N4M.6 A board has one end wedged under a rock having a mass of 380 kg and is supported by another rock that touches the bottom side of the board at a point 85 cm from the end under the rock. The board is 4.5 m long, has a mass of about 22 kg, and projects essentially horizontally out over a river. Is it safe for an adult with a mass of 62 kg to stand at the unsupported end of the board

Answers

Answer 1

Answer:

it is safe to stand at the end of the table

Explanation:

For this exercise we use the rotational equilibrium condition

         Στ = 0

         W x₁ - w x₂ - w_table x₃ = 0

         M x₁ - m x₂ - m_table x₃ = 0

where the mass of the large rock is M = 380 kg and its distance to the pivot point x₁ = 850 cm = 0.85m

the mass of the man is 62 kg and the distance

            x₂ = 4.5 - 0.85

            x₂ = 3.65 m

the mass of the table (m_table = 22 kg) is at its geometric center

            x_{cm} = L/2 = 2.25 m

            x₃ = 2.25 -0.85

            x₃ = 1.4 m

let's look for the maximum mass of man

            m_{maximum} = [tex]\frac{ M x_1 -m_{table} x_3}{ x_2}[/tex]

let's calculate

             m_{maximum} = [tex]\frac{ 380 \ 0.85 - 22 \ 1.4}{3.65}[/tex](380 0.85 - 22 1.4) / 3.65

             m_{maximum} = 80 kg

we can see that the maximum mass that the board supports without turning is greater than the mass of man

             m_{maximum}> m

consequently it is safe to stand at the end of the table


Related Questions

What is a transfer of energy called?
A. Displacement
B. Acceleration
C. Work
D. Torque

Answers

Your answer is c, work

Which of the
following
DECREASES
as you go UP a
mountain?
A. climate
B. altitude
C. amount of oxygen
D. buoyancy

Answers

C. Amount of oxygen

The others either change but don’t decrease or they increase.

Answer:

C. Amount of oxygen

Explanation:

Options A and D are invalid as they aren't affecting factors.

Option B is false as the altitude increases as you go up a mountain.

Option C is true as the air pressure (atmospheric pressure) is inversely proportional to the height/altitude of the mountain.

Which one the answer to this question

Answers

The second bubble is the answer:)

A 20 ft ladder leans against a wall. The bottom of the ladder is 3 ft from the wall at time t=0 and slides away from the wall at a rate of 2ft/sec. Find the velocity of the top of the ladder at time t=1.

Answers

Answer: 0.516 ft/s

Explanation:

Given

Length of ladder L=20 ft

The speed at which the ladder moving away is v=2 ft/s

after 1 sec, the ladder is 5 ft away from the wall

So, the other end of the ladder is at

[tex]\Rightarrow y=\sqrt{20^2-5^2}=19.36\ ft[/tex]

Also, at any instant t

[tex]\Rightarrow l^2=x^2+y^2[/tex]

differentiate w.r.t.

[tex]\Rightarrow 0=2xv+2yv_y\\\\\Rightarrow v_y=-\dfrac{x}{y}\times v\\\\\Rightarrow v_y=-\dfrac{5}{19.36}\times 2=0.516\ ft/s[/tex]

A 64.0 cm long cord is vibrating in such a manner that it forms a standing wave with two antinodes. (The cord is fixed at both ends.) Which harmonic does this wave represent

Answers

Answer:

the wave represents the second harmonic.

Explanation:

Given;

length of the cord, L = 64 cm

The first harmonic of a cord fixed at both ends is given as;

[tex]f_o = \frac{V}{2L}[/tex]

The wavelength of a standing wave with two antinodes is calculated as follows;

L = N---> A -----> N    +   N ----> A -----> N

Where;

N is node

A is antinode

L = N---> A -----> N    +   N ----> A -----> N =  λ/2  + λ/2

L = λ

The harmonic is calculated as;

[tex]f = \frac{V}{\lambda} \\\\f = \frac{V}{L} = 2(\frac{V}{2L} ) = 2(f_o) = 2^{nd} \ harmonic[/tex]

Therefore, the wave represents the second harmonic.

