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3 years ago

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One light-year is the distance light travels in one year. This distance is equal to 9.461 1015 m. After the sun, the star neares

t to Earth is Alpha Centauri, which is about 4.35 light-years from Earth. Express this distance in a. megameters. b. picometers.

1 answer:

Oliga [24]3 years ago

4 0

Answer:

A.) 411.6 Mega metres

B.) 411.6 × 10^18 Picometers

Explanation:

Given that One light-year is the distance light travels in one year. This distance is equal to 9461 1015 m.

After the sun, the star nearest to Earth is Alpha Centauri, which is about 4.35 light-years from Earth.

The distance travelled will be

Distance = 9461 1015 × 4.35

Distance = 411557915.3 m

To Express this distance in

a. megameters, divide the value by 1000000

411557915.3 / 1000000

411.5579153 Mega metre

411.6 Mega metres approximately

b. picometers, multiply by 10^12

That is, 411557915.3 × 10^12

411.6 × 10^18 Picometers

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A plane is flying east when it drops some supplies to a designated target below. The supplies land after falling for 10 seconds.

Diano4ka-milaya [45]

solution:

As Given plane is flying in east direction.

It throws back some supplies to designated target.

Time taken by the supply to reach the target =10 seconds

g = Acceleration due to gravity = - 9.8 m/s²[Taken negative as object is falling Downwards]

As we have to find distance from the ground to plane which is given by d.

d = $\frac{1}{2}\times g\times t^2$

= $\frac{1}{2}\times (9.8) \times(100) =50\times 9.8=490$ meters

Distance from the ground where supplies has to be land to plane = Option B =490 meters

6 0

3 years ago

Read 2 more answers

You toss a ball straight up with an initial speed of 30m/s. How high does it go, and how long is it in the air (neglecting air r

Brut [27]

Explanation:

Given that,

A ball is tossed straight up with an initial speed of 30 m/s

We need to find the height it will go and the time it takes in the air.

At its maximum height, its final speed, v = 0 and it will move under the action of gravity. Using equation of motion :

v = u +at

Here, a = -g

v = u -gt

i.e. u = gt

$t=\dfrac{u}{g}\\\\t=\dfrac{30\ m/s}{9.8\ m/s^2}\\\\t=3.06\ s$

So, the time for upward motion is 3.06 seconds. It means that it will in air for 3.06×2 = 6.12 seconds

Let d is the maximum distance covered by it.

$d=ut-\dfrac{1}{2}gt^2$

Putting all values

$d=30(3.06)-\dfrac{1}{2}\times 9.8\times (3.06)^2\\\\d=45.91\ m$

Hence, it will go to a height of 45.91 m and it will in the air for 6.12 seconds.

8 0

3 years ago

A long, straight metal rod has a radius of 5.75 cm and a charge per unit length of 33.3 nC/m. Find the electric field at the fol

PIT_PIT [208]

Answer:

Explanation:

From the question;

We will make assumptions of certain values since they are not given but the process to achieve the end result will be the same thing.

We are to calculate the following task, i.e. to determine the electric field at the distances:

a) at 4.75 cm

b) at 20.5 cm

c) at 125.0 cm

Given that:

the charge (q) = 33.3 nC/m

= 33.3 × 10⁻⁹ c/m

radius of rod = 5.75 cm

a) from the given information, we will realize that the distance lies inside the rod. Provided that there is no charge distribution inside the rod.

Then, the electric field will be zero.

b) The electric field formula $E = \dfrac{kq }{d}$

$E = \dfrac{9 \times 10^9 \times (33.3 \times 10^{-9}) }{0.205}$

E = 1461.95 N/C

c) The electric field E is calculated as:

$E = \dfrac{9 \times 10^9 \times (33.3 \times 10^{-9}) }{1.25}$

E = 239.76 N/C

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3 years ago

A 7.00 kg bowling ball is held 2.00 m above the ground. Using g- 9.8 m/s^2, how much energy does the bowling ball have due to it

Alexeev081 [22]

It should be B.137 just took the test.

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3 years ago

The opening to a cave is a tall, 30.0-cm-wide crack. A bat is preparing to leave the cave emits a 30.0 kHz ultrasonic chirp. How

Vlada [557]

Answer:

The value is $w = 7.54 \ m$

Explanation:

From the question we are told that

The length of the crack is $a = 0.3 \ m$

The frequency is $f = 30.0 \ kHz = 30 *10^{3} \ Hz$

The distance outside the cave that is being consider is $D = 100 \ m$

The speed of sound is $v_s = 340 \ m/s$

Generally the wavelength of the wave is mathematically represented as

$\lambda = \frac{v}f}$

=> $\lambda = \frac{340 }{30*10^{3}}$

=> $\lambda = 0.0113 \ m/s$

Generally for a single slit the path difference between the interference patterns of the sound wave and the center is mathematically represented as

$y = \frac{ n * \lambda * D}{a}$

=> $y = \frac{ 1 * 0.0113 * 100}{0.3}$

=> $y = 3.77 \ m$

Generally the width of the sound beam is mathematically represented as

$w = 2 * y$

=> $w = 2 * 3.77$

=> $w = 7.54 \ m$

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3 years ago