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3 years ago

What is the weight of a 48 kg girl on Earth? Round the answer to the nearest whole number.

Physics

2 answers:

timama [110]3 years ago

8 0

Answer:

It is 470 N. Thanks for answering your own question, it helped me on the test.

Explanation:

Diano4ka-milaya [45]3 years ago

6 0

Remember, in order to calculate the weight on earth, we need to multiply the mass of the object with the force of Gravity (9.8 m/s^2 on earth)

so, her weight would be:
48 x 9.8 = 470.4 >>>>>>> Will be 470 if we round it to the nearest whole number.

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What are some important factors to consider when choosing a warm-up before your workout?

TEA [102]

Answer:

Prior to exercise, a proper warm-up of 10-15 minutes is extremely important to avoid injuries.

Don't go too hard in the beginning and boost your activity level slowly. A good indication of a proper warm-up is that you feel sweat on your body parts.
Don't overstretch right in the beginning as it can cause sore in your muscles and joints or stress fractures.
Take a break if you feel sick or fatigues and use other drinks along with water to replace electrolytes and body fluids.

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3 years ago

Read 2 more answers

how far can a mother push a 20.0kg baby carriage, using a force of 62.0N at angle of 30.0°to the horizontal, if she can do 2920

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Here is my solution using my app.

4 0

3 years ago

An infinitely long straight wire has a uniform linear charge density of Derive the 4. equation for the electric field a distance

marshall27 [118]

Answer:

$E = \frac{\lambda}{2\pi \epsilon_0 r}$

Explanation:

Let the linear charge density of the charged wire is given as

$\frac{q}{L} = \lambda$

here we can use Gauss law to find the electric field at a distance r from wire

so here we will assume a Gaussian surface of cylinder shape around the wire

so we have

$\int E. dA = \frac{q}{\epsilon_0}$

here we have

$E \int dA = \frac{\lambda L}{\epsilon_0}$

$E. 2\pi r L = \frac{\lambda L}{\epsilon_0}$

so we have

$E = \frac{\lambda}{2\pi \epsilon_0 r}$

4 0

2 years ago

Light with a wavelength of 400 nm strikes the surface of cesium in a photocell, and the maximum kinetic energy of the electrons

Firdavs [7]

Answer:

The longest wavelength of light that is capable of ejecting electrons from that metal is 1292 nm.

Explanation:

Given that,

Wavelength = 400 nm

Energy $E=1.54\times10^{-19}\ J$

We need to calculate the longest wavelength of light that is capable of ejecting electrons from that metal

Using formula of energy

$E = \dfrac{hc}{\lambda}$

$\lambda=\dfrac{hc}{E}$

Put the value into the formula

$\lambda=\dfrac{6.63\times10^{-34}\times3\times10^{8}}{1.54\times10^{-19}}$

$\lambda=1292\times10^{-9}\ m$

$\lambda=1292\ nm$

Hence, The longest wavelength of light that is capable of ejecting electrons from that metal is 1292 nm.

8 0

3 years ago

A crate with a mass of 110 kg glides through a space station with a speed of 4.0 m/s. An astronaut speeds it up by pushing on it

Darina [25.2K]

Answer:

The final speed of the crate after the astronaut push to slow it down is 4.50 m/s

Explanation:

Given:

The crate has mass m = 110 kg and an initial speed vi = 4 m/s.

Solution

We are asked to determine the final speed of the crate. We could apply the steps for energy principle update form as next

Ef=Ei+W (1)

Where Ef and Ei are the find and initial energies of the crate (system) respectively. While W is the work done by the astronaut (surrounding).

The system has two kinds of energy, the kinetic energy which associated with its motion and the rest energy where it has zero speed. The summation of both energies called the particle energy. So, equation (1) will be in the form

(Kf + mc^2) = (KJ+ mc^2) (2)

Where m is the mass of crate, c is the speed of light which equals 3 x 10^8 m/s and the term mc^2 represents the energy at rest and the term K is the kinetic energy.

In this case, the rest energy doesn't change so we can cancel the rest energy in both sides and substitute with the approximate expression of the kinetic energy of the crate at low speeds where K = 1/2 mv^2 and equation (2) will be in the form

(1/2mvf^2+mc^2)=(1/2mvi^2 +mc^2)+W

1/2mvf^2=1/2mvi^2+W (3)

Now we want to calculate the work done on the crate to complete our calculations. Work is the amount of energy transfer between a source of an applied force and the object that experiences this force and equals the force times the displacement of the object. Therefore, the total work done will be given by

W = FΔr (4)

Where F is the force applied by the astronaut and equals 190 N and Δr is the displacement of the crate and equals 6 m. Now we can plug our values for F and Δr to get the work done by the astronaut

W = F Δr= (190N)(6 m) = 1140 J

Now we can plug our values for vi, m and W into equation (3) to get the final speed of the crate

1/2mvf^2=1/2mvi^2+W

vf=5.82 m/s

This is the final speed of the first push when the astronaut applies a positive work done. Then, in the second push, he applies a negative work done on the crate to slow down its speed. Hence, in this case, we could consider the initial speed of the second process to be the final speed of the first process. So,

vi' = vf

In this case, we will apply equation (3) for the second process to be in the

1/2mvf^2=1/2mvi'^2+W' (3*)

The force in the second process is F = 170 N and the displacement is 4 m. The force and the displacement are in the opposite direction, hence the work done is negative and will be calculated by

W'= —F Δr = —(170N)(4 m)= —680J

Now we can plug our values for vi' , m and W' into equation (3*) to get the final speed of the crate

1/2mvf'^2=1/2mvi'^2+W'

vf'=4.50 m/s

The final speed of the crate after the astronaut push to slow it down is 4.50 m/s

7 0

3 years ago