Answer:
T = 0.71 seconds
Explanation:
Given data:
mass m = 1Kg, spring constant K = 78 N/m, time period of oscillation T = 0.71 seconds.
We have to calculate time period when this same spring-mass system oscillates vertically.
As we know
[tex]T = 2\pi \sqrt{\frac{m}{K} }[/tex]
This relation of time period is true under every orientation of the spring-mass system, whether horizontal, vertical, angled or inclined. Therefore, time period of the same spring-mass system oscillating vertically too remains the same.
Therefore, T = 0.71 seconds
Transverse thrusters are used to make large ships fully maneuverable at low speeds without tugboat assistance. A transverse thruster consists of a propeller mounted in a duct; the unit is then mounted below the waterline in the bow or stern of the ship. The duct runs completely across the ship. Calculate the thrust developed by a 1900 kW unit supplied to the propeller if the duct is 2.6 m in diameter and the ship is stationary.
Answer:
Thrust developed = 212.3373 kN
Explanation:
Assuming the ship is stationary
Determine the Thrust developed
power supplied to the propeller ( Punit ) = 1900 KW
Duct distance ( diameter ; D ) = 2.6 m
first step : calculate the area of the duct
A = π/4 * D^2
= π/4 * ( 2.6)^2 = 5.3092 m^2
next : calculate the velocity of propeller
Punit = (A*v*β ) / 2 * V^2 ( assuming β = 999 kg/m^3 ) also given V1 = 0
∴V^3 = Punit * 2 / A*β
= ( 1900 * 10^3 * 2 ) / ( 5.3092 * 999 )
hence V2 = 8.9480 m/s
Finally determine the thrust developed
F = Punit / V2
= (1900 * 10^3) / ( 8.9480)
= 212.3373 kN
If the girl skater has a mass of 30 kg and moves backward at 5 m/s, what is the velocity or the boy skater
his mass is 50 kg?
Question: Two people stand facing each other at a roller-skating rink then push off each other. If the girl skater has a mass of 30 kg and moves backward at 5 m/s, what is the velocity of the boy skater if his mass is 50 kg?
Answer:
3 m/s
Explanation:
Applying,
The Law of conservation of momentum
Momentum of the girl skater = momentum of the boy skater
MV = mv...................... Equation 1
Where M = mass of the girl skater, V = velocity of the girl skater, m = mass of the boy skater, v = velocity of the boy skater
From the question, we were asked to calculate v
v = MV/m.................. Equation 1
Given: M = 30 kg, V = 5 m/s, m = 50 kg
Substitute these values into equation 1
v = (30×5)/50
v = 3 m/s
Hence the velocity of the the boy skater is 3m/s
1.What is the Kinetic energy of a 3 kg object moving at 4 m/s?
Plz help I’ll give points
Answer:
24 J
Explanation:
[tex]K = \frac{1}{2} mv^{2} = \frac{1}{2} (3kg)(4m/s)^{2} = 24 J[/tex]
Plutonium-238 has a half life of 87.7 years. What percentage of a 5 kilogram (kg) sample remains after 50 years?
Answer:
i dont know but i should know try g o o g l e
Explanation:
A block of weight 1200N is on an incline plane of 30° with the horizontal, a force P is applied to the body parallel to the plane, if the coefficient of the static friction is 0.20 and kinetic friction is 0.15 (1) find the value of P to cause motion up the plane (2) find P to prevent motion down the plane. (3) Find P to cause continuous motion up the plane.
Answer:
a) P = 807.85 N, b) P = 392.15 N, c) P = 444.12 N
Explanation:
For this exercise, let's use Newton's second law, let's set a reference frame with the x-axis parallel to the plane and the direction rising as positive, and the y-axis perpendicular to the plane.
Let's use trigonometry to break down the weight
sin θ = Wₓ / W
cos θ = W_y / W
Wₓ = W sin θ
W_y = W cos θ
Wₓ = 1200 sin 30 = 600 N
W_y = 1200 cos 30 = 1039.23 N
Y axis
N- W_y = 0
N = W_y = 1039.23 N
Remember that the friction force always opposes the movement
a) in this case, the system will begin to move upwards, which is why friction is static
P -Wₓ -fr = 0
P = Wₓ + fr
as the system is moving the friction coefficient is dynamic
fr = μ N
fr = 0.20 1039.23
fr = 207.85 N
we substitute
P = 600+ 207.85
P = 807.85 N
b) to avoid downward movement implies that the system is stopped, therefore the friction coefficient is static
P + fr -Wx = 0
fr = μ N
fr = 0.20 1039.23
fr = 207.85 N
we substitute
P = Wₓ -fr
P = 600 - 207,846
P = 392.15 N
c) as the movement is continuous, the friction coefficient is dynamic
P - Wₓ + fr = 0
P = Wₓ - fr
fr = 0.15 1039.23
fr = 155.88 N
P = 600 - 155.88
P = 444.12 N
15 points!
a. Calculate the electric potential energy stored in a 1.4 x 10-7 F capacitor
that stores 3.40 x 10-6 C of charge at 24.0 V.
Answer:
[tex]4.12\times 10^{-5}\ J[/tex].
Explanation:
Given that,
Capacitance, [tex]C=1.4\times 10^{-7}\ F[/tex]
Charge stored in the capacitor, [tex]Q=3.4\times 10^{-6}\ C[/tex]
We need to find the electric potential energy stored in the capacitor. The formula for the electric potential energy stored in the capacitor is given by :
[tex]E=\dfrac{Q^2}{2C}[/tex]
Put all the values,
[tex]E=\dfrac{(3.4\times 10^{-6})^2}{2\times 1.4\times 10^{-7}}\\\\=4.12\times 10^{-5}\ J[/tex]
So, the required electric potential eenergy is equal to [tex]4.12\times 10^{-5}\ J[/tex].
1. A train is moving north at 5 m/s on a straight track. The engine is causing it to accelerate northward at 2 m/s^2.
How far will it go before it is moving at 20 m/s?
A) 83
B) 43
C) 39
D) 94
E) 20
Answer:
It will go up to 93.75 m before it is moving at 20 m/s
Explanation:
As we know that
[tex]v^2 - u^2 = 2aS[/tex]
here v is the final speed i.e 20 m/s
u is the initial speed i.e 5 m/s
a is the acceleration due to gravity i.e 2 m/s^2
Substituting the given values in above equation, we get -
[tex]20^2 - 5^2 = 2*2*S\\S = 93.75[/tex]meters
