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Genrish500 [490]

2 years ago

14

You discover a Cepheid variable star with a 30 day period in the Milky Way. Through careful monitoring for a few years with the

Hubble Space Telescope, you determine that it has a parallax of 0.004 arcseconds. How distant is this Cepheid variable star in parsecs?

1 answer:

soldi70 [24.7K]2 years ago

3 0

Answer:c

Explanation:

Bc it makes sense

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A 56 kg diver runs and dives from the edge of a cliff into the water which is located 4.0 m below. If she is moving at 8.0 m/s t

Reil [10]

Answer:

1) 2197.44 J

2) 0 J

3) 2197.44 J = Constant

4) 2197.44 J

5) Approximately 8.86 m/s

Explanation:

The given parameters are;

The mass of the diver, m = 56 kg

The height of the cliff, h = 4.0 m

The speed with which the diver is moving, vₓ = 8.0 m/s

The gravitational potential energy = Mass, m × Height of the cliff, h × Acceleration due to gravity, g

1) Her gravitational potential energy = 56 × 4.0 × 9.81 = 2197.44 J

2) The kinetic energy = 1/2·m·u²

Where;

u = Her initial velocity = 0 when she just leaves the cliff

Therefore;

Her kinetic energy when she just leaves the cliff = 1/2 × 56 × 0² = 0 J

3) The total mechanical energy = Kinetic energy + Potential energy

The total mechanical energy is constant

Her total mechanical energy relative to the water surface when she leaves the cliff = Her gravitational potential energy = 2197.44 J = Constant

4) Her total mechanical energy relative to the water surface just before she enters the water = 2197.44 J

5) The speed with which she enters the water, v, is given from, v² = u² + 2·g·h

Where;

u = The initial velocity at the top of the cliff before she jumps= 0 m/s

∴ v² = 0² + 2 × 9.81 × 4 = 78.48

v = √78.48 ≈ 8.86 m/s

The speed with which she enters the water, v ≈ 8.86 m/s

7 0

2 years ago

The speed of sound in seawater is 1470 m/s. A dolphin sends out a click that reflects off of an

Nitella [24]

Answer: 0.204 s

Explanation:

The speed of sound $V$ is defined as the distance traveled $d$ in a especific time $t$ :

$V=\frac{d}{t}$

Where:

$V=1470 m/s$ is the speed of sound in seawater

$t$ is the time the sound wave travels from the dolphin and then returns after the reflection

$d=2(150 m)$ is twice the distance between the dolphin and the object to which the sound waves are reflected

Finding $t$ :

$t=\frac{d}{V}$

$t=\frac{2(150 m)}{1470 m/s}$

<u>Finally:</u>

$t=0.204 s$

3 0

2 years ago

Justine is ice-skating at the Lloyd Center what is her final velocity if she accelerates at a rate of 2.0 meters per second for

lilavasa [31]

2*3.5 = 7m/s

You multiply the acceleration per the time (they both are in seconds, otherwise, you should set them in the same units).

7 0

1 year ago

A wave has a frequency of 34 Hz and a wavelength of 2.0 m. What is the speed of the wave? Use . A. 17 m/s B. 36 m/s C. 0.059 m/s

mel-nik [20]

F= (speed)/(wavelength)

Therefore, speed = Frequency x wavelength
V = 68m/s

8 0

2 years ago

Read 2 more answers

Driving your Ferrari through the Italian countryside at a speedy 88 m/s, you approach an opera diva singing a high C (1,046 Hz).

MrRissso [65]

Answer:

You will hear the note E₆

Explanation:

We know that:

Your speed = 88m/s

Original frequency = 1,046 Hz

Sound speed = 340 m/s

The Doppler effect says that:

$f' = \frac{v \pm v0 }{v \mp vs}*f$

Where:

f = original frequency

f' = new frequency

v = velocity of the sound wave

v0 = your velocity

vs = velocity of the source, in this case, the source is the diva, we assume that she does not move, so vs = 0.

Replacing the values that we know in the equation we have:

$f' = \frac{340 m/s + 88m/s}{340 m/s} *1,046 Hz = 1,316.73 Hz$

This frequency is close to the note E₆ (1,318.5 Hz)

7 0

1 year ago