All categories

3 years ago

Which is a result of using a machine?

Physics

1 answer:

kakasveta [241]3 years ago

4 0

Answer:

Yes labor is required therefor jobs like cashier, and other low skill labor jobs I'll not be required

Explanation:

You might be interested in

How can jet streams impact weather

CaHeK987 [17]

Jet streams, which are rapidly moving ribbons of air 6 to 9 miles above the Earth, influence the weather by separating warm and cold air and pushing weather systems around the globe. The movement of a jet stream affects temperatures and precipitation.

HOPE THIS HELPS

6 0

3 years ago

A cannon of mass 6.43 x 103 kg is rigidly bolted to the earth so it can recoil only by a negligible amount. The cannon fires a 7

Nata [24]

Answer:

The velocity of the shell when the cannon is unbolted is 500.14 m/s

Explanation:

Given;

mass of cannon, m₁ = 6430 kg

mass of shell, m₂ = 73.8-kg

initial velocity of the shell, u₂ = 503 m/s

Initial kinetic energy of the shell; when the cannon is rigidly bolted to the earth.

K.E = ¹/₂mv²

K.E = ¹/₂ (73.8)(503)²

K.E = 9336032.1 J

When the cannon is unbolted from the earth, we apply the principle of conservation of linear momentum and kinetic energy

change in initial momentum = change in momentum after

0 = m₁u₁ - m₂u₂

m₁v₁ = m₂v₂

where;

v₁ is the final velocity of cannon

v₂ is the final velocity of shell

$v_1 = \frac{m_2v_2}{m_1}$

Apply the principle of conservation kinetic energy

$K = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2\\\\K = \frac{1}{2}m_1(\frac{m_2v_2}{m_1})^2 + \frac{1}{2}m_2v_2^2\\\\K = \frac{1}{2}m_2v_2^2(\frac{m_2}{m_1}) + \frac{1}{2}m_2v_2^2 \\\\K = \frac{1}{2}m_2v_2^2 (\frac{m_2}{m_1} + 1)\\\\2K = m_2v_2^2 (\frac{m_2}{m_1} + 1)\\\\v_2^2 = \frac{2K}{M_2(\frac{m_2}{m_1} + 1)} \\\\v_2^2 = \frac{2*9336032.1}{73.8(\frac{73.8}{6430} + 1)}\\\\$

$v_2^2 = 250138.173\\\\v_2 = \sqrt{250138.173} \\\\v_2 = 500.14 \ m/s$

Therefore, the velocity of the shell when the cannon is unbolted is 500.14 m/s

3 0

3 years ago

A 4.0 kg box is released from rest on a frictionless ramp. The box slides from the ramp onto a rough horizontal

muminat

Answer:

Assuming air resistance is negligible, all of the potential energy that the object has at the top of the ramp is converted into kinetic energy by the time it gets to the bottom of the ramp. This is because no matter what path the object takes to move the 5m vertically (ie. falling straight down v. sliding on the ramp), gravity does the same amount of work on it.

Thus, calculate the total amount of potential energy at the top of the ramp:

Ep=mgh

Ep=4(9.81)5

Ep=196.2 Joules

Because all of this potential energy is converted into kinetic energy in the object by the bottom of the ramp, the object hits the spring with 196.2J of energy.

By using the formula for elastic potential energy, you can calculate exactly how far the spring compresses.

196.2=(1/2)k(x^2)

392.4=(350)(x^2)

1.1211=x^2

sqrt(1.1211)=x

x=1.059m

As for the last part of the question, after the object compresses the spring fully and stops momentarily, the spring converts it's elastic potential energy back into kinetic energy in the object and pushes it away again.

Explanation:

4 0

3 years ago

Read 2 more answers

To meet the productivity goal, you must move 75 units of product per hour. You have moved 30 units of product in 30 minutes. How

9966 [12]

Answer:

1.5 unit of product per min

Explanation:

30 units of product was moved in 30 minutes.

Number of units left = Total number of units-number of units moved

=75-30 =45 units

45 units is available to be moved for the rest 30 min. To be able to achieve this goal of 75 units of product per hour.

45/30 amount of units must be moved in 1 min

=1.5 unit per min

4 0

3 years ago

An object of mass m is traveling on a horizontal surface. There is a coefficient of kinetic friction  between the object and th

Gnom [1K]

Answer:

8m * (μg/v)^2

Explanation:

k, the spring constant = ?

(k in terms of μ, m, g, and v.)

Frictional force = μmg

Note: lost KE is converted to work done against the friction + PE of the spring

1/2mv2 = μmgx + 1/2kx^2....equation i

Cancel the 1/2 on both sides

mv^2 = μmgx + kx^2

Lets recall that:

Due to frictional effect, further enegy will be lost when the spring recoils backward

Therefore

1/2kx^2 = μmgx..... equation ii

Let's substitute 1/2kx^2 in equation I for ii

So we can say that:

1/2mv^2 = (μmgx)+ μmgx

1/2mv^2 = 2 (μmgx)

1/4mv^2 = μmgx

Cancel out m on both sides

1/4v^2 = μgx

Make x subject of the formula

x = (1/4v^2) / (μg)...... equation iii

substitute x to equation ii

But first make k in equation ii subject of the formula

1/2kx^2 = μmgx

k = 2μmg/x

Now substitute x

k = 2μmg / ((1/4v^2) / (μg))

k = 2μmg * ((μg) / (1/4v^2))

k = 8m * (μg/v)^2

8m * (μg/v)^2

7 0

3 years ago