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sveta [45]
3 years ago
13

Using a simple machine, a student is able to lift a 500N weight by applying only 100N.

Physics
1 answer:
pishuonlain [190]3 years ago
7 0
Hey there!

a \ student  \ is \ able \ to \ lift \ a \ 500(n) \\ weight \ by \ applying \ only  \ 100(n) \\ \\ so, \ this \ info \ here, \ we \ simply \ divide \ by \\ how \ much \ this \ kid \ lifted, \ by \ the \ weight \ he/she \\ \ is \ applying. \\ \\   \left[\begin{array}{ccc}\boxed{\boxed{500(n)/100(n)}}\end{array}\right] \\ \\ and \ from \ this,\ your \ answer \ would \ \\ conclude \ to \ be \ (5) \\ \\ \boxed{5} \ would \ be \ your \ answer!

Hope this helps you!
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A standard 1 kilogram weight is a cylinder 41.5 mm in height and 44.0 mm in diameter. what is the density of the material?
Korvikt [17]

Answer;

=15855.40 kg/m^3

Explanation;

Volume (V) of the cylinder = pi x r^2 x h  

V = 3.14 x (44/2 x 10^-3)^2 x 41.5 x 10^-3  

V = 6.307 x 10^-5 m^3  

By density = m/V  

mass = 1 kg

density = 1/(6.307 x 10^-5) = 15855.40 kg/m^3

6 0
3 years ago
"why are some galaxies' spectra blueshifted rather than redshifted?"
stira [4]
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3 years ago
If the work put into a lever is 930 joules and the work accomplished is 870 joules, the efficiency of the lever is _____.
77julia77 [94]
♥ C) 94%
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5 0
3 years ago
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Two planets A and B, where B has twice the mass of A, orbit the Sun in elliptical orbits. The semi-major axis of the elliptical
lozanna [386]

Answer:

2.83

Explanation:

Kepler's discovered that the square of the orbital period of a planet is proportional to the cube of the semi-major axis of its orbit, that is called Kepler's third law of planet motion and can be expressed as:

T=\frac{2\pi a^{\frac{3}{2}}}{\sqrt{GM}} (1)

with T the orbital period, M the mass of the sun, G the Cavendish constant and a the semi major axis of the elliptical orbit of the planet. By (1) we can see that orbital period is independent of the mass of the planet and depends of the semi major axis, rearranging (1):

\frac{T}{a^{\frac{3}{2}}}=\frac{2\pi}{\sqrt{GM}}

\frac{T^{2}}{a^{3}}=(\frac{2\pi }{\sqrt{GM}})^2 (2)

Because in the right side of the equation (2) we have only constant quantities, that implies the ratio \frac{T^{2}}{a^{3}} is constant for all the planets orbiting the same sun, so we can said that:

\frac{T_{A}^{2}}{a_{A}^{3}}=\frac{T_{B}^{2}}{a_{B}^{3}}

\frac{T_{B}^{2}}{T_{A}^{2}}=\frac{a_{B}^{3}}{a_{A}^{3}}

\frac{T_{B}}{T_{A}}=\sqrt{\frac{a_{B}^{3}}{a_{A}^{3}}}=\sqrt{\frac{(2a_{A})^{3}}{a_{A}^{3}}}

\frac{T_{B}}{T_{A}}=\sqrt{\frac{2^3}{1}}=2.83

6 0
3 years ago
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marissa [1.9K]

The correct answer is B the total velocity is equal at both landing and launch because before your about launch you have 0 velocity then when you have landed you also have 0 velocity. Hope This Helps

3 0
3 years ago
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