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Usimov [2.4K]
3 years ago
15

Use the equation d = st, where d = distance, s = speed, and t = time. If you ride

Physics
1 answer:
eimsori [14]3 years ago
8 0

Answer:

D (50m)

Explanation:

If distance is speed × time

That would be 5m/s × 10s

So the final answer will be 50m.

Blessings

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Kinetic energy is energy of motion while temperature is a measure of that energy in substances.
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How fast would a 0.25 kg football have to be traveling to have the same momentum as a 0.05 kg bullet traveling 500 m/s?
motikmotik
P=mv
0.25v=0.05*500
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4 0
3 years ago
1. Una carga Q1 = + 12 μC se coloca a una distancia r = 0.024 m desde una carga Q2 = + 16 μC. a) Determina la magnitud de la fue
lyudmila [28]

Answer:

1. a. 3,000 N

b. Repulsión

2. 46.875 × 10⁶ N/C

3. a. 81,000 J

b. 6.75 × 10⁹ V

Explanation:

1. Los parámetros dados son;

Q₁ = +12 μC, Q₂ = +16 μC

La distancia entre las cargas, r = 0.024

La magnitud de la fuerza electrostática, F, entre cargas se da como sigue;

F = k \times \dfrac{Q_1 \cdot Q_2}{r^2}

Donde, k = constante de Coulomb = 9.0 × 10⁹ N · m² / C²

Por lo tanto, obtenemos;

F = 9.0 × 10⁹ × 12 × 10⁻⁶ × 16 × 10⁻⁶ / 0.024² = 3.000

La magnitud de la fuerza electrostática, entre las cargas, F = 3000 N

(b) Dado que tanto Q₁ como Q₂ son cargas positivas, y las cargas iguales se repelen entre sí, la fuerza es la repulsión.

2) La intensidad de un campo eléctrico, E, se da como sigue;

E = \dfrac{k \cdot Q}{r^2}

La magnitud de la carga, Q = 24 μC

La distancia donde se mide el campo, r = 48 mm = 0.048 m

Por lo tanto, E = 9.0 × 10⁹ × 12 × 10⁻⁶ / 0.048² = 46,875,000 N / C

La intensidad de un campo eléctrico, E = 46,875,000 N / C = 46.875 × 10⁶ N / C

3. La magnitud de las cargas son;

Q₁ = 24 mC

Q₂ = -12 μC

La distancia entre las cargas, r = 0.032 m

un. El potencial eléctrico de una carga, U_E , se da de la siguiente manera;

U_E = k \times \dfrac{Q_1 \cdot Q_2}{r}

Por lo tanto;

U_E = 9.0×10⁹ × 24 × 10⁻³ × (-12) × 10⁻⁶ /0.032 = -81,000

La energía potencial eléctrica entre la carga, Q₁ y Q₂= -81,000 J

b. El potencial eléctrico de Q₁ en Q₂, V₁ = k \times \dfrac{Q_1 }{r}

Por lo tanto, V₁ = 9.0×10⁹ × 24 × 10⁻³/0.032 = 6.75 × 10⁹

El potencial eléctrico de Q₁ en Q₂, V₁ = 6.75 × 10⁹ V

3 0
3 years ago
Write the equation in slope-intercept form <br> x-2y=4
maxonik [38]

Answer:

y=\frac{1}{2}x-2

Explanation:

Slope-intercept form means we want the y to be by itself in the equation. Every thing we do will be about getting the y alone on the left side of the equation

To start we should move x to the left hand side. We can do this by subtracting x from both sides. That way, there is an x on the right, but not the left.

x-x-2y=4-x

this gives us

-2y=4-x

Great! So now what? Well, the y isn't by itself yet because it still is being multiplied by negative two (-2). In order to move it from the left side to the right side, we have to do the opposite of multiply; divide. So, we will divide both sides by -2

\frac{-2y}{-2} =\frac{4}{-2} -\frac{x}{-2}

-2 divided by -2 is 1, 4 divided by -2 is -2, and -x divided by -2 is \frac{1}{2}x

This gives us the answer:  y=\frac{1}{2}x-2

Tips:

A negative divided by a negative is a positive ex: -4 divided by -2 is positive 2

If you are subtracting by a negative number, you are actually adding by a positive ex: 2-(-2) is actually 2+2

Don't be afraid to have fractions in your equations

Whatever you do to the one side of the equation, you must do it to the other side as well. Multiply the left side by 2? You HAVE TO multiply the right side by two as well. Add 3 to the right side? You HAVE TO add 3 to the left.

For problems like this (and when you have access to the internet), where you need to rewrite an equation, double check your work on desmos, which is an online graphing calculator. Input both the original equation and the equation you rewrote, and if they don't create the same line, you did something wrong.

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Can be absorbed and transformed into heat. i know this because i used to work as a constructor for lights and i had to go over a training on this subject.
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4 years ago
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