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ad-work [718]
4 years ago
12

How could dead skin under fingernails indicate that a struggle occured?

Physics
2 answers:
Sever21 [200]4 years ago
7 0
Cause the skin under the fingernails can be due to the scraping of skin from attacker while the victim is struggling.
Greeley [361]4 years ago
3 0
The skin under the nail shows the victim attempted to fight or struggled getting away.
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An astronaut in deep space is at rest relative to a nearby space station. The astronaut needs to return to the space station. A
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Answer:

what u

Explanation:

know about rolling down in the deep when ue brain goes du m u can call that mental freeze when people talk to much puy that shi t on slow motion

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3 years ago
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To convert minutes per second into kilometre per hour we multiply the speed with​
ankoles [38]

Answer:

To convert m/sec into km/hr, multiply the number by 18 and then divide it by 5.

Explanation:

please mark as brainliest

3 0
3 years ago
Một oto đang chạy với vận tốc là 10m/s thì tăng tốc và chuyển động nhanh dần đều sau 20s thì đạt vận tốc 14m/s
Tasya [4]

a) gia tốc = vf-vi / t

a = 14-10 / 20

a = 0,2ms⁻²

b) dưới dạng a = Δv / t

v = lúc

v = 0,2 × 40

v = 8ms⁻¹

như v = d / t

do đó d = vt

d = 8 × 40

d = 320m

hãy đánh dấu là trí óc nhất

4 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

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4 years ago
Explain how there can be more heat energy in an iceberg than in a pot of boiling water
insens350 [35]
The tension of cold icebergs rubbing creates heat and it can also be how large the quantity is
3 0
4 years ago
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