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2 years ago

What is the definition of input force?

Physics

2 answers:

Len [333]2 years ago

6 0

Answer:

The input force (effort) is the amount of effort used to push down on a rod, or pull on a rope in order to move the weight. In this example, the force the little guy is using to pull the elephant is the input force.

andrezito [222]2 years ago

5 0

Answer:

Explanation:

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A small electric motor produces a force of 5 N that moves a remote-control car 5 m every second. How much power does the motor p

kati45 [8]

Answer:

Explanation:

Given:

Force, f = 5 N

Velocity, v = 5 m/s

Power, p = energy/time

Energy = mass × acceleration × distance

Poer, p = force × velocity

= 5 × 5

= 25 W.

Note 1 watt = 0.00134 horsepower

But 25 watt,

0.00134 hp/1 watt × 25 watt

= 0.0335 hp.

4 0

2 years ago

Read 2 more answers

Which law is associated with inertia

N76 [4]

Answer:

The focus of Lesson 1 is Newton's first law of motion - sometimes referred to as the law of inertia. An object at rest stays at rest and an object in motion stays in motion with the same speed and in the same direction unless acted upon by an unbalanced force.

4 0

3 years ago

Dimension equation of work

kkurt [141]

Answer:

Explanation:

Work

Other units Foot-pound, Erg

In SI base units 1 kg⋅m2⋅s−2

Derivations from other quantities W = F ⋅ s W = τ θ

Dimension M L2 T−2

Idk if this is what u are looking for but i hope this help.:)

3 0

2 years ago

A mercury thermometer is used to measure the temperature of boiling water.<br>Why?

Rama09 [41]

Answer:

It has very high density, so a small bulb of a thermometer can contain much mercury. Mercury remains liquid state over a quite wide range of temperature because it freezes at 39°C and boils at 357°C.

Explanation:

6 0

2 years ago

A 0.560 kg snowball is fired from a cliff 14.2 m high with an initial velocity of 13.3 m/s, directed 26.0° above the horizontal.

enot [183]

Answer:

a) v = 21.34 m/s

b) v = 21.34 m/s

c) v = 21.34 m/s

Explanation:

Mass of the snowball, m = 0.560 kg

Height of the cliff, h = 14.2 m

Initial velocity of the ball, u = 13.3 m/s

θ = 26°

The speed of the slow ball as it reaches the ground, v = ?

The initial Kinetic energy of the snow ball, $KE_{0} = 0.5 mu^{2}$

Potential energy of the snow ball at the given height, PE = mgh

Final Kinetic energy of the ball as it reaches the ground, $KE_{f} = 0.5mv^{2}$

a) Using the principle of energy conservation,

$KE_{0} + PE = KE_{f} \\0.5mu^{2} + mgh = 0.5mv^{2}\\v^{2} =2( 0.5u^{2} + gh)\\v^{2} =u^{2} + 2gh\\v = \sqrt{u^{2} + 2gh} \\v = \sqrt{13.3^{2} + 2*9.8*14.2}\\v = 21.34 m/s$

b) The speed remains v = 21.34 m/s since it is not a function of the angle of launch

c)The principle of energy conservation used cancels out the mass of the object, therefore the speed is not dependent on mass

v = 21. 34 m/s

7 0

3 years ago