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Mars2501 [29]
3 years ago
10

The amount of energy transferred by can be determined by…

Physics
1 answer:
son4ous [18]3 years ago
5 0

Answer:

Power and time

Explanation:

Given the power of an appliance, the energy transferred is given as the product of power and time. Normally, power is defined as the rate of energy transfer per unit time in a circuit and expressed as

P=E/t

Making E the subject of the formula then

E=Pt

Where P is power, E is energy transfer and t is time.

Clearly, energy transfer is determined by power of an appliance and time.

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Suppose you're performing experiments in science class in which you start with 70 bacteria and the amount of bacteria triples ev
barxatty [35]
1h----------------> 70x3=210 bacteria
2h-----------------> 210*3=630 bactaeria
let be y the number of bacteria at the t=0h
it is y=70 3^0
for t= 1h
y=70*3^1=210
for t=2h
y=70*3^2=630

so we can write y=70*3^x, where x is the number of hour


5 0
3 years ago
Read 2 more answers
Two tiny particles having charges of +5.00 μC and +7.00 μC are placed along the x-axis. The +5.00-µC particle is at x = 0.00 cm,
Liula [17]

Answer:

The third charged particle must be placed at x = 0.458 m = 45.8 cm

Explanation:

To solve this problem we apply Coulomb's law:  

Two point charges (q₁, q₂) separated by a distance (d) exert a mutual force (F) whose magnitude is determined by the following formula:  

F = \frac{k*q_1*q_2}{d^2} Formula (1)  

F: Electric force in Newtons (N)

K : Coulomb constant in N*m²/C²

q₁, q₂: Charges in Coulombs (C)  

d: distance between the charges in meters (m)

Equivalence  

1μC= 10⁻⁶C

1m = 100 cm

Data

K = 8.99 * 10⁹ N*m²/C²

q₁ = +5.00 μC = +5.00 * 10⁻⁶ C

q₂= +7.00 μC = +7.00 * 10⁻⁶ C

d₁ = x (m)

d₂ = 1-x (m)

Problem development

Look at the attached graphic.

We assume a positive charge q₃ so F₁₃ and F₂₃ are repulsive forces and must be equal so that the net force is zero:

We use formula (1) to calculate the forces F₁₃ and F₂₃

F_{13} = \frac{k*q_1*q_3}{d_1^2}

F_{23} = \frac{k*q_2*q_3}{d_2^2}

F₁₃ = F₂₃

\frac{k*q_1*q_3}{d_1^2} = \frac{k*q_2*q_3}{d_2^2} We eliminate k and q₃ on both sides

\frac{q_1}{d_1^2}= \frac{q_2}{d_2^2}

\frac{q_1}{x^2}=\frac{q_2}{(1-x)^2}

\frac{5*10^{-6}}{x^2}=\frac{7*10^{-6}}{(1-x)^2} We eliminate 10⁻⁶ on both sides

(1-x)^2 = \frac{7}{5} x^2

1-2x+x^2=\frac{7}{5} x^2

5-10x+5x^2=7 x^2

2x^2+10x-5=0

We solve the quadratic equation:

x_1 = \frac{-b+\sqrt{b^2-4ac} }{2a} = \frac{-10+\sqrt{10^2-4*2*(-5)} }{2*2} = 0.458m

x_2 = \frac{-b-\sqrt{b^2-4ac} }{2a} = \frac{-10-\sqrt{10^2-4*2*(-5)} }{2*2} = -5.458m

In the option x₂, F₁₃ and F₂₃ will go in the same direction and will not be canceled, therefore we take x₁ as the correct option since at that point the forces are in  opposite way .

x = 0.458m = 45.8cm

8 0
3 years ago
An astronaut is standing on the surface of a planetary satellite that has a radius of 1.74 × 10^6 m and a mass of 7.35 × 10^22 k
ExtremeBDS [4]

Answer:

2.87 km/s

Explanation:

radius of planet, R = 1.74 x 10^6 m

Mass of planet, M = 7.35 x 10^22 kg

height, h = 2.55 x 10^6 m

G = 6.67 x 106-11 Nm^2/kg^2

Use teh formula for acceleration due to gravity

g=\frac{GM}{R^{2}}

g=\frac{6.67\times 10^{-11}\times 7.35\times 10^{22}}{1.74^{2}\times 10^{12}}

g = 1.62 m/s^2

initial velocity, u = ?, h = 2.55 x 10^6 m , final velocity, v = 0

Use third equation of motion

v^{2}=u^{2}-2gh

0 = v² - 2 x 1.62 x 2.55 x 10^6

v² = 8262000

v = 2874.37 m/s

v = 2.87 km/s

Thus, the initial speed should be 2.87 km/s.

6 0
3 years ago
A 65-kg ice skater stands facing a wall with his arms bent and then pushes away from the wall by straightening his arms. At the
Marrrta [24]

Our values can be defined like this,

m = 65kg

v = 3.5m / s

d = 0.55m

The problem can be solved for part A, through the Work Theorem that says the following,

W = \Delta KE

Where

KE = Kinetic energy,

Given things like that and replacing we have that the work is given by

W = Fd

and kinetic energy by

\frac {1} {2} mv ^ 2

So,

Fd = \frac {1} {2} m ^ 2

Clearing F,

F = \frac {mv ^ 2} {2d}

Replacing the values

F = \frac {(65) (3.5)} {2 * 0.55}

F = 723.9N

B) The work done by the wall is zero since there was no displacement of the wall, that is d = 0.

6 0
2 years ago
A student puts a beaker of boiling water where it touches a block of ice. Which two statements are true? A. Thermal energy will
rodikova [14]

Answer: C and D

Explanation: I’m pretty sure it’s C and D I got the same question but cold water let me know please, thanks!

6 0
3 years ago
Read 2 more answers
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