = m * g * sin(θ)
where m is the mass of the car, g is the acceleration due to gravity, and θ is the angle of the banked curve (18 degrees).
The friction force can be calculated using the formula:
[tex]F_{friction}[/tex] = μ * N
where μ is the coefficient of friction and N is the normal force.
The normal force can be determined by resolving the gravitational force into its components:
N = m * g * cos(θ)
Combining these equations, we have:
[tex]F_{friction}[/tex] = μ * (m * g * cos(θ))
Since the car is moving in a circle, there is a centripetal force acting towards the center of the curve. This force can be calculated using:
[tex]Fc = (m * v^2) / r[/tex]
where v is the speed of the car (80 m/s) and r is the radius of the curve.
At the minimum radius of the curve, the friction force will be equal to the centripetal force:
μ * (m * g * cos(θ)) = [tex](m * v^2) / r[/tex]
Now we can solve for the minimum radius, r:
r = (μ * [tex]v^2[/tex]) / (g * cos(θ))
Plugging in the given values:
μ = 0.5, v = 80 m/s, g = 9.8 [tex]m/s^2[/tex], and θ = 18 degrees, we can calculate the minimum radius:
[tex]r = (0.5 * 80^2) / (9.8 * cos(18))[/tex]
r ≈ 692.6 meters
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OHANPSE3 5.P.016. A tennis ball nf mass 57 g is initially at rest. While being hit, it experiences an average force of 52 N during a 0.146−s interval. What is its final velocity? m/s
The tennis ball experiences an average force of 52 N over a 0.146-s interval. Using Newton's second law, its final velocity is approximately 133.16 m/s.
To determine the final velocity of the tennis ball, we can use Newton's second law of motion, which states that the force acting on an object is equal to the product of its mass and acceleration: F = m * a.
Rearranging the equation, we can solve for acceleration: a = F / m.
In this case, the force acting on the tennis ball is 52 N, and the mass of the ball is 57 g, which is equivalent to 0.057 kg.
Substituting the values into the equation, we have: a = 52 N / 0.057 kg = 912.28 m/s^2.
We also know that acceleration is the change in velocity divided by the time interval: a = Δv / Δt.
Rearranging the equation, we can solve for the change in velocity: Δv = a * Δt.
Substituting the values of acceleration and time interval into the equation, we have: Δv = 912.28 m/s^2 * 0.146 s = 133.1576 m/s.
Therefore, the final velocity of the tennis ball is approximately 133.16 m/s.
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The speed of a nerve impulse in the human body is about 100 m/s. Estimate the time it takes the nerve impulse to reach your brain when you step on a LEGO piece that is on the floor.
When someone steps on a LEGO piece, the nerve impulse takes about 0.02 seconds to reach the brain.
The speed of nerve impulses in the human body is roughly 100 m/s, which is a rather fast pace. If someone steps on a LEGO piece on the floor.
When someone steps on a LEGO piece, the nerve impulse takes approximately 0.02 seconds (20 milliseconds) to reach the brain. Because the distance traveled by the nerve impulse from the foot to the brain is roughly 2 meters (assuming a typical adult height), we can estimate this.
The formula for time is time = distance/speed.
Using this formula, we can estimate the time it takes a nerve impulse to reach the brain as follows:Time = 2/100
= 0.02 seconds.
Therefore, when someone steps on a LEGO piece, the nerve impulse takes about 0.02 seconds to reach the brain.
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a soccer player was practicing kicking goals, which is kicking the ball such that it leaves the ground with a velocity of a given magnitude (speed) and at a specific angle from the horizontal. They were kicking such that each time the ball got the same magnitude velocity (i.e. speed) but that it had different angles from the horizontal. Odie was coaching the kicker and said that the important thing was to be able to get the ball to land at exactly the same distance from where it was kicked every time. The player could accurately kick the ball at four different angles above the horizontal, each with the same speed: at 30°, 40°, 60°, and 80°. Which two of those angles cause the ball to land the same distance away?
2sin(2theta )
The horizontal and vertical motions are independent so each one gives you information. First find how long it
A) 30° and 80° B) 30° and 60° C) 40° and 80° D) 40° and 60° E) 30° and 40°
The angles those cause the ball to land the same distance is 30° and 60°.
The horizontal distance that the ball travels is given by:
x = v * t * cos(theta)
where:
x is the horizontal distance
v is the initial velocity
t is the time of flight
theta is the angle of the kick
The vertical distance that the ball travels is given by:
y = v * t * sin(theta) - 1/2 * g * t^2
where:
y is the vertical distance
g is the acceleration due to gravity
For the ball to land at the same distance every time, the vertical distance must be the same for each kick. This means that the following equation must be true:
v * t * sin(theta) - 1/2 * g * t^2 = v * t * sin(phi) - 1/2 * g * t^2
where:
phi is the other angle of the kick
Simplifying the equation, we get:
v * sin(theta) = v * sin(phi)
This means that the sine of the two angles must be equal. The only two angles in the problem that satisfy this condition are 30° and 60°.
Therefore, the two angles that cause the ball to land the same distance away are 30° and 60°.
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The speed of an object as a function of position is given by the following equation: v
2
=(7
s
2
1
)x
2
What is the magnitude of the object's acceleration (in m/s
2
) at x=10 m ?
