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maks197457 [2]
4 years ago
15

Which statement best describes the common features of sound waves and light waves

Physics
1 answer:
Viktor [21]4 years ago
7 0
Sound waves or bounces off the wall and light waves are waves of light
You might be interested in
Why are incandescent lights fo inefficient?​
AnnZ [28]

Answer:

Explanation:

Incandescent bulbs are extremely inefficient because 90% of the energy is wasted as heat, with only 10% being converted to visible light. This gives incandescent bulbs a luminous efficiency rating of around 2%. Ultimately, incandescent bulbs perform better as heaters than as light sources.

8 0
3 years ago
A solid block is attached to a spring scale. When the block is suspended in air, the scale reads 21.2 N; when it is completely i
blsea [12.9K]

Answer:

7066kg/m³

Explanation:

The forces in these cases (air and water) are: Fa =mg =ρbVg Fw =(ρb −ρw)Vg where ρw = 1000 kg/m3 is density of water and ρb is density of the block and V is its density. We can find it from this two equations:

Fa /Fw = ρb / (ρb −ρw) ρb = ρw (Fa /Fa −Fw) =1000·(1* 21.2 /21.2 − 18.2)

= 7066kg/m³

Explanation:

8 0
3 years ago
Read 2 more answers
Aconstant current of 3 Afor 4 hours is required to charge an automotive battery. If the terminal voltage is V, where t is in hou
kirza4 [7]

Answer:

(a) 43.2 kC

(b) 0.012V kWh

(c) 0.108V cents

Explanation:

<u>Given:</u>

  • i = current flow = 3 A
  • t = time interval for which the current flow = 4\ h = 4\times 3600\ s = 14400\ s
  • V = terminal voltage of the battery
  • R = rate of energy = 9 cents/kWh

<u>Assume:</u>

  • Q = charge transported as a result of charging
  • E = energy expended
  • C = cost of charging

Part (a):

We know that the charge flow rate is the electric current flow through a wire.

\therefore i = \dfrac{Q}{t}\\\Rightarrow Q =it\\\Rightarrow Q = 3\times 14400\\\Rightarrow Q = 43200\ C\\\Rightarrow Q = 43.200\ kC\\

Hence, 43.2 kC of charge is transported as a result of charging.

Part (b):

We know the electrical energy dissipated due to current flow across a voltage drop for a time interval is given by:

E = Vit\\\Rightarrow E = V\times 3\times 4\\\Rightarrow E = 12V\ Wh\\\Rightarrow E = 0.012V\ kWh\\

Hence, 0.012V kWh is expended in charging the battery.

Part (c):

We know that the energy cost is equal to the product of energy expended and the rate of energy.

\therefore \textrm{Cost}=\textrm{Energy}\times \textrm{Rate}\\\Rightarrow C = ER\\\Rightarrow C = 0.012V\times 9\\\Rightarrow C =0.108V\ cents

Hence, 0.108V cents is the charging cost of the battery.

4 0
4 years ago
A copper cable needs to carry a current of 160 A with a power loss of only 2.0 W/m. What is the required radius of the copper ca
lesantik [10]

Answer:

The radius of the cable is 0.0083 m or 8.3 mm.

Explanation:

The resistance of copper cable of 1 meter length will be given by

R_{cable} = \frac{\rho \times l}{a}    ....     (i)

where the resistivity of copper is \rho = 1.7 \times 10^{-8} \Omega.m , and l is the length of the wire which is considered to be 1m, and a is the cross sectional area of the wire in m^{2}.

From the formula of power we know that, P = I^2 R    ....    (ii)

Therefore 2 W/m  = (160)^2 \times R     ....     (iii)  

where  the resistance,R, actually means the resistance of the cable per meter.

Therefore R ( resistance of cable per meter)

= \frac{2}{160^2}  = 7.812 \times 10^{-5} ohms / meter.       ....    (iv)

Therefore from (i)

7.812 \times 10^{-5} = \frac{1.7 \times 10^{-8} \times 1}{a}  = \frac{1.7 \times 10^{-8} \times 1}{\pi r^{2} }       .....     (v)

where cross sectional area of the cable, a  = \pi r^2,

where r is the radius of the cable, and length of cable,l = 1m.

Therefore r  = \sqrt{\frac{ 1.7 \times 10^{-8}}{\pi \times 7.812 \times 10^{-5} } }  =  0.0083m = 8.3 mm

5 0
4 years ago
On a typical clear day, the atmospheric electric field points downward and has a magnitude of approximately 103 N/C. Compare the
Nina [5.8K]

Answer:

a) FE = 0.764FG

b) a = 2.30 m/s^2

Explanation:

a) To compare the gravitational and electric force over the particle you calculate the following ratio:

\frac{F_E}{F_G}=\frac{qE}{mg}              (1)

FE: electric force

FG: gravitational force

q: charge of the particle = 1.6*10^-19 C

g: gravitational acceleration = 9.8 m/s^2

E: electric field = 103N/C

m: mass of the particle = 2.2*10^-15 g = 2.2*10^-18 kg

You replace the values of all parameters in the equation (1):

\frac{F_E}{F_G}=\frac{(1.6*10^{-19}C)(103N/C)}{(2.2*10^{-18}kg)(9.8m/s^2)}\\\\\frac{F_E}{F_G}=0.764

Then, the gravitational force is 0.764 times the electric force on the particle

b)

The acceleration of the particle is obtained by using the second Newton law:

F_E-F_G=ma\\\\a=\frac{qE-mg}{m}

you replace the values of all variables:

a=\frac{(1.6*10^{-19}C)(103N/C)-(2.2*10^{-18}kg)(9.8m/s^2)}{2.2*10^{-18}kg}\\\\a=-2.30\frac{m}{s^2}

hence, the acceleration of the particle is 2.30m/s^2, the minus sign means that the particle moves downward.

7 0
3 years ago
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