L = λ

A 2.0 kg breadbox on a fric-
tionless incline of angle u 40 is
connected, by a cord that runs over a
pulley, to a light spring of spring con-
stant k 120 N/m, as shown in
Fig. 8-43. The box is released from rest when the spring is unstretched. Assume that the pulley is massless and frictionless. (a) What is the speed of the box when it has moved 10 cm down the in- cline? (b) How far down the incline from its point of release does the box slide before momentarily stopping, and what are the (c) magnitude and (d) direction (up or down the incline) of the box’s acceleration at the instant the box momentarily stops?

Answers

Dhjwbxuzb wm I known kdn wi. Wlzkk n

Calculate the terminal velocity of
the following nain drops faning
through air (a) one with a diameter
of 0.3cm 6 one with a a diameter
of o. Olm. Take the density of
water to be looo Kym3 and the
eis cosity of air to be ixlos pas.
The buoyancy effect of the air
may be ignored)​

Answers

I’m pretty sure it’s a

An enormous thunderstorm covers Dallas-Ft. Worth. Your best friend Clark is a storm chaser and heads to the center of the storm to take some readings while you stay dry at home. While Clark is at the center of the storm, he sees and hears lightning strike a tree that is 150 m from where he is standing. You are 127 km from the tree. How long does it take for the sound to reach Clark

Answers

Answer:

t = 0.437 s

Explanation:

The speed of sound is a constant that is worth v = 343 m / s

           v = d / t

            t = d / v

the time it takes for the sound to reach Clark at d = 150 m is

           t = 150/343

           t = 0.437 s

This same sound takes much longer to reach you

          t₂ = 127 10³/343

          t₂ = 370 s

Find the wavelength of light which is capable of ionizing a hydrogen atom?

Answers

Answer:

The correct answer is -  91.4 nm

Explanation:

According to Bohr's model, the minimum wavelength to ionize Hydrogen atom from n= 1 state is expressed as:

(h×c)/λ=13.6eV

here,

h - Planck constant

c - the speed of light

λ - wavelength

Placing the value in the formula for the wavelength

(6.626×10^−34J.s × 3×10^8 m/s)/λ  =  13.6 ×1.6 × 10^−19 J

λ≈91.4nm

Thus, the correct answer would be = 91.4 nm

When a 20 kg explosive detonates and sends a 5 kilogram piece traveling to the right at 105 m/s
what is the speed and direction of the other 15 kilogram piece of the explosive!

Answers

Answer:

speed: 35m/s

direction: left

Explanation:

Assuming the right side is the positive direction:

before explosion:

P = mv = 0

after explosion:

P' = 15P + 5P

(Set the velocity of the 15kg piece after explosion as v1' and the velocity of the 5kg piece after explosion as v2')

P' = 0.75mv1' + 0.25mv2'

P' = (15kg)v' + (5kg)(105m/s)

P' = 525kg/m/s + (15kg)v1'

P = P'

525kg/m/s + (15kg)v1' = 0

(15kg)v1' = -525kg/m/s

v1' = -35m/s

speed = |-35| = 35m/s

direction is to the left since the right side is the positive direction.

I WILL GIVE BRAINLIEST TO CORRECT ANSWER
Two boxes are being pulled to the right by a force F across a frictionless table. The force is acting directly on M 2 which is connected to M 1 by a massless cord. Which of the following expressions best represents the acceleration of M 1?

Answers

sorry i don’t know, i really need the points

You are at a train yard observing trains (because why not). You see a train car (let's call it car 1) moving to the right ( x direction) towards a stationary train car (let's call this one car 2). Car 1 has an initial velocity of 15.0 m/s. A helpful train employee tells you that Car 1 also has a mass of 1,825 kg and Car 2 has a mass of 2,645 kg. Car 1 gently collides with Car 2, allowing them to connect. After the collision the two train cars stay connected. You can assume that there is no friction in the system. If you have never see train cars connect, you can watch the first 25ish seconds of this video to see two train cars couple. However, these cars have friction, so they stop - unlike our problem. What is the Final Velocity of the system consisting of Car 1 and Car 2

Answers

Answer:

6.12 m/s

Explanation:

