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Ilya [14]
2 years ago
12

keyana is in a plane travelling 800 km/h at 35.0° w of n when the plane encountered the jet stream blowing 200 km/h in a directi

on 82.0° e of n. what is the resultant speed of the plane
Physics
1 answer:
lys-0071 [83]2 years ago
5 0

Answer:

An aeroplane is flying with the velocity of V_(1)=800kmph relative to the air towards south.A wind with velocity of V_(2)=15ms^(-1) is blowing from west to east.What is the velocity of the aeroplane with respect of the earth.

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Which is not true of friction? A. Causes wear and tear of the surfaces B. Helps us to fall easily on roads C. Produces heat D. P
Ganezh [65]

Answer:

B.

Hope this helps

5 0
2 years ago
Read 2 more answers
50g of ice at 0°C is mixed with 50g of water at 80°C, what will be the final temperature of a mixture in
xxTIMURxx [149]

Answer:

0° C

Explanation:

Given that

Mass of ice, m = 50g

Mass of water, m(w) = 50g

Temperature of ice, T(i) = 0° C

Temperature of water, T(w) = 80° C

Also, it is known that

Specific heat of water, c = 1 cal/g/°C

Latent heat of ice, L(w) = 89 cal/g

Let us assume T to be the final temperature of mixture.

This makes the energy balance equation:

Heat gained by ice to change itself into water + heat gained by melted ice(water) to raise its temperature at T° C = heat lost by water to reach at T° C

m(i).L(i) + m(i).c(w)[T - 0] = m(w).c(w)[80 - T], on substituting, we have

50 * 80 + 50 * 1(T - 0) = 50 * 1(80 - T)

4000 + 50T = 4000 - 50T

0 = 100 T

T = 0° C

Thus, the final temperature is 0° C

3 0
2 years ago
How does the height from which you drop the ball relate to the height that the ball bounces back up?
Stels [109]
The higher you go the more potential energy there is, and the lower it is the more kinetic energy there is, so the more kinetic energy there is the higher the ball will bounce.
7 0
3 years ago
An MRI scanner is based on a solenoid magnet that produces a large magnetic field. The magnetic field doesn't stop at the soleno
GaryK [48]

Answer:

The maximum change in  flux is \Delta \o = 0.1404 \ Wb

The average  induced emf     \epsilon =0.11232 V

Explanation:

   From the question we are told that

             The speed of the technician is v = 0.80 m/s

              The distance from the scanner is d = 1.0m

              The  initial magnetic field is  B_i = 0T

               The final magnetic field is B_f = 6.0T

                 The diameter of the loop is  D = 19cm = \frac{19}{100} = 0.19 m

The area of the loop is mathematically represented as

        A  =  \pi [\frac{D}{2} ]^2

             = 3.142 \frac{0.19}{2}

             = 0.02834 m^2

At maximum the change in magnetic field is mathematically represented as

            \Delta \o = (B_f - B_i)A

  =>      \Delta  \o = (6 -0)(0.02834)

                  \Delta \o = 0.1404 \ Wb

The  average induced emf is mathematically represented as

           \epsilon =  \Delta \o v

              = 0.1404 * 0.80

             \epsilon =0.11232 V

7 0
3 years ago
Which example involves the transformation of chemical energy directly into light energy?
zimovet [89]

burning sodium or magnesium

8 0
3 years ago
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