The answer is....
214.28
which would be 214.3 or 214
Answer:
Specific heat at constant pressure is = 1.005 kJ/kg.K
Specific heat at constant volume is = 0.718 kJ/kg.K
Explanation:
given data
temperature T1 = 50°C
temperature T2 = 80°C
solution
we know energy require to heat the air is express as
for constant pressure and volume
Q = m × c × ΔT ........................1
here m is mass of the gas and c is specific heat of the gas and Δ
T is change in temperature of the gas
here both Mass and temperature difference is equal and energy required is dependent on specific heat of air.
and here at constant pressure Specific heat is greater than the specific heat at constant volume,
so the amount of heat required to raise the temperature of one unit mass by one degree at constant pressure is
Specific heat at constant pressure is = 1.005 kJ/kg.K
and
Specific heat at constant volume is = 0.718 kJ/kg.K
Answer:
The distance of the bee from the hive is 740 m.
Explanation:
Given that,
Bee starts fly 500 m due east, 430 m west and 670 m east.
The direction of the bee
500 m in positive direction
430 m in negative direction
670 m in positive direction
We need to calculate the net distance
Using formula of distance
Hence, The distance of the bee from the hive is 740 m.
Given data in the problem :-
- Acceleration (a) = 4.0 m/s^2
- Initial velocity (u) = 0 m/s
- Final velocity (v) = 34 m/s
- Distance travelled by aircraft (S) = ?
From Newton's Laws of Motion we know that ,
v = u + at [t = Time taken by aircraft to cover the distance]
⇒ 34 = 0 + 4t
⇒ t = 34/4 s
∴ t = 8.5 s
From Newton's Laws of Motion we also know that ,
S = u.t + 1/2a.t^2
⇒ S = 0×8.5 + 1/2 × 4 × (8.5)^2 m
∴ S = 144.50 m
Thus the distance travelled by the aircraft while accelerating is 144.50 meter .
<h3>
Answer: The acceleration doubles</h3>
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Explanation:
Consider a mass of 10 kg, so m = 10
Let's say we apply a net force of 20 newtons, so F = 20
The acceleration 'a' is...
F = ma
20 = 10a
20/10 = a
2 = a
a = 2
The acceleration is 2 m/s^2. Every second, the velocity increases by 10 m/s.
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Now let's double the net force on the object
F = 20 goes to F = 40
m = 10 stays the same
F = ma
40 = 10a
10a = 40
a = 40/10
a = 4
The acceleration has also doubled since earlier it was a = 2, but now it's a = 4.
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In summary, if you double the net force applied to the object, then the acceleration doubles as well.
