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expeople1 [14]
3 years ago
12

What is carried by a wave?

Physics
2 answers:
prohojiy [21]3 years ago
6 0

Answer

Energy

Explanation

wave is a disturbance which travels through a medium carrying energy from one point to another without causing any permanent displacement in that medium. So energy is carried by a wave.

Nata [24]3 years ago
3 0
Waves carry energy from one place to another. Because waves carry energy, some waves are used for communication, eg radio and television waves and mobile telephone signals. ... Some types of waves need to be transmitted through matter, either a solid, liquid or a gas. For example, water waves have to travel in water.
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An astronomical telescope has an objective of diameter 20 cm with a focal length of 180 cm. the telescope is used with an eyepie
Bingel [31]
By definition;
M = fo/fe

Where,
M = Angular magnification
fo = Focal length of objective lens
fe = Focal length of eyepiece lens

From the information given;
M = 180/30 = 6
8 0
3 years ago
James Bond is trying to escape his enemy on a speedboat but
SVEN [57.7K]

Answer:

100 m/s

Explanation:

Mass the mass of Bond's boat is m₁. His enemy's boat is twice the mass of Bond's i.e. m₂ = 2 m₁

Initial speed of Bond's boat is 0 as it won't start and remains stationary in the water. The initial speed of enemy's boat is 50 m/s. After the collision, enemy boat is  completely stationary. Let v₁ is speed of bond's boat.

It is the concept of the conservation of momentum. It remains conserved. So,

m_1u_1+m_2u_2=m_1v_1+m_2v_2

Putting all the values, we get :

0+(2m_1)50=m_1v_1+(2m_2)(0)\\\\100m_1=m_1v_1\\\\v_1=100\ m/s

So, Bond's boat is moving with a speed of 100 m/s after the collision.

3 0
3 years ago
A circuit contains four capacitors in parallel (10 F, 3 F, 7 F, and 1 F). What is the equivalent capacitance of this circuit?
Oxana [17]

The equivalent capacitance (C_{eq}) of an electrical circuit containing four capacitors which are connected in parallel is equal to: A. 21 F.

<h3>The types of circuit.</h3>

Basically, the components of an electrical circuit can be connected or arranged in two forms and these are;

  • Series circuit
  • Parallel circuit

<h3>What is a parallel circuit?</h3>

A parallel circuit can be defined as an electrical circuit with the same potential difference (voltage) across its terminals. This ultimately implies that, the equivalent capacitance (C_{eq}) of two (2) capacitors which are connected in parallel is equal to the sum of the individual (each) capacitances.

Mathematically, the equivalent capacitance (C_{eq}) of an electrical circuit containing four capacitors which are connected in parallel is given by this formula:

Ceq = C₁ + C₂ + C₃ + C₄

Substituting the given parameters into the formula, we have;

Ceq = 10 F + 3 F + 7 F + 1 F

Equivalent capacitance, Ceq = 21 F.

Read more equivalent capacitance here: brainly.com/question/27548736

#SPJ1

3 0
2 years ago
25 POINTS FOR CORRECT ANSWER
castortr0y [4]
No.  I do not agree with Stefan.  Quite the contrary.  I disagree
with his description of "<span>angle of incidence" as the angle between
the surface of the mirror and the incoming ray. 

The correct description of "angle of incidence" is </span><span>the angle between
the NORMAL TO the surface of the mirror and the incoming ray. 

Thus, the true angle of incidence is the complement of the angle that
Stefan calculates or measures.</span>
5 0
3 years ago
Read 2 more answers
A 10-cm-long spring is attached to theceiling. When a 2.0 kg mass is hung from it,the spring stretches to a length of 15 cm.a.Wh
alekssr [168]

(a) 392 N/m

Hook's law states that:

F=k\Delta x (1)

where

F is the force exerted on the spring

k is the spring constant

\Delta x is the stretching/compression of the spring

In this problem:

- The force exerted on the spring is equal to the weight of the block attached to the spring:

F=mg=(2.0 kg)(9.8 m/s^2)=19.6 N

- The stretching of the spring is

\Delta x=15 cm-10 cm=5 cm=0.05 m

Solving eq.(1) for k, we find the spring constant:

k=\frac{F}{\Delta x}=\frac{19.6 N}{0.05 m}=392 N/m

(b) 17.5 cm

If a block of m = 3.0 kg is attached to the spring, the new force applied is

F=mg=(3.0 kg)(9.8 m/s^2)=29.4 N

And so, the stretch of the spring is

\Delta x=\frac{F}{k}=\frac{29.4 N}{392 N/m}=0.075 m=7.5 cm

And since the initial lenght of the spring is

x_0 = 10 cm

The final length will be

x_f = x_0 +\Delta x=10 cm+7.5 cm=17.5 cm

8 0
3 years ago
Read 2 more answers
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