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AnnyKZ [126]
3 years ago
13

Can a machine like a lever, produce a greater force than what you put into it? Can it increase the energy that you put into it?

Physics
2 answers:
morpeh [17]3 years ago
7 0

Answer:

yes

Explanation:

the force is multiplied by the levers length of the handle

Ipatiy [6.2K]3 years ago
3 0

Answer:

Yes, a lever can produce greater force than what you put into it, but it does not increase the energy that you put into it. Energy is conserved, it cannot be created or destroyed, just moved.

Explanation:

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Use the graph to answer the question.
Setler79 [48]

Answer:175

Explanation:

5 0
3 years ago
Read 2 more answers
How much work is done against gravity when lowering a 16 kg box 0.50 m? (g = 9.8 m/s2)
leonid [27]

Answer:

The work done against gravity is 78.4 J

Explanation:

The work is calculated by multiplying the force by the distance that the

object moves

W = F × d, where W is the work , F is the force and d is the distance

The SI unit of work is the joule (J)

We need to find the work done against gravity when lowering a

16 kg box 0.50 m

→ F = mg

→ m = 16 kg, and g = 9.8 m/s²

Substitute these value in the rule

→ F = 16 × 9.8 = 156.8 N

→ W = F × d

→ F = 156.8 N and d = 0.50

Substitute these values in the rule

→ W = 78.4 J

<em>The work done against gravity is 78.4 J</em>

6 0
3 years ago
1. A sprinter races in the 100 meter dash. It takes him 10 second to reach the finish line
poizon [28]

Answer:

v = 10 m/s

Explanation:

Given that,

Distance covered by a sprinter, d = 100 m

Time taken by him to reach the finish line, t = 10 s

We need to find his average velocity. We know that velocity is equal to the distance covered divided by time taken. So,

v = d/t

v=\dfrac{100\ m}{10\ s}\\\\v=10\ m/s

Hence, his average velocity is 10 m/s.

6 0
3 years ago
A light rope is attached to a block with mass 3.60 kg that rests on a frictionless, horizontal surface. The horizontal rope pass
Readme [11.4K]

Answer and Explanation:

(a) The fre-body diagrams for each block is shown below. In the block of mass 3.60 kg, there are 3 forces acting on it: horizontal force due to the rope (F_{t}), vertical gravitational force (F_{g}) and vertical normal force (F_{n}), due to the surface. Since there is no vertical movement, F_{g} and F_{n} cancels it out. So, for this block, net force is horizontal due to the rope F_{t}.

The block of mass m is hanging from the pulley, so there is the force of the rope (F_{t}) and the gravitational force (F_{g}). Both are vertical, because there is no surface "holding" block m.

(b) Since both blocks are attached to each other, the acceleration will be the same. To calculate it, we use the Second Law of Motion:

F_{r}=m.a

a=\frac{F_{r}}{m}

a=\frac{18.8}{3.6}

a = 5.22

The acceleration of either block is 5.22 m/s².

(c) Block m has 2 forces acting on it: tension and gravitational force. Gravitational force is the force of attraction the Earth does over an object. It is calculated as the product of mass and gravitational acceleration, which has magnitude g = 9.8 m/s².

Suppose positive referential is going up. To determine mass:

F_{r}=m.a

F_{t}-F_{g}=m.a

F_{t}-m.g=m.a

18.8-9.8m=5.22m

15.02m=18.8

m = 1.25

Block m has 1.25 kg.

(d) Gravitational force is also called weight. So, as described above: F_{g}=m.g.

The weight for the hanging block is

F_{g}=1.25*9.8

F_{g}= 12.25 N

Comparing tension and weight:

\frac{12.25}{18.8} ≈ 0.65

We can see that, weight of the hanging block is almost 0.65 times smaller than the tension on the rope.

4 0
3 years ago
A car and a train move together along straight, parallel paths with the same constant cruising speed v(initial). At t=0 the car
kogti [31]

Answer:

t_1 = \frac{v_i}{a_i}

t_2 = \frac{v_i}{a_i}

Δd = v_it_1 = v_i^2/a_i

Explanation:

As v(t) = v_i + at, when the car is making full stop, v(t_1) = 0 . a = -a_i . Therefore,

0 = v_i - a_it_1\\v_i = a_it_1\\t_1 = \frac{v_i}{a_i}

Apply the same formula above, with v(t_2) = v_i and a = a_i, and the car is starting from 0 speed,  we have

v_i = 0 + a_it_2\\t_2 = \frac{v_i}{a_i}

As s(t) = vt + \frac{at^2}{2}. After t = t_1 + t_2, the car would have traveled a distance of

s(t) = s(t_1) + s(t_2)\\s(t_1) = (v_it_1 - \frac{a_it_1^2}{2})\\ s(t_2) = \frac{a_it_2^2}{2}

Hence s(t) = (v_it_1 - \frac{a_it_1^2}{2}) + \frac{a_it_2^2}{2}

As t_1 = t_2 we can simplify s(t) = v_it_1

After t time, the train would have traveled a distance of s(t) = v_i(t_1 + t_2) = 2v_it_1

Therefore, Δd would be 2v_it_1 - v_it_1 = v_it_1 = v_i^2/a_i

8 0
3 years ago
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