Answer:
f = 7.06 N
Explanation:
The maximum frictional force on the knee joint of the person can be given by the following formula:
[tex]f = \mu R = \mu W \\[/tex]
where,
f = maximum frictional force = ?
μ = static friction coefficient = 0.016
W = Weight load on knee = mg
m = mass supported by knee = 45 kg
g = acceleration due to gravity = 9.81 m/s²
Therefore,
[tex]f = \mu mg\\f = (0.016)(45\ kg)(9.81\ m/s^2)\\[/tex]
f = 7.06 N
Two guitar strings, of equal length and linear density, are tuned such that the second harmonic of the first string has the same frequency as the third harmonic of the second string. The tension of the first string is 510 N. Calculate the tension of the second string.
Answer:
The tension in the second string is 226.7 N.
Explanation:
Length is L, mass per unit length = m
T = 510 N
Let the tension in the second string is T'.
second harmonic of the first string = third harmonic of the second string
[tex]2 f = 3 f'\\\\2\sqrt{\frac{T}{m}} = 3 \sqrt {\frac{T'}{m}}\\\\4 T = 9 T'\\\\4\times 510 = 9 T'\\\\T' = 226.7 N[/tex]
A tennis player receives a shot with the ball (0.0600 kg) traveling horizontally at 59.4 m/s and returns the shot with the ball traveling horizontally at 37.2 m/s in the opposite direction. (Take the direction of the ball's final velocity (toward the net) to be the +x-direction).
(a) What is the impulse delivered to the ball by the racket?
(b) What work does the racket do on the ball?
5 9 . 4
- 3 7 . 2
2 2 . 2
Explanation:
Use the algorithm method.
5 9 . 4
- 3 7 . 2
2 2 . 2
2 Therefore, 59.4-37.2=22.259.4−37.2=22.2.
22.2
22.2
A 0.160kg glider is moving to the right on a frictionless, horizontal air track with a speed of 0.820m/s . It has a head-on collision with a 0.300kg glider that is moving to the left with a speed of 2.27m/s . Suppose the collision is elastic.
Part A
Find the magnitude of the final velocity of the 0.160kg glider. m/s
Part B
Find the direction of the final velocity of the 0.160kg glider.
i. to the right
ii. to the left
Part C
Find the magnitude of the final velocity of the 0.300kg glider. m/s
Part D
Find the direction of the final velocity of the 0.300kg glider.
Answer:
A) v_{f1} = -3.2 m / s, B) LEFT , C) v_{f2} = -0.12 m / s, D) LEFT
Explanation:
This is a collision exercise that can be solved using momentum conservation, for this we define a system formed by gliders, so that the forces during the collision are internal and the moment is conserved.
Let's use the subscript 1 for the lightest glider m1 = 0.160 kg and vo1 = 0.820 m / s
subscript 2 for the heaviest glider me² = 0.820 kg and vo2 = -2.27 m / s
Initial instant. Before the crash
p₀ = m₁ v₀₁ + m₂ v₀₂
Final moment. After the crash
p_f = m₁ v_{f1} + m₂ v_{f2}
p₀ = p_f
m₁ v₀₁ + m₂ v₀₂ = m₁ v_{f1} + m₂ v_{f2}
as the shock is elastic, energy is conserved
K₀ = K_f
½ m₁ v₀₁² + ½ m₂ v₀₂² = ½ m₁ [tex]v_{f1}^2[/tex] + ½ m₂ [tex]v_{f2}^2[/tex]
m₁ (v₀₁² - v_{f1}²) = m₂ (v_{f2}² -v₀₂²)
let's make the relationship
(a + b) (a-b) = a² -b²
m₁ (v₀₁ + v_{f1}) (v₀₁-v+{f1}) = m₂ (v_{f2} + v₀₂) (v_{f2} -v₀₂)
let's write our two equations
m₁ (v₀₁ -v_{f1}) = m₂ (v_(f2) - v₀₂) (1)
m₁ (v₀₁ + v_{f1}) (v₀₁-v_{f1}) = m₂ (v_{f2} + v₀₂) (v_{f2} -v₀₂)
we solve
v₀₁ + v_{f2} = v_{f2} + v₀₂
we substitute in equation 1 and obtain
M = m₁ + m₂
[tex]v_{f1} = \frac{m_1-m_2}{M} v_o_1 + 2 \frac{m_2}{M} v_f_2[/tex]
[tex]v_f_2 = \frac{2m_1}{M} v_o_1 + \frac{m_2-m_1}{M} v_o_2[/tex]vf2 = 2m1 / mm vo1 + m2-m1 / mm vo2
we calculate the values
m₁ + m₂ = 0.160 +0.3000 = 0.46 kg
v_{f1} = [tex]\frac{ 0.160 -0.300} {0.460} \ 0.820 + \frac{2 \ 0300}{0.460} \ (-2.27)[/tex]
v_{f1} = -0,250 - 2,961
v_{f1} = - 3,211 m / s
v_{f2} = [tex]\frac{2 \ 0.160}{0.460} \ 0.820 + \frac{0.300 - 0.160}{0.460 } \ (-2.27)[/tex]
v_{f2} = 0.570 - 0.6909
v_{f2} = -0.12 m / s
now we can answer the different questions
A) v_{f1} = -3.2 m / s
B) the negative sign indicates that it moves to the left
C) v_{f2} = -0.12 m / s
D) the negative sign indicates that it moves to the LEFT
An electric heater is madde of a wire of resistance 100π and connected to a 240v mains supply. Determine the power rating of the heater
Answer:
Power = 576 Watts
Explanation:
The electrical power of an electric circuit can be defined as a measure of the rate at which energy is either produced or absorbed in the circuit.
