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3 years ago

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Ground reaction force acting on carter

1 answer:

mezya [45]3 years ago

8 0

We don't know Carter, and we don't know where he is or what
he's doing, so I'm taking a big chance speculating on an answer.

I'm going to say that if Carter is pretty much just standing there,
or, let's say, lying on the ground taking a nap, then the force of
the ground acting on him is precisely exactly equal to his weight.

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This is a form of energy representing the motion of the molecules which make up an object. A. Thermal Energy B. Kinetic Energy C

bija089 [108]

Answer:

Kinetic energy.

Explanation:

There are many kinds of energy. Some of them are kinetic energy, potential energy, thermal energy etc.
The energy that shows the motion of the object is called its kinetic energy.
Also, the sum of kinetic energy and the gravitational potential energy is called mechanical energy.
Out of the given options, kinetic energy is the form of energy that represents the motion of the molecules which make up an object.
Hence, the correct option is (B).

4 0

3 years ago

by how many times occur in the force of attraction between two bodies change when the distance between then is reduced to one th

xenn [34]

Answer:

<em>The force is now 9 times the original force</em>

Explanation:

<u>Coulomb's Law </u>

The electrostatic force between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

Coulomb's formula is:

$\displaystyle F=k\frac{q_1q_2}{d^2}$

Where:

$k=9\cdot 10^9\ N.m^2/c^2$

q1, q2 = the particles' charge

d= The distance between the particles

Suppose the distance is reduced to d'=d/3, the new force F' is:

$\displaystyle F'=k\frac{q_1q_2}{\left(\frac{d}{3}\right)^2}$

$\displaystyle F'=k\frac{q_1q_2}{\frac{d^2}{9}}$

$\displaystyle F'=9k\frac{q_1q_2}{d^2}$

$\displaystyle F'=9F$

The force is now 9 times the original force

8 0

2 years ago

I. What is the initial velocity of the car?

qaws [65]

Answer:

I. 0 m/s

II. 20 m/s

III. Part BC

Explanation:

I. Determination of the initial velocity.

From the diagram given above,

The motion of the car starts from the origin. This implies that the car start from rest and as such, the initial velocity of the car is 0 m/s

II. Determination of the maximum velocity attained.

From the diagram given above, we can see clearly that the maximum velocity is 20 m/s.

III. Determination of the part of the graph that represents zero acceleration.

It important that we know the meaning of zero acceleration.

Zero acceleration simply means the car is not accelerating. This can only be true when the car is moving with a constant velocity.

From the graph given above, the car has a constant velocity between B and C.

Therefore, part BC illustrates zero acceleration.

6 0

3 years ago

To increase the current in a circuit, which can be increased?

wariber [46]

Answer:

A. Voltage

Explanation:

I got a 100% on the quiz on edg so trust me

4 0

3 years ago

An electron is released from rest at the negative plate of a parallel plate capacitor. The charge per unit area on each plate is

dmitriy555 [2]

Answer:

v = 1.15*10^{7} m/s

Explanation:

given data:

charge/ unit area $= \sigma = 1.99*10^{-7} C/m^2$

plate seperation = 1.69*10^{-2} m

we know that

electric field btwn the plates is $E = \frac{\sigma}{\epsilon}$

force acting on charge is F = q E

Work done by charge q id $\Delta X =\frac{ q\sigma \Delta x}{\epsilon}$

this work done is converted into kinectic enerrgy

$\frac{1}{2}mv^2 =\frac{ q\sigma \Delta x}{\epsilon}$

solving for v

$v = \sqrt{\frac{2q\Delta x}{\epsilon m}$

$\epsilon = 8.85*10^{-12} Nm2/C2$

$v = \sqrt{\frac{2 1.6*10^{-19}1.99*10^{-7}*1.69*10^{-2}}{8.85*10^{-12} *9.1*10^{-31}}$

v = 1.15*10^{7} m/s

8 0

3 years ago