Answer:
6ms^-1
Explanation:
Given that the frequency difference is
( 563- 544) = 19
So alsoThe wavelength of each wave is = v/f = 344 /544
and there are 19 of this waves
So it is assumed that each motorcycle has moved 0.5 of this distance
in one second thus the speed of the motorcycles will be
=> 19/2 x 344/544 = 6.0 m/s
Answer:
d) 1/32 microgram
Explanation:
First half life is the time at which the concentration of the reactant reduced to half.
Second half reaction is the time at which the remaining concentration reduced to half or the initial concentration reduced to 1/4.
Third half life is the time at which the remaining concentration reduced to half or the initial concentration reduced to 1/8.
Forth half life is the time at which the remaining concentration reduced to half or the initial concentration reduced to 1/16.
Fifth half life is the time at which the remaining concentration reduced to half or the initial concentration reduced to 1/32.
The initial mass of the sample = 1 microgram
After 5 half-lives, the mass should reduce to 1/32 of the original.
So the concentration left = 1/32 of 1 microgram = 1/32 microgram
Answer:
The transverse wave will travel with a speed of 25.5 m/s along the cable.
Explanation:
let T = 2.96×10^4 N be the tension in in the steel cable, ρ = 7860 kg/m^3 is the density of the steel and A = 4.49×10^-3 m^2 be the cross-sectional area of the cable.
then, if V is the volume of the cable:
ρ = m/V
m = ρ×V
but V = A×L , where L is the length of the cable.
m = ρ×(A×L)
m/L = ρ×A
then the speed of the wave in the cable is given by:
v = √(T×L/m)
= √(T/A×ρ)
= √[2.96×10^4/(4.49×10^-3×7860)]
= 25.5 m/s
Therefore, the transverse wave will travel with a speed of 25.5 m/s along the cable.
Answer:
34.45m
Explanation:
Magnitude of a vector is equal to the square root of sum of squares of x & y vectors.
Magnitude = 
= 
=34.45m
Answer:
I = 1.875 A
Explanation:
For this exercise we use Ampere's law
∫ B . ds = μ₀ I
We use a circular path around the wire whereby B and ds are parallel, whereby the dot product is reduced to the algebraic product
ds = 2π dr
B (2πr) = μ₀ I
I = B 2π R /μ₀
r= 7.5 cm = 0.075 m
calculate
I = (50 μ₀ /π) 2π 0.075 /μ₀
I = 1.875 A
