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dexar [7]
3 years ago
9

A 75.0 kg ladder that is 3.00 m in length is placed against a wall at an angle theta. The center of gravity of the ladder is at

a point 1.2 m from the base of the ladder. The coefficient of static friction at the base of the ladder is 0.400. There is no friction between the wall and the ladder. What is the minimum angle the ladder makes with the horizontal for the ladder not to slip and fall
Physics
1 answer:
SOVA2 [1]3 years ago
4 0

Answer:

The minimum angle the ladder makes with the horizontal for the ladder not to slip and fall is 45 degrees.

Explanation:

Given that,

Mass of the ladder, m = 75 kg

Length of the ladder, l = 3 m

The center of gravity of the ladder is at a point 1.2 m from the base of the ladder.

The coefficient of static friction at the base of the ladder is 0.4

We need to find the minimum angle the ladder makes with the horizontal for the ladder not to slip and fall.    

At equilibrium, net force and net torque on the ladder is equal to zero. Force is given by :

f=\mu mg

Using condition for torque as :

mg\times 12\cos\theta-fl\sin (180-\theta)=0\\\\mg\times 1.2\cos\theta-\mu mgl\sin (180-\theta)=0\\\\\tan\theta=\dfrac{1.2}{\mu l}\\\\\tan\theta=\dfrac{1.2}{0.4\times 3}\\\\\tan\theta=1\\\\\theta=45^{\circ}

So, the minimum angle the ladder makes with the horizontal for the ladder not to slip and fall is 45 degrees.

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How does the science of heat transfer differ from the science of thermodynamics?
mariarad [96]

Answer:

The science of thermodynamics deals with the fundamental laws that guide how physical processes occur in relation with the energy transfer. When a system or process changes from one state of equilibrium to another, thermodynamics is interested with the amount of heat transfer during the process. On the other hand, the science of heat transfer is simply about the rate of heat and temperature distribution inside a system at a particular point in time.

Explanation:

3 0
3 years ago
Two long, parallel transmission lines, 40.0cm apart, carry 25.0-A and 73.0-A currents.A). Find all locations where the net magne
In-s [12.5K]

Answer:

a) If the currents are in the same direction, the magnetic field is zero at x = 0.298 m = 29.8 cm

That is, in between the wires, 29.8 cm from the 73.0 A wire and 10.2 cm from the 25.0 A wire.

b) If the currents are in opposite directions, the magnetic field is zero at x = 0.608 m = 60.8 cm

That is, along the positive x-axis, 60.8 cm from the 73.0 A wire and 20.8 cm from the 25.0 A wire.

Explanation:

The origin is at the 73.0 A wire and the 25.0 A wire is at x = 0.40 m

The magnetic field in a current carrying wire at a distance r from the wire is given by

B = (μ₀I/2πr)

μ₀ = magnetic constant = (4π × 10⁻⁷) H/m

a) If the currents are in the same direction, at what positions is the magnetic field equal to 0.

According to laws describing the direction.of magnetic fields, this position will be at some point between the two wires.

The magnetic field due to the 73.0 A wire points out of the book, at points along the positive x-axis while the magnetic field due to the 25.0 A wire points into the plane of the book, moving in the negative x-direction.

Hence,

For the 73.0 A wire, I₁ = 73.0 A, r₁ = x

For the 25.0 A wire, I₂ = 25.0 A, r₂ = (0.4 - x)

B = B₁ - B₂ = 0

(μ₀/2π) [(I₁/r₁) - (I₂/r₂)] = 0

(I₁/r₁) = (I₂/r₂)

(I₁/x) = [I₂/(0.4-x)]

(73/x) = [25/(0.4-x)]

73(0.4-x) = 25x

29.2 - 73x = 25x

73x + 25x = 29.2

98x = 29.2

x = (29.2/98) = 0.298 m

b) If the currents are in the opposite directions, at what positions is the magnetic field equal to 0?

According to laws describing the direction.of magnetic fields, this position will be at some point beyond the second wire (since we're initially concerned about the positive x-direction).

The magnetic field due to the 73.0 A wire points out of the book, at points along the positive x-axis while the magnetic field due to the 25.0 A wire (whose direction is now in the opposite direction to the current in the first wire) is also along the positive x-direction.

Hence,

For the 73.0 A wire, I₁ = 73.0 A, r₁ = x

For the 25.0 A wire, I₂ = 25.0 A, r₂ = (x - 0.4)

B = B₁ - B₂ = 0

(μ₀/2π) [(I₁/r₁) - (I₂/r₂)] = 0

(I₁/r₁) = (I₂/r₂)

(I₁/x) = [I₂/(x-0.4)]

(73/x) = [25/(x-0.4)]

73(x-0.4) = 25x

73x - 29.2 = 25x

73x - 25x = 29.2

48x = 29.2

x = (29.2/48) = 0.608 m

Hope this Helps!!!

5 0
3 years ago
LO NECESITO PARA HOY URGENTE!! 2. Exprese en metros las siguientes longitudes a) 48,9 km b) 36,875 cm c) 756,34 hm d) 9876 mm
Vesnalui [34]

Answer:

a)48900 metros

b)0.36875 metros

c)75634 metros

d)9.876 metros

Explanation:

Hola, para resolver debemos convertir unidades utilizando equivalencias

a) 48.9 km  

1 kilometro = 1000 metros

48.9 x 1000 = 48900 metros

b) 36.875 cm  

1 centímetro =0.01 metros

36.875 x 0.01 = 0.36875 metros

c) 756,34 hm

1 hectómetro= 100 metros

756.34 x 100 = 75634 metros

d) 9876 mm

1 milímetro = 0.001 m

9876 x 0.001 = 9.876 metros

4 0
3 years ago
Small, slowly moving spherical particles experience a drag force given by Stokes' law: Fd = 6πηrv where r is the radius of the p
Dominik [7]

Answer:

Explanation:

At the time of a body achieving terminal velocity, the drag force becomes equal to the weight of the body less the buoyant force by the surrounding medium which can be represented by the following equation

\frac{4\pi\times r^3(d-\rho)}{3} =6\pi\times n\times r\times v

Where r is radius of the body , d is density of the material of the body σ is density of the medium and n is coefficient of viscosity of the medium and v is terminal velocity.

Simplifying

v = \frac{2\times r^2(d-\rho)}{9\times n}

Assuming the value of density of air as 1.225 kg/m³ and putting other given values in the formula we get

v = [tex]\frac{2\times (1.2\times10^{-5})^2(2182-1.225)}{9\times 1.8\times10^{-5}}[/tex]

v = 387 x 10⁻⁵ m/s

Terminal velocity = 387 x 10⁻⁵ m/s

Time taken to fall a distance of 100 m

= \frac{100}{387\times10^{-5}}

= 2.6 x 10⁴ s.

5 0
2 years ago
You need to repair a gate on the farm. The gate weighs 100 kg and pivots as indicated. A small diagonal bar supports the gate an
tekilochka [14]

Answer:

The force is  F = 3920 \ N

Explanation:

The diagram for this question is shown on the first uploaded image

   From the question we are told that

          The weight of the gate is G  =  100\  kg

 

The vertical component of F is  F_y =  F\ sin  \theta

   From the diagram , taking moment about the pivot we have  

                W_g  * 2 - F_y  * 1 =  0

Where W_g  is the weight of the gate evaluated as

             W_g  =  m_g * g =  100 * 9.8 =  980 \ N

=>        980 * 2 - Fsin(30)  * 1 =  0

=>         F = \frac{1960}{sin(30)}

=>      F = 3920 \ N

7 0
3 years ago
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