List and describe two ways currents affect life on earth

How many total oxygen atoms are in the compound Molybdenum (V) Dichromate?

alekssr [168]

Answer:

$1.2646\times10^{25}\ atoms$

Explanation:

-The chemical formula for Molybdenum (V) Dichromate is $Mo(Cr_2O_7)_3$

-There are 21 moles of oxygen per one mole of Molybdenum (V) Dichromate

-We apply Avogadro's constant to find the number of atoms of oxygen:

$Avogadro's \ Constant=6.022\times 10^{23} \ mol_1\\\\No\ of \ Atoms=Moles\times Avogadro's \ Constant\\\\=21\times 6.022\times 10^{23} \\\\=1.2646\times10^{25}\ atoms$

Hence, there are $1.2646\times10^{25} \ atoms$

3 0

3 years ago

A 26 foot ladder is lowered down a vertical wall at a rate of 3 feet per minute. The base of the ladder is sliding away from the

lakkis [162]

Answer:

(i) 7.2 feet per minute.

(ii) No, the rate would be different.

(iii) The rate would be always positive.

(iv) the resultant change would be constant.

(v) 0 feet per min

Explanation:

Let the length of ladder is l, x be the height of the top of the ladder from the ground and y be the length of the bottom of the ladder from the wall,

By making the diagram of this situation,

Applying Pythagoras theorem,

$l^2 = x^2 + y^2-----(1)$

Differentiating with respect to t ( time ),

$0=2x\frac{dx}{dt} + 2y\frac{dy}{dt}$ ( l = 26 feet = constant )

$\implies 2y\frac{dy}{dt} = -2x\frac{dx}{dt}$

$\implies \frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt}$

We have,

$y = 10, \frac{dx}{dt}= -3\text{ feet per min}$

$\frac{dy}{dt}=\frac{3x}{10}-----(X)$

(i) From equation (1),

$26^2 = x^2 + 10^2$

$676=x^2 + 100$

$576 = x^2$

$\implies x = 24\text{ feet}$

From equation (X),

$\frac{dy}{dt}=\frac{3\times 24}{10}=7.2\text{ feet per min}$

(ii) From equation (X),

$\frac{dy}{dt}\propto x$

Thus, for different value of x the value of $\frac{dy}{dt}$ would be different.

(iii) Since, distance = Positive number,

So, the value of y will always a positive number.

Thus, from equation (X),

The rate would always be a positive.

(iv) The length of the ladder is constant, so, the resultant change would be constant.

i.e. x = increases ⇒ y = decreases

y = decreases ⇒ y = increases

(v) if ladder hit the ground x = 0,

So, from equation (X),

$\frac{dy}{dt}=0\text{ feet per min}$

3 0

3 years ago

A 3 kg mass object is pushed 0.6 m into a spring with spring constant 210 N/m on a frictionless horizontal surface. Upon release

Svetradugi [14.3K]

Answer with Explanation:

We are given that

Mass of spring,m=3 kg

Distance moved by object,d=0.6 m

Spring constant,k=210N/m

Height,h=1.5 m

a.Work done to compress the spring initially= $\frac{1}{2}kx^2=\frac{1}{2}(210)(0.6)^2=37.8J$

b.

By conservation law of energy

Initial energy of spring=Kinetic energy of object

$37.8=\frac{1}{2}(3)v^2$

$v^2=\frac{37.8\times 2}{3}$

$v=\sqrt{\frac{37.8\times 2}{3}}$

v=5.02 m/s

c.Work done by friction on the incline, $w_{friction}=P.E-spring \;energy$

$W_{friction}=3\times 9.8\times 1.5-37.8=6.3 J$

8 0

3 years ago

Given: Q1 = +10 uc = 1.0 x 10-5C

ser-zykov [4K]

Answer:

-0.038 N

Explanation:

F=K Q1 Q2/r^2 by COULOMB'S LAW

F= 9×10^9×1×10^-5×-1.5×10^-5/(6)^2

F= -0.038 N

5 0

3 years ago

How is an ammeter connected in a circuit to measure current flowing through it?

trasher [3.6K]

Answer:

It is connected in series with the circuit

Explanation:

This is because to measure the current in the circuit, the current in the circuit has to flow through the ammeter. As such, the ammeter must be connected in series with the circuit so as to measure the current flowing through the circuit.

So, to measure the current flowing through a circuit with an ammeter, the ammeter must be connected in series with the circuit.

4 0

3 years ago