The given equation represents the speed of an object as a function of its position. To find the magnitude of the object's acceleration at x=10 m, we need to differentiate the speed equation with respect to time (t) to obtain the acceleration equation.
Differentiating v^2 = (7s^2)/(1+x^2) with respect to t, we get:
2v(dv/dt) = (7s^2)(2x)(dx/dt)
Since we are interested in the magnitude of acceleration, we can rewrite this as:
a = |(7s^2)(x)(dx/dt)/v|
To find the magnitude of acceleration at x=10 m, we need to know the values of s (which is not provided) and also the velocity of the object at x=10 m. Without these additional details, it is not possible to determine the exact magnitude of the object's acceleration at x=10 m.
Therefore, the magnitude of the object's acceleration cannot be determined with the given information.
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A negative point charge −q is fixed at the point (−a,0,0). A positive charge of the same magnitude is fixed at (+a,0,0). Imagine a circular area drawn on the y-z plane, with its center at the origin; the circle subtends angle β at each of the point charges. Calculate the electric flux through this area, in terms of q,β, and ϵ
0
. Discuss the special case when β=π/2.
The expression for the electric flux of a point charge q at distance r from a circular area of radius R isϵ0qR2r2In the present case, the charge is a combination of a positive and a negative charge, and the circle subtends an angle β at each of the charges.
Let the electric field at any point on the circle be E. Since the electric field due to the two charges is in the same direction at every point on the circle, the flux through the area is simply the sum of the fluxes due to the two charges. Therefore, the flux due to the positive charge through the area is ϵ0qR2r2(1+cosβ2), and that due to the negative charge is ϵ0qR2r2(1−cosβ2)Adding the two fluxes gives the total flux through the area as ϵ0qR2r2(2cos2β2)
The special case is when β=π/2. When this is the case, the circle lies in the x-y plane, and the electric field at every point on the circle is parallel to the x-y plane, so the flux through the area is zero. Therefore,ϕ=0. Answer:ϵ0qR2r2(2cos2β2) and ϕ=0.
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A skateboard starts from rest to roll down hill and covers 125 m in 27 s. Find the
acceleration.
The acceleration of the skateboard rolling down the hill is approximately 1.84 m/s².
We can use the formula for acceleration to find the value:
Acceleration = Change in Velocity / Time
Since the skateboard starts from rest, its initial velocity is 0 m/s. The final velocity can be calculated using the formula:
Final Velocity = Distance / Time
Given that the skateboard covers a distance of 125 m in 27 s, we can substitute these values into the formula:
Final Velocity = 125 m / 27 s ≈ 4.63 m/s
Now we can calculate the acceleration using the formula for acceleration:
Acceleration = (Final Velocity - Initial Velocity) / Time
Acceleration = (4.63 m/s - 0 m/s) / 27 s ≈ 0.171 m/s²
Therefore, the acceleration of the skateboard rolling down the hill is approximately 1.84 m/s².
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Water shoots out from a horizontal pipe which is at a height of 52 cm from the floor, if the horizontal distance travelled by the water before it hits the floor is 100 cm, what is the velocity of the water when it leaves the pipe?
The velocity of the water when it leaves the pipe is 125 cm/s if the horizontal distance travelled by the water before it hits the floor is 100 cm.
When water is released from a horizontal pipe at a height of 52 cm from the ground, the time it takes to reach the ground can be determined using the formula t = \sqrt{(2h/g)},
where h is the height of the pipe from the ground and g is the acceleration due to gravity, which is 9.81 m/s².
At a distance of 100 cm from the pipe, the horizontal distance travelled by the water is given. Using this information, we can calculate the velocity of the water when it leaves the pipe using the formula:
v = d/t, where d is the distance travelled by the water horizontally, which is 100 cm in this case.
t can be found by using the formula t = \sqrt{(2h/g)}.
Therefore,
t = [tex]\sqrt{(2(52)/9.81)}[/tex] = 0.8 seconds
Now, using the formula v = d/t, we can find the velocity of the water:
v = 100/0.8
= 125 cm/s
Therefore, the velocity of the water when it leaves the pipe is 125 cm/s.
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The fundamental frequency of an open organ pipe corresponds to the E below middle C (164.8 Hz on the chromatic musical scale). The third resonance (fifth harmonic) of a closed organ pipe has the same frequency. (Assume that the speed of sound in air is 343 m/s.)
The frequency of the fundamental frequency of an open organ pipe corresponding to the E below middle C (164.8 Hz on the chromatic musical scale) can be determined as follows:
Given that f1 is the frequency of the fundamental frequency of the open organ pipe.
Therefore,f1 = (v/2L), where v is the speed of sound in air and L is the length of the open organ pipe.
Using the given values, we have;164.8 = (343/2L)
Multiplying both sides by 2L, we get;2L × 164.8 = 343
Therefore, L = (343/329.6) = 1.04 m
Now we determine the frequency of the third resonance (fifth harmonic) of a closed organ pipe having the same frequency. Given that f2 is the frequency of the third resonance (fifth harmonic) of a closed organ pipe. Therefore, f2 = 5f1.
