An 80-N crate is pushed a distance of 5.0 m upward along a smooth incline that makes an angle of 30o with the horizontal. The force pushing the crate is parallel to the slope. If the speed of the crate increases at a rate of 1.5 m/s2, find the work done by the force. Group of answer choices 260 J 200 J 1 J 200 kJ 140 J

Answers

Answer 1

The work done by the force is 140 J.

Calculation:

It is given that, an 80-N crate is pushed a distance of 5.0 m upward along a smooth incline that makes an angle of [tex]30^\circ[/tex] with the horizontal. The force pushing the crate is parallel to the slope and the acceleration is 1.5 [tex]\text{ m/s}^2[/tex].

It is required to find the work done by the force.

The free-body diagram is shown below,

Here, W=80 N, a=-1.5 [tex]\text{ m/s}^2[/tex]

We can write the net force equation as,

[tex]\begin{gathered}F-W \sin 30^{\circ}=m a \\F-m g \sin 30^{\circ}=\left(\frac{W}{g}\right) a\end{gathered}[/tex]

Here W is the weight of the object, m is the mass of the body, and F is the applied force.

Thus,

[tex]\begin{aligned}F-(80 \mathrm{~N}) \sin 30^{\circ} &=\left(\frac{80 \mathrm{~N}}{9.8 \mathrm{~m} / \mathrm{s}^{2}}\right)\left(-1.5 \mathrm{~m} / \mathrm{s}^{2}\right) \\F &=27.755 \mathrm{~N}\end{aligned}[/tex]

It is known that the work done by the force is calculated as the multiplication of the applied force and the displacement.

Thus the work one is,

[tex]\begin{aligned}W &=(27.755 \mathrm{~N})(5 \mathrm{~m}) \\& \approx 140 \mathrm{~J}\end{aligned}[/tex]

Thus the last option is the correct one.

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An 80-N Crate Is Pushed A Distance Of 5.0 M Upward Along A Smooth Incline That Makes An Angle Of 30o

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Explanation:

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Explanation:

Frequency is the number of occurrences of a repeating event per unit of time.

It is also referred to as temporal frequency, which emphasizes the contrast to spatial frequency and angular frequency.

The period is the duration of time of one cycle in a repeating event, so the period is the reciprocal of the frequency.

Relationship between Period and frequency is as under :

The frequency of a wave describes the number of complete cycles which are completed during a given period of time.

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http://www.sengpielaudio.com/calculator-period.htm

http://www.sengpielaudio.com/calculator-period.htm

https://study.com/academy/lesson/wave-period-definition-formula-quiz.html

An airliner flys 950 miles from S.B. to Denver at 500 mph. How long does it take to fly to
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Answer:

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A hockey puck, with an initial velocity of 65 km/h [W], ricochets off the boards. After 0.76 s in contact with the boards, its final velocity is 47 km/h [E]. Determine the acceleration of the puck.

Answers

Answer:

 a = 40.937 m / s²

Explanation:

For this exercise let's use the relationship between momentum and momentum variation

          I  = Δp

          F t = m v_f - mv₀

          F = m (v_f -v₀) / t

let's reduce the magnitudes to the SI system

          v_f = 47 km / h (1000m / 1 km) (1h / 3600 s) = 13.056 m / s

          v₀ = - 65 km / h = -18.056 m / s

the negative sign is bearing the speed is west

let's calculate

          F = m (13.056 + 18.056) / 0.76

          F = m 40.937

now we can use Newton's second law

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Answers

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Explanation:

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How does energy in the form of light photon travel through the many layers of the Sun?

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Answers

Answer:

The electrostatic force that will result if both charges are doubled and the distance remains the same is 6.0 * 10⁻⁶ N

Explanation:

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In formula; F = KQ₁Q₂/d²

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Hence the electrostatic force is given by  

[tex]\rm F=\frac{Kq_1q_2}{d^2}[/tex]

The given data in the question ,

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F₂ is the  electric force for case 2= ?

Conditions for case 2;

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[tex]\rm F_1=\frac{Kq_1q_2}{d_1^2}[/tex]

For case 2,

[tex]\rm F_2=\frac{K(2q_1)(2q_2)}{d_1^2}[/tex]

[tex]\rm F_2=4\times\frac{Kq_1q_2}{d_1^2}\\\\\rm F_2=4F_1[/tex]

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Answers

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Answers

Answer:

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A ray diagram shows an object placed between 2F and F of a convex lens.

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Answers

Answer:

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Explanation:

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Important Tip:

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Answer:

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Answers

Answer:

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please mark it brainliest

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Answer:

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