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3 years ago

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An auto, starting from rest, undergoes constant acceleration and covers a distance of 1200 meters. The final speed of the auto i

s 60 meters/sec. How long does it take the car to cover the 1200 meters?

1 answer:

Elanso [62]3 years ago

8 0

Answer:

It will take 40 sec

Explanation:

We have given that auto is starting from rest so initial velocity u = 0 m/sec

And final velocity is given as v = 60 m/sec

Distance traveled by car s = 1200 m

According third equation of motion we know that $v^2=u^2+2as$

So $60^2=0^2+2\times a\times 1200$

$a=1.5m/sec^2$

Now according to first equation of motion v = u+at

So $60=0+1.5\times t$

t = 40 sec

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Hoochie [10]

I have a strange hunch that there's some more material or previous work
that goes along with this question, which you haven't included here.

I can't easily find the dates of Mercury's extremes, but here's some of the
other data you're looking for:

Distance at Aphelion (point in it's orbit that's farthest from the sun):
<span><span><span><span><span>69,816,900 km
0. 466 697 AU</span>

</span> </span> </span> <span> Distance at Perihelion (</span></span><span>point in it's orbit that's closest to the sun):</span>
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0.307 499 AU</span> </span>

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3 years ago

Calculate the magnitude of the acceleration due to gravity on the surface of Earth due to the Moon.

Fudgin [204]

Answer:

$g'_h=1.096\times 10^{-5}\ m.s^{-2}$

Explanation:

We know that the gravity on the surface of the moon is,

$g'=\frac{g}{6}$

$g'=1.63\ m.s^{-2}$

<u>Gravity at a height h above the surface of the moon will be given as:</u>

$g'_h=\frac{G.m}{(r+h)^2}$ ..........................(1)

where:

G = universal gravitational constant

m = mass of the moon

r = radius of moon

We have:

$G=6.67\times 10^{-11}\ m^3.s^{-2}.kg^{-1}$

$m=7.35\times 10^{22}\ kg$

$r=1.74\times 10^6\ m$

$h=384.4\times 10^6\ m$ is the distance between the surface of the earth and the moon.

Now put the respective values in eq. (1)

$g'_h=\frac{6.67\times 10^{-11}\times 7.35\times 10^{22}}{(1.74\times 10^6+384.4\times 10^6)^2}$

$g'_h=1.096\times 10^{-5}\ m.s^{-2}$ is the gravity on the moon the earth-surface.

4 0

3 years ago

Little Tammy lines up to tackle Jackson to (unsuccessfully) prove the law of conservation of momentum. Tammy’s mass is 34.0 kg a

Leya [2.2K]

Answer:

So Tammy must move with speed 4.76 m/s in opposite direction of Jackson

Explanation:

As per law of conservation of momentum we know that there is no external force on it

So here we can say that initial momentum of the system must be equal to the final momentum of the system

now we have

$m_1v_1 + m_2v_2 = 0$

final they both comes to rest so here we can say that final momentum must be zero

now we have

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$v = -4.76 m/s$

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3 years ago

SashulF [63]

Answer:

a. Final velocity, V = 2.179 m/s.

b. Final velocity, V = 7.071 m/s.

Explanation:

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

Acceleration = 0.500m/s²

a. To find the velocity of the boat after it has traveled 4.75 m

Since it started from rest, initial velocity is equal to 0m/s.

Now, we would use the third equation of motion to find the final velocity.

$V^{2} = U^{2} + 2aS$

Where;

V represents the final velocity measured in meter per seconds.
U represents the initial velocity measured in meter per seconds.
a represents acceleration measured in meters per seconds square.
S represents the displacement measured in meters.

Substituting into the equation, we have;

$V^{2} = 0^{2} + 2*0.500*4.75$

$V^{2} = 4.75$

Taking the square root, we have;

$V^{2} = \sqrt {4.75}$

<em>Final velocity, V = 2.179 m/s.</em>

b. To find the velocity if the boat has traveled 50 m.

$V^{2} = 0^{2} + 2*0.500*50$

$V^{2} = 50$

Taking the square root, we have;

$V^{2} = \sqrt {50}$

<em>Final velocity, V = 7.071 m/s.</em>

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3 years ago