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raketka [301]
3 years ago
7

Who wanna be my friend im a girl and im 13 and im lightskin and black curly hair

Physics
1 answer:
Dafna11 [192]3 years ago
8 0

Answer:

Would love to be your friend m'lady.

Explanation:

You might be interested in
Acceleration and Force
olga55 [171]

Answer:

I'm pretty sure its 3m/s^2 for the acceleration but I don't know the force part sorry .

Explanation:

15m/s - 0m/s divided by 5 s = 3m/s

I'm no expert or anything so I could be wrong but this is the best I can give you. Sorry

6 0
2 years ago
you are flying on an airplane and you overhear two pasengers comment that flight attendants must be very skiled, one says "when
s344n2d4d5 [400]

How do you get out of a room with only a mirror and a table?

Look in the mirror you see what you saw, use the saw to cut the table, half and half makes a hole, jump in it.

Thought you would like that.

4 0
3 years ago
A sound wave with a frequency of 400 Hz is moving through a solid object. If the wavelength of the sound wave is 8 m, what is th
Paraphin [41]

Answer:

v = 3200 m/s

Explanation:

As we know that the frequency of the sound wave is given as

f = 400 Hz

wavelength of the sound wave is given as

\lambda = 8 m

so now we have

speed = wavelength \times frequency

so we will have

v = (8m) \times (400 Hz)

v = 3200 m/s

4 0
3 years ago
Before colliding, the momentum of block A is -100 kg*m/, and block B is -150 kg*m/s. After, block A has a momentum -200 kg*m/s.
rjkz [21]

Answer:

Momentum of block B after collision =-50\ kg\ ms^{-1}

Explanation:

Given

Before collision:

Momentum of block A = p_{A1}= -100\ kg\ ms^{-1}

Momentum of block B = p_{B1}= -150\ kg\ ms^{-1}

After collision:

Momentum of block A = p_{A2}= -200\ kg\ ms^{-1}

Applying law of conservation of momentum to find momentum of block B after collision p_{B2}.

p_{A1}+p_{B1}=p_{A2}+p_{B2}

Plugging in the given values and simplifying.

-100-150=-200+p_{B2}

-250=-200+p_{B2}

Adding 200 to both sides.

200-250=-200+p_{B2}+200

-50=p_{B2}

∴ p_{B2}=-50\ kg\ ms^{-1}

Momentum of block B after collision =-50\ kg\ ms^{-1}

6 0
3 years ago
A -turn rectangular coil with length and width is in a region with its axis initially aligned to a horizontally directed uniform
madam [21]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The maximum emf is \epsilon_{max}= 26.8 V

The emf induced at t = 1.00 s is \epsilon = 24.1V

The maximum rate of change of magnetic flux is   \frac{d \o}{dt}|_{max}  =26.8V

Explanation:

    From the question we are told that

        The number of turns is N = 44 turns

          The length of the coil is  l = 15.0 cm = \frac{15}{100} = 0.15m

          The width of the coil is  w = 8.50 cm =\frac{8.50}{100} =0.085 m

          The magnetic field is  B = 745 \ mT

          The angular speed is w = 64.0 rad/s

Generally the induced emf is mathematically represented as

        \epsilon = \epsilon_{max} sin (wt)

 Where \epsilon_{max} is the maximum induced emf and this is mathematically represented as

            \epsilon_{max} = N\ B\ A\ w

Where \o is the magnetic flux

            N is the number of turns

             A is the area of the coil which is mathematically evaluated as

             A = l *w

        Substituting values

           A = 0.15 * 0.085

               = 0.01275m^2

substituting values into the equation for  maximum induced emf

         \epsilon_{max} = 44* 745 *10^{-3} * 0.01275 * 64.0

                 \epsilon_{max}= 26.8 V

 given that the time t = 1.0sec

substituting values into the equation for induced emf  \epsilon = \epsilon_{max} sin (wt)

      \epsilon = 26.8 sin (64 * 1)

        \epsilon = 24.1V

   The maximum induced emf can also be represented mathematically as

              \epsilon_{max} = \frac{d \o}{dt}|_{max}

  Where  \o is the magnetic flux and \frac{d \o}{dt}|_{max} is the maximum rate at which magnetic flux changes the value of the maximum rate of change of magnetic flux is

         \frac{d \o}{dt}|_{max}  =26.8V

8 0
3 years ago
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