Answer:
0.266 m
Explanation:
Assuming the lump of patty is 3 Kg then applying the principal of conservation of linear momentum,
P= mv where p is momentum, m is mass and v is the speed of an object. In this case
where sunscripts p and b represent putty and block respectively, c is common velocity.
Substituting the given values then
3*8=v(15+3)
V=24/18=1.33 m/s
The resultant kinetic energy is transferred to spring hence we apply the law of conservation of energy
where k is spring constant and x is the compression of spring. Substituting the given values then
![(3+15)*1.33^{2}=450*x^{2}\\x\approx 0.266 m](https://tex.z-dn.net/?f=%283%2B15%29%2A1.33%5E%7B2%7D%3D450%2Ax%5E%7B2%7D%5C%5Cx%5Capprox%200.266%20m)
Answer:
λ = 2.7608 x 10⁻⁷ m = 276.08 nm
Explanation:
The work function of a metallic surface is the minimum amount of photon energy required to release the photo-electrons from the surface of metal. The work function is given by the following formula:
Work Function = hc/λ
where,
Work Function = (4.5 eV)(1.6 x 10⁻¹⁹ J/1 eV) = 7.2 x 10⁻¹⁹ J
h = Plank's Constant = 6.626 x 10⁻³⁴ J.s
c = speed of light = 3 x 10⁸ m/s
λ = longest wavelength capable of releasing electron.
Therefore,
7.2 x 10⁻¹⁹ J = (6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/λ
λ = (6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(7.2 x 10⁻¹⁹ J)
<u>λ = 2.7608 x 10⁻⁷ m = 276.08 nm</u>
There are two ways to solve this. The longer way is to use those equations to calculate numbers for total distance.
The easier way is to find the area under the graph. That's right, AREA UNDER VELOCITY-TIME graph is the TOTAL DISTANCE travelled!
it's a shortcut.
Let's split up the area into a triangle and rectangle:
Triangle = 0.5(4-0)(10-0) = 20 m
Rectangle = (6-4)(10-0) = 20 m
Total distance = 40 m!
Answer:
Normal force = 8.75 N
Explanation:
given,
frictional force between the steel spatula and the Teflon frying pan=0.350 N
coefficient of friction between material =0.04
normal force = ?
using formula,
Frictional force = coefficient of friction × normal force
![normal\ force = \dfrac{Frictional\ force}{coefficient\ of\ friction}](https://tex.z-dn.net/?f=normal%5C%20force%20%3D%20%5Cdfrac%7BFrictional%5C%20force%7D%7Bcoefficient%5C%20of%5C%20friction%7D)
![normal\ force = \dfrac{0.350}{0.04}](https://tex.z-dn.net/?f=normal%5C%20force%20%3D%20%5Cdfrac%7B0.350%7D%7B0.04%7D)
Normal force = 8.75 N