A 15 m uniform ladder weighing 500 N rests against a frictionless wall. The ladder makes a 60° angle with horizontal. (a) Find the horizontal and vertical forces that the earth exerts on the base of the ladder when an 800 N firefighter is 4 m from the bottom. (b) If the ladder is just on the verge of slipping when the firefighter is 9 m up, what is the coefficient of static friction between ladder and ground?
Answers
Answer:
a) F₁ = 267.3 N, N₁ = 1300 N, b) μ = 0.324
Explanation:
For this exercise we use the rotational equilibrium condition, we have a reference system is the floor and the anticlockwise rotations as positive, in the adjoint we can see a diagram of the forces
let's use subscript 1 for the ladder and 2 for the firefighter
∑ τ = 0
-W₁ x₁ - W₂ x₂ + N₁ y = 0
N₁ = [tex]\frac{W_1 x_1 + W_2 x_2}{y}[/tex] (1)
the center of mass of the ladder is at its geometric center,
d = L / 2 = 15/2 = 7.5 m
cos 60 = x₁ / d₁
x₁ = d₁ cos 60
x₁ = 7.5 cos 60
x₁ = 3.75 m
for the firefighter d₂ = 4 m
cos 60 = x₂ / d₂
x₂ = d₂ cos 60
x₂ = 4 cos 60 = 2 m
for the fulcrum d₃ = 15 m
sin 60 = y / d₃
y = d₃ sin 60
y = 15 sin 60
y = 13 m
we look for the Normal by substituting in equation 1
N₂ = [tex]\frac{500 \ 3.75 \ + 800 \ 2}{13}[/tex]
N₂ = 267.3 N
now let's use the translational equilibrium relations
X axis
F₁ - N₂ = 0
F₁ = N₂
F₁ = 267.3 N
Axis y
N₁ - W₁ -W₂ = 0
N₁ = W₁ + W₂
N₁ = 500 + 800
N₁ = 1300 N
b) for this case change the firefighter's distance d₂ = 9 m
x₂ = 9 cos 60
x₂ = 4.5 m
we substitute in 1
N₂ = \frac{500 \ 3.75 \ + 800 \ 4.5}{13}
N₂ = 421.15 N
of the translational equilibrium equation on the x-axis
fr = F₁ = N₂
fr = 421.15 N
friction force has the expression
fr = μ N
in this case the reaction of the Earth to the support of the ladder is N1 = 1300N
μ = fr / N₁
μ = 421.15 / 1300
μ = 0.324
Related Questions
In July 2015, Oregon State University, the National Oceanic and Atmospheric Administration, and the Coast Guard cooperated to send a hydrophone into Challenger Deep, the deepest part of the Mariana Trench. The titanium shelled recording device withstood the pressure 10,994 meters (nearly 7 miles!) under the ocean's surface. The hydrophone recorded 23 days of audio from the deepest part of the ocean floor. If the spherical hydrophone has a radius of 10 cm, what is the total force exerted on the titanium shell by the ocean water
Answers
Answer:
Explanation:
Pressure due to water column as deep as 10994 meters can be given by the following expression
Pressure = h d g , where h is depth of water , d is density of water and g is acceleration due to gravity .
Pressure = 10994 x 10³ x 9.8
= 10.77 x 10⁷ N / m²
Pressure will act on curved surface of the spherical shell , the effective surface area will be π R² where R is radius of the surface .
Effective surface = 3.14 x 0.1²
= .0314 m²
Total force = pressure due to water column x effective surface
= 10.77 x 10⁷ x .0314 N.
= 33.82 x 10⁵ N .
