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2 years ago

The amount of matter (mass) in a given unit volume

Physics

1 answer:

maxonik [38]2 years ago

6 0

Answer:

Density is the mass of a substance per unit volume. Volume is the amount of space an object occupies.

Explanation:

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What are 3 types of white sugar chemical changes?

iragen [17]

Answer:

glucose ,fructose, sucrose,maltose,lactose etc... are some types of sugar.

4 0

2 years ago

Kiran drove from City A to City B, a distance of 228 mi. She increased her speed by 12 mi/h for the 400-mi trip from City B to C

Degger [83]

Answer:

From city A to city B her speed was 38 mi/h

Explanation:

The traveled distance can be calculated using this equation:

From city A to city B

228 mi = v · t₁

Where:

v = velocity

t₁ = time it took Kiran to travel the 228 mi from city A to city B

From city B to city C

400 mi = (v + 12 mi/h) · t₂

We also know that the entire trip took 14 h, then:

t₁ + t₂ = 14 h

So, we have a system of three equations with three unknwons:

228 mi = v · t₁

400 mi = (v + 12 mi/h) · t₂

t₁ + t₂ = 14 h

Let´s solve the third equation for t₁:

t₁ = 14 h - t₂

Now let´s replace t₁ in the first equation and solve it for t₂

228 mi = v · t₁

228 mi = v · (14 h - t₂)

228 mi/v - 14 h = - t₂

t₂ = 14 h - 228 mi/v

Now let´s replace t₂ in the second equation:

400 mi = (v + 12 mi/h) · t₂

400 mi = (v + 12 mi/h) · (14 h - 228 mi/v)

400 mi = 14 h · v - 228 mi + 168 mi - 2736 mi²/(v · h)

400 mi = 14 h · v - 60 mi - 2736 mi²/(v · h)

460 mi = 14 h · v - 2736 mi²/(v · h)

Multiplicate by v both sides of the equation:

460 mi · v = 14 h · v² - 2736 mi²/h

0 = 14 h · v² - 460 mi · v - 2737 mi²/h

Solving the quadratic equation:

v = 38 mi/h

(The other solution of the equation is negative, and therefore discarded)

From city A to city B her speed was 38 mi/h

4 0

2 years ago

A point charge of -0.70 μC is fixed to one corner of a square. An identical charge is fixed to the diagonally opposite corner. A

MakcuM [25]

Answer:

The magnitude and algebraic sign of q is $14\sqrt{2}\ \mu C$

Explanation:

Given that,

Point charge = -0.70 μC[/tex]

We need to calculate the force for all charges

The electric force at first corner

$F_{1}=\dfrac{-k0.70\times10^{-6}q}{r^2}$

The electric force at opposite corner

$F_{3}=\dfrac{-k0.70\times10^{-6}q}{r^2}$

The net force is

$F=\sqrt{F_{1}^2+F_{2}^2}$

Put the value into the formula

$F=\sqrt{(\dfrac{-k0.70\times10^{-6}q}{r^2})^2+(\dfrac{-k0.70\times10^{-6}q}{r^2})^2}$

The electric force at second corner

$F_{2}=\dfrac{-kq^2}{2r^2}$

The net force acting on either of the charges is zero.

So, $F=F'$

$\sqrt{(\dfrac{k0.70\times10^{-6}q}{r^2})^2+(\dfrac{-k0.70\times10^{-6}q}{r^2})^2}=\dfrac{kq^2}{2r^2}$

$\sqrt{2}\times\dfrac{0.70\times10^{-6}kq}{r^2}=\dfrac{kq^2}{2r^2}$

$q=14\sqrt{2}\ \mu C$

Hence, The magnitude and algebraic sign of q is $14\sqrt{2}\ \mu C$

8 0

2 years ago

Is there any possibility to make 100% efficient system

Ganezh [65]

Answer:

yes

Explanation:

its awesome

4 0

2 years ago

A small charged bead has a mass of 3.6 g. It is held in a uniform electric field E = (200,000 N/C, up). When the bead is release

DiKsa [7]

Answer:

The charge on the bead is $q = 6.084 *10^{-7}\ C$

Explanation:

From the question we are told that

The mass of the bead is $m = 3.6 \ g = 0.0036 \ kg$

The magnitude of the electric field is $E = 200,000 \ N/C$

The acceleration of the bead is $a = 24 m/s^2$

Generally, the electric force on the bead is mathematically represented as

$F_ e = q E$

Where q is the charge on the bead

Now the gravitational force opposing the upward movement of the bead is mathematically represented as

$F_g = mg$

Generally the net force on the bead is mathematically represented as

$F = F_e - F_g = m* a$

=> $qE - mg = ma$

Now substituting values

$q * 200000 - 0.0036 *9.8 = 0.0036 * 24$

$q = 6.084 *10^{-7}\ C$

6 0

2 years ago