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1 year ago

A car starts from rest and accelerates at 4 m/s2 until it reaches 36 m/s. How far did it travel in this time?

Physics

1 answer:

dexar [7]1 year ago

3 0

The distance of a car that starts from rest and accelerates at 4 m/s² until it reaches 36 m/s is 162m.

<h3>How to calculate distance?</h3>

The distance moved by a body can be calculated using the following formula:

S = ut + ½at²

Where;

S = distance
u = initial velocity
t = time
a = acceleration

However, the time taken for the car to complete the acceleration can be calculated as follows:

4 = 36/t

t = 9seconds

S = 0 × t + ½ × 4 × 9²

S = 0 + 162

S = 162m

Therefore, the distance of a car that starts from rest and accelerates at 4 m/s² until it reaches 36 m/s is 162m.

Learn more about distance at: brainly.com/question/15256256

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Some neodymium glass lasers can provide 100TW of power in 1.0 ns pulses at a wavelength of 0.26 micrometers. how much energy is

Aleksandr-060686 [28]

Answer:

E = 10⁵ J

Explanation:

given,

Power, P = 100 TW

= 100 x 10¹² W

time, t = 1 ns

= 1 x 10⁻⁹ s

The energy of a single pulse is:-

Energy = Power x time

E = P t

E = 100 x 10¹² x 1 x 10⁻⁹

E = 10⁵ J

The energy contained in a single pulse is equal to 10⁵ J

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3 years ago

A plastic bag is massed. It is then filled with a gas which is insoluble in water and massed again. The apparent weight of the g

levacccp [35]

The actual weight of the gas = apparent weight + weight.

The actual weight = $W_{A}$ + W

Given that a plastic bag is massed. It is then filled with a gas which is insoluble in water and massed again.

If the apparent weight of the gas is the difference between these two masses, then let the apparent weight = $W_{A}$

The gas is squeezed out of the bag to determine its volume by the displacement of water. Since

density = mass / volume

The density of water is 1000 kg/ $m^{2}$

we can get the mass of the gas by making m the subject of the formula.

W = mg

The actual weight of the gas = apparent weight + weight

That is,

The actual weight = $W_{A}$ + W

Learn more about density here: brainly.com/question/406690

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2 years ago

Why to astronauts appear weightless while they are filmed performing activities inside the orbiting space shuttle?

vagabundo [1.1K]

<h2>Answer: The astronauts are falling at the same rate as the space shuttle as it orbits around earth</h2>

The astronauts seem to float because they are in free fall just like the spacecraft.

However, although they are constantly falling on the Earth, they do not fall because the ship orbits at a sufficient speed (in the same direction of rotation of the Earth) so that the centrifugal force is balanced with the Earth's gravitational pull.

In other words:

The spaccraft and the astronauts are in free fall but the Earth's surface will never be reached as long as they does not decrease the speed.

Then, as they accelerate toward Earth (regardless of their mass), it curves beneath them and never comes close.

That's why astronauts, having the same acceleration as the spacecraft, feel weightless and see themselves floating.

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2 years ago

Read 2 more answers

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:

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