Using the calculated value of f1, we have;f2 = 5(164.8) = 824 Hz
Therefore, the third resonance (fifth harmonic) of a closed organ pipe having the same frequency as the fundamental frequency of an open organ pipe corresponding to the E below middle C is 824 Hz.
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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.
The current through the copper wire can be calculated using Ohm's Law. The resistance is determined based on the length and diameter of the wire, and the voltage is divided by the resistance to find the current.
To calculate the current through the copper wire, we can use Ohm's Law, which states that the current (I) is equal to the voltage (V) divided by the resistance (R). In this case, the resistance can be determined using the formula for the resistance of a wire, which is given by R = (ρ * L) / A, where ρ is the resistivity of copper, L is the length of the wire, and A is the cross-sectional area of the wire.
Given:
Length of wire (L) = 400 m
Diameter of wire (d) = 2 mm = 0.002 m
Voltage (V) = 240 V
First, we need to calculate the cross-sectional area (A) of the wire using the formula A = π * (d/2)^2.
Next, we can find the resistivity (ρ) of copper from the notes or reference materials. The resistivity of copper is approximately 1.7 x 10^-8 Ωm.
Using the obtained values of length (L), cross-sectional area (A), and resistivity (ρ), we can calculate the resistance (R) of the wire.
\Finally, we can calculate the current (I) by dividing the voltage (V) by the resistance (R).
By following these steps and plugging in the appropriate values, we can determine the current through the copper wire.
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The Lamborghini Huracan has an initial Part A acceleration of 0.70 g. Its mass, With a driver, is 1510 kg. If an 80 kg passenger rode along, what would the car's acceleration be? Express your answer with the appropriate units.
The acceleration of the car with the passenger would be 9.81 m/s².
Let's assume that the acceleration of the car with the passenger is a. After adding the passenger, the total mass of the car becomes, Total mass = Mass of the car and driver + Mass of the passenger
M = 1510 kg + 80 kg
= 1590 kg
Now, we can use the formula to find the acceleration;
a = (F_net) / MWhere F_net is the net force acting on the car. The formula can also be written as;
a = (F_applied) / M Where F_applied is the force applied to the car. Initially, when there was no passenger, the net force acting on the car was; F_net = m × g × a Where m is the mass of the car and driver. g is the acceleration due to gravity = 9.8 m/s²a is the initial acceleration
= 0.70 g So, the net force was;
F_net = (1510 kg) × (9.8 m/s²) × (0.70 g)
= 9314 N When the passenger is added, the force applied to the car remains the same. But, the net force changes. Let's find the new net force.F_net2 = (m + m2) × g × a Where m2 is the mass of the passenger
.Now, the net force is;F_net2 = 15582 N Now, we can use the formula;
a = (F_net) / Ma
= 9.81 m/s².The acceleration of the car with the passenger would be 9.81 m/s². Hence, the answer is 9.81 m/s².
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object moves along the x axis according to the equation x=3.60t2−2.00t+3.00, where x is in meters and t is in seconds. (a) Determine the average speed between t=1.50 s and t=3.30 s. m/s (b) Determine the instantaneous speed at t=1.50 s. m/s Determine the instantaneous speed at t=3.30 s. m/s (c) Determine the average acceleration between t=1.50 s and t=3.30 s. m/s2 (d) Determine the instantaneous acceleration at t=1.50 s. m/s2 Determine the instantaneous acceleration at t=3.30 s. m/s2 (e) At what time is the object at rest? s
To solve this problem, we will differentiate the given equation with respect to time to find the velocity and acceleration functions.
Given equation: x = 3.60t^2 - 2.00t + 3.00
(a) Average speed between t = 1.50 s and t = 3.30 s:
To find the average speed, we need to calculate the total distance traveled divided by the total time taken. The total distance traveled is the change in position, which is x(3.30) - x(1.50). Divide this by the time interval, 3.30 s - 1.50 s, to get the average speed.
(b) Instantaneous speed at t = 1.50 s and t = 3.30 s:
To find the instantaneous speed, we need to calculate the magnitude of the velocity at those specific times. Take the derivative of the position equation with respect to time, which will give us the velocity function. Evaluate the velocity function at t = 1.50 s and t = 3.30 s to find the instantaneous speeds.
(c) Average acceleration between t = 1.50 s and t = 3.30 s:
To find the average acceleration, we need to calculate the change in velocity divided by the time interval. The change in velocity is the difference between the instantaneous velocities at t = 3.30 s and t = 1.50 s, divided by the time interval.
(d) Instantaneous acceleration at t = 1.50 s and t = 3.30 s:
To find the instantaneous acceleration, we need to take the derivative of the velocity function with respect to time. Evaluate the resulting acceleration function at t = 1.50 s and t = 3.30 s to find the instantaneous accelerations.
(e) To find when the object is at rest, we need to determine the time(s) when the velocity is zero. Set the velocity function equal to zero and solve for t.
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We have a cube that has a mass of 1.5 kg and each of its edges is 10 cm. Determine
its density in g/cm3 and in kg/m3
A cube has a mass of 1.5 kg and each of its edges is 10 cm. Its density is around 1.5 * 10² kg/m³.
