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blsea
1 month ago
15

A crate with a mass of m = 450 kg rests on the horizontal deck of a ship. The coefficient of static friction between the crate a

nd the deck is μs = 0.73. The coefficient of kinetic friction is μk = 0.59
Write an expression for the force Fv that must be applied to keep the block moving at a constant velocity.
What is the magnitude of the force Fv in newtons?
Physics
1 answer:
Sav [3.1K]1 month ago
4 0

Answer:F_{v} =\mu_{k} mg

The force required has a magnitude of 2601.9 N

Explanation:

m = 450 kg

Static friction coefficient μs = 0.73

Kinetic friction coefficient μk = 0.59

The force necessary to initiate movement of the crate is F_{s} =\mu_{s} mg.

Once the crate begins to move, the frictional force decreases to F_{v} =\mu_{k} mg.

To maintain the motion of the crate at a steady velocity, we must lower the pushing force to F_{v} =\mu_{k} mg.

Subsequently, the pushing force aligns with the frictional force stemming from kinetic friction, enabling balanced forces and consistent velocity.

<pMagnitude of the force

F_{v} =\mu_{k} mg\\F_{v} =0.59 \times 450 \times 9.8\\F_{v} =2601.9 N

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Position or composition
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1 month ago
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Three pendulums with strings of the same length and bobs of the same mass are pulled out to angles θ1, θ2, and θ3, respectively,
Sav [3153]

Answer:

All three pendulums will have the same angular frequencies.

Explanation:

For a simple pendulum, the time period using the approximation sin(\theta )\approx \thetais expressed as:

T=2\pi\sqrt{\frac{l}{g}}

The angular frequency \omega is defined as

\omega =\frac{2\pi }{T}\\\\\omega =\frac{2\pi}{2\pi \sqrt{\frac{l}{g}}}\\\\\therefore \omega =\sqrt{\frac{g}{l}}

Since the angular frequency remains unaffected by the initial angle (valid strictly for small angle approximations), we deduce that the angular frequencies of the three pendulums are identical.

6 0
1 month ago
Two very small 8.55-g spheres, 15.0 cm apart from center to center, are charged by adding equal numbers of electrons to each of
ValentinkaMS [3465]

Answer:

1.43 x 10¹⁷.

They will move away from each other.

Explanation:

The force acting on each charged sphere is determined as F = mass x acceleration

= 8.55 x 10⁻³ x 25 x 9.8

= 2.095 N

Assuming Q is the charge on each sphere

F = \frac{9\times10^9\times Q^2}{(15\times10^{-2})^2}

Using the values, 2.095 = \frac{9\times10^9\times Q^2}{(15\times10^{-2})^2}

We find that Q² = \frac{2.095\times(15)^2\times10^{-4}}{9\times10^9}

Thus, Q = 2.289 X 10⁻⁶

The quantity of electrons = Charge / charge of a single electron

= \frac{2.289\times10^{-6}}{1.6\times10^{-19}}

=1.43 x 10¹³.

They will accelerate away from each other.

4 0
1 month ago
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Starting with only the Balmer series light (visible light), how could we ensure that the solar panels generate a current that Ma
ValentinkaMS [3465]

The right answer is (a).

Solar panels create electric current through the photoelectric effect, which describes how photons strike certain material surfaces, resulting in the release of electrons when light with the correct frequency hits them. A photon will interact with an electron on the panel, causing it to be ejected from the panel's surface.

As the illumination on the panel becomes brighter, the intensity of the light rises, indicating an increase in the number of photons. Each photon has the potential to liberate an electron; thus, as the number of incoming photons rises, so does the quantity of freed electrons. Given that the photoelectric current reflects the rate at which these electrons flow, an increase in light intensity leads to a corresponding rise in the photoelectric current.

If the frequency of the light is increased without a change in brightness, the photoelectric current remains the same because the total number of photons does not increase. Yet, the electrons that are ejected do escape with higher kinetic energy. However, since the total number of electrons liberated stays unchanged, the current remains constant regardless of the electrons' increased energy. Thus, option b is incorrect.

Increasing the wavelength of the light means the energy of the photons decreases. This would cause the emitted electrons to have lower energy. However, if the brightness is consistent, the number of electrons remains the same, and as a result, there would be no change in the photoelectric current. Therefore, choice (c) is also incorrect.

The correct answer is (a). To generate the needed current, the brightness of the incident light must be increased.

8 0
1 month ago
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A particle moves according to a law of motion s = f(t), t ≥ 0, where t is measured in seconds and s in feet. f(t) = 0.01t4 − 0.0
serg [3582]
Since you've completed parts a and b, I will tackle part c.
For part C
To respond to this question, we must identify the zeros of the velocity function:
v(t)=0.04t^3-0.06t^2
This polynomial can be factored:
v(t)=0.04t^3-0.06t^2=t^2(0.04t-0.06)
Finding the zeros now becomes straightforward since the function equals zero when any factor is zero.
t^2=0;\\ 0.04t-0.06=0
By solving these equations, we identify our zeros:
t_1=0; t_2=\frac{3}{2}
The particle remains stationary at t=0 and t=3/2.
For part D
We must discover when the velocity function exceeds zero. We will utilize its factored form.
We will assess when each factor is greater than zero and compile the findings in the following table:
\centering \label{my-label} \begin{tabular}{lllll} Range & -\infty & 0 & 3/2 & +\infty \\ t^2 & - & + & + & + \\ 0.04t-0.06 & - & - & + & + \\ t^2 (0.04t-0.06) & + & - & + & + \end{tabular}
From the table, it's evident that our function is positive when - \infty < t and t>3/2.
This indicates the interval during which the particle moves forward.
For part E
The distance traveled can be represented as:
s(t)=0.01t^4 - 0.02t^3
We simply substitute t=12 to calculate the total distance traveled:
s(12)=0.01(12)^4 - 0.02(12)^3=172.80 ft
For part F
Acceleration is defined as the rate at which velocity changes.
We determine acceleration by deriving the velocity function concerning time.
a(t)=\frac{dv}{dt}=(0.04t^3-0.06t^2)'=0.12t^2-0.12t
To find the acceleration at 1 second, we substitute t=1s into the previous equation:
a(1)=0.12-0.12=0


7 0
1 month ago
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