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3 years ago

If Earth were twice as far from the sun, the force of gravity attracting the Earth to the Sun would be

2 answers:

6 0

If Earth was twice as far from the sun, the force of gravity attracting the Earth to the sun would be only one-quarter as strong. The correct answer will be C.

sveticcg [70]3 years ago

3 0

It would be half as strong

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A ball having a mass of 0.20 kilograms is placed at a height of 3.25 meters. If it is dropped from this height, what will be the

asambeis [7]

EC_1 + EP_1 = EC2 + EP_2

EC_2 = 0

EC_2 = EP_1 - EP_2

EC_2 = mg(H_1 - H_2) = 0.20 kg * 9.8 m/s^2 * (3.25 m - 1.5m) = 3.43 J

7 0

2 years ago

Read 2 more answers

A tennis ball is tossed up off a building with a velocity of 22m/s. It takes 6.4s to reach the ground. How high is the building

Amanda [17]

The height of the tennis ball,relative to the ground is H=h max+h-->h max-the maximum height that the tennis ball reaches relative to the roof of the building; h-the height of the building;h max =v0^2/2g=24,2m(g=10m/s^2).H=gt^2/2=>24,2+h=gt^2/2=>h=gt^2/2-24,2=180,6m

4 0

3 years ago

It has been suggested that rotating cylinders about 10 mi long and 5.9 mi in diameter be placed in space and used as colonies. T

tekilochka [14]

Answer:

ω = 0.05 rad/s

Explanation:

We consider the centripetal force acting as the weight force on the surface of the cylinder. Therefore,

$Centripetal Force = Weight\\\frac{mv^{2}}{r} = mg\\\\here,\\v = linear\ speed = r\omega \\therefore,\\\frac{(r\omega)^{2}}{r} = g\\\\\omega^{2} = \frac{g}{r}\\\\\omega = \sqrt{\frac{g}{r}}\\$

where,

ω = angular velocity of cylinder = ?

g = required acceleration = 9.8 m/s²

r = radius of cylinder = diameter/2 = 5.9 mi/2 = 2.95 mi = 4023.36 m

Therefore,

$\omega = \sqrt{\frac{9.8\ m/s^{2}}{4023.36\ m}}\\\\$

7 0

3 years ago

How did harlow shapley conclude that the sun was not in the center of the milky way galaxy?

timurjin [86]

We learned that We are in the disk of the Galaxy, about 5/8 of the way from the center.

<h3>What is the work of Harlow Shapley?</h3>

Shapley, who was headquartered in Boulder, Colorado, used Cepheid variable stars to estimate the size of the Milky Way Galaxy and its position relative to the Sun. In 1953, he published his "liquid water belt" theory, today known as the concept of a livable zone.

There are many stars, grains of dust, and gas in the Milky Way. It is known as a spiral galaxy because, from the top or bottom, it would appear to be whirling like a pinwheel. About 25,000 light-years from the galaxy's nucleus, the Sun is situated on one of the spiral arms.

Approximately 5/8 of the way from the galaxy's nucleus, we are in the disc. William Herschel believed that the Sun and Earth were about in the middle of the vast cluster of stars known as the Milky Way.

To learn more about Harlow Shapley's original estimate go to - brainly.com/question/28145909

#SPJ4

3 0

1 year ago

A T-shirt cannon can shoot a 0.085 kg T-shirt at nearly 30 m/s. The T-shirt cannon has a mass of 33 kg. If the initial net momen

IgorLugansk [536]

Answer:

Approximately $0.077\; {\rm m\cdot s^{-1}}$ (assuming that external forces on the cannon are negligible.)

Explanation:

If an object of mass $m$ is moving at a velocity of $v$ , the momentum $p$ of that object would be $p = m\, v$ .

Momentum of the t-shirt:

$\begin{aligned} p(\text{t-shirt}) &= m(\text{t-shirt}) \, v(\text{t-shirt}) \\ &= 0.085\; {\rm kg} \times 30\; {\rm m \cdot s^{-1}} \\ &= 2.55 \; {\rm kg \cdot m \cdot s^{-1}} \end{aligned}$ .

If there is no external force (gravity, friction, etc.) on this cannon, the total momentum of this system should be conserved. In other words, if $p(\text{cannon})$ denote the momentum of this cannon:

$p(\text{t-shirt}) + p(\text{cannon}) = 0$ .

$p(\text{cannon}) = -p(\text{t-shirt}) = -2.55\; {\rm kg \cdot m \cdot s^{-1}}$ .

Rewrite $p = m\, v$ to obtain $v = (p / m)$ . Since the mass of this cannon is $m(\text{cannon}) = 33\; {\rm kg}$ , the velocity of this cannon would be:

$\begin{aligned} v(\text{cannon}) &= \frac{p(\text{cannon})}{m(\text{cannon})} \\ &= \frac{-2.55\; {\rm kg \cdot m \cdot s^{-1}}}{33\; {\rm kg}} \\ &\approx 0.077\; {\rm m \cdot s^{-1}}\end{aligned}$ .

8 0

1 year ago