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Solnce55 [7]
2 years ago
4

a container of rocks with a mass of 65.0 kilograms is brought back from the moon's surface where the acceleration due to gravity

is 1.62 meters per second2. what is the weight of the container of rocks on earth's surface?
Physics
1 answer:
den301095 [7]2 years ago
7 0
The mass is an intrinsic property of a substance which means that wherever it is placed, the mass would stay the same. This being said, the mass of the container of rocks in the Earth's surface is still 65 kg. 

To determine the weight of the object on the Earth's surface, we multiply the mass with the acceleration due to gravity, 9.8 m/s². 

           Weight = mg

Substituting with the known values,
   
           Weight = (65 kg)(9.8 m/s²)
            Weight = 637 N

<em>Answer: 637 N</em>
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Charge q1 is distance r from a positive point charge q. charge q2=q1/3 is distance 2r from q. what is the ratio u1/u2 of their p
makvit [3.9K]
We need the power law for the change in potential energy (due to the Coulomb force) in bringing a charge q from infinity to distance r from charge Q. We are only interested in the ratio U₁/U₂, so I'm not going to bother with constants (like the permittivity of space). 

<span>The potential energy of charge q is proportional to </span>
<span>∫[s=r to ∞] qQs⁻²ds = -qQs⁻¹|[s=r to ∞] = qQr⁻¹, </span>

<span>so if r₂ = 3r₁ and q₂ = q₁/4, then </span>
<span>U₁/U₂ = q₁Qr₂/(r₁q₂Q) = (q₁/q₂)(r₂/r₁) </span>
<span>= 4•3 = 12.</span>
5 0
2 years ago
The inclined plane in the figure above has two sections of equal length and different roughness. The dashed line shows where sec
saveliy_v [14]

The static friction exerted on the block by the incline is \mu _ s _1 Mgcos \ \theta.

The given parameters;

  • <em>mass of the block, = M</em>
  • <em>coefficient of static friction in section 1, = </em>\mu_s_1<em />
  • <em>angle of inclination of the plane, = θ</em>

<em />

The normal force on the block is calculated as follows;

Fₙ = Mgcosθ

The static friction exerted on the block by the incline is calculated as follows;

F_s = \mu_s F_n\\\\F_s = \mu _s_1(Mg cos\ \theta)\\\\F_s = \mu _s_1 Mgcos\ \theta

Thus, the static friction exerted on the block by the incline is \mu _ s _1 Mgcos \ \theta

Learn more here:brainly.com/question/17237604

3 0
2 years ago
How does increasing the distance between charged objects affect the electric force between them? the electric force increases be
konstantin123 [22]

the electric force decreases because the distance has an indirect relationship to the force

Explanation:

The electric force between two objects is given by

F=k \frac{q_1 q_2}{r^2}

where

k is the Coulomb's constant

q1 and q2 are the charges of the two objects

r is the distance between the two objects

As we can see from the formula, the magnitude of the force is inversely proportional to the square of the distance: so, when the distance between the object increases, the magnitude of the force decreases.

3 0
2 years ago
Read 2 more answers
A radioactive isotope has a half-life of 2 hours. If a sample of the element contains 600,000 radioactive nuclei at 12 noon, how
storchak [24]

Answer: There will be 75258 nuclei left at 6 pm.

Explanation:

a) half-life of the radioactive substance:

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

t_{\frac{1}{2}}=\frac{0.69}{k}

k=\frac{0.69}{t_{\frac{1}{2}}}=\frac{0.693}{2hours}=0.346hours^{-1}

b) Expression for rate law for first order kinetics is given by:

A=A_0e^{-kt}

where,

k = rate constant  

t = time for decomposition = 6 hours ( from 12 noon to 6 pm)

A = activity at time t = ?

A_0 = initial activity  = 600, 000

A=600000\times e^{-0.346\times 6}

A=75258

Thus there will be 75258 nuclei left at 6 pm.

7 0
2 years ago
The image shows a pendulum that is released from rest at point A. Shari tells her friend that no energy transformation occurs as
Masja [62]
Is  D    the  right  answer
6 0
2 years ago
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