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3 years ago

Their are fewer or more elements than compounds?

2 answers:

4 0

There are 92 elements that are found in nature. Roughly 80 of those are capable of combining with each other to make compounds, so there are a lot more than 92 compounds.

Example: Take carbon and hydrogen. (please)

Carbon and hydrogen combine to make methane, ethane, propane, pentane, hexane, heptane, octane, butane, and several more 'tanes'.

And if you have some oxygen along with the carbon and hydrogen, you can
make all the alcohols there are, plus vinegar.

EleoNora [17]3 years ago

4 0

There are fewer elements than compounds. There are 118 known elements, but millions of compounds and more being made every day

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An engineer in India (standard household voltage = 220 volts) is designing a transformer for use on her

Gnom [1K]

He should a step-up transformer with k=220/120=1.83 so output coil must have 240*1.83=440 turns

5 0

2 years ago

An object is thrown directly up (positive direction) with a velocity (vo) of 20.0 m/s and do= 0. How high does it rise (v = 0 cm

Anit [1.1K]

Answer:

The value of d is 20.4 m.

(C) is correct option.

Explanation:

Given that,

Initial velocity = 20 m/s

Final velocity = 0

We need to calculate the time

Using equation of motion

$v = u+gt$

Where, u = Initial velocity

v = Final velocity

Put the value into the formula

$t = \dfrac{20}{9.8}$

$t= 2.04\ sec$

We need to calculate the distance

Using equation of motion

$s = ut+\dfrac{1}{2}gt^2$

$s=0+\dfrac{1}{2}\times9.8\times(2.04)^2$

$s=20.4\ m$

Hence, The value of d is 20.4 m.

4 0

3 years ago

Calculate the gravitational potential energy of a 2 kg banana hanging 5

Hunter-Best [27]

Answer:

100 J

Explanation:

The potential energy is given by the formula ...

PE = mgh

= (2 kg)(10 m/s^2)(5 m) = 100 J

7 0

3 years ago

A ball is thrown with an initial speed vi at an angle i with the horizontal. The horizontal range of the ball is R, and the ball

adell [148]

Answer:

Part a)

$T = 2\sqrt{\frac{R}{3g}}$

Part b)

$v_x = \frac{\sqrt{3Rg}}{2}$

Part c)

$v_y = \sqrt{Rg/3}$

Part d)

$v = \frac{1}{2}\sqrt{13Rg}$

Part e)

$\theta_i = 33.7 degree$

Part f)

$H = \frac{13R}{8}$

Part g)

$X = \frac{13R}{4}$

Explanation:

Initial speed of the launch is given as

initial speed = $v_i$

angle = $\theta_i$ degree

Now the two components of the velocity

$v_x = v_i cos\theta_i$

similarly we have

$v_y = v_i sin\theta_i$

Part a)

Now we know that horizontal range is given as

$R = \frac{v_i^2 (2sin\theta_icos\theta_i)}{g}$

maximum height is given as

$H = \frac{R}{6} = \frac{v_i^2 sin^2\theta_i}{2g}$

so we have

$v_i sin\theta = \sqrt{Rg/3}$

time of flight is given as

$T = \frac{2v_isin\theta_i}{g}$

$T = \frac{2\sqrt{Rg/3}}{g}$

$T = 2\sqrt{\frac{R}{3g}}$

Part b)

Now the speed of the ball in x direction is always constant

so at the peak of its path the speed of the ball is given as

$R = v_x T$

$R = v_x 2\sqrt{\frac{R}{3g}}$

$v_x = \frac{\sqrt{3Rg}}{2}$

Part c)

Initial vertical velocity is given as

$v_y = v_i sin\theta_i$

$v_i sin\theta = \sqrt{Rg/3}$

Part d)

Initial speed is given as

$v = \sqrt{v_x^2 + v_y^2}$

so we will have

$v = \sqrt{Rg/3 + 3Rg/4}$

$v = \frac{1}{2}\sqrt{13Rg}$

Part e)

Angle of projection is given as

$tan\theta_i = \frac{v_y}{v_x}$

$tan\theta_i = \frac{\sqrt{Rg/3}}{\sqrt{3Rg}/2}$

$\theta_i = 33.7 degree$

Part f)

If we throw at same speed so that it reach maximum height

then the height will be given as

$H = \frac{v^2}{2g}$

$H = \frac{13R}{8}$

Part g)

For maximum range the angle should be 45 degree

so maximum range is

$X = \frac{v^2}{g}$

$X = \frac{13R}{4}$

3 0

3 years ago

A 10 ohms resistor is powered by a 5-V battery. The current flowing<br> through the source is:

mario62 [17]

Resistance=R=10ohm
Voltage=V=5V
Current=I

Applying ohm's law

$\\ \sf\longmapsto \dfrac{V}{I}=R$

$\\ \sf\longmapsto I=\dfrac{V}{R}$

$\\ \sf\longmapsto I=\dfrac{5}{10}$

$\\ \sf\longmapsto I=0.5A$

4 0

1 year ago