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3 years ago

An actor who memorizes his lines word-for-word has demonstrated _____ learning.

Physics

2 answers:

Nana76 [90]3 years ago

8 0

<span>Mechanical association learning used by an actor to memorize his lines</span>

vazorg [7]3 years ago

8 0

Answer:

known as rote memory

Explanation:

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The revolution of the earth around the sun demonstrate what motion?

olga2289 [7]

Answer:

Anticlockwise directions

Please mark me Brainliest to help me

3 0

1 year ago

An electron in the beam of a cathod-ray tube is accelerated by a potential difference of 2.12 kV . Then it passes through a regi

son4ous [18]

Answer:

B=9.1397*10^-4 Tesla

Explanation:

To find the velocity first we put kinetic energy og electron is equal to potential energy of electron

K.E=P.E

$\frac{1}{2}*m*v^{2} =e*V$

where :

m is the mass of electron

v is the velocity

V is the potential difference

$v=\sqrt{\frac{2*e*V}{m} }$ eq 1

Radius of electron moving in magnetic field is given by:

$R=\frac{m*v}{q*B}$ eq 2

where:

m is the mass of electron

v is the velocity

q=e=charge of electron

B is the magnitude of magnetic field

Put v from eq 1 into eq 2

$R=\frac{m*\sqrt{\frac{2*e*V}{m} } }{e B}$

$B=\sqrt{\frac{2*m*V}{e*R^{2} } }$

$B=\sqrt{\frac{2*(9.31*10^{-31})*(2.12*10^{3}) }{(1.60*10^{-19})*(0.170)^{2} } }$

B=9.1397*10^-4 Tesla

3 0

3 years ago

Suppose you could fit 100 dimes, end to end, between your card with the pinhole and your dime-sized sunball. how many suns could

Naddika [18.5K]

Answer: 100 suns

Explanation:

We can solve this with the following relation:

$\frac{d}{x_{sunball-pinhole}}=\frac{D}{x_{sun-pinhole}}$

Where:

$d=17.91 mm =17.91(10)^{-3} m$ is the diameter of a dime

$D$ is the diameter of the Sun

$x_{sun-pinhole}=150,000,000 km=1.5(10)^{11} m$ is the distance between the Sun and the pinhole

$x_{sunball-pinhole}=100 d=1.791 m$ is the amount of dimes that fit in a distance between the sunball and the pinhole

Finding $D$ :

$D=\frac{d}{x_{sunball-pinhole}}x_{sun-pinhole}$

$D=\frac{17.91(10)^{-3} m}{1.791 m} 1.5(10)^{11} m$

$D=1.5(10)^{9} m$ This is roughly the diameter of the Sun

Now, the distance between the Earth and the Sun is one astronomical unit (1 AU), which is equal to:

$1 AU=149,597,870,700 m$

So, we have to divide this distance between $D$ in order to find how many suns could it fit in this distance:

$\frac{149,597,870,700 m}{1.5(10)^{9} m}=99.73 suns \approx 100 suns$

8 0

3 years ago

Gravity pushes objects away from each other TRUE OR FALSE

e-lub [12.9K]

Answer:

false

Explanation:

5 0

3 years ago

The temperature -273°C is —.

ArbitrLikvidat [17]

Answer:

-273°C = Absolute zero

Explanation:

Average Kinetic energy of particles in matter refers to the temperature of matter. This value is not arbitrary zero temperature. It explains the lowest kinetic energy of the particles in matter.

4 0

3 years ago