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melamori03 [73]
3 years ago
14

In the Position versus Time graph of an object moving in a straight line, what is the change in position in the fourth hour?

Physics
1 answer:
Ratling [72]3 years ago
8 0

Answer:

Option B. 15miles.

Explanation:

From the question given, in the 4th hour simply means between 3rd and 4th hour.

Now, let us determine the position of the object at the 3rd and 4th hour. This is illustrated below:

At the 3rd hour:

Position (P3) = 60miles

At the 4th hour:

Position (P4) = 75miles

Finally, we shall determine the change in the position of the object as follow:

Change in the position = P4 – P3

Change in the position = 75 – 60

Change in the position = 15 miles

Therefore, the change in the position of the object in the 4th hour is 15miles.

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When light travels through a small hole, it appears to be an observer that the light spreads out, blurring the outline of the ho
3241004551 [841]

Answer:

support lights as a wave

Explanation:

In the model of light as a particle, the experimenter would expect to see one small hole of light emerging on the wall. However, as the light spreads out, it behaves much like a wave that diffracts when going through a small hole.

7 0
3 years ago
Write newton's second law for an astronaut lying on the floor of one of the habitats
Leviafan [203]
Hey there,
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3 0
3 years ago
Read 2 more answers
A space vehicle is traveling at 5000 km/h relative to the Earth when the exhausted rocket motor is disengaged and sent backward
Art [367]

Answer:

the speed of the command module relative to Earth just after the separation = 4943.2 Km/hr

Explanation:

Given:

speed of space vehicle =5000 km/hr

rocket motor speed = 71 km/hr relative to the command module

mass of module = m

mass of motor = 4m

By conservation of linear momentum

Pi = Pf

Pi= initial momentum

Pf=  final momentum

Since, the motion is only in single direction

MV_i=4mV_{mE}+mV_{cE}

Where M is the mass of the space vehicle which equals the sum of motor's mass and the command's mass, Vi its initial velocity, V_mE is velocity of motor relative to Earth, and V_cE is its velocity of the command relative to Earth.

The velocity of motor relative to Earth equals the velocity of motor relative to command plus the velocity of command relative to Earth.

V_mE = V_mc+V_cE

Where V_mc is the velocity of motor relative to command this yields

5mV_i = 4m(V_{mc}+V_{cE})+mV_{cE}

5V_i = 4V_{mc}+5V_{cE}

substituting  the values we get

V_{cE} = \frac{5V_i-4V_{mc}}{5}

V_{cE} = \frac{5(5000)-4(71)}{5}

= 4943.2 Km/hr

the speed of the command module relative to Earth just after the separation = 4943.2 Km/hr

4 0
3 years ago
Read 2 more answers
Suppose the Sun suddenly stopped emitting light. How long would it take, in seconds, for the Sun’s light to disappear on Earth?
irina [24]

Answer:

480 seconds

Explanation:

It takes the suns light 8 minutes to get to Earth. 8 minutes is 480 seconds.

4 0
4 years ago
After fixing a flat tire on a bicycle you give the wheel a spin. Its initial angular speed was 5.40 rad/s and it rotated 12.3 re
zaharov [31]

To solve this problem it is necessary to apply the concepts related to the kinematic equations of angular motion.

By definition, acceleration can be expressed as the change in angular velocity squared over a given period of distance traveled.

\alpha = \frac{\omega^2}{2\theta}

where,

\omega = Angular velocity

\theta = Angular displacement.

In turn, as a function of time, we can represent it as,

\alpha = \frac{\omega}{t}

For our case we have to,

\omega = 5.4rad/s

\theta = 12.3rev = 12.3rev(\frac{2\pi rad}{1rev})=24.6\pi rad

PART A) In the case of angular acceleration we have to,

\alpha = \frac{\omega^2}{2\theta}

\alpha = \frac{(5.4)^2}{2*24.6\pi}

\alpha = 0.1886rad/s^2

PART B) Through the definition of angular acceleration as a function of time we can calculate it,

\alpha = \frac{\omega}{t}

t = \frac{\omega}{\alpha}

t = \frac{5.4}{0.1886}

t = 28.63s

3 0
4 years ago
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