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mamaluj [8]
3 years ago
14

A force of 210 N is applied to an object and the object accelerates at 14m/s2. Determine the mass of the object in kg.

Physics
1 answer:
lisabon 2012 [21]3 years ago
7 0

Answer:

15 kg

Explanation:

F = m*a

F/a = m

210/14 = 15 kg

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dalvyx [7]

Answer:

46 m at 9.5 degrees east of south

Explanation:

5 0
3 years ago
Four identical masses m are evenly spaced on a frictionless 1D track. The first mass is sent at speed v toward the other three.
SpyIntel [72]

Answer:

The speed decreases 75%.

Explanation:

  • Since no friction present, assuming no external forces acting during the three collisions, total momentum must be conserved.
  • For the first collission, only mass 1 is moving before it, so we can write the following equation:

       p_{i} = p_{f} = m*v_{o}    (1)

  • Since both masses are identical, and they stick together after the collision, we can express the final momentum as follows:

       p_{f1} = 2*m*v_{1}    (2)

  • From (1) and (2) we get:
  • v₁ = v₀/2  (3)
  • Since the masses are moving on a frictionless 1D track, the speed of the set of mass 1 and 2 combined together before colliding with mass 3 is just v₁, so the initial momentum prior the second collision (p₁) can be expressed as follows:

       p_{1} = 2*m*v_{1} = 2*m*\frac{v_{o} }{2}  = m*v (4)

  • Since after the collision the three masses stick together, we can express this final momentum (p₂) as follows:

        p_{2} = 3*m*v_{2}  (5)

  • From (4) and (5) we get:
  • v₂ = v₀/3  (6)
  • Since the masses are moving on a frictionless 1D track, the speed of the set of mass 1, 2 and 3 combined together before colliding with mass 4 is just v₂, so the initial momentum prior the third collision (p₂) can be expressed as follows:

      p_{2} = 3*m*v_{2} = 3*m*\frac{v_{o} }{3}  = m*v (7)

  • Since after the collision the four masses stick together, we can express this final momentum (p₃) as follows:

       p_{3} = 4*m*v_{3}  (8)

  • From (7) and (8) we get:
  • v₃ = v₀/4
  • This means that after the last collision, the speed will have been reduced to a 25% of the initial value, so it will have been reduced in a 75% regarding the initial value of v₀.
5 0
2 years ago
4. A cinder block is sitting on a platform 20 m high. It has a mass of 4 kg. The block has energy. Calculate it.
zavuch27 [327]

Answer:

Explanation:

Since the block is at rest in an elevated position, we can assume that it only has potential energy.

U=mgh is the formula for potential energy where U=potential energy, m= mass, g=acceleration due to gravity, and h=height.

Plug in known variables....

U=4kg*9.8m/s^2*20m

U=784 joules of potential energy or letter A.

4 0
2 years ago
Light of frequency 5*10^14hz liberates electron with energy 2.31*10^-19 joule from a certain surface .what is wave length of ult
stealth61 [152]

Answer:  Light of frequency 5 x 1014 HZ liberates electrons with energy 2.3 x 10-19from a certain metallic surface.

Explanation:

5 0
1 year ago
A stone is dropped from a tower 100 meters above the ground. The stone falls past ground level and into a well. It hits the wate
lilavasa [31]

Take the stone's position at ground level to be the origin, and the downward direction to be negative. Then its position in the air y at time t is given by

y=100\,\mathrm m-\dfrac g2t^2

Let d be the depth of the well. The stone hits the bottom of the well after 5.00 s, so that

-d=100\,\mathrm m-\dfrac g2(5.00\,\mathrm s)^2\implies d=\boxed{22.6\,\mathrm m}

7 0
3 years ago
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