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Helen [10]
3 years ago
10

It would really help if anyone could answer

Physics
1 answer:
ExtremeBDS [4]3 years ago
7 0
B is the answer to this problem
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What happens when a star blows up and it is next another star will it blow up too?
FinnZ [79.3K]

Answer:

Sometimes

Explanation:

Sometimes meaning in occasions. If it is a "vampire star" not likely because they also cause they also make novas happen by sucking the gas from another star.

4 0
4 years ago
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What factors are required for refraction to occur PLEASE HELP!
Cloud [144]
1. Light Travelling from denser medium to less dense medium

2. Light Travelling from less dense medium to denser medium

5 0
3 years ago
What are the energy transfers in a torch?
lapo4ka [179]

Explanation:

The battery is a store of internal energy (shown as chemical energy). The energy is transferred through the wires to the lamp, which then transfers the energy to the surroundings as light. These are the useful energy transfers - we use electric lamps to light up our rooms.

6 0
3 years ago
A toy car runs off the edge of a table that is 2.225 m high. If the car lands 0.400 m away from the base of the table, how fast
ruslelena [56]

Answer:

0.594 m/s

Explanation:

First, find the time it takes to land.

Given, in the y direction:

Δy = 2.225 m

v₀ = 0 m/s

a = 9.8 m/s²

Find: t

Δy = v₀ t + ½ at²

(2.225 m) = (0 m/s) t + ½ (9.8 m/s²) t²

t = 0.674 s

Next, find the horizontal velocity.

Given, in the x direction:

Δx = 0.400 m

a = 0 m/s²

t = 0.674 s

Find: v₀

Δx = v₀ t + ½ at²

(0.400 m) = v₀ (0.674 s) + ½ (0 m/s²) (0.674 s)²

v₀ = 0.594 m/s

8 0
3 years ago
A cube of ice at an initial temperature of -15.00°C weighing 12.5 g total is placed in 85.0 g of water at an initial temperature
Leona [35]

<u>Answer:</u> The specific heat of ice is 2.11 J/g°C

<u>Explanation:</u>

When ice is mixed with water, the amount of heat released by water will be equal to the amount of heat absorbed by ice.

Heat_{\text{absorbed}}=Heat_{\text{released}}

The equation used to calculate heat released or absorbed follows:

Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})

m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]       ......(1)

where,

q = heat absorbed or released

m_1 = mass of ice = 12.5 g

m_2 = mass of water = 85.0 g

T_{final} = final temperature = 22.24°C

T_1 = initial temperature of ice = -15.00°C

T_2 = initial temperature of water = 25.00°C

c_1 = specific heat of ice = ?

c_2 = specific heat of water = 4.186 J/g°C

Putting values in equation 1, we get:

12.5\times c_1\times (22.24-(-15))=-[85.0\times 4.186\times (22.24-25)]

c_1=2.11J/g^oC

Hence, the specific heat of ice is 2.11 J/g°C

3 0
4 years ago
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