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allsm [11]
3 years ago
7

What variables should you keep constant in initial speed

Physics
1 answer:
Tanya [424]3 years ago
6 0
Time and distance I think
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Please help me solve all of them ( a, b, c and d ) thankiew !! <br> I’m also kind of in a rush
Sergio039 [100]

Answer:

a-

V= IR

9V = I ×( 12+6)

I = 9/ 18 A = 0.5 A

b

V=IR

240 = 6 A ×( 20 + R)

40 = 20 + R

R = 20 ohm

c

resultant resistance of the 2 parallel resistances= Ro

1/Ro = 1/ 5 + 1/ 20

1/Ro =( 20+5)/100

= 1/Ro = 1/4

Ro= 4 ohm

V=IR

V = 2A × ( 1+ 4 OHM)

V = 10V

d

equivalent resistance = Ro

1/Ro = 1/(2+8) + 1/(5+5)

1/Ro = 1/10 +1/10

2/10 = 1/ Ro

Ro= 10/2 = 5 ohm

V = IR

12V = I × 5Ohm

I=2.4 A

6 0
2 years ago
Lc circuit consists of a 20.0 mh inductor and a 0.150 µf capacitor. if the maximum instantaneous current is 0.400 a, what is the
slavikrds [6]
The following expression is applicable:

Max. inductor energy = Max. capacitor energy

Where;
Max. inductor energy = LI^2/2, with L = 20.0 mH, I = 0.400 A
Max. capacitor energy = CV_max^2/2, C = 0.150 micro Faraday, V_max = Max. potential difference

Substituting;
LI^2/2 = CV^2/2
LI^2 = CV^2
V^2 = (LI^2)/C
V_max = Sqrt [(LI^2)/C] = Sqrt [(20*10^-3*0.4^2)/(0.15*10^-6)] = 146.06 V
5 0
3 years ago
Suppose I produce radio waves with an antenna that have a peak electric field amplitude E and peak magnetic field amplitude B. I
denis23 [38]

Answer:

Correct answer is 2B

Explanation:

The electric field magnitude and magnetic field magnitude in an electromagnetic waves are related as under

E=cB

where 'c' is velocity of light in the medium of transmission

According to the given question if we double the electric field we have

2E=cB'\\\\2cB=cB'\\\\B'=2B

Thus the magnetic field also doubles

3 0
3 years ago
Which of the following is NOT a force that influences the wind? A. coriolis effect B. magnetic field C. pressure gradient D. fri
Pavlova-9 [17]
A magnetic field does not have any influence on wind. The Coriolis effect, pressure gradient, and friction do affect wind.

The solution is B.
7 0
4 years ago
Read 2 more answers
Determine the centripetal force upon a 40-kg child who makes 10 revolution around the cliffhanger in 29.3 seconds.the radius of
zysi [14]

Answer:

The centripetal force acting on the child is 39400.56 N.

Explanation:

Given:

Mass of the child is, m=40\ kg

Radius of the barrel is, R=2.90\ m

Number of revolutions are, n =10

Time taken for 10 revolutions is, t=29.3\ s

Therefore, the time period of the child is given as:

T=\frac{n}{t}=\frac{10}{29.3}=0.341\ s

Now, angular velocity is related to time period as:

\omega=\frac{2\pi}{T}=\frac{2\pi}{0.341}=18.43\ rad/s

Now, centripetal force acting on the child is given as:

F_{c}=m\omega^2 R\\F_{c}=40\times (18.43)^2\times 2.90\\F_{c}=40\times 339.66\times 2.90\\F_{c}=39400.56\ N

Therefore, the centripetal force acting on the child is 39400.56 N.

8 0
3 years ago
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