Calculate the volume of this regular solid,
1 answer:
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Answer:
question5: F=74312.5N
question6: charge at the end of antenna=0.37N
Explanation:
Coulomb's law: the magnitude of the force of attraction or repulsion due to two charges is proportional to the product of the magnitude of the charges and inversely proportional to the square of distance between the charges.
⇒
∴
where is the force of attraction or repulsion
is Coulumb's constant=
and
are the magnitude of the charges
is the distance between two charges
The force between the two charges is attractive if they are of different polarity
The force between the two charges is repulsive if they are of same polarity
Question5:
Given: q1=0.041 C, q2=0.029 C, r=12 m
therefore by Coulumb's law,
Question6:
Given: q1=, r=5 m, F=
therefore by Coulumb's law,
⇒
Answer:
51.82
Explanation:
First of all, let's convert both vectors to cartesian coordinates:
Va = 36 < 53° = (36*cos(53), 36*sin(53))
Va = (21.67, 28.75)
Vb = 47 < 157° = (47*cos(157), 47*sin(157))
Vb = (-43.26, 18.36)
The sum of both vectors will be:
Va+Vb = (-21.59, 47.11) Now we will calculate the module of this vector:
Answer:
T=1022.42 N
Explanation:
Given that
l = 32 cm ,μ = 1.5 g/cm
L =2 m ,V= 344 m/s
The pipe is closed so n= 3 ,for first over tone
f= 129 Hz
The tension in the string given as
T = f²(4l²) μ
Now by putting the values
T = f²(4l²) μ
T = 129² x (4 x 0.32²) x 1.5 x 10⁻³ x 100
T=1022.42 N