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FinnZ [79.3K]
3 years ago
15

(e) Give an example of a process when electrons behave as: A.particles. B.a wave.

Physics
1 answer:
borishaifa [10]3 years ago
6 0

Answer:

particles

Explanation:

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A white dwarf star has a density of about 1.0 x 10^9 kg/m3. If the earth were to suddenly become as dense as a white dwarf star,
GalinKa [24]

Answer:

R = 98304.75 m = 98.3 km

Explanation:

The density of an object is given as the ratio between the mass of that object and the volume occupied by that object.

Density = Mass/Volume

Now, it is given that the density of Earth has become:

Density = 1 x 10⁹ kg/m³

Mass = Mass of Earth (Constant) = 5.97 x 10²⁴ kg

Volume = 4/3πR³ (Volume of Sphere)

R = Radius of Earth = ?

Therefore,

1 x 10⁹ kg/m³ = (5.97 x 10²⁴ kg)/[4/3πR³]

4/3πR³ = (5.97 x 10²⁴ kg)/(1 x 10⁹ kg/m³)

R³ = (3/4)(5.97 x 10¹⁵ m³)/π

R = ∛[0.95 x 10¹⁵ m³]

<u>R = 98304.75 m = 98.3 km</u>

6 0
3 years ago
I WILL MARK BRAINLIST PLEASE HELP!
Alenkasestr [34]

Answer:

Generally, when thermal energy is transferred to a material, the motion of its particles speeds up and its temperature increases. There are three methods of thermal energy transfer: conduction, convection, and radiation. ... Convection transfers thermal energy through the movement of fluids or gases in circulation cells.

Explanation:

4 0
3 years ago
Read 2 more answers
Importance of choke coil?<br><br>​
Verdich [7]

Answer:The choke coil works because it can act as an inductor. When the current pass through will change as AC currents creates a magnetic field in the coil that works against that current. This is known as inductance and blocks most of the AC current from passing through.

Explanation:

8 0
3 years ago
A 5-cm-high peg is placed in front of a concave mirror with a radius of curvature of 20 cm.
Andrei [34K]

Answer:

Explanation:

Using the magnification formula.

Magnification = Image distance(v)/object distance(u) = Image Height(H1)/Object Height(H2)

M = v/u = H1/H2

v/u = H1/H2...1

3) Given the radius of curvature of the concave lens R = 20cm

Focal length F = R/2

f = 20/2

f = 10cm

Object distance u = 5cm

Object height H2= 5cm

To get the image distance v, we will use the mirror formula

1/f = 1/u+1/v

1/v = 1/10-1/5

1/v = (1-2)/10

1/v =-1/10

v = -10cm

Using the magnification formula

(10)/5 = H1/5

10 = H1

H1 = 10cm

Image height of the peg is 10cm

4) If u = 15cm

1/v = 1/f-1/u

1/v = 1/10-1/15

1/v = 3-2/30

1/v = 1/30

v = 30cm

30/15 = H1/5

15H1 = 150

H1/= 10cm

5) if u = 20cm

1/v = 1/f-1/u

1/v = 1/10-1/20

1/v = 2-1/20

1/v = 1/20

v = 20cm

20/20 = H1/5

20H1 = 100

H1 = 5cm

6) If u = 30cm

1/v = 1/f-1/u

1/v = 1/10-1/30

1/v = 3-1/30

1/v = 2/30

v = 30/2 cm

v =>15cm

15/30 = Hi/5

30H1 = 75

H1 = 75/30

H1 = 2.5cm

4 0
3 years ago
A circuit contains a single 270-pf capacitor hooked across a battery. it is desired to store four times as much energy in a comb
yan [13]
The energy stored by a system of capacitors is given by
U= \frac{1}{2}C_{eq} V^2
where Ceq is the equivalent capacitance of the system, and V is the voltage applied.

In the formula, we can see there is a direct proportionality between U and C. This means that if we want to increase the energy stored by 4 times, we have to increase C by 4 times, if we keep the same voltage.

Calling C_1 = 270 pF the capacitance of the original capacitor, we can solve the problem by asking that, adding a new capacitor with C_x, the new equivalent capacitance of the system C_{eq} must be equal to 4C_1. If we add the new capacitance X in parallel, the equivalent capacitance of the new system is the sum of the two capacitance
C_{eq} = C_1 + C_x
and since Ceq must be equal to 4 C1, we can write
C_1+C_x = 4C_1
from which we find
C_x=3C_1=3 \cdot 270 pF=810 pF
5 0
3 years ago
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