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3 years ago

6

A race car travels 44.3 m/s around a banked (45° with the horizontal) circular (radius = 200 m) track. What is the magnitude of

the resultant force in kN on the 80 kg driver of this car?

1 answer:

djverab [1.8K]3 years ago

3 0

The magnitude of the resultant force is given by the centripetal force, since the car is under a circular motion. So, we have:

$F_c=ma_c$

The centripetal acceleration is given by:

$a_c=\frac{v^2}{r}$

Where v is the linear speed and r the radius of the circular motion. Replacing this and solving:

$F=m\frac{v^2}{r}\\F=80kg\frac{(44.3\frac{m}{s})^2}{200m}\\F=785N*\frac{1kN}{1000N}\\F=0.785kN$

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marishachu [46]

Answer:

Explanation:

Given

mass of automobile $m=1000\ kg$

velocity of automobile $v=10\ m/s$

If the automobile crashes into the brick wall then it's kinetic energy is converted into thermal and demolishing energy to wall.

Car completely stops as it crashes into the wall so Energy converted into demolition and thermal energy

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A car travels at a constant speed around a circular track whose radiu is 2.6 km. The goes once arond the track in 360s . What is

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Answer:

Centripetal acceleration = 0.79 m/s²

Explanation:

<u>Given the following data;</u>

Radius, r = 2.6 km

Time = 360 seconds

<em><u>Conversion:</u></em>

2.6 km to meters = 2.6 * 1000 = 2600 meters

To find the magnitude of centripetal acceleration;

First of all, we would determine the circular speed of the car using the formula;

$Circular \; speed (V) = \frac {2 \pi r}{t}$

Where;

r represents the radius and t is the time.

Substituting into the formula, we have;

$Circular \; speed (V) = \frac {2*3.142*2600}{360}$

$Circular \; speed (V) = \frac {16338.4}{360}$

Circular speed, V = 45.38 m/s

Next, we find the centripetal acceleration;

Mathematically, centripetal acceleration is given by the formula;

$Centripetal \; acceleration = \frac {V^{2}}{r}$

Where;

V is the circular speed (velocity) of an object.
r is the radius of circular path.

Substituting into the formula, we have;

$Centripetal \; acceleration = \frac {45.38^{2}}{2.6}$

$Centripetal \; acceleration = \frac {2059.34}{2600}$

<em>Centripetal acceleration = 0.79 m/s²</em>

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The gravitational acceleration at any distance r is given by

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The Earth's radius is $r_e=6.37 \cdot 10^6 m$ , so the meteoroid is located at a distance of:

$r=r_e+2.60 r_e =3.60 r_e = 2.29 \cdot 10^7 m$

And by substituting this value into the previous formula, we can find the value of g at that altitude:

$g= \frac{GM}{r^2} = \frac{(6.67 \cdot 10^{-11} m^3 kg^{-1} s^{-2})(5.97 \cdot 10^{24} kg)}{(2.29 \cdot 10^7 m)^2} =0.75 m/s^2$

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Answer:

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4 years ago

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Answer:

$1.63\cdot 10^{-24} J$

Explanation:

The energy of a photon is given by:

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In this problem, we have a microwave photon with wavelength

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