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Yuliya22 [10]
3 years ago
9

1. Describe what happens to the gravitational when you increase mass versus when you

Physics
2 answers:
Rudik [331]3 years ago
7 0

Answer: if the mass is doubled, the force of gravity is doubled, meaning it decreases. If the distance is doubled, the force of gravity is 1/4 as strong as before

Explanation:

faust18 [17]3 years ago
6 0

Answer:If one of the masses is doubled, the force of gravity between the objects is doubled. increases, the force of gravity decreases. If the distance is doubled, the force of gravity is one-fourth as strong as before.

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Bromine, a liquid at room temperature, has a boiling point
lukranit [14]

Yes it does !  The so-called "boiling point" is the temperature at which Bromine liquid can change state and become Bromine vapor, if enough additional thermal energy is provided.  The boiling point is higher than room temperature.

3 0
3 years ago
I need help with part D
natka813 [3]
1).  The little projectile is affected by friction all the way through the block.
Friction robs some kinetic energy.

2).  The block is affected by friction as it scrapes along the top of the post.
Friction robs some kinetic energy.

3).  The block is also affected by friction with the air (air resistance) as it
falls to the ground.  Friction robs some kinetic energy.
8 0
3 years ago
two billiard balls moving along the same line hit each other head-on. each has a mass of 0.220 kg; one has an initial velocity o
Tems11 [23]

Hi there!

Since the collision is elastic, we must also satisfy the following condition:

Ei = Ef, or:

KEi = KEf

Begin by writing an expression for momentum. (p = mv) Remember that one ball's direction is negative; in this instance, we can let the second ball be moving LEFT.

mv1 + mv2 = mvf1 + mvf2

0.220(1.84) + 0.220(-.530) = 0.220(vf1 + vf2)

0.2882/0.220 = vf1 + vf2

1.31 = vf1 + vf2

Now, we can express this as a conservation of energy:

1/2mv1² + 1/2mv2² = 1/2mvf1² + 1/2mvf2²

Plug in values and simplify:

0.403315 = 1/2m(vf1² + vf2²)

Simplify further:

3.6665 = vf1² + vf2²

Use the equation derived from momentum above and solve for one variable:

vf2 = 1.31 - vf1

Plug in this expression for vf2:

3.6665 = vf1² + (1.31 - vf1)²

Expand:

3.6665 = vf1² + 1.7161 - 2.62vf1 + vf1²

Simplify:

1.9504 = -2.62vf1 + 2vf1²

Solve for vf1 using a graphing calculator:

vf1 = -0.53 m/s or 1.84 m/s; we must figure out which one is correct.

Since v1 is heading to the right initially with a velocity of 1.84 m/s, we know that the ball's velocity could not have stayed the same in both magnitude and direction, so the final velocity must be -0.53 m/s.

Now, we can solve for the velocity of the other ball (initial of 0.53 m/s):

vf2 = 1.31 - (-0.53) = 1.84 m/s.

Now, you could have also made the connection that when two balls of the SAME MASS experience an ELASTIC collision, the velocities are simply "exchanged" from one to another. I just used this more "extensive" method to prove this.

7 0
3 years ago
A wave with low energy will also have
Aliun [14]
Will also have low frequency 

3 0
3 years ago
Two positive point charges, each of which has a charge of 2.5 × 10−9 C, are located at y = + 0.50m and y = − 0.50m. Find the mag
nadya68 [22]

Answer:

F = 147,78*10⁻⁹ [N]

Explanation:

By symmetry the Fy components of the forces acting on charge in point x = 0,7 m canceled each other, and the total force will be twice Fx ( Fx is x axis component of one of the forces .

The angle β  ( angle between the line running through one of the charges in y axis and the charge in x axis) is

tan β  =  0,5/0,7

tan β  = 0,7142    then   β = arctan 0,7142      ⇒   β = 35 ⁰

cos  β = 0,81

d = √ (0,5)²  +  (0,7)²       d1stance between charges

d = √0,25 + 0,49

d =  √0,74  m

d = 0,86  m

Now Foce between two charges  is:

F  = K* q₁*q₂/ d²      (1)

Where  K  = 9*10⁹ N*m²/C²

q₁  =  2,5* 10⁻⁹C

q₂  = 3,0*10⁻⁹C

d²  = 0,74 m²

Plugging these values in (1)

F  =  9*10⁹* 2,5* 10⁻⁹*3,0*10⁻⁹ / 0,74       [N*m²/C²]*C*C/m²

F = 91,21 * 10⁻⁹   [N]

And  Fx  =  F*cos  β

Fx  =  91,21 * 10⁻⁹ *0,81

Fx =73,89*10⁻⁹  [N]

Then total force acting on charge located at x = 0,7 m is:

F = 2* Fx

F = 2*73,89*10⁻⁹  [N]

F = 147,78*10⁻⁹ [N]

8 0
3 years ago
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