In case A below, a 1 kg solid sphere is released from rest at point S. It rolls without slipping down the ramp shown, and is launched vertically into the air at point T. In case B, a 1 kg block slides without friction down an identically shaped ramp and is also launched vertically at point T. Select the statement that best describes which object will go higher after launch, and why
Answers
Answer:
the block reaches higher than the sphere
\frac{y_{sphere}} {y_block} = 5/7
Explanation:
We are going to solve this interesting problem
A) in this case a sphere rolls on the ramp, let's find the speed of the center of mass at the exit of the ramp
Let's use the concept of conservation of energy
starting point. At the top of the ramp
Em₀ = U = m g y₁
final point. At the exit of the ramp
Em_f = K + U = ½ m v² + ½ I w² + m g y₂
notice that we include the translational and rotational energy, we assume that the height of the exit ramp is y₂
energy is conserved
Em₀ = Em_f
m g y₁ = ½ m v² + ½ I w² + m g y₂
angular and linear velocity are related
v = w r
the moment of inertia of a sphere is
I = [tex]\frac{2}{5}[/tex] m r²
we substitute
m g (y₁ - y₂) = ½ m v² + ½ ([tex]\frac{2}{5}[/tex] m r²) ([tex]\frac{v}{r}[/tex])²
m g h = ½ m v² (1 + [tex]\frac{2}{5}[/tex])
where h is the difference in height between the two sides of the ramp
h = y₂ -y₁
mg h = [tex]\frac{7}{5}[/tex] ([tex]\frac{1}{2}[/tex] m v²)
v = √5/7 √2gh
This is the exit velocity of the vertical movement of the sphere
v_sphere = 0.845 √2gh
B) is the same case, but for a box without friction
starting point
Em₀ = U = mg y₁
final point
Em_f = K + U = ½ m v² + m g y₂
Em₀ = Em_f
mg y₁ = ½ m v² + m g y₂
m g (y₁ -y₂) = ½ m v²
v = √2gh
this is the speed of the box
v_box = √2gh
to know which body reaches higher in the air we can use the kinematic relations
v² = v₀² - 2 g y
at the highest point v = 0
y = vo₀²/ 2g
for the sphere
y_sphere = 5/7 2gh / 2g
y_esfera = 5/7 h
for the block
y_block = 2gh / 2g
y_block = h
therefore the block reaches higher than the sphere
[tex]\frac{y_{sphere}} {y_bolck} = 5/7[/tex]
Could I get help on this question please
Answers
Answer:
124.51 m
Explanation:
From the question given above, the following data were obtained:
Initial velocity (u) = 49.4 m/s
Final velocity (v) = 0 m/s (at maximum height)
Maximum height (h) =?
NOTE: Acceleration due to gravity (g) = 9.8 m/s²
The maximum height to which the cannon ball attained before falling back can be obtained as illustrated below:
v² = u² – 2gh ( since the ball is going against gravity)
0² = 49.4² – (2 × 9.8 × h)
0 = 2440.36 – 19.6h
Collect like terms
0 – 2440.36 = –19.6h
–2440.36 = –19.6h
Divide both side by –19.6
h = –2440.36 / –19.6
h = 124.51 m
Therefore, maximum height to which the cannon ball attained before falling back is 124.51 m
If the solid, purple line represents a prey population
(e.g., rabbits) and the dashed, red line represents a predator population (e.g., coyotes), which of the following statements is a reasonable conclusion based on the graph?
Answers
Answer:
There is no correlation that can be made between the predator and prey populations.
Explanation:
Ermrmrm.
An electric charge at rest produce
Answers
Answer:
Charge at rest only produces electric field. Moving charge produces both electric field and magnetic field.
plz follow me
Based on your graph, predict the length of pendulum that will give you a
period of 2.00s.
Answers
Answer:
About 1 meter
Explanation:
Base on your image, when the period is 2 (horizontal variable), the length of the pendulum is about 1 meter.
