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Triss [41]
1 year ago
14

a bus is moving at 22m/s [E] for 12s. Then the bus driver slows down at 1.2m/s2 [W] until the bus stops. Determine the total dis

placement of the bus. (Answer is 4.7x10^2m)
Physics
1 answer:
KatRina [158]1 year ago
7 0
The total displacement is equal to the total distance. For the east or E direction, the distance is determined using the equation:

d = vt = (22 m/s)(12 s) = 264 m

For the west or W direction, we use the equations:
a = (v - v₀)/t
d = v₀t + 0.5at²

Because the object slows down, the acceleration is negative. So,
-1.2 m/s² = (0 m/s - 22 m/s)/t
t = 18.33 seconds
d = (22 m/s)(18.33 s) + 0.5(-1.2 m/s²)(18.33 s)²
d = 201.67 m

Thus,
Total Displacement = 264 m +  201.67 m = 465.67 or  approximately 4.7×10² m.
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Juan and Kuri are on a carousel. Juan is closer to the center of the carousel than Kuri. Which statement describes their tangent
Licemer1 [7]

Answer:

Juan and Kuri complete one revolution in the same time, but Juan travels a shorter distance and has a lower speed.

Explanation:

Since Juan is closer to the center and Kuri is away from the center so we can say that Juan will move smaller distance in one complete revolution

As we know that the distance moved in one revolution is given as

d = 2\pi r

also the time period of revolution for both will remain same as they move with the time period of carousel

Now we can say that the speed is given as

v = \frac{2\pi r}{T}

so Juan will have less tangential speed. so correct answer will be

Juan and Kuri complete one revolution in the same time, but Juan travels a shorter distance and has a lower speed.

6 0
2 years ago
Read 2 more answers
A spring stretches by 15cm when a mass of 300g hangs down from it,if the spring is then stretched an additional 10cm and release
bearhunter [10]

Answer:

0.1 m

Explanation:

It is given that,

Mass of the object, m = 350 g = 0.35 kg

Spring constant of the spring, k = 5.2 N/m

Amplitude of the oscillation, A = 10 cm = 0.1 m

Frequency of a spring mass system is given by :

Time period:

4 0
2 years ago
A jetboat is drifting with a speed of 5.0\,\dfrac{\text m}{\text s}5.0 s m ​ 5, point, 0, start fraction, start text, m, end tex
love history [14]

The question is incomplete. Here is the entire question.

A jetboat is drifting with a speed of 5.0m/s when the driver turns on the motor. The motor runs for 6.0s causing a constant leftward acceleration of magnitude 4.0m/s². What is the displacement of the boat over the 6.0 seconds time interval?

Answer: Δx = - 42m

Explanation: The jetboat is moving with an acceleration during the time interval, so it is a <u>linear</u> <u>motion</u> <u>with</u> <u>constant</u> <u>acceleration</u>.

For this "type" of motion, displacement (Δx) can be determined by:

\Delta x = v_{i}.t + \frac{a}{2}.t^{2}

v_{i} is the initial velocity

a is acceleration and can be positive or negative, according to the referential.

For Referential, let's assume rightward is positive.

Calculating displacement:

\Delta x = 5(6) - \frac{4}{2}.6^{2}

\Delta x = 30 - 2.36

\Delta x = - 42

Displacement of the boat for t=6.0s interval is \Delta x = - 42m, i.e., 42 m to the left.

8 0
2 years ago
A beam of electrons is sent horizontally down the axis of a tube to strike a fluorescent screen at the end of the tube. On the w
slamgirl [31]

Answer:

The answer is 3.

Explanation:

The answer to this question can be found by applying the right hand rule for which the pointer finger is in the direction of the electron movement, the thumb is pointing in the direction of the magnetic field, so the effect that this will have on the electrons is the direction that the middle finger points in which is right in this example.

So as a result of the magnetic field directed vertically downwards which is at a right angle with the electron beams, the electrons will move to the right and the spot will be deflected to the right of the screen when looking from the electron source.

I hope this answer helps.

4 0
2 years ago
One end of a string is fixed. An object attached to the other end moves on a horizontal plane with uniform circular motion of ra
sveticcg [70]

Answer:

If both the radius and frequency are doubled, then the tension is increased 8 times.

Explanation:

The radial acceleration (a_{r}), measured in meters per square second, experimented by the moving end of the string is determined by the following kinematic formula:

a_{r} = 4\pi^{2}\cdot f^{2}\cdot R (1)

Where:

f - Frequency, measured in hertz.

R - Radius of rotation, measured in meters.

From Second Newton's Law, the centripetal acceleration is due to the existence of tension (T), measured in newtons, through the string, then we derive the following model:

\Sigma F = T = m\cdot a_{r} (2)

Where m is the mass of the object, measured in kilograms.

By applying (1) in (2), we have the following formula:

T = 4\pi^{2}\cdot m\cdot f^{2}\cdot R (3)

From where we conclude that tension is directly proportional to the radius and the square of frequency. Then, if radius and frequency are doubled, then the ratio between tensions is:

\frac{T_{2}}{T_{1}} = \left(\frac{f_{2}}{f_{1}} \right)^{2}\cdot \left(\frac{R_{2}}{R_{1}} \right) (4)

\frac{T_{2}}{T_{1}} = 4\cdot 2

\frac{T_{2}}{T_{1}} = 8

If both the radius and frequency are doubled, then the tension is increased 8 times.

5 0
1 year ago
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