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2 years ago

All of the following devices are used to create an open circuit EXCEPT a:

Physics

1 answer:

Brrunno [24]2 years ago

4 0

Answer:

I think its a fuse because everything else makes sense and is used in electrical circuits and the fuse is the only one that stands out to me ¯\_(ツ)_/¯

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A car is traveling at 60 mph. If the rate of speed increases 4 mph each hour,

Sauron [17]

Answer:

5 hours ................

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2 years ago

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Why is the warm air rising while the cold air is sinking

UNO [17]

Because cold air tends to more dense, and it's therefore heavier, and it sinks.
Hot air though is hot with no density so it's light and rises.

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2 years ago

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Water drops fall from the edge of a roof at a steady rate. a fifth drop starts to fall just as the first drop hits the ground. a

Alchen [17]

The height of the roof is <u>3.57m</u>

Let the drops fall at a rate of 1 drop per t seconds. The first drop takes 5t seconds to reach the ground. The second drop takes 4t seconds to reach the bottom of the 1.00 m window, while the 3rd drop takes 3t s to reach the top of the window.

Calculate the distances traveled by the second and the third drops s₂ and s₃, which start from rest from the roof of the building.

$s_2=\frac{1}{2} g(4t)^2=8gt^2\\ s_3=\frac{1}{2} g(3t)^2=(4.5)gt^2$

The length of the window s is given by,

$s=s_2-s_3\\ (1.00 m)=8gt^2-4.5gt^2=3.5gt^2\\ t^2=\frac{1.00 m}{(3.5)(9.8m/s^2)} =0.02915s^2$

The first drop is at the bottom and it takes 5t seconds to reach down.

The height of the roof h is the distance traveled by the first drop and is given by,

$h=\frac{1}{2} g(5t)^2=\frac{25t^2}{2g} =\frac{25(0.02915s^2)}{2(9.8m/s^2)} =3.57 m$

the height of the roof is 3.57 m

8 0

3 years ago

Read 2 more answers

At the bottom of its path, the ball strikes a 2.30 kg steel block initially at rest on a frictionless surface. The collision is

Triss [41]

Answer:

(a). The speed of the ball after collision is 2.01 m/s.

(b). The speed of the block after collision 1.11 m/s.

Explanation:

Suppose, A steel ball of mass 0.500 kg is fastened to a cord that is 50.0 cm long and fixed at the far end. The ball is then released when the cord is horizontal.

Given that,

Mass of steel block = 2.30 kg

Mass of ball = 0.500 kg

Length of cord = 50.0 cm

We need to calculate the initial speed of the ball

Using conservation of energy

$\dfrac{1}{2}mv^2=mgl$

$v=\sqrt{2gl}$

Put the value into the formula

$u=\sqrt{2\times9.8\times50.0\times10^{-2}}$

$u=3.13\ m/s$

The initial speed of the ball $u_{1}=3.13\ m/s$

The initial speed of the block $u_{2}=0$

(a). We need to calculate the speed of the ball after collision

Using formula of collision

$v_{1}=(\dfrac{m_{1}-m_{2}}{m_{1}+m_{2}})u_{1}+(\dfrac{2m_{2}}{m_{1}+m_{2}})u_{2}$

Put the value into the formula

$v_{1}=(\dfrac{0.5-2.30}{0.5+2.30})\times3.13$

$v_{1}=-2.01\ m/s$

Negative sign shows the opposite direction of initial direction.

(b). We need to calculate the speed of the block after collision

Using formula of collision

$v_{2}=(\dfrac{2m_{1}}{m_{1}+m_{2}})u_{1}+(\dfrac{m_{1}-m_{2}}{m_{1}+m_{2}})u_{2}$

Put the value into the formula

$v_{2}=(\dfrac{2\times0.5}{0.5+2.30})\times3.13+0$

$v_{2}=1.11\ m/s$

Hence, (a). The speed of the ball after collision is 2.01 m/s.

(b). The speed of the block after collision 1.11 m/s.

8 0

3 years ago

A force of 100N moves a body on a horizontal frictionless surface when......

zalisa [80]

Answer:

I. Friction force exerted on the body is less than 100N

Explanation:

For a body to be static, the moving force must be equal to the frictional force. Since the frictional force is a force of opposition. It tends to oppose the moving force acting on an object.

Hence if the moving force is greater than the force of friction, the Force of fiction will not be able to overcome the moving hence the body will tend to move.

Therefore, for a body to move, Fm > Ff or Ff < Ff

Fm is the moving force

Ff is the force of friction

Given

Fm = 100N

For the 100N body to move the frictional force must be less than 100N

4 0

2 years ago