On an ice rink, Manuel pushes off the wall with 540 N of force. If his mass is 72.0kg, what is his

Physics

1 answer:

tangare [24]3 years ago

4 0

Answer: 7.5m/s2

Explanation:

Force applied = 540 N of force.

mass = 72.0kg

acceleration on the skates = ?

Recall that Force is the product of mass of an object and the acceleration by which it moves.

Force = mass x acceleration

540N = 72.0kg x acceleration

Acceleration = 540N/72.0kg

= 7.5m/s2

You might be interested in

Use your data (attachment) from Part 3 and Newton’s laws to explain why the force meter measures a force if the cart is moving a

vazorg [7]

The cart experiences a frictional force which is directly proportional to its weight. This means that there must be a force applied on the car to balance the forces on the car to produce a net force of 0.
This is in accordance to Newton's first law which states that an object at rest will remain at rest and an object in motion will remain in motion unless an external force acts on it. The force must be a resultant force.
Therefore, the force needed increases with the total weight of the cart as well as with the added mass in a linear manner.

4 0

3 years ago

Potential difference is measured in which units?<br> volts<br> amps<br> currents<br> watts

viktelen [127]

It is Volt
First option

4 0

3 years ago

Read 2 more answers

A child of mass 40.0 kg is in a roller coaster car that travels in a loop of radius 7.00 m. at point a the speed of the car is 1

pav-90 [236]

I attached the missing picture.
The force of seat acting on the child is a reaction the force of child pressing down on the seat. This is the third Newton's law. The force of a child pressing down the seat and the force of the seat pushing up on the child are the same.
There two forces acting on the child. The first one is the gravitational force and the second one is centrifugal force. In this example, the force of gravity is always pulling down, but centrifugal force always acts away from the center of circular motion.
Part A
For point A we have:
$F_a=F_cf-F_g$
In this case, the forces are aligned, centrifugal is pointing up and gravitational is pulling down.
$F_a=m\frac{v^2}{r}-mg=179 $N$
Part B
At the point, B situation is a bit more complicated. In this case force of gravity and centrifugal force are not aligned. We have to look at y components of this forces, y-axis, in this case, is just pointing upward.
$F=F_{cf}\cos(30)-mg=m\frac{v^2}{r}\cos(30)-mg=153.2$N$
Part C
The child will stay in place at point A when centrifugal force and force of gravity are in balance:
$F_g=F_{cf}\\
mg=m\frac{v^2}{r}\\
gr=v^2\\
v=\sqrt{gr}=8.29\frac{m}{s}$