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ololo11 [35]
4 years ago
14

g A mass of 2 kg is attached to a spring whose constant is 7 N/m. The mass is initially released from a point 4 m above the equi

librium position with a downward velocity of 10 m/s, and the subsequent motion takes place in a medium that imparts a damping force numerically equal to 10 times the instantaneous velocity. What is the differential equation for the mass-spring system.
Physics
1 answer:
Ann [662]4 years ago
6 0

Answer:

mass 20 times of an amazing and all its motion

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Which statements best describes why science changes?
MrRa [10]

Explanation:

Science changes in so many ways. In my opinion, science changes due to Climatic change.

Hope it helps.

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2 years ago
Can anybody tell me what I'm suppose to do. I click start the numbers comes up to the right ​
Marina CMI [18]
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4 years ago
The magnitude of the force involved in a certain collision (measured in newtons) is equal to the change in momentum (measured in
Vinvika [58]

Answer:

The time-interval of the collision is t=\frac{{\Delta}p}{F}

Explanation:

As given the force is equal to the rate of change of momentum. Mathemticaly this is:

F=\frac{dp}{dt}.

We can rearrange this equation to solve for {\delta}t which gives

t=\frac{{\Delta}p}{F}

which is our answer.

In words this means the time interval is equal to the momentum change in that interval divided by the force applied that caused this momentum change.

4 0
3 years ago
Calcular la aceleración que produce una fuerza de 40 N sobre un cuerpo con 88 Kg de masa. Expresar el resultado en m⁄s^2 *
tigry1 [53]

Answer:

a = 0.45 m/s²

Explanation:

The given question is ''Calculate the acceleration that produces a force of 40 N on a body with 88 kg of mass".

Given that,

Force, F = 40 N

Mass of the body, m = 88 kg

The net force acting on the body is given by :

F = ma

Where

a is the acceleration of the body

a=\dfrac{F}{m}\\\\a=\dfrac{40\ N}{88\ kg}\\\\a=0.45\ m/s^2

So, the required acceleration is 0.45 m/s².

4 0
3 years ago
Suppose you have two identical capacitors. You connect the first capacitor to a battery that has a voltage of 21.2 volts, and yo
HACTEHA [7]

Answer:

r=2.743

Explanation:

The energy stored on a capacitor is of type potencial, therfore depends on the capacity to "store" energy. Inthe case of the capacitor, it stores charge (Q), and the equations you use to calculate it are:

E_p=\frac{Q^2}{2C}=\frac{QV}{2}=\frac{CV^2}{2}

In this case we know V and C, therefore we use the last expression:

E_{p1}=\frac{CV_1^2}{2}

E_{p2}=\frac{CV_2^2}{2}

\frac{E_{p1}}{E_{p2}}=r=\frac{\frac{CV_1^2}{2}}{\frac{CV_2^2}{2}}  \\r=(\frac{V_1}{V_2})^2\\r=(\frac{21.2}{12.8})^2

r=2.743

3 0
3 years ago
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