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2 years ago

Write the relation between horsepower and Watt?

Physics

2 answers:

Vadim26 [7]2 years ago

8 0

Explanation:

Horsepower is related to watt as: 1 horsepower = 746 W = 750 W approximately. 1 horsepower (HP) = 750 watt (approx.) The power is one watt when one joule of work is done in one second.

zvonat [6]2 years ago

5 0

Answer:

1HP = 746 watts

Explanation:

The relation between Horsepower and watt is that One Horse Power is equal to 746 Watts .

$\large{\boxed{ \sf 1 HP = 746 Watts = 0.746 kW }}$

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In the Millikan oil drop experiment the measured charge of any single droplet was always a whole number multiple of -1.60 x 10-1

BaLLatris [955]

Answer:

Number of electrons, n = 6

Explanation:

Total charge in a single droplet, $q=-9.6\times 10^{-19}\ C$

The measured charge of any single droplet, $e=-1.6\times 10^{-19}\ C$

Let n is the number of excess electrons are contained within the drop. According to the quantization of charge :

$q=ne$

$n=\dfrac{q}{e}$

$n=\dfrac{-9.6\times 10^{-19}}{-1.6\times 10^{-19}}$

n = 6

So, there are 6 electrons contained within the drop. Hence, this is the required solution.

7 0

3 years ago

Consider a series RLC circuit where R = 855 Ω and C = 6.25 μF. However, the inductance L of the inductor is unknown. To find its

sashaice [31]

Answer:

L= 0.059 mH

Explanation:

Given that

R = 855 Ω and C = 6.25 μF

V= 84 V

Frequency

ω = 51900 1/s

We know that

$\omega=\sqrt{\dfrac{1}{LC}}$

L=Inductance

C=Capacitance

ω =angular Frequency

ω² L C =1

(51900)² x L x 6.25 x 10⁻⁶ = 1

L= 5.99 x 10⁻⁵ H

L= 0.059 mH

6 0

3 years ago

PLEASE ASAP ILL GIVE BRAINLIEST.

taurus [48]

Answer:

applied force

Explanation:

any force where you push or pull is always applied force.

4 0

2 years ago

A point charge is placed at each corner of square with side leanth a. The charges all have same magnitude q. My question, What i

nexus9112 [7]

Answer:

E = k q / a² (1.3535) (- i ^ + j ^)

E = k q / a² 1.914 , θ’= 135

Explanation:

For this exercise we will use Newton's second law where we must add as vectors

E_total = E₁₂ i ^ + E₁₄ j ^ + E₁₃

Let's look for the value of each term

On the x axis

E₁₂ = k q / a²

On the y axis

E₁₄ = k q / a²

For the charge in the opposite corner we look for the distance

d = √ (a² + a²) = a √2

let's look for the field

E₁₃ = k q / d²

E₁₃ = k q / 2a²

let's use trigonometry to find the two components of this field

cos 45 = E₁₃ₓ / E₁₃

E₁₃ₓ = E₁₃ cos 45

sin 45 = E_{13y} / E₁₃

E_{13y} = E₁₃ sin 45

E₁₃ₓ = k q / 2a² cos 45

E_{13y} = k q / 2a² sin 45

let's find each component of the electric field

X axis

Eₓ = -E₁₂ - E₁₃ₓ

Eₓ = - k q / a² - k q / 2a² cos 45

Eₓ = - k q / a² (1 + cos 45/2)

cos 45 = sin 45 = 0.707

Eₓ = - k q / a² (1 + 0.707 / 2)

Eₓ = - k q / a² (1.3535)

Y axis

E_y = E₁₄ + E_{13y}

E_y = k q / a² + k q / 2a² sin 45

E_y = k q / a² (1 + sin 45/2)

E_y = k q / a² (1.3535)

we can give the results in two ways

E = k q / a² (1.3535) (- i ^ + j ^)

In modulus and angle form, let's use Pythagoras' theorem for the angle

E = √ (Eₓ² + E_y²)

E = k q / a² 1.3535 √2

E = k q / a² 1.914

we use trigonometry for the angle

tan θ = E_y / Eₓ

θ = tan⁻¹ (E_y / Eₓ)

θ = tan⁻¹ (1 / -1)

θ = 45

in the third quadrant, if we measure the angle of the positive side of the x-axis

θ‘= 90 + 45

θ’= 135

4 0

3 years ago

A boy whirls a stone in a horizontal circle of radius 1.1 m and at height 2.1 m above ground level. The string breaks, and the s

DedPeter [7]

Answer:

$212.8 m/s^{2}$

Explanation:

Time taken by stone to cover horizontal distance

$t=\sqrt{\frac {2h}{g}}$ where t is time, h is height of whirling the stone in horizontal circle, g is gravitational constant, Substituting h for 2.1 m and g for 9.81