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A 75 kg man starts climbing a ladder that leans against a wall. If the weight of the ladder is negligible, determine how far up the ladder the man can climb before the ladders starts to slip. The coefficient of friction on both surfaces is μS=0.25

Answers

Answer 1

Answer:

The man can climb $\bold { X (max) = 0.25* L* tan \alpha }$ , before the ladders starts to slip.

A - point at the top of the ladder

B - point at the bottom of the ladder

C - point where the man is positioned in the ladder

L- the length of the ladder

α - angle between ladder and ground

x - distance between C and B

The forces act on the ladder,

Horizontal reaction force (T) of the wall against the ladder

Vertical (upward) reaction force (R) of ground against the ladder.

Frictionalhorizontal ( to the left ) force (F)

Vertical( downwards) of the man,

mg = 75 Kg x 9.8 m/s² = 735 N

in static conditions,

∑Fx = T - F = 0 Since, T = F

∑Fy = mg - R = 0 Since, 735 - R = 0, R = 735

∑ Torques(b) = 0

In point B the torque produced by forces R and F is Zero

Then:

∑Torques(b) = 0

And the arm lever for each force,

mg = 735

Since, ∑Torques(b) = 0

$\bold {735* x* cos\alpha = F* L* sin\alpha }$ Since,T = F

$\bold {F = \frac {735* x* cos\apha }{L* sin\alpha }}$ $\bold { \frac {cos \alpha } { sin\alpha }= cot\alpha =\frac 1{tan\alpha}}$

$\bold {F = \frac {735* x* cos\apha }{L }}$

$\bold {F = 735* x* tan\alpha }}$

F < μR the ladder will starts slipping over the ground

μ(s) = 0.25

$\bold { X (max) = 0.25* L* tan \alpha }$

Therefore, the man can climb $\bold { X (max) = 0.25* L* tan \alpha }$ , before the ladders starts to slip. \

To know more about Torque,

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Answer 2

Answer:

Answer:

x (max) = 0,25*L*tanα

Explanation:

Letá call

A: point at the top of the ladder

B: the point at the bottom of the ladder

C: point where the man is up the ladder

L the length of the ladder

α angle between ladder and ground

"x" distance between C and B

Description

The following forces act on the ladder

Point A: horizontal (to the right) reaction (T) of the wall against the ladder

Point B : Vertical (upwards) reaction (R) of ground against the ladder

frictional horizontal ( to the left ) force (F)

Point C : Weight (vertical downwards)) of the man mg

mg = 75 Kg * 9,8 m/s² mg = 735 [N]

Then in static conditions:

∑Fx = T - F = 0 ⇒ T = F

∑Fy = mg - R = 0 ⇒ 735 - R = 0 ⇒ R = 735

∑Torques(b) = 0

Note: In point B the torque produced by forces R and F are equal to 0

Then:

∑Torques(b) = 0

And the arm lever for each force is:

mg = 735

d₁ for mg and d₂ for T

cos α = d₁/x then d₁ = x*cosα

sin α = d₂ / L then d₂ = L*sinα

Then:

∑Torques(b) = 0 ⇒ 735*x*cosα - T*L*sinα = 0

735*x*cosα = T*L*sinα

T = F then 735*x*cosα = F*L*sinα

F = (735)*x*cosα/L*sinα cos α / sinα = cotgα = 1/tanα

F = (735)*x*cotanα/L or F = (735)*x/L*tanα

When F < μ* R the ladder will stars slippering over the ground

μ(s) = 0,25 0,25*R = 735*x/L*tanα

x = 0,25*R*tanα*L/735

But R = mg = 735 then

0,25*L*tanα = x

Then x (max) = 0,25*L*tanα

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The rms speed of the molecules in 1.3 g of hydrogen gas is 1600 m/s.Part A. What is the total translational kinetic energy of the gas molecules?
Part B. What is the thermal energy of the gas?
Part C. 500J of work are done to compress the gas while, in the same process, 2000J of heat energy are transferred from the gas to the environment. Afterward, what is the rms speed of the molecules?

Answers

a. The total translational kinetic energy of the gas molecules is 1672 Joules.

b. The thermal energy of a gas molecule is equal to 1672 Joules.

c. The rms speed of the gas molecules is equal to 512.83 m/s.

Given the following data:

Mass of hydrogen gas = 1.3 gram.

Speed (rms), c = 1600 m/s.

Work done = 500 Joules.

Quantity of energy = 2000 Joules.

Scientific data:

Mass of proton = $1.67 * 10^(-27)$ kg.

Avogadro constant = $6.02 * 10^(23)$

a. To calculate the total translational kinetic energy of the gas molecules:

How to calculate translational kinetic energy.

First of all, we would determine the number of moles of hydrogen gas contained in 1.3 grams:

Note:Molar mass of hydrogen gas = 2 g/mol.

$Number \;of \;moles = \frac {mass}{molar\;mass}\n\nNumber \;of \;moles = \frac {1.3}{2}$

Number of moles = 0.65 moles.

Next, we would determine the number of molecules in 0.65 moles of hydrogen gas:

By stoichiometry:

1 mole = $6.02 * 10^(23)$ molecules.

0.65 mole = X molecules.

Cross-multiplying, we have:

X = $0.65 * 6.02 * 10^(23)$ = $3.913 * 10^(23)$ molecules.

Mathematically, total translational kinetic energy is given by this formula:

$T = (1)/(2) mc^2$

Substituting the given parameters into the formula, we have;

$T = (1)/(2) * 2 * 1.67 * 10^(-27) * 3.913 * 10^(23) * (1600)^2\n\nT = 6.53 * 10^(-4) * 2560000$

T = 1,671.68 ≈ 1672 Joules.

b. In Science, the total translational kinetic energy is equal to the thermal energy of a gas molecule.