To determine the density of the cube, we need to use the formula:
Density = Mass / Volume
First, let's calculate the volume of the cube. Since each of its edges is 10 cm, the volume can be calculated as:
Volume = (Edge length)³
Converting the edge length from centimeters to meters:
Edge length = 10 cm = 0.1 m
Now, we can calculate the volume:
Volume = (0.1 m)³ = 0.001 m³
Next, we substitute the given mass into the formula:
Density = Mass / Volume = 1.5 kg / 0.001 m³ = 1500 kg/m³
Density = 1.5 * 10⁸ g/cm³ * (1 kg / 1000 g) * (1 m / 100 cm)³
Density = 1.5 * 10⁸ / (1000 * 100³) kg/m³
Density = 1.5 * 10⁸ / (10^6) kg/m³
Density = 1.5 * 10² kg/m³
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Consider a portion of a cell membrane that has a thickness of 7.50 nm and 1.1 μm x 1.1 μm in area. A measurement of the potential difference across the inner and outer surfaces of the membrane gives a reading of 91.3 mV. The resistivity of the membrane material is 1.30 x 107 Ω·m. (a) Determine the amount of current that flows through this portion of the membrane. 1.1 What is the resistance of the membrane? Which side of the membrane contributes to its resistance? Which sides of the membrane contribute to the cross sectional area of the membrane?A (b) By what factor does the current change if the side dimensions of the membrane portion is doubled? The other values do not change. O increase by a factor of 2 O increase by a factor of 8 O decrease by a factor of 2 o decrease by a factor of 4 increase by a factor of 4
Therefore, the current flowing through the portion of the membrane is I = 91.3 mV / R .In this case, the current will increase by a factor of 4.
The amount of current flowing through the portion of the cell membrane can be determined using Ohm's Law, which states that current (I) is equal to the potential difference (V) divided by the resistance (R). Given that the potential difference across the membrane is 91.3 mV and the resistivity of the membrane material is 1.30 x 10^7 Ω·m,
We can calculate the resistance of the membrane by rearranging the formula: R = V/I. However, to determine the resistance, we first need to find the current. Since the current is equal to the potential difference divided by the resistance, we rearrange the formula to solve for current: I = V/R. Therefore, the current flowing through the portion of the membrane is I = 91.3 mV / R.
To calculate the resistance of the membrane, we need to consider the dimensions and the resistivity of the material. The resistivity represents the intrinsic property of the material, while the dimensions affect the cross-sectional area through which the current flows. Given that the membrane has a thickness of 7.50 nm and an area of 1.1 μm x 1.1 μm
We can determine the resistance using the formula: R = ρ * (L/A), where ρ is the resistivity, L is the thickness, and A is the cross-sectional area. In this case, R = (1.30 x 10^7 Ω·m) * (7.50 nm / (1.1 μm x 1.1 μm)). The resulting resistance of the membrane is the resistance of the portion of the membrane through which the current flows.
The resistance is determined by the resistivity and the thickness of the membrane. In this case, the thickness of the membrane is given as 7.50 nm, so it is the side of the membrane that contributes to its resistance. As for the cross-sectional area, the membrane has dimensions of 1.1 μm x 1.1 μm, indicating that both sides of the membrane contribute to the cross-sectional area.
Regarding the second part of the question, if the side dimensions of the membrane portion are doubled while keeping the other values constant, the cross-sectional area of the membrane will increase by a factor of 4. This is because the area is proportional to the square of the dimensions.
According to Ohm's Law, if the resistance remains the same while the cross-sectional area increases, the current will also increase by the same factor. Therefore, in this case, the current will increase by a factor of 4.
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A ball is thrown vertically down with an initial speed of v
0
from a height h. What is the speed of the ball when it hits the ground? How long does it take the ball to reach the ground? (b) How do your answers change if instead the ball is thrown vertically upwards with speed v
0
? Show it.
The speed of the ball when it hits the ground is √[v0² + 2gh] and the time taken to reach the ground is t = (√[2h/g]). The speed of the ball when it hits the ground is √[2gh] and the time taken to reach the ground is t = (√[2h/g]).
The problem statement is as follows:
A ball is thrown vertically down with an initial speed of v0 from a height h. What is the speed of the ball when it hits the ground?
How long does it take the ball to reach the ground?
(b) How do your answers change if instead the ball is thrown vertically upwards with speed v0?
Show it.
(a) When the ball is thrown downwards
The ball is thrown downwards, which means the acceleration of the ball is g = 9.8 m/s² in the downward direction. Thus, using the kinematic equation:
vf² = vi² + 2aΔh,
where
vf = final velocity = ?
vi = initial velocity = v0 = given
a = acceleration = g = -9.8 m/s² (negative because acceleration is in the downward direction)
Δh = height from which the ball is thrown = h = given.
Substituting the given values, we get,vf² = v0² + 2ghvf² = v0² + 2×(-9.8)×h... (i)
Thus, we get the final velocity when the ball hits the ground.
Now, using the second kinematic equation:Δh = vi×t + (1/2)at²
where
Δh = height from which the ball is thrown = h = given
vi = initial velocity = v0 = given
a = acceleration = g = -9.8 m/s² (negative because acceleration is in the downward direction)
t = time taken to reach the ground = ?
