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3 years ago

The force of gravity on Mercury is less than the Earth because:

2 answers:

8 0

The answer is B.

Earth's mass is <span>5.972 × 10^24 kg
Mercury's mass is </span><span>3.285 × 10^23 kg

Hope i helped</span>

Vinvika [58]3 years ago

7 0

Mercury has less mass than earth. So the answer is B

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The Pasig Revolving Tower has an estimated height of 45.23 meters. How high is it in yard?

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3 years ago

Suppose the initial kinetic energy and final potential energy in an experiment are both zero. What can you conclude? A. The fina

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Answer:

C. The final kinetic energy is equal to the initial potential energy.

Explanation:

Based on the Principle of energy conservation:

Sum of the Initial Energy = Sum of the Final Energy

Initial Kinetic Energy + Initial Potential Energy = Final Kinetic Energy + Final Potential Energy..........(1)

Since according to the question:

Initial Kinetic Energy = 0

Final Potential Energy = 0

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2 years ago

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3 years ago

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3 years ago

The addition of 9.0×105 J is required to convert a block of ice at -10 ∘C to water at 11 ∘C. i need help this is due in less tha

Law Incorporation [45]

The mass of the block of ice is 2.24 kg

Explanation:

The amount of heat needed for the whole process consists of three different amounts of heat:

$Q_1$ : the amount of heat needed to raise the temperature of the block of ice from $-10^{\circ}C$ to $0^{\circ}C$

$Q_2$ : the amount of heat needed to melt the block of ice at melting point

$Q_3$ : the amount of heat needed to raise the temperature of the water from $0^{\circ}C$ to $11^{\circ}C$

The total amount of heat needed can be written as

$Q=Q_1+Q_2+Q_3=mC_i\Delta T_1 + m\lambda_f + mC_w\Delta T_3$

where we have:

$Q=9.0 \cdot 10^5 J$ (total amount of heat required)

m is the mass of the block of ice

$C_i = 2108 J/kg^{\circ}C$ is the specific heat of ice

$\lambda_f=3.34\cdot 10^5 J/kg$ is the latent heat of fusion of ice

$C_w=4186 J/kg^{\circ}C$ is the specific heat capacity of water

$\Delta T_1 = 0-(-10)=10^{\circ}C$ is the change in temperature in the 1st process

$\Delta T_3 = 11-0=11^{\circ}C$ is the change in temperature in the 3rd process

Solving the equation for m, we find the mass of the block of ice:

$m=\frac{Q}{C_i\Delta T_1 + \lambda_f+C_w\Delta T_3}=\frac{9.0\cdot 10^5}{(2108)(10)+3.34\cdot 10^5+(4186)(11)}=2.24 kg$

Learn more about specific heat capacity:

brainly.com/question/3032746

brainly.com/question/4759369

#LearnwithBrainly

8 0

3 years ago