Thermal energy = 1672 Joules.

c. To calculate the rms speed of the gas molecules:

$Net\;energy = 500 + 1672 -2000$

Net energy = 172 Joules.

For the rms speed:

$172 = (1)/(2) * 2 * 1.67 * 10^(-27) * 3.913 * 10^(23) * c^2\n\n172 = 6.54 * 10^(-4) c^2\n\nc = \sqrt{(172)/(6.54 * 10^(-4)) } \n\nc=√(262996.95)$

c = 512.83 m/s.

Read more on rms speed here: brainly.com/question/7427089

Final answer:

The total translational kinetic energy and thermal energy of 1.3g of hydrogen gas with rms speed of 1600 m/s is 5.01x10^25 Joules. After work of 500 Joules is done to compress the gas and 2000 Joules of heat energy are transferred out, the kinetic and thermal energy remains the same, thus the rms speed remains largely the same (with a negligible change due to roundoff errors).

Explanation:

You're asking about the behavior of a hydrogen gas in terms of its kinetic and thermal energy, as well as changes in its root mean square (rms) speed as work is done to compress the gas and heat is transferred out of it.

Part A: The total translational kinetic energy can be calculated using the formula 1/2*m*v^2, where m is the mass and v is the speed. For hydrogen in monoatomic gas, 1.3g of hydrogen is about 0.65 moles. 1 mole's mass is about 1g, so 0.65 moles would be about 0.65g. Convert this to kilograms: 0.65g = 0.00065kg. To find the individual molecule's kinetic energy, you multiply by Avogadro's number (6.02*10^23) as there are that many molecules in a mole. Therefore, the Total translational kinetic energy = 1/2 * 0.00065 kg * (1600 m/s)^2 * 6.02x10^23 = 5.01x10^25 Joules.

Part B: At equilibrium, the thermal energy of a gas is equal to its kinetic energy, so the thermal energy would also be 5.01x10^25 Joules.

Part C: According to the principle of energy conservation, the final kinetic (and thus, thermal) energy of the gas will be its initial energy plus the work done on it minus the heat transferred out of it. Therefore, the final energy = 5.01x10^25 Joules + 500 Joules - 2000 Joules = 5.01x10^25 Joules. To find the new rms speed, you set this equal to the kinetic energy formula and solve for v. Doing so gets you a modulus change in the root mean square speed. Please note that this involves some simplifying assumptions and may not reflect what would happen in a more complex system.

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Tonya picks up a leaf from the ground and holds it at arm’s length. She lets go, and the leaf falls to the ground. What force pulled the leaf to the ground?

Answers

Answer:

Gravity

Explanation:

Gravity is constantly pulling objects downward. Without it, everything would float out into space.

I hope this answer helps :)

Answer:

The answer for the given question above would be option C. GRAVITATIONAL FORCE. Based on the given scenario above of a leaf that falls to the ground when Tonya let it go, the force that pulled the leaf to the ground is the gravitational force. This kind of force is a force that attracts any object with mass.

Hope this helps!!!

A trombone can produce pitches ranging from 85 Hz to 660 Hz approximately. When the trombone is producing a 357 Hz tone, what is the wavelength of that tone in air at standard conditions?

Answers

Answer:

The wavelength of that tone in air at standard condition is 0.96 m.

Explanation:

Given that, a trombone can produce pitches ranging from 85 Hz to 660 Hz approximately. We need to find the wavelength of that tone in air when the trombone is producing a 357 Hz tone.

We know that the speed of sound in air is approximately 343 m/s. Speed of a wave is given by :

$v=f\lambda\n\n\lambda=(v)/(f)\n\n\lambda=(343\ m/s)/(357\ Hz)\n\n\lambda=0.96\ m$

So, the wavelength of that tone in air at standard condition is 0.96 m. Hence, this is the required solution.

A straight trail with a uniform inclination of 16° leads from a lodge at an elevation of 600 feet to a mountain lake at an elevation of 7,000 feet. What is the length of thetrail (to the nearest foot)?
O A. 6,658 ft
OB. 25,396 ft
OC. 7,282 ft
OD. 23,219 ft

Answers

I believed the answer is d

What are the density, specific gravity and mass of the air in a room whose dimensions are 4 m * 6 m * 8 m at 100 kPa and 25 C.

Answers

Answer:

Density = 1.1839 kg/m³

Mass = 227.3088 kg

Specific Gravity = 0.00118746 kg/m³

Explanation:

Room dimensions are 4 m, 6 m & 8 m. Thus, volume = 4 × 6 × 8 = 192 m³

Now, from tables, density of air at 25°C is 1.1839 kg/m³

Now formula for density is;

ρ = mass(m)/volume(v)

Plugging in the relevant values to give;

1.1839 = m/192

m = 227.3088 kg

Formula for specific gravity of air is;

S.G_air = density of air/density of water

From tables, density of water at 25°C is 997 kg/m³

S.G_air = 1.1839/997 = 0.00118746 kg/m³

Please show steps as to how to solve this problem
Thank you!

Answers

Answer:

Torques must balance

F1 * X1 = F2 * Y2

or M1 g X1 = M2 g X2

X2 = M1 / M2 * X1 = 130.4 / 62.3 * 10.7

X2 = 22.4 cm

Torque = F1 * X2 =

62.3 gm* 980 cm/sec^2 * 22.4 cm = 137,000 gm cm^2 / sec^2

Normally x cross y will be out of the page

r X F for F1 will be into the page so the torque must be negative

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