Explanation:
Rate X Time = Distance
Distance / Time = Rate
225 km / 3.35 hr = 67.2 km/hr
A satellite weighing 5,400 kg is launched into orbit 3.6400 x 107 m above the center of the earth.
The mass of Earth is 6.0 × 1024 kg. The gravitational constant is 6.673 × 10–11 N•m2/kg2.
The gravitational force of Earth on the satellite is ___
Group of answer choices
9.1 x 10^4
1.6 x 10^3
2.1 x 10^6
Answer:
[tex]\tt F=1.63*10^3 N[/tex]
Explanation:
Gravitational force is defined as the force of attraction between two objects with mass. It is a fundamental force of nature, and it is what keeps us on the ground and what keeps the planets in orbit around the Sun.
The gravitational force between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers
For the Question:
We can use the following formula to calculate the gravitational force between the Earth and the satellite:
[tex]\boxed{\tt F =\frac{ G * M * m }{ r^2}}[/tex]
Where:
F is the gravitational force
G is the gravitational constant[tex]\tt (6.673 * 10^{-11} Nm^2/kg^2)[/tex]
M is the mass of the Earth [tex]\tt (6.0 * 10^24 kg)[/tex]
m is the mass of the satellite[tex]\tt (5,400 kg)[/tex]
r is the distance between the satellite and the center of the Earth [tex]\tt (3.6400 * 10^7 m)[/tex]
Plugging in these values, we get the following:
[tex]\tt F = \frac{6.673 * 10^{-11} * 6.0 * 10^{24}* 5,400 }{ (3.6400 * 10^7 )^2}[/tex]
[tex]\tt F=1.63*10^3 N[/tex]
Therefore, answer is [tex]\tt F=1.63*10^3 N[/tex]
The temperature of a aluminum bar rises by 10.0°C when it absorbs 4.73 kJ of energy by heat. The mass of the bar is 525 g. Determine the specific heat of aluminum from these data. Answer is in kJ/kg · °C.
Answer:
Certainly! We can use the formula:
q = mcΔT
where q is the amount of heat absorbed, m is the mass of the aluminum bar, c is the specific heat capacity of aluminum, and ΔT is the change in temperature.
Substituting the given values, we get:
4.73 kJ = (0.525 kg) x c x (10.0°C)
Solving for c, we get:
c = 0.901 kJ/kg · °C
Therefore, the specific heat of aluminum is 0.901 kJ/kg · °C.
Explanation: