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2 years ago

HELP WILL MARK BRAINLIEST

Physics

1 answer:

zvonat [6]2 years ago

6 0

Answer is D

Explanation:

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3 years ago

Help, im not sure how to do this at all

antiseptic1488 [7]

Answer:

60 m

Explanation:

The boat has two separate motions:

1- A motion due north, with constant speed of 10 m/s

1- A motion due east, due to the current, at speed of 2 m/s

We know that the river is 300 m wide, so we can consider first motion 1) to find how much does it take for the boat to cross the river:

$t=\frac{d}{v}=\frac{300 m}{10 m/s}=30s$

Now we can find how far downstream the boat moved by calculating the distance that the boat covered moving east during this time interval:

$d'=vt=(2 m/s)(30 s)=60 m$

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2 years ago

What is the final temperature when a 3.0 kg gold bar at 99 0C is dropped into 0.22 kg of water at 25oC?

liq [111]

Answer:

46.9 C

Explanation:

The heat released by the gold bar is equal to the heat absorbed by the water:

$m_g C_g (T_g-T_f)=m_w C_w (T_f-T_w)$

where:

$m_g = 3.0 kg$ is the mass of the gold bar

$C_g=129 J/kg C$ is the specific heat of gold

$T_g=99 C$ is the initial temperature of the gold bar

$m_w = 0.22 kg$ is the mass of the water

$C_w=4186 J/kg C$ is the specific heat of water

$T_w=25 C$ is the initial temperature of the water

$T_f$ is the final temperature of both gold and water at equilibrium

We can re-arrange the formula and solve for T_f, so we find:

$m_g C_g T_g -m_g C_g T_f = m_w C_w T_f - m_w C_w T_w\\m_g C_g T_g +m_w C_w T_w= m_w C_w T_f +m_g C_g T_f \\T_f=\frac{m_g C_g T_g +m_w C_w T_w}{m_w C_w + m_g C_g}=\\=\frac{(3.0)(129)(99)+(0.22)(4186)(25)}{(0.22)(4186)+(3.0)(129)}=\frac{38313+23023}{921+387}=\frac{61336}{1308}=46.9 C$

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2 years ago