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Natali [406]
2 years ago
14

A block of mass M is attached to a spring of negligible mass and can slide on a horizontal surface along the x-direction, as sho

wn above. There is friction present between the block and the surface. The spring exerts no force on the block when the center of the block is located at x=0. At the instant shown above, the block is located at x=0 and moving toward the right with speed v=v0. (a) For the instant shown above, with the block located at x=0 and moving to the right, predict the direction of the net force on the block. If the net force is zero, select “The net force is zero.”
_ To the left _ The net force is zero _ To the right
Briefly justify your prediction.
Physics
1 answer:
Hoochie [10]2 years ago
6 0

Answer:

Because of the frictional force, the net force will oppose direction of the block and be directed towards the left even tho the spring exerts no force at this point

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We are going to focus on the gravitational force between Earth and our Moon. The Earth has a mass of about 6 x 10/24 kg and the
Basile [38]

Answer

Gravity is what holds us down on the earth's (or moon's) surface. If you were to weigh yourself on a scale on Earth and then on the moon, the weight read on the moon would be 1/6 your earth weight

7 0
2 years ago
An inelastic collision of two objects is characterized by the following.
serious [3.7K]

Options:

(a) Total kinetic energy of the system remains constant.

(b) Total momentum of the system is conserved.

(c) Both A and B are true.

(d) Neither A nor B are true.

Answer:

(b) Total momentum of the system is conserved.

Explanation:

An inelastic collision is a type of collision in which momentum is conserved and kinetic energy is not conserved. That is, there is loss of kinetic energy.

In an inelastic collision:

Total momentum before collision = Total momentum after collision

An example of inelastic collision is seen in the ballistic pendulum, The ballistic pendulum is a device in which a projectile such as a bullet is fired into a suspended heavy wooden stationary block.

8 0
3 years ago
A fluid flows along the x axis with a velocity given by V = ( x / t ) ˆ i , where x is in feet and t in seconds. (a) Plot the sp
umka21 [38]

Answer:

c)

 V_local = -x/t^2

 V_convec = x/t^2

d)

a =  V_local +  V_convec = 0

e) When a particle moves towards postive x direction its convective velocity increases, but at the same time the local velocity deacreases (at the same rate) when time increases

Explanation:

Hi!

You can see plots for a) and b) attached on this document

c)

The local acceleration is just teh aprtial derivative of the velocity with respect to t:

\frac{dV}{dt} = \frac{d}{dt} \frac{x}{t}=- \frac{x}{t^2}

And the convective acceleration is given by the product of the velocity times the gradient of the velocity, that is:

\vec{v} \cdot \nabla \vec{v} = v ( \frac{dv}{dx} ) =\frac{x}{t} \frac{1}{t} = \frac{x}{t^2}

d)

Since the acceleration of any fluid particle is the sum of the local and convective accelerations, we can easily see that it is equal to zero, since they are equal but with opposit sign

e)

This is because of teh particular form of the velocity. A particle will move towards areas of higher velocities (convectice acceleration), but as time increases,  the velocity is also decreasing (local acceleration), and the sum of these quantities adds up to zero

3 0
3 years ago
A 30.0 kg child, initially at rest, slides down a 2.0 m tall slide. The child reaches the bottom of the slide with a speed of 6
HACTEHA [7]

Answer:

Explanation:

Total energy is constant

E = mgh + ½mv² + Fd

At the top of the slide, all energy is potential

E = mgh + 0 + 0

At the bottom of the slide, all potential energy has converted to kinetic and work of friction.

mgh = ½mv² + W

W = mgh - ½mv²

W = 30.0[(9.81)(2.0) - ½6²]

W = 48.6 J

5 0
2 years ago
A hollow, uniformly charged sphere has an inner radius of r1 = 0.105 m and an outer radius of r2 = 0.31 m. The sphere has a net
Sliva [168]

Answer:

E = 77532.42N/C

Explanation:

In order to find the magnitude of the electric field for a point that is in between the inner radius and outer radius, you take into account the Gauss' law for the electric flux trough a spherical surface with radius r:

\int E\cdot dS=\frac{Q}{\epsilon_o}       (1)

Q: net charge of the hollow sphere = 1.9*10-6C

ε0: dielectric permittivity of vacuum = 8.85*10^-12 C^2/Nm^2

Furthermore, you have that the net charge contained in a sphere of radius r is:

Q=\rho V=\rho \frac{4\pi (r^3-r_1^3)}{3}      (2)

with the charge density is:

\rho=\frac{Q}{\frac{4}{3}\pi(r_2^3-r_1^3)}          (3)

r2: outer radius = 0.31m

r1: inner radius = 0.105m

The electric field trough the Gaussian surface is parallel to the normal to the surface, the, you have in the integral of the equation (1):

\int E\cdot dS=E(4\pi r^2)      (4)

where you have used the expression for a surface of a sphere.

Next, you replace the expressions of equations (2), (3) and (4) in the equation (1) and solve for E:

E(4\pi r^2)=\frac{1}{\epsilon_o}\frac{Q}{\frac{4}{3}\pi(r_2^3-r_1^3)}(\frac{4\pi (r^3-r_1^3)}{3})\\\\E=\frac{1}{\epsilon_o}\frac{Q(r^3-r_1^3)}{4\pi r^2(r_2^3-r_1^3)}

you replace the values of all parameters, and with r = 0.17m

E=\frac{(1.6*10^{-6}C)((0.17m)^3-(0.105m)^3)}{4\pi(8.85*10^{-12}C^2/Nm^2)(0.17m)^2((0.31m)^3-(0.105m)^3)}\\\\E=77532.42\frac{N}{C}

The magnitude of the electric field at a distance r=0.17m to the center of the hollow sphere is 77532.42N/C

5 0
3 years ago
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