a student is pushing a 50 kilogram cart with a force of 500 newtons another students measures the speed of the cart and finds th
1 answer:
The force of friction is 300 N
Explanation:
We can solve the problem by applying Newton's second law of motion: in fact, the net force acting on an object is equal to the product between the mass of the object and its acceleration. So we can write
where
is the net force acting on the object
m is its mass
a is its acceleration
For the cart in this problem, we have two forces acting on it:
- The force of push, F = 500 N, forward
- The force of friction, , backward
So Newton's second law can be rewritten as
where
m = 50 kg
is the acceleration of the cart
And solving for , we find the force of friction:
Learn more about force of friction:
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Answer:
a) The potential energy in the system is greatest at X.
Explanation:
Let be X the point where a ball rests at the top of a hill. By applying the Principle of Energy Conservation, the total energy in the physical system remains constant and gravitational potential energy at the top of the hill is equal to the sum of kinetic energy, a lower gravitational energy and dissipated work due to nonconservative forces (friction, dragging).
Conclusions are showed as follows:
a) The potential energy in the system is greatest at X.
b) The kinetic energy is the lowest at X and Z.
c) Total energy remains constant as the ball moves from X to Y.
Hence, the correct answer is A.
Answer:
= 7.02 ° C
Explanation:
The liquid water gives heat to melt the ice (Q₁) maintaining the temperature of 0 ° C and then the two waters are equilibrated to a final temperature.
Let's start by calculating the heat needed to melt the ice
Q₁ = m L
Q₁ = 0.090 3.33 10⁵
Q₁ = 2997 10⁴ J
This is the heat needed to melt all the ice
Now let's calculate at what temperature the water reaches when it releases this heat
Q = M (T₀ -
)
Q₁ = Q
= T₀ - Q₁ / M
= 20.0 - 2997 104 / (0.600 4186)
= 20.0 - 11.93
= 8.07 ° C
This is the temperature of the water when all the ice is melted
Now the two bodies of water exchange heat until they reach an equilibrium temperature
Temperatures are
Water of greater mass T₀₂ = 8.07ºC
Melted ice T₀₁ = 0ºC
M (T₀₂ -
) = m
(
- T₀₁)
M T₀₂ + m T₀₁ = m + M
= (M T₀₂ + 0) / (m + M)
= M / (m + M) T₀₂
let's calculate
= 0.600 / (0.600 + 0.090) 8.07
= 7.02 ° C
The RC time constant, also called tau, the time constant (in seconds) of an RC circuit, is equal to the product of the circuit resistance (in ohms) and the circuit capacitance (in farads), i.e.
Here,
R = Resistance
C = Capacitance
Replacing we have that
Therefore the time constant of this circuit is