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Zanzabum
2 years ago
8

Describe kepler's contribution to the current structure of the solar system.

Physics
1 answer:
notka56 [123]2 years ago
7 0
Kepler used mathematics to determine the paths of planets around the sun
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Scientists believe the universe follows a set of “rules” known as?
g100num [7]
I believe the answer is Natural Laws
3 0
3 years ago
A whale swims due east for a distance of 6.9 km, turns around and goes due west for 1.8 km, and finally turns around again and h
xeze [42]

The question is incomplete. Here is the complete question:

A whale swims due east for a  distance of 6.9 km, turns around and  goes due west for 1.8 km and finally  turns around again and heads 3.7 km  due east.  (a) What is the total distance  traveled by the whale? (b) What are the  magnitude and direction of the displacement of the whale?

Answer:

(a) Distance = 12.4 km

(b) Displacement = 8.8 km due east

Explanation:

Consider east direction as positive and west direction as negative.

Given:

The motion is along the east-west line.

The whale first swims 6.9 km due east, then 1.8 km due west and again 3.7 km due east.

(a)

Distance traveled by the whale is equal to the sum of the lengths of all the distances traveled. Therefore,

Distance traveled = 6.9 km + 1.8 km + 3.7 km = 12.4 km

Therefore, the distance traveled by the whale is 12.4 km.

(b)

Displacement of the whale is given by considering the sign of each of the individual displacements. Therefore,

Displacement of the whale = (+6.9 km) + (-1.8 km) + (+3.7 km)

Displacement of the whale = 6.9 km - 1.8 km + 3.7 km = 8.8 km

The answer is positive. So, the direction is due east.

Therefore, the displacement of the whale is <u>8.8 km due east.</u>

6 0
3 years ago
Which of these is not a layer of the skin?
lina2011 [118]
4. hyperdermis is not a layer of skin
3 0
3 years ago
Read 2 more answers
1. A student lifts a box of books that weighs 185 N. The box is
aksik [14]

1)  148 J

When lifting an object, the work done on the object is equal to its change in gravitational potential energy. Mathematically:

W = \Delta U = (mg) \Delta h

where

mg is the weight of the object

\Delta h is the change in height

For the box in this problem,

mg = 185 N

\Delta h = 0.800 m

Substituting into the equation, we find:

W=(185)(0.800)=148 J

2) (a) 28875 J

The work done by a force applied parallel to the direction of motion of the object is given by

W=Fd

where

F is the magnitude of the force

d is the displacement

In this problem,

F = 825 N is the force applied by the two students together

d = 35 m is the displacement of the car

Substituting,

W=(825)(35)=28875 J

2) (b) 57750 J

As seen previously, the equation that gives the work done by the force is

W=Fd

We see that the work done is proportional to the magnitude of the force: therefore, if the force is doubled, then the work done is also doubled.

The work done previously was

W = 28875 J

Now the force is doubled, so the new work done will be

W' = 2(28875)=57750 J

3) 4.4 J

In this case, the force acting on the ball is the force of gravity, whose magnitude is:

F = mg

where

m = 0.180 kg is the mass of the ball

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

Solving the equation,

F=(0.180)(9.8)=1.76 N

Now we find the work done by gravity using the same formula applied before:

W=Fd

where d = 2.5 m is the displacement of the ball. We can apply this version of the formula since the force is parallel to the displacement. Substituting,

W=(1.76)(2.5)=4.4 J

4) 595.2 kg

In this case, we have the work done on the box:

W = 7.0 kJ = 7000 J

And we also know the change in height of the box:

\Delta h = 1.2 m

As we stated in part a), the work done on the box is equal to its change in gravitational potential energy:

W=mg \Delta h

Solving for m, we find

m=\frac{W}{g \Delta h}

And substituting the numerical values, we find the mass of the box:

m=\frac{7000}{(9.8)(1.2)}=595.2 kg

5) They do the same work

In fact, the net work done by each person on the box is equal to the change in gravitational potential energy of the box:

W=mg \Delta h

Where \Delta h is the difference in height between the final position and the initial position of the box.

This means that the work done on the box depends only on its initial and final position, not on the path taken. The two men carry the box along different paths, however the reach at the end the same position, and they started from the same position: this means that the value of \Delta h is the same for both of them, so the work they have done is exactly the same.

5 0
3 years ago
A vertical scale on a spring balance reads from 0 to 155 N . The scale has a length of 10.0 cm from the 0 to 155 N reading. A fi
Harrizon [31]

Answer:

mass of the fish is 8.11 kg

Explanation:

As we know that the frequency of oscillation of spring block system is given as

f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}

here we know that the reading of scale varies from 0 to 155 N from length varies from x = 0 to x = 10 cm

Now we have

k = \frac{155}{0.10} N/m

k = 1550 N/m

so now we have

2.20 = \frac{1}{2\pi}\sqrt{\frac{1550}{m}}

m = 8.11 kg

so mass of the fish is 8.11 kg

4 0
2 years ago
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