Answer:
Energy is essentially work done by an object or on object.
From,
W = Fd
It's directly proportional to mass.
from,
K. E = 1/2mv²
Energy is directly proportional to mass.
P. E = mgh
Energy is directly proportional to mass.
H = mc∆T
Energy is directly proportional to mass.
Thus increasing mass will increase the energy also imparted on another object since all the above eqns show that relationship.
And for 2 moving bodies
K.Ei = K.Ef(energy conservation)
m1u²1 + m2u²2 = m1v²1 + m2v²2
The relationship is the same that the greater mass the greater the impact.
Given:
The thermal energy added to the system is Q = 90 J
The work done by the system on the surroundings is W = 30 J
To find the change in internal energy.
Explanation:
According to the first law of thermodynamics, the change in internal energy can be calculated by the formula
On substituting the values, the change in internal energy will be
Final Answer: The chage in internal energy is 60 J (option D)
Answer:
New pressure is 0.534 atm
Explanation:
Given:
Initial volume of the gas, V₁ = 250 mL
Initial pressure of the gas, P₁ = 1.00 atm
Initial temperature of the gas, T₁ = 20° C = 293 K
Final volume of the gas, V₂ = 500 mL
Final pressure of the gas = P₂
Final temperature of the gas, T₁ = 40° C = 313 K
now,
we know for a gas
PV = nRT
where,
n is the moles
R is the ideal gas constant
also, for a constant gas
we have
(P₁V₁/T₁) = (P₂V₂/T₂)
on substituting the values in the above equation, we get
(1.00 × 250)/293 = (P₂ × 500)/313
or
P₂ = 0.534 atm
Hence, the <u>new pressure is 0.534 atm</u>
By definition we have that the final speed is:
Vf² = Vo² + 2 * a * d
Where,
Vo: Final speed
a: acceleration
d: distance.
We cleared this expression the acceleration:
a = (Vf²-Vo²) / (2 * d)
Substituting the values:
a = ((0) ^ 2- (60) ^ 2) / ((2) * (123) * (1/5280))
a = -77268 mi / h ^ 2
its stopping distance on a roadway sloping downward at an angle of 17.0 ° is:
First you must make a free body diagram and see the acceleration of the car:
g = 32.2 feet / sec ^ 2
a = -77268 (mi / h ^ 2) * (5280/1) (feet / mi) * (1/3600) ^ 2 (h / s) ^ 2
a = -31.48 feet / sec ^ 2
A = a + g * sin (θ) = -31.48 + 32.2 * sin17.0
A = -22.07 feet / sec ^ 2
Clearing the braking distance:
Vf² = Vo² + 2 * a * d
d = (Vf²-Vo²) / (2 * a)
Substituting the values:
d = ((0) ^ 2- (60 * (5280/3600)) ^ 2) / (2 * (- 22.07))
d = 175.44 feet
answer:
its stopping distance on a roadway sloping downward at an angle of 17.0 ° is 175.44 feet
The answer is: Salt! :)
Have a great day!
