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melamori03 [73]
3 years ago
9

What is an electron

Physics
2 answers:
Sunny_sXe [5.5K]3 years ago
5 0

Answer:

A stable subatomic particle with a charge of negative electricity that is found in all atoms.

Explanation:

Fiesta28 [93]3 years ago
3 0

Answer:

C. A negative particle outside the nucleus

Explanation:

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A train travels 85 kilometers in 5 hours, and then 63 kilometers in 5 hours what is its average speed?
Lesechka [4]
We know average speed =total distance/time taken
So avg speed=(85+63)/(5+5)=14.8km/hr
3 0
3 years ago
Everyone can be hypnotized?
MAXImum [283]

Answer:

false

Explanation:

8 0
3 years ago
A mass free to vibrate on a level, frictionless surface at the end of a horizontal spring is pulled 35 cm from its equilibrium p
saul85 [17]

Answer:

0.67 s

Explanation:

This is a simple harmonic motion (SHM).

The displacement, x, of an SHM is given by

x = A\cos(\omega t)

A is the amplitude and \omega is the angular frequency.

We could use a sine function, in which case we will include a phase angle, to indicate that the oscillation began from a non-equilibrium point. We are using the cosine function for this particular case because the oscillation began from an extreme end, which is one-quarter of a single oscillation, when measured from the equilibrium point. One-quarter of an oscillation corresponds to a phase angle of 90° or \frac{\pi}{4} radian.

From trigonometry, \sin A =\cos B if A and B are complementary.

At t = 0, x = 3.5

3.5 = A\cos(\omega \times0)

A =3.5

So

x = 3.5\cos(\omega t)

At t = 0.12, x = 1.5

1.5 = 3.5\cos(0.12\omega)

\cos(0.12\omega)=\dfrac{1.5}{3.5}=0.4286

0.12\omega =\cos^{-1}0.4286

0.12\omega = 1.13

\omega = 9.4

The period, T, is related to \omega by

T = \dfrac{2\pi}{\omega} = \dfrac{2\times3.14}{9.4}=0.67

5 0
3 years ago
If you get too many fouls, you may not be able to play in the rest of the game.
trapecia [35]

Answer:

In the NBA that's true

4 0
3 years ago
Read 2 more answers
A 235.0 g metal block absorbs 2.44 × 103 J of heat to raise its temperature by 35 K. What is the specific heat of the metal? Sho
andrew-mc [135]
When an object absorbs an amount of energy equal to Q, its temperature raises by \Delta T following the formula
Q=m C_s \Delta T
where m is the mass of the object and C_s is the specific heat capacity of the material.

In our problem, we have Q=2.44 \cdot 10^3 J, m=235.0 g and \Delta T=35 K, so we can re-arrange the formula and substitute the numbers to find the specific heat capacity of the metal:
C_s =  \frac{Q}{m \Delta T}= \frac{2.44 \cdot 10^3 J}{(235.0 g)(35 K)}=0.297 J g^{-1} K^{-1}
3 0
3 years ago
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