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mel-nik [20]
3 years ago
7

Why can't we implement the center tapped full wave rectifier without center-tapped transformer

Physics
1 answer:
maksim [4K]3 years ago
6 0
For a full wave bridge you don't want a center tap
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A projectile is shot at an angle 45 degrees to the horizontalnear the surface of the earth but in the absence of air resistance.
juin [17]

Answer:

Explanation:

Given

For first case

launch angle \theta =45^{\circ}C

at highest point h=150 m/s

150=u\cos 45

u=\frac{150}{\cos 45}=212.132 m/s

For second case

\theta _2=37^{\circ}C

at highest Point velocity is u\cos \theta _2

=212.132\times \cos 37

=169.41 m/s

as there is no acceleration in x direction therefore horizontal velocity is same          

7 0
3 years ago
Read 2 more answers
An electron travels with speed 6.0×10^6 m/s between the two parallel charged plates shown in the figure. The plates are separate
svlad2 [7]

Answer:

B= 3.33 m T

Explanation:

Given that

Speed ,C= 6 x 10⁶  m/s

d= 1 cm = 0.01 m

V= 200 V

The electric field E given as

V= E .d

E=Electric field

d=Distance

V=Voltage

200 = 0.01 x E

E=20000 V/m

The relationship between magnetic and electric field given as

E= C x B

20000 = 6 x 10⁶   x B

B =3333.333 x 10⁻⁶ T

B= 3.33 x 10⁻³ T

B= 3.33 m T

Therefore the magnetic filed will be 3.33 m T.

5 0
2 years ago
A student walks 4 block east, 7 blocks west, 1 blocks east then 2 blocks west in an hour her average speed is____ block/hour
S_A_V [24]

Answer:

2 m/s

Explanation:

The total time = 1 hour

The vertical displacement = 1 - 1

Vertical displacement = 0

Horizontal displacement = 4 - 2

Horizontal displacement = 2

Total displacement = sqrt (2^2 - 0^2)

Displacement - 2

Average velocity is displacement/time

= 2x1

=  2 m/s

The average velocity is 2 metres per second.

7 0
3 years ago
A person is pushing a box. The net external force on the 60-kg box is stated to be 90 N. If the force of friction opposing the m
Tems11 [23]

Answer:

b.1.5m/s^2

Explanation:

We are given that

Mass of box=60kg

Net external force applied on the box=90 N

Friction force =30N

We have to find the acceleration of the box.

We know that

Net external force=ma

Substitute the values then we get

90=60a

a=\frac{90}{60}=1.5m/s^2

Hence, option b is true.

5 0
2 years ago
How fast will it be traveling after it goes 87 m ?
swat32

To solve this we can use one of the main four kinematic equations, preferably we want something that we can use directly without solving for other values.

So in this problem we know that:

Inital velocity or Vi = 0 (started from rest)

The acceleration of the object = 7.5

The distance the object travels Δx = 87m

So, our best equation here with what we have would be:

Vf² = Vi² + 2aΔx

Now we now that Vi is 0 so:

Vf² = 2aΔx

We need final velocity so lets take the square root of Vf to isolate it:

Vf = √2aΔx

Now we plug in

Vf = √2(7.5)(87)

Vf = 36.124 m/s or 36.1 m/s

6 0
2 years ago
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