Answer:
a) 0 metres
b) From time 0 s to 10 s , the car was accelerated. Its velocity accelerated from 0m/s to 20 m/s
c) 20 m/s
Explanation:
a) <em>Formula of displacement= velocity x time</em>
time=40 s
velocity =0 m/s
∴ displacement= 0 x 40 = 0 m
Magnitude of displacement is 0 m
b) The increase in velocity shows that there has been acceleration.
c) The average velocity of the car is =
{initial velocity + final velocity}
=
=20
Therefore, the magnitude of the average velocity of the car is 20 m/s
The minimum speed of the water must be 3.4 m/s
Explanation:
There are two forces acting on the water in the pail when it is at the top of its circular motion:
- The force of gravity, mg, acting downward (where m is the mass of the water and g the acceleration of gravity)
- The normal reaction, N also acting downward
Since the water is in circular motion, the net force must be equal to the centripetal force, so:
Where:
v is the speed of the pail
r = 1.2 m is the radius of the circle
The water starts to spill out when the normal reaction of the pail becomes zero:
N = 0
When this occurs, the equation becomes:
And substitutin the values of g and r, we find the minimum speed that the water must have in order not to spill out:
Learn more about circular motion:
brainly.com/question/2562955
brainly.com/question/6372960
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Answer:
My scenario would be A Car vs. a guard rail on a road. You have a car that is coming down a Highway at a speed of 43 Mph Miles per hour (69.2018 Kmh)
And it hits a steel guardrail and the car smashes in at the front and the guardrail is only bent while the car has the bumper and the hood along with the headlights and windshield along with the passenger side window break.
Explanation:
This is caused by so much force reacting from one object to another but also depends on molecular density.
Answer:
a) x = v₀² sin 2θ / g
b) t_total = 2 v₀ sin θ / g
c) x = 16.7 m
Explanation:
This is a projectile launching exercise, let's use trigonometry to find the components of the initial velocity
sin θ = 
/ vo
cos θ = v₀ₓ / vo
v_{oy} = v_{o} sin θ
v₀ₓ = v₀ cos θ
v_{oy} = 13.5 sin 32 = 7.15 m / s
v₀ₓ = 13.5 cos 32 = 11.45 m / s
a) In the x axis there is no acceleration so the velocity is constant
v₀ₓ = x / t
x = v₀ₓ t
the time the ball is in the air is twice the time to reach the maximum height, where the vertical speed is zero
v_{y} = v_{oy} - gt
0 = v₀ sin θ - gt
t = v_{o} sin θ / g
we substitute
x = v₀ cos θ (2 v_{o} sin θ / g)
x = v₀² /g 2 cos θ sin θ
x = v₀² sin 2θ / g
at the point where the receiver receives the ball is at the same height, so this coincides with the range of the projectile launch,
b) The acceleration to which the ball is subjected is equal in the rise and fall, therefore it takes the same time for both parties, let's find the rise time
at the highest point the vertical speed is zero
v_{y} = v_{oy} - gt
v_{y} = 0
t = v_{oy} / g
t = v₀ sin θ / g
as the time to get on and off is the same the total time or flight time is
t_total = 2 t
t_total = 2 v₀ sin θ / g
c) we calculate
x = 13.5 2 sin (2 32) / 9.8
x = 16.7 m
