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inna [77]
3 years ago
14

How much power should a braked 1.5t car have to be braked to reduce its speed from 30m / s to 10m / s in 5s?​

Physics
1 answer:
grandymaker [24]3 years ago
7 0

Answer:

-120000 W

Explanation:

Power = change in energy / time

P = ΔE / t

P = (½ mv₂² − ½ mv₁²) / t

P = m (v₂² − v₁²) / (2t)

Given m = 1.5 t = 1500 kg, v₂ = 10 m/s, v₁ = 30 m/s, and t = 5 s:

P = (1500 kg) ((10 m/s)² − (30 m/s)²) / (2 × 5 s)

P = -120000 W

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Elza [17]

Answer:

1. W = F d = 20 N * 6 m = 120 J

2. F = W / d = 60 J / 2 m = 30 N

3. d = W / F = 350 J / 85 N = 4.12 m

4. P = W / t = F d / t = 45 N * 9 m / 10 s = 40.5 Watts

5. W = P t = 500 W * 120 sec = 60,000 J

6. t = W / P = 550 J / 310 W = 1.77 sec

5 0
2 years ago
Identical spheres are dropped from a height of 100 m above the surfaces of Planet X and Planet Y. The speed of the spheres as a
Anna007 [38]

Answer:B

Explanation:

8 0
3 years ago
The position of an electron is given by , with t in seconds and in meters. At t = 3.99 s, what are (a) the x-component, (b) the
egoroff_w [7]

Answer:

A. Vx = 3.63 m/s

B. Vy = -45.73 m/s

C. |V| = 45.87 m/s

D. θ = -85.46°

Explanation:

Given that position, r, is given as:

r = 3.63tˆi − 5.73t^2ˆj + 8.16ˆk

Velocity is the derivative of position, r:

V = dr/dt = 3.63 - 11.46t^j

A. x component of velocity, Vx = 3.63 m/s

B. y component of velocity, Vy = -11.46t

t = 3.99 secs,

Vy = - 11.46 * 3.99 = -45.73 m/s

C. Magnitude of velocity, |V| = √[(-45.73)² + 3.63²]

|V| = √(2091.2329 + 13.1769)

|V| = √(2104.4098)

|V| = 45.87 m/s

D. Angle of the velocity relative to the x axis, θ is given as:

tanθ = Vy/Vx

tanθ = -45.73/3.63

tanθ = -12.6

θ = -85.46°

7 0
3 years ago
Waves transport
Brilliant_brown [7]

Answer:

d)energy

Explanation:

Waves can transfer energy over distance without moving matter the entire distance. For example, an ocean wave can travel many kilometers without the water itself moving many kilometers. The water moves up and down—a motion known as a disturbance. It is the disturbance that travels in a wave, transferring energy.

5 0
3 years ago
A child of mass 40.0 kg is in a roller coaster car that travels in a loop of radius 7.00 m. at point a the speed of the car is 1
pav-90 [236]
I attached the missing picture.
The force of seat acting on the child is a reaction the force of child pressing down on the seat. This is the third Newton's law. The force of a child pressing down the seat and the force of the seat pushing up on the child are the same.
There two forces acting on the child. The first one is the gravitational force and the second one is centrifugal force. In this example, the force of gravity is always pulling down, but centrifugal force always acts away from the center of circular motion.
Part A
For point A we have:
F_a=F_cf-F_g
In this case, the forces are aligned, centrifugal is pointing up and gravitational is pulling down.
F_a=m\frac{v^2}{r}-mg=179 $N
Part B
At the point, B situation is a bit more complicated. In this case force of gravity and centrifugal force are not aligned. We have to look at y components of this forces, y-axis, in this case, is just pointing upward.
F=F_{cf}\cos(30)-mg=m\frac{v^2}{r}\cos(30)-mg=153.2$N
Part C
The child will stay in place at point A when centrifugal force and force of gravity are in balance:
F_g=F_{cf}\\
mg=m\frac{v^2}{r}\\
gr=v^2\\
v=\sqrt{gr}=8.29\frac{m}{s}

6 0
3 years ago
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