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larisa86 [58]
3 years ago
7

A straight wire carries a current of 10 A at an angle of 30° with respect to the direction of a uniform 0.30-T magnetic field. F

ind the magnitude of the magnetic force on a 0.50-m length of the wire.
Physics
2 answers:
Dimas [21]3 years ago
6 0

Answer:

Explanation:

Given that,

Current in wire is

I = 10A

And the current makes an angle of 30° with respect to the magnetic field

Then, θ = 30°

And the magnetic field is

B = 0.3 T

Length of the wire is

L = 0.5m

Force on the wire F?

The force on the wire in calculated using

F = iL × B

Where

The magnitude of the cross produce of L and B is

L × B = LB•Sinθ

Then, force becomes

F = iLB•Sinθ

F = 10 × 0.5 × 0.3 × Sin30

F = 0.75 N

The force on the wire is 0.75 Newton

Mariana [72]3 years ago
3 0

Answer:

The magnitude of the magnetic force on the wire is 0.75 N

Explanation:

Given;

current in the wire, I = 10 A

angle of inclination to magnetic field, θ = 30°

magnetic field strength, B = 0.30-T

length of the wire, L = 0.50-m

The magnitude of magnetic force acting on the wire is given as;

F = BILSinθ

where;

B is magnetic field strength

I is the current in the wire

L is length of the wire

θ is the angle of inclination

F = 0.3 x 10 x 0.5 x sin(30)

F = 0.3 x 10 x 0.5 x 0.5

F = 0.75 N

Therefore,  the magnitude of the magnetic force on the wire is 0.75 N

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A basketball is held over head at a height of 2.4 m. The ball is lobbed to a teammate at 8 m/s at an angle of 40'. If the ball i
cupoosta [38]

Explanation:

since both the teammates are of the same height, their height won't matter. Because now the basketball won't cover any vertical distance.

We have to calculate its range the horizontal distance covered by it when tossed from one teammate to the other.

range can be calculated by the formula :-

\boxed{\mathfrak{range =  \frac{  u  {}^{2}   \sin 2\theta }{g} }}

u is the velocity during its take off and \theta is the angle at which its thrown

Given that

  • u = 8m/ s
  • \theta = 40°

calculating range using the above formula

= \frac{ {8}^{2} \sin2(40)  }{10}

=  \frac{64 \times  \sin(80) }{10}

value of sin 80 = 0. 985

=  \frac{64 \times 0.985}{10}

=  \frac{63.027}{10}

= 6.3027

Hence,

\mathfrak { \blue{the \: teammate \: is \:  \red{\underline{6.3027 \: meters} }\: away } }

7 0
3 years ago
Two sealed 1 l containers full of gas are at room temperature. Container a has a pressure of 4 atm and container b has a pressur
loris [4]

Answer:

The number density of the gas in container A is twice the number density of the gas in container B.

Explanation:

Here we have

P·V =n·R·T

n = P·V/(RT)

Therefore since V₁ = V₂ and T₁ = T₂

n₁ = P₁V₁/(RT₁)

n₂ = P₂V₂/(RT₂)

P₁ = 4 atm

P₂ = 2 atm

n₁ = 4V₁/(RT₁)

n₂ =2·V₁/(RT₁)

∴ n₁ = 2 × n₂

Therefore, the number of moles in container A is two times that in container B and the number density of the gas in container A is two times the number density in container B.

This can be shown based on the fact that the pressure  of the container is due to the collision of the gas molecules on the walls of the container, with a kinetic energy that is dependent on temperature and mass, and since the temperature is constant, then the mass of container B is twice that of A and therefore, the number density of container A is twice that of B.

5 0
3 years ago
Can someone help me ?
hammer [34]

Answer:

1)    Time interval                 Blue Car                      Red Car

          0 - 2 s                Constant Velocity           Increasing Velocity

          2 - 3 s                Constant Velocity           Constant Velocity

          3 - 5 s                Constant Velocity           Increasing Velocity

          5 - 6 s                Constant Velocity           Decreasing Velocity

2) For Red and Blue car y₂  = 120       v = \frac{y_{2}-y_{1}}{t_{2}-t_{1}} = \frac{120-0}{6-0} = 20 m/s

     We get the same velocity for two cars because it is the average velocity of the car at the given interval of time. It is measured for initial and final position.

3)   At t = 2s, the cars are the same position, and are moving at the same rate

                    Position - same

                    Velocity - same

The position-time graph shares the same spot for two cars.

4 0
3 years ago
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Tom [10]

Answer:

+ 24 N

Explanation:

the computation is shown below:

Given that

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x = 0.15m

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F = kx

= (160) (0.15)

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As the instant block is released so the acting of the force on the block is positive and it would be in a positive direction i.e. right direction

Therefore the third option is correct

3 0
3 years ago
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