Using the law of conservation of momentum

momentum before collision = momentum after collision

m₁v₁ + m₂v₂ = (m₁ + m₂)V    (since the train cars become attached to each other) where m₁ = mass of car 1 = 1,825 kg, m₂ = mass of car 2 = 2,645 kg, v₁ = initial velocity of car 1 = + 15.0 m/s (positive since it is moving in the positive x direction), v₂ = initial velocity of car 2 = 0 m/s (since it is initially stationary) and V = velocity of both cars after collision,

So, m₁v₁ + m₂v₂ = (m₁ + m₂)V  

m₁v₁ + m₂(0 m/s) = (m₁ + m₂)V  

m₁v₁ + 0 = (m₁ + m₂)V  

V = m₁v₁/(m₁ + m₂)

substituting the values of the other variables into the equation, we have

V = 1,825 kg × 15.0 m/s/(1,825 kg + 2,645 kg)

V = 27375 kgm/s/ 4470kg

V = 6.124 m/s

V ≅ 6.12 m/s

Your boss asks you to design a drone that begins its flight near the surface and rises to 9600 m. At the surface it will fly through air having a density of 1.23 kg per cubic meter and at its highest altitude the air density will become 0.62 kg per cubic meter. If the flight velocity near sea level is 45 mph, then how fast will in need to go at its highest altitude to maintain the same lift. Assume the coefficient of lift remains constant.

Answers

Answer:

[tex]63.38\ \text{mph}[/tex]

Explanation:

L = Lift force

[tex]\rho[/tex] = Density of air

A = Surface area

v = Velocity

[tex]v_1[/tex] = 45 mph

[tex]\rho_1=1.23\ \text{kg/m}^3[/tex]

[tex]\rho_2=0.62\ \text{kg/m}^3[/tex]

Coefficient of lift is given by

[tex]CL=\dfrac{2L}{\rho v^2A}\\\Rightarrow \rho=\dfrac{2L}{CL v^2A}[/tex]

So

[tex]\rho\propto \dfrac{1}{v^2}[/tex]

[tex]\dfrac{\rho_1}{\rho_2}=\dfrac{v_2^2}{v_1^2}\\\Rightarrow v_2=\sqrt{\dfrac{\rho_1}{\rho_2}}\times v_1\\\Rightarrow v_2=\sqrt{\dfrac{1.23}{0.62}}\times 45\\\Rightarrow v_2=63.38\ \text{mph}[/tex]

The velocity at the required altitude should be [tex]63.38\ \text{mph}[/tex] to maintain the same lift.

A hand dryer blows heated air downwards out of the exit duct at a velocity of 4 m/s. The temperature and density of the ambient air at the inlet are 15 C and 1.23 kg/m3, while at the outlet it has temperature 35 C and density 1.15 kg/m3 The blower power is 10.0 W and the heater power is 715 W. Consider the inlet to be at the large mass of ambient air which has negligible velocity.
a) What is the pressure at the outlet? 4 m/s, 35 C
b) You will be applying the energy equation. Why can you ignore any height differences in this situation?
c) If the specific heat of air C-1000 J/(kg K), where Δυ-C Δ T, find the change in internal energy per unit mass from the inlet to outlet.
d) Find the mass flow rate through the dryer.
e) What is the power loss in the system?
f) What is the loss in the system?
g) What is the head loss in the system?
h) What is the total loss coefficient of the system, referred to the outlet velocity?
i) If there were no heater, would the temperature of gas at the outlet be higher, the same, or lower than the inlet? Explain why.

Answers

Answer:

nzkdjdksishdjsdjjdjnzkskejeoueeuieeoyrie ryrhdhcksodopdncndnszdkhfoeosheiwuef wokxkzdkjdoeehxjbxbdkeiehdhdhddjjddjdhhdhdhhhjdhfjdjjfjddhdheudiwiehdbejwowud

Explanation:

isos

2. Plastic is a great conductor of charge so it moves quicker.

True
False

Answers

Answer:

the answer is false

Explanation:

plastic doesnt conduct anything

Which layer of the atmosphere is the least dense?

