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maria [59]
3 years ago
6

11) A man is on a 1/4 on a bridge. A train is coming the same direction he is going. The man can run across the bridge in the sa

me direction and make in barely in time. He can also run backwards towards the train and also barely make it. How fast is the train going relative to the man?
Physics
1 answer:
lesya692 [45]3 years ago
7 0

15mph

If the man turns and runs toward point A, he will cover

3/8 of the length of the bridge in the time that it takes

the train to reach A.

If the man runs forward toward point B, what part of the bridge

will he cover before the train reaches A? Well, he will cover

3/8 of the bridge, only heading forward toward B. This will put

him 3/8 + 3/8 = 6/8 = 3/4 of the way across the bridge by the

time the train reaches A.

since we know that the man and the train will meet at B, this

means that in the time it takes the man to run the remaining

1/4 of the bridge, the train will cover the entire length of

the bridge.

If it takes the man the same time to cover 1/4 of the bridge

that it takes the train to cover the whole bridge, then the train

must be going four times as fast as the man. Another way of saying

this is that the man runs at 1/4 the speed of the train.  

Since the train's speed is known to be 60 mph, this means that

the man runs at (1/4) 60 = 15 mph.

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The plank is 5.1 meters long and has a mass of 9.2 kg, ack has a mass of 93.6 kg, and Diane has a mass of 64.2 kg. Assuming Dian
kodGreya [7K]

Answer:

The answer is 0.8 m

Explanation:

According the attached diagram, if we take the momentum in the point 0:

\frac{m_{D}*5.1 }{2} =m_{J} *X

Where

mD = mass of Diane = 64.2 kg

mJ = mass of Jack = 93.6 kg

Replacing:

X=\frac{\frac{64.2*5.1}{2} }{93.6} =1.75m

\frac{L}{2} -X=Y=\frac{5.1}{2} -1.75\\Y=0.8m

8 0
3 years ago
calculate the period of a wave whose frequency is 5 Hertz and whose wavelength is one centimeter give your answer in a decimal f
olga2289 [7]
The period of the wave is the reciprocal of its frequency.

       1 / (5 per second)  =  0.2 second .

The wavelength is irrelevant to the period.  But since you
gave it to us, we can also calculate the speed of the wave.

Wave speed = (frequency) x (wavelength)

                   = (5 per second) x (1cm)  =  5 cm per second
4 0
3 years ago
A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.6 m/s at ground level.
galina1969 [7]

Answer:

44.64 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.8 m/s²

v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 4.2\times 1180+80.6^2}\\\Rightarrow v=128.01\ m/s

v=u+at\\\Rightarrow 128.01=80.6+4.2t\\\Rightarrow t=\frac{128.01-80.6}{4.2}=11.29\ s

<u>Time taken to reach 1180 m is 11.29 seconds</u>

v=u+at\\\Rightarrow 0=128.01-9.8t\\\Rightarrow t=\frac{128.01}{9.8}=13.06\ s

<u>Time the rocket will keep going up after the engines shut off is 13.06 seconds.</u>

v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-128.01^2}{2\times -9.8}\\\Rightarrow s=836.05\ m

The distance the rocket will keep going up after the engines shut off is 836.05 m

Total distance traveled by the rocket in the upward direction is 1180+836.05 = 2016.05 m

The rocket will fall from this height

s=ut+\frac{1}{2}at^2\\\Rightarrow 2016.05=0t+\frac{1}{2}\times 9.8\times t^2\\\Rightarrow t=\sqrt{\frac{2016.05\times 2}{9.8}}\\\Rightarrow t=20.29\ s

<u>Time taken by the rocket to fall from maximum height is 20.29 seconds</u>

Time the rocket will stay in the air is 11.29+13.06+20.29 = 44.64 seconds

5 0
3 years ago
What percentage of the earth's surface is covered in water
Alla [95]

Answer:

71 % of the earth's surface is covered in water

8 0
3 years ago
Read 2 more answers
Please Please Please help on these 3 questions :) 20 POINTS!! - NO LINKS PLEASE
GenaCL600 [577]

Answer:

Catapult on the ground: Normal, gravity

Catapult (I'm assuming launching marshmallow): Reaction of Force Applied

Marshmallow: Force Applied

Explanation:

This is the forces that act on a stationary object and a launched object. The catapult may also experience a force friction if your teacher is taking a more practical sense.

3 0
3 years ago
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