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3 years ago

When several radio telescopes are wired together, the resulting network is called a radio

Physics

1 answer:

WINSTONCH [101]3 years ago

4 0

Answer:

Interferometer

Explanation:

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A 225 kg block is pulled by two horizontal forces. The first force is 178 N at a 41.7-degree angle and the second is 259 N at a

yawa3891 [41]

Answer:

52.9 N, 364.7 N

Explanation:

First of all, we need to resolve both forces along the x- and y- direction. We have:

- Force A (178 N)

$A_x = (178 N)(cos 41.7^{\circ})=132.9 N\\A_y = (178 N)(sin 41.7^{\circ})=118.4 N$

- Force B (259 N)

$B_x = (259 N)(cos 108^{\circ})=-80.0 N\\B_y = (259 N)(sin 108^{\circ})=246.3 N$

So the x- and y- component of the total force acting on the block are:

$R_x = A_x + B_x = 132.9 N - 80.0 N =52.9 N\\R_y = A_y + B_y = 118.4 N +246.3 N = 364.7 N$

7 0

3 years ago

If the final speed of an object as is strikes the ground is 77 m/s and it was in the air for 6.5 seconds. What was the initial d

azamat

Answer:

The answer to your question is: 13.2 m/s

Explanation:

final speed (fs) = 77 m/s

t = 6.5 s

gravity (g) = 9.81 m/s2

initial speed (is) = ?

Formula

fs = is + gt from this equation we clear "is" = fs - gt

Substitution is = 77 - (9,81)(6.5)

Process is = 77 - 63.8

is = 13.2 m/s

8 0

3 years ago

Which platform has touch controls?

Crazy boy [7]

I think C but not sure

8 0

3 years ago

Read 2 more answers

Please Please Please help on these 3 questions :) 20 POINTS!! - NO LINKS PLEASE

GenaCL600 [577]

Answer:

Catapult on the ground: Normal, gravity

Catapult (I'm assuming launching marshmallow): Reaction of Force Applied

Marshmallow: Force Applied

Explanation:

This is the forces that act on a stationary object and a launched object. The catapult may also experience a force friction if your teacher is taking a more practical sense.

3 0

3 years ago

The specific heat of water in its solid phase (ice) is 2090 J/(kg K), while in the liquid phase (water) its specific heat is 419

oee [108]

Answer:

d. 149 ⁰C.

Explanation:

Given;

mass of the block of ice, m = 2 kg

specific heat capacity of the ice, C = 2090 J/(kgK)

initial temperature of the ice, t₁ = -90 ⁰C

heat added to the ice, H = 1,000,000 J

let the final temperature of the ice = t₂

The final temperature of the ice after adding the heat is calculated as follows;

$H = mC_{ice} \Delta t\\\\H = mC_{ice} (t_2 - t_1)\\\\1,000,000 = 2 \times 2090 \times (t_2 - (-90))\\\\1,000,000 = 4,180(t_2 +90)\\\\1,000,000 = 4,180t_2 + 376,200\\\\1,000,000 - 376,200 = 4,180t_2\\\\623,800 = 4,180 t_2\\\\t_2 = \frac{623,800}{4,180} \\\\t_2 = 149 \ ^0C$

Therefore, the new temperature of the water is 149 ⁰C.

4 0

2 years ago