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5 days ago

How many turns should a 10-cm long solenoid have if it is to generate a 1.5 x 10-3 t magnetic field on 1.0 a of current?

Physics

1 answer:

kicyunya [2.2K]5 days ago

3 0

This problem can be solved using Ampere’s Law:

Bh = μoNI

In this equation:

B = Magnetic Field

h = length of the coil

μo = permeability = 4π*10^-7 T·m/A

N = number of coil turns

I = current

Given values are B = 0.0015T, I = 1.0A, h = 10 cm = 0.1m

Utilizing Ampere's law to determine the number of turns:
This can be rearranged to:
N = Bh/μoI
N = (0.0015)(0.1)/(4π*10^-7)(1.0)
N = 119.4

Final answer: 119.4 turns

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10 days ago

A 7.5 kg cannon ball leaves a canon with a speed of 185 m/s. Find the average net force applied to the ball if the cannon muzzle

Keith_Richards [2256]

To determine the average net force, we can calculate acceleration using:

x = 0.5*a*t^2

v = a*t

where x=3.6m and v=185 m/s.

Thus,

t=v/a and therefore x = 0.5*a*(v/a)^2 = 0.5 * (v^2)/a

which gives us a= (0.5*v^2)/x

Since we have the known values of v and x, we can compute a by substituting these numbers.

The average net force is then given as:

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22 days ago

Compressed air is used to fire a 60 g ball vertically upward from a 0.70-m-tall tube. The air exerts an upward force of 3.0 N on

Yuliya22 [2420]

Answer:

2.87 m

Explanation:

Given parameters:

Mass of the ball (m) = 60 g = 0.06 kg

Height of the tube (h) = 0.70 m

Force applied on the ball by compressed air (F) = 3.0 N

Initial velocity of the ball (u) = 0 m/s (Assumed)

Final velocity of the ball at the tube's exit (v) =?

Acceleration of the ball (a) =?

The ball's weight is derived from multiplying mass and gravity. Therefore,

Weight (W) = $mg=0.06\times 9.8=0.588\ N$

Thus, the total force acting on the ball equals the net of upward force minus the weight.

Net force = Air force - Weight

$F_{net}=F-mg\\F_{net}=3.0-0.588 = 2.412\ N$

According to Newton's second law, net force equals the mass multiplied by acceleration.

$F_{net}=ma\\\\a=\frac{F_{net}}{m}=\frac{2.412\ N}{0.06\ kg}=40.2\ m/s^2$

Acceleration (a) is calculated as 40.2 m/s².

Using the motion equation, we find:

$v^2=u^2+2ah\\\\v^2=0+2\times 40.2\times 0.7\\\\v=\sqrt{56.28}=7.5\ m/s$

Let’s denote the maximum height achieved as 'H'.

Next, we apply the principle of energy conservation from the pipe's peak to the maximum height.

A decrease in kinetic energy equals an increase in potential energy.

$\frac{1}{2}mv^2=mgH\\\\H=\frac{v^2}{2g}$

Substituting the values, we solve for 'H', yielding:

$H=\frac{56.28}{2\times 9.8}\\\\H=\frac{56.28}{19.6}=2.87\ m$

Hence, the ball ascends to a height of 2.87 m above the top of the tube.

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24 days ago

A 0.050-kg lump of clay moving horizontally at 12 m/s strikes and sticks to a stationary 0.15-kg cart that can move on a frictio

Ostrovityanka [2204]

Answer:

Explanation:

According to the parameters provided,

mass of the clay lump, m₁ = 0.05 kg

initial velocity of the lump, u₁ = 12 m/s

mass of the cart, m₂ = 0.15 kg

initial speed of the cart, u₂ = 0

As the clay adheres to the cart, we have an inelastic collision scenario. Let v represent the combined speed of both the cart and lump post-collision. Given that momentum is conserved, we have:

$m_1u_1+m_2u_2=(m_1+m_2)v$

$v=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}$

$v=\dfrac{0.05\ kg\times 12\ m/s+0}{0.05\ kg+0.15\ kg}$

The resultant speed is v = 3 m/s.

Thus, the final speed of both cart and lump following the collision is 3 m/s. This concludes the solution.

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19 days ago

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