Answer:
a= 4.4×10 m/s^2
Explanation:
pressure P = E/c
Where, E = 100 W/m^2 intensity of light
c= speed of light = 3×10^8 m/s
P = 1000/ 3×10^8
P = 3.33×10^(-6) Pa
Force F = P×A
- P is the pressure and c= speed of light
F = 3.33×10^{-6}×6.65×10(-29)
= 2.22×10^{-6}
acceleration a = F/m = 2.22×10^{-6}/ 5.10×10^{-27}
a= 4.4×10 m/s^2
Answer:
6.88 mA
Explanation:
Given:
Resistance, R = 594 Ω
Capacitance = 1.3 μF
emf, V = 6.53 V
Time, t = 1 time constant
Now,
The initial current, I₀ = 
or
I₀ = 
or
I₀ = 0.0109 A
also,
I = ![I_0[1-e^{-\frac{t}{\tau}}]](https://tex.z-dn.net/?f=I_0%5B1-e%5E%7B-%5Cfrac%7Bt%7D%7B%5Ctau%7D%7D%5D)
here,
τ = time constant
e = 2.717
on substituting the respective values, we get
I = ![0.0109[1-e^{-\frac{\tau}{\tau}}]](https://tex.z-dn.net/?f=0.0109%5B1-e%5E%7B-%5Cfrac%7B%5Ctau%7D%7B%5Ctau%7D%7D%5D)
or
I = ![0.0109[1-2.717^{-1}]](https://tex.z-dn.net/?f=0.0109%5B1-2.717%5E%7B-1%7D%5D)
or
I = 0.00688 A
or
I = 6.88 mA
Answer:
A u = 0.36c B u = 0.961c
Explanation:
In special relativity the transformation of velocities is carried out using the Lorentz equations, if the movement in the x direction remains
u ’= (u-v) / (1- uv / c²)
Where u’ is the speed with respect to the mobile system, in this case the initial nucleus of uranium, u the speed with respect to the fixed system (the observer in the laboratory) and v the speed of the mobile system with respect to the laboratory
The data give is u ’= 0.43c and the initial core velocity v = 0.94c
Let's clear the speed with respect to the observer (u)
u’ (1- u v / c²) = u -v
u + u ’uv / c² = v - u’
u (1 + u ’v / c²) = v - u’
u = (v-u ’) / (1+ u’ v / c²)
Let's calculate
u = (0.94 c - 0.43c) / (1+ 0.43c 0.94 c / c²)
u = 0.51c / (1 + 0.4042)
u = 0.36c
We repeat the calculation for the other piece
In this case u ’= - 0.35c
We calculate
u = (0.94c + 0.35c) / (1 - 0.35c 0.94c / c²)
u = 1.29c / (1- 0.329)
u = 0.961c
Answer:
mb = 3.75 kg
Explanation:
System of forces in balance
ΣFx =0
ΣFy = 0
Forces acting on the box
T₁ : Tension in string 1 ,at angle of 50° with the horizontal on the left
T₂ = 40 N : Tension in string 2, at angle of 75° with the horizontal on the right.
Wb :Weightt of the box (vertical downward)
x-y T₁ and T₂ components
T₁x= T₁cos50°
T₁y= T₁sin50°
T₂x= 30*cos75° = 7.76 N
T₂y= 30*sin75° = 28.98 N
Calculation of the Wb
ΣFx = 0
T₂x-T₁x = 0
T₂x=T₁x
7.76 = T₁cos50°
T₁ = 7.76 /cos50° = 12.07 N
ΣFy = 0
T₂y+T₁y-Wb = 0
28.98 + 12.07(cos50°) = Wb
Wb = 36.74 N
Calculation of the mb ( mass of the box)
Wb = mb* g
g: acceleration due to gravity = 9.8 m/s²
mb = Wb/g
mb = 36.74 /9.8
mb = 3.75 kg
4.Use Ohm’s Law to determine the resistance in a circuit if the voltage is 12.0 volts and the current is 4.0 amps.
A. 8.0 ohms B. 48 ohms C. 3.0 ohms D. 12 ohms
Ohm's law is V=IR, or I=V/R, or R=V/I. (I= current, V= voltage, R= resistance.) Let's plug in our variables: V=12.0, I=4.0, R=? into the equation R=V/I. 12.0/4.0=3.0, so the resistance is 3.0 ohms.
