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3 years ago

The orbit of mars is never far from the ecliptic. why?

1 answer:

7 0

The plane of the ecliptic is an imaginary plane that passes from the Sun through the Earth. So Earth has an inclination of zero degrees from the plane.

All planets of our Solar System do not lie in this imaginary plane.

Mars is less than 2 degrees off which is why it appears to be centralized. Jupiter is also less than 2 degrees off. But some such as the Plutoid or Dwarf Planet Pluto can be off as much as 17 degrees.

Hope this gives you a bit understanding!

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A box is pushed 40 m by a mover. The amount of work done was 2,240 j. How much force was exerted on the box

Georgia [21]

The force exerted on the box is 56 N

Explanation:

The work done by a force on an object 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 of the displacement

For the box in this problem, we have:

W = 2240 J is the work done

d = 40 m is the displacement of the box

Assuming that the force is parallel to the displacement, $\theta=0$

Solving the equation for F, we find the force exerted on the box:

$F=\frac{W}{d cos \theta}=\frac{2240}{(40)(cos 0)}=56 N$

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

3 0

3 years ago

Calculate the net force on the right charge due to the other two. Enter a positive value if the force is directed to the right a

lbvjy [14]

Answer:

Answer:

A. - 0.017N. It acts to the left.

B. - 0.043N. It acts to the left.

C. 0.060N. It acts to the right.

Explanation:

A. For the +65μC charge, we consider it to be the origin. Hence, the two other charges are on the +x axis.

The net coulombs force on the charge is

F = [KQ(1)Q(2)]/(r^2) + [KQ(1)Q(3)]/(r^2)

Where K = Coloumbs constant =

Q(1) = charge on the leftmost side.

Q(2) = charge in the middle.

Q(3) = charge on the rightmost side.

F = [(8.988 × 10^9)×(65×10^-6)×(48×10^-6)]/(40^2) + [(8.988 × 10^9)×(-95×10^-6)×(65×10^-6)]/(40^2)

F = 0.01753 - 0.03469

F = -0.017N

It has a negative sign, hence, it acts to the left.

B. For the +48μC charge, we consider it to be the origin. Hence, the leftmost charge is on the - x axis and the rightmost charge is on the +x axis.

The net coulombs force on the charge is

F = [-KQ(1)Q(3)]/(r^2) + [KQ(2)Q(3)]/(r^2)

F = [-(8.988×10^9)×(65×10^-6)×(48×10^-6)]/(40^2) + [(8.988 × 10^9)×(48×10^-6)×(-95×10^-6)]/(40^2)

F = -0.017 - 0.02562

F = - 0.043N

It has a negative sign, hence, it acts to the left.

C. For the -95μC charge, we consider it to be the origin. Hence, the two other charges are on the - x axis.

The net coulombs force on the charge is

F = [-KQ(1)Q(3)]/(r^2) - [KQ(2)Q(3)]/(r^2)

F = [-(8.988×10^9)×(65×10^-6)×(-95×10^-6)]/(40^2) - [(8.988 × 10^9)×(48×10^-6)×(-95×10^-6)]/(40^2)

F = +0.03469 + 0.02562

F = +0.060N

It has a positive sign, hence, it acts to the right.

Read more on Brainly.com - brainly.com/question/14592748#readmore

Explanation:

5 0

4 years ago

Question 4 (18 marks) (a) During a Physics Lab experiment, 1 st year SFY students analyzed the behavior of capacitors by connect

Nataly_w [17]

Answer:

1.) 274.5v

2.) 206.8v

Explanation:

1.) Given that In one part of the lab activities, students connected a 2.50 µF capacitor to a 746 V power source, whilst connected a second 6.80 µF capacitor to a 562 V source.

The potential difference and charge across EACH capacitor will be

V = Voe

Where Vo = initial voltage

e = natural logarithm = 2.718

For the first capacitor 2.50 µF,

V = Vo × 2.718

746 = Vo × 2.718

Vo = 746/2.718

Vo = 274.5v

To calculate the charge, use the below formula.

Q = CV

Q = 2.5 × 10^-6 × 274.5

Q = 6.86 × 10^-4 C

For the second capacitor 6.80 µF

V = Voe

562 = Vo × 2.718

Vo = 562/2.718

Vo = 206.77v

The charge on it will be

Q = CV

Q = 6.8 × 10^-6 × 206.77

Q = 1.41 × 10^-3 C

B.) Using the formula V = Voe again

165 = Vo × 2.718

Vo = 165 /2.718

Vo = 60.71v

Q = C × 60.71

Q = C

4 0

3 years ago

Help plzzzzzzzzzzzzzzzz!

aksik [14]

concave <span>ray diagrams were constructed in order to determine the general location, size, orientation, and type of image formed by concave mirrors. Perhaps you noticed that there is a definite relationship between the image characteristics and the location where an object placed in front of a concave mirror. but, convex</span><span>ray diagrams were constructed in order to determine the location, size, orientation, and type of image formed by concave mirrors. The ray diagram constructed earlier for a convex mirror revealed that the image of the object was virtual, upright, reduced in size and located behind the mirror. </span>

3 0

3 years ago

A person drives 70 km/h in 1 hour to the east, then 80 km/h for another hour to the east. What

andre [41]

Answer: The average velocity is 150 km/h

Explanation: 70+80=150

6 0

3 years ago