Answer:
Option D.
Explanation:
From the question given above, the following data were obtained:
Force applied (F) = 5 N
Extention (e) = 0.075 m
Spring constant (K) =?
The spring constant for the spring can be obtained as follow:
F = Ke
5 = K × 0.075
Divide both side by 0.075
K = 5 / 0.075
K = 67 N/m
Thus, the spring constant for the spring is 67 N/m
Current = charge per second
2 Coulombs per second = 2 Amperes
Potential difference = (current)x(resistance) in volts.
That's (2 Amperes) x (2 ohms).
That's how to do it.
I think you can find the answer now.
The work done by friction to move the sled is - 1,323 J.
<h3>
What is Coefficient of friction?</h3>
- The friction coefficient is the ratio of the normal force pressing two surfaces together to the frictional force preventing motion between them.
- Typically, it is represented by the Greek letter µ. In terms of math, is equal to F/N, where F stands for frictional force and N for normal force.
- The coefficient of friction has no dimensions because both F and N are measured in units of force (such as newtons or pounds). For both static and kinetic friction, the coefficient of friction has a range of values.
- When an object experiences static friction, the frictional force resists any applied force, causing the object to stay at rest until the static frictional force is removed. The frictional force opposes an object's motion in kinetic friction.
Solution:
Given that
Coefficient of friction (µ) = 0.10
Mass (m) = 90kg
distance covered (d) = 30m
We use the formula:
friction work = -µmgdcos∅
friction work = -0.100 × 90 kg × 9.8 m/s² × 30 m × cos 60°
friction work = - 1,323 J
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Answer:
Check Explanation.
Explanation:
(1). The force that will continue to move the sled toward the right and speeding up at steady rate(constant acceleration) = option C => The force is toward the right and is constant strength (magnitude).
(2). The force that will continue to move the sled toward the right and at Constant velocity = option G => No applied force is needed. This is because acceleration = 0,and force is also = 0.
(3). The force that will continue to move at steady rate and an acceleration to the right = option A => The force is toward the right and is decreasing in strength (magnitude).
(4). The force that will slow the sled moving to the right at steady rate(constant acceleration) = option F => The force is toward the left and is of constant strength (magnitude).
(5). The force that moves the sled rest toward the a steady (velocity) towards the left = option G => No applied force is needed.
Since the train travels 30 miles per hour, you can divide the total distance by the miles per hour to get the time, so it takes 240mi/30 mph, which is 8 hours, since the mile units cancel themselves out