All categories

3 years ago

Star Characteristic Questions What are the two things that stars are emitting into space?

Physics

1 answer:

katovenus [111]3 years ago

6 0

Answer:

1- Light , 2-Heat , ( and Nuclear energy)

Explanation:

You might be interested in

Can Anyone help me with this?

agasfer [191]

1. Velocity

2. Time

3. Idk

4. Idk

5. D. I think it may be A. but I think D.

7 0

3 years ago

Read 2 more answers

A delivery truck leaves a warehouse and travels 2.60 km north. The truck makes a left turn and travels 1.25 km west before makin

yulyashka [42]

Answer:

4.19 km and 107.35 degrees north of east

Explanation:

So in the end, the truck is (2.6 + 1.4 = 4km) north and 1.25 km west from the warehouse. We can use the Pythagorean formula to calculate the magnitude and direction α of the truck displacement from the warehouse:

$s = \sqrt{s_n^2 + s_w^2} = \sqrt{4^2 + 1.25^2} = \sqrt{16 + 1.5625} = \sqrt{17.5625} = 4.19$ km

$tan\alpha = \frac{s_n}{s_w} = \frac{4}{1.25} = 3.2$

$\alpha = tan^{-1}3.2 = 1.27 rad \approx 72.65 degrees$ north or west or (180 - 72.65) = 107.35 degrees north of east

3 0

3 years ago

The slope of the line on a speed-time graph tells the speed , true or false ?

KengaRu [80]

True I hope this helps you out

8 0

3 years ago

Read 2 more answers

A small charged bead has a mass of 1.9 g . it is held in a uniform electric field e⃗ = (200,000 n/c, up). when the bead is relea

dmitriy555 [2]

As we know that when charge is released in electric field

It will have two forces on it

1. electrostatic force

2. gravitational force

now if the ball will accelerate upwards so we can say

net upward force = mass * acceleration

$qE - mg = ma$

$q*200000 - 1.9 * 10^{-3}* 9.8 = 1.9 * 10^{-3}* 15$

now we can find charge q on it by above equation

$q = 0.2356 * 10^{-6} C$

So above is the charge on the particle

7 0

3 years ago

The barrel of a rifle has a length of 0.89 m. A bullet leaves the muzzle of a rifle with a speed of 620 m/s. What is the acceler

guapka [62]

Answer:

215955.06 m/s^2

Explanation:

length of barrel, s = 0.89 m

initial velocity of the bullet, u = 0 m/s

Final velocity of the bullet, v = 620 m/s

Let a be the acceleration of the bullet in the barrel

Use third equation of motion, we get

$v^{2}=u^{2}+ 2as$

$620^{2}=0^{2}+ 2\times a \times 0.89$

a = 215955.06 m/s^2

Thus, the acceleration of the bullet inside the barrel is 215955.06 m/s^2.

6 0

3 years ago