Answer:
f1 = 12.90 Hz
Explanation:
To calculate the first harmonic frequency you use the following formula for n = 1:
( 1 )
It is necessary that the unist are in meters, then you have:
L: length of the string = 60cm = 0.6m
M: mass of the string = 0.05kg
T: tension on the string = 20 N
you replace the values of L, M and T in the expression (1) for getting f1:
Hence, the first harmonic has a frequency of 12.90 Hz
Answer:
The maximum radius the asteroid can have for her to be able to leave it entirely simply by jumping straight up is approximately 1782.45 meters
Explanation:
Whereby the height the astronaut can jump on Earth = 0.500 m, we have the following kinematic equation;
v² = u² - 2·g·h
Where;
v = The final velocity
u = The initial velocity
g = The acceleration due to gravity ≈ 9.8 m/s²
h = The height she jumps
At the maximum height, = 0.500 m, she jumps, v = 0, therefore, we have;
0² = u² - 2·g·
u² = 2 × 9.8 × 0.5 = 9.8
u = √9.8 ≈ 3.13
u = 3.13 m/s
Her initial jumping velocity ≈ 3.13 m/s
Escape velocity,
Where;
M = The mass of the asteroid
G = The Universal gravitational constant = 6.67408 × 10⁻¹¹ m³/(kg·s²)
r = The radius of the asteroid
The average density of the Earth = 5515 kg/m³
The mass of the asteroid, M = Density × Volume = 5515 kg/m³× 4/3 × π × r³
The escape velocity, she has, ≈ 3.13 m/s is therefore;
Therefore, the maximum radius of the asteroid can have for her jumping velocity to be equal to the escape velocity for her to be able to leave it entirely simply by jumping straight up = r ≈ 1782.45 meters.
Answer:
B=9.1397*10^-4 Tesla
Explanation:
To find the velocity first we put kinetic energy og electron is equal to potential energy of electron
K.E=P.E
where :
m is the mass of electron
v is the velocity
V is the potential difference
eq 1
Radius of electron moving in magnetic field is given by:
eq 2
where:
m is the mass of electron
v is the velocity
q=e=charge of electron
B is the magnitude of magnetic field
Put v from eq 1 into eq 2
B=9.1397*10^-4 Tesla