Answer:
The answer is below
Explanation:
Given that:
mass (m) = 86 kg, distance (L) = 2.75 m, θ = 31°, force (F) = 595 N, initial velocity (
) = 2.4 m/s, g = acceleration due to gravity = 9.8 m/s²
The net work can be gotten from the equation:

From the work-energy theorem equation, we can get her speed at the top of the ramp (
)
Hence:

Answer:
Radius of the solenoid is 0.93 meters.
Explanation:
It is given that,
The magnetic field strength within the solenoid is given by the equation,
, t is time in seconds

The induced electric field outside the solenoid is 1.1 V/m at a distance of 2.0 m from the axis of the solenoid, x = 2 m
The electric field due to changing magnetic field is given by :

x is the distance from the axis of the solenoid
, r is the radius of the solenoid


r = 0.93 meters
So, the radius of the solenoid is 0.93 meters. Hence, this is the required solution.
Answer:
99.95%
Explanation:
A double pulsar system named PSR J0737-3039A/B in Puppis constellation was discovered in the year 2003. Pulsars are second most densest object in the universe after black holes and they emit radio waves at regular intervals. This pair presented a great and natural setup to test the Theory of General Relativity presented by Einstein in 1915. In this theory Einstein had presented a set of equations on how the space-time fabric will be curved because of the very dense objects such as Neutron stars. It also predicted how the gravitational waves are created because of stars orbiting each other.
A team of astrophysicists led by Michael Kramer, conducted a study on how these gravitational waves will impact the time in which the radio waves emitted by pulsars will reach Earth. The result of the study proved the theory of General Relativity to be accurate up to 99.95%.
Answer:
w = √ 1 / CL
This does not violate energy conservation because the voltage of the power source is equal to the voltage drop in the resistence
Explanation:
This problem refers to electrical circuits, the circuits where this phenomenon occurs are series RLC circuits, where the resistor, the capacitor and the inductance are placed in series.
In these circuits the impedance is
X = √ (R² + (
-
)² )
where Xc and XL is the capacitive and inductive impedance, respectively
X_{C} = 1 / wC
X_{L} = wL
From this expression we can see that for the resonance frequency
X_{C} = X_{L}
the impedance of the circuit is minimal, therefore the current and voltage are maximum and an increase in signal intensity is observed.
This does not violate energy conservation because the voltage of the power source is equal to the voltage drop in the resistence
V = IR
Since the contribution of the two other components is canceled, this occurs for
X_{C} = X_{L}
1 / wC = w L
w = √ 1 / CL
1 watt = 1 joule/sec
2,000 watts = 2,000 joules/sec
(2,000 joule/sec) x (120 sec)
= (2,000 x 120) (joule-sec/sec)
= 240,000 joules .