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2 years ago

A machine does 50 J of Work on a 10 Kg Object. What is the Change in the Objects Total Energy?

Physics

1 answer:

Otrada [13]2 years ago

3 0

The change in the total energy of the object is zero (0).

The given parameters:

work done by the machine, W = 50 J

mass of the object, m = 10 kg

To find:

the change in the total energy of the object

The change in the total energy of the object is the difference between the objects initial energy due to its position and the work done on the object.

Based on work energy-theory, the work done on the object is equal to the energy of the object.

The energy of the object = work-done on the object

The change in total energy = 50 J - 50 J = 0

Thus, the change in the total energy of the object is zero (0).

Learn more here: brainly.com/question/20377140

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You are driving your 1700 kg car at 21 m/s down a hill with a 5.0∘ slope when a deer suddenly jumps out onto the roadway. You sl

podryga [215]

Answer:

d = 27.7 m

Explanation:

Here the car is driving on the inclined plane

So here we can say that work done by the gravity and work done by friction is equal to change in kinetic energy of the system

So here we can write it as

$mgsin\theta \times d - F_f \times d = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$

now we have

m = 1700 kg

$v_f = 0$

$v_i = 21 m/s$

$F_f = 1.5 \times 10^4 N$

$(1700)(9.81)sin5 (d) - (1.5\times 10^4)d = 0 - \frac{1}{2}(1700)(21^2)$

$1453.5 d - (1.5\times 10^4)d = -374850$

$d = 27.7 m$

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3 years ago

How much positive electric charge is in 10 moles of carbon? NA=6.022×1023mol−1, e=1.60×10−19C.

levacccp [35]

Answer:

4.75 x 10^-25 kg

Explanation:

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3 years ago

What is the energy per photon absorbed during the transition from n = 2 to n = 3 in the hydrogen atom?

adelina 88 [10]

Answer : The energy of one photon of hydrogen atom is, $3.03\times 10^{-19}J$

Explanation :

First we have to calculate the wavelength of hydrogen atom.

Using Rydberg's Equation:

$\frac{1}{\lambda}=R_H\left(\frac{1}{n_i^2}-\frac{1}{n_f^2} \right )$

Where,

$\lambda$ = Wavelength of radiation

$R_H$ = Rydberg's Constant = 10973731.6 m⁻¹

$n_f$ = Higher energy level = 3

$n_i$ = Lower energy level = 2

Putting the values, in above equation, we get:

$\frac{1}{\lambda}=(10973731.6)\left(\frac{1}{2^2}-\frac{1}{3^2} \right )$

$\lambda=6.56\times 10^{-7}m$

Now we have to calculate the energy.

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

where,

h = Planck's constant = $6.626\times 10^{-34}Js$

c = speed of light = $3\times 10^8m/s$

$\lambda$ = wavelength = $6.56\times 10^{-7}m$

Putting the values, in this formula, we get:

$E=\frac{(6.626\times 10^{-34}Js)\times (3\times 10^8m/s)}{6.56\times 10^{-7}m}$

$E=3.03\times 10^{-19}J$

Therefore, the energy of one photon of hydrogen atom is, $3.03\times 10^{-19}J$

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3 years ago

At the periphery of a hurricane the air is ____, and several kilometers above the surface, in the eye, the air is ____.

tatuchka [14]

<span>At the periphery of a hurricane the air is sinking, and several kilometers above the surface, in the eye, the air is sinking. </span>

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3 years ago

A rocket, initially at rest on the ground, accelerates straight upward from rest with constant acceleration 58.8m/s2 . The accel

Zanzabum

Split the operation in two parts. Part A) constant acceleration 58.8m/s^2, Part B) free fall.

Part A)
Height reached, y = a*[t^2] / 2 = 58.8 m/s^2 * [7.00 s]^2 / 2 = 1440.6 m

Now you need the final speed to use it as initial speed of the next part.

Vf = Vo + at = 0 + 58.8m/s^2 * 7.00 s = 411.6 m/s

Part B) Free fall

Maximum height, y max ==> Vf = 0

Vf = Vo - gt ==> t = [Vo - Vf]/g = 411.6 m/s / 9.8 m/s^2 = 42 s

ymax = yo + Vo*t - g[t^2] / 2

ymax = 1440.6 m + 411.6m/s * 42 s - 9.8m/s^2 * [42s]^2 /2
ymax = 1440.6 m + 17287.2m - 8643.6m = 10084.2 m

Answer: ymax = 10084.2m

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3 years ago