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3 years ago

Red shift data shows that galaxies are O expanding O shrinking O moving away O moving closer

Physics

1 answer:

Anon25 [30]3 years ago

7 0

Answer:

Should be moving away

Explanation:

Red is a longer wavelength therefore further away. Wavelength is stretched out more and on the red end. I hope this is right. I decided to research and answer since you didn’t have other answers. Are you taking this on edg? I hope I helped!

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Four identical metallic objects carry the following charges 1.82 6.65 4.80 and 9.30 C The objects are brought simultaneously int

GuDViN [60]

Answer:

a) 5.64 C

b) 3.5*10¹⁹ protons

Explanation:

Since the four metallic objects are identical, and total charge must be conserved, this means that after brought simultaneously into contact so that each touches the others, once separated, total charge must be the same than before being brought in contact.
But due they are identical, after charges were able to transfer freely between them, the four objects must have the same final charge, i.e. the fourth part of the total charge, as follows:

$Q_{n} = \frac{Q_{tot}}{4} = \frac{22.57C}{4} = 5.64 C (1)$

This charge will be divided between n protons, since the charge is positive.
Since each proton carries a charge equal to the elementary charge e, which value is 1.6*10⁻¹⁹ C, we can find the number of protons in excess, doing the following calculation:
$n_{p} =\frac{Q_{n}}{e} = \frac{5.64C}{1.6e-19C} = 3.5 e19 C (2)$

7 0

3 years ago

Which is longer, 10 cm or .01 m?

motikmotik

Answer:

They´re the same.

Explanation:

Someone deleted my answer. And my brainly is gone..

Have a good night ma´am/sir.

Be safe!

6 0

3 years ago

When playing a game of disc golf, each throw to your target is considered to be a O Point O Stroke O Hit O Toss

Hunter-Best [27]

Answer:

an o point i think

Explanation:

5 0

3 years ago

A 6.00 kg object is lifted vertically through a distance of 5.25 m by a light string under a tension of 80.0 N. Find: (2 marks)

jeka57 [31]

Explanation:

It is given that,

Mas of the object, m = 6 kg

It is lifted through a distance, h = 5.25 m

Tension in the string, T = 80 N

(a) By considering the free body diagram of the object, the forces can be equated as :

$T-mg=ma$

$a=\dfrac{T-mg}{m}$

$a=\dfrac{80-6\times 9.8}{6}$

$a=3.33\ m/s^2$

Work done by tension, $W_t=F\times h$

$W_t=80\times 5.25$

$W_t=420\ J$

(b) Work done by gravity, $W_g=mgh$

$W_g=6\times 9.8\times 5.25$

$W_g=308.7\ J$

(c) Let v is the final speed of the object and u = 0

$v=\sqrt{2ah}$

$v=\sqrt{2\times 3.33\times 5.25}$

v = 5.91 m/s

Hence, this is the required solution.

7 0

4 years ago

1. Is the collision between the ball and the pendulum elastic or inelastic? Justify your answer by calculating the kinetic energ

Nitella [24]

Answer:

So energy is not conserved and inelastic shock

Explanation:

In the collision between a bullet and a ballistic pendulum, characterized in that the bullet is embedded in the block, if the kinetic energy is conserved the shock is elastic and if it is not inelastic.

Let's find the kinetic energy just before the crash

K₀ = ½ m vₓ₀²

After the crash we can use the law of conservation of energy

Starting point. Right after the crash, before starting to climb

Em₀ = K = ½ (m + M) v₂²

Final point. At the maximum height of the pendulum

$Em_{f}$ = U = (m + M) g h

Where m is the mass of the bullet and M is the mass of the pendulum

Em₀ = Em_{f}

½ (m + M) v₂² = (m + M) g h

v₂ = √ 2g h

Now we can calculate the final kinetic energy

K_{f} = ½ (m + M) v₂²²

K_{f} = ½ (m + M) (2gh)

The relationship between these two kinetic energies is

K₀ / K_{f} = ½ m vₓ₀² / (½ (m + M) 2 g h)

K₀ / K_{f} = m / (m + M) vₓ₀² / 2 g h

We can see that in this relationship the Ko> Kf

So energy is not conserved and inelastic shock

5 0

4 years ago