Answer:
(a): a = 0.4m/s²
(b): α = 8 radians/s²
Explanation:
First we propose an equation to determine the linear acceleration and an equation to determine the space traveled in the ramp (5m):
a= (Vf-Vi)/t = (2m/s)/t
a: linear acceleration.
Vf: speed at the end of the ramp.
Vi: speed at the beginning of the ramp (zero).
d= (1/2)×a×t² = 5m
d: distance of the ramp (5m).
We replace the first equation in the second to determine the travel time on the ramp:
d = 5m = (1/2)×( (2m/s)/t)×t² = (1m/s)×t ⇒ t = 5s
And the linear acceleration will be:
a = (2m/s)/5s = 0.4m/s²
Now we determine the perimeter of the cylinder to know the linear distance traveled on the ramp in a revolution:
perimeter = π×diameter = π×0.1m = 0.3142m
To determine the angular acceleration we divide the linear acceleration by the radius of the cylinder:
α = (0.4m/s²)/(0.05m) = 8 radians/s²
α: angular aceleration.
Answer:
at t=46/22, x=24 699/1210 ≈ 24.56m
Explanation:
The general equation for location is:
x(t) = x₀ + v₀·t + 1/2 a·t²
Where:
x(t) is the location at time t. Let's say this is the height above the base of the cliff.
x₀ is the starting position. At the base of the cliff we'll take x₀=0 and at the top x₀=46.0
v₀ is the initial velocity. For the ball it is 0, for the stone it is 22.0.
a is the standard gravity. In this example it is pointed downwards at -9.8 m/s².
Now that we have this formula, we have to write it two times, once for the ball and once for the stone, and then figure out for which t they are equal, which is the point of collision.
Ball: x(t) = 46.0 + 0 - 1/2*9.8 t²
Stone: x(t) = 0 + 22·t - 1/2*9.8 t²
Since both objects are subject to the same gravity, the 1/2 a·t² term cancels out on both side, and what we're left with is actually quite a simple equation:
46 = 22·t
so t = 46/22 ≈ 2.09
Put this t back into either original (i.e., with the quadratic term) equation and get:
x(46/22) = 46 - 1/2 * 9.806 * (46/22)² ≈ 24.56 m
Answer:
Work= -7.68×10⁻¹⁴J
Explanation:
Given data
q₁=q₂=1.6×10⁻¹⁹C
r₁=2.00×10⁻¹⁰m
r₂=3.00×10⁻¹⁵m
To find
Work
Solution
The work done on the charge is equal to difference in potential energy
W=ΔU
![Work=U_{1}-U_{2}\\ Work=-kq_{1}q_{2}[\frac{1}{r_{2}}-\frac{1}{r_{1}} ]\\Work=(-9*10^{9})*(1.6*10^{-19} )^{2}[\frac{1}{3.0*10^{-15} }-\frac{1}{2*10^{-10} } ]\\ Work=-7.68*10^{-14}J](https://tex.z-dn.net/?f=Work%3DU_%7B1%7D-U_%7B2%7D%5C%5C%20Work%3D-kq_%7B1%7Dq_%7B2%7D%5B%5Cfrac%7B1%7D%7Br_%7B2%7D%7D-%5Cfrac%7B1%7D%7Br_%7B1%7D%7D%20%5D%5C%5CWork%3D%28-9%2A10%5E%7B9%7D%29%2A%281.6%2A10%5E%7B-19%7D%20%29%5E%7B2%7D%5B%5Cfrac%7B1%7D%7B3.0%2A10%5E%7B-15%7D%20%7D-%5Cfrac%7B1%7D%7B2%2A10%5E%7B-10%7D%20%7D%20%5D%5C%5C%20%20Work%3D-7.68%2A10%5E%7B-14%7DJ)
Answer:
271cm^2
Explanation:
volume 1= 15^3 =3375
temp. 1. = 20+273 = 293
temp. 2. = 50+273 = 353
volume 2 =?
According to Charles law
volume is proportional to temperature
v2 = v1 * t2 / t1
v2 = 3375 * 353 / 293
v2 = 4066cm^3
v = area * length
4066 = area * 15
area = 4066/15 = 271cm^2
Answer:
The equipment to use is: a beaker, a fixed amount of water, a thermometer.
The mass of water, the time, the temperature for each time should be noted and a graph of Temperature versus time should be made
Explanation:
The design of an experiment is to place the beaker in the microwave, with a good amount of water (approximately ⅔ of its capacity) and turn it on for small periods of time, generally the minimum is 30 s, quickly open the microwave, place a thermometer or better yet an infrared thermometer to measure the temperature of the water; repeat this several times.
The advantage of the infrared thermometer is that it reduces the transfer of heat between the water and the thermometer.
The mass of water, the time, the temperature for each time should be noted and a graph of Temperature versus time should be made.
The equipment to use is: a beaker, a fixed amount of water, a thermometer.
The main precaution that must be taken is not to open the microwave while it is on.
