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sveta [45]
3 years ago
8

In the diagram, each charge is

Physics
1 answer:
Jlenok [28]3 years ago
3 0

Answer:

Quick maths

First you find the fafarick and the lalickc and the caprisum and the joinnt

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How long will it take a plane to fly 1256km<br> if it travels 500km/hr?
expeople1 [14]

Answer:

<h3>The answer is 2.51 s</h3>

Explanation:

The time taken can be found by using the formula

t =  \frac{d}{v}  \\

d is the distance

v is the velocity

From the question we have

t =  \frac{1256}{500}  \\  = 2.512

We have the final answer as

<h3>2.51 s</h3>

Hope this helps you

4 0
3 years ago
I dnt know how to do it
NNADVOKAT [17]

Here's what you need to know about waves:

Wavelength = (speed) / (frequency)

Now ... The question gives you the speed and the frequency,
but they're stated in unusual ways, with complicated numbers.

Frequency:  How many each second ?
The thing that's making the waves is vibrating 47 times in 26.9 seconds.
Frequency = (47) / (46.9 s) =  1.747... per second.  (1.747... Hz)

Speed:  How far a point on a wave travels in 1 second.
The crest of one wave travels 4.16 meters in 13.7 seconds.
Speed = (4.16 m / 13.7 sec) = 0.304... m/s

Wavelength = (speed) / (frequency)

Wavelength = (0.304 m/s) / (1.747 Hz)  =  0.174 meter per second


4 0
3 years ago
Two lightbulbs both operate on 120V . One has a power of 25W and the other 100W. (ii) Which lightbulb carries more current? Choo
Vikki [24]

The lightbulb that carries more current will be the 25W bulb.

<h3>How to explain the information?</h3>

It should be noted that an electric current simply means the stream of charged particles that move through an electrical conductor or space.

In this case, it should be noted that the power is the same for both bulbs. Therefore, the 25W bulb will have the higher resistance so that it will have lower power.

Therefore, the lightbulb that carries more current will be the 25W bulb.

Learn more about current on:

brainly.com/question/1100341

#SPJ4

7 0
1 year ago
One of the waste products of a nuclear reactor is plutonium-239 . This nucleus is radioactive and decays by splitting into a hel
Gekata [30.6K]

Answer:

a) v_{U-235} = 2.68 \cdot 10^{5} m/s

v_{He-4} = -1.57 \cdot 10^{7} m/s  

b) E_{He-4} = 8.23 \cdot 10^{-13} J

E_{U-235} = 1.41 \cdot 10^{-14} J

 

Explanation:

Searching the missed information we have:                                        

E: is the energy emitted in the plutonium decay = 8.40x10⁻¹³ J

m(⁴He): is the mass of the helium nucleus = 6.68x10⁻²⁷ kg  

m(²³⁵U): is the mass of the helium U-235 nucleus = 3.92x10⁻²⁵ kg            

a) We can find the velocities of the two nuclei by conservation of linear momentum and kinetic energy:

Linear momentum:

p_{i} = p_{f}

m_{Pu-239}v_{Pu-239} = m_{He-4}v_{He-4} + m_{U-235}v_{U-235}

Since the plutonium nucleus is originally at rest, v_{Pu-239} = 0:

0 = m_{He-4}v_{He-4} + m_{U-235}v_{U-235}  

v_{He-4} = -\frac{m_{U-235}v_{U-235}}{m_{He-4}}    (1)

Kinetic Energy:

E_{Pu-239} = \frac{1}{2}m_{He-4}v_{He-4}^{2} + \frac{1}{2}m_{U-235}v_{U-235}^{2}

2*8.40 \cdot 10^{-13} J = m_{He-4}v_{He-4}^{2} + m_{U-235}v_{U-235}^{2}    

1.68\cdot 10^{-12} J = m_{He-4}v_{He-4}^{2} + m_{U-235}v_{U-235}^{2}   (2)    

By entering equation (1) into (2) we have:

1.68\cdot 10^{-12} J = m_{He-4}(-\frac{m_{U-235}v_{U-235}}{m_{He-4}})^{2} + m_{U-235}v_{U-235}^{2}  

1.68\cdot 10^{-12} J = 6.68 \cdot 10^{-27} kg*(-\frac{3.92 \cdot 10^{-25} kg*v_{U-235}}{6.68 \cdot 10^{-27} kg})^{2} +3.92 \cdot 10^{-25} kg*v_{U-235}^{2}  

Solving the above equation for v_{U-235} we have:

v_{U-235} = 2.68 \cdot 10^{5} m/s

And by entering that value into equation (1):

v_{He-4} = -\frac{3.92 \cdot 10^{-25} kg*2.68 \cdot 10^{5} m/s}{6.68 \cdot 10^{-27} kg} = -1.57 \cdot 10^{7} m/s                        

The minus sign means that the helium-4 nucleus is moving in the opposite direction to the uranium-235 nucleus.

b) Now, the kinetic energy of each nucleus is:

For He-4:

E_{He-4} = \frac{1}{2}m_{He-4}v_{He-4}^{2} = \frac{1}{2} 6.68 \cdot 10^{-27} kg*(-1.57 \cdot 10^{7} m/s)^{2} = 8.23 \cdot 10^{-13} J

For U-235:

E_{U-235} = \frac{1}{2}m_{U-235}v_{U-235}^{2} = \frac{1}{2} 3.92 \cdot 10^{-25} kg*(2.68 \cdot 10^{5} m/s)^{2} = 1.41 \cdot 10^{-14} J

 

I hope it helps you!                                                                                    

3 0
3 years ago
1. An EM wave representing 650 nm laser light is traveling in the +zˆ direction. At one point in space and time, the electric fi
ivann1987 [24]

Answer:

Explanation:

a ) Direction of the magnetic field will be in positive x direction.

The direction of the vector E X B gives the direction of motion of wave.

b ) Magnitude of magnetic field is given by the relation

E₀ / B₀ = c , c is velocity of light

B₀ = E₀ / c

= 20 / (3 x 10⁸)

= 6.67 x 10⁻⁸ T

c ) Average power flowing per unit area by this wave is called Poynting vector

c ε₀E₀² , ε₀ = 8.85X10⁻¹²

= 3 X 10⁸ X 8.85 X 10⁻¹² X 20²

= 1.062 W m⁻²

5 0
3 years ago
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