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2 years ago

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a car is moving with a velocity of 25m/s for 15s. calculate the displacement of the car. The acceleration of the car over the 15

s

1 answer:

SpyIntel [72]2 years ago

8 0

Explanation:

this might help you i think so

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If you know that the universe is estimated to be 14 billion years old . what is the age of the universe in seconds ???? (knowing

Svetlanka [38]

Answer:

<em>The estimated age of the universe is</em> $4.42*10^{17}\ seconds$

Explanation:

<u>Time Units Conversion</u>

Sometimes we need to express magnitudes in different units, depending on the specific situation handled in the question or problem at hand. Magnitudes like length, time, mass, temperature, pressure, etc., can be expressed in several different units.

The most common units for time are years, months, days, hours, minutes, and seconds. There are 3,600 seconds in one hour and 24 hours in one day.

We are given the estimated age of the universe as 14 billion solar years. If one solar year is 365.25 days long, then the age, expressed in days is

14,000,000,000 * 365.25 days

Some quantities need to be expressed in scientific notation. This is the case since the numbers are too high

$14,000,000,000 * 365.25 = 5.1135*10^{12}\ days$

There are 24*3600=86400 seconds in one day, so

$5.1135*10^{12}\ days=5.1135*10^{12} * 86400 =4.42*10^{17}\ seconds$

The estimated age of the universe is $4.42*10^{17}\ seconds$

6 0

2 years ago

Read 2 more answers

A box has a mass of 5.8kg. The box is lifted from the garage floor and placed on a shelf 2.5m off the ground. How much gravitati

Damm [24]

Gravitational potential energy =

(mass) x (gravity) x (height)

= (5.8 kg) x (9.8 m/s²) x (2.5 m)

= 142.1 Joules (C)

5 0

3 years ago

A rocket travels in the x-direction at speed 0.70c with respect to the earth. An experimenter on the rocket observes a collision

marishachu [46]

Answer:

A) The space time coordinate x of the collision in Earth's reference frame is

$x \approx 103,46x10^{9}m$ .

B) The space time coordinate t of the collision in Earth's reference frame is

$t=377,29s$

Explanation:

We are told a rocket travels in the x-direction at speed v=0,70 c (c=299792458 m/s is the exact value of the speed of light) with respect to the Earth. A collision between two comets is observed from the rocket and it is determined that the space time coordinates of the collision are (x',t') = (3.4 x 10¹⁰ m, 190 s).

An event indicates something that occurs at a given location in space and time, in this case the event is the collision between the two comets. We know the space time coordinates of the collision seen from the reference frame of the rocket and we want to find out the space time coordinates in Earth's reference frame.

<em>Lorentz transformation</em>

The Lorentz transformation relates things between two reference frames when one of them is moving with constant velocity with respect to the other. In this case the two reference frames are the Earth and the rocket that is moving with speed v=0,70 c in the x axis.

The Lorentz transformation is

$x'=\frac{x-vt}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

$y'=y$

$z'=z$

$t'=\frac{t-\frac{v}{c^{2}}x}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

prime coordinates are the ones from the rocket reference frame and unprimed variables are from the Earth's reference frame. Since we want position x and time t in the Earth's frame we need the inverse Lorentz transformation. This can be obtained by replacing v by -v and swapping primed an unprimed variables in the first set of equations

$x=\frac{x'+vt'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

$y=y'$

$z=z'$

$t=\frac{t'+\frac{v}{c^{2}}x'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

First we calculate the expression in the denominator

$\frac{v^{2}}{c^{2}}=\frac{(0,70)^{2}c^{2}}{c^{2}} =(0,70)^{2}$

$\sqrt{1-\frac{v^{2}}{c^{2}}} =0,714$

then we calculate t

$t=\frac{t'+\frac{v}{c^{2}}x'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

$t=\frac{190s+\frac{0,70c}{c^{2}}.3,4x10^{10}m}{0,714}$

$t=\frac{190s+\frac{0,70c .3,4x10^{10}m}{299792458\frac{m}{s}}}{0,714}$

$t=\frac{190s+79,388s}{0,714}$

finally we get that

$t=377,29s$

then we calculate x

$x=\frac{x'+vt'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

$x=\frac{3,4x10^{10}m+0,70c.190s}{0,714}}$

$x=\frac{3,4x10^{10}m+0,70.299792458\frac{m}{s}.190s}{0,714}}$

$x=\frac{3,4x10^{10}m+39872396914m}{0,714}}$

$x=\frac{73872396914m}{0,714}}$

$x=103462740775,91m$

finally we get that

$x \approx 103,46x10^{9} m$

5 0

3 years ago

How far will a 600 kg boat travel in 10 s if there is a constant 900 N force on it and it starts from rest?

natta225 [31]

Answer:

here

Explanation:

There are two forces acting upon the skydiver - gravity (down) and air resistance (up). The force of gravity has a magnitude of m•g = (72 kg) •(9.8 m/s/s) = 706 N. ... a 3.25-kg object rightward with a constant acceleration of 1.20 m/s/s if the force of ... of 33.8 kg, how far (in meters) will it move in 1.31 seconds, starting from rest?

4 0

2 years ago

I don't get it at all

andrezito [222]

The most i can help with is Saliva glands or salivary glands

3 0

3 years ago

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