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2 years ago

7

A student compresses the spring in a pop up toy .020 meter if the sprinf has a spring constant of 340 newtons per meter how much

energy is being stored in the spring

1 answer:

vodomira [7]2 years ago

5 0

The answer is letter c-0.068 j

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Hacks in real life are tricks to help solve a problem with a simple solution.

Say for example, add a pinch of salt to whip egg white faster.

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2 years ago

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How does the density of fluid affect the magnitude of buoyancy acting on an object immersed in it

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3 years ago

A battery has a potential resistance of 12 V and a current of 1280 mA. What is the resistance? PLS HELP ME ASAPPPPPPPPPP

Romashka-Z-Leto [24]

Resistance (R) = <u>9.375 ohm (Ω)</u>

Steps:

$R = \frac{V}{I}$

$= \frac{12 \: volt}{1280 \: milliampere}$

$= \frac{12 \: volt}{1.28 \: ampare}$

$= 9.375 \: ohm (Ω)$

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2 years ago

An open train car moves with speed 18.5 m/s on a flat frictionless railroad track, with no engine pulling the car. It begins to

olga2289 [7]

Answer:

The speed decreases.

Explanation:

This can be explained using the conservation of linear momentum.

Since there is no friction, the initial moment of the train must be equal to its linear moment after it is filled with water.

the initial linear momentum is

$m_{1}v_{1}$

where $m_{1}$ is the initial mass of the train, and $v_{1}$ the initial speed of the train.

And linear momentum after the water filled the train car is

$m_{2}v_{2}$

where $m_{2}$ is mass of the train after the rain, and $v_{2}$ the speed of the train after the rain

<u>the equality must be fulfilled:</u>

$m_{1}v_{1}=m_{2}v_{2}$

We know that if water is added to the train, $m_{2}$ that is the mass after the water is added, is greater than $m_{1}$ which is the mass of the train without the water.

Therefore, in order for the conservation of the linear momentum to be fulfilled: $m_{1}v_{1}=m_{2}v_{2}$

the speed after the water is added ( $v_{2}$ ) must be smaller than the initial train speed ( $v_{1}$ ) . So the speed of the car decreases.

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2 years ago

An airliner arrives at the terminal, and its engines are shut off. The rotor of one of the engines has an initial clockwise angu

Ilia_Sergeevich [38]

(a) 1200 rad/s

The angular acceleration of the rotor is given by:

$\alpha = \frac{\omega_f - \omega_i}{t}$

where we have

$\alpha = -80.0 rad/s^2$ is the angular acceleration (negative since the rotor is slowing down)

$\omega_f$ is the final angular speed

$\omega_i = 2000 rad/s$ is the initial angular speed

t = 10.0 s is the time interval

Solving for $\omega_f$ , we find the final angular speed after 10.0 s:

$\omega_f = \omega_i + \alpha t = 2000 rad/s + (-80.0 rad/s^2)(10.0 s)=1200 rad/s$

(b) 25 s

We can calculate the time needed for the rotor to come to rest, by using again the same formula:

$\alpha = \frac{\omega_f - \omega_i}{t}$

If we re-arrange it for t, we get:

$t = \frac{\omega_f - \omega_i}{\alpha}$

where here we have

$\omega_i = 2000 rad/s$ is the initial angular speed

$\omega_f=0$ is the final angular speed

$\alpha = -80.0 rad/s^2$ is the angular acceleration

Solving the equation,

$t=\frac{0-2000 rad/s}{-80.0 rad/s^2}=25 s$

6 0

2 years ago