Explanation:
(a) We have,
Length of solenoid, l = 55 cm = 0.55 m
Diameter of the solenoid, d = 10 cm
Radius, r = 5 cm = 0.05 m
Number of loops in the solenoid is 1000.
(a) The self inductance in the solenoid is given by :
A is area
(b) The energy stored in the inductor is given by :
Hence, this is the required solution.
7. solar flare: f.
8. core: h.
9. chromosphere: b.
10. sunspot: d.
11. corona: c.
12. nuclear fusion: j.
13. photosphere: a.
14. solar wind: g.
15. prominence: e.
16. radiation zone: k.
17. convection zone: i.
The average force on the ball by the racket is 98 N. The correct option is the third option - 98 N
From the question, we are to determine the average force on the ball by the racket.
From the formula,
Where F is the force
m is the mass
v is the velocity
and t is the time
From the given information
m = 0.07 kg
v = 56 m/s
t = 0.04 s
Putting the parameters into the formula,
we get
F = 98 N
Hence, the average force on the ball by the racket is 98 N. The correct option is the third option - 98 N
Learn more on calculating force exerted on an object here: brainly.com/question/13590154
Answer:
a = 1.152s
b = 0.817 m
c = 7.29m/s
Explanation: let the following
From the first equation of linear motion
V = u+at..........1
parameters be represented as :
t = Time taken
v = Final velocity
a = Acceleration due to gravity = 9.8m/s²
u = Initial velocity = 4 m/s
s = Displacement
V = 0
Substitute the values into equation 1
0 = 4-9.8(t)
-4 = -9.8t
t = 4/9.8
t = 0.408s
From : s = ut+1/2at^2.........2
S = 4×0.408+0.5(-9.8)×0.408^2
S= 1.632-4.9(0.166)
S = 1.632-0.815
S = 0.817m
Her highest height above the board is 0.817 m
Total height she would fall is 0.817+1.90 = 2.717 m
From equation 2
s = ut+1/2at^2
2.717 m = 0t+0.5(9.8)t^2
2.717 m = 0+4.9t^2
2.717 m = 4.9t^2
2.717/4.9 = t^2
0.554 =t^2
t =√0.554
t = 0.744s
Hence, her feet were in the air for 0.744+0.408seconds
= 1.152s
Also recall from equation 1
V= u+at
V = 0+9.8(0.744)
V = 7.29m/s
Hence, the velocity when she hits the water is 7.29m/s
Finally,
a = 1.152s
b = 0.817 m
c = 7.29m/s
(a) The required magnitude of the electric field when the point charge is an electron is 5.57 x 10⁻¹¹ N/C.
(b) The required magnitude of the electric field when the point charge is an proton is 1.02 x 10⁻⁷ N/C.
<h3>
Magnitude of electric field </h3>
The magnitude of electric field is given by the following equation.
F = qE
But F = mg
mg = qE
E = mg/q
where;
- E is the electric field
- m is mass of the particle
- g is acceleration due to gravity
- q is charge of the particle
<h3>For an electron</h3>
E = (9.11 x 10⁻³¹ x 9.8)/(1.602 x 10⁻¹⁹)
E = 5.57 x 10⁻¹¹ N/C
<h3>For proton</h3>
E = (1.67 x 10⁻²⁷ x 9.8)/(1.602 x 10⁻¹⁹)
E = 1.02 x 10⁻⁷ N/C
Thus, the required vertical electric field is greater when the charge is proton.
Learn more about electric field here: brainly.com/question/14372859
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