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4 years ago

12

The distance between Earth and Mars is 225 million km. When converted using the conversion factor 1 AU = 1.5 × 108 km, the dista

nce between the planets is . The astronomical unit (AU) is a light-year.

1 answer:

noname [10]4 years ago

4 0

To convert km to AU, we divide 225,000,000 km by the factor of 1.5 x 10^8 = 150,000,000 km. This gives us 225,000,000 / 150,000,000 = 1.5 AU. Therefore, the distance between Earth and Mars in AU is 1.5 AU.
The AU is not equivalent to a light-year. A light-year is equivalent to around 9.5 x 10^12 kilometers.

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A kid is on a stationary sled, on

gregori [183]

The mass of the kid and the sled is 50.0 kg

Explanation:

The sled starts to move when the force applied becomes larger than the maximum force of static friction, which is given by

$F_{max} = \mu_s mg$

where

$\mu_s$ is the coefficient of static friction

m is the mass of the kid + the sled

$g=9.8 m/s^2$ is the acceleration of gravity

Here the sled starts moving when the force applied is 61.2 N, so

$F_{max}=61.2 N$

And we also know

$\mu_s = 0.125$

Therefore, we can find the mass of the kid and the sled by re-arranging the equation:

$m=\frac{F_{max}}{\mu_s g}=\frac{61.2}{(0.125)(9.8)}=50.0 kg$

Learn more about friction:

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6 0

4 years ago

Consider a steel guitar string of initial length L=1.00L=1.00 meter and cross-sectional area A=0.500A=0.500 square millimeters.

erma4kov [3.2K]

Answer:

The extent to which it would stretch is $\Delta L = 0.015 \ m$

Explanation:

From the question we are told that

The initial length is $L = 1.00m$

The area is $A = 0.500 mm^2 = \frac{0.500}{1 *10^6} = 0.500*10^6 \ m^2$

The Young modulus of the steel is $Y = 2.0*10^{11} Pa$

The tension is $T =1500 N$

The Young modulus is mathematically represented as

$Y = \frac{\sigma}{e}$

Where $\sigma$ is the stress which is mathematically represented as

$\sigma = \frac{F}{A}$

Substituting values

$\sigma = \frac{1500}{0.500*10^{-6}}$

$\sigma = 3.0*10^9 N/m^2$

And e is the strain which is mathematically represented as

$e = \frac{\Delta L}{L }$

Where $\Delta L$ The extension of the steel string

Substituting these into the equation above

$Y = \frac{3.0*10^9}{\frac{\Delta L}{L} }$

Substituting values

$2.0 *10^{11} = \frac{3.0*10^9}{\frac{\Delta L}{L} }$

$\Delta L = \frac{3.0*10^9 * 1}{2.0 *10^{11}}$

$\Delta L = 0.015 \ m$

7 0

3 years ago

A boxer punches a sheet of paper in midair and brings it from rest up to a speed of 30 m/s in 0.060 s .

zimovet [89]

Answer:

Force exerted, F = 1.5 N

Explanation:

It is given that, a boxer punches a sheet of paper in midair and brings it from rest up to a speed of 30 m/s in 0.060 s.

i.e. u = 0

v = 30 m/s

Time taken, t = 0.06 s

Mass of the paper, m = 0.003 kg

We need to find the force the boxer exert on it. The force can be calculated using second law of motion as :

$F=m\times a$

$F=m\times (\dfrac{v-u}{t})$

$F=0.003\times (\dfrac{30}{0.06})$

F = 1.5 N

So, the force the boxer exert on the paper is 1.5 N. Hence, this is the required solution.

6 0

3 years ago

A bus traveling at 24 m/s slows down to 12 m/s in 5.0 seconds. What is the acceleration?

Contact [7]

Answer:

a = -2.4 m/s²

Explanation:

Given,

The initial speed of the bus, u = 24 m/s

The final speed of bus, v = 12 m/s

Time taken to reach final speed is, t = 5.0 s

The acceleration of the body is given by the change in velocity by time

a = (v - u) / t

= (12 - 24) / 5

= -2.4 m/s²

The negative sign in the acceleration indicates that the bus is decelerating.

Therefore, the acceleration of the bus is, a = -2.4 m/s²

8 0

3 years ago

How much force does it take to bring a 1,050 N car from rest to a velocity of 42 m/s in 13 seconds?

frozen [14]

Answer:

F = 339.23 N

Explanation:

Weight of a car, W = 1050 N

Initial velocity, u = 0

Final velocity, v = 42 m/s

Time, t = 13 s

The weight of an object is given by :

W = mg

g is the acceleration due to gravity

$m=\dfrac{W}{g}\\\\m=\dfrac{1050}{10}\\\\m=105\ kg$

The force acting on car is :

$F=ma\\\\F=\dfrac{m(v-u)}{t}\\\\F=\dfrac{105\times (42-0)}{13}\\\\F=339.23\ N$

So, the force acting on the car is 339.23 N.

5 0

3 years ago