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yarga [219]
3 years ago
10

A desk was moved 12 m using 280 j of work. how much force was used to move the desk?

Physics
1 answer:
Oksana_A [137]3 years ago
4 0
Let F =  required force, N

Given:
d = 12 m, distance
W = 280 J, work done

By definition,
W = F*d,
therefore
(F N)*(12 m) = (280 J)
F = 280/12 = 23.33 N

Answer: The force is 23.3 N (nearest tenth)
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A particle executes simple harmonic motion with an amplitude of 1.69 cm. At what positive displacement from the midpoint of its
m_a_m_a [10]

Answer: 0.0146m

Explanation: The formula that defines the velocity of a simple harmonic motion is given as

v = ω√A² - x²

Where v = linear velocity, A = amplitude = 1.69cm = 0.0169m, x = displacement.

The maximum speed of a simple harmonic motion is derived when x = A, hence v = ωA

One half of maximum speed = speed of motion

3ωA/2 = ω√A² - x²

ω cancels out on both sides of the equation, hence we have that

A/2 = √A² - x²

(0.0169)/2 = √(0.0169² - x²)

0.00845 = √(0.0169² - x²)

By squaring both sides, we have that

0.00845² = 0.0169² - x²

x² = 0.0169² - 0.00845²

x² = 0.0002142

x = √0.0002142

x = 0.0146m

5 0
3 years ago
Help Please!!!!!!!!!!!!!!!!!!
julia-pushkina [17]
The answer is 59.4 degrees.
4 0
2 years ago
If the wagon travels 18.75 m, what is the work done on the wagon
Schach [20]

If the wagon travels 18.75 m, then the work done on the wagon is

(18.75 m) x (the steady force applied to the wagon all the way, in Newtons) .

The unit is Joules .


4 0
3 years ago
If you shout on the moon,will the sound travel faster or slower than on earth?why?
labwork [276]
As long as the sound is inside the helmet of your space suit, it will travel
at the same speed as it would on Earth, through the same mixture of gases
at the same pressure.  Once it passes through the visor of your space helmet,
its 'speed' has no meaning, since there's nothing for sound to travel through on
the moon, and it doesn't travel at all.
7 0
3 years ago
A 5.0 A electric current passes through an aluminum wire of 4.0~\times~10^{-6}~m^2 cross-sectional area. Aluminum has one free e
Serhud [2]

Answer: The electron number density (the number of electrons per unit volume) in the wire is 6.0 \times 10^{28} m^{-3}.

Explanation:

Given: Current = 5.0 A

Area = 4.0 \times 10^{-6} m^{2}

Density = 2.7 g/cm^{3}, Molar mass = 27 g

The electron density is calculated as follows.

n = \frac{density}{mass per atom}\\= \frac{\rho}{\frac{M}{N_{A}}}\\

where,

\rho = density

M = molar mass

N_{A} = Avogadro's number

Substitute the values into above formula as follows.

n = \frac{\rho \times N_{A}}{M}\\= \frac{2.7 g/cm^{3} \times 6.02 \times 10^{23}/mol}{27 g/mol}\\= \frac{16.254 \times 10^{23}}{27} cm^{3}\\= 0.602 \times 10^{23} \times \frac{10^{6} cm^{3}}{1 m^{3}}\\= 6.0 \times 10^{28} m^{-3}

Thus, we can conclude that the electron number density (the number of electrons per unit volume) in the wire is 6.0 \times 10^{28} m^{-3}.

8 0
2 years ago
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