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Zanzabum
3 years ago
9

Suddenly a worker picks up the bag of gravel. Use energy conservation to find the speed of the bucket after it has descended 2.3

0 m from rest. (You can check your answer by solving this problem using Newton's laws.)
Physics
1 answer:
fiasKO [112]3 years ago
8 0

Explanation:

A worker picks up the bag of gravel. We need to find the speed of the bucket after it has descended 2.30 m from rest. It is case of conservation of energy. So,

\dfrac{1}{2}mv^2=mgh\\\\v=\sqrt{2gh}

h = 2.3 m

v=\sqrt{2\times 9.8\times 2.3} \\\\v=6.71\ m/s

So, the speed of the bucket after it has descended 2.30 m from rest is 6.71 m/s.

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Is your increase in gravitational potential energy the same in both cases? When Climbing a mountain on a zigzag path and on a st
baherus [9]

Answer:

The increase in gravitational potential energy is the same in both cases

Explanation:

It is easier to climb a mountain in a zigzag way rather than climbing on a straight line but since the distance is the same ( vertical height ) , mass and gravity is the same. Hence the increase in gravitational potential energy is the same in both cases.

gravitational potential energy = mgh ( same in both cases )

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Write the ratio of the following?<br><br>CaCO3<br>C2H6<br>Fe(NO3)3​
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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

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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

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