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3 years ago

Find the following answer based on the image.

Physics

1 answer:

saul85 [17]3 years ago

5 0

<u>We are given:</u>

Mass of Neptune = 1.03 * 10²⁶ kg

Distance from the center of Neptune (r) = 2.27 * 10⁷

now, computing the value of the acceleration due to gravity (g)

<u>Finding g:</u>

We know the formula:

g = G(mass of planet) / (r)²

g = [6.67 * 10⁻¹¹ * 1.03*10²⁶] / (2.27*10⁷) [since G is 6.67*10⁻¹¹]

g = (6.87 * 10¹⁵) / (5.15 * 10¹⁴)

which can be rewritten as:

g = (6.87 * 10¹⁵ * 10⁻¹⁴) / 5.15

g = (6.87 * 10¹⁵⁻¹⁴) / 5.15

g = (6.87/5.15) * 10

g = 1.34 * 10

g = 13.4 m/s² <em>(approx)</em>

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Which of the following methods has led to the earliest discoveries of massive planets orbiting near their parent stars a. detect

Mrac [35]

Answer:

c. detecting the gravitational effect of an orbiting planet (The Wobble"") by looking for the Doppler shifts in the star's spectrum

Explanation:

In a solar system the mass of the star and planets affect each other's orbital movements. The center of gravity of a star and a planet is inside the star. This causes the star to be closer and farther from the Earth at different times. Due to this wobble the star appears to be red shifted when it is farther and blue shifted when it is closer.

When the mass of the planet is high, like a hot Jupiter it causes more wobble i.e., change in radial velocity. This makes it easier to detect the planet. The earliest hot Jupiter found by this method is the planet 51 Pegasi b.

3 0

3 years ago

A car is traveling in a race. The car went from the initial velocity of 35 m/s to the final velocity of 65 m/s in 5 seconds

melisa1 [442]

Answer:

Explanation:

65-35 = 30

30 divided by 5 = 6

The answer will be 6 m/s

6 0

2 years ago

Two horizontal metal plates, each 10.0 cm square, are aligned 1.00 cm apart with one above the other. They are given equal-magni

Ber [7]

Answer:

a) motion is PARABOLIC, b) positive particle is accelerated towards the negative plate, c) x = 6.19 10⁹ m

Explanation:

This exercise looks at the motion of a positively charged particle in an electric field.

a) Since the field is vertical the acceleration in this direction is

F = m a

the electric force is

F = q E

we substitute

q E = m a

a = qE / m

the mass of the particle is m = 2.00 10-16 kg

a = 1.6 10⁻¹⁹ 2.02 10³ / 2.00 10⁻¹⁶ kg

a = 1,616 m / s²

on the x-axis there are no relationships because there are no forces.

Since the particle has velocities in both axes, its motion is PARABOLIC,

b) the positive particle is accelerated towards the negative plate,

The field is descending, for which the event is down

c) where hit the particle on the x-axis

they indicate that the particle leaves the center of the negative plate, for which we will fix our reference system at this point.

Let's find the components of the initial velocity.

sin θ = v_{oy} / v

cos θ = v₀ₓ / v

v_{oy} = v₀ sin θ

v₀ₓ = v₀ cos θ

v_{oy) = 1.02 10⁵ sin 37 = 0.6139 10⁵ m / s

v₀ₓ = 1.02 10⁵ cos 37 = 0.8146 10⁵ m / s

Let's find the time it takes to hit the negative plate

y = y₀I + v_{oy} t + ½ a and t2

in this case the positions are y = y₀ = 0 and the accelerations

a = - 1,616m/s2,

we substitute

0 = 0 + v_{oy} t - ½ a_y t²

v_{oy}= ½ a_y t

t = 2v_{oy} / a_y

let's calculate

t = 2 0.6139 10⁵ / 1.616

t = 7.597 10⁴s

in this time the particle travels a horizontal distance

x = v₀ₓ t

x = 0.8145 10⁵ 7.597 10⁴

x = 6.19 10⁹ m

the particle falls off the plate

4 0

2 years ago

5. A family leaves for an evening walk around the neighborhood. They leave at 5 pm and return at 6 pm. They traveled a total of

vlabodo [156]

This question involves the concepts of average speed, distance, and time.

The average speed of the family is "2.5 mi/hr".

<h3>AVERAGE SPEED</h3>

The average speed is defined as the ratio of the total distance covered with the total time taken to cover it.

$v=\frac{s}{t}$

where,

s = distance covered = 2.5 miles
t = total time taken = 1 hour
v = average speed = ?

Therefore,

$v=\frac{2.5\ mi}{1\ hr}$

v = 2.5 mi/hr

Learn more about average speed here:

brainly.com/question/12322912

4 0

2 years ago

3.00 textbook rests on a frictionless, horizontal tabletop surface. A cord attached to the book passes over a pulley whose diame

sammy [17]

Answer:

a1 = 3.56 m/s²

Explanation:

We are given;

Mass of book on horizontal surface; m1 = 3 kg

Mass of hanging book; m2 = 4 kg

Diameter of pulley; D = 0.15 m

Radius of pulley; r = D/2 = 0.15/2 = 0.075 m

Change in displacement; Δx = Δy = 1 m

Time; t = 0.75

I've drawn a free body diagram to depict this question.

Since we want to find the tension of the cord on 3.00 kg book, it means we are looking for T1 as depicted in the FBD attached. T1 is calculated from taking moments about the x-axis to give;

ΣF_x = T1 = m1 × a1

a1 is acceleration and can be calculated from Newton's 2nd equation of motion.

s = ut + ½at²

our s is now Δx and a1 is a.

Thus;

Δx = ut + ½a1(t²)

u is initial velocity and equal to zero because the 3 kg book was at rest initially.

Thus, plugging in the relevant values;

1 = 0 + ½a1(0.75²)

Multiply through by 2;

2 = 0.75²a1

a1 = 2/0.75²

a1 = 3.56 m/s²

6 0

3 years ago