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3 years ago

11

which statement about waves is false? a. waves transfer matter b. waves can change direction c. waves can interact with each oth

er d. waves can transfer energy to matter

1 answer:

satela [25.4K]3 years ago

8 0

A.

Waves can transfer energy but they can’t transfer matter

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A professor designing a class demonstration connects a parallel-plate capacitor to a battery, so that the potential difference b

Lesechka [4]

Answer:

a) Q = 397.57 pC , Q = 3.18 104 pC , b) C = 1.157 10⁻¹⁰ F , V = 3.4375 V ,

c) U = 54.7 nJ , d) ΔU = 54 nJ,

Explanation:

a) The capacity of a capacitor is defined

C = Q / V

Q = C V

can also be calculated using geometry consideration

C = e or A / d

we reduce to the SI system

A = 25.0 cm² (1 m / 10² cm) 2 = 25.0 10⁻⁴ m²

d = 1.53 cm = 1.53 10⁻² m

we substitute

Q = eo A / d V

Q = 8.85 10⁻¹² 25 10⁻⁴ / 1.53 10⁻² 275

Q = 3.9757 10⁻¹⁰ C

let's reduce to pC

Q = 3.9757 10⁻¹⁰ C (10¹² pC / 1 C)

Q = 397.57 pC

when the capacitor is introduced into the water the dielectric constant is different

Q = k Q₀

Q = 80 397.57

Q = 3.18 104 pC

b) Find capacitance and voltage after submerged in water

C = k C₀

C = 80 8.85 10⁻¹² 25 10⁻⁴ / 1.53 10⁻²

C = 1.157 10⁻¹⁰ F

V = Vo / k

V = 275/80

V = 3.4375 V

c) The stored energy is

U = ½ C V²

U = ½, 85 10⁻¹² 25 10⁻⁴ / 1.53 10⁻² 275²

U = 5.47 10⁻⁸ J

let's reduce to nJ

109 nJ = 1 J

U = 54.7 nJ

d) energy after submerging

U = ½ (kCo) (Vo / k) 2

U = ½ Co Vo2 / k

U = U₀ / k

U = 54.7 / 80 nJ

U = 0.68375 nJ

the energy change is

ΔU = U₀ -U

ΔU = 54.7 - 0.687375

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3 years ago

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Answer:

A, and D are the answers

Explanation:

The pulley. It is located where the bicycle chain and gears are. The chain is wrapped around the pulley which turns and causes the wheel to turn on its axle.

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3 years 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

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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

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The amount of heat required to melt lead at it melting temperature is 23.0 kJ/kg. The amount of heat required to melt mercury at

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The addition of 24 kJ of energy will allow all of the mercury and lead to change from solid to liquid. The temperature of each substance will also increase.

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