All categories

3 years ago

I NEED HELP LIKE RIGHT NOW!!!!... please

Physics

2 answers:

Maslowich3 years ago

8 0

1.) TRUE

7.) I believe it is FALSE

that's all I got for now

harkovskaia [24]3 years ago

6 0

In my humble opinion . . . . .

1). true

2). a poor question, fishing for "False"

3). true

4). False

5). poor question, fishing for "true"

6). false

7). false

8). false (not totally sure)

9). false

10). true

11). true

12). 600m

13). 0.05 L

14). 400 g

15). kilogram

16). folding

17). solar system

18). matter

19). free-fall

20). smallpox

21). (idk)

22). the continents were once one large land mass

23). the same as her mass on Earth, or anywhere else

24). (idk)

25). magma is one, idk the other

26). a force

27). small

28). formed slowly

29). (idk)

30). distance

31). by collision with a high-speed particle

32). I agree that there is still habitable land remaining on Earth. But that's not a "therefore" that follows from the first statement.

You might be interested in

A student is walking at a constant speed of 5 km/h. A car passes the student

Ulleksa [173]

Answer:

5 km/hr because he thinks it moves in the same speed of his...

8 0

3 years ago

How can you tell the difference between halite, calcite, and gypsum

ehidna [41]

Halite is transparent, but it can be died many different colors calcite is a white or colorless mineral consisting of calcium carbonate. It is a major constituent of sedimentary rocks such as limestone, marble, and chalk, can occur in crystalline form gypsum is a soft white or gray mineral consisting of hydrated calcium sulfate.

6 0

3 years ago

The motivation for Isaac Newton to discover his laws of motion was to explain the properties of planetary orbits that were obser

Dafna1 [17]

A) Orbital speed: $v=\sqrt{\frac{GM}{R}}$

B) Kinetic energy: $K= \frac{GmM}{2R}$

D) The orbital period is $T=\frac{2\pi}{\sqrt{GM}}R^{3/2}$

F) The angular momentum is $L=m\sqrt{GMR}$

G) Exponent of radial dependence:

Speed: -1/2

Kinetic energy: -1

Orbital period: 3/2

Angular momentum: 1/2

Explanation:

We know that for a satellite in circular orbit around a planet of mass M, the gravitational force between the satellite and the planet is

$F=G\frac{mM}{R^2}$

where m is the mass of the satellite.

This force provides the centripetal force needed for the circular motion, which is

$F=m\frac{v^2}{R}$

where v is the orbital speed.

Since the gravitational force provides the centripetal force, we can equate the two expressions:

$G\frac{mM}{R^2}=m\frac{v^2}{r}$

And solving for v, we find

$v=\sqrt{\frac{GM}{R}}$

The kinetic energy of an object is given by

$K=\frac{1}{2}mv^2$

where

m is the mass of the object

v is its speed

In this problem,

m is the mass of the satellite

$v=\sqrt{\frac{GM}{R}}$ is the speed of the satellite (found in part A)

Substituting, we find an expression for the kinetic energy of the satellite:

$K=\frac{1}{2}m(\sqrt{\frac{GM}{R}})^2 = \frac{GmM}{2R}$

The orbital speed of the satellite can be rewritten as the ratio between the distance covered during one orbit (the circumference of the orbit) divided by the period of revolution:

$v=\frac{2\pi R}{T}$

where

$2\pi R$ is the circumference of the orbit

T is the orbital period

We already found that the orbital speed is

$v=\sqrt{\frac{GM}{R}}$

Substituting into the equation,

$\sqrt{\frac{GM}{R}}=\frac{2\pi R}{T}$

And making T the subject,

$T=\frac{2\pi R}{\sqrt{\frac{GM}{R}}}=\frac{2\pi}{\sqrt{GM}}R^{3/2}$

The angular momentum of an object is defined as

$L=mvr$

where

m is the mass of the object

v is its speed

r is the radius of the orbit

For the satellite here we have

m (mass of the satellite)

$v=\sqrt{\frac{GM}{R}}$ (orbital speed)

R (orbital radius)

Substituting,

$L=m\sqrt{\frac{GM}{R}}R=m\sqrt{GMR}$

First, we rewrite the list of expressions for the different quantities that we found:

Orbital speed: $v=\sqrt{\frac{GM}{R}}$

Kinetic energy: $K= \frac{GmM}{2R}$

Orbital period: $T=\frac{2\pi}{\sqrt{GM}}R^{3/2}$

Angular momentum: $L=m\sqrt{GMR}$

Now we observed the dependence of each quantity from R:

Orbital speed: $v\propto R^{-1/2}$

Kinetic energy: $K \propto R^{-1}$

Orbital period: $T \propto R^{3/2}$

Angular momentum: $L \propto R^{1/2}$

So the exponent of the radial dependence of each quantity is:

Speed: -1/2

Kinetic energy: -1

Orbital period: 3/2

Angular momentum: 1/2

Learn more about circular motion:

brainly.com/question/2562955

brainly.com/question/6372960

#LearnwithBrainly

7 0

3 years ago

The cushion is picked up and put on the couch. A work is being done. B no work is being done

Goshia [24]

Work is being done because picking up a cushion and putting it somewhere is the motion of doing something (work).

5 0

4 years ago

Read 2 more answers

A plastic rod of length d = 1.5 m lies along the x-axis with its midpoint at the origin. The rod carries a uniform linear charge

Serga [27]

Answer:

Explanation:

Let the plastic rod extends from - L to + L .

consider a small length of dx on the rod on the positive x axis at distance x . charge on it = λ dx where λ is linear charge density .

It will create a field at point P on y -axis . Distance of point P

= √ x² + .15²

electric field at P due to small charged length

dE = k λ dx x / (x² + .15² )

Its component along Y - axis

= dE cosθ where θ is angle between direction of field dE and y axis

= dE x .15 / √ x² + .15²

= k λ dx .15 / (x² + .15² )³/²

If we consider the same strip along the x axis at the same position on negative x axis , same result will be found . It is to be noted that the component of field in perpendicular to y axis will cancel out each other . Now for electric field due to whole rod at point p , we shall have to integrate the above expression from - L to + L

E = ∫ k λ .15 / (x² + .15² )³/² dx

= k λ x L / .15 √( L² / 4 + .15² )

6 0

4 years ago