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3 years ago

A 1kg ball is dropped (from rest) 100m onto a spring with spring constant 125N/m. How much does the spring compress?

Physics

1 answer:

Anastasy [175]3 years ago

7 0

Answer:

3.95 m

Explanation:

m = 1 kg, h = 100 m, k = 125 N/m

Let the spring is compressed by y.

Use the conservation of energy

potential energy of the mass is equal to the energy stored in the spring

m x g x h = 1/2 x ky^2

1 x 9.8 x 100 = 0.5 x 125 x y^2

y^2 = 15.68

y = 3.95 m

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svetoff [14.1K]

Answer:

F(Mars) = 2 G m M / (4 R)^2 force of Sun on Mars

F(Merc) = G m M / R^2 force of force of Sun on Mercury

R = distance of Sun from Mercury, m = mass of Mercury

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2 years ago

4. A car accelerates, from rest, at 2.4 m/s?. How fast is the car traveling

pantera1 [17]

Answer:

19.2m/s

Explanation:

Assuming that 2.4m/s^2 was the acceleration and not a typo, we can use the equation v=at, where v=velocity, a=acceleration, and t=time,

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Part complete How long must a simple pendulum be if it is to make exactly one swing per five seconds?

shtirl [24]

Answer:

$L=6.21m$

Explanation:

For the simple pendulum problem we need to remember that:

$\frac{d^{2}\theta}{dt^{2}}+\frac{g}{L}sin(\theta)=0$ ,

where $\theta$ is the angular position, t is time, g is the gravity, and L is the length of the pendulum. We also need to remember that there is a relationship between the angular frequency and the length of the pendulum:

$\omega^{2}=\frac{g}{L}$ ,

where $\omega$ is the angular frequency.

There is also an equation that relates the oscillation period and the angular frequeny:

$\omega=\frac{2\pi}{T}$ ,

where T is the oscillation period. Now, we can easily solve for L:

$(\frac{2\pi}{T})^{2}=\frac{g}{L}\\\\L=g(\frac{T}{2\pi})^{2}\\\\L=9.8(\frac{5}{2\pi})^{2}\\\\L=6.21m$

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3 years ago

How is a projectile different from an object in free fall

aksik [14]

Answer:

Explanation:

Projectile Motion. Projectile motion is different than free fall: it involves two dimensions instead of one. ... Balls traveling in two dimensions, only one of which experiences acceleration, require two sets of equations: one set for the x-direction and the other for the y-direction.

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3 years ago

There is a uniform magnetic field of magnitude B, pervading all space, perpendicular to the plane of rod and rails. The rod is r

Charra [1.4K]

The right hand rule to find the direction of the magnetic field for a falling bar is:

The charge is positive the magnetic field is outgoing, horizontally and towards us.

The charge of the bar is negative, the magnetic field is incoming, that is horizontal away from us.

The magnetic force is given by the vector product of the velocity and the magnetic field.

F = q v x B

Where the bolds indicate vectors, F is the force, q the charge on the particle, v the velocity and B the magnetic field.

In the vector product, the vectors are perpendicular, which is why the right-hand rule has been established, see attached:

The thumb points in the direction of speed.

Fingers extended in the direction of the magnetic field.

The palm is in the direction of the force if the charge is positive and in the opposite direction if the charge is negative.

They indicate that the bar is dropped, therefore its speed is vertical and downwards, it moves to the left therefore this is the direction of the force, we use the right hand rule, the magnetic field must be horizontal, we have two possibilities:

If the charge is positive the magnetic field is outgoing, horizontally and towards us.

If the charge of the bar is negative, the magnetic field is incoming, that is, horizontal away from us

In conclusion using the right hand rule we can find the direction of the magnetic field for a falling bar is:

The charge of the bar is negative, the magnetic field is incoming, that is horizontal away from us.

The charge is positive the magnetic field is outgoing, horizontally and towards us.

Learn more about the right hand rule here: brainly.com/question/12847190

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2 years ago