the friction force provided by the brakes is 30000 N.
<h3>What is friction force?</h3>
Friction force is the force that opposes the motion between two bodies in contact.
To calculate the average friction force provided by the brakes, we apply the formula below.
Formula:
- K.E = F'd............. Equation 1
Where:
- K.E = Kinetic energy of the train
- F' = Friction force provided by the brakes
- d = distance
Make F' the subject of the equation
- F' = K.E/d............ Equation 2
From the question,
Given:
Substitute these values into equation 2
- F' = (8.1 ×10⁶)/270
- F' = 30000 N
Hence, the friction force provided by the brakes is 30000 N
Learn more about friction force here: brainly.com/question/13680415
I’m assuming that’s m^3? If so then simply divide 160,000 by 20 and you get the answer.
8,000 kg/m^3
Answer:
R= 5.4 ohms
Explanation:
Given that
V= 9 V
Power ,P= 15 W
Lets take resistor resistance = R
We know that Power given as
V=Voltage
P=Power
R=Resistance
Now by putting the all the values in the above equation
R= 5.4 ohms
Therefore the resistance of the resister will be 5.4 ohms.
Before swinging, T has only potential energy, (no speed)
Ui = mgh
Where h is the vertical displacement of T
From the laws of geometry,
cos45 = (L-h)/L
cos45 = 1-h/L
h/L = 1-cos45
h = L(1-cos45)
Therefore
Ui = mgL(1-cos45)
Proceeding the same way,
Twill raise to aheight of h' due to swing
h' = L(1-cos30)
The PE of T after swing is
Uf = mgh'
Uf = mgL(1-cos30)
Along with the PE , T has some kinetic energy results due to the moment.
Tf = 0.5*mv^2
According to the law of conservation of energy,
Ui = Uf+Tf
mgL(1-cos45) = mgL(1-cos30) + 0.5*mv^2
gL(co30-cos45) = 0.5*v^2
9.8*20*(co30-cos45) = 0.5*V^2
v = 7.89 m/s
<span>The speed f T after swing is 7.89 m/s</span>
Answer:
a) 79.7rad/s
b) -18.7rad/s^2
c) 53m
Explanation:
We will use the MKS system of unit, so:
now, The angular speed is given by:
in order to obtain the angular acceleration we have to apply the following formula:
The linear displacement is given by:
