<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:

east of south
Explanation:
Given:
- distance of the person form the initial position,

- direction of the person from the initial position,
north of east
- distance supposed to travel form the initial position,

- direction supposed to travel from the initial position, due North
<u>Now refer the schematic for visualization of situation:</u>

...............(1)

.................(2)
<u>Now the direction of the desired position with respect to south:</u>


east of south
<u>Now the distance from the current position to the desired position:</u>



Answer:
false
Explanation:
sound travels slower than light. that is why we see lightning before we hear the thunder
AS
work done =W = F.d = F d cosФ (Ф is angle between force F and displacement d) If a body/object is moving on a smooth surface (friction-less surface ) .There is no force acting on that body. F=0 so W=FdcosФ= (0)dcosФ ⇒ W=0
Now if a body is facing some amount of force but under the action of force there is no displacement covered. d=0 so W =FdcosФ= F(0)cosФ ⇒W=0
example: A person is applying a force on rigid wall but wall remains at rest there is no displacement occurs in wall.
The third term upon which work done dependent is angle between force and displacement i.e Ф. If Ф=90° then W= FdcosФ= Fdcos90⇒ W=0 ( as cos 90°=0)
Answer:
Magnitude of magnetic field is 1.29 x 10⁻⁴ T
Explanation:
Given :
Current flowing through the wire, I = 16.9 A
Length of wire. L = 0.69 m
Magnetic force experienced by the wire, F = 1.5 x 10⁻³ N
Consider B be the applied magnetic field.
The relation to determine the magnetic force experienced by current carrying wire is:
F = ILBsinθ
Here θ is the angle between magnetic field and current carrying wire.
According to the problem, the magnetic field and current carrying wire are perpendicular to each other, that means θ = 90⁰. So, the above equation becomes:
F = ILB

Substitute the suitable values in the above equation.

B = 1.29 x 10⁻⁴ T