Question 5 options:

exosphere


thermosphere


mesosphere


troposphere

Answers

the exosphere is the least dense

Answer:

exosphere

Explanation: Just took the quiz and got an A, 100%

Light containing two different wavelengths passes through a diffraction grating with 1,250 slits/cm. On a screen 17.5 cm from the grating, the third-order maximum of the shorter wavelength falls midway between the central maximum and the first side maximum for the longer wavelength. If the neighboring maxima of the longer wavelength are 8.44 mm apart on the screen, what are the wavelengths in the light

Answers

Answer:

[tex]\lambda_s =6.43*10^-4m[/tex]

Explanation:

From the question we are told that:

Diffraction grating [tex]N=1250slits/cm[/tex]

Distance b/w Screen and grating length [tex]d_{sg}=17.5 cm[/tex]

Distance b/w neighboring maxima and Screen [tex]d_{ms}=8.44[/tex]

 

Generally the equation for grating space is mathematically given by

[tex]d(g)=\frac{1}{N}[/tex]

[tex]d(g)=\frac{100}{1250}[/tex]

[tex]d(g)=0.08[/tex]

Generally the equation for small angle approximation is mathematically given by

[tex]\triangle y=\frac{\lambda d}{L}[/tex]

Therefore for longest wavelength

[tex]\lambda _l=\frac{8.44*10^{-3}*(0.08)}{0.175m}[/tex]

[tex]\lambda _l=3.858*10^{-3}[/tex]

Therefore the third order maximum equation for the shorter wavelength as

[tex]\lambda_s =\frac{1}{6} \lambda_l[/tex]

[tex]\lambda_s =\frac{1}{6} (3.858*10^-^3)[/tex]

[tex]\lambda_s =6.43*10^-4m[/tex]

The wavelengths in the light is given as

[tex]\lambda_s =6.43*10^-4m[/tex]

What is the importance of using locally available resources in creating art?​

Answers

Answer:

please give me brainlist and follow

Explanation:

Using locally available resources for art help in the preservation of environment. A significant and practical aspects of art is material significance. The items used by artists while making an art piece affects both the form and the material. Every material delivers something special in the creative process.

Why is it harder to breathe on a
mountain?
A. The air pressure is so high the lungs can't expand.
B. The air is denser and oxygen can't flow easily into the
lungs.
C. The denser oxygen molecules sink below the
surrounding air.
D. The air is less dense so there are fewer oxygen
molecules.

Answers

I think it’s d but I’m not sure

A copper wire of resistivity 2.6 × 10-8 Ω m, has a cross sectional area of 35 × 10-4 cm2
. Calculate
the length of this wire required to make a 10 Ω coil.

Answers

Answer:

the length of the wire is 134.62 m.

Explanation:

Given;

resistivity of the copper wire, ρ = 2.6 x 10⁻⁸ Ωm

cross-sectional area of the wire, A  = 35 x 10⁻⁴ cm² = ( 35 x 10⁻⁴) x 10⁻⁴ m²

resistance of the wire, R = 10Ω

The length of the wire is calculated as follows;

[tex]R = \frac{\rho L}{A} \\\\L = \frac{RA}{\rho} \\\\L= \frac{10 \times (35\times 10^{-4}) \times 10^{-4}}{2.6 \times 10^{-8}} \\\\L = 134.62 \ m[/tex]

Therefore, the length of the wire is 134.62 m.

A 1.65-m-long wire having a mass of 0.100 kg is fixed at both ends. The tension in the wire is maintained at 16.0 N. (a) What are the frequencies of the first three allowed modes of vibration

Answers

Answer:

Explanation:

mass per unit length ρ = .100 / 1.65 = .0606 . kg /m

length of wire L = 1.65 m

For fundamental frequency , the expression is as follows

n = [tex]\frac{1}{2L} \sqrt{\frac{T}{m} }[/tex]

L = 1.65 , T = 16 n and m = .0606

n = [tex]\frac{1}{2\times 1.65} \sqrt{\frac{16}{.0606} }[/tex]

= 4.9 /s .

This is fundamental frequency .

other mode of vibration ( first three ) will be as follows

4.9 x 2 = 9.8 /s ,

4.9 x 3 = 14.7 /s .

