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3 years ago

What is the potential difference across a 15Ω resistor that has a current of 3.0 A?

Physics

2 answers:

ololo11 [35]3 years ago

7 0

Answer:

45 V

Explanation:

sesenic [268]3 years ago

6 0

V = I • R

V = (15 ohms) x (3 A)

V = 45 volts

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In creating his definition of horsepower, James Watt, the inventor of the steam engine, calculated the power output of a horse o

Studentka2010 [4]

Answer:

Part a)

$F = 182.3 Lb$

Part b)

$P = 0.55 HP$

Explanation:

Diameter of the circle = 24 ft

Diameter = 731.52 cm = 7.3152 m

now the horse complete 144 trips in one hour

so time to complete one trip is given as

$t = \frac{3600}{144} s$

$t = 25 s$

now the speed of the horse is given as

$v = \frac{2\pi r}{t}$

$v = \frac{\pi(7.3152)}{25}$

$v = 0.92 m/s$

Part a)

Now we know that the power is defined as rate of work done

it is given as

$P = F v$

$746 = F(0.92)$

$F = 810.9 N$

$F = 182.3 Lb$

Part b)

Work done to climb up to 3 m height is given by

$W = mgh$

now we have

$Power = \frac{Work}{time}$

$P = \frac{mgh}{t}$

$P = \frac{(70kg)(9.81)(3)}{5.0s}$

$P = 412.02 Watt$

now we know that 1 HP = 746 Watt

so we have

$P = \frac{412}{746} = 0.55 HP$

8 0

2 years ago

The model below shows a calcium atom. An image has a mix of red and blue balls in its center and 4 concentric black rings around

zubka84 [21]

Answer:

8 electrons in the third energy level

Explanation:

From the description,the third energy level has 8 electron (represented by the small green balls you describe)

4 0

3 years ago

Match: The graph below shows the potential and kinetic energy curves for a pendulum. Label each pendulum image with the correspo

Mariana [72]

Top right = B
bottom left = A
bottom right = C

4 0

3 years ago

The electric field 1.5 cm from a very small charged object points toward the object with a magnitude of 180,000 N/C. What is the

Ray Of Light [21]

Answer:

q = 4.5 nC

Explanation:

given,

electric field of small charged object, E = 180000 N/C

distance between them, r = 1.5 cm = 0.015 m

using equation of electric field

$E = \dfrac{kq}{r^2}$

k = 9 x 10⁹ N.m²/C²

q is the charge of the object

$q= \dfrac{Er^2}{k}$

now,

$q= \dfrac{180000\times 0.015^2}{9\times 10^9}$

q = 4.5 x 10⁻⁹ C

q = 4.5 nC

the charge on the object is equal to 4.5 nC

8 0

3 years ago

Read 2 more answers

Problem 1. Cylinders and a pendulum A uniform solid cylinder of radius r and mass m can roll inside a hollow cylinder of radius

Nataliya [291]

Answer:

The two answers are in the explanation

Explanation:

Please find the attached files for the solution

7 0

3 years ago