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Nikitich [7]
3 years ago
9

Newton's law of gravitation states that two masses will ______.

Physics
2 answers:
hjlf3 years ago
8 0

Answer:

The correct option is B

Explanation:

Newton's law of gravitation states that every particle of an object/matter attract every other particle of another object/matter in the universe with a force directly proportional to the product of the particle masses but inversely proportional to the square the of the distance between them.        

The formula for this law is

F = G × Mm/r²

Where F = Force (in newtons)

M and m = the two masses (in kg)

r = distance

spin [16.1K]3 years ago
4 0
Pretty sure the answer is B. Attract each other
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If your car is moving at 100 km/hr, how many kilometers will your car travel in 15 minutes?
steposvetlana [31]
It will be 25 if the car moves at that speed for 15 minutes
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large amounts of energy are converted to small amounts of mass. small amounts of energy are converted to large amounts of mass.
Norma-Jean [14]

Answer:

small amounts of mass are converted to large amounts of energy

Explanation:

According to the mass-energy equivalence, which Albert Einstein initially proposed as a general principle. It was revealed that mass and energy are connected and that a "small amount of mass can be converted into enormous amounts of energy."

Using the formula E=mc^2. This means Energy equals mass times the speed of light squared.

Hence, it is true that "small amounts of mass are converted to large amounts of energy."

3 0
2 years ago
Scientists treat the number of stars in a given volume of space as a Poisson random variable. The density of our galaxy in the v
vladimir1956 [14]

Answer:

P(X\ge 1) = 0.9502

Explanation:

Given

Density = 3 starts in 10 cubic light years.

Required

Determine the probability of 1 or more in 10 cubic light years

Since the number of stars follow a Poisson distribution, we make use of:

P(X=k) = f(x) = (\lambda T)^k\frac{ e^{-\lambda T}}{k!}

\lambda = density

\lambda = \frac{3}{10}

\lambda = 0.3

T = the light years

T = 10

Calculating P(X \ge 1)

In probability:

P(X \ge 1) = 1 - P(X = 0)

Calculating P(X=0)

Substitute 0 for k and the values for \lambda and T in

P(X=k) = f(x) = (\lambda T)^k\frac{ e^{-\lambda T}}{k!}

P(X=0) = (0.3* 10)^0 * \frac{ e^{-0.3 * 10}}{0!}

P(X=0) = (3)^0 * \frac{ e^{-0.3 * 10}}{1}

P(X=0) = (3)^0 *  e^{-0.3 * 10}

P(X=0) = 1 *  e^{-0.3 * 10}

P(X=0) = 1 *  e^{-3}

P(X=0) = e^{-3}

P(X=0) = 0.04979

Substitute 0.04979 for P(X=0) in P(X \ge 1) = 1 - P(X = 0)

P(X\ge 1) = 1 - 0.04979

P(X\ge 1) = 0.95021

P(X\ge 1) = 0.9502 ---  approximated

<em>Hence, the required probability is 0.9502</em>

3 0
3 years ago
How many significant digits are in the measurement 589.040 newtons
Neporo4naja [7]
5... 589. 40....even though the zeros don't really count
5 0
3 years ago
Read 2 more answers
A 0.0550-kg ice cube at −30.0°C is placed in 0.400 kg of 35.0°C water in a very well-insulated container. What is the final temp
KatRina [158]

Answer:

19.34°C

Explanation:

When the ice cube is placed in the water, heat will be transferred from the hot water to it such that the heat gained (Q₁) by the ice is equal to the heat lost(Q₂) by the hot water and a final equilibrium temperature is reached between the melted ice and the cooling/cooled hot water. i.e

Q₁ = -Q₂                  ----------------------(i)

{A} Q₁ is the heat gained by the ice and it is given by the sum of ;

(i) the heat required to raise the temperature of the ice from -30°C to 0°C. This is given by [m₁ x c₁ x ΔT]

<em>Where;</em>

m₁ = mass of ice = 0.0550kg

c₁ = a constant called specific heat capacity of ice = 2108J/kg°C

ΔT₁ = change in the temperature of ice as it melts from -30°C to 0°C = [0 - (-30)]°C = [0 + 30]°C = 30°C

