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viktelen [127]
3 years ago
6

How much work can be done in 30 seconds by a 1000-watt microwave oven?

Physics
2 answers:
iris [78.8K]3 years ago
7 0
Half of it can be be at 1000 watt microwave oven!!!                                                                                                                                                                                                                                                                                                            I hope it help!!!!!!
gayaneshka [121]3 years ago
6 0

Answer:

30000 J

Explanation:

Power is defined as the amount of work done per unit time:

P=\frac{W}{t}

where

P is the power

W is the work done

t is the time taken

In this problem, we know the power of the microwave oven: P=1000 W, and we are asked to find the work done in a time of t=30 s, so by re-arranging the equation we find:

W=Pt=(1000 W)(30 s)=30,000 J=30 kJ

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The interference pattern seen when light passes through narrow, closely spaced slits, is due to
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<h2>Answer: Diffraction</h2><h2 />

Diffraction  is a characteristic phenomenon that occurs in all types of waves .

In this sense,  <u>diffraction</u> happens when a wave (the light in this case) meets an obstacle or a slit .When this occurs, the light bends around the corners of the obstacle or passes through the opening of the slit that acts as an obstacle, forming <u><em>multiple patterns</em></u> with the shape of the aperture of the slit.

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8 0
4 years ago
a soccer player kicks a ball. According to Newton's third law of motion, the player's foot exerts a force on the ball. The ball
Leto [7]

Answer:

The forces are exerted on different objects so they are not balanced forces.

Explanation:

8 0
3 years ago
A parallel-plate capacitor with plates of area 600 cm^2 is charged to a potential difference V and is then disconnected from the
Julli [10]

Answer:

<h2>a) Q = 0.759µC</h2><h2>b) E = 39.5µJ</h2>

Explanation:

a) The charge Q on the positive charge capacitor can be gotten using the formula Q = CV

C = capacitance of the capacitor (in Farads )

V = voltage (in volts) = 100V

C = ∈A/d

∈ = permittivity of free space = 8.85 × 10^-12 F/m

A = cross sectional area = 600 cm²

d= distance between the plates = 0.7cm

C = 8.85 × 10^-12 * 600/0.7

C = 7.59*10^-9Farads

Q = 7.59*10^-9 * 100

Q = 7.59*10^-7Coulombs

Q = 0.759*10^-6C

Q = 0.759µC

b) Energy stored in a capacitor is expressed as E = 1/2CV²

E = 1/2 * 7.59*10^-9 * 100²

E = 0.0000395Joules

E = 39.5*10^-6Joules

E = 39.5µJ

7 0
3 years ago
Janet ran 30 meters in 4 seconds,what is her speed.
lubasha [3.4K]

Answer:

7.5 meters per second

Explanation:

30 / 4 = 7.5

7.5 * 1 = 7.5

7.5 * 2 = 15

7.5 * 3 = 22.5

7.5 * 4 = 30

8 0
4 years ago
Read 2 more answers
A bobsledder pushes her sled across horizontal snow to get it going, then jumps in. After she jumps in, the sled gradually slows
anastassius [24]

Answer:

In the vertical direction the acting forces are the normal force and the weight of the bobsleder plus the sled. In the horizontal direction the acting force is the friciton force.

Explanation:

Hi there!

Please, see the attached figure for a graphic representation of the forces acting on the sled after the bobsleder jumped in.

In the vertical direction, the acting forces are the normal force (N) and the weight of the sled plus the bobsledder (W).

Since the sled is not being accelerated in the vertical direction, the sum of forces in that direction is zero:

∑Fy = W + N = 0 ⇒ W = N

The weight is calculated as follows:

W = (mb + ms) · g

Where:

mb = mass of the bobsleder.

ms = mass of the sled.

g = acceleration due to gravity.

In the horizontal direction the only acting force is the friction force (Fr). The friction force is calculated a follows:

Fr = N · μ

Where:

N = normal force.

μ = kinetic friction coefficient.

Since N = W = (mb + ms) · g

Fr = (mb + ms) · g · μ

If we want to find the acceleration of the sled after the bobsleder jumps in, we can apply Newton's second law:

∑F = m · a

Where "a" is the acceleration and "m" is the mass of the object (in this case, the mass of bobsleder plus the mass of the sled).

∑F = Fr =  (mb + ms) · g · μ =  (mb + ms) · a

(mb + ms) · g · μ =  (mb + ms) · a

Solving for "a":

g · μ = a

3 0
3 years ago
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