A warm hockey puck has a coefficient of restitution of 0.50, while a frozen hockey puck has a coefficient of restitution of only 0.35. In the NHL, the pucks to be used in games are kept frozen. During a game, the referee retrieves a puck from the cooler to restart play but is told by the equipment manager that several warm pucks were just put into the cooler. To check to make sure he has a game-ready puck, the referee drops the puck on its side from a height of 2 m. How high should the puck bounce if it is a frozen puck
Answers
Answer:
the required height is 0.2449 m only
Explanation:
Given the data in the question;
Initial height = 2m
so speed of the puck before hitting the ground will be;
u² = 2gh
Initial speed u_ball = √2gh
u_ball = √( 2 × 9.8 × 2 )
u_ball = √39.2
u_ball = 6.26 m/s
given that; FOR THE FROZEN PUCK, coefficient of restitution = 0.35 only
R = - (v_ball - v_ground / u_ball - u_ ground)
so
0.35 = - (v_ball - 0 / 6.26 - 0)
0.35 = -v_ball / - 6.26
-v_ball = 0.35 × (- 6.26)
-v_ball = -2.191 m/s
v_ball = 2.191 m/s
to get the height;
v² = 2gh
h = v² / 2g
we substitute
h = (2.191)² / 2×9.8
h = 4.800481 / 19.6
h = 0.2449 m
Therefore, the required height is 0.2449 m only
If a skaters mass increases how does that effect kinetic energy
Answers
Answer:
By paying close attention to the formula for average kinetic energy, we can see that by increasing the mass by a proportional amount will lead to an increase in the total average kinetic energy. There is a direct relationship being observed between the values.
a point charge q1 = 2.40 uC is held stationary at the origin. A second point charge q2 = -4.30uC moves from the point x= .150 m, y= 0.0 m, to the point x = .250 m, y= 0.0m
a) what is the charge in potential energy of the pair of charges?
b) How much work is done by the electric force on q2
Answers
Answer:150M
Explanation:
If a 500-pound object is moved 200 feet how much work is being done?
a. 200 FT LB
b. 500 FT LB
c. 1000 FT LB
d. 100,000 FT LB
Answers
Answer:
D
Explanation:
Work = Distance x Mass
work done = 100,000 FT LB
What is work done ?
Work is done whenever a force moves something over a distance or The work done by a force is defined to be the product of component of the force in the direction of the displacement and the magnitude of this displacement.
Work done = force * displacement
given :
force = 500 pound
displacement = 200 feet
work done = 500 * 200 = 100,000 FT LB
correct option is d. 100,000 FT LB
learn more about work done
https://brainly.com/question/13662169?referrer=searchResults
#SPJ2
True or False:
Some stars appear dimmer than others. Dim stars are always further
away from us than bright stars.
Answers
Answer:
A star's brightness also depends on its proximity to us. The more distant an object is, the dimmer it appears.
Steam at 6 MPa, 400C is flowing in a pipe. Connected to this pipe through a valve is a tank of volume 0.4 m 3 . This tank initially contains saturated water vapor at 0.1MPa. The valve is opened and the tank fills with steam until the pressure is 6MPa, and then the valve is closed. The process takes place adiabatically. Determine the temperature in the tank right as the valve is closed.
Answers
Answer:
2400°C
Explanation:
Volume of tank = 0.4 m^3
steam pressure = 6 Mpa
Steam temperature = 40°C
Initial pressure of tank = 0.1 MPa
Final pressure of Tank = 6 Mpa
Calculate the temperature in the tank when the Pressure in Tank = 6Mpa
since the volume of the Tank is constant = 0.4 m^3
we will apply Gay-Lussac's Law
= [tex]\frac{T1}{P1} = \frac{T2}{P2}[/tex] ------ ( 1 )
T1 = 40°c
P1 = 0.1 MPa
P2 = 6 Mpa
T2 = ?
From equation 1 above
T2 = ( T1 * P2 ) / P1
= ( 40 * 6 ) / 0.1
= 2400°C
WILL GIVE BRAINLIEST!!
What is a way to transfer charge in which an object becomes polarized?
Answers
Answer:
I answered Number 4 (Solids and Elasticity)
Explanation:
solids and elasticity
The standard test to determine the maximum lateral acceleration of a car is to drive it around a 200-ft diameter circle painted on a level asphalt surface. The driver slowly increases the vehicle speed until he is no longer able to keep both wheel pairs straddling the line. If the maximum speed is 35 mi/hr for a 3000-lb car, compute the magnitude F of the total friction force exerted by the pavement on the car tires.
