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3 years ago

A box with the mass of 20 kg at 5 m is lifted to 20 m. How much work was 7 points done?

Physics

1 answer:

kirza4 [7]3 years ago

3 0

Answer:

$2,900\: \mathrm{J}$

Explanation:

Work is given by the equation $W=F\Delta x$ where $F$ is force and $\Delta x$ is displacement.

Displacement is defined as change in position. In this case, the box moves from 5m to 20m. Its displacement is $20-5=15\:\mathrm{m}$ .

The force acting on the box is the force of gravity, given as $F_g=mg$ .

Therefore, the total work done is:

$W=F\Delta x=mg\Delta x = 20\cdot 9.81\cdot 15=2,943=\fbox{$2,900\:\mathrm{J}$}$ (two significant figures).

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Energy Calculations

MaRussiya [10]

Answer: 20,734.69 N/m

Explanation:

The elastic potential energy (ELPE) of the rubber band is given by

$E=\frac{1}{2}kx^2$

where

k is the spring constant

x = 0.035 m is the stretching of the rubber band

E = 12.7 J is the ELPE of the rubber band

Substituting the numbers and re-arranging the equation, we find

$k=\frac{2E}{x^2}=\frac{2\cdot 12.7 J}{(0.035 m)^2}=20,734.69 N/m$

6 0

3 years ago

Scientists use laser range-finding to measure the distance to the moon with great accuracy. A brief laser pulse is fired at the

Fofino [41]

Answer:

d = 2,042 10-3 m

Explanation:

The laser diffracts in the circular slit, so the process equation is

d sin θ= m λ

The first diffraction minimum occurs for m = 1

We can use trigonometry in the mirror

tan θ = Y / L

Where L is the distance from the Moon to Earth

Since the angle is extremely small

tan θ = sin θ / cos θ

Cos θ = 1

tant θ = sin θ = y / L

We replace

d y / L = λ

d = λ L / y

Let's calculate

d = 532 10⁻⁹ 3.84 10⁶/1 10³

d = 2,042 10-3 m

5 0

3 years ago

2. Kevin works as a janitor, and he is pushing a fully-

dybincka [34]

The time taken for him to move the bin 6.5 m is 2.30 s.

The given parameters;

weight of the load, w = 557 N
force applied , F = 410 N
angle of force, = 15°
coefficient of kinetic friction = 0.46
distance moved, d = 6.5 m

The net horizontal force on the recycling bin is calculated as follows;

$Fcos\theta - F_k = ma$

where;

m is the mass of the recycling bin
$F_k$ is the frictional force

W = mg

$557 = 9.8m\\\\m = \frac{557}{9.8} \\\\m = 56.84 \ kg$

The net horizontal force on the recycling bin is calculated as;

$Fcos \theta - F_k = ma\\\\Fcos\theta - \mu_kF_n = ma\\\\410\times cos(15) \ - \ 0.46(557) = 56.84 a\\\\139.8 = 56.84a\\\\a = \frac{139.8}{56.84} \\\\a = 2.46 \ m/s^2$

The time taken for him to move the bin 6.5 m is calculated as follows;

$s = v_0t + \frac{1}{2} at^2\\\\6.5 = 0 + \frac{1}{2} \times 2.46\times t^2\\\\6.5 = 1.23 t^2\\\\t^2 = \frac{6.5 }{1.23} \\\\t^2 = 5.285\\\\t = \sqrt{5.285} \\\\t = 2.30 \ s$

Thus, the time taken for him to move the bin 6.5 m is 2.30 s.

Learn more here:brainly.com/question/21684583

7 0

3 years ago

Two charges (q1 = 3.8*10-6C, q2 = 3.2*10-6C) are separated by a distance of d = 3.25 m. Consider q1 to be located at the origin.

Sergio039 [100]

Answer:

The distance is 1.69 m.

Explanation:

Given that,

First charge $q_{1}= 3.8\times10^{-6}\ C$

Second charge $q_{2}=3.2\times10^{-6}\ C$

Distance = 3.25 m

We need to calculate the distance

Using formula of electric field

$E_{1}=E_{2}$

$\dfrac{kq_{1}}{x^2}=\dfrac{kq_{2}}{(d-x)^2}$

$\dfrac{q_{1}}{q_{2}}=\dfrac{(x)^2}{(d-x)^2}$

$\sqrt{\dfrac{q_{1}}{q_{2}}}=\dfrac{x}{d-x}$

$x=(d-x)\times\sqrt{\dfrac{q_{1}}{q_{2}}}$

Put the value into the formula

$x=(3.25-x)\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}$

$x+x\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}=3.25\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}$

$x(1+\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}})=3.25\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}$

$x=\dfrac{3.25\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}}{(1+\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}})}$

$x=1.69\ m$

Hence, The distance is 1.69 m.

5 0

3 years ago

Why must mine tailings be stored and disposed of carefully?

gavmur [86]

After thorough researching, the mine tailings must be stored and disposed of carefully because they have lots of chemical and various toxic materials. They can also leach to the aquifers. The correct answer to the following given statement above is they have chemicals which are dangerous.

7 0

3 years ago