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N76 [4]
3 years ago
10

Where is the center of gravity if the 9.00 kg mass of the barbell itself is taken into account?

Physics
1 answer:
Pavel [41]3 years ago
3 0

The problem is missing some parts but nevertheless here is the answer:

 

Given:

1.7 m long barbell with 23 kg weight on the left and 34 in the right end.

 

Solution:

 Calculating from the left side:

Center of gravity = ΣM/ΣW = [23*0 + 9*(1.7/2) + 34*1.7]/[23+9+34] 

Center of gravity = .992 m from the left end

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How many joules of heat are needed to raise the temperature of 50.0 g of aluminum from 10°C to 110°C, if the specific heat of al
dusya [7]

Answer:

Heat required to raise the temperature of the aluminium is 4750 J

Explanation:

As we know that the heat energy required to raise the temperature of the aluminium is given as

Q = ms\Delta T

here we know that

m = 50 g

\Delta T = 110 - 10

\Delta T = 100 ^oC

so we have

Q = 50(0.95)(100)

Q = 4750 J

5 0
3 years ago
If two vectors are perpendicular to each other, how should you add them?
Aleksandr [31]

Let us consider two vectors A and B.

As per the question, the two vectors are perpendicular to each other.

Hence the angle between them  \theta =90 degree

We are asked to calculate the resultant of these two vectors.

As per parallelogram law of vector addition, the resultant of two vectors are-

                      R=\sqrt{A^{2}+ B^{2}+2ABcos\theta

                                =\sqrt{ A^{2}+ B^{2}+2AB*cos90}    [cos90=0]

                                =\sqrt{ A^{2}+ B^{2}

This is the way by which we can add two perpendicular vectors.


8 0
3 years ago
Read 2 more answers
What is the energy of the photon emitted when an electron in a mercury atom drops from energy level f to energy level b?
xeze [42]

The energy of the photon emitted when an electron in a mercury atom drops from energy level f to energy level b is 3.06 eV.

<h3>Change in energy level of the electron</h3>

When photons jump from a higher energy level to a lower level, they emit or radiate energy.

The change in energy level of the electrons is calculated as follows;

ΔE = Eb - Ef

ΔE = -2.68 eV - (-5.74 eV)

ΔE = 3.06 eV

Thus, the energy of the photon emitted when an electron in a mercury atom drops from energy level f to energy level b is 3.06 eV.

Learn more about energy level here: brainly.com/question/14287666

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7 0
2 years ago
A rock is thrown upward with an initial velocity of 16 ft/s from an initial height of 5 ft. write a quadratic function equation
Andrei [34K]
During upward projection the final velocity is zero, and the gravitational acceleration is -10 m/s² (against the gravity).
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S = 1/2gt² + ut
Where s is the height h, g is gravitational acceleration, and t is the time and u is the initial velocity u, is 16 ft/s.
Thus; h= 1/2(-10)t² + 16t
We get; h = -5t² + 16t
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5 0
3 years ago
Which transformation of energy takes place when a slingshot launches a stone? (PE stands for potential energy.) A. elastic PE to
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B. Elastic potential to kinetic energy

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