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ollegr [7]
2 years ago
8

You are throwing a stone straight-up in the absence of air friction. The stone is caught at the same height from which it was th

rown. the speed of the stone when thrown is equal to the speed when caught.A. TrueB. False
Physics
1 answer:
balu736 [363]2 years ago
7 0

Answer:

A. True

Explanation:

When a stone is thrown straight-up, it has an initial velocity which decreases gradually as the stone move to maximum height due to constant acceleration due to gravity acting downward on the stone, at the maximum height the final velocity of the stone is zero. As the stone descends the velocity starts to increase and becomes maximum  before it hits the ground.

Height of the motion is given by;

H = \frac{u^2}{2g}

g is acceleration due to gravity which is constant

H is height traveled

u is the speed of throw, which determines the value of height traveled.

Therefore, when the stone is caught at the same height from which it was thrown in the absence of air resistance, the speed of the stone when thrown will be equal to the speed when caught.

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2 years ago
Four copper wires of equal length are connected in series. Their cross-sectional areas are 0.7 cm2 , 2.5 cm2 , 2.2 cm2 , and 3 c
Travka [436]

Answer:

22.1 V

Explanation:

We are given that

A_1=0.7 cm^2=0.7\times 10^{-4} m^2

A_2=2.5 cm^2=2.5\times 10^{-4} m^2

A_3=2.2 cm^2=2.2\times 10^{-4} m^2

A_4=3 cm^2=3\times 10^{-4} m^2

Using 1cm^2=10^{-4} m^2

We know that

R=\frac{\rho l}{A}

In series

R=R_1+R_2+R_3+R_4

R=\frac{\rho l}{A_1}+\frac{\rho l}{A_2}+\frac{\rho l}{A_3}+\frac{\rho l}{A_4}

R=\frac{\rho l}{\frac{1}{A_1}+\frac{1}{A_2}+\frac{1}{A_3}+\frac{1}{A_4}}

Substitute the values

R=\rho A(\frac{1}{0.7\times 10^{-4}}+\frac{1}{2.5\times 10^{-4}}+\frac{1}{2.2\times 10^{-4}}+\frac{1}{3\times 10^{-4}})

R=\rho l(2.62\times 10^4)

V=145 V

I=\frac{V}{R}=\frac{145}{\rho l(2.62\times 10^4)}

Voltage across the 2.5 square cm wire=IR=I\times \frac{\rho l}{A_2}

Voltage across the 2.5 square cm wire=\frac{145}{\rho l(2.62\times 10^4)}\times \frac{\rho l}{2.5\times 10^{-4}}=22.1 V

Voltage across the 2.5 square cm wire=22.1 V

6 0
3 years ago
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