Answer:
0.8 seconds
Explanation:
F=ma
Let x be the seconds the force is applied.
m = 20kg
F = 50 Newtons (kg*m/sec^2)
acceleration, a, is provided for x seconds to increase the speed from 1 m/s to 3 m/s, an increase of 2m/s
Let's calculate the acceleration of the cart:
F=ma
(50 kg*m/s^2) = (20kg)*a
a = 2.5 m/s^2
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The acceleration is 2.5 m/s^2. The cart increases speed by 2.5 m/s every second.
We want the number of seconds it takes to add 2.0 m/sec to the speed:
(2.5 m/s^2)*x = 2.0 m/s
x = (2.0/2.5) sec
x = 0.8 seconds
Answer:
ΔX = λ = 0.68 m
Explanation:
Wave speed is related to wavelength and frequency by the equation
v = λ f
where the speed of sound is 340 m / s
λ = v / f
λ = 340/500
λ = 0.68 m
this is the wavelength, it is the minimum distance for which the wave epitates its movement, which is equal to the distance between two consecutive compressions of the sound
ΔX = λ = 0.68 m
Answer:
the electroscope separate by the presence of charge carriers
Explanation:
Metal bodies are characterized by having free (mobile) electrons. In the electroscope the plates are in balance; when the external metal ball is touched, a charge is introduced into the device, when the body that touched the ball is separated, an excess charge remains. This charge, being a metal, is distributed over the entire surface, giving a uniform density and an electric force of repulsion is created between the two charged sheets, which tends to separate the sheets. This force is counteracted by the tension component as the sheets are separated at a given angle, the separation reaches the point where
Fe - Tx = 0
Fe = Tx
In summary, the electroscope separate its leaves by the presence of charge carriers
Answer
given,
mass of copper rod = 1 kg
horizontal rails = 1 m
Current (I) = 50 A
coefficient of static friction = 0.6
magnetic force acting on a current carrying wire is
F = B i L
Rod is not necessarily vertical
the normal reaction N = mg-F y
static friction f = μ_s (mg-F y )
horizontal acceleration is zero
B_w = B sinθ
B_d = B cosθ
iLB cosθ= μ_s (mg- iLB sinθ)
B = 0.1 T
Raising the temperature results in the radiator giving off photons of high-energy ultraviolet light. As heat is added, the radiator emits photons across a wide range of visible-light frequencies
E
(9.35) = 1057.46 lumens (lm)