All categories

2 years ago

A force of 30.0 N is applied to a 3.00 kg object for 3.00 seconds. Calculate the velocity experienced by the object.

Physics

1 answer:

olganol [36]2 years ago

6 0

Answer:

Explanation:

F = ma and

$a=\frac{v}{t}$

We have F, we have m, but in order to solve for v, we need a.

30.0 = 3.00a so

a = 10.0 m/s/s. Plug that in for a in the second equation and solve for v:

$10.0=\frac{v}{3.00}$ so

v = 10.0(3.00) so

v = 30.0 m/s

You might be interested in

What is friction and how does it affect objects in motion?

lesantik [10]

Friction is a force. A resistance force that tries to slow down and stop objects in motion. The difference between friction and air resistance is that friction acts on surfaces. For example, if I were skateboarding on a rough surface, more friction would occur so I would travel slower. If I were skareboarding on ice, I would go faster because there is much less friction to slow me down. Hope that helps:)

3 0

3 years ago

Read 2 more answers

A thin block of soft wood with a mass of 0.078 kg rests on a horizontal frictionless surface. A bullet with a mass of 4.67 g is

natali 33 [55]

Answer:

Final speed of the bullet is 228.3 m/s

Explanation:

As we know that there is no external force on the system of wooden block and the bullet

so we can say momentum of the system is conserved here

so here we can say

$P_i = P_f$

$m_1v_1 = m_1v_{1f} + m_2v_{2f}$

$4.67\times 10^{-3}(613) = 4.67 \times 10^{-3}v_{1f} + (0.078)(23)$

so we will have

$2.86 = 4.67\times 10^{-3} v_{1f} + 1.794$

$v_{1f} = 228.3 m/s$

7 0

2 years ago

The force of gravity is an inverse square law. this means that, if you double the distance between two large masses, the gravita

likoan [24]

F~1/r²
doubling the distance r, Decreases the force by ¼

4 0

3 years ago

A heat engine has a maximum possible efficiency of 0.780. If it operates between a deep lake with a constant temperature of-24.8

Volgvan

Answer : The temperature of the hot reservoir (in Kelvins) is 1128.18 K

Explanation :

Efficiency of carnot heat engine : It is the ratio of work done by the system to the system to the amount of heat transferred to the system at the higher temperature.

Formula used for efficiency of the heat engine.

$\eta =1-\frac{T_c}{T_h}$