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Darya [45]
3 years ago
10

A temperature reading of zero on the Celsius scale is equal to absolute zero.

Physics
2 answers:
GrogVix [38]3 years ago
8 0
No, that's false.  'Zero' on the Celsius scale
is 273 on the absolute scale.
zhannawk [14.2K]3 years ago
5 0
False.

Absolute zero is actually 0 Kelvin.

Absolute zero is = -273°C, and not 0°C.

0 k = -273°C<span />
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Why would physics be used to study the movement of ocean waves?
Lunna [17]
I think it’s c because the other ones are just options not facts
5 0
3 years ago
In a generator, as the magnet spins, opposite poles of the magnet push the electrons in opposite directions. This back-and-forth
Natali [406]
C) alternating current .
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B)direct current </span>
6 0
3 years ago
Read 2 more answers
Verify that the SI unit of impulse is the same as the SI unit of momentum.
lys-0071 [83]

Maybe this will help you out:

Momentum is calculate by the formula:

P = mv

Where:

P = momentum

m = mass      

v = velocity

The SI unit:

mass = kg\\ velocity = \dfrac{m}{s}

So the unit of momentum would be:

kg.\dfrac{m}{s}

Impulse is defined as the change in momentum or how much force changes momentum. It can be calculate with the formula:

I = FΔt

where:

I = impulse

F = Force

Δt = change in time

The SI unit:

F = Newtons (N) or kg.\dfrac{m}{s^{2} }

t = Seconds (s)

So the unit of impulse would be derived this way:

I = FΔt

I = kg.\dfrac{m}{s^{2} } x s

or

\dfrac{kg.m.s}{s^{2}} = \dfrac{kg.m.s}{s.s}

You can then cancel out one s each from the numerator and denominator and you'll be left with:

kg.\dfrac{m}{s}

So then:

Momentum:                             Impulse

kg.\dfrac{m}{s}                                       kg.\dfrac{m}{s}

4 0
3 years ago
In an attempt to reduce the extraordinarily long travel times for voyaging to distant stars, some people have suggested travelin
alexandr402 [8]

Answer:

a) v=0.999124c

b) E=7.566*10^{22}

c) E_a=760 times\ larger

Explanation:

From the question we are told that

Distance to Betelgeuse d_b=430ly

Mass of Rocket M_r=20000

Total Time in years traveled T_d=36years

Total energy used by the United States in the year 2000 E_{2000}=1.0*10^20

Generally the equation of speed of rocket v mathematically given by

v=\frac{2d}{\triangle t}

v=860ly/ \triangle t

where

\triangle t=\frac{\triangle t'}{(\sqrt{1-860/ \triangle t)^2}}

\triangle t=\frac{36}{(\sqrt{1-860/ \triangle t)^2}}

\triangle t=\sqrt{(860)^2+(36)^2}

\triangle t=860.7532

Therefore

v=\frac{860ly}{ 860.7532}

v=0.999124c

b)

Generally the equation of the energy E required to attain prior speed mathematically given by

E=\frac{1}{\sqrt{1-(v/c)^2} }-1(20000kg)(3*10^8m/s)^2

E=7.566*10^{22}

c)Generally the equation of the energy E_a required to accelerate the rocket mathematically given by

E_a=\frac{E}{E_{2000}}

E_a=\frac{7.566*10^{22}}{1.0*10^{20}}

E_a=760 times\ larger

8 0
3 years ago
A bullet is fired horizontally at a height of 1.3 meters and a velocity of 950 m/s. How long was the bullet in the air?
seropon [69]

Answer:

<em>The bullet was 0.52 seconds in the air.</em>

Explanation:

<u>Horizontal Motion </u>

It occurs when an object is thrown horizontally with a speed v from a height h.

The object describes a curved path ruled exclusively by gravity until it hits the ground.

To calculate the time the object takes to hit the ground, we use the following equation:

\displaystyle t=\sqrt{\frac{2y}{g}}

Note it doesn't depend on the initial velocity but on the height.

The bullet is fired horizontally at h=1.3 m, thus:

\displaystyle t=\sqrt{\frac{2\cdot 1.3}{9.8}}

\displaystyle t=\sqrt{\frac{2.6}{9.8}}

t = 0.52 s

The bullet was 0.52 seconds in the air.

3 0
2 years ago
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