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QveST [7]
2 years ago
9

Peak to peak value of voltage is

Physics
2 answers:
Mnenie [13.5K]2 years ago
6 0
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hram777 [196]2 years ago
3 0

Answer: Peak-to-peak voltage is the distance from the lowest negative amplitude, or trough, to the highest positive amplitude, or crest, of the AC voltage waveform. In other words, peak-to-peak voltage is equal to the full height of the waveform. Peak-to-peak voltage can be found using peak voltage or RMS voltage.

Explanation:hope tht gave u a clue have a wonderful Christmas Eve time with your family!❄️

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Two long, straight parallel wires are placed 38 cm apart, one above the other. The top and bottom wires are carrying currents 4.
ElenaW [278]

Answer:

The force per unit length (N/m) on the top wire is 16.842 N/m

Explanation:

Given;

distance between the two parallel wire, d = 38 cm = 0.38 m

current in the first wire, I₁ = 4.0 kA

current in the second wire, I₂ = 8.0 kA

Force per unit length, between two parallel wires is given as;

\frac{F}{L} = \frac{\mu_oI_1I_2 }{2\pi d }

where;

μ₀ is constant = 4π x 10⁻⁷ T.m/A

Substitute the given values in the above equation and calculate the force per unit length

\frac{F}{L} = \frac{\mu_oI_1I_2 }{2\pi d } = \frac{4\pi *10^{-7}*4000*8000 }{2\pi *0.38} = 16.842 \ N/m

Therefore, the force per unit length (N/m) on the top wire is 16.842 N/m

4 0
3 years ago
A group of students prepare for a robotic competition and build a robot that can launch large spheres of mass M in the horizonta
jeyben [28]

Answer:

V_0

Explanation:

Given that, the range covered by the sphere, M, when released by the robot from the height, H, with the horizontal speed V_0 is D as shown in the figure.

The initial velocity in the vertical direction is 0.

Let g be the acceleration due to gravity, which always acts vertically downwards, so, it will not change the horizontal direction of the speed, i.e. V_0 will remain constant throughout the projectile motion.

So, if the time of flight is t, then

D=V_0t\; \cdots (i)

Now, from the equation of motion

s=ut+\frac 1 2 at^2\;\cdots (ii)

Where s is the displacement is the direction of force, u is the initial velocity, a is the constant acceleration and t is time.

Here, s= -H, u=0, and a=-g (negative sign is for taking the sigh convention positive in +y direction as shown in the figure.)

So, from equation (ii),

-H=0\times t +\frac 1 2 (-g)t^2

\Rightarrow H=\frac 1 2 gt^2

\Rightarrow t=\sqrt {\frac {2H}{g}}\;\cdots (iii)

Similarly, for the launched height 2H, the new time of flight, t', is

t'=\sqrt {\frac {4H}{g}}

From equation (iii), we have

\Rightarrow t'=\sqrt 2 t\;\cdots (iv)

Now, the spheres may be launched at speed V_0 or 2V_0.

Let, the distance covered in the x-direction be D_1 for V_0 and D_2 for 2V_0, we have

D_1=V_0t'

D_1=V_0\times \sqrt 2 t [from equation (iv)]

\Rightarrow D_1=\sqrt 2 D [from equation (i)]

\Rightarrow D_1=1.41 D (approximately)

This is in the 3 points range as given in the figure.

Similarly, D_2=2V_0t'

D_2=2V_0\times \sqrt 2 t [from equation (iv)]

\Rightarrow D_2=2\sqrt 2 D [from equation (i)]

\Rightarrow D_2=2.82 D (approximately)

This is out of range, so there is no point for 2V_0.

Hence, students must choose the speed V_0 to launch the sphere to get the maximum number of points.

7 0
3 years ago
Is this right? plzz anwser soon
-Dominant- [34]

Answer:

yes

Explanation:

6 0
3 years ago
Read 2 more answers
In which situation is a person doing work on an object?
Marianna [84]

Answer : When a person applies force and covers distance then the person is doing work on an object.

6 0
3 years ago
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If you move 50 meters in 10 seconds, what is your speed ​
iogann1982 [59]

Explanation:

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8 0
3 years ago
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