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3 years ago

What is the The most appropriate SI unit for measuring the length of an automobile

1 answer:

7 0

If you're talking about S.I. Unit then it must be "meter" (nothing else). Meter is the only S.I. Unit of length and it also uses in measuring the length of an automobile 'cause it varies from few meters to several thousand meters.

In short, Your Correct answer would be "meter" (m)

Hope this helps!

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PLEASE HELP!!

leva [86]

I think the answer is B

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3 years ago

How does beam spreading affect the temperature on earth and the seasons

navik [9.2K]

Get to school and learn boi

8 0

3 years ago

Find the velocity and distance travelled of a car that is accelerating at 3 m/s2 for 4 seconds.

marta [7]

Explanation:

Given:

v₀ = 0 m/s

a = 3 m/s²

t = 4 s

Find: Δx and v

Δx = v₀ t + ½ at²

Δx = (0 m/s) (4 s) + ½ (3 m/s²) (4 s)²

Δx = 24 m

v = at + v₀

v = (3 m/s²) (4 s) + 0 m/s

v = 12 m/s

6 0

3 years ago

Question is attatched!

lina2011 [118]

All it does is lets him pull in a more convenient direction to raise the load. It has no effect on the required force.

4 0

3 years ago

Two satellites are in circular orbits around the earth. The orbit for satellite A is at a height of 403 km above the earth’s sur

BARSIC [14]

Answer:

$v_A=7667m/s\\\\v_B=7487m/s$

Explanation:

The gravitational force exerted on the satellites is given by the Newton's Law of Universal Gravitation:

$F_g=\frac{GMm}{R^{2} }$

Where M is the mass of the earth, m is the mass of a satellite, R the radius of its orbit and G is the gravitational constant.

Also, we know that the centripetal force of an object describing a circular motion is given by:

$F_c=m\frac{v^{2}}{R}$

Where m is the mass of the object, v is its speed and R is its distance to the center of the circle.

Then, since the gravitational force is the centripetal force in this case, we can equalize the two expressions and solve for v:

$\frac{GMm}{R^2}=m\frac{v^2}{R}\\ \\\implies v=\sqrt{\frac{GM}{R}}$

Finally, we plug in the values for G (6.67*10^-11Nm^2/kg^2), M (5.97*10^24kg) and R for each satellite. Take in account that R is the radius of the orbit, not the distance to the planet's surface. So $R_A=6774km=6.774*10^6m$ and $R_B=7103km=7.103*10^6m$ (Since $R_{earth}=6371km$ ). Then, we get:

$v_A=\sqrt{\frac{(6.67*10^{-11}Nm^2/kg^2)(5.97*10^{24}kg)}{6.774*10^6m} }=7667m/s\\\\v_B=\sqrt{\frac{(6.67*10^{-11}Nm^2/kg^2)(5.97*10^{24}kg)}{7.103*10^6m} }=7487m/s$

In words, the orbital speed for satellite A is 7667m/s (a) and for satellite B is 7487m/s (b).

7 0

3 years ago