Answers
Answer:
the magnitude F of the total friction force is 2456.7 lb
Explanation:
Given the data in the question;
maximum speed = 35 mi/hr = ( 35×5280 / 60×60) = 51.3333 ft/s
diameter = 200ft
radius = 200/2 = 100 ft
First we calculate the normal component of the acceleration;
[tex]a_{n}[/tex] = v² / p
where v is the velocity of the car( 51.3333 ft/s)
p is the radius of the curvature( 100 ft)
so we substitute
[tex]a_{n}[/tex] = (51.3333 ft/s)² / 100ft
[tex]a_{n}[/tex] = (2635.1076 ft²/s²) / 100ft
[tex]a_{n}[/tex] = 26.35 ft/s²
we convert Feet Per Second Squared (ft/s²) to Standard Gravity (g)
1 ft/s² = 0.0310809502 g
[tex]a_{n}[/tex] = 26.35 ft/s² × 0.0310809502 g
[tex]a_{n}[/tex] = 0.8189g
Now consider the dynamic equilibrium of forces in the Normal Direction;
∑[tex]F_{n}[/tex] = m[tex]a_{n}[/tex]
F = m[tex]a_{n}[/tex]
we know that mass of the car is 3000-lb = 3000lb([tex]\frac{1}{g}slug[/tex]/1 lb)
so
we substitute
F = 3000lb([tex]\frac{1}{g}slug[/tex]/1 lb) × 0.8189g
F = 2456.7 lb
Therefore; the magnitude F of the total friction force is 2456.7 lb
6.
ribbon
AA
SON
120 N
Two teams of students are competing in a tug-o-war contest, as shown in the
picture above. How does the ribbon move?
Answers
Answer:
The ribbon will move to the right.
Explanation:
To know the the correct answer to the question, we shall determine the net force and direction. This can be obtained as follow:
Force to the right (Fᵣ) = 120 N
Force to the left (Fₗ) = 80 N
Net force (Fₙ) =?
Fₙ = Fᵣ – Fₗ
Fₙ = 120 – 80
Fₙ = 40 N to the right.
From the calculation made above, the net force is 40 N to the right. Thus, the ribbon will move to the right.
You are working as a letter sorter in a U.S. Post Office. Postal regulations require that employees' footwear must have a minimum coefficient of static friction of 0.5 on a specified tile surface. You are wearing athletic shoes for which you do not know the coefficient of static friction. In order to determine the coefficient, you imagine that there is an emergency and start running across the room. You have a coworker time you, and find that you can begin at rest and move 4.38 m in 1.21 s. If you try to move faster than this, your feet slip. Assuming your acceleration is constant, does your footwear qualify for the postal regulation?
Answers
Answer:
μ = 0.66, therefore if it compiesy with the regulations
Explanation:
Let's solve this exercise in part, let's start by finding with kinematics the acceleration of man
y = v₀ t + ½ a t²
as it starts from rest the initial velocity is zero
y = ½ a t²
a = [tex]\frac{2y}{t^2}[/tex]
a =\frac{2 \ 4.38}{1.21^2}
a = 6.46 m / s²
Now let's use Newton's second law,
Axis y
N- W = 0
N = W
N = m g
X axis
on this axis the man exerts a backward force and by the law of action and reaction the floor exerts a forward force of the same magnitude, this forward force is the friction force.
fr = m a
the friction force has an expression
fr = my N
let's substitute
μ mg = m a
μ = a / g
let's calculate
μ = 6.46 / 9.8
μ = 0.66
therefore if you comply with the regulations
6th grade science I mark as brainliest !
Answers
Answer:
first is Atoms
4) is True
A swimmer pushing off from the wall of a pool exerts a force of 1 newton on the wall. What is the reaction force of the wall on the swimmer?
