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2 years ago

2 ways to change frictional force between 2 objects

2 answers:

8 0

It can be changed by shorter distances and by the amount of weight it has or the amount of force that is pushing that object to go however distance it can.

Zanzabum2 years ago

5 0

In order to change the frictional force between two solid surfaces, it can be changed by shorter distances and by the amount of weight it has or the amount of force that is pushing that object to go however distance it can.

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Energy is converted from solar to chemical in Process A and then from one form of chemical to another in Process B. Process A is

Pepsi [2]

I think the answer is photosynthis, when plants turn light into food and energy.

9 0

3 years ago

Read 2 more answers

Which formula correctly shows how to calculat the time taken, given the average velocity and the displacement

Alisiya [41]

We're so good here on Brainly, we can answer it
even WITHOUT seeing the choices.

Time = (displacement) / (magnitude of average velocity) .

8 0

3 years ago

A space probe is fired as a projectile from the Earth's surface with an initial speed of 2.05 104 m/s. What will its speed be wh

Elanso [62]

Answer:

The value is $v = 2.3359 *10^{4} \ m/s$

Explanation:

From the question we are told that

The initial speed is $u = 2.05 *10^{4} \ m/s$

Generally the total energy possessed by the space probe when on earth is mathematically represented as

$T__{E}} = KE__{i}} + KE__{e}}$

Here $KE_i$ is the kinetic energy of the space probe due to its initial speed which is mathematically represented as

$KE_i = \frac{1}{2} * m * u^2$

=> $KE_i = \frac{1}{2} * m * (2.05 *10^{4})^2$

=> $KE_i = 2.101 *10^{8} \ \ m \ \ J$

And $KE_e$ is the kinetic energy that the space probe requires to escape the Earth's gravitational pull , this is mathematically represented as

$KE_e = \frac{1}{2} * m * v_e^2$

Here $v_e$ is the escape velocity from earth which has a value $v_e = 11.2 *10^{3} \ m/s$

=> $KE_e = \frac{1}{2} * m * (11.3 *10^{3})^2$

=> $KE_e = 6.272 *10^{7} \ \ m \ \ J$

Generally given that at a position that is very far from the earth that the is Zero, the kinetic energy at that position is mathematically represented as

$KE_p = \frac{1}{2} * m * v^2$

Generally from the law energy conservation we have that

$T__{E}} = KE_p$

$2.101 *10^{8} m + 6.272 *10^{7} m = \frac{1}{2} * m * v^2$

=> $5.4564 *10^{8} = v^2$

=> $v = \sqrt{5.4564 *10^{8}}$

=> $v = 2.3359 *10^{4} \ m/s$

4 0

2 years ago

A 6.0 m section of wire carries a current of 5.2 A from east to west in the earth's magnetic field of 1.0 × 105 T at a location

NNADVOKAT [17]

Well, since you only want direction, ignore the numbers. Use the right hand rule.
Current (pointer finger) points west (left).
Magnetic field (middle finger) points south (towards you).
Force (thumb) then points up (away from the earth)

4 0

3 years ago

PLEASE HELP<br><br> TOPIC - CONVECTION

Alex777 [14]

Answer:

Dry air burns out and gets warm
It's smoke and smoke always rises
Not ours
It makes sure no air is inside bc air gets cold an freezes

6 0

2 years ago