The asteroid 234 Ida has a mass of about 4 × 1016 kg and an average radius of about 16 km. What is the acceleration due to gravity on 234 Ida? Assume that the asteroid is spherical; use G = 6.67 × 10–11 Nm2/kg2.A. 1 cm/s2
B. 2 cm/s2
C. 5 cm/s2
D. 6 cm/s2

Answers

Answer 1

Answer:

The asteroid 234 Ida has a mass of about 4×1016 kg and an average radius of about 16 km. The acceleration due to gravity will be 1.04 cm/s². Hence, option A is correct.

What is the acceleration due to gravity?

The acceleration an object experiences as a result of gravitational force is known as acceleration due to gravity. M/s² is its SI unit. Its vector nature—which includes both magnitude and direction—makes it a quantity. The unit g stands for gravitational acceleration. At sea level, the standard value of g on the earth's surface is 9.8 m/s².

The formula for the acceleration due to gravity is g=GM/r².

According to the question, the given values are :

Mass, M = 4 × 1016 kg or

M = 4 × 10¹⁶.

Radius, r = 16 km or,

r = 16000 meter.

G = 6.67 × 10⁻¹¹ Nm²/kg²

g = (6.67 × 10⁻¹¹ ) (4 × 10¹⁶) / 16000²

g = 0.0104 m/s² or,

g = 1.04 cm/s².

Hence, the acceleration due to gravity will be 1.04 m/s²

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Answer 2

Answer:

Answer:

1 cm/s²

Explanation:

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The sun does not normally affect the tides to a noticeable degree. however, under certain circumstances the gravitational pull of the sun does affect the tides. under what two (2) conditions.the sun does not normally affect the tides to a noticeable degree. however, under certain circumstances the gravitational pull of the sun does affect the tides. under what two (2) conditions.

Answers

The answers are :
1) when the sun, moon, and earth are in a line only
2) when the gravitational forces of the Moon and the Sun are
perpendicular to one another with respect to the Earth.

Answer:

High and low tides are result of combined effect of gravitational pull of the sun and the moon. When the two align in a straight line, the range of tides is maximum. This happens on new moon and full moon day.

On the other hand, when the sun and the moon align at right angles, the effect of gravity is minimum and the range of the tides is minimum.

In an experiment to measure the acceleration of g due to gravity, two values, 9.96m/s (s is squared) and 9.72m/s (s is squared), are determined. Find (1) the percent difference of the measurements, (2) the percent error of each measurement, and (3) the percent error of their mean. (Accepted value:g=9.80m/s (s is squared))

Answers

Answer:

a)2.46 %

b)For 1 :101.52 %

For 2 : 99.08 %

c)100..4 %

Explanation:

Given that

g₁ = 9.96 m/s²

g₂ = 9.72 m/s²

The actual value of g = 9.8 m/s²

The difference Δ g = 9.96 -9.72 =0.24 m/s²

$The\ percentage\ difference=(0.24)/(9.72)* 100=2.46\ percentage\n$

For first one :

$Error\ in\ the\ percentage =(9.96)/(9.81)* 100 =101.52\ perncetage$

For second :

$Error\ in\ the\ percentage =(9.72)/(9.81)* 100 =99.08\ perncetage$

The mean g(mean )

$g(mean )=(9.96+9.72)/(2)\ m/s^2\ng(mean)=9.84\ m/s^2$

$The\ percentage=(9.84)/(9.8)* 100=100.40\ percentage$

a)2.46 %

b)For 1 :101.52 %

For 2 : 99.08 %

c)100..4 %

Final answer:

The percent difference between the two measurements is 2.44%. The percent error for the first measurement is 1.63%, and for the second measurement is 0.82%. The percent error of their mean is 0.41%.

Explanation:

In physics, the percent difference is calculated by subtracting the two values, taking the absolute value, dividing by the average of the two values, and then multiplying by 100. Therefore, the percent difference between the two measurements 9.96m/s² and 9.72m/s² is:

|(9.96-9.72)|/((9.96+9.72)/2)*100 = 2.44%

The percent error involves taking the absolute difference between the experimental value and the accepted value, divided by the accepted value, then multiplied by 100. So, the percent error for the two measurements with accepted value of 9.80m/s² are:

For 9.96m/s²: |(9.96-9.80)|/9.80*100 = 1.63%

For 9.72m/s²: |(9.72-9.8)|/9.8*100 = 0.82%

The percent error of the mean involves doing the above but using the mean of the experimental measurements:

|(Mean of measurements - Accepted value)|/Accepted value * 100 |(9.96+9.72)/2-9.8|/9.8*100 = 0.41%

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A car turns from a road into a parking lot and into an available parking space. The car’s initial velocity is 4 m/s [E 45° N]. The car’s velocity just before the driver decreases speed is 4 m/s [E 10° N]. The turn takes 3s. What's the average acceleration of the car during the turn? The answer should have directions with an angle.