Answers
Answer: 1 Newton
Explanation:
"Every action has an equal and opposite reaction."
Please mark as Brainliest if it is correct.
Force is an action-reaction principle. It stated that the force always exists in a pair. The reaction force of the wall on the swimmer will be 1N.
What is Newton's third law of motion?Newton's third law of motion state that every action has an equal and opposite reaction. It is an action-reaction principle. It stated that the force always exists in a pair.
Both occur in an action-reaction form. Hence for every action, there is an equal and opposite reaction.
[tex]\rm F_{action}=F_{reaction} \\\\ \rm F_{action= 1N[/tex]
[tex]\rm F_{reaction} = 1N[/tex]
Hence the reaction force of the wall on the swimmer will be 1N.
To learn more about Newton's third law refer to the link;
https://brainly.com/question/1077877
A place kicker must kick the football from a point [06] m from the goal and clear a bar 3.00 m above the ground. The ball leaves the ground with a speed of 20.0 m/s at an angle of 53.0 degrees above the horizontal. (a) By how much (m) does the ball clear (positive value) or fall short (negative value) of the cross bar? (This is vertical distance above or below the cross bar.) (b) When it gets to the cross bar, what is the vertical component of the ball’s velocity (m/s)? (Is it rising or falling-pay attention to the sign?)
Answers
Answer:
Explanation:
Horizontal component of initial velocity of throw = 20 cos 53 = 12 m /s
Vertical component = 20 sin 53 = 15.97 m /s
Distance to be travelled horizontally = 6 m .
time taken by ball to travel this distance = 6 / 12 = 0.5 s
vertical displacement during this period can be calculated as follows .
Initial vertical velocity = 15.97 m /s
time of travel = .5 s
acceleration = - 9.8 m /s²
s = ut - 1/2 g t²
= 15.97 x .5 - .5 x 9.8 x 0.5²
= 7.985 - 1.225
= 6.76 m
Goal post is 6 m high , so ball will cross the goal post .76 m or 76 cm above cross bar .
b ) vertical component of ball when it crosses the goal post . Let it be v .
v = u - gt
Applying this formula for vertical movement ,
v = 15.97 - 9.8 x .5
= 15.97 - 4.9
= 11.07 m /s .
Calculate the electric field associated to an electric dipole for two charges separated 10-8 m with a dipole moment of 10-33 C m. Do not use unit of measure, just a whole number. Give the result in standard notation, not in scientific notation. Use for the Coulomb constant the value k
Answers
Answer:
18 N/C
Explanation:
Given that:
Electric field constant, k = 9*10^9 N/c
Distance, r = 10^-8 m
Dipole moment, p = 10^-33
Using the relation for electric field due to dipole :
E = [2KP / r³]
E = (2 * (9*10^9) * 10^-33) ÷ (10^-8)^3
E = (18 * 10^9 * 10^-33) ÷ 10^-24
E = [18 * 10^(9-33)] ÷ 10^-24
E = (18 * 10^-24) / 10^-24
E = 18 * 10^-24+24
E = 18 * 10^0
E = 18 N/C
Help please ! Ill give brainliest !! ☁️✨
Answers
Force Meter > used to measure force
Milli > Prefix that means 1/1,000
Centi > Prefix that means 1/100
Kilo > Prefixes that means 1,000
Thermometer > used to measure Temperature
Explanation: uh just trust
Your favorite golfer taps the golf ball with just enough force that it rolls into the ninth hole is an example of what law of motion???
Answers
Second law definition: The acceleration of an object is dependent upon the force acting on the object & the mass of the object.
And the golf ball does so when it’s hit by enough force.
Answer:
mass and on the net force acting on it. ... Tap again to see term ... Newton's second law of motion states that an object's acceleration depends on its ... You hit a ping-pong ball & a tennis ball with a tennis rack
True Or False weather conditions in the atmosphere can be recognized through direct observation.
