Answer:
T = 764.41 N
Explanation:
In this case the tension of the string is determined by the centripetal force. The formula to calculate the centripetal force is given by:
(1)
m: mass object = 2.3 kg
r: radius of the circular orbit = 0.034 m
v: tangential speed of the object
However, it is necessary to calculate the velocity v first. To find v you use the formula for the kinetic energy:
You have the value of the kinetic energy (13.0 J), then, you replace the values of K and m, and solve for v^2:
you replace this value of v in the equation (1). Also, you replace the values of r and m:
hence, the tension in the string must be T = Fc = 764.41 N
Answer:
Speed at which the ball passes the window’s top = 10.89 m/s
Explanation:
Height of window = 3.3 m
Time took to cover window = 0.27 s
Initial velocity, u = 0m/s
We have equation of motion s = ut + 0.5at²
For the top of window (position A)
For the bottom of window (position B)
We also have
Solving
So after 1.11 seconds ball reaches at top of window,
We have equation of motion v = u + at
Speed at which the ball passes the window’s top = 10.89 m/s
Answer:
The drift speed of the electrons in the wire is 2.12x10⁻⁴ m/s.
Explanation:
We can find the drift speed by using the following equation:
Where:
I: is the current = 4.50 A
n: is the number of electrons
q: is the modulus of the electron's charge = 1.6x10⁻¹⁹ C
A: is the cross-sectional area = 2.20x10⁻⁶ m²
We need to find the number of electrons:
Now, we can find the drift speed:
Therefore, the drift speed of the electrons in the wire is 2.12x10⁻⁴ m/s.
I hope it helps you!
Answer:
The answer is a for Plato users.
Explanation:
Since the angle of the refracted ray moves away from the normal, it must be traveling in a faster medium.
Answer:
a) the maximum transverse speed of a point on the string at an antinode is 5.9899 m/s
b) the maximum transverse speed of a point on the string at x = 0.075 m is 4.2338 m/s
Explanation:
Given the data in the question;
as the equation of standing wave on a string is fixed at both ends
y = 2AsinKx cosωt
but k = 2π/λ and ω = 2πf
λ = 4 × 0.150 = 0.6 m
and f = v/λ = 260 / 0.6 = 433.33 Hz
ω = 2πf = 2π × 433.33 = 2722.69
given that A = 2.20 mm = 2.2×10⁻³
so 
= A × ω
= 2.2×10⁻³ × 2722.69 m/s
= 5.9899 m/s
therefore, the maximum transverse speed of a point on the string at an antinode is 5.9899 m/s
b)
A' = 2AsinKx
= 2.20sin( 2π/0.6 ( 0.075) rad )
= 2.20 sin( 0.7853 rad ) mm
= 2.20 × 0.706825 mm
A' = 1.555 mm = 1.555×10⁻³
so
= A' × ω
= 1.555×10⁻³ × 2722.69
= 4.2338 m/s
Therefore, the maximum transverse speed of a point on the string at x = 0.075 m is 4.2338 m/s
