Ask question

Ask question

All categories

3 years ago

11

A carpenter lifts a 10 kg piece of wood to his shoulder 1.5 m above the ground. What is the wood's potential energy on the carpe

nter's shoulder? (*Hint: "g" is the acceleration due to gravity of an object on Earth. G = 10 m/s^2)

1 answer:

Burka [1]3 years ago

8 0

Answer:

The wood's potential energy on the carpenter's shoulder is 150 J.

Explanation:

Given;

mass of the wood, m = 10 kg

height through which the wood was raised, h = 1.5 m

acceleration due to gravity, g = 10 m/s²

The wood's potential energy on the carpenter's shoulder is calculated as;

P.E = mgh

P.E = 10 x 10 x 1.5

P.E = 150 J

Therefore, the wood's potential energy on the carpenter's shoulder is 150 J.

You might be interested in

The discharge of a stream is Choose one: likely to decrease downstream in arid regions and increase downstream in temperate regi

Sauron [17]

Answer:

<em>likely to decrease downstream in arid regions and increase downstream in temperate regions</em>

<em></em>

Explanation:

Arid regions are is a region with a severe lack of water, usually to the extent that affect the organisms living in the region. Arid regions are characterized by a very low depth of rainfall per year. Temperate region on the other hand experience more distinct seasonal change and wider temperature change. Temperate regions get a fairly large amount of rainfall per year.

In arid regions, the soil is very dry, and the rate of infiltration and percolation is high relative to the amount of rainfall available. The effect is that more water is infiltrated into the soil as you move downstream, leading to a decrease in the discharge of a stream as you move downstream. Most temperate region have soils that are usually saturated in the peak of the rainfall season, leading to a greater stream discharge as you move downstream.

3 0

3 years ago

A) Charge q1 = +5.60 nC is on the x-axis at x = 0 and an unknown charge q2 is on the x-axis at x = -4.00 cm. The total electric

jeka94

Answer:

a) F₃₁ = 63.0 μN

b) F₃₂ = - 14.0 μN

c) q₂ = - 5.0 nC

Explanation:

a)

Assuming that the three charges can be taken as point charges, the forces between them must obey Coulomb's Law, and can be found independent each other, applying the superposition principle.
So, we can find the force that q₁ exerts along the x-axis on q₃, as follows:

$F_{31} =\frac{k*q_{1}*q_{3} }{r_{13}^{2}} = \frac{9e9Nm2/C2*5.6e-9C*2.0e-9C}{(0.04m)^{2}} = 63.0 \mu N (1)$

b)

Since total force exerted by q₁ and q₂ on q₃ is 49.0 μN, we can find the force exerted only by q₂ (which is along the x-axis only too) just by difference, as follows:

$F_{32} = F_{3} - F_{31} = 49.0\mu N - 63.0\mu N = -14.0 \mu N (2)$

c)

Finally, in order to find the value of q₂, as we know the value and sign of F₃₂, we can apply again the Coulomb's Law, solving for q₂, as follows:

$q_{2} = \frac{F_{32} * r_{23}^{2} }{k*q_{3}} = \frac{(-14\mu N)*(0.08m)^{2}}{9e9Nm2/C2* 2 nC} = - 5 nC (3)$

6 0

3 years ago

A NASA satellite has just observed an asteroid that is on a collision course with the Earth. The asteroid has an estimated mass,

Citrus2011 [14]

Answer:

v = 7934.2 m/s

Explanation:

Here the total energy of the Asteroid and the Earth system will remains conserved

So we will have

$-\frac{GMm}{r} + \frac{1}{2}mv_0^2 = -\frac{GMm}{R} + \frac{1}{2}mv^2$

now we know that

$v_0 = 660 m/s$

$M = 5.98 \times 10^{24} kg$

$m = 5 \times 10^9 kg$

$r = 4 \times 10^9 m$

$R = 6.37 \times 10^6 m$

now from above formula

$GMm(\frac{1}{R} - \frac{1}{r}) + \frac{1}{2}mv_0^2 = \frac{1}{2}mv^2$

now we have

$2GM(\frac{1}{R} - \frac{1}{r}) + v_0^2 = v^2$

now plug in all data

$2(6.67 \times 10^{-11})(5.98 \times 10^{24})(\frac{1}{6.37 \times 10^6} - \frac{1}{4 \times 10^9}) + (660)^2 = v^2$

$v = 7934.2 m/s$

5 0

3 years ago

The Balmer series in hydrogen atom produces only infrared emission. O True O False

lorasvet [3.4K]

Answer:

False

Explanation:

As we know that, the Balmer series gives the n values as,

$n_{i}=2$ .

$[tex]n_{f}=3,4,5,.....\infty$ .

Now the value of wavelength can be calculated as,

$\frac{1}{\lambda}=R(\frac{1}{n_{i} }-\frac{1}{n_{f} } )z^{2}$ .

Here, $R=109677 cm^{-1}$ .

And $n_{f}=3$ .

Now,

$\frac{1}{\lambda}=109677 cm^{-1}(\frac{1}{2}-\frac{1}{3})$ .

Therefore,

$\lambda=\frac{6}{109677} cm\\\lambda=547\times 10^{-9} m\\ \lambda=547 nm$

Therefore, the wavelength of Balmer series lies in visible region which is 547 nm.

8 0

3 years ago

Read 2 more answers

a seismic wave has an amplitude of 0.012 Meters.If the amplitude of this wave reduces to 0.006 meters, what happens to the energ

irina1246 [14]

Answer:The energy of the wave by a factor of 4

Explanation:

5 0

3 years ago