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o-na [289]
2 years ago
11

You have two beakers of different types of oil, which have different specific heat capacities. You put each beaker in a water ba

th at 50 °C. The water is heated and stays at 50 °C. Which statement is true?
A: If you put the same amount of oil in each beaker, the temperature of the oil in each beaker will be the same at all times.

B:If the mass of oil in the two beakers is the same, the oil with the lower specific heat capacity will heat up more slowly.

C:If the mass of oil in the two beakers is the same, the oil with the lower specific heat capacity will heat up more quickly.

D:The oil with the higher specific heat capacity will heat up more quickly regardless of the mass of oil.
Physics
1 answer:
eimsori [14]2 years ago
8 0

Answer:

C

Explanation:

C:If the mass of oil in the two beakers is the same, the oil with the lower specific heat capacity will heat up more quickly.

The answer is C

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An object has a kinetic energy of 275 j and a momentum of magnitude 25.0 kg · m/s. find the (a) speed and (b) mass of the object
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A helium atom (mass 4.0 u) moving at 598 m/s to the right collides with an oxygen molecule (mass 32 u) moving in the same direct
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Answer:

Speed of the helium after collision = 246 m/s

Explanation:

Given that

Mass of helium ,m₁ = 4 u

u₁=598 m/s

Mass of oxygen ,m₂ = 32 u

u₂  = 401 m/s

v₂ =445 m/s

Given that initially both are moving in the same direction and lets take they are moving in the right direction.

Speed of the helium after collision = v₁

There is no any external force on the masses that is why the linear momentum will be conserve.

Initial linear momentum = Final linear momentum

P = m v

m₁u₁+m₂u₂ = m₁v₁+m₂v₂

598 x 4 + 32 x 401 = 4 x v₁+ 32 x 445

v₁ = 246 m/s

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A 170 kg astronaut (including space suit) acquires a speed of 2.25 m/s by pushing off with his legs from a 2600 kg space capsule
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Explanation:

Mass of the astronaut, m₁ = 170 kg

Speed of astronaut, v₁ = 2.25 m/s

mass of space capsule, m₂ = 2600 kg

Let v₂ is the speed of the space capsule. It can be calculated using the conservation of momentum as :

initial momentum = final momentum

Since, initial momentum is zero. So,

m_1v_1+m_2v_2=0

170\ kg\times 2.25\ m/s+2600\ kg\times v_2=0

v_2=-0.17\ m/s

So, the change in speed of the space capsule is 0.17 m/s. Hence, this is the required solution.

8 0
3 years ago
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