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Mumz [18]
2 years ago
15

What are the units for specific charge? A. C/kg B. C*kg C. kg/C D. 1/C*kg

Physics
1 answer:
Stolb23 [73]2 years ago
6 0

Answer:

Option A

Explanation:

Your answer is the first option or "C/kg." C/kg is a ratio for specific charge. C for the charge of the electron, kg for the mass of the electron. To find the specific charge we divide the two, which usually involves scientific notations and simplifying.

<u>Example:</u>
\frac{C}{kg}
\frac{3.4\times10^3C}{2.7\times10^5kg}

Hope this helps.

You might be interested in
The speed of a box traveling on a horizontal friction surface changes from vi = 13 m/s to vf = 11.5 m/s in a distance of d = 8.5
KiRa [710]

Answer:

0.68 s

Explanation:

We are given that

Initial velocity of box=u=13m/s

Final velocity of box=v=11.5 m/s

Distance=d=8.5 m

We have to find the time taken by box to slow by this amount.

We know that

v^2-u^2=2as

Substitute the values

(11.5)^2-(13)^2=2a(8.5)

132.25-169=17a

-36.75=17a

a=\frac{-36.75}{17}=-2.2m/s^2

We know that

Acceleration=a=\frac{v-u}{t}

Substitute the values

-2.2=\frac{11.5-13}{t}

-2.2=\frac{-1.5}{t}

t=\frac{1.5}{2.2}=0.68 s

Hence, the time taken by box to slow by this amount=0.68 s

8 0
3 years ago
8. The legs of a young man are each 0.650 meters long. What is his maximum walking speed?
Norma-Jean [14]

Answer:

2.52 m/s

Explanation:

When the man takes a step, his foot is stationary while his body revolves around it.  At the point when his body is directly above his foot, there will be no normal force at his maximum speed.

Sum of the forces in the radial direction:

∑F = ma

mg = m v² / r

g = v² / r

v = √(gr)

Given that r = 0.650 m:

v = √(9.8 m/s² × 0.650 m)

v = 2.52 m/s

8 0
3 years ago
What is the relationship between balancing equations and the law of conservation of matter
harkovskaia [24]
Every chemical equation adheres to the law of conservation of mass, which states that matter cannot be created or destroyed. Therefore, there must be the same number of atoms of each element on each side of a chemical equation.
7 0
2 years ago
to move a resting box of 100 Newton on the ground with kinetic friction coeficient of 0,250 is applied a force of 60 N horizonta
krok68 [10]
Work is calculated by multiplying force by the distance that the object had moved. The applied force is 60 N, moving the object by 10 m. Thus, the work does is 600 J. For the friction force which is equal to,
                          100N x 0.250 = 25.0 N
the work done is,
                        W = (60 N - 25 N) x 10 m = 350 J
The kinetic energy of the box can be equated to this force. Thus, the answer is also 350 J. 
6 0
3 years ago
A wedge with an inclination of angle θ rests next to a wall. A block of mass m is sliding down the plane. There is no friction b
Softa [21]

Answer:

  The net force on the block  F(net)  = mgsinθ).

   Fw =mg(cosθ)(sinθ)

Explanation:

(a)

Here, m is the mass of the block, n is the normal force, \thetaθ is the wedge angle, and Fw  is the force exerted by the wall on the wedge.

Since the block sliding down, the net force on the block is along the plane of the wedge that is equal to horizontal component of weight of the block.

                    F(net)  = mgsinθ

The net force on the block  F(net)  = mgsinθ).

The direction of motion of the block is along the direction of net force acting on the block. Since there is no frictional force between the wedge and block, the only force acting on the block along the direction of motion is mgsinθ.

(b)

From the free body diagram, the normal force n is equal to mgcosθ .

                           n=mgcosθ

The horizontal component of normal force on the block is equal to force

                           Fw=n*sin(θ) that exerted by the wall on the wedge.

Substitute mgcosθ for n in the above equation;

                           Fw =mg(cosθ)(sinθ)

Since, there is no friction between the wedge and the wall, there is component force acting on the wall to restrict the motion of the wedge on the surface and that force is arises from the horizontal component for normal force on the block.

6 0
3 years ago
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