Answer:
1+1=2
Explanation:
You just need to add and voila the answer
Answer:
C
Explanation:
Because everything on Earth falls at the same speed, the masses of the balls do not matter. Since the acceleration due to gravity is constant, their speeds will both be increasing at the same rate, and therefore the difference in speeds would remain constant until they hit the ground. Hope this helps!
Answer: a) 16Hz, 3m b) 48Hz, 1mc) 80Hz, 0.6m
Explanation:
a) Fundamental frequency in string is represented as Fo = V/2L where;
Fo is the fundamental frequency
V is the speed of the transverse wave = 48m/s
L is the length of the wire. = 1.50m
Substituting this values in the formula given we have;
Fo = 48/2(1.5)
Fo = 48/3
Fo = 16Hz
The fundamental tone is therefore 16Hz
Using v =f¶
Where f is the frequency and ¶ is the wavelength, the wavelength of the fundamental note will be;
¶ = v/fo
¶ = 48/16 = 3m
b) Overtones or harmonics is the multiple integral of the fundamental frequency. The multiples are I'm arithmetical progression.
First overtone f1 = 2fo
Second overtone f2 = 3fo etc.
Since fo = 16Hz
Second overtone f2 = 3×16 = 48Hz
¶ = v/f2 = 48/48
¶ = 1m
c) Fourth harmonic or overtone will be f4 = 5fo
F4 = 5×16 = 80Hz
The fourth harmonic is therefore 80Hz
¶ = v/f4 = 48/80
¶ = 0.6m
Answer:
9) a = 25 [m/s^2], t = 4 [s]
10) a = 0.0875 [m/s^2], t = 34.3 [s]
11) t = 32 [s]
Explanation:
To solve this problem we must use kinematics equations. In this way we have:
9)
a)

where:
Vf = final velocity = 0
Vi = initial velocity = 100 [m/s]
a = acceleration [m/s^2]
x = distance = 200 [m]
Note: the final speed is zero, as the car stops completely when it stops. The negative sign of the equation means that the car loses speed or slows down as it stops.
0 = (100)^2 - (2*a*200)
a = 25 [m/s^2]
b)
Now using the following equation:

0 = 100 - (25*t)
t = 4 [s]
10)
a)
To solve this problem we must use kinematics equations. In this way we have:

Note: The positive sign of the equation means that the car increases his speed.
5^2 = 2^2 + 2*a*(125 - 5)
25 - 4 = 2*a* (120)
a = 0.0875 [m/s^2]
b)
Now using the following equation:

5 = 2 + 0.0875*t
3 = 0.0875*t
t = 34.3 [s]
11)
To solve this problem we must use kinematics equations. In this way we have:

10^2 = 2^2 + 2*a*(200 - 10)
100 - 4 = 2*a* (190)
a = 0.25 [m/s^2]
Now using the following equation:

10 = 2 + 0.25*t
8 = 0.25*t
t = 32 [s]
Given Information:
Current = I = 20 A
Diameter = d = 0.205 cm = 0.00205 m
Length of wire = L = 1 m
Required Information:
Energy produced = P = ?
Answer:
P = 2.03 J/s
Explanation:
We know that power required in a wire is
P = I²R
and R = ρL/A
Where ρ is the resistivity of the copper wire 1.68x10⁻⁸ Ω.m
L is the length of the wire and A is the area of the cross-section and is given by
A = πr²
A = π(d/2)²
A = π(0.00205/2)²
A = 3.3x10⁻⁶ m²
R = ρL/A
R = 1.68x10⁻⁸*(1)/3.3x10⁻⁶
R = 5.09x10⁻³ Ω
P = I²R
P = (20)²*5.09x10⁻³
P = 2.03 Watts or P = 2.03 J/s
Therefore, 2.03 J/s of energy is produced in 1.00 m of 12-gauge copper wire carrying a current of 20 A