What is all matter made up of?
2 answers:
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Explanation:
It is given that,
Frequency of vibration, f = 215 Hz
Amplitude, A = 0.832 mm
(a) Let T is the period of this motion. It is given by the following relation as :
(b) Speed of sound in air, v = 343 m/s
It can be given by :
Hence, this is the required solution.
Answer:
Distance between peak height (vertically) of projectile and mountain height = (2975.2 - 1800) = 1175.2 m
Distance between where the projectile lands and ship B = (3188.8 - 3110) = 8.8 m
Explanation:
Given the velocity and angle of shot of the projectile, one can calculate the range and maximum height attained by the projectile.
H = (v₀² Sin²θ)/2g
v₀ = initial velocity of projectile = 2.50 × 10² m/s = 250 m/s
θ = 75°, g = 9.8 m/s²
H = 250² (Sin² 75)/(2 × 9.8) = 2975.2 m
Range of projectile
R = v₀² (sin2θ)/g
R = 250² (sin2×75)/9.8
R = 250² (sin 150)/9.8 = 3188.8 m
Height of mountain = 1.80 × 10³ = 1800 m
Maximum height of projectile = 2975.2 m
Distance between peak height (vertically) of projectile and mountain height = 2975.2 - 1800 = 1175.2 m
Distance of ship B from ship A = 2.5 × 10³ + 6.1 × 10² = 2500 + 610 = 3110 m
Range of projectile = 3188.8 m
Distance between where the projectile lands and ship B = 3188.8 - 3110 = 8.8 m
Answer:
(a)Magnitude of the electric field at 0 cm from the center of the sphere:
E=0
(b) Magnitude of the electric field at 10.0 cm from the center of the sphere :
E= 244.7 KN/C
(c) Magnitude of the electric field at 40.0 cm from the center of the sphere :
E=978.8 KN/C
(d) Magnitude of the electric field at 71 cm from the center of the sphere :
E=310.7 KN/C
Explanation:
If we have a uniform charge sphere we can use the following formulas to calculate the Electric field due to the charge of the sphere:
: Formula (1) To calculate the electric field in the region outside the sphere r ≥ a
:Formula (2) To calculate the electric field in the inner region of the sphere. r ≤ a
Where:
K: coulomb constant
a: sphere radius
Q: Total sphere charge
r : Distance from the center of the sphere to the region where the electric field is calculated
Equivalences
1μC=10⁻⁶C
1cm= 10⁻²m
Data
k= 9*10⁹ N*m²/C²
Q=17.4 μC=17.4 *10⁻⁶C
a= 40 cm = 40*10⁻²m = 0.4m
Problem development
(a)Magnitude of the electric field at 0 cm :
We replace r=0 in the formula (2) , then, E=0
(b) Magnitude of the electric field at 10.0 cm from the center of the sphere
r<a , We apply the Formula (2):
E= 244.7 KN/C
(c) Magnitude of the electric field at 40.0 cm from the center of the sphere
r=a, We apply the Formula (1) :
E=978.8 KN/C
(d) Magnitude of the electric field at 71 cm from the center of the sphere
r>a , We apply the Formula (1) :
E=310.7 KN/C