We will solve this question using the second law of motion which states that force is directly equal to the product of mass and acceleration.
![\sf \: F=ma](https://tex.z-dn.net/?f=%20%5Csf%20%5C%3A%20F%3Dma)
Where,
- F is force
- m is mass
- a is acceleration
In our case,
- F = ?
- m = 2500 kg
- a = 20m/s
![\tt \: F_{net} = 2500 \times 20 \\ \tt= 50000](https://tex.z-dn.net/?f=%20%5Ctt%20%5C%3A%20F_%7Bnet%7D%20%20%3D%202500%20%5Ctimes%2020%20%5C%5C%20%20%20%5Ctt%3D%2050000)
<em>Thus, The force of 50000 Newton is required to accelerate a car of 2500 kg...~</em>
Answer:
11
Explanation:
1. You are going to be rounding down.
2. change the metric ton to kg.
1.000 * 10^3 kg = 1000 kg
1000 / 87 = 11.49 = 11 people
Answer: 16.3 seconds
Explanation: Given that the
Initial velocity U = 80 ft/s
Let's first calculate the maximum height reached by using third equation of motion.
V^2 = U^2 - 2gH
Where V = final velocity and H = maximum height.
Since the toy is moving against the gravity, g will be negative.
At maximum height, V = 0
0 = 80^2 - 2 × 9.81 × H
6400 = 19.62H
H = 6400/19.62
H = 326.2
Let's us second equation of motion to find time.
H = Ut - 1/2gt^2
Let assume that the ball is dropped from the maximum height. Then,
U = 0. The equation will be reduced to
H = 1/2gt^2
326.2 = 1/2 × 9.81 × t^2
326.2 = 4.905t^2
t^2 = 326.2/4.905
t = sqrt( 66.5 )
t = 8.15 seconds
The time it will take for the rocket to return to ground level will be 2t.
That is, 2 × 8.15 = 16.3 seconds
Answer:
Acceleration of the car will be ![a=0.1375m/sec^2](https://tex.z-dn.net/?f=a%3D0.1375m%2Fsec%5E2)
Explanation:
We have given mass of the ball m = 1600 kg
Force in north direction F= 7560 N
Resistance force which opposes the movement of car ![F_R=7340N](https://tex.z-dn.net/?f=F_R%3D7340N)
So net force on the car ![F_{net}=F-F_R=7560-7340=220N](https://tex.z-dn.net/?f=F_%7Bnet%7D%3DF-F_R%3D7560-7340%3D220N)
According to second law of motion we know that F=ma
So ![220=1600\times a](https://tex.z-dn.net/?f=220%3D1600%5Ctimes%20a)
![a=0.1375m/sec^2](https://tex.z-dn.net/?f=a%3D0.1375m%2Fsec%5E2)