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Neko [114]
2 years ago
6

A 55.0 kg runner who weighs 539.0 N is accelerating at 3.2 m/s2. After 2

Physics
2 answers:
Darina [25.2K]2 years ago
8 0

Answer:

The answer is C. 352.0 kg-m/s

drek231 [11]2 years ago
4 0

Answer:

C. 352.0 kg-m/s

Explanation: Took the test on APEX

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A spaceship which is 50,000 kilometers from the center of Earth has a mass of 3,000 kilograms. What is the magnitude of the forc
Anna11 [10]
Fg=Gx(M1M2/r^2)
Fg=478N
6 0
2 years ago
Read 2 more answers
A certain rigid aluminum container contains a liquid at a gauge pressure of P0 = 2.02 × 105 Pa at sea level where the atmospheri
MaRussiya [10]

Answer:

dz=19217687.07\ m

Explanation:

Given:

  • initial gauge pressure in the container, P_0=2.02\times 10^{5}\ Pa
  • atmospheric pressure at sea level, P_a=1.01\times 10^5\ Pa
  • initial volume, V_0=4.4\times 10^{-4}\ m^3
  • maximum pressure difference bearable by the container, dP_{max}=2.26\times 10^{5}\ Pa
  • density of the air, \rho_a=1.2\ kg.m^{-3}
  • density of sea water, \rho_s=1.2\ kg.m^{-3}

<u>The relation between the change in pressure with height is given as:</u>

\frac{dP_{max}}{dz} =\rho_a.g_n

where:

dz = height in the atmosphere

g_n= standard value of gravity

<em>Now putting the respective values:</em>

\frac{2.26\times 10^{5}}{dz} =1.2\times 9.8

dz=19217.687\ km

dz=19217687.07\ m

Is the maximum height above the ground that the container can be lifted before bursting. (<em>Since the density of air and the density of sea water are assumed to be constant.</em>)

7 0
2 years ago
A small ball of mass 2.00 kilograms is moving at a velocity 1.50 meters/second. It hits a larger, stationary ball of mass 5.00 k
rewona [7]

The kinetic energy of the small ball before the collision is

                             KE  =  (1/2) (mass) (speed)²

                                     = (1/2) (2 kg) (1.5 m/s)

                                     =    (1 kg)  (2.25 m²/s²)

                                     =        2.25 joules.

Now is a good time to review the Law of Conservation of Energy:

                     Energy is never created or destroyed. 
                     If it seems that some energy disappeared,
                     it actually had to go somewhere.
                     And if it seems like some energy magically appeared,
                     it actually had to come from somewhere.

The small ball has 2.25 joules of kinetic energy before the collision.
If the small ball doesn't have a jet engine on it or a hamster inside,
and does not stop briefly to eat spinach, then there won't be any
more kinetic energy than that after the collision.  The large ball
and the small ball will just have to share the same 2.25 joules.

3 0
2 years ago
For which of the following problems would a scientist most likely use carbon-14?
spayn [35]

Answer:

To calculate the age of a piece of bone

Explanation:

Carbon 14 is an isotope of carbon that is unstable and decays into Nitrogen 14 by emitting an electron. The decay rate of radioactive material is  normally expressed in terms of its "half-life" (the time required by half the radioactive nuclei of a sample to undergo radioactive decay). The nice thing about carbon 14 is that its "half-life" is about 5730 years, which gives a nice reference to measure the age of fossils that are some thousand years old.

Carbon 14 dating is used to determine the age of objects that have been living organisms long ago. They measure how much carbon 14 is left in the object after years of decaying without having exchange with the ambient via respiration, ingestion, absorption, etc. and therefore having renewed the normal amount of carbon 14 that is in the ambient.

A rock is not a living organism, so its age cannot be determined by carbon 14 dating.

3 0
2 years ago
A 3.0-kg mass and a 5.0-kg mass hang vertically at the opposite ends of a very light rope that goes over an ideal pulley. If the
AleksAgata [21]

Answer:

acceleration = 2.4525‬ m/s²

Explanation:

Data: Let m1 = 3.0 Kg, m2 = 5.0 Kg, g = 9.81 m/s²

Tension in the rope = T

Sol: m2 > m1

i) for downward motion of m2:

m2 a = m2 g - T

5 a = 5 × 9.81 m/s² - T  

⇒ T = 49.05‬ m/s² - 5 a     Eqn (a)‬

ii) for upward motion of m1

m a = T - m1 g

3 a = T - 3 × 9.8 m/s²

⇒ T =  3 a + 29.43‬ m/s²   Eqn (b)

Equating Eqn (a) and(b)

49.05‬ m/s² - 5 a = T =  3 a + 29.43‬ m/s²

49.05‬ m/s² - 29.43‬ m/s² = 3 a + 5 a

19.62 m/s² = 8 a

⇒ a = 2.4525‬ m/s²

5 0
2 years ago
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