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3 years ago

You accelerate from 2m/s to 6m/s while traveling a distance of 2m; what was your acceleration?

Physics

1 answer:

nata0808 [166]3 years ago

7 0

Answer:

The acceleration is 8 m/s²

Explanation:

The given parameters are;

The initial velocity, u = 2 m/s

The final velocity, v = 6 m/s

The distance the acceleration took place, s = 2 m

The acceleration, a, can be found from the following kinematic equation;

v² = u² + 2·a·s

By substituting the values, we have;

6² = 2² + 2 × a × 2

6² - 2² = 2 × a × 2

32 = 4·a

a = 32/4 = 8 m/s²

The acceleration, a, of the given motion = 8 m/s².

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Question 2

1. FORCE

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4. ELECTROAGNETIC FORCE

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2 years ago

Why is a human powered generator better than the sun (solar cells)?

zheka24 [161]

Answer:

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3 years ago

A car with a constant velocity of 22 m/s is driven for 6.8 s. How far did it travel? (V =A d/AT)

SVEN [57.7K]

Answer: 150 m

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option C

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3 years ago

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

Luden [163]

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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