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Hitman42 [59]
3 years ago
8

You are traveling at 100 km/hr to make it to work on time. You have to be to work in .25 hr or you will be late. You have 16 km

to travel. Will you make it to work on time?
Physics
1 answer:
Murrr4er [49]3 years ago
8 0

Answer:

You are able to travel 25 km in the

remaining 1/4 hour, so you will make it.

Explanation:

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MrRa [10]

Wave A has a higher amplitude than Wave B.

8 0
4 years ago
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Analyze how most volcanic mountains formed in California
Burka [1]
Los angelos fault got eem coach
5 0
3 years ago
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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
Solve these problems on charges by finding out Q1 and Q2. Q1Q2= 4x10^6, Q1 + Q2 =8.00×10^-6​
inna [77]

Answer:

Answer:

Q_1 = 7Q

1

=7

Q_2 = 10Q

2

=10

Q_3 = 13.5Q

3

=13.5

Step-by-step explanation:

Given

5, 7, 7, 8, 10, 11, 12, 15, 17.

Required

Determine Q1, Q2 and Q3

The number of data is 9

Calculating Q1:

Q1 is calculated as:

Q_1 = \frac{1}{4}(N + 1)Q

1

=

4

1

(N+1)

Substitute 9 for N

Q_1 = \frac{1}{4}(9 + 1)Q

1

=

4

1

(9+1)

Q_1 = \frac{1}{4}*10Q

1

=

4

1

∗10

Q_1 = 2.5th\ itemQ

1

=2.5th item

This means that the Q1 is the mean of the 2nd and 3rd data.

So:

Q_1 = \frac{1}{2}(7+7)Q

1

=

2

1

(7+7)

Q_1 = \frac{1}{2}*14Q

1

=

2

1

∗14

Q_1 = 7Q

1

=7

Calculating Q2:

Q2 is calculated as:

Q_2 = \frac{1}{2}(N + 1)Q

2

=

2

1

(N+1)

Substitute 9 for N

Q_2 = \frac{1}{2}(9 + 1)Q

2

=

2

1

(9+1)

Q_2 = \frac{1}{2}*10Q

2

=

2

1

∗10

Q_2 = 5th\ itemQ

2

=5th item

Q_2 = 10Q

2

=10

Calculating Q3:

Q3 is calculated as:

Q_3 = \frac{3}{4}(N + 1)Q

3

=

4

3

(N+1)

Substitute 9 for N

Q_3 = \frac{3}{4}(9 + 1)Q

3

=

4

3

(9+1)

Q_3 = \frac{3}{4}*10Q

3

=

4

3

∗10

Q_3 = 7.5th\ itemQ

3

=7.5th item

This means that the Q3 is the mean of the 7th and 8th data.

So:

Q_3 = \frac{1}{2}(12+15)Q

3

=

2

1

(12+15)

Q_3 = \frac{1}{2}*27Q

3

=

2

1

∗27

Q_3 = 13.5Q

3

=13.5

4 0
3 years ago
A truck covers 40.0 m in 7.80 s while uniformly slowing down to a final velocity of 1.70 m/s. Please provide with explanation an
Karolina [17]

Answer:

a = -0.223 m/s

Explanation:

4 0
3 years ago
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