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2 years ago

What is her velocity? 1.5 m/s 2 m/s 2.5 m/s 5 m/s

Physics

2 answers:

fiasKO [112]2 years ago

7 0

Answer:

There is no actual question attached to this, to get a real answer be sure to include the documents/question that is provided on your work.

Explanation:

Xelga [282]2 years ago

6 0

Answer:

C. 2.5

Explanation:

This is the answer to that one question on Edge nuity, I got it right.

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2 years ago

What is solar energy I need lots of detail

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2 years ago

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The formula K2S indicates that

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There are two atoms of potassium bonded to one atom of sulfur.

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3 years ago

A converging lens of focal length 20 cm is placed in contact with, and to the left of, a diverging lens of focal length 30 cm. I

scZoUnD [109]

Answer:

Magnification will be equal to 3

Explanation:

We have given focal length of the converging lens $F_1=20cm$

Focal length of the diverging lens $F_2=30cm$

Object is placed 40 cm to the length of the converging lens d = 40 cm

Combination of the focal length will be equal to

$\frac{1}{F}=\frac{1}{F_1}+\frac{1}{F_2}$

$\frac{1}{F}=\frac{1}{20}+\frac{1}{-30}=\frac{1}{20}-\frac{1}{30}=\frac{1}{60}$

F = 60 cm

So combination of the focal length will be 60 cm

Magnification is given by

$M=\frac{F}{F-d}=\frac{60}{60-40}=3$

So magnification will be equal to 3

3 0

2 years ago

A ball is projected with an initial velocity of 40 meter per second and reached maximum height of 160 meters calculate tge angle

Andru [333]

There's a problem with the question as given. Even with a maximum projection angle of θ = 90°, the initial velocity is not large enough to get the ball up in the air 160 m. With angle 90°, the ball's height y at time t would be

y = (40 m/s) t - 1/2 g t ²

Set y = 160 m, and you'll find that there is no (real) solution for t, so the ball never attains the given maximum height.

From another perspective: recall that

v ² - v₀² = 2a ∆y

where

• v₀ = initial velocity

• v = final velocity

• a = acceleration

• ∆y = displacement

At its maximum height, the ball has zero vertical velocity, and ∆y = maximum height = 160 m. The ball is in free fall once it's launched, so a = -g.

So we have

0² - (40 m/s)² = -2g (160 m)

but this reduces to

(40 m/s)² = 2 (9.8 m/s²) (160 m)

1600 m²/s² ≠ 3136 m²/s²

7 0

2 years ago