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Hatshy [7]
3 years ago
6

Two astronauts are 2.00 m apart in their spaceship. One speaks to the other. The conversation is transmitted to earth via electr

omagnetic waves. The time it takes for sound waves to travel at 354 m/s through the air between the astronauts equals the time it takes for the electromagnetic waves to travel to the earth. How far away from the earth is the spaceship?
Physics
1 answer:
Eduardwww [97]3 years ago
5 0

Answer:

D=1693742.7m

Explanation:

For sound waves we have v=d/t where v is the speed of sound and d the distance between the astronauts, while for electromagnetic waves we have c=D/t where c is the speed of light and D the distance between the spaceship and Earth. <em>We have written both times as the same</em> because is what is imposed by the problem, so we have t=d/v=D/c, which means:

D=\frac{dc}{v}

And for our values:

D=\frac{(2m)(299792458m/s)}{354m/s}=1693742.7m

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Cheese attributes its sour taste because of its lactic acid content. However, this should not be an indication to whether the cheese is spoiling already. We can tell its expiry date through the manufacturing details of the product. The product should be good to consume before the manufacturing expiry date. Every package must include expiry date in order to inform the consumer about the range of its consumability attributes. In the case where molds are present on the cheese already, cheese on some parts may not be edible already. Molds are due to the oxidation of glucose through the action of specific bacteria.
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4 years ago
How is the frequency of an electromagnetic wave related to energy
aivan3 [116]

Answer: The energy carried by an electromagnetic wave is proportional to the frequency of the wave.  

Explanation:

Electromagnetic waves that are of higher energy than visible light (higher frequency, shorter wavelength) include ultraviolet light, X-rays, and gamma rays.

7 0
3 years ago
Can someone please help me with these physics problems? I just don’t even know where to start.
KIM [24]

#1

for the block of mass 5 kg normal force is given as

F_n = mg

F_n = 5*9.8 = 49 N

friction force is given as

F_f = \mu F_n

F_f = 0.1*49 = 4.9 N

Net force is given as

F_{net} = ma

F_{net} = 5*2 = 10 N

now we know that

F_{net} = F_{app} - F_f

10 = F_{app} - 4.9

F_{app} = 14.9 N

#2

Normal force is given as

F_n = mg

F_n = 6*9.8

F_n = 58.8 N

now we know that

F_{net} = F_{app} - F_f

F_{net} = 0

as object moves with constant velocity

F_{app} = F_f = 15 N

now for coefficient of friction we can use

F_f = \mu F_n

15 = \mu * 58.8

\mu = 0.255

#3

net force upwards is given as

F = 1.2 * 10^{-4} N

mass is given as

m = 7 * 10^{-5} kg

now as per newton's law we can say

F = ma

1.2 * 10^{-4} = 7 * 10^{-5} * a

a = 1.71 m/s^2

#4

As we know that when block is sliding on rough surface

part a)

net force = applied force - frictional force

F_{net} = F_{app} - F_f

ma = F_{app} - F_f

5*6 = 40 - F_{f}

F_f = 40 - 30 = 10 N

part b)

for coefficient of friction we can use

F_f = \mu F_n

10 = \mu * F_n

here normal force is given as

F_n = mg = 5*9.8 = 49 N

now we have

\mu = \frac{10}{49} = 0.204

#5

if an object is initially at rest and moves 20 m in 5 s

so we can use kinematics to find out the acceleration

d = v_i*t + \frac{1}{2}at^2

20 = 0 + \frac{1}{2}a(5^2)

a = 1.6 m/s^2

now net force is given as

F_{net} = ma

F_{net} = 10*1.6 = 16 N

#6

an object travelling with speed 25 m/s comes to stop in 1.5 s

so here acceleration of object is given as

a = \frac{v_f - v_i}{t}

a = \frac{0 - 25}{1.5} = -16.67 m/s^2

now the force is gievn as

F = ma

F = 5*16.67 = 83.3 N

3 0
4 years ago
How can magnets cause objects to have kinetic energy? ​
Damm [24]

Answer:

this has to do with the drop in voltage, or potential as electrons traverse the material .

4 0
3 years ago
Will mark as BRAINLIEST.......
irinina [24]

Explanation:

It is given that,

The Displacement x of particle moving in one dimension under the action of constant force is related to the time by equation as:

x=4t^3+3t^2-5t+2

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x is in meters and t is in sec

We know that,

Velocity,

v=\dfrac{dx}{dt}\\\\v=\dfrac{d(4t^3+3t^2-5t+2)}{dt}\\\\v=12t^2+6t-5

(a) i. t = 2 s

v=12(2)^2+6(2)-5=55\ m/s

At t = 4 s

v=12(4)^2+6(4)-5=211\ m/s

(b) Acceleration,

a=\dfrac{dv}{dt}\\\\a=\dfrac{d(12t^2+6t-5)}{dt}\\\\a=24t+6

Pu t = 3 s in above equation

So,

a=24(3)+6\\\\a=78\ m/s^2

Hence, (a) (i) v = 55 m/s (ii) v = 211 m/s and (b) 78 m/s²

6 0
3 years ago
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