Solution: From the given question, we shall find the vector quantity among the
(A) Time , (B) Velocity, (C) Distance , (D) Speed
Concept: <u>Vector Quantity: </u>All those physical quantities which have magnitude as well as specific directions, are called Vector Quantities.
Here, Time, Distance and Speed have only magnitude but have no directions so they will be scalar quantities.
Now, <u>Velocity:</u> It is defined as the change in displacement per unit time. Since the change in the displacement will be in particular direction only. Hence, velocity will be the vector quantity.
Hence, the option (B) Velocity will be the correct option.
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Answer:
The shortest distance in which you can stop the automobile by locking the brakes is 53.64 m
Explanation:
Given;
coefficient of kinetic friction, μ = 0.84
speed of the automobile, u = 29.0 m/s
To determine the the shortest distance in which you can stop an automobile by locking the brakes, we apply the following equation;
v² = u² + 2ax
where;
v is the final velocity
u is the initial velocity
a is the acceleration
x is the shortest distance
First we determine a;
From Newton's second law of motion
∑F = ma
F is the kinetic friction that opposes the motion of the car
-Fk = ma
but, -Fk = -μN
-μN = ma
-μmg = ma
-μg = a
- 0.8 x 9.8 = a
-7.84 m/s² = a
Now, substitute in the value of a in the equation above
v² = u² + 2ax
when the automobile stops, the final velocity, v = 0
0 = 29² + 2(-7.84)x
0 = 841 - 15.68x
15.68x = 841
x = 841 / 15.68
x = 53.64 m
Thus, the shortest distance in which you can stop the automobile by locking the brakes is 53.64 m
The acceleration due to gravity is g/4
The acceleration above the earth surface is given by the relation
g^'=gr^2/〖(h+r)〗^2
Since the satellite orbits the earth in a orbit of radius equal to earth radius, therefore
g^'=(gr^2)/〖(r+r)〗^2 =g/4
Thus the acceleration due to gravity on the satellite is g/4.
Answer:
Explanation:
A mass of 700 kg will exert a force of
700 x 9.8
= 6860 N.
Amount of compression x = 4 cm
= 4 x 10⁻² m
Force constant K = force of compression / compression
= 6860 / 4 x 10⁻²
= 1715 x 10² Nm⁻¹.
Let us take compression of r at any moment
Restoring force by spring
= k r
Force required to compress = kr
Let it is compressed by small length dr during which force will remain constant.
Work done
dW = Force x displacement
= -kr -dr
= kr dr
Work done to compress by length d
for it r ranges from 0 to -d
Integrating on both sides
W = 
= [ kr²/2]₀^-4
= 1/2 kX16X10⁻⁴
= .5 x 1715 x 10² x 16 x 10⁻⁴
= 137.20 J