Answer:
See the explanation below.
Explanation:
We know that density is defined as the relationship between mass and volume.
where:
m = mass [kg]
V = volume [m³]
Therefore Ro is given in:
Indicate whether the statement is true or false. The amount of matter in an object is its weight.
2 answers:
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Answer:
The answer to the question is;
The total potential energy of the mass on the spring when the mass is at either endpoint of its motion is 5.0255 Joules.
Explanation:
To answer the question, we note that the maximum speed is 2.30 m/s and the mass is 1.90 kg
Therefore the maximum kinetic energy of motion is given by
Kinetic Energy, KE =
Where,
m = Attached vibrating mass = 1.90 kg
v = velocity of the string = 2.3 m/s
Therefore Kinetic Energy, KE = ×1.9×2.3² = 5.0255 J
From the law of conservation of energy, we have the kinetic energy, during the cause of the vibration is converted to potential energy when the mass is at either endpoint of its motion
Therefore Potential Energy PE at end point = Kinetic Energy, KE at the middle of the motion
That is the total potential energy of the mass on the spring when the mass is at either endpoint of its motion is equal to the maximum kinetic energy.
Total PE = Maximum KE = 5.0255 J.
Answer:
We conclude that the kinetic energy of a 1.75 kg ball traveling at a speed of 54 m/s is 2551.5 J.
Explanation:
Given
- Mass m = 1.75 kg
- Velocity v = 54 m/s
To determine
Kinetic Energy (K.E) = ?
We know that a body can possess energy due to its movement — Kinetic Energy.
Kinetic Energy (K.E) can be determined using the formula
where
- m is the mass (kg)
- v is the velocity (m/s)
- K.E is the Kinetic Energy (J)
now substituting m = 1.75, and v = 54 in the formula
J
Therefore, the kinetic energy of a 1.75 kg ball traveling at a speed of 54 m/s is 2551.5 J.
Answer:
Equilibrium temperature will be
Explanation:
We have given weight of the lead m = 2.61 gram
Let the final temperature is T
Specific heat of the lead c = 0.128
Initial temperature of the lead = 11°C
So heat gain by the lead = 2.61×0.128×(T-11°C)
Mass of the water m = 7.67 gram
Specific heat = 4.184
Temperature of the water = 52.6°C
So heat lost by water = 7.67×4.184×(T-52.6)
We know that heat lost = heat gained
So