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3 years ago

8

Make a list of five materials or substances you use every day that are made from organic compounds. Write a minimum of one parag

raph per material about its benefits and uses in your everyday life.

1 answer:

Vaselesa [24]3 years ago

6 0

1.)Baking powder NaHCO3; sodium bicarbonate Used for baking for cooking, releases CO2 in reaction with other ingredients.

2.)Detergent Sodium sulphate, sodium hydroxide and phosphatecompounds Used for washing clothes.

3.)Toothpaste Calciuym carbonate, sodium flouride Used for cleaning teeth.

4.)Salt NaCl; Sodium Chloride Used in seasoning for cooking; also used as preservative.

5.)Vinegar C2H4O2

acetic acid, ethanoic acid

Used as preservative and for seasoning foods. Utilized for various household cleaning uses.

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A bowling ball moving with a velocity of 5V to the right collides elastically with a beach ball moving at a velocity 2V to the l

katen-ka-za [31]

Answer:

$v'_2=3V$

Explanation:

From the question we are told that:

Bowling ball Speed $v_1=5 m/s$

Beach ball Speed $v_2=2 m/s$

Let The Mass be equal i.e

$M_1=M_2$

Therefore

Generally the equation for Velocity of beach ball after collision $v'_2$ is mathematically given by

Since Velocity is Vector Quantity

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$v'_2=5-2$

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3 years ago

A 10-turn coil of wire having a diameter of 1.0 cm and a resistance of 0.50 Ω is in a 1.0 mT magnetic field, with the coil orien

n200080 [17]

Answer:

The voltage across the capacitor is 1.57 V.

Explanation:

Given that,

Number of turns = 10

Diameter = 1.0 cm

Resistance = 0.50 Ω

Capacitor = 1.0μ F

Magnetic field = 1.0 mT

We need to calculate the flux

Using formula of flux

$\phi=NBA$

Put the value into the formula

$\phi=10\times1.0\times10^{-3}\times\pi\times(0.5\times10^{-2})^2$

$\phi=7.85\times10^{-7}\ Tm^2$

We need to calculate the induced emf

Using formula of induced emf

$\epsilon=\dfrac{d\phi}{dt}$

Put the value into the formula

$\epsilon=\dfrac{7.85\times10^{-7}}{dt}$

Put the value of emf from ohm's law

$\epsilon =IR$

$IR=\dfrac{7.85\times10^{-7}}{dt}$

$Idt=\dfrac{7.85\times10^{-7}}{R}$

$Idt=\dfrac{7.85\times10^{-7}}{0.50}$

$Idt=0.00000157=1.57\times10^{-6}\ C$

We know that,

$Idt=dq$

$dq=1.57\times10^{-6}\ C$

We need to calculate the voltage across the capacitor

Using formula of charge

$dq=C dV$

$dV=\dfrac{dq}{C}$

Put the value into the formula

$dV=\dfrac{1.57\times10^{-6}}{1.0\times10^{-6}}$

$dV=1.57\ V$

Hence, The voltage across the capacitor is 1.57 V.

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