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AURORKA [14]
3 years ago
12

A 5 kg ball travels at a velocity of 10 m/s, what is the momentum of the ball

Physics
2 answers:
Sholpan [36]3 years ago
8 0

Answer:

50 kg·m/s.

Explanation:

Momentum can be calculated using p = mv. (Let p represent momentum, m represent mass, and v represent velocity.)

Plugging our terms in, we have p = 5 kg * 10 m/s. This gives us p = 50 kg·m/s as an answer.

Zina [86]3 years ago
4 0
Answer= 59kg•m/s

you do 5kg x 10m/s
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Which planet has the closest gravitational pull to earth
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Answer:

venus

Explanation:

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What would be your estimate of the age of the universe if you measured a value for Hubble's constant of H0 = 30 km/s/Mly ? You c
Maksim231197 [3]

Answer:

The age of the universe would be 9.9 billion years

Explanation:

We can calculate an estimate for the age of the Universe from Hubble's Law. Let's suppose the distance between two galaxies is D and the apparent velocity with which they are separating from each other is v. At some point, the galaxies were touching, and we can consider that time the moment of the Big Bang.

Thus, the time it has taken for the galaxies to reach their current separations is:

\displaystyle{t=D/v}

and from Hubble's Law:

v =H_0D

Therefore:

\displaystyle{t=D/v=D/(H_0\times D)=1/H_0}

With the given value for the Hubble's constant we have:

H_0=(30\ km/s/Mly) \times (1 Mly/ 9.461 \times 10^{18} km) = 3.17\times 10^{-18}\ 1/s

and thus,

t=1/H_0 = 1/(3.17\times 10^{-18} 1/s) = 0.315 \times 10^{18}\ s \approx 9988584474.8858\ years \approx 9.9\ billion\ years

6 0
3 years ago
Let us remember your previous lesson on Physical Press Components! Directions: Analyze the folawna fitness components. Put a che
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1. Encontrar la densidad de una sustancia, Si pesa 459 g y ocupa un volumen de 34.5 cmɜ.
ryzh [129]

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5 0
3 years ago
An LC circuit consists of a 3.4-µF capacitor and a coil with a self-inductance 0.080 H and no appreciable resistance. At t = 0 t
alexira [117]

Answer

given,

capacitance = C = 3.4-µF

inductance = L = 0.08 H

frequency is expressed as

f = \dfrac{1}{2\pi\sqrt{LC}}

time period

T = \dfrac{1}{f}=2\pi\sqrt{LC}

after time T/4 current reach maximum

 t = \dfrac{T}{4}

 t = \dfrac{2\pi\sqrt{LC}}{4}

 t = \dfrac{2\pi\sqrt{0.08 \times 3.4 \times 10^{-6}}}{4}

        t = 8.2 x 10⁻⁴ s

        t = 0.82 ms

b) using law of conservation

  \dfrac{1}{2}CV^2=\dfrac{1}{2}LI^2

  I^2 = \dfrac{CV^2}{L}

  I^2 = \dfrac{C}{L}\dfrac{Q^2}{C^2}

  I =\sqrt{\dfrac{Q^2}{CL}}

  I =\sqrt{\dfrac{(5.4 \times 10^{-6})^2}{0.08 \times 3.4 \times 10^{-6}}}

       I = 0.010 A

       I = 10 mA

4 0
3 years ago
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