The frictional force exerted by the road on the car is 3000N
Given the mass of the car is 1000 kg , the velocity of the car is 20m/s
and the time is 6.6s
We need to find the frictional force exerted by the road on the car
We know that Force = mass * acceleration
Now here Mass is given but acceleration is not given
So, we will find acceleration by using the formula v = u+at
Where u = 0
v = 20m/s
a = ?
t = 6.6
Substitute the values in the formula We get
v = u+at
20 = 0+(a)(6.6)
20/6.6 = a
∴ a = 3.03 m/s^2
Rounding to nearest tenth is 3m/s^2
Hence the acceleration is 3m/s^2
Now substituting the value of acceleration in F = ma
Where m = mass
a = acceleration
F = 1000*3
∴ F = 3000N
Hence the frictional force exerted by the road on the car is 3000 N
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Answer:
9. A 1500kg car traveling +6m/s with a 2000kg truck at rest. The vehicles collide, but do not stick together. The car has a velocity -3m/s after the collision. What is the velocity of the truck? a. What type of collision occurred above?
Answer:
D = 2.828 m
Explanation:
given,
distance of source of light = 1.24 m below surface of pool
refractive index of the water = n₁ = 1.33
refractive index of air = n₂ = 1
refraction angle be = 90°
let C be the critical angle
Radius = d tan C
d is the depth of the source
Using Snell's law
n₁ sin C = n₂ sin R
1.33 x sin C = 1 x sin 90°
C = 48.75°
hence,
R = 1.24 x tan 48.75°
R = 1.414 m
Diameter = 2 x R
D = 2 x 1.414
D = 2.828 m
Answer:
19. down
20. True
Explanation:
The examples I saw were projectiles launched at different angles and speeds. In each case, the only acceleration was ...
19: down
20: due to gravity (true)