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Zinaida [17]
3 years ago
15

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

Physics
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

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Starting from rest, a car undergoes a constant acceleration of 6 m/s^2. How far will the car travel in the two seconds?
oksano4ka [1.4K]

If time is specified, the distance may be estimated in constant acceleration using the formula: X=(at2)/2 if the beginning velocity is 0. (A automobile begins from a stop...) As a result, X=(6*10*10)/2=600/2 = 300 m.

4 0
2 years ago
You pull with a force of 295 N on a rope that is attached to a block of mass 22 kg, and the block slides across the floor at a c
Sergeeva-Olga [200]

Answer:

Fnet = 0

Explanation:

  • Since the block slides across the floor at constant speed, this means that it's not accelerated.
  • According Newton's 2nd Law, if the acceleration is zero, the net force on the sliding mass must be zero.
  • This means that there must be a friction force opposing to the horizontal component of the applied force, equal in magnitude to it:

       F_{appx} = F_{app} * cos \theta = 295 N * cos 35 = 242 N  (1)

  • In the vertical direction, the block is not accelerated either, so the sum of the normal force and the vertical component of the applied force, must be equal in magnitude to the force of gravity on the block:

      F_{appy} = F_{app} * cos \theta = 295 N * sin 35 = 169 N  (2)

⇒    169 N + Fn = Fg = 216 N  (3)

  • This means that there must be a normal force equal to the difference between Fappy and Fg, as follows:
  • Fn = 216 N - 169 N = 47  N (4)

6 0
3 years ago
Underground water is being pumped into a pool whose cross section is 3 m x 4 m while water is discharged through a 0.076m-diamet
Svetllana [295]
Given:

Area of pool = 3m×4m
Diameter of orifice = 0.076m
Outlet Velocity = 6.3m/s
Accumulation velocity = 1.5cm/min

Required:

Inlet flowrate

Solution:

The problem can be solved by this general formula.

Accumulation = Inlet flowrate - Outlet flowrate
Accumulation velocity × Area of pool = Inlet flowrate - Outlet velocity × Area of orifice

First, we need to convert the units of the accumulation velocity into m/s to be consistent.

Accumulation velocity = 1.5cm/min × (1min/60s)×(1m/100cm)
Accumulation velocity = 0.00025 m/s

We then calculate the area of the pool and the area of the orifice by:

Area of pool = 3 × 4 m²
Area of pool = 12m²

Area of orifice = πd²/4 = π(0.076m)²/4
Area of orifice = 0.00454m²

Since we have all we need, we plug in the values to the general equation earlier

Accumulation velocity × Area of pool = Inlet flowrate - Outlet velocity × Area of orifice

0.00025 m/s × 12m² = Inlet flowrate - 6.3m/s × 0.00454m²

Transposing terms,

Inlet flowrate = 0.316 m³/s
6 0
3 years ago
what is the percent efficiency of a light bulb that emits 14 joules of light for every 60 joules of electric energy thst enters
jasenka [17]
If you're using the bulb as a source of light, then it's. 14/60 = 23.3% efficient.
If you're using it to heat a bird nest or a hamster cage, then it's. 46/60 = 76.7% efficient !
It just depends on your point of view, and what you consider 'useful' output.
5 0
3 years ago
A house brick has a volume of 1900 cm³ and a weight in air of 80N.What is its apparent weight in water?The density of water is 1
Shtirlitz [24]

Answer:

61 N

Explanation:

We'll begin by calculating the mass of the brick when placed in water. This can be obtained as follow:

Volume of brick = 1900 cm³

Density of water = 1 g/cm³

Mass of brick in water =…?

Density = mass / volume

1 = mass of brick in water / 1900

Cross multiply

Mass of brick in water = 1 × 1900

Mass of brick in water = 1900 g

Next, we shall convert 1900 g to Kg.

1000 g = 1 Kg

Therefore,

1900 g = 1900 g × 1 Kg / 1000 g

1900 g = 1.9 Kg

Next, we shall determine the weight in water. This can be obtained as follow:

Mass (m) = 1.9 Kg

Acceleration due to gravity (g) = 10 m/s²

Weight (W) =?

W = m × g

W = 1.9 × 10

W = 19 N

Thus, the weight of the brick in water is 19 N.

Finally, we shall determine the apparent weight of the brick in water. This can be obtained as follow:

Weight in air = 80 N

Weight in water = 19 N

Apparent weight =?

Apparent weight = weight in air – weight in water

Apparent weight = 80 – 19

Apparent weight = 61 N

3 0
3 years ago
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