Answers

Write the velocity vectors in component form.

• initial velocity:

v₁ = 4 m/s at 45º N of E

v₁ = (4 m/s) (cos(45º) i + sin(45º) j)

v₁ ≈ (2.83 m/s) i + (2.83 m/s) j

• final velocity:

v₂ = 4 m/s at 10º N of E

v₂ = (4 m/s) (cos(10º) i + sin(10º) j)

v₂ ≈ (3.94 m/s) i + (0.695 m/s) j

The average acceleration over this 3-second interval is then

a = (v₂ - v₁) / (3 s)

a ≈ (0.370 m/s²) + (-0.711 m/s²)

with magnitude

||a|| = √[(0.370 m/s²)² + (-0.711 m/s²)²] ≈ 0.802 m/s²

and direction θ such that

tan(θ) = (-0.711 m/s²) / (0.370 m/s²) ≈ -1.92

→ θ ≈ -62.5º

which corresponds to an angle of about 62.5º S of E, or 27.5º E of S. To use the notation in the question, you could say it's E 62.5º S or S 27.5º E.

Imagine two billiard balls on a pool table. Ball A has a mass of 2 kilograms and ballB has a mass of 3 kilograms. The initial velocity of ball A is 9 meters per second to
the right, and the initial velocity of the ball B is 6 meters per second to the left. The
final velocity of ball A is 9 meters per second to the left, while the final velocity of
ball B is 6 meters per second to the right.

1. Explain what happens to each ball after the collision. Why do you think this
occurs? Which of Newton’s laws does this represent?

Answers

This is an example of an elastic collision. The two objects collide and return to their original shapes and move separately. In such a collision, kinetic energy is conserved. I think we can agree that this represents Newton's third law by demonstrating conservation of momentum.

Answer:

Yes, the law of conservation of momentum is satisfied. The total momentum before the collision is 1.5 kg • m/s and the total momentum after the collision is 1.5 kg • m/s. The momentum before and after the collision is the same.

Explanation:

A uniform, 4.5 kg, square, solid wooden gate 2.0 m on each side hangs vertically from a frictionless pivot at the center of its upper edge. A 1.2 kg raven flying horizontally at 4.5 m/s flies into this door at its center and bounces back at 1.5 m/s in the opposite direction. What is the angular speed of the gate just after it is struck by the unfortunate raven?

Answers

Answer:

Explanation:

Mass of the gate, $m_1 = 4.5 kg$

Mass of the raven, $m_2 = 1.2 kg$

Initial speed of raven, $v_1 = 4.5 m/s$

Final speed of raven, $v_2 = - 1.5 m/s$

Moment of Inertia of the gate about the axis passing through one end:

$I = (1)/(3) m_1 a^2\nI = (1)/(3) *4.5 * 2^2\nI = 6 kg m^2$

Angular momentum of the gate, $L = I \omega$

$L = 5.33 \omega$

Using the law of conservation of angular momentum:

$m_2 v_f (a/2) + I\omega = m_2v_i (a/2)\nI\omega = m_2 (a/2)(v_i - v_f)\n$

Which of the following statements correctly compares the relationship between the earth, its atmosphere and radiation?1. The earth is cooled and its atmosphere is heated by solar radiation.

2. The earth is heated and its atmosphere is cooled by terrestrial radiation.

3. The earth is cooled and its atmosphere is heated by terrestrial radiation.

4. The earth is heated and its atmosphere is cooled by solar radiation.

Don't answer unless you know for sure. Thank you so much!

Answers

Answer: The option 4 is correct answer.

Explanation:

Terrestrial radiation is a long wave electromagnetic radiation. It originates from the earth and its atmosphere.

The sun emits a huge amount of energy. It travels across the space. The atmosphere is not directly heated by the solar radiation. It is heated by the terrestrial radiation that the planet itself emits.

When the land is heated then it emits radiation which heats up the atmosphere.

The earth is cooled and its atmosphere is heated by terrestrial radiation.

Therefore, the relationship between the earth, its atmosphere and radiation is correctly compared by statement 4.

The earth is cooled and its atmosphere is heated by terrestrial radiation.

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