Work is calculated by multiplying force by the distance that the object had moved. The applied force is 60 N, moving the object by 10 m. Thus, the work does is 600 J. For the friction force which is equal to,
100N x 0.250 = 25.0 N
the work done is,
W = (60 N - 25 N) x 10 m = 350 J
The kinetic energy of the box can be equated to this force. Thus, the answer is also 350 J.
Explanation:
Check out the picture I drew for a minute before reading this...
B. Distance [the red line] is a scalar quantity reflecting how far an object has traveled. Displacement [the green line] is a vector quantity reflecting how far an object has moved from a point. The key difference is that distance can be any sort of path while displacement is always a vector (or a straight line) between a starting point and a finishing point. Sometimes distance and displacement are equal to one another. Sometimes you have a distance traveled, but zero displacement overall; which is what's going on in your question.
A. The distance that the racecar traveled is indeed 500m. But at the end of the lap, it is right back where it started. So overall, it has been displaced 0m.
Answer:
x= 9.53 ounces
Explanation:
Given that
Mean ,μ= 9 ounces
Standard deviation ,σ=0.8 ounces
He wants to sell only those potatoes that are among the heaviest 25%.
P=25% = 0.25
When P= 0.25 then Z=0.674
Lest take x is the the minimum weight required to be brought to the farmer's market.
We know that
x = Z . σ + μ
x= 0.674 ₓ 0.8 + 9 ounces
x= 9.53 ounces
Atomic mass= number of protons + number of neutrons
hope this helps
Answer:
True The grid with more slits gives more angle separation increases
True. The grating with 10 slits produces better-defined (narrower) peaks
Explanation:
Such a system can be seen as a diffraction network in this case with different number of lines per unit length, the expression for the constructive interference of a diffraction network is
d sin θ = m λ
where d is the distance between slits or lines, m the order of diffraction and λ the wavelength.
For network with 5 slits
d = 1/5 = 0.2
For the network with 10 slits
d = 1/10 = 0.1
let's calculate the separation (teat) for each one
θ = sin⁻¹ (m λ / d)
for 5 slits
θ₅ = sin⁻¹ (m λ 5)
for 10 slits
θ₁₀ = sin⁻¹ (m λ 10)
we can appreciate that for more slits the angle increases
the intensity of a series of slits is
I = I₀ sin²2 (N d/2) / sin² d/2)
when there are more slits (N) the peaks have greater intensity and are more acute (half width decreases)
let's analyze the claims
False
True The grid with more slits gives more angle separation increases
False
True The expression for the intensity of the diffraction peaks the intensity of the peaks increases with the number of slits as well as their spectral width decreases
False
