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2 years ago

5) The recoil momentum of a gun

Physics

1 answer:

prisoha [69]2 years ago

4 0

Explanation:

c) is equal to the momentum of the fired bullet

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A spearfisher stands in shallow water and sees a fish a few feet in front of her. She throws her spear directly toward the posit

DENIUS [597]

Answer:

the spear will end up above the fish relative to the actual position of the fish.

Explanation:

due to refraction of light coming from the fish the fish will appear slightly above from its real position

So due to this refraction the spearfisher will throw the spear directly at the image of the fish due to which it will not reach the position of fish but it will reach the position above the fish.

So here we can say that the spear will end up above the fish relative to the actual position of the fish

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3 years ago

How did the magnet’s density measurement using the Archimedes’ Principle compare to the density measurement using the calculated

guapka [62]

Answer:

The two methods will yield different results as one is subject to experimental errors that us the Archimedes method of measurement, the the density measurement method will be more accurate

Explanation:

This is because the density method using the calculated volume will huve room for less errors that's occur in practical method i.e Archimedes method due to human error

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3 years ago

Help me please!!!

lilavasa [31]

Both planets are similar in shape and have a rocky surface. Not sure about the phases though

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2 years ago

Read 2 more answers

Two trains A and B of length 400 m each are moving on two parallel tracks with

Vinvika [58]

Answer:

The initial distance between the trains is 1450 m. 

Explanation:

In the question two trains are of equal length 400 m and moves at a uniform speed of 72 km/h. train A is moving ahead of train B. If the train B has to overtake train A it should accelerate.

Train B’s acceleration is $1m/s^2$ and it accelerated for 50 seconds.

 $a=1 m/s^2$ 

t=50 s 

initial speed u=72km/h 

we have to convert this speed into m/s 

 $u=72 \times 5/18=20 m/s$ 

Distance covered in accelerating phase $S=ut+1/2 at^2$ 

 $=20 \times 50+1/2 \times 1 \times 50^2$ 

 $=1000+1250=2250 m$ 

If a train is just behind another, the distance covered by the train located behind during overtaking phase will be equal to the sum of the lengths of the trains.

Here length of train A+length of train $B=400+400=800 m$ 