Answer:
T = 712.9 N
Explanation:
First, we will find the speed of the wave:
v = fλ
where,
v = speed of the wave = ?
f = frequency = 890 Hz
λ = wavelength = 0.1 m
Therefore,
v = (890 Hz)(0.1 m)
v = 89 m/s
Now, we will find the linear mass density of the wire:

where,
μ = linear mass density of wie = ?
m = mass of wire = 90 g = 0.09 kg
L = length of wire = 1 m
Therefore,

μ = 0.09 kg/m
Now, the tension in wire (T) will be:
T = μv² = (0.09 kg/m)(89 m/s)²
<u>T = 712.9 N</u>
Electric field = potential difference
-----------------------------
distance between plates
Distance between plates = 45
----------
500
= 0.09 meters.
Answer:
E. Student 1 is correct, because as θ is increased, h is the same.
Explanation:
Here we have the object of a certain mass falling under gravity so the force acting on the it will depend on mass of the object and the acceleration due to gravity.
Mathematically:

As we know that the work done is evaluated as the force applied on a body and the displacement of the body in the direction of the force.
And for work we have:

where:
displacement of the object
angle between the force and displacement vectors
Given that the height of the object is same in each trail of falling object under the gravity be it a free-fall or the incline plane.
- In case of free-fall the angle between the force is and the displacement is zero.
- In case when the body moves along the inclined plane the force applied by the gravity is same because it depends upon the mass of the object. And the net displacement in the direction of the gravitational force is the height of the object which is constant in both the cases.
So, the work done by the gravitational force is same in the two cases.
<span>160 Joules
For this problem, we can ignore the vertical component of the applied force and focus on only the horizontal component of 80 N and since work is defined as force over distance, let's multiply the force by the distance:
80 N * 2.0 m = 160 Nm = 160 kg*m^2/s^2 = 160 Joules.
So the cart has a final kinetic energy of 160 Joules.</span>
Answer:
110.87 dB
Explanation:
(I got it right on Acellus)
I= P/4(pi)r^2 = 60/4(pi)6.25^2
60/4(pi)6.25^2=0.12223
B=10log(I/Io)
B=10log(0.12223/1*10^-12) = 110.87 dB
111 in sigfigs