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3 years ago

Which of the following causes objects to orbit the Sun?

Physics

2 answers:

algol [13]3 years ago

7 0

B. The suns gravity

hope this helps

Gekata [30.6K]3 years ago

4 0

In order for an object to be in orbit around a larger object. The larger object my exert its gravity on the smaller object. Like a planet around the sun, or a satellite around the Earth or Moon. This answer is B

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Does an eclipse occur every time the moon is in the new or full moon phase?

harina [27]

Answer:

solar eclipse mostly occurs when the moon gets between Earth and the sun, and the moon casts a shadow over Earth. A solar eclipse can only take place at the phase of new moon, when the moon passes directly between the sun and Earth and its shadows fall upon Earth's surface.

7 0

3 years ago

Read 2 more answers

Two electrons exert a force of repulsion of 1.2 N on each other. How far apart are they? The elementary charge is 1.602 × 10−19

aleksandr82 [10.1K]

Answer:

The answer to your question is distance between these electrons

= 1.386 x 10⁻¹⁴ m

Explanation:

Data

Force = F = 1.2 N

distance = d = ?

charge = q₁ = q₂ = 1.602 x 10⁻¹⁹ C

K = 8.987 x 10⁹ Nm²/C²

Formula

-To solve this problem use the Coulomb's equation

F = kq₁q₂ / r²

-Solve for r²

r² = kq₁q₂ / F

-Substitution

r² = (8.987 x 10⁹)(1.602 x 10⁻¹⁹)(1.602 x 10⁻¹⁹) / 1.2

- Simplification

r² = 2.306 x 10⁻²⁸ / 1.2

r² = 1.922 x 10⁻²⁸

-Result

r = 1.386 x 10⁻¹⁴ m

7 0

4 years ago

A luggage handler pulls a suitcase of mass 19.6 kg up a ramp inclined at an angle 24.0 ∘ above the horizontal by a force F⃗ of m

Dvinal [7]

(a) 638.4 J

The work done by a force is given by

$W=Fd cos \theta$

where

F is the magnitude of the force

d is the displacement of the object

$\theta$ is the angle between the direction of the force and the displacement

Here we want to calculate the work done by the force F, of magnitude

F = 152 N

The displacement of the suitcase is

d = 4.20 m along the ramp

And the force is parallel to the displacement, so $\theta=0^{\circ}$ . Therefore, the work done by this force is

$W_F=(152)(4.2)(cos 0)=638.4 J$

b) -328.2 J

The magnitude of the gravitational force is

W = mg

where

m = 19.6 kg is the mass of the suitcase

$g=9.8 m/s^2$ is the acceleration of gravity

Substituting,

$W=(19.6)(9.8)=192.1 N$

Again, the displacement is

d = 4.20 m

The gravitational force acts vertically downward, so the angle between the displacement and the force is

$\theta= 90^{\circ} - \alpha = 90+24=114^{\circ}$

Where $\alpha = 24^{\circ}$ is the angle between the incline and the horizontal.

Therefore, the work done by gravity is

$W_g=(192.1)(4.20)(cos 114^{\circ})=-328.2 J$

c) 0

The magnitude of the normal force is equal to the component of the weight perpendicular to the ramp, therefore:

$R=mg cos \alpha$

And substituting

m = 19.6 kg

g = 9.8 m/s^2

$\alpha=24^{\circ}$

We find

$R=(19.6)(9.8)(cos 24)=175.5 N$

Now: the angle between the direction of the normal force and the displacement of the suitcase is 90 degrees:

$\theta=90^{\circ}$

Therefore, the work done by the normal force is

$W_R=R d cos \theta =(175.4)(4.20)(cos 90)=0$

d) -194.5 J

The magnitude of the force of friction is

$F_f = \mu R$

where

$\mu = 0.264$ is the coefficient of kinetic friction

R = 175.5 N is the normal force

Substituting,

$F_f = (0.264)(175.5)=46.3 N$

The displacement is still

d = 4.20 m

And the friction force points down along the slope, so the angle between the friction and the displacement is

$\theta=180^{\circ}$

Therefore, the work done by friction is

$W_f = F_f d cos \theta =(46.3)(4.20)(cos 180)=-194.5 J$

e) 115.7 J

The total work done on the suitcase is simply equal to the sum of the work done by each force,therefore:

$W=W_F + W_g + W_R +W_f = 638.4 +(-328.2)+0+(-194.5)=115.7 J$

f) 3.3 m/s

First of all, we have to find the work done by each force on the suitcase while it has travelled a distance of

d = 3.80 m

Using the same procedure as in part a-d, we find:

$W_F=(152)(3.80)(cos 0)=577.6 J$

$W_g=(192.1)(3.80)(cos 114^{\circ})=-296.9 J$

$W_R=(175.4)(3.80)(cos 90)=0$

$W_f =(46.3)(3.80)(cos 180)=-175.9 J$

So the total work done is

$W=577.6+(-296.9)+0+(-175.9)=104.8 J$

Now we can use the work-energy theorem to find the final speed of the suitcase: in fact, the total work done is equal to the gain in kinetic energy of the suitcase, therefore

$W=\Delta K = K_f - K_i\\W=\frac{1}{2}mv^2\\v=\sqrt{\frac{2W}{m}}=\sqrt{\frac{2(104.8)}{19.6}}=3.3 m/s$

6 0

3 years ago

The sources of marine debris and explain why it is a problem with global origins

Lera25 [3.4K]

Marine debris is any trash or litter that ends up in a marine environment. This harms wildlife through entanglement, ingestion, and disturbs the habitat of many species of animals. This becomes a global problem when it is a danger to human health. Nails, syringes, and glass can cause harm to people on the beach. On top of that, trash in the waterway increases the amount of chemicals destroying our water quality.
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3 years ago

1. Write the goal of the lab or the question you tried to answer.

oksano4ka [1.4K]

Answer:The goal of the lab was to collect and transfer data including the tennis ball, football, and other objects.

Explanation: Edgenuity kid

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3 years ago

Read 2 more answers