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Feliz [49]
3 years ago
14

Object A has a mass of 10 kg and is moving at a velocity of 2.0

Physics
1 answer:
luda_lava [24]3 years ago
4 0

Answer:20kgm/s

Explanation:

Mass=10kg

Velocity=2m/s

Momentum=mass x velocity

Momentum=10 x 2

Momentum=20

Momentum=20kgm/s

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A force of 10 N causes a spring to extend by 20 mm. Find a) the spring constant of the spring in N/m b) the extension of the spr
suter [353]

(a) The spring constant is 500 N/m.

(b) The extension of the spring when 25 N force is applied is 0.05 m.

(c) The applied force to cause an extension of 5 mm is 2.5 N.

The given parameters:

  • Applied force, F = 10 N
  • Extension of the spring, x = 20 mm

The spring constant is calculated as follows;

F = kx\\\\k = \frac{F}{x} \\\\k = \frac{10}{20 \times 10^{-3}} \\\\k = 500 \ N/m

The extension of the spring when 25 N force is applied is calculated as follows;

F = kx\\\\x = \frac{F}{k} \\\\x = \frac{25}{500} \\\\x = 0.05 \ m

The applied force to cause an extension of 5 mm is calculated as follows;

F = kx\\\\F = 500 \times 5 \times 10^{-3}\\\\F = 2.5 \ N

Learn more about Hook's law here: brainly.com/question/12253978

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2 years ago
What is the purpose of oil used in a car's engine?
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B) to reduce friction
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Thermal energy transfers from a cup of tea at 350 K to the hand holding it. <br> A)True<br> B)False
Alexandra [31]

Answer:

true

Explanation:

it is an energy transfer

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WHAT IS THE DENSTY OF ALL THE LAYER IN THE ATMOSPHERE
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The troposphere starts at the Earth's surface and extends 8 to 14.5 kilometers high (5 to 9 miles). This part of the atmosphere is the most dense. Almost all weather is in this region.

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8 0
3 years ago
Read 2 more answers
A capacitor with initial charge q0 is discharged through a resistor. a) In terms of the time constant τ, how long is required fo
-BARSIC- [3]

Answer:

It would take \tau(\ln 9 - \ln 8) time for the capacitor to discharge from q_0 to \displaystyle \frac{8}{9} \, q_0.

It would take \tau(\ln 9 - \ln 7) time for the capacitor to discharge from q_0 to \displaystyle \frac{7}{9}\, q_0.

Note that \ln 9 = 2\,\ln 3, and that\ln 8 = 3\, \ln 2.

Explanation:

In an RC circuit, a capacitor is connected directly to a resistor. Let the time constant of this circuit is \tau, and the initial charge of the capacitor be q_0. Then at time t, the charge stored in the capacitor would be:

\displaystyle q(t) = q_0 \, e^{-t / \tau}.

<h3>a)</h3>

\displaystyle q(t) = \left(1 - \frac{1}{9}\right) \, q_0 = \frac{8}{9}\, q_0.

Apply the equation \displaystyle q(t) = q_0 \, e^{-t / \tau}:

\displaystyle \frac{8}{9}\, q_0 = q_0 \, e^{-t/\tau}.

The goal is to solve for t in terms of \tau. Rearrange the equation:

\displaystyle e^{-t/\tau} = \frac{8}{9}.

Take the natural logarithm of both sides:

\displaystyle \ln\, e^{-t/\tau} = \ln \frac{8}{9}.

\displaystyle -\frac{t}{\tau} = \ln 8 - \ln 9.

t = - \tau \, \left(\ln 8 - \ln 9\right) = \tau(\ln 9 - \ln 8).

<h3>b)</h3>

\displaystyle q(t) = \left(1 - \frac{1}{9}\right) \, q_0 = \frac{7}{9}\, q_0.

Apply the equation \displaystyle q(t) = q_0 \, e^{-t / \tau}:

\displaystyle \frac{7}{9}\, q_0 = q_0 \, e^{-t/\tau}.

The goal is to solve for t in terms of \tau. Rearrange the equation:

\displaystyle e^{-t/\tau} = \frac{7}{9}.

Take the natural logarithm of both sides:

\displaystyle \ln\, e^{-t/\tau} = \ln \frac{7}{9}.

\displaystyle -\frac{t}{\tau} = \ln 7 - \ln 9.

t = - \tau \, \left(\ln 7 - \ln 9\right) = \tau(\ln 9 - \ln 7).

7 0
3 years ago
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