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3 years ago

How much work will a machine with a power rating of 1.1 × 103 watts do in 2.0 hours?

Physics

2 answers:

lyudmila [28]3 years ago

7 0

Answer:

Explanation:

8.1X10 to the power of 6 Joules.

Pie3 years ago

4 0

Answer:

7.92×10^6Joules

Explanation:

Power is defined as the rate of change in work done. Mathematically, Power = Workdone/Time

Given power = 1.1×10³watts

Time = 2.0hours = 2×60×60

= 7,200seconds

Workdone = power × time

Work done = 1.1×10³×7200

Workdone = 7,920,000Joules

= 7.92×10^6Joules

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When light waves hit ice, most of them bounce off and radiate back into space. Because of this, it can be said that ice is a(n)

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3 years ago

Read 2 more answers

A proton starts from rest near the surface of a sheet of charge. It experiences a force of 14.0 microNewtons towards the sheet.

Naddika [18.5K]

Answer:

The surface charge density of the sheet is 1547.516 C/m²

Explanation:

Given;

force experienced by the proton, F = 14.0 μN

Electric field near an infinite sheet with surface charge density (σ) is given as;

E = σ/2ε₀

The force of attraction the proton experienced towards the sheet;

F = Eq

F = q(σ/2ε₀) $= \frac{q \sigma}{2 \epsilon_o}$

where;

q is the charge of the proton

σ is the surface charge density of the sheet

ε₀ is the permittivity of free space

$F = \frac{q \sigma}{2 \epsilon_o} \\\\\sigma = \frac{2F \epsilon_o}{q} \\\\\sigma = \frac{2*14*10^{-6}*8.854*10^{-12}}{1.602*10^{-19}} = 1547.516 \ C/m^2$

Therefore, the surface charge density of the sheet is 1547.516 C/m²

6 0

3 years ago

What is the gravitational potential energy of a two-particle system with masses 4.5 kg and 6.3 kg, if they are separated by 1.7

zepelin [54]

Answer:

7.41 x 10^-10 J

Explanation:

m = 4.5 kg

M = 6.3 kg

d = 1.7 m

The formula for the gravitational potential energy is given by

$U = \frac{-GMm}{d}$

Where, G is the Universal gravitational constant

G = 6.67 x 10^-11 Nm^2/kg^2

$U = \frac{-6.67 \times 10^{-11}\times 6.3 \times 4.5}{1.7}$

U = - 1.112 x 10^-9 J

Now the separation is tripled, d' = 3 x d = 3 x 1.7 m = 5.1 m

Let the potential energy is U'

The formula for the gravitational potential energy is given by

$U' = \frac{-GMm}{d'}$

Where, G is the Universal gravitational constant

G = 6.67 x 10^-11 Nm^2/kg^2

$U' = \frac{-6.67 \times 10^{-11}\times 6.3 \times 4.5}{5.1}$

U' = - 3.71 x 10^-10 J

the work done is equal to the change in potential energy

W = U' - U

W = - 3.71 x 10^-10 + 1.112 x 10^-9

W = 7.41 x 10^-10 J

7 0

3 years ago

A ball having mass 2 kg is connected by a string of length 2 m to a pivot point and held in place in a vertical position. A cons

Virty [35]

Complete Question

A ball having mass 2 kg is connected by a string of length 2 m to a pivot point and held in place in a vertical position. A constant wind force of magnitude 13.2 N blows from left to right. Pivot Pivot F F (a) (b) H m m L L If the mass is released from the vertical position, what maximum height above its initial position will it attain? Assume that the string does not break in the process. The acceleration of gravity is 9.8 m/s 2 . Answer in units of m.What will be the equilibrium height of the mass?

Answer:

$H_m=1.65m$

$H_E=1.16307m$

Explanation:

From the question we are told that

Mass of ball $M=2kg$

Length of string $L= 2m$

Wind force $F=13.2N$

Generally the equation for $\angle \theta$ is mathematically given as

$tan\theta=\frac{F}{mg}$

$\theta=tan^-^1\frac{F}{mg}$

$\theta=tan^-^1\frac{13.2}{2*2}$

$\theta=73.14\textdegree$

Max angle = $2*\theta= 2*73.14=>146.28\textdegree$

Generally the equation for max Height $H_m$ is mathematically given as

$H_m=L(1-cos146.28)$

$H_m=0.9(1+0.8318)$

$H_m=1.65m$

Generally the equation for Equilibrium Height $H_E$ is mathematically given as

$H_E=L(1-cos73.14)$

$H_E=0.9(1+0.2923)$

$H_E=1.16307m$

8 0

2 years ago

A simple harmonic oscillator completes 1550 cycles in 30 min. (a) Calculate the period. s (b) Calculate the frequency of the mot

nordsb [41]

Answer:

(a) 1.16 s

(b)0.861 Hz

Explanation:

(a) Period : The period of a simple harmonic motion is the time in seconds, required for a object undergoing oscillation to complete one cycle.

From the question,

If 1550 cycles is completed in (30×60) seconds,

1 cycle is completed in x seconds

x = 30×60/1550

x = 1.16 s

Hence the period is 1.16 seconds.

(b) Frequency : This can be defined as the number of cycles that is completed in one seconds, by an oscillating body. The S.I unit of frequency is Hertz (Hz).

Mathematically, Frequency is given as

F = 1/T ........................... Equation 1

Where F = frequency, T = period.

Given: T = 1.16 s.

Substitute into equation 1

F = 1/1.16

F = 0.862 Hz

Hence thee frequency = 0.862 Hz

6 0

3 years ago