Can you use a machine to gain both force and speed at the same time? explain.

1 answer:

8 0

Well, It rather depends on your definition of "machine." The normal physics set of simple machines - levers, pulleys, ramps all give you increased the force at the expense of reduced speed or increased the rate at the cost of reduced force. So, no - by definition a machine is an arrangement for multiplying one while paying the cost by reducing the other. You are looking at an example of the Conservation of Energy. One of the giant rules we are pretty sure cannot be violated.<span>
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The magnitude of the magnetic force on a current-carrying wire held in a magnetic

Kamila [148]

Answer:

All of the above

Explanation:

The magnitude of the magnetic force on a current-carrying wire held in a magnetic is given by the equation $F = BIlsin \theta$

Where B = Strength of the magnetic field

I = The current carried by the wire

l = length of the wire in the magnetic field

θ = Angle between the wire and the magnetic field

Based on the relationship written above, the magnitude of the magnetic force on the current - carrying wire in the magnetic field depends on the strength of the magnetic field (B), length of the wire(l), current in the wire (I).

All the options are correct.

3 0

4 years ago

A ball is projected upward at time t = 0.0 s, from a point on a roof 20 m above the ground. The ball rises, then falls and strik

enot [183]

Answer:

$9.8\ m/s^2$

Explanation:

Given:

x = initial vertical height of the ball = 20 m
y = final height of the ball = 0 m
a = acceleration of the ball in air = $-g = -9.8\ m/s^2$
u = initial vertical velocity of the ball = 37.9 m/s

Assume:

t = time interval

When an object is in air, the gravitational force due to earth in the downward direction and air resistance opposite to the motion of the object acts on it. But, air resistance is neglected unless it is mentioned in the question. This means only gravitational force due to the earth acts on it in the downward direction.

Let us first find out the time interval for which the ball remains in the air i.e., the time instant from which the ball was projected vertically up to the time instant at which the ball strikes the ground.

Since the ball in the air remains at constant acceleration motion, we have the following relation:

$y-x = ut+\dfrac{1}{2}at^2\\\Rightarrow 0-20=37.9t+\dfrac{1}{2}(-9.8)t^2\\\Rightarrow 4.9t^2-37.9t-20=0\\$

On solving the above quadratic equation, we have

t = -0.50 s or t = 8.23 s

Since t = -0.50 s is the time in the past. So, it is wrong and t = 8.23 s is the desirable time instant.

This means the ball is in the air for 8.23 s. Since the ball in the air remains in constant acceleration motion in the downward direction. So, the ball has a constant acceleration closest to $9.8\ m/s^2$ at t = 6.2 s. This is because the ball remains in the air for 8.23 s.