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Sveta_85 [38]
2 years ago
9

1.20 Newton force is working on a 250 gram object. What is the acceleration?

Physics
1 answer:
Leokris [45]2 years ago
7 0

Answer:

The answer is B

Explanation:

250g = 0.25kg

F = m × a

a = F/m

= 1.2/0.25

= 4.8m/s²

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A particle moving in the x direction is being acted upon by a net force f(x)=cx2, for some constant
makkiz [27]
The work-energy theorem states that the change in kinetic energy of the particle is equal to the work done on the particle:
\Delta K = W
The work done on the particle is the integral of the force on dx:
W= \int\limits^{3L}_L {F(x)} \, dx = \int\limits^{3L}_L {cx^2} \, dx = \frac{26}{3}cL^3
So, this corresponds to the change in kinetic energy of the particle.
6 0
3 years ago
Calculate the work done by a force of 30 N in lifting a load of 2 kg to a height of 10 m (g = 10 m/s2)
Mars2501 [29]

Answer:

<u>300 J</u>

Explanation:

Given :

  1. Applied force = 30 N
  2. Mass = 2 kg
  3. Height = 10 m
  4. g = 10 m/s²

Work done by the force :

  • Work done = Force x Displacement
  • Work done = mg x h
  • Work done = 30 N x 10 m
  • Work done = <u>300 J</u>

<u></u>

<u><em>Note</em></u> :

  • What you have calculated is the work done by gravitational force on the object (that too, incorrectly)
  • But in the end, it asks for work done by the force of 30N
  • Hence, the given answer ~
3 0
2 years ago
An air-conditioning system is to be filled from a rigid container that initially contains 5 kg of liquid R-134a at 24°C. The val
OverLord2011 [107]

Answer:

Answer

The Final Quality of teh R-134a in the container  is  0.5056

The Total Heat transfer is Q_{in} = 22.62 KJ

Explanation:

Explanation is  in the following  attachments

3 0
3 years ago
A piece of string 2 meters and has a mass of 5g. On one end of the string hangs a 200 g mass. Find the tension of the string and
viktelen [127]

Explanation:

It is given that,

Length of the string, l = 2 m

Mass of the string, m=5\ g=5\times 10^{-3}\ kg

Hanged mass in the string, m'=200\ g=0.2\ kg

1. The tension in the string is given by :

T=m'g

T=0.2\times 9.8

T = 1.96 N

2. Velocity of the transverse wave in the string is given by :

v=\sqrt{\dfrac{T}{M}}

m = M/l

v=\sqrt{\dfrac{Tl}{m}}

v=\sqrt{\dfrac{1.96\times 2}{5\times 10^{-3}}}

v = 28 m/s

Hence, this is the required solution.

4 0
3 years ago
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