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Svet_ta [14]
3 years ago
10

What is the maximum speed with which a 1200-kg car can round a turn of radius 90.0 m on a flat road if the coefficient of fricti

on between tires and road is 0.80?
Physics
1 answer:
RideAnS [48]3 years ago
5 0

Answer:

Maximum speed of the car is 26.56 m/s.

Explanation:

Given that,

Mass of the car, m = 1200 kg

Radius of the curve, r = 90 m

The coefficient of friction between tires and the road is 0.8.

We need to find the maximum speed of the car. On the circular curve, the centripetal force is balanced by the force of friction. So,

\dfrac{v^2}{r}=\mu g\\\\v=\sqrt{\mu gr} \\\\v=\sqrt{0.8\times 9.8\times 90} \\\\v=26.56\ m/s

So, the maximum speed of the car is 26.56 m/s.

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svlad2 [7]
The answer is actually 36 J.

Hope this helped :)

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7 0
2 years ago
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A force of 250 N is applied to a hydraulic jack piston that is 0.02 m in diameter. If the piston that supports the load has a di
Vikentia [17]

Answer:

option B

Explanation:

given,

Force exerted by the hydraulic jack piston = F₁ = 250 N

diameter of piston, d₁ = 0.02 m

                                r₁ = 0.01 m

diameter of second piston,  d₂ = 0.15 m

                                r₂ = 0.075 m

mass of the jack to lift = ?

now,

    \dfrac{F_1}{A_1} =\dfrac{F_2}{A_2}

    \dfrac{250}{\pi r_1^2} =\dfrac{F_2}{\pi r_2^2}

    \dfrac{250}{0.01^2} =\dfrac{F_2}{0.075^2}

    F_2= \dfrac{250}{0.01^2}\times {0.075^2}

               F₂ = 14062.5 N

F = m g

m = \dfrac{F}{g}

m = \dfrac{14062.5}{9.8}

m = 1435 Kg

hence, the correct answer is option B

5 0
3 years ago
How can Newton's third law describe the forces affecting a rocket as it
rusak2 [61]

Answer:D

Explanation:

Apex

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3 years ago
A circular grill of diameter 0.25 m has an emissivity of 0.8. If the surface temperature is maintained at 150°C, determine the r
Schach [20]

Answer:

the required electrical power when the room air and surroundings are at 30°C.= 52.51822 Watt

Explanation:

Power required to maintain the surface temperature at 150°C from 20°C

P= εσA(T^4-t^4)

P= power in watt

ε= emissivity

A=  area of surface

T= 150°C= 423 K

t= 20°C= 303K

/sigma= 5.67×10^{-8} watt/m^2K^4

putting vales we get

= 0.8\times5.67\times10^{-8} \pi\frac{0.25^4}{4}(423^4-303^4)

P=52.51822 Watt

the required electrical power when the room air and surroundings are at 30°C.= 52.51822 Watt

4 0
2 years ago
. In a solid, particles do not move around each
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Answer:

I am sure the awnser to this is

B. vibrate

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