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Zolol [24]
1 year ago
6

One particular descent goes from 2100m to 1600m. Assuming work done against friction is 90% of the potential energy change of th

e cyclist and the cycle, what increase in speed in Km/h can a rider attain by the end of the
descent?
Physics
1 answer:
alisha [4.7K]1 year ago
7 0

Answer:

1/2 M V^2 = .1 M g H       where 10% of PE goes into KE

V^2 = .2 g H = .2 * 9.8 * (2100 - 1600) = 980 m^2 / s^2

V = 31.1 m/s       increase in speed during descent

1 km / hr = 1000 m / 3600 sec = .278 m/s

V = 31.1 m/s / (.278 m/s / km /hr)= 112 km/hr

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The force needed to accelerate a stationary 2000 racing car to a speed of 144 km/h in a distance of 80 meters
CaHeK987 [17]

Answer:

518400 Newton

Explanation:

Force F = mass m * Acceleration a

a = velocity v / time t

v = distance d / t

Therefore;

F = m * v² / d

with m = 2000 (unit not defined but say g)

v = 144 Km/h

d = 80 meters =0.08Km

F = 2Kg * 144Km/h * 144Km/h / 0.08Km

F = 518400 Newton

7 0
3 years ago
1.The lunch lady pushes a 100 kg zombie with 300 N of force. How much is the zombie accelerated?
Molodets [167]

Answer:

1. A=3.00m/s  2.m=50kg

Explanation:

1. Use the formula a=f/m

a=300/100

a=3

2.Use the formula m=f/a

m=1000/20

m=50kg

5 0
3 years ago
A 2.0-ohm resistor is connected in a series with a 20.0 -V battery and a three-branch parallel network with branches whose resis
Llana [10]

Answer:

Explanation:

Resolving the parallel resistor branches

\frac{1}{Rtotal} =\frac{1}{8.0ohm} +\frac{1}{8.0ohm} +\frac{1}{8.0ohm} =2.67Ohm

The equivalent resistor is now in series with the 2.0 Ohm resistor

so, by using Ohm's Law

I=\frac{V}{Rtotal+R} \\I=\frac{20.0V}{2.67Ohm+2.0Ohm} \\\\I= 4.28A

4 0
3 years ago
How to work out this problem??
tensa zangetsu [6.8K]
Veo en youtube es buena explicacion hablame por mensaje y te doy el link
8 0
3 years ago
A 2.0-kg mass is oscillating about the origin at 24 rad/s. The amplitude of the oscillations is 0.040 m. At what position is the
Darya [45]

Answer:

0.0327 m

Explanation:

m = 2 kg

ω = 24 rad/s

A = 0.040 m

Let at position y, the potential energy is twice the kinetic energy.

The potential energy is given by

U = 1/2 m x ω² x y²

The kinetic energy is given by

K = 1/2 m x ω² x (A² - y²)

Equate both the energies as according to the question

1/2 m x ω² x y² = 2 x 1/2 m x ω² x (A² - y²)

y² = 2 A² - 2 y²

3y² = 2A²

y² = 2/3 A²

y = 0.82 A = 0.82 x 0.040 = 0.0327 m

4 0
3 years ago
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