Mathematically, electrical power is given by the formula;
[tex] Electrical \; power = current * voltage [/tex]
This ultimately implies that, the quantity (current times voltage ) is electrical power and it is measured (S.I units) in Watt (W).
Given the following data;
Resistance = 100 ohms
Voltage = 240 V
To find the power rating of the heater;
Power = V²/R
Where;
V is the voltage.
R is the resistance.
Substituting into the formula, we have;
Power = 240²/100
Power = 57600/100
Power = 576 Watts
When an automobile moves with constant velocity the power developed is used to overcome the frictional forces exerted by the air and the road. If the power developed in an engine is 50.0 hp, what total frictional force acts on the car at 55 mph (24.6 m/s)
P = F v
where P is power, F is the magnitude of force, and v is speed. So
50.0 hp = 37,280 W = F (24.6 m/s)
==> F = (37,280 W) / (24.6 m/s) ≈ 1520 N
chemical kinetics half lives
A mass is attached to the end of a spring and set into oscillation on a horizontal frictionless surface by releasing it from a compressed position. The record of time is started when the oscillating mass first passes through the equilibrium position, and the position of the mass at any time is described by
The question is incomplete. The complete question is :
A mass is attached to the end of a spring and set into oscillation on a horizontal frictionless surface by releasing it from a compressed position. The record of time is started when the oscillating mass first passes through the equilibrium position, and the position of the mass at any time is described by x = (4.7 cm)sin[(7.9 rad/s)πt].
Determine the following:
(a) frequency of the motion
(b) period of the motion
(c) amplitude of the motion
(d) first time after t = 0 that the object reaches the position x = 2.6 cm
Solution :
Given equation : x = (4.7 cm)sin[(7.9 rad/s)πt].
Comparing it with the general equation of simple harmonic motion,
x = A sin (ωt + Φ)
A = 4.7 cm
ω = 7.9 π
a). Therefore, frequency, [tex]$f=\frac{\omega}{2 \pi}$[/tex]
[tex]$=\frac{7.9 \pi}{2 \pi}$[/tex]
= 3.95 Hz
b). The period, [tex]$T=\frac{1}{f}$[/tex]
[tex]$T=\frac{1}{3.95}[/tex]
= 0.253 seconds
c). Amplitude is A = 4.7 cm
d). We have,
x = A sin (ωt + Φ)
[tex]$x_t=4.7 \sin (7.9 \pi t)$[/tex]
[tex]$2.6 = 4.7 \sin (7.9 \pi t)$[/tex]
[tex]$\sin (7.9 \pi t) = \frac{26}{47}$[/tex]
[tex]$7.9 \pi t = \sin^{-1}\left(\frac{26}{47}\right)$[/tex]
Hence, t = 0.0236 seconds.
Cold air rises because it is denser than water, is this true?
Answer:
true
Explanation:
im not sure please dont attack me
You are driving to the grocery store at 20 m/s. You are 150 m from an intersection when the traffic light turns red. Assume that your reaction time is 0.50 s and that your car brakes with constant acceleration.
Required:
a. How far are you from the intersection when you begin to apply the brakes?
b. What acceleration will bring you to rest right at the intersection?
c. How long does it take you to stop?
