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3 years ago

What rate do things fall to Earth?

Physics

1 answer:

Dima020 [189]3 years ago

3 0

Answer:

9.8 meters per square second

Explanation:

Free Falling Object. the value of g is 9.8 meters per square second on the surface of the earth. The gravitational acceleration g decreases with the square of the distance from the center of the earth. But for many practical problems, we can assume this factor to be a constant.

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Two satellites are in circular orbits around the earth. The orbit for satellite A is at a height of 403 km above the earth’s sur

BARSIC [14]

Answer:

$v_A=7667m/s\\\\v_B=7487m/s$

Explanation:

The gravitational force exerted on the satellites is given by the Newton's Law of Universal Gravitation:

$F_g=\frac{GMm}{R^{2} }$

Where M is the mass of the earth, m is the mass of a satellite, R the radius of its orbit and G is the gravitational constant.

Also, we know that the centripetal force of an object describing a circular motion is given by:

$F_c=m\frac{v^{2}}{R}$

Where m is the mass of the object, v is its speed and R is its distance to the center of the circle.

Then, since the gravitational force is the centripetal force in this case, we can equalize the two expressions and solve for v:

$\frac{GMm}{R^2}=m\frac{v^2}{R}\\ \\\implies v=\sqrt{\frac{GM}{R}}$

Finally, we plug in the values for G (6.67*10^-11Nm^2/kg^2), M (5.97*10^24kg) and R for each satellite. Take in account that R is the radius of the orbit, not the distance to the planet's surface. So $R_A=6774km=6.774*10^6m$ and $R_B=7103km=7.103*10^6m$ (Since $R_{earth}=6371km$ ). Then, we get:

$v_A=\sqrt{\frac{(6.67*10^{-11}Nm^2/kg^2)(5.97*10^{24}kg)}{6.774*10^6m} }=7667m/s\\\\v_B=\sqrt{\frac{(6.67*10^{-11}Nm^2/kg^2)(5.97*10^{24}kg)}{7.103*10^6m} }=7487m/s$

In words, the orbital speed for satellite A is 7667m/s (a) and for satellite B is 7487m/s (b).

7 0

2 years ago

At a hot air balloon race, a person on the ground shots a ball of confetti from a cannon to start the race. One of the balloons

Sati [7]

Answer:

a) The balloon and the ball will meet after 1.43 s.

b) The ball will reach the balloon at 15.7 m above the ground.

Explanation:

The height of the confetti ball is given by the following equation:

y = y0 + v0 · t + 1/2 · g · t²

Where:

y = height of the ball at time t

y0 = initial height

v0 = initial velocity

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

The height of the ball is given by this equation:

y = y0 + v · t

Where v is the constant velocity.

When the ball and the ballon meet, both heights are equal. Let´s consider the ground as the origin of the frame of reference so that y0 = 0:

y balloon = y ball

y0 + v · t = y0 + v0 · t + 1/2 · g · t² (y0 = 0)

11 m/s · t = 18 m/s · t -1/2 · 9.8 m/s² · t²

0 = -4.9 m/s² · t² + 18 m/s · t - 11 m/s · t

0 = -4.9 m/s² · t² + 7 m/s · t

0 = t( -4.9 m/s² · t + 7 m/s)

t = 0 and

0 = -4.9 m/s² · t + 7 m/s

-7 m/s / - 4.9 m/s² = t

t = 1.43 s

They will meet after 1.43 s

b) Now let´s calculate the height of the balloon after 1.43 s

y = v · t

y = 11 m/s · 1.43 s = 15.7 m

The ball will reach the balloon at 15.7 m above the ground.

7 0

2 years ago

a crowbar of 2 meter is used to lift an object of 800N if the effort arm is 160cm , calculste the effort applied

Vitek1552 [10]

Answer:

200 N

Explanation:

The crowbar is 2 meter, or 200 cm. The effort arm is 160 cm, so the moment arm of the object is 40 cm.

(800 N) (40 cm) = F (160 cm)

F = 200 N

5 0

3 years ago

AGREE OR DISAGREE: All light can be seen.

Tema [17]

Answer:

Disagree

Explanation:

We cannot see infared lights.

3 0

2 years ago

Scientists want to place a telescope on the moon to improve their view of distant planets. The telescope weighs 200 pounds on Ea

Maru [420]

Answer:

33,02 lb

Explanation:

g_m ≈ 1,62 m/s2

g ≈ 9,81 m/s2

m = 200 lb

m_m = m * g_m / m = 200 * 1,62 / 9,81 = 33,02 lb

7 0

3 years ago

Read 2 more answers