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3 years ago

Based on the law of conservation of energy, how can we reasonably improve a machine’s ability to do work?

Physics

2 answers:

marshall27 [118]3 years ago

8 0

Answer:

Reduce the friction between its moving parts.

Explanation:

Sholpan [36]3 years ago

6 0

You will have to redefine the machine's boundaries.

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my choice1A child pushes a toy truck up an inclined plane. The lifting force is...Select one:ut ofO a. the weight of the truck.O

mash [69]

When pushing a toy truck up an inclined plane, the force that makes the truck goes up is the force that the child uses to push the truck. (The force that pushes

Therefore the correct option is C.

4 0

1 year ago

Calculate the final velocity right after a 117 kg rugby player who is initially running at 7.45 m/s collides head‑on with a padd

Free_Kalibri [48]

Answer:

v_f = 0.87 m/s

Explanation:

We are given;

F_avg = -17700 N (negative because it's backward)

m = 117 kg

Δt = 5.50 × 10^(−2) s

v_i = 7.45 m/s

Now, formula for impulse is given by;

I = F•Δt = - 17700 x 5.50 × 10^(−2) = - 973.5 kg.m/s

From impulse momentum theory, we know that;

Change in momentum of particle is equal to impulse.

Thus,

Δp = I = m•v_f - m•v_i

Thus,

-973.5= 117(v_f - 7.45)

Thus,

-973.5/117 = (v_f - 7.45)

-8.3205 + 7.45 = v_f

v_f = - 0.87 m/s

We'll take absolute value as;

v_f = 0.87 m/s

5 0

3 years ago

Ignoring air resistance and the little friction from the plastic tube, the magnet was a freely-falling object in each trial. If

horsena [70]

Answer:

emf will also be 10 times less as compared to when it has fallen $40 \mathrm{m}$

Explanation:

We know, from faraday's law-

$e m f=-N \frac{\Delta \Phi}{\Delta T}$

and $\Phi=B . A$

So, as the height increases the velocity with which it will cross the ring will also increase. $(v=\sqrt{2 g h})$

Given

$\mathrm{V} 1(\text { Speed at } 40 \mathrm{m})=2 \mathrm{x} \mathrm{V} 2(\text { speed at } 10 \mathrm{m})$

$\sqrt{2 g h_{2}}=2 \times \sqrt{2 g h_{1}}=28.28 \mathrm{m} / \mathrm{s}$

Now, from $40 \mathrm{cm}$

$V_{3}=\sqrt{2 g h_{3}}=\sqrt{2 \times 10 \times 0.4}=2.82 \mathrm{m} / \mathrm{s}$

From equation a and b we see that velocity when dropped from $40 \mathrm{m}$ is 10 times greater when height is 40 $\mathrm{cm}$ so, emf will also be 10 times less as compared to when it has fallen $40 \mathrm{m}$

4 0

3 years ago

Please see the picture

Tpy6a [65]

Answer:

t.f. are you sure that's english? it looks like not a real thing

Explanation:

3 0

3 years ago

A 120 g, 8.0-cm-diameter gyroscope is spun at 1000 rpm and allowed to precess. What is the precession period?

dolphi86 [110]

To solve this problem we will derive the expression of the precession period from the moment of inertia of the given object. We will convert the units that are not in SI, and finally we will find the precession period with the variables found. Let's start defining the moment of inertia.

$I = MR^2$