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stepladder
1 month ago
8

According to Newton's Law of Universal Gravitation, which of the following would cause the attractive force between a planet and

the Sun to be doubled?
A. Doubling the mass of the planet
B. Quadrupling the distance from the sun
C. Doubling the distance from the sun
D. Squaring the mass of the planet oxford
Physics
1 answer:
Maru [3.3K]1 month ago
4 0
The correct options could be either a or c.
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A student with a mass of 66.0 kg climbs a staircase in 44.0 s. If the distance between the base and the top of the staircase is
serg [3582]

power = 205.8 \: watt \\ solution \\ mass = 66 \: kg \\ time = 44 \: sec \\ distance = 14 \\ now \\ power = \frac{w}{t} \\ \: \: \: \: \: \: \: \: = \frac{f \times d}{t} \\ \: \: \: \: \: = \frac{m \times g \times d}{t} \\ \: \: \: \: \: \: = \frac{66 \times 9.8 \times 14}{44} \\ \: \: \: \: = \frac{9055.2}{44} \\ \: \: \: \: \: = 205.8 \: watt \\ hope \: it \: helps

4 0
1 month ago
A projectile is fired from ground level with a speed of 150 m/s at an angle 30.° above the horizontal on an airless planet where
Yuliya22 [3333]

Answer:

130 m/s (to two significant figures)

Explanation:

In projectile motion, the launching velocity and launch angle help to determine both the horizontal and vertical velocity components.

u represents the initial projectile velocity = 150 m/s

uₓ = u cos θ = 150 cos 30° = 129.9 m/s

uᵧ = u sin θ = 150 sin 30° = 75.0 m/s

A projectile's motion can be viewed as made up of independent vertical and horizontal elements.

The vertical motion is affected by gravitational acceleration (which pulls down on the projectile), altering the vertical velocity component due to this acting force.

Conversely, there is no acting force in the horizontal direction, which means the horizontal component maintains a steady velocity throughout the projectile's flight.

Thus, at t = 4 s, the horizontal component of the projectile's speed remains equal to the initial horizontal velocity component.

At t = 4 s, the horizontal component of velocity is uₓ = u cos θ = 150 cos 30° = 129.9 m/s ≈ 130 m/s

6 0
1 month ago
Read 2 more answers
How, if at all, would the equations written in Parts C and E change if the projectile was thrown from the cliff at an angle abov
Keith_Richards [3271]

Answer:

x = v₀ cos θ   t,   y = y₀ + v₀ sin θ t - ½ g t2

Explanation:

This pertains to a projectile motion scenario. Here, we will express the equations for both the x and y dimensions.

Now, we will apply trigonometry to determine the initial velocity components.

              sin θ = v_{oy} / v₀

              cos θ = v₀ₓ / v₀

              v_{y} = v_{oy} sin θ

              v₀ₓ = v₀ cos θ

Next, let's formulate the equations of motion.

X axis

         x = v₀ₓ t

         x = v₀ cos θ   t

        vₓ = v₀ cos θ

Y axis

        y = y₀ + v_{oy} t - ½ g t2

        y = y₀ + v₀ sin θ t - ½ g t2

        v_{y} = v₀ - g t

       v_{y} = v₀  sin θ - gt

        v_{y}^{2} = v_{oy}^2 sin² θ - 2 g y

It is evident that the major distinction lies in the fact that in an inclined launch compared to a horizontal one, the velocity comprises different components

7 0
1 month ago
In a fluorescent tube of diameter 3 cm, 3 1018 electrons and 0.75 1018 positive ions (with a charge of e) flow through a cross-s
Softa [3030]

Answer:

The current flowing through the tube is 0.601 A

Explanation:

Given data;

the diameter of the fluorescent tube is d = 3 cm

the incoming negative charge in the tube is -e = 3 x 10¹⁸ electrons/second

the outgoing positive charge equals +e = 0.75 x 10¹⁸ electrons/ second

The current within the fluorescent tube results from both positive and negative charges which contribute to maintaining electrical neutrality in the conductor (fluorescent tube).

Q = It

I = Q/t

where;

I signifies current in Amperes (A)

Q represents charge measured in Coulombs (C)

t denotes time which is in seconds (s)

1 electron (e) accounts for 1.602 x 10⁻¹⁹ C

Thus, 3 x 10¹⁸ e/s can be computed as

= (3 x 10¹⁸ e/s  x 1.602 x 10⁻¹⁹ C) / 1e

= 0.4806 C/s

This reflects the negative charge per second (Q/t) = 0.4806 C/s

Meanwhile, the positive charge per second amounts to

(0.75 x 10¹⁸ e/s  x 1.602 x 10⁻¹⁹ C) / 1e

Thus, the positive charge per second is equal to 0.12015 C/s

The total charge per second in the tube is then obtained by summing both charges: Q / t = (0.4806 C/s + 0.12015 C/s)

                                                                I = 0.601 A

Therefore, the current within the tube computes to 0.601 A

7 0
1 month ago
Calculate the weight of a 4.5 kg rabbit.
Yuliya22 [3333]
The answer is 13900589.
3 0
1 month ago
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