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4 years ago

This is the question with the options

Physics

1 answer:

melamori03 [73]4 years ago

4 0

Answer:

Explanation:

Average speed is calculated as (total distance)/(total time). We have three segments in the journey, indexed by 0, 1, and 2:

$v = \frac{s}{t}=\frac{v_0t_0+v_1t_1+v_2t_2}{t_0+t_1+t_2}\\t_1=t_2\implies\\v=\frac{v_0t_0+v_1t_1+v_2t_1}{t_0+t_1+t_1}=\frac{v_0t_0+v_1t_1+v_2t_1}{\frac{v_0t_0}{v_0}+\frac{v_1t_1}{v_1}+\frac{v_2t_1}{v_2}}$

We also know that the distance of the first segment is the same as one of segment 2 and 3 together:

$v_0t_0=v_1t_1+v_2t_1\\v_0\frac{t_0}{t_1}=v_1+v_2$

Going back to the average speed expression, divide by t_1:

$\frac{v_0t_0+v_1t_1+v_2t_1}{\frac{v_0t_0}{v_0}+\frac{v_1t_1}{v_1}+\frac{v_2t_1}{v_2}}=\frac{\frac{v_0t_0}{t_1}+v_1+v_2}{\frac{v_0t_0}{v_0t_1}+2}$

and combine the two equations:

$\frac{\frac{v_0t_0}{t_1}+v_1+v_2}{\frac{v_0t_0}{v_0t_1}+2}=\frac{2(v_1+v_2)}{\frac{v_1+v_2}{v_0}+2} = \frac{2v_0(v_1+v_2)}{v_1+v_2+2v_0}$

The last form matches your choice (c).

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A 99.5 N grocery cart is pushed 12.9 m along an aisle by a shopper who exerts a constant horizontal force of 34.6 N. The acceler

Romashka [77]

1) 9.4 m/s

First of all, we can calculate the work done by the horizontal force, given by

W = Fd

where

F = 34.6 N is the magnitude of the force

d = 12.9 m is the displacement of the cart

Solving ,

W = (34.6 N)(12.9 m) = 446.3 J

According to the work-energy theorem, this is also equal to the kinetic energy gained by the cart:

$W=K_f - K_i$

Since the cart was initially at rest, $K_i = 0$ , so

$W=K_f = \frac{1}{2}mv^2$ (1)

where

m is the of the cart

v is the final speed

The mass of the cart can be found starting from its weight, $F_g = 99.5 N$ :

$m=\frac{F_g}{g}=\frac{99.5 N}{9.8 m/s^2}=10.2 kg$

So solving eq.(1) for v, we find the final speed of the cart:

$v=\sqrt{\frac{2W}{m}}=\sqrt{\frac{2(446.3 J)}{10.2 kg}}=9.4 m/s$

2) $2.51\cdot 10^7 J$

The work done on the train is given by

W = Fd

where

F is the magnitude of the force

d is the displacement of the train

In this problem,

$F=4.28 \cdot 10^5 N$

$d=586 m$

So the work done is

$W=(4.28\cdot 10^5 N)(586 m)=2.51\cdot 10^7 J$

3) $2.51\cdot 10^7 J$

According to the work-energy theorem, the change in kinetic energy of the train is equal to the work done on it:

$W=\Delta K = K_f - K_i$

where

W is the work done

$\Delta K$ is the change in kinetic energy

Therefore, the change in kinetic energy is

$\Delta K = W = 2.51\cdot 10^7 J$

4) 37.2 m/s

According to the work-energy theorem,

$W=\Delta K = K_f - K_i$

where

$K_f$ is the final kinetic energy of the train

$K_i = 0$ is the initial kinetic energy of the train, which is zero since the train started from rest

Re-writing the equation,

$W=K_f = \frac{1}{2}mv^2$

where

m = 36300 kg is the mass of the train

v is the final speed of the train

Solving for v, we find

$v=\sqrt{\frac{2W}{m}}=\sqrt{\frac{2(2.51\cdot 10^7 J)}{36300 kg}}=37.2 m/s$

7 0

4 years ago

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olya-2409 [2.1K]

Explanation:

4.he invented the atomic model with all different electrons proton and neutrons

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3 years ago

How did Archimedes measure the mass of the block of gold?

MA_775_DIABLO [31]

Archimedes found a piece of gold and a piece of silver with exactly the same mass. He dropped the gold into a bowl filled to the brim with water and measured the volume of water that spilled out. Then he did the same thing with the piece of solver. Although both metals had the same mass, the silver gad a larger volume; therefore, it displaced more water than the gold did. That's because the silver was less dense than gold. Afterwards he applied the same method to the crown for the king he served who had got a new crown from a jeweler who gave it to him. Archimedes found a piece of pure gold that had the same mass as the crown. He placed the pure gold chuck and the crown in water, one at a time. The crown displaced more water the piece of gold. Therefore, its density was less than pure gold.

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Lapatulllka [165]

Answer:

95.5 m

Explanation:

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In this case, the magnitude of the displacement is the diameter of the circular track.

d = 300 m / π

d ≈ 95.5 m

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3 years ago

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True.

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4 years ago