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3 years ago

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A 40.0 kg wagon is towed up a hill inclined at 18.5 degrees with respect to the horizontal. The tow rope is parallel to the incl

ine and has a tension of 140 N in it. Assume that the wagon starts from rest at the bottom of the hill, and neglect friction. How fast is the wagon going after moving 80 m up the hill? (THe Answer my teacher gives us is 7.90 m/s, but I have no idea how you get it)

1 answer:

sp2606 [1]3 years ago

5 0

Answer:

7.9m/s

Explanation:

We are given that

Mass of wagon=40 kg

$\theta=18.5^{\circ}$

Tension= $140 N$

Initial velocity of wagon= $u=0$

Displacement=s=80 m

Net force applied on wagon= $F=T-mgsin\theta=140-40(9.8)sin18.5=15.7 N$

By using $g=9.8m/s^2$

$a=\frac{F}{a}=\frac{15.7}{40}=0.39m/s^2$

We know that

$v^2-u^2=2as$

Using the formula

$v^2=2\times 0.39\times 80$

$v=\sqrt{2\times 0.39\times 80}=7.9m/s$

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Estimate the radiation pressure due to a bulb that emits 25 W of EM radiation at a distance of 6.5 cm from the center of the bul

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A small loop of area A = 5.5 mm2 is inside a long solenoid that has n = 919 turns/cm and carries a sinusoidally varying current

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Answer: $410.90\times 10^{-6} V$

Explanation:

Given

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$\phi _{B}=1.818\times 10^{-6}\sin (226t)$

$E=-\frac{\mathrm{d} \phi _B}{\mathrm{d} t}$

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3 years ago

A force F with arrow applied to an object of mass m1 produces an acceleration of 3.10 m/s2. The same force applied to a second o

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Answer:

(a) The value of the ratio m₁/m₂ is 0.581

(b) the acceleration of the combined masses is 1.139 m/s²

Explanation:

Given;

The acceleration of force applied to M₁, a₁ = 3.10 m/s²

The same force applied to M₂ has acceleration, a₂ = 1.80 m/s²

Let this force = F

According Newton's second law of motion;

F = ma

(a) the value of the ratio m₁/m₂

since the applied force is same in both cases, M₁a₁ = M₂a₂

$\frac{m_1}{m_2} = \frac{a_2}{a_1} \\\\\frac{m_1}{m_2} = \frac{1.8}{3.1} \\\\\frac{m_1}{m_2} = 0.581$

(b) the acceleration of m₁ and m₂ combined as one object under the action force F

F = ma

$a = \frac{F}{M} \\\\a = \frac{F}{m_1 + m_2} \\\\a = \frac{F}{0.581m_2 + m_2}\\\\a = \frac{F}{1.581m_2}$

$But, m_2 = \frac{F}{a_2} \\\\a = \frac{F}{1.581m_2} = \frac{F*a_2}{1.581F} \\\\a = \frac{a_2}{1.581} \\\\a = \frac{1.8}{1.581} = 1.139 \ m/s^2$

Therefore, the acceleration of the combined masses is 1.139 m/s²

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3 years ago

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galina1969 [7]

Answer:

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Explanation:

F = 20N, m = 4.0kg, a = ?

Using the formula for Force

F = ma

Then making acceleration the subject of the formula

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a = 5ms^-2

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2 years ago