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sukhopar [10]
2 years ago
12

What is the peak power consumption of a 120-v ac microwave oven that draws 10. 0 a.

Physics
1 answer:
Lina20 [59]2 years ago
7 0
The peak power consumption of a 120-V AC microwave oven that draws 10.0 A? is 2400 W.
You might be interested in
An object experiences an acceleration of -6.8 m/s​2.​ As a result, it accelerates from 54 m/s to a complete stop. How much dista
inessss [21]

Answer:

The distance traveled during its acceleration, d = 214.38 m

Explanation:

Given,

The object's acceleration, a = -6.8 m/s²

The initial speed of the object, u = 54 m/s

The final speed of the object, v = 0

The acceleration of the object is given by the formula,

                                      a = (v - u) / t   m/s²

       ∴                              t = (v - u) / a

                                         = (0 - 54) / (-6.8)

                                         = 7.94 s

The average velocity of the object,

                                       V = (54 + 0)/2

                                           = 27 m/s

The displacement of the object,

                                 d = V x t   meter

                                    = 27 x 7.94

                                    = 214.38 m

Hence, the distance the object traveled during that acceleration is, a = 214.38 m

3 0
3 years ago
ate around its central axis. A rope wrapped around the drum of radius 1.24 m exerts a force of 4.56 N to the right on the cylind
Mashcka [7]

Answer:

Magnitude the net torque about its axis of rotation is 1.3338 Nm

Explanation:

The radius of the wrapped rope around the drum, r = 1.24 m

Force applied to the right side of the drum, F = 4.56 N

The radius of the rope wrapped around the core, r' = 0.57 m

Force on the cylinder in the downward direction, F' = 7.58 N

Now, the magnitude of the net torque is given by:

\tau_{net} = \tau + \tau'

where

\tau = Torque due to Force, F

\tau' = Torque due to Force, F'

\tau = F\times r\tau' = F'\times r'

Now,

\tau_{net} = - F\times r + F'\times r'\tau_{net} = - 4.56\times 1.24 + 7.58\times 0.57 \\\\= - 1.3338\ Nm

The net torque comes out to be negative, this shows that rotation of cylinder is in the clockwise direction from its stationary position.

Now, the magnitude of the net torque:

|\tau_{net}| = 1.3338\ Nm

8 0
3 years ago
True or False. Can Metalloids conduct electricity under certain conditions?
Phoenix [80]

can metalloids conduct electricity under certain conditions?

true

5 0
3 years ago
UN BLOQUE TIENE DE DIMENSIONES AB = 4 cm, BC = 2.5 cm, BF = 6 cm Y UNA MASA DE 300 gr  DETERMINE SU VOLUMEN  HALLE SU DENSIDAD
torisob [31]

Responder:

- Volumen del bloque = 60cm³

- Densidad = 5g / cm³

Explicación:

Dada la dimensión del bloque

AB = 4 cm, BC = 2.5 cm y BF = 6 cm

Volumen del bloque = Largo × Ancho × Alto

Volumen del bloque = AB × BC × BF

Volumen del bloque = 4 × 2.5 × 6

Volumen del bloque = 60cm³

Dada la masa del bloque = 300 gramos

Densidad = Masa / Volumen

Densidad = 300g / 60cm³

Densidad = 5g / cm³

Algunos materiales que tienen una densidad de 5g / cm³ o cercanos son radio, germanio y europio con densidades de 5g / cm³, 5.3g / cm³ y 5.24g / cm³ respectivamente.

6 0
4 years ago
An object has the acceleration graph shown in (Figure 1). Its velocity at t=0s is vx=2.0m/s. Draw the object's velocity graph fo
timama [110]

Answer:

Explanation:

We may notice that change in velocity can be obtained by calculating areas between acceleration lines and horizontal axis ("Time"). Mathematically, we know that:

v_{b}-v_{a} = \int\limits^{t_{b}}_{t_{a}} {a(t)} \, dt

v_{b} = v_{a}+ \int\limits^{t_{b}}_{t_{a}} {a(t)} \, dt

Where:

v_{a}, v_{b} - Initial and final velocities, measured in meters per second.

t_{a}, t_{b} - Initial and final times, measured in seconds.

a(t) - Acceleration, measured in meters per square second.

Acceleration is the slope of velocity, as we know that each line is an horizontal one, then, velocity curves are lines with slopes different of zero. There are three region where velocities should be found:

Region I (t = 0 s to t = 4 s)

v_{4} = 2\,\frac{m}{s}  +\int\limits^{4\,s}_{0\,s} {\left(-2\,\frac{m}{s^{2}} \right)} \, dt

v_{4} = 2\,\frac{m}{s}+\left(-2\,\frac{m}{s^{2}} \right) \cdot (4\,s-0\,s)

v_{4} = -6\,\frac{m}{s}

Region II (t = 4 s to t = 6 s)

v_{6} = -6\,\frac{m}{s}  +\int\limits^{6\,s}_{4\,s} {\left(1\,\frac{m}{s^{2}} \right)} \, dt

v_{6} = -6\,\frac{m}{s}+\left(1\,\frac{m}{s^{2}} \right) \cdot (6\,s-4\,s)

v_{6} = -4\,\frac{m}{s}

Region III (t = 6 s to t = 10 s)

v_{10} = -4\,\frac{m}{s}  +\int\limits^{10\,s}_{6\,s} {\left(2\,\frac{m}{s^{2}} \right)} \, dt

v_{10} = -4\,\frac{m}{s}+\left(2\,\frac{m}{s^{2}} \right) \cdot (10\,s-6\,s)

v_{10} = 4\,\frac{m}{s}

Finally, we draw the object's velocity graph as follows. Graphic is attached below.

3 0
4 years ago
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