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Licemer1 [7]
3 years ago
8

A train rolls past a stationary observer. To him, the train is moving at a speed of 23m/s west, and a woman on the train is movi

ng at a speed of 22.4m/s west. How long does it take the women to move 13 m relative to the train?
A. 24.1s
B.21.7s
C. 23.2s
D. 25.6s
Physics
1 answer:
GREYUIT [131]3 years ago
4 0

Answer:

21.7 seconds.

Explanation:

Woman's velocity relative to train (23 m/s - 22.4 m/s) = 0.6 m/s

Distance woman wants to travel = 13m

To find how long she will take to move 13m relative to the train, take the distance she wants to travel divided by her velocity relative to the train.

(13m)/(0.6 m/s) = 21.6667 seconds or 21.7 seconds.  

Therefore, it will take the woman 21.7 seconds to move 13m.

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Which of the following best describes a property of water?
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C. weak cohesive forces exist between its molecules.

Explanation:

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A compound whose molecules contain one boron atom and three fluorine atoms would be named monoboron fluoride. Please select the
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Two charged objects have a repulsive force of 0.040 N. If the distance separating the objects is doubled, then what is the new f
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Two charged objects have a repulsive force of 0.080 N. If the charge of both of the objects is doubled, then what is the new force? Explanation: Electrostatic force is directly related to the charge of each object. So if the charge of both objects is doubled, then the force will become four times greater.

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7 0
3 years ago
An airplane is moving at 350 km/hr. If a bomb is
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Answers:

a) -171.402 m/s

b) 17.49 s

c) 1700.99 m

Explanation:

We can solve this problem with the following equations:

y=y_{o}+V_{oy}t-\frac{1}{2}gt^{2} (1)

x=V_{ox}t (2)

V_{f}=V_{oy}-gt (3)

Where:

y=0 m is the bomb's final jeight

y_{o}=1.5 km \frac{1000 m}{1 km}=1500 m is the bomb'e initial height

V_{oy}=0 m/s is the bomb's initial vertical velocity, since the airplane was moving horizontally

t is the time

g=9.8 m/s^{2} is the acceleration due gravity

x is the bomb's range

V_{ox}=350 \frac{km}{h} \frac{1000 m}{1 km} \frac{1 h}{3600 s}=97.22 m/s is the bomb's initial horizontal velocity

V_{f} is the bomb's fina velocity

Knowing this, let's begin with the answers:

<h3>b) Time</h3>

With the conditions given above, equation (1) is now written as:

y_{o}=\frac{1}{2}gt^{2} (4)

Isolating t:

t=\sqrt{\frac{2 y_{o}}{g}} (5)

t=\sqrt{\frac{2 (1500 m)}{9.8 m/s^{2}}} (6)

t=17.49 s (7)

<h3>a) Final velocity</h3>

Since V_{oy}=0 m/s, equation (3) is written as:

V_{f}=-gt (8)

V_{f}=-(97.22)(17.49 s) (9)

V_{f}=-171.402 m/s (10) The negative sign ony indicates the direction is downwards

<h3>c) Range</h3>

Substituting (7) in (2):

x=(97.22 m/s)(17.49 s) (11)

x=1700.99 m (12)

5 0
3 years ago
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