1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Marat540 [252]
3 years ago
11

A 10.2-kg mass is located at the origin, and a 4.6-kg mass is located at x = 8.1 cm. Assuming g is constant, what is the locatio

n of the center of mass xcom and the location of the center of gravity xcogof the two masses? Are the locations the same? Why or why not?
Physics
1 answer:
goldfiish [28.3K]3 years ago
3 0

Answer:

center of mass of the two masses will lie at x = 2.52 cm

center of gravity of the two masses will lie at x = 2.52 cm

So center of mass is same as center of gravity because value of gravity is constant here

Explanation:

Position of centre of mass is given as

r_{cm} = \frac{m_1r_1 + m_2r_2}{m_1 + m_2}

here we have

m_1 = 10.2 kg

m_2 = 4.6 kg

r_1 = (0, 0)

r_2 = (8.1cm, 0)

now we have

r_{cm} = \frac{10.2 (0,0) + 4.6 (8.1 , 0)}{10.2 + 4.6}

r_{cm} = {(37.26, 0)}{14.8}

r_{cm} = (2.52 cm, 0)

so center of mass of the two masses will lie at x = 2.52 cm

now for center of gravity we can use

r_g_{cm} = \frac{m_1gr_1 + m_2gr_2}{m_1g + m_2g}

here we have

m_1 = 10.2 kg

m_2 = 4.6 kg

r_1 = (0, 0)

r_2 = (8.1cm, 0)

now we have

r_g_{cm} = \frac{10.2(9.8) (0,0) + 4.6(9.8) (8.1 , 0)}{10.2(9.8) + 4.6(9.8)}

r_g_{cm} = {(37.26, 0)}{14.8}

r_g_{cm} = (2.52 cm, 0)

So center of mass is same as center of gravity because value of gravity is constant here

You might be interested in
HEY CAN ANYONE ANSWER DIS RQ!!!!!!
emmasim [6.3K]

Answer:

Density is the mass to volume ratio of an object.

> It tells you how compact the mass is.

> Density = mass/volume

The density of water is 1 g/ml or 1 kg/L or 1000 kg/m3

• If an object has a density less than that of water, it

will float.

• If an object has a density more than that of water, it

will sink

Explanation:

hope this helps some

7 0
3 years ago
Read 2 more answers
PLS HELP. WILL MARK AS THE BRAINLIEST. PLS ANSWER STEP BY STEP.
USPshnik [31]

Answer:12

Explanation:

its the way u read it

3 0
3 years ago
Read 2 more answers
A sample contains radioactive atoms of two types, A and B. Initially there are five times as many A atoms as there are B atoms.
victus00 [196]

Answer:

Explanation:

Initially no of atoms of A = N₀(A)

Initially no of atoms of B = N₀(B)

5 X N₀(A)  = N₀(B)

N = N₀ e^{-\lambda t}

N is no of atoms after time t , λ is decay constant and t is time .

For A

N(A) = N(A)₀ e^{-\lambda_1 t}

For B

N(B) = N(B)₀ e^{-\lambda_2 t}

N(A) = N(B) , for t = 2 h

N(A)₀ e^{-\lambda_1 t} = N(B)₀ e^{-\lambda_2 t}

N(A)₀ e^{-\lambda_1 t} = 5 x N₀(A)  e^{-\lambda_2 t}

e^{-\lambda_1 t} = 5  e^{-\lambda_2 t}

e^{\lambda_2 t} = 5  e^{\lambda_1 t}

half life = .693 / λ

For A

.77 =  .693 / λ₁

λ₁ = .9 h⁻¹

e^{\lambda_2 t} = 5  e^{\lambda_1 t}

Putting t = 2 h , λ₁ = .9 h⁻¹

e^{\lambda_2\times  2} = 5  e^{.9\times  2}

e^{\lambda_2\times  2} = 30.25

2 x λ₂ = 3.41

λ₂ = 1.7047

Half life of B = .693 / 1.7047

= .4065 hours .

= .41 hours .

6 0
3 years ago
An electron emitted from a filament is travelling at 1.5 x 105 m/s when it enters an acceleration of an electron gun in a televi
Crank

Answer:

The acceleration of the electron is 1.457 x 10¹⁵ m/s².

Explanation:

Given;

initial velocity of the emitted electron, u = 1.5 x 10⁵ m/s

distance traveled by the electron, d = 0.01 m

final velocity of the electron, v = 5.4 x 10⁶ m/s

The acceleration of the electron is calculated as;

v² = u² + 2ad

(5.4 x 10⁶)² = (1.5 x 10⁵)² + (2 x 0.01)a

(2 x 0.01)a = (5.4 x 10⁶)² - (1.5 x 10⁵)²

(2 x 0.01)a = 2.91375 x 10¹³

a = \frac{2.91375 \ \times \ 10^{13}}{2 \ \times \ 0.01} \\\\a = 1.457 \ \times \ 10^{15} \ m/s^2

Therefore, the acceleration of the electron is 1.457 x 10¹⁵ m/s².

7 0
3 years ago
The specific heat of gold is 0.126 J/g°C. What mass of gold would change its temperature from 25.0 °C to 60.0 °C with the additi
fiasKO [112]
<span>The amount of heat energy needed to increase the temperature of a substance by </span>\Delta T<span> is given by:
</span>Q=m C_s \Delta T<span>
where m is the mass of the substance, Cs is its specific heat capacity and </span>\Delta T<span> is the increase in temperature of the substance.

In this problem, we have a certain mass m of gold, with specific heat capacity </span>C_s=0.126 J/g^{\circ}C<span>, to which we add Q=2825 J of energy. Its temperature increases by </span>\Delta T=60-25=35 ^{\circ}C<span>. Therefore, if we re-arrange the previous equation, we can find the mass of the block of gold:
</span>m= \frac{Q}{C_s \Delta T} = \frac{2825J}{0.126\cdot 35}} =641 g<span>
So, the correct answer is B.</span>
3 0
4 years ago
Read 2 more answers
Other questions:
  • MARKING BRAINLIEST AND I NEED WORK SHOWN ! i’m a junior in high school
    7·1 answer
  • A car moving in a straight line starts at x=0 at t=0. it passes the point x=25.0m with a speed of 11.0m/s at t=3.0s. it passesth
    7·1 answer
  • Quiz Active 1 2 3 4 5 6 7 8 10 TIME REMAINING 52:33 Which conclusion is best supported by the information? Rhea kicks a soccer b
    7·1 answer
  • What are some ways in which body fat can be measured
    6·1 answer
  • a toboggan loaded with vacationing students (total weight 1300 N) slides down a slope at 30 degrees and there is no friction. wh
    5·1 answer
  • You are a member of a citizen's committee investigating safety in the high school sports program. You are interested in knee dam
    6·1 answer
  • You want to move a heavy box with mass 30.0 across a carpeted floor. You pull hard on one of the edges of the box at an angle 30
    10·1 answer
  • Hello can someone please help me with this.
    9·1 answer
  • I need help ASAP
    12·1 answer
  • Vin Diesel jumps out of a plane. Gravity pulls on him with 184N of force and his parachute pushes him up with 82 N of force.
    8·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!