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julsineya [31]
2 years ago
13

What is the power through a device with a resistance of 100 ohms if a current of 8 A is running through it?

Physics
1 answer:
Kazeer [188]2 years ago
5 0

Answer:

6400 W (or) 6.4 KW

Explanation:

Formula we use,

→ P = I²R

Let's solve for the power of device,

→ P = I²R

→ P = (8)² × 100

→ P = 64 × 100

→ [ P = 6400 W ]

Hence, the power is 6400 W.

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A girl delivering newspapers covers her route by traveling 3.00 blocks west, 4.00 blocks north, and then 6.00 blocks east. What
Harrizon [31]

The concepts used to solve this problem are those related to the Pythagorean theorem for which we will calculate the distance and the pitch.

According to the attached diagram we have that the expression of the resulting displacement is

R = \sqrt{(3b)^2+(4b)^2}

Therefore the resultant displacement of the girl is

R = \sqrt{(9b^2+16b^2)}

R = \sqrt{25b^2}

R = 5b

Therefore the girl has displaced around of 5 blocks

4 0
3 years ago
The tub of a washer goes into its spin-dry cycle, starting from rest and reaching an angular speed of 2.0 rev/s in 10.0 s. At th
Arte-miy333 [17]

Answer:

22 revolutions

Explanation:

2 rev/s = 2*(2π rad/rev) = 12.57 rad/s

The angular acceleration when it starting

\alpha_a = \frac{\Delta \omega}{\Delta t} = \frac{12.57}{10} = 1.257 rad/s^2

The angular acceleration when it stopping:

\alpha_o = \frac{\Delta \omega}{\Delta t} = \frac{-12.57}{12} = -1.05 rad/s^2

The angular distance it covers when starting from rest:

\omega^2 - 0^2 = 2\alpha_a\theta_a

\theta_a = \frac{\omega^2}{2\alpha_a} = \frac{12.57^2}{2*1.257} = 62.8 rad

The angular distance it covers when coming to complete stop:

0 - \omega^2 = 2\alpha_o\theta_o

\theta_o = \frac{-\omega^2}{2\alpha_o} = \frac{-12.57^2}{2*(-1.05)} = 75.4 rad

So the total angular distance it covers within 22 s is 62.8 + 75.4 = 138.23 rad or 138.23 / (2π) = 22 revolutions

6 0
3 years ago
The charge of a single electron
vlabodo [156]
B. Electrons are negative so one single 1 would be negative 1
7 0
3 years ago
The electron gun in an old TV picture tube accelerates electrons between two parallel plates 1.4cm apart with a 28kV potential d
yulyashka [42]

Answer:

a)   F = 3.2 10⁻¹⁰ N , b)       v = 9.9 10⁷ m / s

Explanation:

a) The electric force is

       F = q E

The electric field is related to the potential reference

     V = E d

     E = V / d

Let's replace

    F = e V / d

Let's calculate

    F = 1.6 10⁻¹⁹ 28 10³ / 1.4 10⁻²

    F = 3.2 10⁻¹⁰ N

b) For this part we can use kinematics

          v² = v₀ + 2 a d

          v = √ 2 ad

Acceleration can be found with Newton's second law

        e V / d = m a

        a = e / m V / d

        a = 1.6 10⁻¹⁹ / 9.1 10⁻³¹ 28 10³ / 1.4 10⁻²

        a = 3,516 10⁻¹⁷ m / s²

Let's calculate the speed

       v = √ (2 3,516 10¹⁷ 1.4 10⁻²)

       v = √ (98,448 10¹⁴)

       v = 9.9 10⁷ m / s

3 0
3 years ago
Vector ~A has a negative x-component 3.07 units in length and a positive y-component 3.17 units in length. When a vector ~B = b1
luda_lava [24]

Answer:

a. 3.07 b. 1.26

Explanation:

Given that A = -3.07i + 3.17j and B = b1i + b1j and C = A + B = 0i + 4.43j

Since A + B = -3.07i + 3.17j + b1i + b2j

= (-3.07 + b1)i + (3.17 + b2)j

So,(-3.07 + b1)i + (3.17 + b2)j = 0i + 4.43j

Comparing components,

-3.07 + b1 = 0 (1) and 3.17 + b2 = 4.43 (2)

a. From (1), b1 = 3.07

b. From(2) b2 = 4.43 - 3.17 = 1.26

4 0
3 years ago
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