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Bond [772]
3 years ago
7

A force is applied to a 30-kg object, resulting in an acceleration of 15 m/s. What is the force being applied to the object?

Physics
2 answers:
scoundrel [369]3 years ago
6 0
From Second Newton's Law:

F=a*m
a=15 \frac{m}{ s^{2} }
m=30kg

F=15 \frac{m}{ s^{2} }*30kg=450N
telo118 [61]3 years ago
5 0
The sum of all forces applied to the object = (mass) x (acceleration)

The sum of all forces applied to the object = (30) x (15)

= <u>450 newtons</u>, in the direction of the acceleration.
 
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hoa [83]

Answer: D

Explanation: I have my answers VS. Go0gle answers (go0gle answers are pics)

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(C) Both have ample water. No. Venus doesn't have much water >_>

(D) Both have a solid inner core and a liquid outer core. Yesh. In my lessons (before), Venus has a solid inner core and liquid outer core.

5 0
3 years ago
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How are the climates of coastal regions affected by the specific heat capacity of water?
Katen [24]

Answer:

The specific heat capacity can be defined as the amount of heat required to raise the temperature of 1 unit of mass by 1 unit temperature. The specific heat capacity of water is 4.186 joule/gram °C which is higher than common substances. The land has lower specific heat capacity. Thus, the land gets hot quickly than water.

This results in warming up air near the land which creates a difference in pressure across the coastal region. Sea breeze blows from sea towards landmass. Opposite happens at night, when water is still warm and land gets cooled down quickly. Then land breeze blows  from landmass towards the sea. This breeze maintains a moderate temperature and windy and humid weather in the coastal regions.

5 0
4 years ago
to move a resting box of 100 Newton on the ground with kinetic friction coeficient of 0,250 is applied a force of 60 N horizonta
krok68 [10]
Work is calculated by multiplying force by the distance that the object had moved. The applied force is 60 N, moving the object by 10 m. Thus, the work does is 600 J. For the friction force which is equal to,
                          100N x 0.250 = 25.0 N
the work done is,
                        W = (60 N - 25 N) x 10 m = 350 J
The kinetic energy of the box can be equated to this force. Thus, the answer is also 350 J. 
6 0
4 years ago
How much power is needed to lift a 20-kg object to a height of 4.0 m in 2.5 seconds?
4vir4ik [10]

Answer:

313.92w

Explanation:

Formula for power:

P=W/∆t = Fv

Givens:

m=20kg

∆y=4.0m

∆t=2.5s

a=9.81m/s²

In order to find power, we first need to solve for work.

W=Fd (force*displacement), f=mg

W=mg∆y

W=(20kg)(9.81m/s²)(4.0m)

W=784.8J

P=W/∆t

P=784.8J/2.5s

P=313.92 watts

5 0
3 years ago
A particle of mass 4.00 kg is attached to a spring with a force constant of 100 N/m. It is oscillating on a frictionless, horizo
zloy xaker [14]

Solution :

Given :

Mass attached to the spring = 4 kg

Mass dropped = 6 kg

Force constant = 100 N/m

Initial amplitude = 2 m

Therefore,

a). $v_{initial} = A w$

          $= 2 \times \sqrt{\frac{100}{4}}$

          = 10 m/s

Final velocity, v at equilibrium position, v = 5 m/s

Now, $\frac{1}{2}(4+4)5^2 = \frac{1}{2} kA'$

A' = amplitude = 1.4142 m

b). $T=2 \pi \sqrt{\frac{m}{k}}$

    m' = 2m

    Hence, $T'=\sqrt2 T$

c). $\frac{\frac{1}{2}(4+4)5^2 + \frac{1}{2}\times 4 \times 10^2}{\frac{1}{2} \times 4 \times 10^2}$

  $=\frac{1}{2}$

Therefore, factor $=\frac{1}{2}$

Thus, the energy will change half times as the result of the collision.

7 0
3 years ago
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