(ii) and the heat required to melt the ice completely - This is called the heat of fusion. This is given by [m₁ x L₁]

Where;

m₁ = mass of ice = 0.0550kg

L₁ = a constant called latent heat of fusion of ice = 334 x 10³J/kg

Therefore,

Q₁ = [m₁ x c₁ x ΔT₁] + [m₁ x L₁]        ------------------(ii)

Substitute the values of m₁, c₁, ΔT₁ and  L₁ into equation (ii) as follows;

Q₁ = [0.0550 x 2108 x 30] + [0.0550 x 334 x 10³]

Q₁ = [3478.2] + [18370]

Q₁ = 21848.2 J

{B} Q₂ is the heat lost by the hot water and is given by

Q₂ = m₂ x c₂ x ΔT₂                -----------------(iii)

Where;

m₂ = mass of water = 0.400kg

c₂ = a constant called specific heat capacity of water = 4200J/Kg°C

ΔT₂ = change in the temperature of water as it cools from 35°C to the final temperature of the hot water (T) = (T - 35)°C

Substitute these values into equation (iii) as follows;

Q₂ = 0.400 x 4200 x (T - 35)

Q₂ = 1680 x (T-35) J

{C} Now to get the final temperature, substitute the values of Q₁ and Q₂ into equation (i) as follows;

Q₁ = -Q₂

=> 21848.2 = - 1680 x (T-35)

=> 35 - T  = 21848.2 / 1680

=> 35 - T  = 13

=> T  = 35 - 13

=> T  = 22

Therefore the final temperature of the hot water is 22°C.

Now let's find the final temperature of the mixture.

The mixture contains hot water at 22°C and melted ice at 0°C

At this temperature, the heat (Q_{W}) due to the hot water will be equal to the negative of the one (Q_{I}) due to the melted ice.

i.e

Q_{W} = -Q_{I}             -----------------(a)

Where;

Q_{I} = m_{I} x c_{I} x ΔT_{I}         [m_{I} = mass of ice, c_{I} = specific heat capacity of melted ice which is now water and ΔT_{I} = change in temperature of the melted ice]

and

Q_{W} = m_{W} x c_{W} x ΔT_{W}    

[m_{W} = mass of water, c_{W} = specific heat capacity of water and ΔT_{W} = change in temperature of the water]

Substitute the values of Q_{W} and Q_{I} into equation (a) as follows

m_{W} x c_{W} x ΔT_{W}   =  - m_{I} x c_{I} x ΔT_{I}

Note that c_{W} and c_{I} are the same since they are both specific heat capacities of water. Therefore, the equation above becomes;

m_{W} x ΔT_{W}   = -m_{I} x ΔT_{I}   -----------------------(b)

Now, let's analyse ΔT_{W} and ΔT_{I}. The final temperature (T_{F}) of the two kinds of water(melted ice and cooled water) are now the same.

=> ΔT_{W} = change in temperature of water = final temperature of water(T_{F}) - initial temperature of water(T_{IW})

ΔT_{W} = T_{F} - T_{IW}

Where;

T_{IW} = 22°C           [which is the final temperature of water before mixture]

=> ΔT_{I} = change in temperature of melted ice = final temperature of water(T_{F}) - initial temperature of melted ice (T_{II})

ΔT_{I} = T_{F} - T_{II}

T_{II} = 0°C     (Initial temperature of the melted ice)

Substitute these values into equation (b) as follows;

m_{W} x ΔT_{W}   =  - m_{I} x ΔT_{I}

0.400 x (T_{F} - T_{IW}) = -0.0550 x (T_{F} - T_{II})

0.400 x (T_{F} - 22) = -0.0550 x (T_{F} - 0)

0.400 x (T_{F} - 22) = -0.0550 x (T_{F})

0.400T_{F} - 8.8 = -0.0550T_{F}

0.400T_{F} + 0.0550T_{F} =  8.8  

0.455T_{F} = 8.8

T_{F} = 19.34°C

Therefore, the final temperature of the mixture is 19.34°C

8 0
2 years ago
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