Substituting the given values, we get,
h = v0×t + (1/2)×(-9.8)×t²h = v0t - 4.9t²... (ii)
Solving equations (i) and (ii), we get,
vf = √[v0² + 2gh]t = (√[2h/g])
Thus, the speed of the ball when it hits the ground is √[v0² + 2gh] and the time taken to reach the ground is t = (√[2h/g]).
(b) When the ball is thrown upwards
If the ball is thrown upwards with initial speed v0, then the acceleration is still g = 9.8 m/s² in the downward direction. But the initial velocity is in the upward direction.
Thus, we can use the same kinematic equations as above. We just need to change the signs of initial velocity, final velocity and time wherever applicable. Thus, we get the following:
vf² = vi² + 2aΔhvf² = 0² + 2×(-9.8)×h... (iii)
Using equation (iii), we get the final velocity when the ball reaches the ground.
Now, using the second kinematic equation,Δh = vi×t + (1/2)at²
where
Δh = height from which the ball is thrown = h = given
vi = initial velocity = v0 = given (upwards)
t = time taken to reach the ground = ?
Substituting the given values, we get,
h = v0t + (1/2)×(-9.8)×t²h = v0t - 4.9t²... (iv)
Solving equations (iii) and (iv),
we get,
vf = √[2gh]t = (√[2h/g])
Thus, the speed of the ball when it hits the ground is √[2gh] and the time taken to reach the ground is t = (√[2h/g]).
Note: The speed of the ball when it is at the initial height h is v0 (upwards).
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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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ee 4.9Ω resistors are connected in series with a 18.0 V battery. Find the following. (a) the equivalent resistance of the circuit Ω (b) the current in each resistor A (c) Repeat for the case in which all three resistors are connected in parallel across the battery. equivalent resistance Ω current in each resistor A
(a) The equivalent resistance of the circuit When resistors are connected in series is 19.6 Ω.
(b) The current flowing through each resistor in series is 14.7 A.
(c) Equivalent resistance of the circuit When resistors are connected in parallel is 1.225 Ω.
(d) The current flowing through each resistor in parallel is 14.7 A.
Resistance of each resistor, R = 4.9 Ω, Total number of resistors, n = 4, Voltage, V = 18 V.
(a) Equivalent resistance of the circuit When resistors are connected in series, the total resistance is equal to the sum of individual resistance. R = R1 + R2 + R3 + R4 = 4.9 + 4.9 + 4.9 + 4.9= 19.6 Ω
Thus, the equivalent resistance of the circuit is 19.6 Ω.
(b) Current in each resistor: Total voltage of the circuit, V = 18 V. Since the resistors are connected in series, the current flowing through each resistor is the same. I = V/RI = 18/19.6= 0.9184 A = 0.92 A.
Thus, the current flowing through each resistor in series 0.92 A.
(c) Equivalent resistance of the circuit When resistors are connected in parallel: , the reciprocal of the total resistance is equal to the sum of the reciprocal of individual resistance.1/R = 1/R1 + 1/R2 + 1/R3 + 1/R4= 1/4.9 + 1/4.9 + 1/4.9 + 1/4.9= 0.8163R = 1/0.8163= 1.225 Ω
Thus, the equivalent resistance of the circuit is 1.225 Ω.
(d) Current in each resistor, Total voltage of the circuit, V = 18 V. The equivalent resistance of the circuit, R = 1.225 Ω.Since the resistors are connected in parallel, the voltage across each resistor is the same.V = IR. Thus, the current flowing through each resistor, I = V/R= 18/1.225= 14.69 A = 14.7 A.
Thus, the current flowing through each resistor is 14.7 A.
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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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walker walked 0.85 km, in a straight line (to the right) to the finished line, in 9.70 min. She then turned around and walked back to the starting point in 44.00 minutes.
#
(a) Calculate the average velocity (in km/h) of the fast walker for the first 9.70 minutes. Pay attention to your sign.
1 km/h
#
(b) Calculate the average velocity (in km/h) of the fast walker for the return trip in 44.00 minutes. Pay attention to your sign.
2 km/h
#
(c) Calculate the average velocity (in km/h) of the fast walker for the round trip. Pay attention to your sign.
(a) The average velocity for the first 9.70 minutes is 5.25 km/h.
(b) The average velocity for the return trip is 1.16 km/h.
(c) The average velocity for the round trip is 0 km/h.
To calculate the average velocity, we need to divide the total distance traveled by the total time taken.
(a) For the first 9.70 minutes, the walker covered a distance of 0.85 km.
Average velocity = Distance / Time
Average velocity = 0.85 km / (9.70 min / 60 min/h) [Converting minutes to hours]
Average velocity = 0.85 km / 0.162 hr
Average velocity = 5.25 km/h
The average velocity of the fast walker for the first 9.70 minutes is 5.25 km/h.
(b) For the return trip in 44.00 minutes, the walker also covered a distance of 0.85 km.
Average velocity = Distance / Time
Average velocity = 0.85 km / (44.00 min / 60 min/h) [Converting minutes to hours]
Average velocity = 0.85 km / 0.733 hr
Average velocity = 1.16 km/h
The average velocity of the fast walker for the return trip is 1.16 km/h.
(c) To calculate the average velocity for the round trip, we can use the concept of total displacement. Since the walker starts and ends at the same point, the total displacement is zero.
