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Rina8888 [55]
4 years ago
13

In what direction must the current I2 flow in the straight wire in order to create a field at the center of the loop in the dire

ction opposite to that created by the loop?
Physics
1 answer:
guajiro [1.7K]4 years ago
6 0

Answer:

Right to left

Explanation:

From right hand thumb rule, the magnetic field at the centre of loop points out of the plane of paper. Thus the straight wire current should create magnetic field into the plane of paper at the centre. Hence the direction of current in long wire should be from right to left

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The key to making a concise mathematical definition of escape velocity is to consider the energy. If an object is launched at it
aleksklad [387]

Answer:

The total Mechanical energy will be zero

Explanation: Escape velocity is the velocity required by a free object in order to overcome the impact of the force of gravity. The total mechanical energy of an object is the total energy possessed by an object which includes its kinectic and potential energy.

since the object is moving at an escape velocity which is 11.2m/s the object will be assumed to be weightless

Etotal = kinetic energy + potential energy

kinetic energy= 1/2*M*V*V

Potential energy=MGH

Etotal=1/2*0*11.2*11.2+0*0*0

Etotal=0+0

Etotal=0.

3 0
3 years ago
Which material is very strong and tough but shows very little elongation as it absorbs energy? A. spider silk B.rubber C.Kevlar®
Furkat [3]

C. Kevlar

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5 0
3 years ago
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If it takes 15 minutes to dry your hair with a 1.200 kW hair drier, how much energy is used in drying your hair?
Elodia [21]
1.2 kW * 0.25 h = 0.300 kWh
6 0
3 years ago
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A large sheet of charge has a uniform charge density of 9  μCm2. What is the electric field due to this charge at a point just
Alex73 [517]

Answer:

Explanation:

Surface charge density, σ = 9 μC/m² = 9 x 10^-6 C/m²

According to the Gauss theorem,

Electric field due to the sheet is given by

E = \frac {\sigma }{2\epsilon _{0}}

E = \frac{9\times 10^{-6}}{2\times 8.854\times 10^{-12}}

E = 5.08 x 10^5 N/C

7 0
3 years ago
A quarterback passes a football from height h = 2.1 m above the field, with initial velocity v0 = 13.5 m/s at an angle θ = 32° a
SOVA2 [1]

Answer:

a)    x = v₀² sin 2θ / g

b)    t_total = 2 v₀ sin θ / g

c)    x = 16.7 m

Explanation:

This is a projectile launching exercise, let's use trigonometry to find the components of the initial velocity

        sin θ = v_{oy} / vo

        cos θ = v₀ₓ / vo

         v_{oy} = v_{o} sin θ

         v₀ₓ = v₀ cos θ

         v_{oy} = 13.5 sin 32 = 7.15 m / s

         v₀ₓ = 13.5 cos 32 = 11.45 m / s

a) In the x axis there is no acceleration so the velocity is constant

         v₀ₓ = x / t

          x = v₀ₓ t

the time the ball is in the air is twice the time to reach the maximum height, where the vertical speed is zero

          v_{y} = v_{oy} - gt

          0 = v₀ sin θ - gt

          t = v_{o} sin θ / g

         

we substitute

       x = v₀ cos θ (2 v_{o} sin θ / g)

       x = v₀² /g      2 cos θ sin θ

       x = v₀² sin 2θ / g

at the point where the receiver receives the ball is at the same height, so this coincides with the range of the projectile launch,

b) The acceleration to which the ball is subjected is equal in the rise and fall, therefore it takes the same time for both parties, let's find the rise time

at the highest point the vertical speed is zero

          v_{y} = v_{oy} - gt

          v_{y} = 0

           t = v_{oy} / g

           t = v₀ sin θ / g

as the time to get on and off is the same the total time or flight time is

           t_total = 2 t

           t_total = 2 v₀ sin θ / g

c) we calculate

          x = 13.5 2 sin (2 32) / 9.8

          x = 16.7 m

5 0
3 years ago
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