All categories

2 years ago

Tim is pushing a box with a 7 N force. The weight of the box is 5 N. A frictional force

1 answer:

5 0

Friction and the applied force are not balanced. The 7 N push is countered by the frictional force of 3 N to push the box with a net force of 4 N.

You might be interested in

Lukalu is rappelling off a cliff. The parametric equations that describe her horizontal and vertical position as a function of t

andre [41]

Answer:

2.5 s, 5 m

Explanation:

The equations for the horizontal and vertical position of Lukalu are:

$x(t) = 8t\\y(t) = -16t^2 + 100$

we can find the time it takes her to reach the ground by requiring that the vertical position becomes zero:

y(t) = 0

So we find:

$0=-16t^2 +100\\16t^2 = 100\\t=\sqrt{\frac{100}{16}}=2.5 s$

The horizontal distance of Lukalu instead will be given by the equation for the horizontal position, substituting t = 2.5 s:

$x=8t = 8 \cdot 2.5 s =5 m$

4 0

3 years ago

When an average force F is exerted over a certain distance on a shopping cart of mass m, its kinetic energy increases by 12mv2.

VMariaS [17]

Answer:

A) $d=\dfrac{1}{2F}mv^2$

B) $\Delta KE'=2\times \dfrac{1}{2}mv^2$

Explanation:

Given that

Force = F

Increase in Kinetic energy = $\dfrac{1}{2}mv^2$

$\Delta KE=\dfrac{1}{2}mv^2$

we know that

Work done by all the forces =change in the kinetic energy

Lets distance = d

We know work done by force F

W= F .d

F.d=ΔKE

$F.d=\dfrac{1}{2}mv^2$

$d=\dfrac{1}{2F}mv^2$

If the force become twice

F' = 2 F

F'.d=ΔKE'

2 F .d = ΔKE' ( F.d =Δ KE)

2ΔKE = ΔKE'

$\Delta KE'=2\times \dfrac{1}{2}mv^2$

Therefore the final kinetic energy will become the twice if the force become twice.

8 0

3 years ago

Which word is not a good way to describe exercise?

Karolina [17]

In my opinion, calling exercise "work" wouldn't be a good way of describing it, I don't think people would exercise if its called work

3 0

3 years ago

5 ways in which friction can be useful

saul85 [17]

Friction can be useful when you trying to stop a car, also it can be useful in energy losses, increase wear and tear and produces heat

6 0

3 years ago

Lana71 [14]

Answer:

A. Condensation B. Freezing. C. condensation (not sure) D. Sublimation E. Evaporation. F Evaporation (not sure)

Hope it heps:)

6 0

3 years ago