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ruslelena [56]
3 years ago
14

The team monitoring a space probe exploring the outer solar system fi nds that radio transmissions from the probe take 2.53 hours

to reach earth. how distant (in meters) is the probe
Physics
1 answer:
Aleksandr-060686 [28]3 years ago
3 0
If you can observe, we are only given one parameter here which is the time. If you want to compute for the distance, you have to know the speed. The hint here is the 'radio transmissions'. All the information gathered by the probe from the space, is sent back to the Earth by electromagnetic waves. Hence, we must know the speed of electromagnetic waves. Since they are as fast as light, their speed is equal to 300 million meters per second. Then, we can finally determine the distance.

d = speed*time
d = (300×10⁶ m/s)(2.53 hours)

Since 1 hour = 3,600 seconds,

d = (300×10⁶ m/s)(2.53 hours)(3,600 seconds/1 hour)
d = 2.73×10⁻¹² m
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Explanation:

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5 0
3 years ago
Which of these pairs of atoms are isotpoes? (Physical Science) Pair A Pair B Pair C # protons 6 8 5 2 12 12 # neutrons 8 8 5 3 1
Aleksandr-060686 [28]

Answer:

I guess that the atoms are:

Protons:     6     8    5    2    12    12

Neutrons:  8     8     5   3     13    14

Now, two atoms are isotopes if they share the same number of protons (so both atoms are the same element) but they have a different number of neutrons.

From the given options, the only two that have the same number of protons but a different number of neutrons are:

Protons 12, neutrons 13

and

Protons 12, neutrons 14.

These two are isiotopes.

4 0
3 years ago
How much work is done in holding a 10 newton sack of potatoes while waiting in line at the grocery store for 3 minutes.​
hodyreva [135]

Answer:

Zero

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W = F × s

F = 10 N,

t = 3min = 180sec

s = 0( no change in postion)

W = 10 ×0

W = 0

4 0
3 years ago
A ball, which has a mass of 1.25 kg, is thrown straight up from the top of a building 225 meters tall with a velocity of 52.0 m/
Elena-2011 [213]

First we will find the speed of the ball just before it will hit the floor

so in order to find the speed of the cart we will first use energy conservation

KE_i + PE_i = KE_f + PE_f

\frac{1}{2}mv_i^2 + mgh = \frac{1}{2}mv_f^2 + 0

\frac{1}{2}(1.25)(52)^2 + 1.25(9.8)(225) = \frac{1}{2}(1.25)v_f^2

So by solving above equation we will have

v_f = 84.3 m/s

now in order to find the momentum we can use

P = mv

P = 1.25 \times 84.3

P = 105.4 kg m/s

6 0
3 years ago
2.- Si una cámara fotográfica emite un pulso de sonido para enfocar un objeto, determinar
uranmaximum [27]

Answer:

a. El tiempo de recorrido es 5.882\times 10^{-3} segundos para un objeto localizado a un metro de distancia de la cámara fotográfica.

b. El tiempo de recorrido es 0.118 segundos para un objeto localizado a un metro de distancia de la cámara fotográfica.

Explanation:

El sonido es un tipo de onda mecánica, que es un tipo de onda que necesita de un medio material para propagarse. En este caso, entendemos que el sonido se propaga a través del aire atmosférico hasta llegar a su destino y devolverse a rapidez constante. Entonces, podemos estimar el tiempo (t), medido en segundos, a partir de la siguiente fórmula:

t = \frac{2\cdot x_{s}}{v_{s}}

Donde:

x_{s} - Distancia entre la cámara fotográfica y el objeto, medida en metros.

v_{s} - Rapidez del sonido en el aire atmosférico, medida en metros por segundo.

A continuación, calculamos el tiempo de recorrido:

a. (x_{s} = 1\,m, v_{s} = 340\,\frac{m}{s})

t = \frac{2\cdot (1\,m)}{340\,\frac{m}{s} }

t = 5.882\times 10^{-3}\,s

El tiempo de recorrido es 5.882\times 10^{-3} segundos para un objeto localizado a un metro de distancia de la cámara fotográfica.

b. (x_{s} = 20\,m, v_{s} = 340\,\frac{m}{s})

t = \frac{2\cdot (20\,m)}{340\,\frac{m}{s} }

t = 0.118\,s

El tiempo de recorrido es 0.118 segundos para un objeto localizado a un metro de distancia de la cámara fotográfica.

8 0
3 years ago
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