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3 years ago

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15. For waves moving at a constant speed, if the wavelength is doubled, the frequency is

1 answer:

faltersainse [42]3 years ago

5 0

Answer:

halved

.............

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What factors affect potential energy

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Mass ,gravity and height

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3 years ago

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When atoms are split, they release energy. This concept applies to (2 points)

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This applies to nuclear reactions, specifically nuclear fission.

This huge release of energy has been used in atomic bombs and in the nuclear reactors that generate electricity.

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6 0

3 years ago

A 5.00 kilogram mass is traveling at 100. meters per second. Determine the speed of the mass after an impulse with a magnitude o

faltersainse [42]

m = mass = 5 kg

$v_{i}$ = initial velocity = 100 m/s

$v_{f}$ = final velocity = ?

I = impulse = 30 Ns

Using the impulse-change in momentum equation

I = m( $v_{f}$ - $v_{i}$ )

30 = 5 ( $v_{f}$ - 100)

$v_{f}$ = 106 m/s

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3 years ago

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Six artificial satellites circle a space station at constant speed. The mass m of each satellite, distance L from the space stat

nikklg [1K]

Answers:

a) $T_{2}>T_{5}>T_{1}>T_{3}=T_{6}>T_{4}$

b) $a_{4}>a_{6}>a_{1}>a_{3}>a_{5}>a_{2}$

Explanation:

a) Since we are told the satellites circle the space station at constant speed, we can assume they follow a uniform circular motion and their tangential speeds $V$ are given by:

$V=\omega L=\frac{2\pi}{T} L$ (1)

Where:

$\omega$ is the angular frequency

$L$ is the radius of the orbit of each satellite

$T$ is the period of the orbit of each satellite

Isolating $T$ :

$T=\frac{2 \pi L}{V}$ (2)

Applying this equation to each satellite:

$T_{1}=\frac{2 \pi L}{V_{1}}=261.79 s$ (3)

$T_{2}=\frac{2 \pi L}{V_{2}}=1570.79 s$ (4)

$T_{3}=\frac{2 \pi L}{V_{3}}=196.349 s$ (5)

$T_{4}=\frac{2 \pi L}{V_{4}}=98.174 s$ (6)

$T_{5}=\frac{2 \pi L}{V_{5}}=785.398 s$ (7)

$T_{6}=\frac{2 \pi L}{V_{6}}=196.349 s$ (8)

Ordering this periods from largest to smallest:

$T_{2}>T_{5}>T_{1}>T_{3}=T_{6}>T_{4}$

b) Acceleration $a$ is defined as the variation of velocity in time:

$a=\frac{V}{T}$ (9)

Applying this equation to each satellite:

$a_{1}=\frac{V_{1}}{T_{1}}=0.458 m/s^{2}$ (10)

$a_{2}=\frac{V_{2}}{T_{2}}=0.0254 m/s^{2}$ (11)

$a_{3}=\frac{V_{3}}{T_{3}}=0.4074 m/s^{2}$ (12)

$a_{4}=\frac{V_{4}}{T_{4}}=1.629 m/s^{2}$ (13)

$a_{5}=\frac{V_{5}}{T_{5}}=0.101 m/s^{2}$ (14)

$a_{6}=\frac{V_{6}}{T_{6}}=0.814 m/s^{2}$ (15)

Ordering this acceerations from largest to smallest:

$a_{4}>a_{6}>a_{1}>a_{3}>a_{5}>a_{2}$

4 0

3 years ago

A large sheet of charge has a uniform charge density of 9  μCm2. What is the electric field due to this charge at a point just

Alex73 [517]

Answer:

Explanation:

Surface charge density, σ = 9 μC/m² = 9 x 10^-6 C/m²

According to the Gauss theorem,

Electric field due to the sheet is given by

$E = \frac {\sigma }{2\epsilon _{0}}$

$E = \frac{9\times 10^{-6}}{2\times 8.854\times 10^{-12}}$

E = 5.08 x 10^5 N/C

7 0

3 years ago