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Angelina_Jolie [31]

3 years ago

8

1.03 Transformation of energy flvs science question

1 answer:

FinnZ [79.3K]3 years ago

7 0

Explanation:

the process of conversion of energy from one form to another is called transformation of energy.

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3. In the wave models, you imagined adding energy into the wave by

castortr0y [4]

Answer:

by moving your hands up and down your we're creating transverse wave which travel in a right angle shape towards your friend

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3 years ago

Aman tosses a jart upward with a velocity of 14.1 m/s a 60° angle

crimeas [40]

Answer:

hold on let me check my answer

Explanation:

8 0

4 years ago

An ordinary workshop grindstone has a radius of 7.50 cm and rotates at 6500 rev/min. (a) Calculate the magnitude of the centripe

jonny [76]

To solve this problem it is necessary to apply the concepts related to the relationship between tangential velocity and centripetal velocity, as well as the kinematic equations of angular motion. By definition we know that the direction of centripetal acceleration is perpendicular to the direction of tangential velocity, therefore:

$a_c= \frac{v^2}{r}=r\omega^2$

Where,

V = the linear speed

r = Radius

$\omega =$ Angular speed

The angular speed is given by

$\omega =6500\frac{rev}{min}(\frac{2\pi rad}{1 rev})(\frac{1 min}{60s})$

$\omega = 680.6784 rad/s.$

Replacing at our first equation we have that the centripetal acceleration would be

$a_c = r\omega^2$

$a_c = (0.0750)(680.6784)^2$

$a_c = 34 749.23 m/s^2$

To transform it into multiples of the earth's gravity which is given as $9.8m / s$ the equivalent of 1g.

$a_c =34749.23 \frac{m}{s^2} (1\frac{g}{9.8m/s^2})$

$a_c = 3545.84g$

PART B) Now the linear speed would be subject to:

$v = \omega r$

$v= (680.6784)(0.0750)$

$v=51.05 m/s$

Therefore the linear speed of a point on its edge is 51.05m/s

8 0

3 years ago

A circular loop of radius 11.7 cm is placed in a uniform magnetic field. (a) If the field is directed perpendicular to the plane

Mama L [17]

Answer:

Magnetic field, B = 0.199 T

Explanation:

It is given that,

Radius of circular loop, r = 11.7 cm = 0.117 m

Magnetic flux through the loop, $\phi=8.6\times 10^{-3}\ T/m^2$

The magnetic flux linked through the loop is :

$\phi=B.A$

$\phi=BA\ cos\theta$

Here, $\theta=0$

$B=\dfrac{\phi}{A}$

or

$B=\dfrac{\phi}{\pi r^2}$

$B=\dfrac{8.6\times 10^{-3}}{\pi (0.117 )^2}$

B = 0.199 T

So, the strength of the magnetic field is 0.199 T. Hence, this is the required solution.

4 0

3 years ago

All of these are types of electromagnetic waves except for:

dimaraw [331]

C- Sigma Waves, This is what my teacher used to get us to remember https://www.youtube.com/watch?v=bjOGNVH3D4Y

3 0

3 years ago