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3 years ago

Is the facceleration due to gravity the same on Earth and Earth's moon?

Physics

1 answer:

Alchen [17]3 years ago

3 0

Answer:

Earth's average surface gravity is about 9.8 meters per second per second. ... The Moon's surface gravity is weaker because it is far less massive than Earth. A body's surface gravity is proportional to its mass but inversely proportional to the square of its radius.

Explanation:

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3730 watts equals about how many horsepower A.5 B.10 C.20 D.30

ikadub [295]

3,730 W is equal to about 5 horsepower. (4.9982 hp)

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3 years ago

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Are electrons lower in mass than neutrons

tamaranim1 [39]

Answer:

ya I just looked it up on Google

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3 years ago

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A rocket initially at rest accelerates at a rate of 99.0 meters/second ^2.Calculate the distance covered by the rocket if it. At

Pani-rosa [81]

The distance covered is 1000 m

Explanation:

The rocket is moving by uniformly accelerated motion, so we can find the distance it covers by using the following suvat equation:

$s=vt-\frac{1}{2}at^2$

where

s is the distance covered

v is the final velocity

t is the time

a is the acceleration

For the rocket in this problem, we have:

v = 445 m/s is the final velocity

$a=99.0 m/s^2$ is the acceleration

t = 4.50 s is the time

Substituting, we find the distance covered:

$s=(445)(4.50)-\frac{1}{2}(99.0)(4.50)^2=1000 m$

Learn more about accelerated motion:

brainly.com/question/9527152

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#LearnwithBrainly

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4 years ago

Another term for "phone to pocket" is:

Gemiola [76]

Answer:

Explanation:

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3 years ago

The Tevatron acceleator at the Fermi National Accelerator Laboratory (Fermilab) outside Chicago boosts protons to 1 TeV (1000 Ge

Eva8 [605]

Answer:

a) $v = c \cdot 0.04 = 1.2\cdot 10^{7} m/s$

b) $v = c \cdot 0.71 = 2.1\cdot 10^{8} m/s$

c) $v = c \cdot 0.994 = 2.97\cdot 10^{8} m/s$

d) $v = c \cdot 0.999 = 2.997\cdot 10^{8} m/s$

e) $v = c \cdot 0.9999 = 2.999\cdot 10^{8} m/s$

Explanation:

At that energies, the speed of proton is in the relativistic theory field, so we need to use the relativistic kinetic energy equation.

$KE=mc^{2}(\gamma -1) = mc^{2}(\frac{1}{\sqrt{1-\beta^{2}}} -1)$ (1)

Here β = v/c, when v is the speed of the particle and c is the speed of light in vacuum.

Let's solve (1) for β.

$\beta = \sqrt{1-\frac{1}{\left (\frac{KE}{mc^{2}}+1 \right )^{2}}}$

We can write the mass of a proton in MeV/c².

$m_{p}=938.28 MeV/c^{2}$

Now we can calculate the speed in each stage.

a) Cockcroft-Walton (750 keV)

$\beta = \sqrt{1-\frac{1}{\left (\frac{0.75 MeV}{938.28 MeV}+1 \right )^{2}}}$

$\beta = 0.04$

$v = c \cdot 0.04 = 1.2\cdot 10^{7} m/s$

b) Linac (400 MeV)

$\beta = \sqrt{1-\frac{1}{\left (\frac{400 MeV}{938.28 MeV}+1 \right )^{2}}}$

$\beta = 0.71$

$v = c \cdot 0.71 = 2.1\cdot 10^{8} m/s$

c) Booster (8 GeV)

$\beta = \sqrt{1-\frac{1}{\left (\frac{8000 MeV}{938.28 MeV}+1 \right )^{2}}}$

$\beta = 0.994$

$v = c \cdot 0.994 = 2.97\cdot 10^{8} m/s$

d) Main ring or injector (150 Gev)

$\beta = \sqrt{1-\frac{1}{\left (\frac{150000 MeV}{938.28 MeV}+1 \right )^{2}}}$

$\beta = 0.999$

$v = c \cdot 0.999 = 2.997\cdot 10^{8} m/s$

e) Tevatron (1 TeV)

$\beta = \sqrt{1-\frac{1}{\left (\frac{1000000 MeV}{938.28 MeV}+1 \right )^{2}}}$

$\beta = 0.9999$

$v = c \cdot 0.9999 = 2.999\cdot 10^{8} m/s$

Have a nice day!

4 0

4 years ago