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lozanna [386]
3 years ago
13

The wasted energy in a charger is:

Physics
1 answer:
kirza4 [7]3 years ago
4 0
Thermal is the wasted energy in a charger
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HELP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!pls
barxatty [35]
I think this is the solution:

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5 0
2 years ago
Un tubo cilindrico hueco de cobre mide 3 m de longitud tienen un diametro exterior de 4cm y un diametro interior de 2 cm¿cuanto
Irina-Kira [14]

Answer:

 W = 9.93 10² N

Explanation:

To solve this exercise we must use the concept of density

           ρ = m / V

the tabulated density of copper is rho = 8966 kg / m³

let's find the volume of the cylindrical tube

           V = A L

           V = π (R_ext  ² - R_int ²) L

let's calculate

          V = π (4² - 2²) 10⁻⁴  3

          V = 1.13 10⁻²  m³

         m = ρ V

        m = 8966 1.13 10⁻²

        m = 1.01 10² kg

the weight of the tube

        W = mg

         W = 1.01 10² 9.8

         W = 9.93 10² N

4 0
2 years ago
The forces in (Figure 1) are acting on a 1.0 kg object.What is ax , the x -component of the object's acceleration
a_sh-v [17]

The x -component of the object's acceleration is 2 m/s².

<h3>What's the resultant force along x- direction?</h3>
  • Forces along x axis direction are as follows
  1. 4N along +x axis, so it's taken as +4 N
  2. 2N along -x axis , so it's taken as -2N.
  • Resultant force along x direction = 4N - 2N = 2 N which is along + ve x direction.

<h3>What's the acceleration along x axis direction?</h3>
  • As per Newton's second law, Force = mass × acceleration of the object
  • Force along x axis= mass × acceleration along x axis= 2N
  • Acceleration = 2/ mass = 2/1 = 2 m/s²

Thus, we can conclude that the acceleration along x axis is 2 m/s².

Disclaimer: The question was given incomplete on the portal. Here is the complete question.

Question: The forces in (Figure 1) are acting on a 1.0 kg object. What is ax, the x-component of the object's acceleration?

Learn more about the acceleration here:

brainly.com/question/460763

#SPJ1

3 0
2 years ago
Calculate the net force and the acceleration on the block
Darina [25.2K]

Answer:

Net Force = 10N

Acceleration = 2m/s^2

Explanation:

calculate the net force and the acceleration on the block

Net force on the block F = mass * acceleration

Net force acting in the positive direction = 4N + 6N = 10N

Mass = 5kg

According to newton's second law;

a = F/m

a = 10N/5

a = 2m/s^2

hence the acceleration on the block is 2m/s^2

7 0
3 years ago
A 0.274 kg block is pushed up a frictionless ramp inclined at angle of 32° above the horizon. The push force is a constant 2.55
SpyIntel [72]

Answer:

The answer is 2,416 m/s. Let's jump in.

Explanation:

We do work with the amount of energy we can transfer to objects. According to energy theory:

W = ΔE

Also as we know W = F.x

We choose our reference point as a horizontal line at the block's rest point.<u> At the rest, block doesn't have kinetic energy</u> and <u>since it is on the reference point(as we decided) it also has no potential energy.</u>

Under the force block gains;

W = F.x → W=2,55.0,71=1,8105\frac{N}{m}

In the second position block has both kinetic and potential energy. Following the law of conservation of energy;

W = ΔE = Kinetic energy + Potantial Energy

W = ΔE = \frac{1}{2} mV^{2} + mgh

Here we can find h in the triangle i draw in the picture using sine theorem;

In a triangle \frac{a}{sinA}=\frac{b}{sinB}=\frac{c}{sinC}

In our situation

\frac{0,71}{sin90} =\frac{h}{sin32} → h=0,376

Therefore

1,8105=\frac{1}{2} 0,274V^{2} +0,274.9,81.0,376

→ V=2,416

7 0
3 years ago
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