A scientist notices that an oil slick floating on water when viewed from above has many different colors reflecting off the surface, making it look rainbow-like (an effect known as iridescence). She aims a spectrometer at a particular spot and measures the wavelength to be 750 nm (in air). The index of refraction of water is 1.33.
Part A: The index of refraction of the oil is 1.20. What is the minimum thickness of the oil slick at that spot? t= 313nm
Part B: Suppose the oil had an index of refraction of 1.50. What would the minimum thickness be now? t=125nm

Answers

Answer:

a) The minimum thickness of the oil slick at the spot is 313 nm

b) the minimum thickness be now will be 125 nm

Explanation:  

Given the data in the question;

a) The index of refraction of the oil is 1.20. What is the minimum thickness of the oil slick at that spot?

t[tex]_{min[/tex] = λ/2n

given that; wavelength λ = 750 nm and  index of refraction of the oil n = 1.20

we substitute

t[tex]_{min[/tex] = 750 / 2(1.20)

t[tex]_{min[/tex] = 750 / 2.4

t[tex]_{min[/tex] = 312.5 ≈ 313 nm

Therefore, The minimum thickness of the oil slick at the spot is 313 nm

b)

Suppose the oil had an index of refraction of 1.50. What would the minimum thickness be now?

minimum thickness of the oil slick at the spot will be;

t[tex]_{min[/tex] = λ/4n

given that; wavelength λ = 750 nm and  index of refraction of the oil n = 1.50

we substitute

t[tex]_{min[/tex] = 750 / 4(1.50)

t[tex]_{min[/tex] = 750 / 6

t[tex]_{min[/tex] = 125 nm

Therefore, the minimum thickness be now will be 125 nm

Please answer this for 15 points please don’t put in a link.

Answers

It’s b I took the test

Answer:

c. Double Replacement

Explanation:

As in Double Replacement reaction exchanges the cations (or the anions) of two ionic compounds.

Here, in BaCl2 , Ba has replaced with NO3 to form Ba(NO3)2

and in 2AgNo3 , Ag has replaced with Cl to form 2AgCl.

How many gallons of water does it take to produce the following:
a. Cheeseburger
b. Pound of butter
c. A pair of jeans

Answers

Answer:

a. 660 gallons

b.665 gallons

c. 1,800

A bullet of mass 4.00 g is fired horizontally into a wooden block of mass 1.30 kg resting on a horizontal surface. The coefficient of kinetic friction between block and surface is 0.170. The bullet remains embedded in the block, which is observed to slide a distance 0.240 m along the surface before stopping. Part A What was the initial speed of the bullet

Answers

Answer:

[tex]291.67\ \text{m/s}[/tex]

Explanation:

[tex]m_1[/tex] = Mass of bullet = 4 g

[tex]m_2[/tex] = Mass of block = 1.3 kg

[tex]\mu[/tex] = Coefficient of friction = 0.17

[tex]s[/tex] = Displacement of block = 0.24 m

[tex]v_1[/tex] = Velocity of bullet

[tex]v[/tex] = Velocity of combined mass

[tex]g[/tex] = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]

The energy balance of the system is given by

[tex]\dfrac{1}{2}(m_1+m_2)v^2=\mu(m_1+m_2)gs\\\Rightarrow v=\sqrt{2\mu gs}[/tex]

As the momentum is conserved in the system we have

[tex]m_1v_1=(m_1+m_2)v\\\Rightarrow m_1v_1=(m_1+m_2)\sqrt{2\mu gs}\\\Rightarrow v_1=\dfrac{(m_1+m_2)\sqrt{2\mu gs}}{m_1}\\\Rightarrow v_1=\dfrac{(4\times 10^{-3}+1.3)\times \sqrt{2\times 0.17\times 9.81\times 0.24}}{4\times 10^{-3}}\\\Rightarrow v_1=291.67\ \text{m/s}[/tex]

The initial speed of the bullet is [tex]291.67\ \text{m/s}[/tex].

Scientists have concluded that the uppermost part of the mantle is partially-molten. Which observation helped them reach this conclusion?

