The answer would be 187.95 kg.m/s.
To get the momentum, all you have to do is multiply the mass of the moving object by the velocity.
p = mv
Where:
P = momentum
m = mass
v = velocity
Not the question is asking what is the total momentum of the football player and uniform. So we need to first get the combined mass of the football player and the uniform.
Mass of football player = 85.0 kg
Mass of the uniform = <u> 4.5 kg</u>
TOTAL MASS 89.5 kg
So now we have the mass. So let us get the momentum of the combined masses.
p = mv
= (89.5kg)(2.1m/s)
= 187.95 kg.m/s
Answer:
0.63 s
Explanation:
From the question given above, the following data were obtained:
Mass (m) = 50 g
Extention (e) = 10 cm
Period (T) =?
Next, we obtained 50 g to Kg. This can be obtained as follow:
1000 g = 1 Kg
Therefore,
50 g = 50 g × 1 Kg / 1000 g
50 g = 0.05 kg
Next, we shall convert 10 cm to m. This is illustrated below:
100 cm = 1 m
Therefore,
10 cm = 10 cm × 1 m / 100 cm
10 cm = 0.1 m
Next, we shall determine the force exerted on the spring. This can be obtained as follow:
Mass = 0.05 Kg
Acceleration due to gravity (g) = 9.8 m/s²
Force (F) =?
F = mg
F = 0.05 × 9.8
F = 0.49 N
Next, we shall determine the spring constant of the spring.
Extention (e) = 0.1 m
Force (F) = 0.49 N
Spring constant (K) =?
F = Ke
0.49 = K × 0.1
Divide both side by 0.1
K = 0.49 /0.1
K = 4.9 N/m
Finally, we shall determine the period as follow:
Mass = 0.05 Kg
Spring constant (K) = 4.9 N/m
Pi (π) = 3.14
Period (T) =?
T = 2π√(m/k)
T = 2 × 3.14 × √(0.05 / 4.9)
T = 6.28 × √(0.05 / 4.9)
T = 0.63 s
Thus, the period of oscillation is 0.63 s
Answer:
v = 14 m/s
= 31.3 mph
The answer would be the same if the mass of the car were 2000 kg
Explanation:
Let V be the final velocity of the car after skidding, and v be the initial velocity of the car. Let a be the acceleration of the car and Δx be the distance the car travels after applying brakes (length of the skid marks). Let Fk be the force of friction between the tyres and the road. Let N be the normal force exerted on the car and μ be the co efficient of kinetic friction.
V^2 = v^2 + 2×a×Δx
Now V, the final velocity is zero as the car stops
0 = v^2 + 2×a×Δx
v^2 = -2×a×Δx
v =√-2×a×Δx .....*
Now applying Newton's Second Law
Fnet = m×a
-Fk = m×a
-μ×N = m×a
-μ×m×g = m×a (The mass cancels out)
a = -μ×g
Substituting the value of a back to equation *
v = √-2×(-μ×g)×Δx
v = √-2×(-0.5×9.8)×20
v = 14 m/s
Therefore the speed the car was travelling with v = 14 m/s
which is equal to 31.3 mph
Now if you were to change the mass of the car to 2000 kg the value for v would still be the same. As it is seen above mass cancels out so it does not influence or affect the value of the velocity obtained.
Show transcribed image text. Compare the work done by the electric field when the particle travels<span> from point W to point X to that </span>done<span> when the </span>particle travels<span> from point Z to point Y. ... Is the </span>work done<span> on the </span>particle<span> by the </span>electric field<span> positive, negative, or zero? Explain using force and displacement vectors.</span>
