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IgorLugansk [536]
2 years ago
6

Our textbook describes a common model for friction (both static and kinetic) in which the area of contact between interacting su

rfaces is irrelevant to the size of the frictional force. However, we can't help noticing how wide the tires are on drag racing cars! One explanation for the large size of the tires would be to give a larger contact patch between the tire and the track in search of more traction. But our model says contact area doesn't matter. Clarify this situation in a mini essay of 2 or 3 sentences. Sort out whether contact area matters or not. Can you think of other advantages of really wide tires -- in addition to how cool they look.
Physics
1 answer:
Mkey [24]2 years ago
7 0

Contact area is not a factor of the amount of friction force acting between two bodies in contact, preventing rolling or sliding

Reasons:

  • Friction force is the product of the normal reaction and the coefficient of friction.
  • The coefficient of friction is higher in softer tires than hard tires which are required to be wider to reduce the pressure that can cause deformation withstood by hard tires.
  • The wide tires in drag racing cars can be thought of being made from the soft compound tire material having a higher coefficient of friction that increases the friction force.

Other advantages of wide tires are;

  • Increased traction
  • Shorter stopping distance

Learn more here:

brainly.com/question/17209968

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How does a distance-time graph show you which direction the runner is moving? Use your own words and science vocabulary.
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In a distance time graph, when the x and y values are positive (in first quadrant), the runner is moving forward.

While, if the distance value ( in the y axis) is negative, the runner is moving backwards ( towards the start) .

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How do waves shape Earth’s surface? HELP PLS 10 POINTS PLUS BRAINLEST
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Answer:Waves shape the earths surface because they change the form of the coastal land form, meaning that when a wave is formed and moves they are also moving sand and rocks which can change the shape of the surface

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3 years ago
Calculate the Schwarzschild radius (in kilometers) for each of the following.1.) A 1 ×108MSun black hole in the center of a quas
Westkost [7]

Answer:

(I). The Schwarzschild radius is 2.94\times10^{8}\ km

(II). The Schwarzschild radius is 17.7 km.

(III). The Schwarzschild radius is 1.1\times10^{-7}\ km

(IV). The Schwarzschild radius is 7.4\times10^{-29}\ km

Explanation:

Given that,

Mass of black hole m= 1\times10^{8} M_{sun}

(I). We need to calculate the Schwarzschild radius

Using formula of radius

R_{g}=\dfrac{2MG}{c^2}

Where, G = gravitational constant

M = mass

c = speed of light

Put the value into the formula

R_{g}=\dfrac{2\times6.67\times10^{-11}\times1\times10^{8}\times1.989\times10^{30}}{(3\times10^{8})^2}

R_{g}=2.94\times10^{8}\ km

(II). Mass of block hole m= 6 M_{sun}

We need to calculate the Schwarzschild radius

Using formula of radius

R_{g}=\dfrac{2MG}{c^2}

Put the value into the formula

R_{g}=\dfrac{2\times6.67\times10^{-11}\times6\times1.989\times10^{30}}{(3\times10^{8})^2}

R_{g}=17.7\ km

(III). Mass of block hole m= mass of moon

We need to calculate the Schwarzschild radius

Using formula of radius

R_{g}=\dfrac{2MG}{c^2}

Put the value into the formula

R_{g}=\dfrac{2\times6.67\times10^{-11}\times7.35\times10^{22}}{(3\times10^{8})^2}

R_{g}=1.1\times10^{-7}\ km

(IV). Mass = 50 kg

We need to calculate the Schwarzschild radius

Using formula of radius

R_{g}=\dfrac{2MG}{c^2}

Put the value into the formula

R_{g}=\dfrac{2\times6.67\times10^{-11}\times50}{(3\times10^{8})^2}

R_{g}=7.4\times10^{-29}\ km

Hence, (I). The Schwarzschild radius is 2.94\times10^{8}\ km

(II). The Schwarzschild radius is 17.7 km.

(III). The Schwarzschild radius is 1.1\times10^{-7}\ km

(IV). The Schwarzschild radius is 7.4\times10^{-29}\ km

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Which property increases as an electromagnetic waves energy decreases
Rufina [12.5K]

Answer:

Wavelength

Explanation:

3 0
3 years ago
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