Ask question

Ask question

All categories

3 years ago

15

Which characteristics can be used to differentiate star systems? Select three options

2 answers:

Veseljchak [2.6K]3 years ago

7 0

Answer: 1,3, and 4 on EDGE science just did the assignment

Explanation:

Lady bird [3.3K]3 years ago

5 0

Answer:

Explanation:

the answer is a,c,d because the stars are characterized by the color temperature, size, composition, and brightness

You might be interested in

What would happen to the moon if there was no sun and earth ?

iragen [17]

Answer:

High tides would be much smaller than they are now, and low tides would be even lower. This is because the sun would be influencing the tides, not the moon; however, the sun has a weaker pull, which would decrease the tides. ... Winds could become much faster and much stronger without the moon.

Explanation:

google

4 0

3 years ago

Suppose a particle moves along a straight line with velocity v(t)=t2e−2tv(t)=t2e−2t meters per second after t seconds. How many

dimulka [17.4K]

Explanation:

It is given that,

Velocity of the particle moving in straight line is :

$v(t)=t^2e^{-2t}\ m/s$

We need to find the distance (x) traveled by the particle during the first t seconds. It is given by :

$x=\int\limits {v.dt}$

$x=\int\limits {t^2e^{-2t}dt}$

Using by parts integration, we get the value of x as :

$x=\dfrac{-(2t^2+2t+1)e^{-2t}}{4}\ meters$

Hence, this is the required solution.

6 0

3 years ago

A sled is moving down a steep hill. The mass of the sled is 50 kg and the net force acting on it is 20 N. What must be done to f

Rashid [163]

I believe the answer is #4. u can always ask google if u believe that's the wrong answer :)

4 0

3 years ago

Read 2 more answers

For a brass alloy, the stress at which plastic deformation begins is 345 MPa (50,000 psi), and the modulus of elasticity is 103

Alona [7]

Answer:

a) P = 44850 N

b) $\delta l =0.254\ mm$

Explanation:

Given:

Cross-section area of the specimen, A = 130 mm² = 0.00013 m²

stress, σ = 345 MPa = 345 × 10⁶ Pa

Modulus of elasticity, E = 103 GPa = 103 × 10⁹ Pa

Initial length, L = 76 mm = 0.076 m

a) The stress is given as:

$\sigma=\frac{\textup{Load}}{\textup{Area}}$

on substituting the values, we get

$345\times10^6=\frac{\textup{Load}}{0.00013}$

or

Load, P = 44850 N

Hence<u> the maximum load that can be applied is 44850 N = 44.85 KN</u>

b)The deformation ( $\delta l$ ) due to an axial load is given as:

$\delta l =\frac{PL}{AE}$

on substituting the values, we get

$\delta l =\frac{44850\times0.076}{0.00013\times103\times 10^9}$

or

$\delta l =0.254\ mm$

3 0

3 years ago

The water waves below are traveling along the surface of the ocean at a speed of 2.5 m/s and

gavmur [86]

Answer:2 seconds

Time= Distance/speed

=50/25

=2seconds

3 0

3 years ago