Answer:

Explanation:
First of all let's define the specific molar heat capacity.
(1)
Where:
Q is the released heat by the system
n is the number of moles
ΔT is the difference of temperature of the system
Now, we can find n with the molar mass (M) the mass of the compound (m).
Using (1) we have:


I hope it helps!
Answer:
mechanical advantage!
Explanation:
The Mechanical advantage of a machine is the factor by which the machine changes the input force.
When a a machine multiplies an input force, that's called a mechanical advantage.
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Defenition of Mechanical Advantage
brainly.com/question/16617083?referrer=searchResults
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Hope this helps! <3
It was about 9:30 p.m. sorry if the answer is wrong
Answer:
The mutual speed immediately after the touchdown-saving tackle is 4.80 m/s
Explanation:
Given that,
Mass of halfback = 98 kg
Speed of halfback= 4.2 m/s
Mass of corner back = 85 kg
Speed of corner back = 5.5 m/s
We need to calculate their mutual speed immediately after the touchdown-saving tackle
Using conservation of momentum

Where,
= mass of halfback
=mass of corner back
= velocity of halfback
= velocity of corner back
Put the value into the formula



Hence, The mutual speed immediately after the touchdown-saving tackle is 4.80 m/s
Answer:
The puck moves a vertical height of 2.6 cm before stopping
Explanation:
As the puck is accelerated by the spring, the kinetic energy of the puck equals the elastic potential energy of the spring.
So, 1/2mv² = 1/2kx² where m = mass of puck = 39.2 g = 0.0392 g, v = velocity of puck, k = spring constant = 59 N/m and x = compression of spring = 1.3 cm = 0.013 cm.
Now, since the puck has an initial velocity, v before it slides up the inclined surface, its loss in kinetic energy equals its gain in potential energy before it stops. So
1/2mv² = mgh where h = vertical height puck moves and g = acceleration due to gravity = 9.8 m/s².
Substituting the kinetic energy of the puck for the potential energy of the spring, we have
1/2kx² = mgh
h = kx²/2mg
= 59 N/m × (0.013 m)²/(0.0392 kg × 9.8 m/s²)
= 0.009971 Nm/0.38416 N
= 0.0259 m
= 2.59 cm
≅ 2.6 cm
So the puck moves a vertical height of 2.6 cm before stopping