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2 years ago

Please help me on questions 3 and 6, thank you! :D I'll give brainliest!

Physics

2 answers:

KIM [24]2 years ago

6 0

3. To increase gravitational pull to the earth, the more mass an object has, the stronger the gravitational pull, so increase mass.

6. While sound energy depends on a medium, radiant energy uses it's own sources to stay independent from other mediums.

Anarel [89]2 years ago

4 0

Answer:

3. if you increase your mass you also increase the gravitational pull

6. Radiant energy doesn't depend on a medium and sound energy is dependent on a medium.

Explanation:

i hope this helps-

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How are the planets sizes related to their surface gravity

dexar [7]

Answer:

The surface gravity is inversely proportional to the square of the radius of the planet

Explanation:

The gravity at the surface of a planet is given by:

$g=\frac{GM}{R^2}$

where

G is the gravitational constant

M is the mass of the planet

R is the radius of the planet

We see from the formula that the surface gravity is inversely proportional to the square of the radius of the planet, R.

At the Earth's surface, the value of the surface gravity is approximately 9.81 m/s^2.

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2 years ago

How much force is needed to stop a body of mass 10kg

almond37 [142]

Answer:

Force of 100 N will be needed.

Explanation:

$Force, F = m \times g$

g is acceleration due to gravity

m is the mass

$F = (10 \times 10) \\ F = 100 \: newtons$

3 0

2 years ago

Read 2 more answers

Three identical resistors have an equivalent resistance 15.6 ohm when they are connected in series. What is the resistance for e

attashe74 [19]

Answer:

$R=5.2\ \Omega$

Explanation:

In series combination, the equivalent resistance is given by :

$R_{eq}=R_1+R_2+R_3+...$

Let the identical resistors be R. We have, $R_{eq}=15.6\ \Omega$

So,

$R+R+R=15.6\\\\3R=15.6\\\\R=\dfrac{15.6}{3}\\\\R=5.2\ \Omega$

So, the resistance of each resistor is $5.2\ \Omega$ .

4 0

2 years ago

An object of 4 kg has a speed of 6 m/s and moves a distance of 8 m. What is its kinetic energy in joules.

11Alexandr11 [23.1K]

Answer:

The answer to your question is Ke = 72 J

Explanation:

Kinetic energy depends on the speed of and object and its mass.

Data

mass = m = 4 kg

speed = v = 6 m/s

distance = d = 8 m

Kinetic energy = ke = ?

Formula

Ke = (1/2) mv²

Substitution

Ke = (1/2) (4)(6)²

Simplification

Ke = (1/2)(4)(36)

Ke = (1/2)(144)

Ke = 72 Joules

Result

Ke = 72 J

3 0

2 years ago

In a historical movie, two knights on horseback start from rest 86 m apart and ride directly toward each other to do battle. Sir

Harlamova29_29 [7]

Answer:

Relative to Sir George's starting point, the knights collide at a distance of 38.43 m from Sir George's starting point.

Explanation:

Let the distance covered by Sir George be $S_{1}$

and the distance covered by Sir Alfred be $S_{2}$

Since the knights collide, hence they must have traveled for the same amount of time just before collision

From one of the equations of motion for linear motion

$S = ut + \frac{1}{2}at^{2}$

Where $S$ is the distance traveled

$u$ is the initial velocity

$a$ is the acceleration

and $t$ is the time

For Sir George,

$S = S_{1}$

$u$ = 0 m/s (Since they start from rest)

$a =$ 0.21 m/s²

Hence,

$S = ut + \frac{1}{2}at^{2}$ becomes

$S_{1} = (0)t + \frac{1}{2}(0.21)t^{2}\\S_{1} = 0.105 t^{2}\\$

$t^{2} = \frac{S_{1}}{0.105}$

Now, for Sir Alfred

$S = S_{2}$

$u$ = 0 m/s (Since they start from rest)

$a =$ 0.26 m/s²

Hence,

$S = ut + \frac{1}{2}at^{2}$ becomes

$S_{2} = (0)t + \frac{1}{2}(0.26)t^{2}\\S_{2} = 0.13 t^{2}\\$

$t^{2} = \frac{S_{2}}{0.13}$

Since, they traveled for the same time, $t$ just before collision, we can write

$\frac{S_{1}}{0.105}= \frac{S_{2}}{0.13}$

Since, the two nights are 86 m apart, that is, the sum of the distances covered by the knights just before collision is 86 m. Then we can write that

$S_{1} + S_{2} = 86$ m

Then, $S_{2} = 86 - S_{1}$

Then,

$\frac{S_{1}}{0.105}= \frac{S_{2}}{0.13}$ becomes

$\frac{S_{1}}{0.105}= \frac{86 -S_{1}}{0.13}$

$0.13{S_{1}}= 0.105({86 -S_{1}})\\0.13{S_{1}}= 9.03 - 0.105S_{1}}\\0.13{S_{1}} + 0.105S_{1}}= 9.03 \\0.235{S_{1}} = 9.03\\{S_{1}} =\frac{9.03}{0.235}$

$S_{1} = 38.43$ m

∴ Sir George covered a distance of 38.43 m just before collision.

Hence, relative to Sir George's starting point, the knights collide at a distance of 38.43 m from Sir George's starting point.

6 0

3 years ago