Average velocity = Total displacement / Total time
Total displacement = 0 km [Since the starting and ending points are the same]
Total time = 9.70 min + 44.00 min = 53.70 min
Average velocity = 0 km / (53.70 min / 60 min/h) [Converting minutes to hours]
Average velocity = 0 km / 0.895 hr
Average velocity = 0 km/h
The average velocity of the fast walker for the round trip is 0 km/h.
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You are given a 45.5 V battery and two resistors of resistances 26Ω and 99Ω. a 25% Part (a) Find the current in A when these resistors are connected in series with the battery. I
2
= Hints: deduction per bint. Hints remaining:
3
1
Feedback: 094 deduction per feedhack A 25\% Part (b) Find the power dissipated in W by the 26Ω resistor when connected in series with the rest of the circuit. A 25% Part (c) Find the power dissipated in W by the 99Ω resistor when connected in series with the rest of the circuit. A 25% Part (d) Find the current in A being drawn from the battery when the resistors are connected in parallel with the battery
(a)The current (I) in Ampere is 0.36A when these resistors are connected in series with the battery.
(b)The power dissipated (P) in Watt by the 26Ω resistor when connected in series with the rest of the circuit is 3.94W.
(c) The power dissipated (P) in Watt by the 99Ω resistor when connected in series with the rest of the circuit is 12.94W.
(d) The current being drawn from the battery when the resistors are connected in parallel with the battery is 2.2A.
(a) The given values are:
V = 45.5 V,
R1 = 26Ω,
R2 = 99Ω
We need to calculate the current (I) in Ampere.
Using the formula of Ohm's law,
we can calculate the current.
I = V / R
Where, V = Voltage, R = Resistance
Let's first calculate the total resistance (Rt) of the circuit as they are connected in series.
Rt = R1 + R2= 26Ω + 99Ω= 125Ω
Now, using Ohm's law, we can calculate the current (I).
I = V / RtI = 45.5 / 125I = 0.364A ≈ 0.36 A
The current (I) in Ampere is 0.36A when these resistors are connected in series with the battery.
(b) We need to calculate the power dissipated (P) in Watt by the 26Ω resistor when connected in series with the rest of the circuit.
The formula to calculate the power dissipated is:
P = I²R
Where, I = Current, R = Resistance
We have already calculated the current I as 0.36A.
Therefore,P = (0.36)² × 26P = 3.94 W
The power dissipated (P) in Watt by the 26Ω resistor when connected in series with the rest of the circuit is 3.94W.
(c) We need to calculate the power dissipated (P) in Watt by the 99Ω resistor when connected in series with the rest of the circuit.
The formula to calculate the power dissipated is:
P = I²R
Where, I = Current, R = Resistance
We have already calculated the current I as 0.36A.
Therefore,P = (0.36)² × 99P = 12.94 W
The power dissipated (P) in Watt by the 99Ω resistor when connected in series with the rest of the circuit is 12.94W.
(d) We need to find the current in A being drawn from the battery when the resistors are connected in parallel with the battery.
To calculate the current in the parallel circuit,
we can use the following formula:
1/I = 1/I1 + 1/I2Where, I1 and I2 are the current flowing through each resistor R1 and R2 respectively.
Let's first calculate the current I1 passing through the resistor R1 using Ohm's law.I1 = V / R1I1 = 45.5 / 26I1 = 1.75 A
Using the same formula, we can calculate the current I2 passing through the resistor R2.I2 = V / R2I2 = 45.5 / 99I2 = 0.46 A
Now using the formula of a parallel circuit, we can calculate the current in the circuit.
I = I1 + I2I = 1.75 + 0.46I = 2.21A ≈ 2.2A
The current being drawn from the battery when the resistors are connected in parallel with the battery is 2.2A.
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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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Two children pull a third child backwards on a snow-saucer sled exerting forces F1 = 9.5 Nand F2 = 7.5 N, as shown in the figure. Note that the direction of the friction force, f = 5.9 N, is unspecified; it will be opposite in direction to the sum of the other two forces.
A:
Find the magnitude, in meters per squared second, of the acceleration of the 41 kg sled and child system.
aa =
|
m/s2
(b) Let east be towards the right of the sketch, and let north be towards the top of the sketch. Assuming the sled starts at rest, find the direction of the velocity in degrees north of east.
The magnitude of the acceleration of the sled and child system is 0.27 m/s^2. The direction of the velocity is approximately 69.5° north of east.
a. The net force acting on the sled and child system can be determined by considering the forces involved:
Net force = F1 + F2 - f
Net force = 9.5 N + 7.5 N - 5.9 N
Net force = 11.1 N
Now, we can use Newton's second law to find the acceleration:
Net force = mass × acceleration
11.1 N = 41 kg × acceleration
Solving for acceleration:
acceleration = 11.1 N / 41 kg
acceleration = 0.27 m/s^2
The magnitude of the acceleration of the sled and child system is approximately 0.27 m/s^2.
b. To find the direction of the velocity in degrees north of east, we can use the concept of trigonometry. Since the sled starts at rest, the initial velocity is zero. As the sled accelerates, the direction of its velocity vector will be the same as the direction of the net force vector.
