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2 years ago

what type of simple machine is shown in the diagram?write the length and height of slope in it. Please help ASAP!!!

Physics

2 answers:

Akimi4 [234]2 years ago

7 0

the machine shown in the diagram is called a tramp.

the height of the tramp is 5 and the length is 12

vesna_86 [32]2 years ago

5 0

Answer:

Incline plane

5/12 rise/run

Explanation:

Incline plane

5/12 rise/run

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Radiation from the Sun reaching Earth (just outside the atmosphere) has an intensity of 1.39 kW/m2. (a) Assuming that Earth (and

Zigmanuir [339]

Answer:

$F=5.8\times 10^{8}\ N$

$F=35.57\times 10^{21}\ N$

Explanation:

Given that

Intensity I

$I= 1.39\ KW/m^2$

$Speed\ of \ light = 2.99\times 10^8\ m/s$

Radius of earth,R = 6370 Km

We know that surface area of earth, A

$A=\pi R^2$

$A=\pi (6370\times 10^3)^2\ m^2$

$A=1.27\times 10^{14}\ m^2$

As we know that pressure due to intensity given as

$P=\dfrac{I}{V}$

V =Velocity of light

$V=3\times 10^8\ m/s$

$P=\dfrac{1.39\times 1000}{=3\times 10^8}$

$P=4.6\times 10^{-6}\ Pa$

We know that force F

F = P .A

$F=4.6\times 10^{-6}\times 1.27\times 10^{14}\ N$

$F=5.8\times 10^{8}\ N$

b)Gravitational force F

$F=G\dfrac{m.M}{r^2}$

$M = mass\ of\ sun = 2\times 10^{30} kg\\m = mass\ of\ earth = 6\times 10^{24}kg$

$r =1.5\times 10^{11}\ m$

$G =6.67\times 10^{-11}Nm^2/kg^2$

So F

$F=6.67\times 10^{-11}\times \dfrac{2\times 10^{30}\times 6\times 10^{24}kg}{1.5\times 10^{11}}$

$F=35.57\times 10^{21}\ N$

7 0

3 years ago

Read 2 more answers

Which of the following describes density only in terms of metric units?

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A) Kilograms per cubic meter. Every other option either contains pounds or feet, which are both units of measurement from the standard system, not the metric system.

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3 years ago

So far, you’ve been working with an "ideal" pulley system. How do you think real pulley systems are different, and how would tha

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Answer:

In an ideal pulley system is assumed as a perfect system, and the efficiency of the pulley system is taken as 100% such that there are no losses of the energy input to the system through the system's component

However, in a real pulley system, there are several means through which energy is lost from the system through friction, which is converted into heat, sound, as well as other forms of energy

Given that the mechanical advantage = Force output/(Force input), and that the input force is known, the energy loss comes from the output force which is then reduced, and therefore, the Actual Mechanical Advantage (AMA) is less than the Ideal Mechanical Advantage of an "ideal" pulley system

The relationship between the actual and ideal mechanical advantage is given by the efficiency of the pulley system as follows;

$Efficiency \, \% = \dfrac{AMA}{IMA} \times 100$