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STatiana [176]
3 years ago
15

A car racing on a flat track travels at 22 m/s around a curve with a 56-m radius. Find the car’s centripetal acceleration. What

minimum coefficient of static friction between the tires and road is necessary for the car to round the curve without slipping?
Physics
1 answer:
Temka [501]3 years ago
8 0

Answer:

0.88

Explanation:

velocity (v) = 22 m/s

radius (r) = 56 m

acceleration due to gravity (g) = 9.8 m/s^{2}

What minimum coefficient of static friction between the tires and road is necessary for the car to round the curve without slipping?

for the car to successfully go round the curve, the frictional force must be equal to or greater than the normal force, therefore

Normal force = frictional force

ma = μmg

where

  • m = mass of the car
  • μ = coefficient of friction
  • g = acceleration due to gravity
  • a = \frac{v^{2} }{r} = \frac{22^{2} }{56} = 8.6
  • putting in all required values the equation now becomes

        8.6m = μm x 9.8

        8.6 = 9.8 x  μ

          μ = 8.6/9.8 = 0.88

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1,000 grams = 1 kilogram
20 grams = 0.02 kilogram

Kinetic energy = (1/2) (mass) x (speed)²

                           (1/2) (0.02) x (15)² =

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4 0
3 years ago
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This is about the magnet fields. thanks in advance.​
Wewaii [24]
The answer is A. The outer lines change as it moves
8 0
2 years ago
A radar for tracking aircraft broadcasts a 12 GHz microwave beam from a 2.0-m-diameter circular radar antenna. From a wave persp
NISA [10]

A) 750 m

First of all, let's find the wavelength of the microwave. We have

f=12GHz=12\cdot 10^9 Hz is the frequency

c=3.0\cdot 10^8 m/s is the speed of light

So the wavelength of the beam is

\lambda=\frac{c}{f}=\frac{3\cdot 10^8 m/s}{12\cdot 10^9 Hz}=0.025 m

Now we can use the formula of the single-slit diffraction to find the radius of aperture of the beam:

y=\frac{m\lambda D}{a}

where

m = 1 since we are interested only in the central fringe

D = 30 km = 30,000 m

a = 2.0 m is the aperture of the antenna (which corresponds to the width of the slit)

Substituting, we find

y=\frac{(1)(0.025 m)(30000 m)}{2.0 m}=375 m

and so, the diameter is

d=2y = 750 m

B) 0.23 W/m^2

First we calculate the area of the surface of the microwave at a distance of 30 km. Since the diameter of the circle is 750 m, the radius is

r=\frac{750 m}{2}=375 m

So the area is

A=\pi r^2 = \pi (375 m)^2=4.42\cdot 10^5 m^2

And since the power is

P=100 kW = 1\cdot 10^5 W

The average intensity is

I=\frac{P}{A}=\frac{1\cdot 10^5 W}{4.42\cdot 10^5 m^2}=0.23 W/m^2

4 0
3 years ago
Please help me anyone i will mark brainliest
Mashcka [7]

Answer:

for 4.567 I can't tell if it x 10^{3} or x 10^{5} so:

answer using x 10^{3} = 0.0000644697N

answer using x 10^{5} = 0.00644697

Explanation:

use equation F = GMm/R^{2}

3 0
3 years ago
A 1.0 kg piece of copper with a specific heat of cCu=390J/(kg⋅K) is placed in 1.0 kg of water with a specific heat of cw=4190J/(
Valentin [98]

Answer:

The temperature change of the copper is greater than the temperature change of the water.

Explanation:

deltaQ = mc(deltaT)

Where,

delta T = change in the temperature

m =mass

c = heat capacity

\frac{(deltaT)_{Cu}}{(deltaT)_{w}}=\frac{4190J/kg.K}{390J/kg.K}\\ \\(deltaT)_{Cu}=10.74(deltaT)_{w}

The temperature change in the copper is nearly 11 times the temperature change in the water.

So, the correct option is,

The temperature change of the copper is greater than the temperature change of the water.

Hope this helps!

7 0
3 years ago
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