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2 years ago

An object kept in front of lens, which is thick at the edges, produces a _________,

Physics

1 answer:

Wewaii [24]2 years ago

4 0

Answer:

Explanation:

A diverging lens forms a virtual image that is as the same size as the object

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What happens to acceleration if forces are balanced

Vikentia [17]

Answer:

No acceleration.

Explanation:

If the forces are Balanced it means they are in equilibrium and there will be no net force, therefore the object will not accelerate and the velocity will remain constant.

5 0

3 years ago

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Which of the following is an example of a physical change but not a chemical change?

vazorg [7]

A water pipe freezes and cracks on a cold night. <em> (D)</em>

There are three physical changes going on in this scenario:

1). The air gets cold and dark at night.

2). The water in the pipe freezes and expands.

3). The pipe cracks.

There are no chemical changes in the description.

7 0

3 years ago

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A 0.15 kg baseball moving at 20 m/s is stopped by a player in 0.010 s. What is the average force of the ball?

Vlad [161]

Answer: 300N

Explanation:

Impulse= Mass * Velocity

F.T = M * V

F= MV/T

F= (0.15*20)/ 0.01

F= 300N

8 0

2 years ago

In the middle of the night you are standing a horizontal distance of 14.0 m from the high fence that surrounds the estate of you

Oksi-84 [34.3K]

Answer:

Explanation:

a ) Height to be cleared = 5 - 1.6 = 3.4 m

Horizontal distance to be cleared = 5 m .

angle of throw = 56°

here y = 3.4 , x = 5 , θ = 56

equation of trajectory

y = x tanθ - 1/2 g ( x/ucosθ)²

3.4 = 5 tan56 - 1/2 g ( 5/ucos56)²

3.4 = 7.4 - 122.5 / .3125u²

122.5 / .3125u² = 4

u² = 98

u = 9.9 m /s

Range = u² sin 2 x 56 / g

= 9.9 x 9.9 x .927 / 9.8

= 9.27 m

horizontal distance be-yond the fence will the rock land on the ground

= 9.27 - 5

= 4.27 m

6 0

3 years ago

Read 2 more answers

Cannonball a is launched across level ground with speed v₀ at an angle of 45. Cannonball b is launched from the same location wi

postnew [5]

Answer:

Cannonball b spends more time in the air than cannonball a.

Explanation:

Starting with the definition of acceleration, we have that:

$a=\frac{\Delta v}{\Delta t}\\\\\Delta t= \frac{\Delta v}{a}$

Since both cannonballs will stop in their maximum height, their final velocity is zero. And since the acceleration in the y-axis is g, we have:

$\Delta t= -\frac{v_{oy}}{g}$

Now, this time interval is from the moment the cannonballs are launched to the moment of their maximum height, exactly the half of their time in the air. So their flying time t_f is (the minus sign is ignored since we are interested in the magnitudes only):

$t_f=2\frac{v_{oy}}{g}$

Then, we can see that the time the cannonballs spend in the air is proportional to the vertical component of the initial velocity. And we know that:

$v_{oy}=v_o\sin\theta\\\\\implies t_f=2\frac{v_o\sin\theta}{g}$

Finally, since $\sin60\°=\frac{\sqrt{3} }{2}$ and $\sin45\°=\frac{\sqrt{2} }{2}$ , we can conclude that:

$t_{fa}=\sqrt{2}\frac{v_o }{g} \\\\t_{fb}=\sqrt{3}\frac{v_o }{g}\\\\\implies t_{fb}>t_{fa}$

In words, the cannonball b spends more time in the air than cannonball a.

5 0

3 years ago