Recall that the force on an object is related to the mass and acceleration of that object by the formula F = ma, where m is the mass of the object and a is its acceleration. What happens when we double F? Well, you might remember from algebra that, in order to keep our equality true, if we double one side, we must also double the other, so our equation becomes 2F = 2ma. Now, this means one of two things: either the mass has doubled, or the acceleration has doubled.
We can tell right away that it'd be absurd if a race car doubled in mass every time it hit the gas, so the quantity doubling must be the <em>acceleration. </em>If we call the car's current velocity v1, we'll be adding the doubled acceleration to get its new velocity. Mathematically, v = v1 + 2a.
We can now conclude that, by doubling the force:
- The acceleration of the car will double,
- The mass of the car will stay the same, and
- The velocity of the car will increase by double the original acceleration
Answer:
1. Installing solar panels on the roof. This is an example of renewable energy, and solar panels help provide electricity for a home.
2. Using LED lights instead of traditional lightbulbs. The modern LED lights use much less electricity and function up to 90% more efficiently than the traditional lightbulbs.
3. Using rain barrels to collect water from the roof. Installing rain barrels to collect rainwater to water your plants and garden is a great way to save water and cut down on your water bill.
Hope this helps! :)
Answer:
5-The force on A is exactly equal to the force on B.
Explanation:
This is proven mathematically:
The force on A by B is given as:
F(A, B) = [K * Q(A) * Q(B)] / R²
The force on B by A is given as:
F(B, A) = [K * Q(B) * Q(A)] / R²
Where K = Coulombs constant
R is the distance between them
Examining the two formulas closely show that they yield the same result.
Answer:
75 N
Explanation:
t = Time taken = 2 seconds
u = Initial velocity
v = Final velocity
s = Displacement = 30 m
a = Acceleration
![s=ut+\frac{1}{2}at^2\\\Rightarrow 30=0\times 2+\frac{1}{2}\times a\times 2^2\\\Rightarrow a=\frac{30\times 2}{2^2}\\\Rightarrow a=15\ m/s^2](https://tex.z-dn.net/?f=s%3Dut%2B%5Cfrac%7B1%7D%7B2%7Dat%5E2%5C%5C%5CRightarrow%2030%3D0%5Ctimes%202%2B%5Cfrac%7B1%7D%7B2%7D%5Ctimes%20a%5Ctimes%202%5E2%5C%5C%5CRightarrow%20a%3D%5Cfrac%7B30%5Ctimes%202%7D%7B2%5E2%7D%5C%5C%5CRightarrow%20a%3D15%5C%20m%2Fs%5E2)
The acceleration due to gravity on the planet is 15 m/s²
Force
F = ma
![F=5\times 15\\\Rightarrow F=75\ N](https://tex.z-dn.net/?f=F%3D5%5Ctimes%2015%5C%5C%5CRightarrow%20F%3D75%5C%20N)
The gravitational force exerted on the object near the planet’s surface is 75 N
Answer:
v = 6.315 cm
Explanation:
given,
R₁ = 4 cm = 0.04 m
R₂ = 15 cm = 0.15 m
n =1.5
![\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}=(n-1)(\dfrac{1}{R_1}-\dfrac{1}{R_2})](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7Bf%7D%3D%5Cdfrac%7B1%7D%7Bv%7D-%5Cdfrac%7B1%7D%7Bu%7D%3D%28n-1%29%28%5Cdfrac%7B1%7D%7BR_1%7D-%5Cdfrac%7B1%7D%7BR_2%7D%29)
![\dfrac{1}{v}-\dfrac{1}{-1}=(1.5-1)(\dfrac{1}{0.04}+\dfrac{1}{0.15})](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7Bv%7D-%5Cdfrac%7B1%7D%7B-1%7D%3D%281.5-1%29%28%5Cdfrac%7B1%7D%7B0.04%7D%2B%5Cdfrac%7B1%7D%7B0.15%7D%29)
![\dfrac{1}{v}+1 = 0.5 \times 31.66](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7Bv%7D%2B1%20%3D%200.5%20%5Ctimes%2031.66)
![\dfrac{1}{v} = 15.833](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7Bv%7D%20%3D%2015.833)
v = 0.06315 m
v = 6.315 cm
hence, the distance of the image from the back surface is v = 6.315 cm