In an earthquake the point where a fault first slips is called the ?

Compared to the mass and charge of a proton, the electron has

maxonik [38]

B. opposite charge and smaller mass

8 0

3 years ago

An 85.0-N crate of apples sits at rest on a ramp that runs from the ground to the bed of a truck. The ramp is inclined at 28.0°

irina [24]

Answer:

A 75.1 N and a direction of 152° to the vertical.

B 85.0 N at 0° to the vertical.

Explanation:

A) The interaction partner of this normal force has what magnitude and direction?

The interaction partner of this normal force is the component of the weight of the crate perpendicular to the ramp. It has a magnitude of 85cos28° = 75.1 N and a direction of 180° - 28° = 152° to the vertical(since it is directed downwards perpendicular to the ramp).

B) The normal and frictional forces are perpendicular components of the contact force exerted on the crate by the ramp. What is the magnitude and direction of the contact force?

Since this force has to balance the weight of the crate, its magnitude is 85.0 N. Its direction has to be vertically opposite to that of the weight.

Since the weight is 180° to the vertical (since it is directed downwards), this force is 0° to the vertical.

So, this force has a magnitude of 85.0 N and a direction of 0° to the vertical.

8 0

2 years ago

Machines A, B, and C do the same amount of work, but they each take a different amount of time. Machine A takes more time as com

Amiraneli [1.4K]

So power is considered as the rate of doing work. Base on the problem given, my analysis is that the machine who finish the work faster is machine C. Therefore, in order to finish the same amount of work in a short period of time you are going to expend the most power. My answer is Machine C.

6 0

2 years ago

Read 2 more answers

What is the daily use of the deltoid?

Ghella [55]

Answer:

In our everyday life, the deltoids take on most of the work in rotating our arms, but besides that the deltoid muscle also allows us to carry objects at a safe distance from the body. It is also responsible from stopping a dislocation or injury to the humerus when we carry something heavy.

6 0

2 years ago

Read 2 more answers

Suppose you have a sphere of radius 1.00 m that carries a charge of +1.00 µC at its center. The radius of the sphere is changed

Gekata [30.6K]

Answer:

The flux remains the same and the field increases.

Explanation:

According to the Gauss' law, the surface integral of the electric field over a closed surface, called the Gaussian surface, is equal to $\dfrac{1}{\epsilon_o}$ times the net charge $q$ enclosed by the surface.

$\oint \vec E \cdot d\vec A=\dfrac{q}{\epsilon_o}\ \ \ \ ............\ (1).$

where,

$\epsilon_o$ = electrical permititivity of free space.

The term on the LHS of Gauss law' $\oint \vec E \cdot d\vec A$ is the electric flux $\phi$ through the Gaussian surface.

Therefore,

$\phi=\dfrac{q}{\epsilon_o}\ \ \ .....................\ (2).$

From equation (2), the electric flux through the sphere depends only on the charge enclosed by the sphere and not on the size of the sphere, therefore, when the radius of the sphere is changed to 0.500 m from 1.00 m, the electric flux through the surface will not change.

Consider the Gaussian sphere of radius $r$ same as that of the given sphere and concentric with the given sphere, then,

$\oint \vec E \cdot d\vec A=\oint E\ dA$

This is because the direction of the electric field through a Gaussian spherical surface and its surface area element $d\vec A$ is always normal to it.

Therefore,

$\oint \vec E \cdot d\vec A = E\oint dA=E\ 4\pi r^2$

Using equation (1),

$E\ 4\pi r^2=\dfrac{q}{\epsilon_o}\\E=\dfrac{q}{4\pi r^2 \epsilon_o}\\E\propto \dfrac{q}{r^2}$

The electric field is inversely proportional to the square of the radius of the sphere, therefore, on decreasing the radius from 1.00 m to 0.500 m, the electric field increases.

Thus, the correct answer is "The flux remains the same and the field increases".

7 0

3 years ago