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3 years ago

Khalad is measuring the amplitude of a wave. What can be known about this wave?

Physics

2 answers:

GalinKa [24]3 years ago

6 0

It is B) transverse wave

ZanzabumX [31]3 years ago

6 0

The answer is B. Transverse wave. I just took my quiz.

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Is this the actual answer??

Nastasia [14]

It probably is the actual answer.

3 0

3 years ago

A wire carrying a current of 26.9 A is bent into a circular arc with a radius of 0.6 cm that sweeps out 0.900 radians. What is t

melomori [17]

The magnetic field at the center of the arc is 4 × 10^(-4) T.

To find the answer, we need to know about the magnetic field due to a circular arc.

<h3>What's the mathematical expression of magnetic field at the center of a circular arc?</h3>

According to Biot savert's law, magnetic field at the center of a circular arc is

B=(μ₀ I/4π)× (arc/radius²)

As arc is given as angle × radius, so

B=( μ₀I/4π)×(angle/radius)

<h3>What will be the magnetic field at the center of a circular arc, if the arc has current 26.9 A, radius 0.6 cm and angle 0.9 radian?</h3>

B=(μ₀ I/4π)× (0.9/0.006)

= (10^(-7)× 26.9)× (0.9/0.006)

= 4 × 10^(-4) T

Thus, we can conclude that the magnitude of magnetic field at the center of the circular arc is 4 × 10^(-4) T.

Learn more about the magnetic field of a circular arc here:

brainly.com/question/15259752

#SPJ4

5 0

2 years ago

A dentist causes the bit of a high-speed drill to accelerate from an angular speed of 1.10 104 rad/s to an angular speed of 3.14

Anestetic [448]

Answer:

3.63 s

Explanation:

We can solve the problem by using the equivalent SUVAT equations for the angular motion.

To find the angular acceleration, we can use the following equation:

$\omega_f^2 - \omega_i ^2 =2 \alpha \theta$

where

$\omega_f = 3.14\cdot 10^4 rad/s$ is the final angular speed

$\omega_i = 1.10 \cdot 10^4 rad/s$ is the initial angular speed

$\theta= 2.00 \cdot 10^4 rad$ is the angular distance covered

$\alpha$ is the angular acceleration

Re-arranging the formula, we can find $\alpha$ :

$\alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}=\frac{(3.14\cdot 10^4 rad/s)^2-(1.10\cdot 10^4 rad/s)^2}{2(2.00\cdot 10^4 rad)}=2.16\cdot 10^4 rad/s^2$

Now we want to know the time the bit takes starting from rest to reach a speed of $\omega_f=7.85\cdot 10^4 rad/s$ . So, we can use the following equation:

$\alpha = \frac{\omega_f-\omega_i}{t}$

where:

$\alpha=2.16\cdot 10^4 rad/s^2$ is the angular acceleration

$\omega_f = 7.85\cdot 10^4 rad/s$ is the final speed

$\omega_i = 0$ is the initial speed

t is the time

Re-arranging the equation, we can find the time:

$t=\frac{\omega_f-\omega_i}{\alpha}=\frac{7.85\cdot 10^4 rad/s-0}{2.16\cdot 10^4 rad/s^2}=3.63 s$

4 0

3 years ago

Read 2 more answers

A lead bar 12 cm long, 2 cm wide, and 2 cm talk has a density of 11.3g/cm3. If you cut the bar in half, what is the density of o

VMariaS [17]

Answer:

The density of one halves 11.3 g / cm cube

Explanation:

Density remain same because cutting the bar in half , mass and volume will decrease to half so density will not change .

density = $\frac{mass}{volume}$

5 0

3 years ago

You need to push a heavy box across a rough floor, and you want to minimize the average force applied to the box during the time

7nadin3 [17]

When pushing the body it is necessary to break the frictional force generated by the floor. Once this frictional force is overcome, the body will begin to move. Ideally, if a constant velocity is maintained or close to this value, the acceleration that will be exerted will tend to be zero and therefore, by Newton's second law the value of the Force will also tend to minimum values.

Remember that this law tells us that

$F= ma$

$F= m \frac{\Delta v}{t}$

Therefore the best strategy is A. keep pushing the box forward at a steady speed

7 0

3 years ago