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kondor19780726 [428]
3 years ago
7

A particle could have a charge of

Physics
1 answer:
tekilochka [14]3 years ago
7 0
The correct answer is 
<span> (3) 3.2 × 10^−19 C

In fact, electric charge is quantized: the elementary charge is the charge of the electron, </span>e=1.6 \cdot 10^{-19} C, and every particle can only have an electric charge which is an integer multiple of this value. Of the options listed above, only option (3) is an integer multiple of the elementary charge: in fact, it corresponds to 2 times the electron charge:
2e=2 \cdot 1.6 \cdot 10^{-19} C=3.2 \cdot 10^{-19} C
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A car goes around in a circular curve on a horizontal road at constant speed. what is the direction of the friction force on the
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3 years ago
In the graph above, what is the instantaneous speed of the object after the first five seconds?
alexdok [17]

Answer:

D) 25 m/s

Explanation:

In order to solve this problem we must use the following kinematics equation.

v_{f} =v_{i} +(a*t)\\

where:

Vf = final speed [m/s]

Vi = initial speed = 0

a = acceleration = 5[m/s^2]

t = time = 5[s]

After 5 seconds the acceleration is equal to 5 [m/s^2]

Now replacing the values in the equation:

Vf = 0 + (5*5)

Vf = 25[m/s]

8 0
3 years ago
The tension in the rope is constant and equal to 40 N as the block is pulled. What is the instantaneous power (in W) supplied by
Eduardwww [97]

Complete Question:

A 10 kg block is pulled across a horizontal surface by a rope that is oriented at 60° relative to the horizontal surface.

The tension in the rope is constant and equal to 40 N as the block is pulled. What is the instantaneous power (in W) supplied by the tension in the rope if the block when the block is 5 m away from its starting point? The coefficient of kinetic friction between the block and the floor is 0.2 and you may assume that the block starting at rest.

Answer:

Power = 54.07 W

Explanation:

Mass of the block = 10 kg

Angle made with the horizontal, θ = 60°

Distance covered, d = 5 m

Tension in the rope, T = 40 N

Coefficient of kinetic friction, \mu = 0.2

Let the Normal reaction = N

The weight of the block acting downwards = mg

The vertical resolution of the 40 N force, f_{y} = 40sin \theta

\sum f(y) = 0

N + 40 sin \theta - mg = 0\\N = -40sin60 + 10*9.81 = 0\\N = 63.46 N

\sum f(x) = 0

40 cos 60 - f_{r} - ma = 0\\ f_{r} = \mu N\\ f_{r} = 0.2 * 63.46\\ f_{r} = 12.69 N\\40cos 60 - 12.69-10a = 0\\7.31 = 10a\\a = 0.731 m/s^{2}

v^{2}  = u^{2} + 2as\\u = 0 m/s\\v^{2}  =  2 * 0.731 * 5\\v^{2}  = 7.31\\v = \sqrt{7.31} \\v = 2.704 m/s

Power, P = Fvcos \theta

P = 40 *2.704 cos60\\P = 54.074 W

7 0
3 years ago
A 1,000 kg car is travelling at 6.5 m/s to the North. A 3,500 kg truck is travelling South at the same velocity. What is the tot
Molodets [167]

Answer:

16250 kgm/s due south

Explanation:

Applying,

M = mv................. Equation 1

Where M = momentum, m = mass, v = velocity.

From the car,

Given: m = 1000 kg, v = 6.5 m/s

Substitute these values into equation 1

M = 1000(6.5)

M = 6500 kgm/s

For the truck,

Given: m = 3500 kg, v = 6.5 m/s

Substitute these values into equation 1

M' = 3500(6.5)

M' = 22750 kgm/s.

Assuming South to be negative direction,

From the question,

Total momentum of the two vehicles = (6500-22750)

Total momentum of the two vehicles = -16250 kgm/s

Hence the total momentum of the two vehicles is 16250 kgm/s due south

3 0
3 years ago
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