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frutty [35]
3 years ago
9

A baseball with a mass of 0.15 kg is moving at a speed of 40 m/s. What is

Physics
2 answers:
Afina-wow [57]3 years ago
6 0

Answer:

120 J

Explanation:

KE = mv²/2 = (0.15 kg * [40 m/s]²)/2 = 120 J

aev [14]3 years ago
3 0

Answer:

120 J

Explanation:

KE = 1/2 mv²

KE = 1/2 (0.15)(40)²

KE = 1/2 (0.15)(1600)

KE = 1/2 (240)

KE = 120 J

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QUICK!! What class of leer is shown below?
wlad13 [49]

Answer:

3rd class lever

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5 0
2 years ago
Read 2 more answers
If someone looks far enough into space, they should be able to see the beginning of the universe true of false
kodGreya [7K]
The answer is no, it would be impossible to see the beginning of the universe
8 0
2 years ago
A black hole in the universe is a An empty region in space b- A massive collapsed star c- A moon that is always turned to its da
marusya05 [52]

Answer:

b. a massive collapsed star

Explanation:

A black hole in the universe is nothing but a massive collapsed star. When the size of the star crosses a particular limit it cannot holds its mass and it collapses under it own self. This is called supernova. A black hole is actually a region in space where gravity is so strong that even light cannot escape through it. Gravity so strong because the matter has been pressed into a tiny space. hence option b is correct

4 0
3 years ago
What is the energy in joules of a photon with a frequency of 3.16e 12 s-1?
erica [24]
We have: Energy(E) = Planck's constant(h) × Frequency(∨)
Here, Planck's constant(h) = 6.626 × 10⁻³⁴ J/s
Frequency (∨) = 3.16 × 10¹² /s

Substitute the values into the expression:
E = (6.626 × 10⁻³⁴)(3.16 × 10¹²) J
E = 2.093 × 10⁻²¹ Joules

In short, Your Final answer would be 2.093 × 10⁻²¹ J

Hope this helps!
5 0
3 years ago
Read 2 more answers
A circular parallel plate capacitor is constructed with a radius of 0.52 mm, a plate separation of 0.013 mm, and filled with an
olya-2409 [2.1K]

Answer:

289282

Explanation:

r = Radius of plate = 0.52 mm

d = Plate separation = 0.013 mm

A = Area = \pi r^2

V = Potential applied = 2 mV

k = Dielectric constant = 40

\epsilon_0 = Electric constant = 8.854\times 10^{-12}\ \text{F/m}

Capacitance is given by

C=\dfrac{k\epsilon_0A}{d}

Charge is given by

Q=CV\\\Rightarrow Q=\dfrac{k\epsilon_0AV}{d}\\\Rightarrow Q=\dfrac{40\times 8.854\times 10^{-12}\times\pi \times (0.52\times 10^{-3})^2\times 2\times 10^{-3}}{0.013\times 10^{-3}}\\\Rightarrow Q=4.6285\times 10^{-14}\ \text{C}

Number of electron is given by

n=\dfrac{Q}{e}\\\Rightarrow n=\dfrac{4.6285\times 10^{-14}}{1.6\times10^{-19}}\\\Rightarrow n=289281.25\ \text{electrons}

The number of charge carriers that will accumulate on this capacitor is approximately 289282.

6 0
2 years ago
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