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kaheart [24]
3 years ago
15

Hell it’s due soon??

Physics
2 answers:
aleksandr82 [10.1K]3 years ago
5 0

Explanation:

don't know don't know don't know

Lapatulllka [165]3 years ago
4 0
Outward.................
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What would Hubble's constant be if we found one galaxy moving away at 30,000 km/s at a distance of 600 Mpc?
lora16 [44]

Answer:

H₀ = 1.6 x 10⁻¹⁸ s⁻¹

Explanation:

The Hubble's Constant can be found by the following formula:

v = H_o D\\\\H_o = \frac{v}{D}

where,

H₀ = Hubble's Constant = ?

v = speed of galaxy = 30000 km/s = 3 x 10⁷ m/s

D = Distacance = 600 Mpc = (6 x 10⁸ pc)(3.086 x 10¹⁶ m/1 pc)

D = 18.52 x 10²⁴ m

Therefore,

H_o = \frac{3\ x\ 10^7\ m/s}{18.52\ x\ 10^{24}\ m}

<u>H₀ = 1.6 x 10⁻¹⁸ s⁻¹</u>

3 0
3 years ago
Two astronauts are playing catch with a ball in space. The first astronaut throws the ball;and
seraphim [82]

recoils and must be tethered or he's gone.

3 0
4 years ago
En un experimento de calorimetría, 0.50 kg de un metal a 100°C se añaden a 0.50 kg de agua a 20°C en un vaso de calorímetro de a
Maru [420]

Answer:

c=0.14J/gC

Explanation:

A.

2) The specific heat will be the same because it is a property of the substance and does not depend on the medium.

B.

We can use the expression for heat transmission

Q=mc(T_2-T_1)

In this case the heat given by the metal (which is at a higher temperature) is equal to that gained by the water, that is to say

Q_1=-Q_2

for water we have to

c = 4.18J / g ° C

replacing we have

c_{metal}*(500g)(100\°C-25\°C)=-(250g)(4.18\frac{J}{g\°C})(20\°C-25\°C)\\c_{metal}=0.14\frac{J}{g\°C}

I hope this is useful for you

A.

2) El calor específico será igual porque es una propiedad de la sustancia y no depende del medio.

B.

Podemos usar la expresión para la transmisión de calor

Q=mc(T_2-T_1)

En este caso el calor cedido por el metal (que está a mayor temperatura) es igual al ganado por el agua, es decir

Q_1=-Q_2

para el agua tenemos que

c=4.18J/g°C

reemplazando tenemos

c_{metal}*(500g)(100\°C-25\°C)=-(250g)(4.18\frac{J}{g\°C})(20\°C-25\°C)\\c_{metal}=0.14\frac{J}{g\°C}

7 0
3 years ago
A snowboarder goes down the hill with a slope of 28° if friction acts on him as he slides down which of the following is the cor
xxMikexx [17]

Answer:

A

Explanation:

All of the frictions are the same, but weight always goes straight down so it can only be A or B. Since they are going down a slope, then the normal force must be sloped. A is the only one out of A and B with a sloped normal force, so it has to be A

6 0
3 years ago
Write a short description of how the motion of the racers might change from the start of the race to the finish line
Ad libitum [116K]
The motion of the racers might change from the start because the pressure goes up so all the racer wants is to speed up and win, so when the racer first starts he or she is calm because he's not driving yet and when he or she is on his/hers way to he finish line he/she just wants to win and gets under pressure so he speeds up even more and drifts. Your welcome
6 0
3 years ago
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