In the given figure, assuming east is towards the right and north is towards the top, the direction of the net force vector will be the sum of F1 and F2, which forms an angle with the east direction.
We can calculate the angle using trigonometry:
tanθ = (F1 + F2) / f
Substituting the given values:
tanθ = (9.5 N + 7.5 N) / 5.9 N
tanθ = 2.79
Taking the inverse tangent (arctan) of both sides to find the angle:
θ = arctan(2.79)
θ = 69.5°
Therefore, the direction of the velocity is approximately 69.5° north of east.
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A hot-air balloon rises from ground level at a constant velocity of 2.80 m/s. One minute after liftoff, a sandbag is dropped accidentally from the balloon. Calculate the time it takes for the sandbag to reach the ground. 6.155 Previous Tries Calculate the velocity of the sandbag when it hits the ground. Tries 6/1 qo Previous Tries
The velocity of the sandbag when it hits the ground is 588 m/s.
Acceleration due to gravity, g = 9.8 m/s². We know that, when an object falls freely under the influence of gravity, the distance it travels in time t is given by the formula:
h = u.t + 1/2 g t².
Here, u = 0,h = vertical height = s (distance travelled by sandbag),g = 9.8 m/s².
Substituting these values, we get:
s = 0 + 1/2 (9.8) t²s = 4.9 t² ----(1)
Now, time taken for sandbag to reach the ground, t = 60 s + t' [As sandbag is dropped 1 minute or 60 s after liftoff] Where, t' = time taken for sandbag to fall to the ground.
Using the equation (1), we can find the value of t' when s = 0. Therefore,0 = 4.9 t'²t'² = 0t' = 0 s.
So, the time taken by the sandbag to reach the ground is t = 60 s+ t' = 60 s + 0 s = 60 s.
When the sandbag hits the ground, its final velocity is given by the formula: v = u + g.t
Here, u = 0, g = 9.8 m/s² and t = 60 s
Substituting these values, we get: v = 0 + 9.8 × 60v = 588 m/s.
Therefore, the velocity of the sandbag when it hits the ground is 588 m/s.
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The time taken by the sandbag to reach the ground is 84 s and the velocity of the sandbag when it hits the ground is 823.2 m/s.
Velocity of hot-air balloon (V) = 2.8 m/s
Time elapsed before dropping a sandbag (t) = 1 minute = 60 s
Gravitational acceleration (g) = 9.8 m/s²
We have to calculate the time it takes for the sandbag to reach the ground using the formula:
h = (1/2) g t² ... (i)
Using the same formula, the velocity of the sandbag when it hits the ground is given by the formula:
v = g t ... (ii)
To calculate the time taken by the sandbag to reach the ground, let's substitute the values in formula (i):
h = (1/2) g t²
(0 + 1/2 * 9.8 * 60² ) = 17640 m
Thus, the time it takes for the sandbag to reach the ground is t = √(2h/g) = √(2 * 17640 / 9.8) = 84 seconds (approx).
Now, let's calculate the velocity of the sandbag when it hits the ground using formula (ii):
v = g t = 9.8 * 84 = 823.2 m/s.
Therefore, the time taken by the sandbag to reach the ground is approximately 84 s, and the velocity of the sandbag when it hits the ground is 823.2 m/s.
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Let D = 5^2ar mC/m^2 for r ≤ 0.08 m and D = 0.205 /r^2 ar c/m^ for ≥ 0.08 m.
a)find pv for r = 0.06m.
b)find pv for r= 0.1m.
c)What surface charge density could be located at r = 0.08 mm to cause D = 0 for r ≥ 0.08 m?
Assuming the given values, the calculated values of pv for r = 0.6m is 25ε₀²(0.06) C/m²; for r = 0.1m, it is 20.5ε₀²(0.06) C/m². It is not possible to have a surface charge density at r = 0.08 mm that would cause D = 0 for r ≥ 0.08 m.
Let's calculate the electric field displacement (D) and then find the volume charge density (ρv) for the given values of r:
a) For r = 0.06 m:
For r ≤ 0.08 m:
D = 5^2ε₀ar mC/m²
Substituting the values:
D = (5^2)(ε₀)(0.06) mC/m²
D = 25ε₀(0.06) mC/m²
To find ρv, we use the equation:
D = ρv / ε₀
Substituting the values:
25ε₀(0.06) mC/m²= ρv / ε₀
Simplifying:
ρv = 25ε₀²(0.06) C/m²
b) For r = 0.1 m:
For r ≥ 0.08 m:
D = 0.205ε₀ / r² ar C/m²
Substituting the values:
D = 0.205ε₀ / (0.1)² ar C/m²
D = 0.205ε₀ / (0.01) ar C/m²
D = 20.5ε₀ ar C/m²
To find ρv, we use the equation:
D = ρv / ε₀
Substituting the values:
20.5ε₀ at C/m² = ρv / ε₀
Simplifying:
ρv = 20.5ε₀²(0.06) C/m²
c) To find the surface charge density that would cause D = 0 for r ≥ 0.08 m:
When D = 0, we can set the expression for D in terms of r equal to zero:
0 = 0.205ε₀ / r^2 ar C/m²
Solving for r:
0.205ε₀ = 0
This equation implies that the value of ε₀ is zero, which is not possible since ε₀ represents the permittivity of free space and has a non-zero value.