Answers
Answer:
yes, we can tell if there will be rain by the swelling in rain clouds, we can also see if a tornado is forming based on the look of the wall cloud. we can use satellites to predict the amount of precipitation or wind.
Explanation:
thank you for ur generosity
Given a volume of 1000. Cm^3 of an ideal gas at 300 k, what volume would iy occupy at a temperature of 600 k
Answers
Answer:2000 cm³
Here, pressure remains constant.
So, b the gas law
V/V' = T/ T'
1000 / V' = 300 / 600
V' = 2000 cm³
Explanation:also pls mark brainliest
The head of a rattlesnake can accelerate at 50 m/s2 in striking a victim. If a car could do as well, how long would it take to reach a speed of 100 km/h from rest
Answers
Answer:
the time for the car to reach the final velocity is 0.56 s.
Explanation:
Given;
acceleration of the car, a = 50 m/s²
final velocity of the car, v = 100 km/h = 27.778 m/s
the initial velocity of the car, u = 0
The time for the car to reach the final velocity is calculated as;
v = u + at
27.778 = 0 + 50t
27.778 = 50t
t = 27.778 / 50
t = 0.56 s
Therefore, the time for the car to reach the final velocity is 0.56 s.
20 points!!!! A 2,00ON steel rod that is 5 meters long is placed in a corner between the floor and a wall, and balanced at an angle using a cord attached to the wall The rod is balanced such that its top end is 2.38 meters away from the wall, The cord is 40 cm long, and it is attached to the wall at a height of 75 cm above the floor. The diagram to the right shows the situation If the lower end of the rod does not slip from the corner, what is the tension in the cord?
Answers
Answer:
WE NEED TO ADD ALL 40+2.38 +75+5
Explanation:
PLSE GIVE SOME POINTS DUDE
Can someone please help me get this right pleaseee I’ll mark brainless .
Answers
Answer:i think it is c
Explanation:
Answer:
Explanation: i think its c to try it
A tortoise and hare start from rest and have a race. As the race begins, both accelerate forward. The hare accelerates uniformly at a rate of 1 m/s2 for 4 seconds. It then continues at a constant speed for 12.9 seconds, before getting tired and slowing down with constant acceleration coming to rest 66 meters from where it started. The tortoise accelerates uniformly for the entire distance, finally catching the hare just as the hare comes to a stop. 1)How fast is the hare going 1.6 seconds after it starts
Answers
Answer:
v = 4 m/s
Explanation:
Given that,
Initial speed of hare, u = 0
The hare accelerates uniformly at a rate of 1 m/s² for 4 seconds.
We need to find how fast is the hare going 1.6 seconds after it starts. Let the speed be v. So,
v = u+at
Substitute all the values,
v = 0+1×4
v = 4 m/s
So, the required speed of the hare is 4 m/s after it starts.
the water behind hoover dam in nevada is 206 m higher than the colorado river below it. at what rate must water pass through the hydraulic turbines of this dam to produce 100 mw of power if the turbines are 100 percent efficient
Answers
Answer:
the required mass flow rate is 49484.37 kg/s
Explanation:
Given the data in the question;
we first determine the relation for mass flow rate of water that passes through the turbine;
so the relation for net work on the turbine due to the change in potential energy considering 100% efficiency is;
[tex]W_{net}[/tex] = m ( Δ P.E )
so we substitute (gh) for ( Δ P.E );
[tex]W_{net}[/tex] = m (gh)
m = [tex]W_{net}[/tex] / gh
so we substitute our given values into the equation
m = 100 MW / ( 9.81 m/s²) × 206 m
m = ( 100 MW × 10⁶W/MW) / ( 9.81 m/s²) × 206 m
m = 10 × 10⁷ / 2020.86
m = 49484.37 kg/s
Therefore, the required mass flow rate is 49484.37 kg/s
Which of the following variables can be measured in joules?
A. momentum
B. Energy
C. Power
D. Work
Answers
Answer:
The variables that can be measured in joules are
B. Energy
D. Work
Hope it will help :)