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3 years ago

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

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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

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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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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 $s$

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

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

$v=\frac{860ly}{ 860.7532}$

$v=0.999124c$

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:

The bullet was 0.52 seconds in the air.

Explanation:

Horizontal Motion

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