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maw [93]
3 years ago
6

Velocity of a machine is always greater than mechanical advantage? why​

Physics
1 answer:
MrRa [10]3 years ago
3 0

Answer:The mechanical advantage of a machine is always less than its velocity ratio.It is because mechanical advantage decreases due to the friction and weight of moving parts of the machine, but the velocity ratio remains constant.

Explanation: hope this helps

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A 2.0-kg block is on a perfectly smooth (frictionless) ramp that makes an angle of 30^\circ30 ​∘ ​​ with the horizontal. What is
Nastasia [14]

Answer:

Explanation:

Since the surface is frictionless therefore there will be no friction force on block but there will be weight of block which we can divide in to two components i.e. mgcosθ &mgsinθ which is perpendicular and parallel to the surface respectively.

In response to mgcosθ ramp will apply a normal force to the block which will be of equal magnitude to that of mgcosθ.

Therefore Ramp will apply a Force of mgcosθ on block where m is the mass of block.

7 0
3 years ago
Water flows through a garden hose which is attached to a nozzle. The water flows through hose with a speed of 1.81 m/s and throu
Serggg [28]

Answer:

a) 17.086m

b) 0.1671 m

Explanation:

Given data: speed of water through the hose  = 1.81 m/s

through the nozzle = 18.3 m/s

We know that maximum height of an object with upward velocity v is given by,

a) H = v^2/2g

where H is the maximum height water emerges  

= 18.3^2/(2×9.8) = 17.086 m answer

b) Again,  

H = v^2/2g

= 1.81^2/(2×9.8) = 0.1671 m

6 0
2 years ago
8. A sprinter on a school track team is running north at a velocity of 6.0 m/s. After 5.0 s, she
Marysya12 [62]

Answer:

acc. = 4-(-6) /5= 10/5=2 m/s^2

6 0
3 years ago
The earth and the moon exert forces on each other which forces is greater? explain
Helen [10]

Thank you for your question, what you say is true, the gravitational force exerted by the Earth on the Moon has to be equal to the centripetal force.

An interesting application of this principle is that it allows you to determine a relation between the period of an orbit and its size. Let us assume for simplicity the Moon's orbit as circular (it is not, but this is a good approximation for our purposes).

The gravitational acceleration that the Moon experience due to the gravitational attraction from the Earth is given by:

ag=G(MEarth+MMoon)/r2

Where G is the gravitational constant, M stands for mass, and r is the radius of the orbit. The centripetal acceleration is given by:

acentr=(4 pi2 r)/T2

Where T is the period. Since the two accelerations have to be equal, we obtain:

(4 pi2 r) /T2=G(MEarth+MMoon)/r2

Which implies:

r3/T2=G(MEarth+MMoon)/4 pi2=const.

This is the so-called third Kepler law, that states that the cube of the radius of the orbit is proportional to the square of the period.

This has interesting applications. In the Solar System, for example, if you know the period and the radius of one planet orbit, by knowing another planet's period you can determine its orbit radius. I hope that this answers your question.


8 0
3 years ago
Which process do hydrogen atoms use to make the Sun's energy?
Maru [420]
Nuclear fusion, converts hydrogen atoms into helium
8 0
3 years ago
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