A 12 kg bowling ball would require what force to accelerate it down an alley at a rate of 2.5 m/s ^ 2

(ik it says physics but astronomy is a field of physics sooo) A recently discovered planet in a different solar system is locate

marusya05 [52]

The distance, to the nearest tenth, is 314.9 light-years.

<em>The given data is:</em>

A recently discovered planet is located 1.85*10^5 miles from Earth.

Now we want to transform this distance to light-years.

Remember that a light-year is defined as "the distance that the light would travel in one year".

using the relation:

distance = speed*time

The speed of light is:

speed = 6.706*10^8 mi/h

And in one year has 8760 hours, then we have:

time = 8760 h

replacing these in the equation we get:

distance = speed*time

distance = (6.706*10^8 mi/h)*(8760 h) = 5,874,456,000,000 miles

Son one light-year is equivalent to 5,874,456,000,000 miles

1 light-year = 5,874,456,000,000 miles

So to transform a distance in miles to light-years, we just need to divide that distance by 5,874,456,000,000 miles:

The distance between the new planet and Earth was:

D = 1.85*10^15 mi = ( 1.85*10^15)/(5,874,456,000,000) = 314.9 light-years.

if you want to learn more about this, you can read:

brainly.com/question/1302132

3 0

3 years ago

The components of vector A are Ax and Ay (both positive), and the angle that it makes with respect to the positive x axis is θ.

DIA [1.3K]

Answer:

a) $\theta = tan^{-1} (\frac{11}{11})= tan^{-1} (1)=45$ degrees

b) $\theta = tan^{-1} (\frac{11}{19})= tan^{-1} (0.579)=30.07$ degrees

c) $\theta = tan^{-1} (\frac{19}{11})= 59.93$ degrees

Explanation:

If we have a vector $A= (A_x ,A_y)$ in a two dimensional space. The angle respect the x axis can be founded from the following expression:

$tan (\theta)= \frac{A_y}{A_x}$

And then the angle is given by:

$\theta = tan^{-1} (\frac{A_y}{A_x})$

Part a

Ax = 11 m and Ay = 11 m

For this case the angle would be:

$\theta = tan^{-1} (\frac{11}{11})= tan^{-1} (1)=45$ degrees

Part b

Ax = 19 m and Ay = 11 m

For this case the angle would be:

$\theta = tan^{-1} (\frac{11}{19})= tan^{-1} (0.579)=30.07$ degrees

Part c

Ax = 11 m and Ay = 19 m

For this case the angle would be:

$\theta = tan^{-1} (\frac{19}{11})= 59.93$ degrees

8 0

3 years ago

Read 2 more answers

Relative to the ground, a car has a velocity of 15.3 m/s, directed due north. Relative to this car, a truck has a velocity of 22

bogdanovich [222]

Answer:

The magnitude of the truck's velocity relative to the ground is 35.82 m/s.

Explanation:

Given that,

Velocity of car relative to ground = 15.3 m/s

Velocity of truck relative to car = 22.5 m/s

We need to calculate the magnitude of the truck's velocity relative to the ground

We need to calculate the x component of the velocity

$v_{x}=22.5\cos\theta$

$v_{x}=22.5\cos52^{\circ}$

$v_{x}=13.852\ m/s$

We need to calculate the y component of the velocity

$v_{y}=15.3+22.5\sin\theta$

$v_{y}=15.3+22.5\sin52^{\circ}$

$v_{y}=33.030\ m/s$

Using Pythagorean theorem

$|v|=\sqrt{v_{x}^2+v_{y}^2}$

$|v|=\sqrt{(13.852)^2+(33.030)^2}$

$|v|=35.82\ m/s$

Hence, The magnitude of the truck's velocity relative to the ground is 35.82 m/s.

5 0

3 years ago

You stand on a straight desert road at night and observe a vehicle approaching. This vehicle is equipped with two small headligh

zhuklara [117]

Complete Question

You stand on a straight desert road at night and observe a vehicle approaching. This vehicle is equipped with two small headlights that are 0.681 m apart. At what distance, in kilometres, are you marginally able to discern that there are two headlights rather than a single light source?Take the wavelength of the light to be 549 nm and your pupil diameter as 4.63 mm.

Answer:

The distance is $z = 4707.6 \ m$