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GenaCL600 [577]
3 years ago
9

A race car has a centripetal acceleration of 13.33 m/s^2 as it goes around a curve. if the curve is a circle with a radius 30 m

what is the speed of the car?
Physics
1 answer:
anzhelika [568]3 years ago
3 0

Answer:

The speed of the car, v = 19.997 m/s

Explanation:

Given,

The centripetal acceleration of the car, a = 13.33 m/s²

The radius of the curve, r = 30 m

The centripetal force acting on the car is given by the formula

                                   F = mv²/r

Where    v²/r is the acceleration component of the force

                                       a = v²/r

Substituting the values in the above equation

                                        13.33 = v²/30

                                         v² = 13.33 x 30

                                         v² = 399.9

                                         v = 19.997 m/s

Hence, the speed of the car, v = 19.997 m/s

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kompoz [17]

the earth exerts a gravitational force

4 0
3 years ago
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An air hockey game has a puck of mass 30 grams and a diameter of 100 mm. The air film under the puck is 0.1 mm thick. Calculate
OverLord2011 [107]

Answer:

time required after impact for a puck is 2.18 seconds

Explanation:

given data

mass = 30 g = 0.03 kg

diameter = 100 mm = 0.1 m

thick = 0.1 mm = 1 ×10^{-4} m

dynamic viscosity = 1.75 ×10^{-5} Ns/m²

air temperature = 15°C

to find out

time required after impact for a puck to lose 10%

solution

we know velocity varies here 0 to v

we consider here initial velocity = v

so final velocity = 0.9v

so change in velocity is du = v

and clearance dy = h

and shear stress acting on surface is here express as

= µ \frac{du}{dy}

so

= µ  \frac{v}{h}   ............1

put here value

= 1.75×10^{-5} × \frac{v}{10^{-4}}

= 0.175 v

and

area between air and puck is given by

Area = \frac{\pi }{4} d^{2}

area  =  \frac{\pi }{4} 0.1^{2}

area = 7.85 × \frac{v}{10^{-3}} m²

so

force on puck is express as

Force = × area

force = 0.175 v × 7.85 × 10^{-3}

force = 1.374 × 10^{-3} v    

and now apply newton second law

force = mass × acceleration

- force = mass \frac{dv}{dt}

- 1.374 × 10^{-3} v = 0.03 \frac{0.9v - v }{t}

t =  \frac{0.1 v * 0.03}{1.37*10^{-3} v}

time = 2.18

so time required after impact for a puck is 2.18 seconds

3 0
3 years ago
You are given four resistors, 2 ohms, 3 ohms, 5 ohms, and 10 ohms. Your friend say you can connect them so you obtain an equival
AlexFokin [52]

If we will connect the resistors 2ohms, 3ohms, 5ohms in series and the 10ohms resistance parallel then we get equivalent resistance of 5 ohms.

The equivalent circuit is,

R equivalent for the series connection is,

\begin{gathered} Req(S)=2+3+5=10ohms \\ Now,\text{ } \\ Req\text{ }for\text{ }10\text{ }ohms\text{ }and\text{ }10\text{ }ohms\text{ }is, \\ \frac{1}{Req}=\frac{1}{10}+\frac{1}{10}=\frac{2}{10}=\frac{1}{5} \\ So, \\ Req=5\text{ ohms} \end{gathered}

The equivalent resistance is 5 ohms.

So your friend is saying true.

6 0
1 year ago
The sun emits electromagnetic waves with a power of 4.0 × 10²⁶ W. Determine the intensity of electromagnetic waves from the sun
Sphinxa [80]

Answer:

I_v = 2,700 W / m^2

I_m = 610 W / m^2

I_s = 16 W / m^2

Explanation:

Given:

- The Power of EM waves emitted by Sun P_s = 4.0*10^26 W

- Radius of Venus r_v = 1.08 * 10^11 m

- Radius of Mars r_m = 2.28 * 10^11 m

- Radius of Saturn r_s = 1.43 * 10^12 m

Find:

Determine the intensity of electromagnetic waves from the sun just outside the atmospheres of (a) Venus, (b) Mars, and (c) Saturn.

Solution:

- We know that Power is related to intensity and surface area of an object follows:

                                        I = P / 4*pi*r^2

Where, A is the surface area of a sphere models the atmosphere around the planets.

a)

- The intensity at the surface of Venus is calculated as:

                                       I_v = P_s / 4*pi*r^2_v

                                       I_v = 4.0*10^26 / 4*pi*(1.08*10^11)^2

                                       I_v = 2,700 W / m^2

b)

- The intensity at the surface of Mars is calculated as:

                                       I_m = P_s / 4*pi*r^2_m

                                       I_m = 4.0*10^26 / 4*pi*(2.28*10^11)^2

                                      I_m = 610 W / m^2

c)

- The intensity at the surface of Saturn is calculated as:

                                       I_s = P_s / 4*pi*r^2_s

                                       I_s = 4.0*10^26 / 4*pi*(1.43*10^12)^2

                                      I_s = 16 W / m^2

7 0
3 years ago
What is electricity?
ASHA 777 [7]

Answer:

The flow of electric charge

Explanation:

Electricity:It is defined flow of electric charge.

It is defined as the flow of charge per unit time.

Electric charge can be negative or positive .

Mathematical representation :

If charge Q flowing through the conductor and time taken in flow of charge is t seconds.

Then, current=I=\frac{Q}{t}

S.I unit of current is Ampere.It is scalar quantity.

Answer:The flow of electric charge.

8 0
3 years ago
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