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Kipish [7]
3 years ago
12

A 57 g tennis ball is traveling at 45 m/s to the right

Physics
1 answer:
andrey2020 [161]3 years ago
8 0

What is the question asking

You might be interested in
alculate the kinetic energies of (a) a 2.00×103-kg automobile moving at 100.0 km/h; (b) an 80.0-kg runner sprinting at 10.0 m/s;
zzz [600]

Answer:

(a) 7.72×10⁵ J

(b) 4000 J

(c) 1.82×10⁻¹⁶ J

Explanation:

Kinetic Energy: This can be defined energy of a body due to its motion. The expression for kinetic energy is given as,

Ek = 1/2mv²................... Equation 1

Where Ek = Kinetic energy, m = mass, v = velocity

(a)

For a moving automobile,

Ek = 1/2mv².

Given: m = 2.0×10³ kg, v = 100 km/h = 100(1000/3600) m/s = 27.78 m/s

Substitute into equation 1

Ek = 1/2(2.0×10³)(27.78²)

Ek = 7.72×10⁵ J

(b)

For a sprinting runner,

Given: m = 80 kg, v = 10 m/s

Substitute into equation 1 above,

Ek = 1/2(80)(10²)

Ek = 40(100)

Ek = 4000 J

(c)

For a moving electron,

Given: m = 9.10×10⁻³¹ kg, v = 2.0×10⁷ m/s

Substitute into equation 1 above,

Ek = 1/2(9.10×10⁻³¹)(2.0×10⁷)²

Ek = 1.82×10⁻¹⁶ J

8 0
3 years ago
If you can answer all of this then your a legend (I'm giving you all my points)
navik [9.2K]

Answer:

1. B

2. B

3. D

4. A

Hope this was correct! A lot of the answers are already in the article itself and the wording is just different. I suggest now that for information retainment, you read the article again with the correct points in mind and see if you can spot the points where the answers are stated!

3 0
2 years ago
The semi-major axis of this ellipse is 8.8 cm, and the distance from one of the foci to the
scoundrel [369]

The eccentricity is 0.5

8 0
3 years ago
A car traveling initially at 9.49 m/s accelerates at the rate of 0.988 m/s^2 for 3.05s. What is it’s velocity at the end of the
weeeeeb [17]

Answer:

12.50 m/s

Explanation:

Vi = 9.49 m/s

a = 0.988 m/s²

t = 3.05 s

Vf = ?

Vf = Vi + at

Vf = 9.49 + (0.988)(3.05)

Vf = 12.50 m/s

6 0
3 years ago
Read 2 more answers
In which mechanical test is a specimen deformed with a gradually increasing load that is applied uniaxially along the long axis
IRINA_888 [86]

Answer:

Compression Test

Explanation:

The Specimen is undergoing a compression test. It is similar to tensile test with the difference that the force is compressive and applied along the direction of stress. Both Tensile and compression tests are performed on Universal Testing machine. Compression test is done to determine the product's reaction when it is compressed, squashed and crushed.

7 0
3 years ago
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