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2 years ago

A student walks 4 km in 30 minutes. What is the student’s average speed in km/h? Be sure to include your units. Please help!!!!!

Physics

1 answer:

VladimirAG [237]2 years ago

8 0

4 km in 30 min then 8 km in 60min or an hr. so 8km/hr.

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A cyclist is riding his bike up a mountain trail. When he starts up the trail, he is going 8 m/s. As the trail gets steeper, he

shusha [124]

Answer:

a) a = - 0.0833 m / s², b) t = 4.4 s

Explanation:

a) this is a kinematics exercise where the acceleration is along the inclined plane

v = v₀ - a t

a = v₀ - v / t

a = 3 - 8/60

a = - 0.0833 m / s²

b) in this case the final velocity is zero

v = v₀ - a t

0 = v₀ - at

t = v₀ / a

t = 28 / 6.4

t = 4.375 s

t = 4.4 s

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2 years ago

Water is boiled at sea level in a coffeemaker equipped with an immersion-type electric heating element. The coffee maker contain

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Answer:

$P=1362\ W$

$t'=251.659\ s$ is time required to heat to boiling point form initial temperature.

Explanation:

Given:

initial temperature of water, $T_i=18^{\circ}C$

time taken to vapourize half a liter of water, $t=18\ min=1080\ s$

desity of water, $\rho=1\ kg.L^{-1}$

So, the givne mass of water, $m=1\ kg$

enthalpy of vaporization of water, $h_{fg}=2256.4\times 10^{-3}\ J.kg^{-1}$

specific heat of water, $c=4180\ J.kg^{-1}.K^{-1}$

Amount of heat required to raise the temperature of given water mass to 100°C:

$Q_s=m.c.\Delta T$

$Q_s=1\times 4180\times (100-18)$

$Q_s=342760\ J$

Now the amount of heat required to vaporize 0.5 kg of water:

$Q_v=m'\times h_{fg}$

where:

$m'=0.5\ kg=$ mass of water vaporized due to boiling

$Q_v=0.5\times 2256.4$

$Q_v=1.1282\times 10^{6}\ J$

Now the power rating of the boiler:

$P=\frac{Q_s+Q_v}{t}$

$P=\frac{342760+1128200}{1080}$

$P=1362\ W$

Now the time required to heat to boiling point form initial temperature:

$t'=\frac{Q_s}{P}$

$t'=\frac{342760}{1362}$

$t'=251.659\ s$

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3 years ago

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Sliva [168]

Answer:

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Explanation:

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satela [25.4K]

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I could be wrong but I believe it’s 1/2

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