Answers

Answer:

 P and S waves slow down when they reach this layer. The asthenosphere, also known as the magma chamber, is the uppermost component of the mantle. This layer is partially molten and is a ductile zone in a tectonically poor state.

It's almost hard and seismic waves move through the asthenosphere at a slow rate. The fragile lithosphere and the uppermost portion of the asthenosphere are assumed to be rigid.

seismic waves travel more quickly through denser materials and therefore generally travel more quickly with the depth it moves more slowly through a liquid than a solid. Molten areas within the Earth slow down P waves and stop S waves because their shearing motion cannot be transmitted through a liquid. Partially molten areas may slow down the P waves and attenuate or weaken S waves.

hope this helps...

S and P wave slow down and stop in  the uppermost part of the mantle. - For this, scientists have concluded that the uppermost part of the mantle is partially-molten.

What is mantle?

A planetary body's mantle is a layer that is surrounded by the crust on top and the core underneath. The largest and most substantial layer of a planetary body, mantles are often comprised of rock or ice. Planetary bodies that have undergone density differentiation typically have mantles. Mantles are found on all terrestrial planets (including Earth), many asteroids, and a few planetary moons.

Between the crust and the outer core, there is a silicate rock layer known as the Earth's mantle. Despite being mostly solid, it behaves like a viscous fluid over geological time. Oceanic crust is created by the partial melting of the mantle at mid-ocean ridges, and continental crust is created by the partial melting of the mantle at subduction zones.

Learn more about mantle here:

https://brainly.com/question/28827790

#SPJ2

A solenoid that is 93.9 cm long has a cross-sectional area of 17.3 cm2. There are 1270 turns of wire carrying a current of 7.80 A. (a) Calculate the energy density of the magnetic field inside the solenoid. (b) Find the total energy in joules stored in the magnetic field there (neglect end effects).

Answers

Answer:

[tex]65.6\ \text{J/m}^3[/tex]

[tex]0.11\ \text{J}[/tex]

Explanation:

B = Magnetic field = [tex]\mu_0 \dfrac{N}{l}I[/tex]

[tex]\mu_0[/tex] = Vacuum permeability = [tex]4\pi10^{-7}\ \text{H/m}[/tex]

N = Number of turns = 1270

[tex]l[/tex] = Length of solenoid = 93.9 cm = 0.939 m

[tex]I[/tex] = Current = 7.8 A

A = Area of solenoid = [tex]17.3\ \text{cm}^2[/tex]

Energy density of a solenoid is given by

[tex]u_m=\dfrac{B^2}{2\mu_0}\\\Rightarrow u_m=\dfrac{(\mu_0 \dfrac{N}{l}I)^2}{2\mu_0}\\\Rightarrow u_m=\dfrac{\mu_0N^2I^2}{2l^2}\\\Rightarrow u_m=\dfrac{4\pi\times 10^{-7}\times 1230^2\times 7.8^2}{2\times 0.939^2}\\\Rightarrow u_m=65.6\ \text{J/m}^3[/tex]

The energy density of the magnetic field inside the solenoid is [tex]65.6\ \text{J/m}^3[/tex]

Energy is given by

[tex]U_m=u_mAl\\\Rightarrow U_m=65.6\times 17.3\times 10^{-4}\times 0.939\\\Rightarrow U_m=0.11\ \text{J}[/tex]

The total energy in joules stored in the magnetic field is [tex]0.11\ \text{J}[/tex].

The low pressure area near Earth's equator is filled by cool air moving in from
А
Europe and South America
B
the North and South Pole
с
the Prime Meridian
D
the Atlantic and Pacific Ocean

Answers

The answer is A! Europe and South America

Light rays enter a transparent material. Which description best describes what happens to the light rays

Answers

When light encounters transparent materials, almost all of it passes directly through them. Glass, for example, is transparent to all visible light. ... Most of the light is either reflected by the object or absorbed and converted to thermal energy. Materials such as wood, stone, and metals are opaque to visible light.

Which device converts electric energy into mechanical energy?
O A. An electromagnet
O B. A motor
O C. A transformer
O D. A generator

Answers

Answer:

B motor

Explanation:

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