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3 years ago

A biker to move from speed of 0 m/s to final speed of 25 m/s in 10 s. What is the acceleration of the bike?

Physics

1 answer:

Rzqust [24]3 years ago

4 0

Answer:

a = 2.5 m/s²

Explanation:

given,

initial speed = 0 m/s

final speed = 25 m/s

time = 10 s

acceleration of bike is equal to change in velocity per unit time

$a = \dfrac{\Delta v}{t}$

$a = \dfrac{v_f-v_i}{t}$

$a = \dfrac{25-0}{10}$

a = 2.5 m/s²

hence, acceleration of bike is equal to a = 2.5 m/s²

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3 years ago

Fluid originally flows through atube at a rate of 200 cm3/s. Toillustrate the sensitivity of the Poiseuille flow rate to various

Alexxx [7]

Answer:

$Q_{2}=1200cm^{3}/s$

Explanation:

Given data

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$F=n\frac{vA}{l}\\$

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3 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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3 years ago