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3 years ago

A coil of wire that is carrying a current and produces a magnetic field is

Physics

2 answers:

labwork [276]3 years ago

8 0

The correct choice is

a solenoid

a solenoid is a sequence of a number of loops in the form of a helix in which when current flows, a magnetic field is produced along the axis of the solenoid. The magnetic field thus produced by a solenoid is given as

B = μ₀ n i

where B = magnetic field , n = number of turns per unit length, i = current in the solenoid.

snow_tiger [21]3 years ago

4 0

Answer:

Solenoid is the correct answer

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NEED ANSWER ASAP!!!

xenn [34]

Answer:

Explanation:

it has high pressure of speed

3 0

3 years ago

A jet lands on an aircraft carrier at 140 mi/h. What is its acceleration if it stops in 2.0 seconds?

sergejj [24]

Answer:

Acceleration, $a=-31.29\ m/s^2$

Explanation:

It is given that,

Initial speed of the aircraft, u = 140 mi/h = 62.58 m/s

Finally, it stops, v = 0

Time taken, t = 2 s

Let a is the acceleration of the aircraft. We know that the rate of change of velocity is called acceleration of the object. It is given by :

$a=\dfrac{v-u}{t}$

$a=\dfrac{0-62.58}{2}$

$a=-31.29\ m/s^2$

So, the acceleration of the aircraft is $31.29\ m/s^2$ and the car is decelerating. Hence, this is the required solution.

4 0

4 years ago

You have been hired to design a spring-launched roller coaster that will carry two passengers per car. The car goes up a 13-m-hi

Andrew [12]

Answer:

22.15 N/m

Explanation:

As we know potential energy = m*g*h

Potential energy of spring = (1/2)kx^2

m*g*h = (1/2)kx^2

Substituting the given values, we get -

(400)*(9.8)*(10) = (0.5)*(k)*(2.0^2)

k = 39200/2.645

k = 19600 N/m

For safety reasons, this spring constant is increased by 13 % So the new spring constant is

k = 19600 * 1.13 = 22148 N/m = 22.15 N/m

3 0

3 years ago

Are momentum and kinetic energy conserved during an inelastic collision?

nevsk [136]

Energy and momentum are always conserved. Kinetic energy is not conserved in an inelastic collision though. And that is because it is converted to another form of energy

8 0

3 years ago

An electron that has a velocity with x component 2.4 x 106 m/s and y component 3.6 x 106 m/s moves through a uniform magnetic fi

likoan [24]

Answer:

(a) 7.315 x 10^(-14) N

(b) - 7.315 x 10^(-14) N

Explanation:

As you referred at the final remark, the electron and proton undergo a magnetic force of same magnitude but opposite direction. Using the definition of magnetic force, a cross product must be done. One technique is either calculate the magnitude of the velocity and magnetic field and multiplying by sin (90°), but it is necessary to assure both vectors are perpendicular between each other ( which is not the case) or do directly the cross product dealing with a determinant (which is the most convenient approach), thus,

(a) The electron has a velocity defined as: $\overrightarrow{v}=(2.4x10^{6} i + 3.6x10^{6} j) \frac{[m]}{[s]}\\\\$

In respect to the magnetic field; $\overrightarrow{B}=(0.027 i - 0.15 j) [T]$

The magnetic force can be written as;

$\overrightarrow{F} = q(\overrightarrow{v} x \overrightarrow{B})\\ \\\\\overrightarrow{F}= q \left[\begin{array}{ccc}i&j&k\\2.4x10^{6}&3.6x10^{6}&0\\0.027&-0.15&0\end{array}\right]$

Bear in mind $q =-1.6021x10^{-19} [C]$

thus,

$\overrightarrow{F}= q \left[\begin{array}{ccc}i&j&k\\2.4x10^{6}&3.6x10^{6}&0\\0.027&-0.15&0\end{array}\right]\\\\\\\overrightarrow{F}= q(2.4x10^{6}* (-0.15)- (0.027*3.6x10^{6}))\\\\\\\overrightarrow{F}= -1.6021x10^{-19} [C](-457200) [T]\frac{m}{s}\\\\\overrightarrow{F}=(7.3152x10^{-14}) k [\frac{N*m/s}{C*m/s}]\\\\|F|= \sqrt{ (7.3152x10^{-14})^{2}[N]^2 *k^{2}}\\\\F=7.3152x10^{-14} [N]$

Note: The cross product is operated as a determinant. Likewise, the product of the unit vector k is squared and that is operated as dot product whose value is equal to one, i.e, $k^{2}=k\cdot k = 1$

(b) Considering the proton charge has the same magnitude as electron does, but the sign is positive, thus

$\overrightarrow{F}= q \left[\begin{array}{ccc}i&j&k\\2.4x10^{6}&3.6x10^{6}&0\\0.027&-0.15&0\end{array}\right]\\\\\\\overrightarrow{F}= q(2.4x10^{6}* (-0.15)- (0.027*3.6x10^{6}))\\\\\\\overrightarrow{F}= 1.6021x10^{-19} [C](-457200) [T]\frac{m}{s}\\\\\overrightarrow{F}=(-7.3152x10^{-14}) k [\frac{N*m/s}{C*m/s}]\\\\|F|= \sqrt{ (-7.3152x10^{-14})^{2}[N]^2 *k^{2}}\\\\F=-7.3152x10^{-14} [N]$

Note: The cross product is operated as a determinant. Likewise, the product of the unit vector k is squared and that is operated as dot product whose value is equal to one, i.e, $k^{2}=k\cdot k = 1$

Final remarks: The cross product was performed in R3 due to the geometrical conditions of the problem.

6 0

4 years ago