the velocity of a car traveling in the positive direction decreases from 32 m/s to 24 m/s in 4 seconds. what is the average acce
1 answer:
Answer:
Negative sign shows that velocity of the car is decreases at a constant rate
Explanation:
We have given velocity of the car is decreases from 32 m /sec to 24 m/sec in 4 second
So initial velocity of the car u = 32 m /sec
And finally car reaches to a velocity of 24 m/sec
Time taken to change in velocity = 4 sec
So final velocity v = 24 m/sec
From first equation of motion v = u+at
So
Negative sign shows that velocity of the car is decreases at a constant rate
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Given:
v = 50.0 m/s, the launch velocity
θ = 36.9°, the launch angle above the horizontal
Assume g = 9.8 m/s² and ignore air resistance.
The vertical component of the launch velocity is
Vy = (50 m/s)*sin(50°) = 30.02 m/s
The time, t, to reach maximum height is given by
(30.02 m/s) - (9.8 m/s²)*(t s) = 0
t = 3.0634 s
The time fo flight is 2*t = 6.1268 s
The horizontal velocity is
u = (50 m/s)cos(36.9°) = 39.9842 m/s
The horizontal distance traveled at time t is given in the table below.
Answer:
t, s x, m
------ --------
0 0
1 39.98
2 79.79
3 112.68
4 159.58
5 199.47
6 239.37
v = 50.0 m/s, the launch velocity
θ = 36.9°, the launch angle above the horizontal
Assume g = 9.8 m/s² and ignore air resistance.
The vertical component of the launch velocity is
Vy = (50 m/s)*sin(50°) = 30.02 m/s
The time, t, to reach maximum height is given by
(30.02 m/s) - (9.8 m/s²)*(t s) = 0
t = 3.0634 s
The time fo flight is 2*t = 6.1268 s
The horizontal velocity is
u = (50 m/s)cos(36.9°) = 39.9842 m/s
The horizontal distance traveled at time t is given in the table below.
Answer:
t, s x, m
------ --------
0 0
1 39.98
2 79.79
3 112.68
4 159.58
5 199.47
6 239.37
12) The weight of the plane is
13) The lift provided by the wing is
Explanation:
12)
The weight of an object is equal to the force of gravity acting on the object. Near the Earth's surface, it can be calculated as
where
W is the weight
m is the mass of the object
g is the acceleration of gravity
In this problem, we know that the mass of the plane is
m = 75 tonnes
Since 1 ton = 1000 kg, the mass in kg is
m = 75,000 kg
Also, the acceleration of gravity is
Therefore, the weight of the plane is:
13)
We can solve this question by applying Newton's second law along the vertical direction of motion. In fact, the net force acting along the vertical direction must be equal to the product between the mass of the plane and the vertical acceleration:
where
is the net force in the y-direction
m is the mass of the plane
is the acceleration in the y-direction
There are two forces acting in the vertical direction:
- The lift L, acting upward
- The weight W, acting downward
So the vertical net force is
And the equation becomes
Also, we know that the plane is travelling in level flight, which means that the vertical acceleration is zero:
Therefore, we get
And so the lift is
Learn more about forces:
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