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3 years ago

A father uses a rope to pull a child on a sled at a constant speed. of the ones listed which forces are present?

Physics

1 answer:

Amanda [17]3 years ago

4 0

All the forces may be present.

The child on the sled has a weight- which is due to the force of gravity on the child and the sled.

The sled and the child exert a force on the ground equal to the combined weight of the sled and the child. The ground exerts a normal force on the sled.

The force used by the father to pull the sled is the applied force.

The sled slides on the ground and as a result, the force of friction exists between the ground and the sled, directed opposite to the direction of the motion of the sled.

The father pulls the sled using a rope. As a result, the rope is under Tension.

As the sled moves it also experiences a force of air resistance, which is dependent on the sled's speed.

However, since the father pulls the sled along with constant speed, the sum of all the forces acting on the sled is zero.

Since there is no movement in the upward or downward directions, the weight of the child and the sled is equal to the normal force acted upon the sled by the ground.

The force applied by the father on the rope is equal to the tension in the rope.

Friction and air resistance act opposite to the direction of motion of the sled. for the sled to move at constant speed, the tension in the rope must be equal to the sum of the forces due to friction and air resistance.

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A small object of mass 3.82 g and charge -16.5 µC is suspended motionless above the ground when immersed in a uniform electric f

horrorfan [7]

Answer:

2271.16N/C upward

Explanation:

The diagram well illustrate all the forces acting on the mass. The weight is acting downward and the force is acting upward in other to balance the weight.since the question says it is motionless, then indeed the forces are balanced.

First we determine the downward weight using

$W=mg\\g=9.81m/s^{2}$

Hence for a mass of 3.82g 0r 0.00382kg we have the weight to be

$W=0.00382kg*9.81m/s^{2}$

$W=0.0375N$

To calculate the electric field,

$E=f/q\\E=0.0375/16.5*10^{-6} \\E=2271.16N/C$

Since the charge on the mass is negative, in order to generate upward force, there must be a like charge below it that is repelling it, Hebce we can conclude that the electric field lines are upward.

Hence the magnitude of the electric force is 2271.16N/C and the direction is upward

4 0

3 years ago

Suppose the magnitude of the proton charge differs from the magnitude of the electron charge by a mere 1 part in 109

Oksana_A [137]

Answer:

A) F = 1.09 10 5 N, b) Yes

Explanation:

Part A

For this exercise we need the number of free electrons in copper, as the valence of copper +1 there is a free electron for each atom. Let's use the concept of density to find the mass of copper in the sphere

ρ = m / V

.m = ρ V = ρ 4/3 π r³

The radius is half the diameter

r = 1.9 10⁻² / 2 = 0.95 10⁻² m

ρ = 8960 kg / m3

m = 8960 4/3 π (0.95 10⁻²)³

m = 3.2179 10⁻² kg

The molecular weight of copper is 63,546 g / mol which has 6,022 10²³ atoms

With this we can use a rule of proportions to enter the number of atom is this mass

#_atom = 6.022 10²³ 3.2179 10⁻² / 63.546 10⁻³

#_atom = 3,049 10²³

Therefore there is the same number of electrons, as they indicate that the charge of the protone and the electon differs by 1/10⁹ the total charge for each spherical is

q = e / 10⁹ #_atom

q = e / 10⁹ 3,049 1023

q = 3,049 10⁴ (-1.6 10⁻¹⁹)

q = -4,878 10-5 C

Electric force is

F = k q₁q₂ / r²

F = k q² / r²

Let's calculate

F = 8.99 10⁹ (4.878 10⁻⁵)²2 / (1.4 10⁻²)²

F = 1.09 10 5 N

This is a force of repulsion.

Part B