An object's momentum is the product of its mass and its velocity:
p = mv
p is its momentum, m is its mass, and v is its velocity.
Given values:
p = -80kg×m/s
m = 8kg
Plug in these values and solve for v:
-80 = 8v
v = -10m/s
Choice D
Answer:
F = ⅔ F₀
Explanation:
For this exercise we use Coulomb's law
F = k q₁q₂ / r²
let's use the subscript "o" for the initial conditions
F₀ = k q² / r²
now the charge changes q₁ = q₂ = 2q and the new distance is r = 3 r
we substitute
F = k 4q² / 9 r²
F = k q² r² 4/9
F = ⅔ F₀
Due to attraction ... of opposite charges
The work done by the normal force n when the box slides down a frictionless incline and gaining speed is zero.
<h3>What is normal force?</h3>
The force of contact is called the normal force. When the two surfaces are in contact with each other, then the normal force acts.
This force is applied by the solid bodies on each other in order to prevent the passing through each other.
A box slides down a frictionless incline, gaining speed. For this box, the value of work done by normal force has to be found out. Let's analyze the given condition.
- The body is gaining the speed, which means there is a change in kinetic energy.
- The change in kinetic energy is equal to the work done.
- The friction force is the product of coefficient of the friction and normal force.
- The friction force for the given case is zero. Thus, the normal force must be equal to the zero.
Thus, the work done by the normal force n when the box slides down a frictionless incline and gaining speed is zero.
Learn more about the normal force here;
brainly.com/question/10941832
The answer is C.
The question says the potential difference is what is changing, which means we're solving for V.
It tells us that potential difference increases by a factor of two, which just means V doubles.
With this info, we can pick some numbers, plug it into Ohms law and see what happens.
Here's an example where I just picked random numbers that are easy to work with:
V=I*R
10=I*5
I=2
Lets increase the potential difference (V) by a factor of two and see what happens to current:
V=I*R
20=I*5 (all I've done is double the potential difference from 10 to 20)
I=4
When we increase V by a factor of 2, I increases by a factor of 2. We went from I=2 to I=4.
We can increase V by factor of 2 again and see:
V=I*R
40=I*5
I=8
Okay, current just increased by a factor of 2 again when we increased the potential difference by a factor of 2.
It's always good to check work with alternate numbers, so here's one more set:
V=I*R
16=I*4
(remember, we know we're solving for V, so I'm just plugging in random numbers for I and R)
I=4
Increase V by factor of 2:
32=I*4
I=8
So, when we increase V (the potential difference) by a factor of 2, I (current) always increases by a factor of 2 as well.
Hope this helps!
