Average speed = (total distance covered) / (time to cover the distance)
total distance covered = (4km + 2km + 1km) = 7 km
time to cover the distance = (32min + 22min + 16min) = 70 min
Average speed = (7 km) / (70 min)
Average speed = 0.1 km/minute
Answer:
kinetic friction may be greater than 400 N or smaller than 400 N
Explanation:
As we know that maximum value of static friction on the rough surface is known as limiting friction and the formula of this limiting friction is known as

now when object is sliding on the rough surface then the friction force on that surface is known as kinetic friction and the formula of kinetic friction is known as

now we know that

so here value of limiting static friction force is always more than kinetic friction
also we know that
initially when body is at rest then static friction value will lie from 0 N to maximum limiting friction
and hence kinetic friction may be greater than static friction or if the static friction is maximum limiting friction then kinetic friction is smaller than static friction
so kinetic friction may be greater than 400 N or smaller than 400 N
Answer:
1. 20.54m/s
2. 1.52s
Explanation:
QUESTION 1:
The speed the stone impact the ground is the final speed/velocity, which can be calculated using the formula:
v² = u² + 2as
Where;
v = final velocity (m/s)
u = initial velocity (m/s)
a = acceleration due to gravity (m/s²)
s = distance (m)
From the provided information, u = 5.65m/s, v = ?, s = 19.9m, a = 9.8m/s²
v² = 5.65² + 2 (9.8 × 19.9)
v² = 31.9225 + 2 (195.02)
v² = 31.9225 + 390.04
v² = 421.9625
v = √421.9625
v = 20.5417
v = 20.54m/s
QUESTION 2:
Using v = u + at
Where v = final velocity (m/s) = 20.54m/s
t = time (s)
u = initial velocity (m/s) = 5.65m/s
a = acceleration due to gravity (m/s²)
v = u + at
20.54 = 5.65 + 9.8t
20.54 - 5.65 = 9.8t
14.89 = 9.8t
t = 14.89/9.8
t = 1.519
t = 1.52s
The correct answer to the question is : 9375 N.
CALCULATION:
As per the question, the mass of the car m = 1500 Kg.
The diametre of the circular track D = 200 m.
Hence, the radius of the circular path R = 
= 
= 100 m.
The velocity of the truck v = 25 m/s.
When a body moves in a circular path, the body needs a centripetal force which helps the body stick to the orbit. It acts along the radius and towards the centre.
Hence, the force acting on the car is centripetal force.
The magnitude of the centripetal force is calculated as -
Force F = 
= 
= 9375 N. [ANS}
The centripetal force is provided to the car in two ways. It is the friction which provides the necessary centripetal force. Sometimes friction is not sufficient. At that time, the road is banked to some extent which provides the necessary centripetal force.
Answer:
C the baseball began at rest and rolls at a rate of 14.7 m/s after 1.5 seconds