Answer:
no it would not. that is an open circuit and it would need to be closed at the switch for current to flow.
Answer:
270 m/s²
Explanation:
Given:
α = 150 rad/s²
ω = 12.0 rad/s
r = 1.30 m
Find:
a
The acceleration will have two components: a radial component and a tangential component.
The tangential component is:
at = αr
at = (150 rad/s²)(1.30 m)
at = 195 m/s²
The radial component is:
ar = v² / r
ar = ω² r
ar = (12.0 rad/s)² (1.30 m)
ar = 187.2 m/s²
So the magnitude of the total acceleration is:
a² = at² + ar²
a² = (195 m/s²)² + (187.2 m/s²)²
a = 270 m/s²
The solution for this problem is:
For 1st minimum, let m be equal to 1.
d = slit width
D = screen distance.
Θ = arcsin (m * lambda/ (d))
= 0.13934 rad, 7.9836 deg
y = D*tan (Θ)
y = 6.50 * tan (7.9836)
= 0.91161 m is the distance from the central maximum to the first-order minimum
Answer: C
Reflection
Explanation: Light travels in a straight line. Reflection is one of the properties of light. And this is the property in which mirror make use of. The ability of light to bounce back. It's this bouncing back characteristics of light ray that eventually produce the image of an object by the mirror.
If the light ray is absorbed, no image will be produced.
Answer:
66.2 sec
Explanation:
C₁ = 1.0 F
C₂ = 1.0 F
ΔV = Potential difference across the capacitor = 6.0 V
C = parallel combination of capacitors
Parallel combination of capacitors is given as
C = C₁ + C₂
C = 1.0 + 1.0
C = 2.0 F
R = resistance = 33 Ω
Time constant is given as
T = RC
T = 33 x 2
T = 66 sec
V₀ = initial potential difference across the combination = 6.0 Volts
V = final potential difference = 2.2 volts
Using the equation
t = 66.2 sec
