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3 years ago

10

. Calculate the efficiency of a bicycle if the input work to turn the pedal is 45J and the output work is 20J. * 1 point 2.25 2.

25% 44 44.4%

1 answer:

cestrela7 [59]3 years ago

8 0

20/45=0.4*100= 44.4 so the answer is..................................................

Answer: 44.4%

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A proud new Jaguar owner drives her car at a speed of 25 m/s into a corner. The coefficients of friction between the road and th

ehidna [41]

Answer:

ac = 3.92 m/s²

Explanation:

In this case the frictional force must balance the centripetal force for the car not to skid. Therefore,

Frictional Force = Centripetal Force

where,

Frictional Force = μ(Normal Force) = μ(weight) = μmg

Centripetal Force = (m)(ac)

Therefore,

μmg = (m)(ac)

ac = μg

where,

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μ = coefficient of friction of tires (kinetic) = 0.4

g = 9.8 m/s²

Therefore,

ac = (0.4)(9.8 m/s²)

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2 years ago

Which of the following is formed by constructive erosion? 1. delta 2. gully 3. rill 4. glacier

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1. Delta, is formed by constructive erosion.

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3 years ago

How much power is required to lift a 60 kg bed to the top of a 3 meter flight of stairs in 10 seconds?

Zigmanuir [339]

Answer:

176.58Watts

Explanation:

Power= work done /time

Where mass(m)=60kg

Height (h) =3m

Time(s)=10s

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Power=mgh/t

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3 years ago

Questions E6 a& b and E7 a&b?

erica [24]

Explanation:

6a) Work = force × distance

W = Fd

W = (60 N) (10 m)

W = 600 J

6b) Change in energy = work

ΔKE = 600 J

7a) Kinetic energy is half the mass times the square of the velocity.

KE = ½ mv²

KE = ½ (0.4 kg) (25 m/s)²

KE = 125 J

7b) Work = change in energy. When the ball is stopped, it has zero kinetic energy.

W = ΔKE

W = 0 J − 125 J

W = -125 J

8 0

2 years ago

A 39-foot ladder is leaning against a vertical wall. If the bottom of the ladder is being pulled away from the wall at the rate

Viefleur [7K]

Answer:

The rate of change of the area when the bottom of the ladder (denoted by $b$ ) is at 36 ft. from the wall is the following:

$\frac{dA}{dt}|_{b=36}=-571.2\, ft^2/s$

Explanation:

The Area of the triangle is given by $A=h\times b$ where $h=\sqrt{l^2-b^2}$ (by using the Pythagoras' Theorem) and $b$ is the length of the base of the triangle or the distance between the bottom of the ladder and the wall.

The area is then

$A=\sqrt{l^2-b^2}b$

The rate of change of the area is given by its time derivative

$\frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\cdot b\right)$

$\implies \frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\right)\cdot b+\frac{db}{dt}\cdot\sqrt{l^2-b^2}$

$\implies\frac{dA}{dt}=\frac{1}{2\sqrt{l^2-b^2}}\frac{d}{dt}(l^2-b^2)\cdot b+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}$ Product rule

$\implies\frac{dA}{dt}=-\frac{1}{2\sqrt{l^2-b^2}}\cdot 2\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}$ Chain rule

$\implies\frac{dA}{dt}=-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}$

$\implies\frac{dA}{dt}=\frac{db}{dt}\left(-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2+\sqrt{l^2-b^2}}\right)$

In here we can identify $b=36\, ft$ , $l=39$ and $\frac{db}{dt}=8\,ft/s$ .

The result is then

$\frac{dA}{dt}=8\left(-\frac{1}{\sqrt{39^2-36^2}}\cdot 36^2+\sqrt{39^2-36^2}}\right)=-571.2\, ft^2/s$

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2 years ago