As per the question the initial speed of the car [ u] is 42 m/s.
The car applied its brake and comes to rest after 5.5 second.
The final velocity [v] of the car will be zero.
From the equation of kinematics we know that
[ here a stands for acceleration]
Here a is taken negative as it the car is decelerating uniformly.
We are asked to calculate the stopping distance .
From equation of kinematics we know that
[here S is the distance]
![= 42*5.5 +\frac{1}{2} [-7.64] [5.5]^2 m](https://tex.z-dn.net/?f=%3D%2042%2A5.5%20%2B%5Cfrac%7B1%7D%7B2%7D%20%5B-7.64%5D%20%5B5.5%5D%5E2%20m)
[ans]
Answer:
The acceleration of the collar is 10 m/s²
Explanation:
Given;
mass of the collar, m = 1 kg
applied force on the bar, F = 10 N
The acceleration of the collar can be calculated by applying Newton's second law of motion;
F = ma
where;
F is the applied force
m is mass of the object
a is the acceleration
a = F / m
a = 10 / 1
a = 10 m/s²
Therefore, the acceleration of the collar is 10 m/s²
Answer:
The Gravitational Force is reduced 4 times
Explanation:
The equation of Gravitational force follows:
F = (G*m1*m2)/r^2
Assume that G*m1*m2 = 1 and r = 1:
F = 1/1^2 = 1 N
Multiply the radius by 2
F = 1/2^2 = 1/4 N
So doubling the distance reduces the force 4 times.
A is the correct answer !!!
Given :- A resistor of 150 ohm, hence Resistance (R) = 150 ohm
Potential Difference (v) = 24 V
Current (I) = ?
V = IR
24 = I × 150
I = 24/150
I = 0.16 ampere
hope it helps!
