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4 years ago

The average distance from the sun to Pluto is approximately 6.10 × 109 km. How long does it take light from

1 answer:

3 0

$V= \frac{S}{t}$
$t= \frac{S}{V}$ <u />
$t= \frac{S}{c}$
$t= \frac{6.1*10^{12}}{299792458}$
$t=20347.4098071s$

It takes 20347.4098071s for light from the sun to reach Pluto.
The 6.1*10^9 is replaced by 6.1*10^12 on line 4 because we convert the distance from km to m.
c = speed of light. If a different value was given in the previous question then use that instead of the value I used to do the final calculation.

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Bird bones have air pockets in them to reduce their weight. This also gives them an average density significantly less than that

mixas84 [53]

Answer:

mas of water displaced = 41.4 g

Explanation:

Weight in air = True weight = 45 g

Apparent weight = 3.6 g

Apparent weight = True weight - Buoyant force

Buoyant force = 45 g - 3.6 g = 41.4 g

Weight of water displaced = Buoyant force

Weight of water displaced = 41.4 g dyne

mas of water displaced = 41.4 g

8 0

3 years ago

How are earthquake epicenters active volcanoes and mountain ranges distributed on the map?

Sati [7]

Answer:

Volcanoes and earthquakes are not randomly distributed around the globe.

Explanation:

Instead they tend to occur along limited zones or belts. With the understanding of plate tectonics, scientists recognized that these belts occur along plate boundaries.

6 0

2 years ago

In a drill during basketball practice, a player runs the length of the 30-meter court and back. The player does this three times

Sergeeva-Olga [200]

Answer:

0 m/s

Explanation:

Average velocity of an object is given by the net displacement divided by time taken. Displacement is equal to the shortest path covered by the object.

In this problem, a player runs the length of the 30-meter court and back. The player does this three times in 60 seconds.

As the player runs the court and returns to the original point. It would mean that the shortest path covered is 0.

Average velocity = displacement/time

v=0/30

v = 0 m/s

Hence, the correct option is (1).

6 0

4 years ago

A refrigerator draws 4.5 A of current while operating on a 120-V power line. If the refrigerator runs 50% of the time and electr

Murljashka [212]

Answer:

$ 29.16

Explanation:

The following data were obtained from the question:

Current (I) = 4.5 A

Voltage (V) = 120 V

Time (t) = 30 days

1 KWh = $ 0.15

Cost for 30 days =.?

From the question given above, we were told that the refrigerator runs 50% of the time.

Thus, the time of operation of the refrigerator will be:

Time (t) = 30 × 50% = 30 × 50/100

= 15 days.

Next, we shall convert 15 days to hours. This can be obtained as follow:

1 day = 24 h

15 days = 15 days × 24 h / 1 day

15 days = 360 h

Next, we shall determine the energy consumed in KWh This can be obtained as follow:

Current (I) = 4.5 A

Voltage (V) = 120 V

Time (t) = 360 h

Energy (E) =?

E = ivt

E = 4.5 × 120 × 360

E = 194400 Wh

Next, we shall convert 194400 Wh to KWh. This can be obtained as follow:

1000 Wh = 1 KWh

Therefore,

194400 Wh = 194400 Wh × 1 KWh/ 1000 Wh

194400 Wh = 194.4 KWh

Thus the energy consumed is 194.4 KWh

Finally, we shall determine the cost of energy consumed. This can be obtained as follow:

From the question given above,

1 KWh cost $ 0.15

Therefore, 194.4 KWh will cost = 194.4 KWh × $ 0.15 = $ 29.16

Therefore, the cost of running the refrigerator for 30 days is $ 29.16

3 0

3 years ago

An electron starts from rest near a negatively charged metal plate, and is accelerated towards a positive plate through a potent

Greeley [361]

Answer:

The magnetic field at the location is $5.375\times10^{-15}\ T$

Explanation:

Given that,

Potential difference = 1320 V

Distance = 0.008 m

We need to calculate the velocity

Using conservation of energy

Electric potential energy lost = kinetic energy gained

$qV=\dfrac{1}{2}mv^2$

$v^2=\dfrac{2qV}{m}$

Put the value into the formula

$v^2=\dfrac{2\times1.6\times10^{-19}\times1320}{9.1\times10^{-31}}$

$v=\sqrt{\dfrac{2\times1.6\times10^{-19}\times1320}{9.1\times10^{-31}}}$

$v=2.15\times10^{7}\ m/s$

We need to calculate the magnetic field at the location

Using formula of magnetic field

$B=\dfrac{\mu qv}{4\pi r^2}$

Put the value into the formula

$B=\dfrac{4\pi\times10^{-7}\times1.6\times10^{-19}\times2.15\times10^{7}}{4\pi\times(0.008)^2}$

$B=5.375\times10^{-15}\ T$

Hence, The magnetic field at the location is $5.375\times10^{-15}\ T$

5 0

3 years ago