A 4 kilogram box is at rest on a frictionless floor. A net force of 12 newtons acts on it. What is the acceleration of the box?

Physics

1 answer:

Rashid [163]3 years ago

4 0

Answer:

Explanation:

Step one:

given

Mass m= 4kg

Force F= 12N

Required

Acceleration the relation between force, acceleration, and mass is Newton's first equation of motion, which says a body will continue to be at rest or uniform motion unless acted upon by an external force

F=ma

a=F/m

a=12/4

a=3 m/s^2

You might be interested in

Suppose that when you move the north pole of a bar magnetic toward a coil it induces a positive current in the coil. The strengt

dolphi86 [110]

Answer:

<em>a) A positive current will be induced in the coil</em>

Explanation:

Electromagnetic induction is the induction of an electric field on a conductor due to a changing magnetic field flux. The change in the flux can be by moving the magnet relative to the conductor, or by changing the intensity of the magnetic field of the magnet. In the case of this electromagnets<em>, the gradual increase in the the electromagnet's field strength will cause a flux change, which will in turn induce an electric current on the coil.</em>

According to Lenz law, the induced current acts in such a way as to negate the motion or action that is producing it. <em>A positive current will be induced on the coil so as to repel any form of attraction between the north pole of the electromagnet and the coil</em>. This law obeys the law of conservation of energy, since work has to be done to move the move them closer to themselves.

3 0

3 years ago

Which algebraic rule describes the transformation?

Paraphin [41]

Answer:

(y) - (x - 2, y + 5)

Explanation:

7 0

3 years ago

A kangaroo jumps up with an initial velocity of 36 feet persecond from the ground (assume its starting height is 0 feet).Use the

kkurt [141]

Given

Initial velocity:

36 ft/s

Initial height:

0 ft

Vertical motion model:

h(t) = -16t^2 + ut + s

v = initial velocity

s = is the height

Procedure

We are going to use the model provided for the vertical motion.

$\begin{gathered} h(t)=-16t^2+36t+0 \\ h(t)=-16t^2+36t \end{gathered}$

We know that at the maximum height the final velocity is 0.

Then we will use the following expression to calculate the maximum height:

$\begin{gathered} v^2_f=v^2_o-2ah_{\max } \\ 0=v^2_o-2ah_{\max } \\ 2ah_{\max }=v^2_o \\ h_{\max }=\frac{v^2_o}{2a} \\ h_{\max }=\frac{(36ft/s)^2}{2\cdot32ft/s^2} \\ h_{\max }=20.25\text{ ft} \end{gathered}$

Now for time: