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3 years ago

9

The components and functionality of the event horizon telescope.

1 answer:

andrew11 [14]3 years ago

6 0

Answer:

The Event Horizon Telescope comprises of a an interconnection of radio observatories. These radio observatories work in unison and helps in efficient observation of radio sources associated with black holes with angular resolution in comparison to their event horizons. This equipment functions by ensuring that anything or light which goes through the event horizon cannot escape from the black hole. Instead there will be a continual falling towards the center of the black hole.

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Two tugboats pull a disabled supertanker. Each tug exerts a constant force of 1.60×106 N, one at an angle 13.0 west of north, an

Juliette [100K]

Answer: $W=1.93\times 10^9 J$

Explanation:

Given

Force $F=1.6\times 10^{6} N$

one at an angle of $13^{\circ}$ East of North and another at $13^{\circ}$ West of North

Net Force is in North Direction

$F_{net}=2F\cos 13$

Forces in horizontal direction will cancel out each other

thus Work done will be by north direction forces

$W=2F\cdot \cos 30\cdot s$

here $s=0.7 km$

$W=2\times 1.6\times 10^{6}\cdot \cos 30\cdot 700$

$W=1.93\times 10^9 J$

3 0

4 years ago

A concert loudspeaker suspended high off the ground emits 39 w of sound power. a small microphone with a 1.0 cm2 area is 60 m fr

Gelneren [198K]

As we know that intensity of sound is defined as power received per unit area

so here the power of source is given as

$P = 39 W$

distance of microphone is given as

$d = 60 m$

now if loudspeaker is considered as spherical source then we will have

$Intensity = \frac{power}{area}$

$Intensity = \frac{39}{4\pi r^2$

here

r = d = 60 m

$Intensity = \frac{39}{4\pi (60^2)}$

$Intensity = 0.86 \times 10^{-3} W/m^2$

7 0

4 years ago

A trumpet player on a moving railroad flatcar moves toward a second trumpet player standing alongside the track both play a 490

labwork [276]

Answer:

The value is $v_s = 1.394 \ m/s$

Explanation:

From the question we are told that

The frequency of the second player is $f_2 = 490 \ Hz$

The beat frequency is $f_b = 2.0 \ Hz$

The speed of sound is $v_s = 343 \ m/s$

Generally the frequency of the note played by the first player is mathematically represented as

$f_1 = f_2 + f_b$

=> $f_1 = 490 + 2.0$

=> $f_1 = 492 Hz$

From the relation of Doppler Shift we have that

$f_1 = \frac{ f_2 (v+ v_o )}{v-v_s }$

Here $v_o\ is\ the\ velocity\ of\ the\ observer\ with\ value\ 0 \ m/s$

So

$492 = \frac{ 490 (343+0 )}{343 -v_s }$

=> $v_s = 1.394 \ m/s$

4 0

3 years ago

Your camera's zoom lens has an adjustable focal length ranging from 77 to 224 mm. What is its range of powers

LiRa [457]

Answer: $0.446\ D\ \text{to}\ 1.298\ D$

Explanation:

Given

Camera has an adjustable focal length ranging from $77$ to $224$ mm

Power is given by Reciprocal of focal length in meter

Power corresponding to 77 cm

$\Rightarrow P=\dfrac{1}{0.77}\\\\\Rightarrow P=1.298\ D$

Power corresponding to 224 cm

$\Rightarrow P=\dfrac{1}{2.24}\\\\\Rightarrow P=0.446\ D$

Therefore, the range of the power is $0.446\ D\ \text{to}\ 1.298\ D$

7 0

3 years ago

Find the mass of a truck moving at 30m/s with 425,000 joules of energy

Leno4ka [110]

Answer:944.4kg

Explanation:

Kinetic energy(ke)=425000 joules

Velocity(v)=30m/s

Mass=(2 x ke) ➗(v x v)

Mass=(2 x 425000) ➗ (30 x 30)

Mass=(850000) ➗ 900

Mass=944.4

Mass=944.4kg

4 0

4 years ago