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Fynjy0 [20]
2 years ago
6

What type of stars will potentially one day

Physics
1 answer:
xz_007 [3.2K]2 years ago
4 0

Answer:

2, High mass stars.

Add-on:

i hope this helped at all. im sorry if its incorrect.

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In a game of tug of war, a rope is pulled by a force of 182 N to the right and by a force of 108 N to the left. Calculate the ma
Mrac [35]

Answer:

74 N to the right

Explanation:

the forces are going in opposite horizontal directions, meaning that they are directly opposing each other. this means that you can subtract the force applied in the direction that is greater from the direction that is less to get the net force for the greater direction

this means 182 N - 108 N = 74 N to the right

6 0
3 years ago
Which of these is an example of climate? A. a weather forecast B. last month's total rainfall C. the current temperature D. an a
Monica [59]

Answer:

The answer to your question is D An annual weather pattern

Explanation:

5 0
3 years ago
Read 2 more answers
A canoe floats in a lake. Inside the canoe is a 25 kg steel solid ball. If the ball is thrown into the lake, does the level of t
Rus_ich [418]

Answer:

Explanation:

volume of ball bearing = 4/3 π r³

= 4/3 x 3.14 x 1.5³

= 14.13 cc.

if D  be the density of steel , weight of the ball bearing

= 14.13 x D x g

For the first case , water will be displaced to keep it floating

weight of displaced water will be equal to weight of steel

weight of displaced water = 14.13 Dg

mass of displaced water = 14.13 D

volume of displaced water = mass / density of water

= 14.13 D / d                             ; (where d is density of water) .

Now when the steel ball bearing is dipped in water , it will displace water equal to its volume only and it will be drowned

its volume = 14 .13 cc

14.13 D / d  >  14.13  ( because D/d is more than one , since D > d )

volume of water displaced in first case is greater

water level will go up higher in first case .

Hence in the second case water level will go down .

Same will happen in case of 25 kg steel .

4 0
3 years ago
A gold bar 20.0kg at 35.0°c is placed in a large insulated 0.8kg glass container at 15°c and 2.0kg of water at 25°c.. calculate
Oksanka [162]

Answer:

The final equilibrium temperature is approximately 26.69 °C

Explanation:

The heat transferred, ΔQ, from a hot body to a cold one is given by the following formula;

ΔQ = m·c·ΔT

Where;

m = The mass of the body

c = The specific heat capacity of the body

ΔT = The temperature change of the body

The given mass of the gold bar, m₁ = 20.0 kg

The initial temperature of the gold bar, T₁ = 35.0 °C

The specific heat capacity of gold, c₁ = 0.13 kJ/(kg·K)

The mass of the glass container, m₂ = 0.8 kg

The initial temperature of the glass container, T₂ = 15°C

The specific heat capacity of glass, c₂ = 0.792 kJ/(kg·K)

The mass of the added water, m₃ = 2.0 kg

The initial temperature of the added water, T₃ = 25°C

The specific heat capacity of water, c₃ = 4.2 kJ/(kg·K)

The heat lost by the gold = The heat gained by the glass and the water

Let 'T' represent the temperature at the final equilibrium, we have;

m₁·c₁·ΔT₁ = m₂·c₂·ΔT₂ + m₃·c₃·ΔT₃

Where;

ΔT₁ = T₁ - T

ΔT₂ = T - T₂

ΔT₃ = T - T₃

∴ 20.0 × 0.13 × (35 - T) = 0.8 × 0.792 × (T - 15) + 2.0 × 4.2 × (T - 25)

Expanding and collecting like terms (using a graphing calculator) gives;

91 - 2.6·T = 9.0336·T - 219.504

9.0336·T + 2.6·T = 219.504 + 91 = 310.504

11.6336·T = 310.504

T = 310.504/11.6336 ≈ 26.69

The final equilibrium temperature, T ≈ 26.69 °C.

4 0
3 years ago
You are writing a science report and want to find accurate trustworthy information. Which would be the best resource?
fgiga [73]
<span>an encyclopedia

Happy studying!</span>
8 0
4 years ago
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