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bekas [8.4K]
3 years ago
14

The magnetic field produced by the solenoid in a magnetic resonance imaging (MRI) system designed for measurements on whole huma

n bodies has a field strength of 7.0 T, and the current in the solenoid is 2.0 \times 10^{2} A. What is the number of turns per meter of length of the solenoid
Physics
2 answers:
Svet_ta [14]3 years ago
8 0

Answer:

Explanation:

Magnetic field, B = 7 T

Current , i = 2 x 100 A

let n be the number of turns per unit length.

The magnetic field due to a solenoid is given by

B=\mu _{0}ni

7=4\times 31.4\times 10^{-7}\times n\times 200

n = 27866.24 turns per metre

Vesna [10]3 years ago
7 0

Answer:

Number of turns per unit length will be n=2.786\times 10^4turns/m

Explanation:

We have given that strength of the magnetic field produced by the solenoid B = 7 T

Current in the solenoid i = 200 A

Let the number of turns per unit length is n

Magnetic field due to solenoid is given as B=\mu ni here \mu is permeability of free space n is number of turns per unit length and i is current

So 7=4\pi \times 10^{-7}\times  n\times 200

n=2.786\times 10^4turns/m

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A projectile is launched at an angle of 36.7 degrees above the horizontal with an initial speed of 175 m/s and lands at the same
Softa [21]

Answer:

a) The maximum height reached by the projectile is 558 m.

b) The projectile was 21.3 s in the air.

Explanation:

The position and velocity of the projectile at any time "t" is given by the following vectors:

r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)

v = (v0 · cos α, v0 · sin α + g · t)

Where:

r = position vector at time "t"

x0 = initial horizontal position

v0 = initial velocity

t = time

α = launching angle

y0 = initial vertical position

g = acceleration due to gravity (-9.80 m/s² considering the upward direction as positive).

v = velocity vector at time t

a) Notice in the figure that at maximum height the velocity vector is horizontal. That means that the y-component of the velocity (vy) at that time is 0. Using this, we can find the time at which the projectile is at maximum height:

vy = v0 · sin α + g · t

0 = 175 m/s · sin 36.7° - 9.80 m/s² · t

-  175 m/s · sin 36.7° /  - 9.80 m/s² = t

t = 10.7 s

Now, we have to find the magnitude of the y-component of the vector position at that time to obtain the maximum height (In the figure, the vector position at t = 10.7 s is r1 and its y-component is r1y).

Notice in the figure that the frame of reference is located at the launching point, so that y0 = 0.

y = y0 + v0 · t · sin α + 1/2 · g · t²

y = 175 m/s · 10.7 s · sin 36.7° - 1/2 · 9.8 m/s² · (10.7 s)²

y = 558 m

The maximum height reached by the projectile is 558 m

b) Since the motion of the projectile is parabolic and the acceleration is the same during all the trajectory, the time of flight will be twice the time it takes the projectile to reach the maximum height. Then, the time of flight of the projectile will be (2 · 10.7 s) 21.4 s. However, let´s calculate it using the equation for the position of the projectile.

We know that at final time the y-component of the vector position (r final in the figure) is 0 (because the vector is horizontal, see figure). Then:

y = y0 + v0 · t · sin α + 1/2 · g · t²

0 = 175 m/s · t · sin 36.7° - 1/2 · 9.8 m/s² · t²

0 = t (175 m/s ·  sin 36.7 - 1/2 · 9.8 m/s² · t)

0 = 175 m/s ·  sin 36.7 - 1/2 · 9.8 m/s² · t

-  175 m/s ·  sin 36.7 / -(1/2 · 9.8 m/s²) = t

t = 21.3 s

The projectile was 21.3 s in the air.

7 0
3 years ago
A child drops a ball from a window. The ball strikes the ground in 3.0 seconds. What is the velocity ofthe ball the instant befo
Sergeeva-Olga [200]

Answer: The velocity of the ball is 30.0 m/s

This can be calculated by using the value of acceleration as 10.0 m/s2 in free fall and the given time of 3.0 seconds. To get the velocity, one will have to multiply the acceleration with the given time and the quotient would result to 30.0 m/s. Mostly all object regardless of their mass, fall to earth with the same acceleration in the absence of air resistance and as the child drops the ball from a window, it gains speed as it falls.

5 0
3 years ago
A pool ball moving 1.83 m/s strikes an identical ball at rest. Afterward, the first ball moves 1.15 m/s at a 23.3 degrees angle.
chubhunter [2.5K]

Answer:

 v_{1fy} = - 0.4549 m / s

Explanation:

This is an exercise of conservation of the momentum, for this we must define a system formed by the two balls, so that the forces during the collision have internal and the momentum is conserved

initial. Before the crash

      p₀ = m v₁₀

final. After the crash

      p_{f} = m v_{1f} + m v_{2f}

Recall that velocities are a vector so it has x and y components

       p₀ = p_{f}

we write this equation for each axis

X axis

       m v₁₀ = m v_{1fx} + m v_{2fx}

       

Y Axis  

       0 = -m v_{1fy} + m v_{2fy}

the exercise tells us the initial velocity v₁₀ = 1.83 m / s, the final velocity v_{2f} = 1.15, let's use trigonometry to find its components

      sin 23.3 = v_{2fy} / v_{2f}

      cos 23.3 = v_{2fx} / v_{2f}

      v_{2fy} = v_{2f} sin 23.3

      v_{2fx} = v_{2f} cos 23.3

we substitute in the momentum conservation equation

       m v₁₀ = m v_{1f} cos θ + m v_{2f} cos 23.3

       0 = - m v_{1f} sin θ + m v_{2f} sin 23.3

      1.83 = v_{1f} cos θ + 1.15 cos 23.3

       0 = - v_{1f} sin θ + 1.15 sin 23.3

      1.83 = v_{1f} cos θ + 1.0562

        0 = - v_{1f} sin θ + 0.4549

     v_{1f} sin θ = 0.4549

     v_{1f}  cos θ = -0.7738

we divide these two equations

      tan θ = - 0.5878

      θ = tan-1 (-0.5878)

       θ = -30.45º

we substitute in one of the two and find the final velocity of the incident ball

        v_{1f} cos (-30.45) = - 0.7738

        v_{1f} = -0.7738 / cos 30.45

        v_{1f} = -0.8976 m / s

the component and this speed is

       v_{1fy} = v1f sin θ

       v_{1fy} = 0.8976 sin (30.45)

       v_{1fy} = - 0.4549 m / s

8 0
3 years ago
The chemical energy in your food is
ale4655 [162]

Answer:

Electromanetic

Explanation:ESPERO TE AYUDE

8 0
3 years ago
How would i solve for force when there are multiple accelerations?
damaskus [11]

"Multiple accelerations" is a puzzling phrase, and I'd be curious to understand it
better.  Sadly however, you haven't explained it at all.

If the multiple accelerations are the accelerations of multiple objects, then
the net force on each object is the product of (its mass) x (its acceleration).

If the multiple accelerations are the acceleration of one object at different times,
then at any instant of time, the net force on the object is the product of (its mass) x
(its acceleration at that instant).

8 0
3 years ago
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