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Alexeev081 [22]
2 years ago
9

Date: 12-3-21 4. The momentum of a 5-kilogram object moving at 6 meters per second is

Physics
2 answers:
Dafna11 [192]2 years ago
6 0
30 kg•m/s

p = mv
p = 5kg • 6m/s
p = 30 kg•m/s
ASHA 777 [7]2 years ago
5 0

Answer

30 kg . m/sec

Explanation:

mark me as brainliest!

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You charge an initially uncharged 65.7-mf capacitor through a 39.1-Ï resistor by means of a 9.00-v battery having negligible int
uysha [10]
In a RC-circuit, with the capacitor initially uncharged,  when we connect the battery to the circuit the charge on the capacitor starts to increase following the law:
Q(t) = Q_0 (1-e^{-t/\tau})
where t is the time, Q_0 = CV is the maximum charge on the capacitor at voltage V, and \tau = RC is the time constant of the circuit.
Using this law, we can answer all the three questions of the problem.

1) Using R=39.1 \Omega and C= 65.7 mF=65.7\cdot 10^{-3}F, the time constant of the circuit is:
\tau = RC=(39.1 \Omega)(65.7 \cdot 10^{-3}F)=2.57 s

2) To find the charge on the capacitor at time t=1.95 \tau, we must find before the maximum charge on the capacitor, which is
Q_0 = CV=(65.7 \cdot 10^{-3}F)(9 V)=0.59 C
And then, the charge at time t=1.95 \tau is equal to
Q(1.95 \tau) = Q_0 (1-e^{-t/\tau})=(0.59 C)(1-e^{-1.95})=0.51 C

3) After a long time (let's say much larger than the time constant of the circuit), the capacitor will be fully charged, this means its charge will be Q_0 = 0.59 C. We can see this also from the previous formule, by using t=\infty:
Q(t) = Q_0 (1-e^{-\infty})=Q_0(1-0) = 0.59 C

4 0
3 years ago
Why do stars appear so steady when viewed from the surface of moon or<br> by an astronaut in space?
Nutka1998 [239]

Answer:

Explanation:

The light from these little disks is also refracted by Earth's atmosphere, as it travels toward our eyes. That's because, in the direction of any horizon, you're looking through more atmosphere than when you look overhead. If you could see stars and planets from outer space, both would shine steadily.

Answer From GauthMath If you like it then please heart it and comment thanks.

4 0
3 years ago
Is it possible to be moving but not be in motion?
GREYUIT [131]

No. Motion is the thing that when you're moving, you're in it.

But it IS possible for one person to say you're moving and another person to say you're not moving, both at the same time, and both of them are correct !

4 0
3 years ago
Read 2 more answers
What is the mathematical relationship between wavelength and velocity? Inverse, horizontal, linear or quadratic? I NEED MAJOR HE
Tju [1.3M]

Answer: Wavelength is the measure of the length of a complete wave cycle. The velocity of a wave is the distance traveled by a point on the wave. In general, for any wave the relation between Velocity and Wavelength is proportionate. It is expressed through the wave velocity formula.

Explanation: For any given wave, the product of wavelength and frequency gives the velocity. It is mathematically given by wave velocity formula written as-

V=f×λ

Where,

V is the velocity of the wave measure using m/s.

f is the frequency of the wave measured using Hz.

λ is the wavelength of the wave measured using m. Velocity and Wavelength Relation

Amplitude, Frequency, wavelength, and velocity are the characteristic of a wave. For a constant frequency, the wavelength is directly proportional to velocity.

Given by:

V∝λ

Example:

For a constant frequency, If the wavelength is doubled. The velocity of the wave will also double.

For a constant frequency, If the wavelength is made four times. The velocity of the wave will also be increased by four times.

Hope you understood the relation between wavelength and velocity of a wave. I truely hope this helps you out tho! Goodluck!

5 0
3 years ago
Read 2 more answers
A particle with charge − 2.74 × 10 − 6 C −2.74×10−6 C is released at rest in a region of constant, uniform electric field. Assum
s2008m [1.1K]

Answer:

241.7 s

Explanation:

We are given that

Charge of particle=q=-2.74\times 10^{-6} C

Kinetic energy of particle=K_E=6.65\times 10^{-10} J

Initial time=t_1=6.36 s

Final potential difference=V_2=0.351 V

We have to find the time t after that the particle is released and traveled through a potential difference 0.351 V.

We know that

qV=K.E

Using the formula

2.74\times 10^{-6}V_1=6.65\times 10^{-10} J

V_1=\frac{6.65\times 10^{-10}}{2.74\times 10^{-6}}=2.43\times 10^{-4} V

Initial voltage=V_1=2.43\times 10^{-4} V

\frac{\initial\;voltage}{final\;voltage}=(\frac{initial\;time}{final\;time})^2

Using the formula

\frac{V_1}{V_2}=(\frac{6.36}{t})^2

\frac{2.43\times 10^{-4}}{0.351}=\frac{(6.36)^2}{t^2}

t^2=\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}

t=\sqrt{\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}}

t=241.7 s

Hence, after 241.7 s the particle is released has it traveled through a potential difference of 0.351 V.

6 0
3 years ago
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