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Answer: 6s
Explanation:
Vs=32m/s speed at beginning of slowing down
Vf=0m/s stop speed
a= -6 m/s² acceleration
----------------
Use equation for acceleration :
a=(Vf-Vs)/t
a*t=Vf-Vs
t=(Vf-Vs)/a
t=(0-36)/-6
t=-36/-6
t=6 s
<u>Answer</u>:
The coefficient of static friction between the tires and the road is 1.987
<u>Explanation</u>:
<u>Given</u>:
Radius of the track, r = 516 m
Tangential Acceleration
= 3.89 m/s^2
Speed,v = 32.8 m/s
<u>To Find:</u>
The coefficient of static friction between the tires and the road = ?
<u>Solution</u>:
The radial Acceleration is given by,




Now the total acceleration is
=>
=>
=>
=>
The frictional force on the car will be f = ma------------(1)
And the force due to gravity is W = mg--------------------(2)
Now the coefficient of static friction is

From (1) and (2)


Substituting the values, we get


Answer:
option (E) is correct.
Explanation:
Work done is defined as the product of force and the distance in the direction of force.
force, f = 100 N
Coefficient of friction, = 0.25
distance = 15 m
So, net force F = f - friction force
F = 100 - 0.25 x m g
Work = (100 - 0.25 mg) x d cosθ
For minimum work, the angle should be maximum.
So, the value of θ is 76°.
thus, option (E) is correct.
Answer:
40.0⁰
Explanation:
The formula for calculating the magnetic flux is expressed as:
where:
is the magnetic flux
B is the magnetic field
A is the cross sectional area
is the angle that the normal to the plane of the loop make with the direction of the magnetic field.
Given
A = 0.250m²
B = 0.020T
= 3.83 × 10⁻³T· m²
3.83 × 10⁻³ = 0.020*0.250cosθ
3.83 × 10⁻³ = 0.005cosθ
cosθ = 0.00383/0.005
cosθ = 0.766
θ = cos⁻¹0.766
θ = 40.0⁰
<em>Hence the angle normal to the plane of the loop make with the direction of the magnetic field is 40.0⁰</em>