The total thermal energy needed to raise the temperature of this ice cube to 0.0 °C and completely melt the ice cube is 5848 J

During a solar eclipse, the Moon, Earth, and Sun all lie on the same line, with the Moon between the Earth and the Sun.

lord [1]

Answer:

(a) $F_{sm} = 4.327\times 10^{20}\ N$

(b) $F_{em} = 1.983\times 10^{20}\ N$

(c) $F_{se} = 3.521\times 10^{20}\ N$

Solution:

As per the question:

Mass of Earth, $M_{e} = 5.972\times 10^{24}\ kg$

Mass of Moon, $M_{m} = 7.34\times 10^{22}\ kg$

Mass of Sun, $M_{s} = 1.989\times 10^{30}\ kg$

Distance between the earth and the moon, $R_{em} = 3.84\times 10^{8}\ m$

Distance between the earth and the sun, $R_{es} = 1.5\times 10^{11}\ m$

Distance between the sun and the moon, $R_{sm} = 1.5\times 10^{11}\ m$

Now,

We know that the gravitational force between two bodies of mass m and m' separated by a distance 'r' is given y:

$F_{G} = \frac{Gmm'_{2}}{r^{2}}$ (1)

Now,

(a) The force exerted by the Sun on the Moon is given by eqn (1):

$F_{sm} = \frac{GM_{s}M_{m}}{R_{sm}^{2}}$

$F_{sm} = \frac{6.67\times 10^{- 11}\times 1.989\times 10^{30}\times 7.34\times 10^{22}}{(1.5\times 10^{11})^{2}}$

$F_{sm} = 4.327\times 10^{20}\ N$

(b) The force exerted by the Earth on the Moon is given by eqn (1):

$F_{em} = \frac{GM_{s}M_{m}}{R_{em}^{2}}$

$F_{em} = \frac{6.67\times 10^{- 11}\times 5.972\times 10^{24}\times 7.34\times 10^{22}}{(3.84\times 10^{8})^{2}}$

$F_{em} = 1.983\times 10^{20}\ N$

(c) The force exerted by the Sun on the Earth is given by eqn (1):

$F_{se} = \frac{GM_{s}M_{m}}{R_{es}^{2}}$

$F_{se} = \frac{6.67\times 10^{- 11}\times 1.989\times 10^{30}\times 5.972\times 10^{24}}{((1.5\times 10^{11}))^{2}}$

$F_{se} = 3.521\times 10^{20}\ N$

7 0

3 years ago

The upper body primal movement patterns are

kramer

Answer:

Bend to extend. Bending with a hips back movement, back straight, feet flat and forward. ...

Squat. This is a hips down motion. ...

Lunge. This is a long, linear stride, lowering your back knee to just above the ground, with a completely upright torso. ...

Rotate. ...

Push. ...

Pull. ...

Gait.

8 0

3 years ago

How much magnification is needed to see a nanometer?.

ArbitrLikvidat [17]

20,000,000

To see things at a nanometer, which is a trillionth of a meter, you would need to increase magnification nearly 20,000,000 times those are ionizing atoms

6 0

2 years ago

On the Em spectrum visible light between

Nesterboy [21]

Hello, Thank you for posting your question here on brainly! feel free to ask as many as needed!

 the portion of the electromagnetic spectrum that is visible to the human eye. Electromagnetic radiation in this range of wavelengths is called visible light or simply light. A typical human eye will respond to wavelengths from about 390 to 700 nm.
I hope i helped you.

7 0

3 years ago

A 0.250-kg wooden rod is 1.35 m long and pivots at one end. It is held horizontally and then released. Part A What is the angula

Ivan

We assume that the rod's weight is evenly distributed, making its center of gravity 0.675 m from the end.
First, we calculate the moment present on the rod:
τ = F*d
τ = m*g*d
τ = 0.25 * 9.81 * 0.675
τ = 1.66

Next, in the case of rotational motion, Newton's second law is:
τ = Iα, where I is moment of inertia and α is the angular acceleration
The moment of inertia for a rod is:
I = (ML²)/12
I = (0.25*1.35²)/12
I = 0.038

Now, we use the formula given by Newton's law:
α = τ / I
α = 1.66 / 0.038
α = 43.7 rad/s²

The angular acceleration is 43.7 radians per seconds squared.

5 0

3 years ago