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4 years ago

For a concave mirror, when the object is at focus the image is formed at

Physics

1 answer:

Sergeeva-Olga [200]4 years ago

8 0

Answer:

When the object is placed at the focus the image is formed at infinity.

Explanation:

When a ray passes through focus and incident on a concave mirror then it will travel parallel to principal axis after reflection.Hence the image is formed at infinity.

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A difference in electric potential is commonly known as voltage, and is provided by standard batteries. If you built two identic

Lapatulllka [165]

Answer:

higher

Explanation:

A bigger voltage will result in a faster movement of charge.

8 0

4 years ago

In the first stage of a two-stage rocket, the rocket is fired from the launch pad starting from rest but with a constant acceler

AnnZ [28]

Answer:

3091.56

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s² (positive downward and negative upward)

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=0\times 25+\frac{1}{2}\times 3.5\times 25^2\\\Rightarrow s=1093.75\ m$

Distance traveled in the first stage is 1093.75 m

$v=u+at\\\Rightarrow v=0+3.5\times 25\\\Rightarrow v=87.5\ m/s$

Velocity at the end of first stage is 87.5 m/s

$v=u+at\\\Rightarrow a=\frac{v-u}{t}\\\Rightarrow a=\frac{132.5-87.5}{10}\\\Rightarrow a=4.5\ m/s^2$

Acceleration of the second stage is 4.5 m/s²

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=87.5\times 10+\frac{1}{2}\times 4.5\times 10^2\\\Rightarrow s=1100\ m$

Distance traveled in the second stage is 1100 m

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-132.5^2}{2\times -9.81}\\\Rightarrow s=894.81\ m$

Distance traveled after the second stage has stopped firing is 894.81 m

Total height the rocket reached = 1093.75+1100+897.81 = 3091.56 m

3 0

4 years ago

In your daily life you come across a range of motion in which acceleration is in the direction of motion

irakobra [83]

That makes no sense to me somehow

3 0

3 years ago

An arrow is shot vertically upward at a rate of 250ft/s. Use the projectile formula h=−16t2+v0t to determine at what time(s), in

SIZIF [17.4K]

Answer:

The arrow is at a height of 500 feet at time t = 2.35 seconds.

Explanation:

It is given that,

An arrow is shot vertically upward at a rate of 250 ft/s, v₀ = 250 ft/s

The projectile formula is given by :

$h=-16t^2+v_ot$

We need to find the time(s), in seconds, the arrow is at a height of 500 ft. So,

$-16t^2+250t=500$

On solving the above quadratic equation, we get the value of t as, t = 2.35 seconds

So, the arrow is at a height of 500 feet at time t = 2.35 seconds. Hence, this is the required solution.

6 0

4 years ago

Read 2 more answers

Which of the following best describes the shape of Earth's orbit?

nikklg [1K]

Elliptical, as shown by most projections

4 0

3 years ago