Answer:
(a) 89 m/s
(b) 11000 N
Explanation:
Note that answers are given to 2 significant figures which is what we have in the values in the question.
(a) Speed is given by the ratio of distance to time. In the question, the time given was the time it took the pulse to travel the length of the cable twice. Thus, the distance travelled is twice the length of the cable.
(b) The tension, 
, is given by
where 
is the speed, is the tension and 
is the mass per unit length.
Hence,
To determine , we need to know the mass of the cable. We use the density formula:
where 
is the mass and 
is the volume.
If the length is denoted by 
, then
The density of steel = 8050 kg/m3
The cable is approximately a cylinder with diameter 1.5 cm and length or height of 620 m. Its volume is
Answer:
Δv = 12 m/s, but we are not given the direction, so there are really an infinite number of potential solutions.
Maximum initial speed is 40.6 m/s
Minimum initial speed is 16.6 m/s
Explanation:
Assume this is a NET impulse so we can ignore friction.
An impulse results in a change of momentum
The impulse applied was
p = Ft = 1400(6.0) = 8400 N•s
p = mΔv
Δv = 8400 / 700 = 12 m/s
If the impulse was applied in the direction the car was already moving, the initial velocity was
vi = 28.6 - 12 = 16.6 m/s
if the impulse was applied in the direction opposite of the original velocity, the initial velocity was
vi = 28.6 + 12 = 40.6 m/s
Other angles of Net force would result in various initial velocities.
Answer:
Explanation:
Moment of inertia of a disc = 1/2 M R²
Since mass is same for both and radius are r and 2r, their moment of inertia can be in the ratio of 1: 4 . Let them be I and 4I . Angular speed are ω₀ and - ω₀ .
We shall apply law of conservation of angular momentum .
initial total angular momentum
I x ω₀ - 4I x ω₀ = - 3Iω₀
Let final common angular momentum be ω
total final angular momentum = ( I + 4I ) ω
Applying law of conservation of angular momentum
( I + 4I ) ω = - 3Iω₀
ω = - 3 / 5 ω₀ .
b )
Initial total rotational K E
= 1/2 I ω₀² + 1/2 4I ω₀²
= 1/2 x5I ω₀²
Final total rotational K E
= 1/2 ( I + 4I ) ( - 3 / 5 ω₀ )²
= 1/2 x 9 / 5 I ω₀²
= 9 / 10I ω₀²
change in rotational kinetic energy = 9 / 10I ω₀² - 1/2 x5I ω₀²
(9/10 - 5/2) xI ω₀²
=( .9 - 2.5 )I ω₀²
= - 1.6 I ω₀² Ans
Answer:
Increases
Explanation:
The expression for the capacitance is as follows as;
Here, C is the capacitance, 
is the permittivity of free space, A is the area and d is the distance between the parallel plate capacitor.
It can be concluded from the above expression, the capacitance is inversely proportional to the distance. According to the given problem, the capacitor is disconnected from the battery and the distance between the plates is increased. Then, the capacitance of the given capacitor will decrease in this case.
The expression for the energy stored in the parallel plate capacitor is as follows;
Here, E is the energy stored in the capacitor, C is the capacitance and Q is the charge.
Energy stored in the given capacitor is inversely proportional to the capacitor. The charge on the capacitor is constant. In the given problem, as the distance between the parallel plates is being separated, the energy stored in this capacitor increases.
Therefore, the option (c) is correct.
The coefficient of static friction is 0.234.
Answer:
Explanation:
Frictional force is equal to the product of coefficient of friction and normal force acting on any object.
So here the mass of the object is given as 2 kg, so the normal force will be acting under the influence of acceleration due to gravity.
Normal force = mass * acceleration due to gravity
Normal force = 2 * 9.8 = 19.6 N.
And the frictional force is given as 4.6 N, then
Coefficient of static friction = 4.6 N / 19.6 N = 0.234
So the coefficient of static friction is 0.234.