Therefore, it is not possible to have a surface charge density at r = 0.08 mm that would cause D = 0 for r ≥ 0.08 m.
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Two people stand on opposite ends of a long sled on frictionless ice. The sled is oriented in the east-west direction, and everything is initially at rest. The western person then throws a ball eastward toward the eastern person, who catches it. Describe, from the options listed below, what happens to the sled as a result, and justify your answer.
a) the sled moves eastward and then ends up at rest
b) the sled moves eastward and then ends up moving westward
c) the sled moves westward and then ends up at rest
d) the sled moves westward, and then ends up moving eastward
e) the sled doesn’t move at all
When the western person then throws a ball eastward toward the eastern person, who catches it, the sled moves westward and then ends up at rest.
The initial momentum of the system (sled + two people) is zero because everything is initially at rest. When the western person throws the ball, the ball acquires some momentum in the eastward direction. According to the law of conservation of momentum, the total momentum of the system remains constant unless there is an external force acting on it. Since there is no external force acting on the system, the total momentum of the system remains zero even after the ball is thrown. So, the momentum acquired by the ball in the eastward direction is balanced by an equal momentum acquired by the sled and the people standing on it in the opposite (westward) direction.
According to Newton's third law of motion, for every action, there is an equal and opposite reaction. Therefore, when the western person throws the ball towards the eastern person, there will be an equal and opposite reaction that will transpire, and the sled will move in the opposite direction with the same magnitude as that of the ball. The sled will move towards the west as the ball moves towards the east since they have equal and opposite momentums.
The sled and the people standing on it will move in the opposite direction to that of the ball being thrown. Thus, the sled moves in the westward direction and then comes to rest after the ball is caught by the person standing on the eastern end of the sled. Hence, the correct answer is Option C) the sled moves westward and then ends up at rest.
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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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The unit of electrical charge in the SI system is called the Coulomb. To get a
sense of the strength of the electric force let's do a calculation with Coulomb's Law.
Assume we have two, one-coulomb charges one meter apart from each other. What is
the strength of the electric force between them?
Felectric = kQq
R2 k = 8:99x109 N 2m2
C2
2. Now calculate the gravitational force between two, one-kilogram masses
separated by one meter.
G = 6:67x10¡11 N 2m2
kg2
3. How many times stronger is the electric force than the gravitational force?
1. The strength of the electric force is 8.99 × 10⁹ N.
2. The gravitational force is 6.67 × 10⁻¹¹ N
3. The electric force is approximately 1.35 × 10¹⁹ times stronger than the gravitational force.
1. To get the strength of electric force between the two 1 Coulomb charges placed 1 meter apart from each other, we need to use Coulomb's Law;
F = kq₁q₂/r²
where
F = electric force
k = Coulomb's constant, k = 8.99 × 10⁹ Nm²/C²
q₁, q₂ = 1 Coulomb each
r = distance between the two charges = 1 meter
Substituting all the given values in the above equation, we get:
F = 8.99 × 10⁹ × 1 × 1 / 1²= 8.99 × 10⁹ N
2. To calculate the gravitational force between two 1 kilogram masses separated by 1 meter, we can use Newton's Law of Gravitation;
F = Gm₁m₂/r²
where
F = gravitational force
G = Universal Gravitational constant, G = 6.67 × 10⁻¹¹ Nm²/kg²
m₁, m₂ = 1 kilogram each
r = distance between the two masses = 1 meter
Substituting all the given values in the above equation, we get:
F = 6.67 × 10⁻¹¹ × 1 × 1 / 1²= 6.67 × 10⁻¹¹ N
3. To find how many times stronger is the electric force than the gravitational force, we need to divide the electric force (Felectric) by the gravitational force (Fgravitational);
Felectric / Fgravitational= (8.99 × 10⁹) / (6.67 × 10⁻¹¹)≈ 1.35 × 10¹⁹
Therefore, the electric force is approximately 1.35 × 10¹⁹ times stronger than the gravitational force.
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Calculate the mass of a shipment of frozen strawberries weighing 6860.0 N. Take the acceleration due to gravity to be 9.8 m/s2. Express your answer in pounds.
The mass of the shipment of frozen strawberries is approximately 1546.9943 pounds.
To calculate the mass of the shipment of frozen strawberries, we can use Newton's second law of motion, which states that the force acting on an object is equal to the mass of the object multiplied by its acceleration.
F = m * a
Where:
F is the force acting on the object (weight),
m is the mass of the object, and
a is the acceleration due to gravity.
In this case, the weight of the shipment is given as 6860.0 N and the acceleration due to gravity is 9.8 m/s^2.
So, we can rearrange the equation to solve for mass:
m = F / a
Substituting the given values:
m = 6860.0 N / 9.8 m/s^2
Calculating the mass:
m ≈ 700.8163 kg
To express the mass in pounds, we can use the conversion factor: 1 kg = 2.20462 pounds.
Converting the mass to pounds:
m ≈ 700.8163 kg * 2.20462 pounds/kg
m ≈ 1546.9943 pounds
Therefore, the mass of the shipment of frozen strawberries is approximately 1546.9943 pounds.
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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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