Answer:
(i) The wavelength is 0.985 m
(ii) The frequency of the wave is 36.84 Hz
Explanation:
Given;
mass of the string, m = 0.0133 kg
tensional force on the string, T = 8.89 N
length of the string, L = 1.97 m
Velocity of the wave is:
(i) The wavelength:
Fourth harmonic of a string with two nodes, the wavelength is given as,
L = 2λ
λ = L/2
λ = 1.97 / 2
λ = 0.985 m
(ii) Frequency of the wave is:
v = fλ
f = v / λ
f = 36.29 / 0.985
f = 36.84 Hz
Answer:
The work done by friction was
Explanation:
Given that,
Mass of car = 1000 kg
Initial speed of car =108 km/h =30 m/s
When the car is stop by brakes.
Then, final speed of car will be zero.
We need to calculate the work done by friction
Using formula of work done
Put the value of m and v
Hence, The work done by friction was
Answer:
Below
Explanation:
First draw the vectors that represent both electric fields.
E1 is the elictric field created by q1, E2 is the one created by q2.
● q1 is negative so E1 will point from P.
● q2 is positive so E2 will point out of P
(Picture below)
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The resulting electric field is equal to the sum of the two fields since both vectors are colinear.
Let E be the total field.
● E = E1 + E2
The formula of the electric field intensity is:
● E = K ×(q/d^2)
-K is Coulomb's constant
-d is the distance between the charge and the object ( here P)
-q is the charge
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● E1 = K × (q1/d1^2)
The distance between q1 and P is the qum of 0.15 m 0.25 m. (0.4 m)
Coulombs constant is 9×10^9 m^2/C^2
● E1 = 9×10^9 ×[-6.39 × 10^(-9)/ 0.4^2]
● E1 = -359.43 N/C
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● E2 = K ×(q2/d^2)
The distance between q2 and P is 0.25 m.
● E2 = 9×10^9×[3.22×10^(-9) /0.25^2]
● E2 = 463.68 N/C
■■■■■■■■■■■■■■■■■■■■■■■■■■
● E = E1 + E2
● E = -359.43+463.68
● E = 105.25 N/C
