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3 years ago

A rope hangs between two cliffs, edge A and edge B. There is a 250 kg hiker sliding across and is closer to

Physics

1 answer:

iVinArrow [24]3 years ago

4 0

Answer:

1655.5 N

Explanation:

We resolve the forces on the rope into horizontal and vertical components. Since the only vertical force exerted on the rope is the weight of the man, this weight equals the vertical components of the tensions on both sides of the rope .

The only horizontal forces acting on the rope are horizontal components of both tensions on both sides. Since the hiker doesn't move sideways, they cancel out and he is balanced. So, the net horizontal force is zero.

Let T be the tension in the rope. So, each vertical component on side A is Tsin35° and that on side B is Tsin65°. Let the weight of the man be W = mg where m = mass of hiker = 250 kg and g = acceleration due to gravity = 9.8 m/s².

So, since the hiker balances, Tsin35° + Tsin65° = mg

0.5736T + 0.9063T = 250 kg × 9.8 m/s²

1.4799T = 2450 N

dividing both sides by 1.4799, we have

T = 2450 N/1.4799

T = 1655.52 N

T ≅ 1655.5 N

So, the tension in the rope equals T = 1655.5 N

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3 years ago

If you need 40.0 Nm of torque in order to loosen a nut on a wn

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3 years ago

g determine what frequency is required of a source powering a 100 uf capacitor a 500 ohm resistor and a s50 mH inductor in serie

polet [3.4K]

Answer: $71.16\ Hz$

Explanation:

Given

Capacitance $C=100\ \mu F$

Resistance $R=500\ \Omega$

Inductance $L=50\ mH$

In LCR circuit, current is maximum at resonance frequency i.e.

$X_L=X_C\ \text{and}\ \omega_o=\dfrac{1}{\sqrt{LC}}$

Insert the values

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Also, frequency is given by

$\Rightarrow 2\pi f=\omega_o\\\\\Rightarrow f=\frac{\omega_o}{2\pi}$

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3 years ago

Silver has a work function of 4.5 eV . Part A What is the longest wavelength of light that will release an electron from a silve

uranmaximum [27]

Answer:

λ = 2.7608 x 10⁻⁷ m = 276.08 nm

Explanation:

The work function of a metallic surface is the minimum amount of photon energy required to release the photo-electrons from the surface of metal. The work function is given by the following formula:

Work Function = hc/λ

where,

Work Function = (4.5 eV)(1.6 x 10⁻¹⁹ J/1 eV) = 7.2 x 10⁻¹⁹ J

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c = speed of light = 3 x 10⁸ m/s

λ = longest wavelength capable of releasing electron.

Therefore,

7.2 x 10⁻¹⁹ J = (6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/λ

λ = (6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(7.2 x 10⁻¹⁹ J)

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