Hi there!
a.
Use the formula d = st to solve:
d = 20 × 0.5 = 10m
150 - 10 = 140m away when brakes are applied
b.
Use the following kinematic equation to solve:
vf² = vi² + 2ad
Plug in known values:
0 = 20² + 2(150)(a)
Solve:
0 = 400 + 300a
-300a = 400
a = -4/3 (≈ -1.33) m/s² required
c.
Use the following kinematic equation to solve:
vf = vi + at
0 = 20 - 4/3t
Solve:
4/3t = 20
Multiply both sides by 3/4 for ease of solving:
t = 15 sec
You place an 8 kg ball on the top of your 2 cm^2 finger tip. Calculate the
PRESSURE. Show MATH, answer and unit.
Answer:
the pressure exerted by the object is 392,000 N/m²
Explanation:
Given;
mass of the object, m = 8 kg
area of your finger, A = 2 cm² = 2.0 x 10⁻⁴ m²
acceleration due to gravity, g = 9.8 m/s²
The pressure exerted by the object is calculated as;
[tex]Pressure = \frac{F}{A} = \frac{mg}{A} = \frac{8 \times 9.8}{2\times 10^{-4}} = 392,000 \ N/m^2[/tex]
Therefore, the pressure exerted by the object is 392,000 N/m²
(c) The ball leaves the tennis player's racket at a speed of 50 m/s and travels a
distance of 20 m before bouncing.
(i) Calculate how long it takes the ball to travel this distance.
(1 mark)
Answer:
t=0.417s
Explanation:
After the ball hits the racket it is in freefall(assume air resistance as negligible)
so a=-g
use
x-x0=v0t+1/2at^2
Plug in givens
20=50t-4.9t^2
Solve quadratic equation using quadratic formula
t= 0.417 seconds, (the other answer is extraneous because it is too big because in 1 second, the ball travels 50 meters)
A 56.0 kg bungee jumper jumps off a bridge and undergoes simple harmonic motion. If the period of oscillation is 11.2 s, what is the spring constant of the bungee cord, assuming it has negligible mass compared to that of the jumper in N/m
Answer:
2.80N/m
Explanation:
Given data
mass m= 56kg
perios T= 11.2s
The expression for the period is given as
T=2π√m/k
Substitute
11.2= 2*3.142*√56/k
square both sides
11.2^2= 2*3.142*56/k
125.44= 351.904/k
k=351.904/125.44
k= 2.80N/m
Hence the spring constant is 2.80N/m
The force an ideal spring exerts on an object is given by , where measures the displacement of the object from its equilibrium position. If , how much work is done by this force as the object moves from to
Answer:
The correct answer is "1.2 J".
Explanation:
Seems that the given question is incomplete. Find the attachment of the complete query.
According to the question,
x₁ = -0.20 mx₂ = 0 mk = 60 N/mNow,
⇒ [tex]W=\int_{x_1}^{x_2}F \ dx[/tex]
⇒ [tex]=\int_{x_1}^{x_2}-kx \ dx[/tex]
⇒ [tex]=-k \int_{-0.20}^{0}x \ dx[/tex]
By putting the values, we get
⇒ [tex]=-(60)[\frac{x^2}{2} ]^0_{-0.20}[/tex]
⇒ [tex]=-60[\frac{0}{2}-\frac{0.04}{2} ][/tex]
⇒ [tex]=1.2 \ J[/tex]
write the formulae of magnesium chloride and sodium sulfate
Answer:
Magnesium Chloride: MgCl2
Sodium Sulfate: Na2SO4
A rigid tank contains 10 lbm of air at 30 psia and 60 F. Find the volume of the tank in ft3. The tank is now heated until the pressure doubles. Find the heat transfer in Btu.
Answer:
Hence the amount of heat transfer is 918.75 Btu.
Explanation:
Now,
A study finds that the metabolic rate of mammals is proportional to m^3/4 , where m is the total body mass. By what factor does the metabolic rate of a 70.0-kg human exceed that of a 4.91-kg cat?
Answer:
The mass of human is 2898 times of the mass of cat.
Explanation:
A study finds that the metabolic rate of mammals is proportional to m^3/4 i.e.
[tex]M=\dfrac{km^3}{4}[/tex]
Where
k is constant
If m = 70 kg, the mass of human
[tex]M=\dfrac{70^3}{4}\\\\=85750[/tex]
If m = 4.91 kg, the mass of cat
[tex]M'=\dfrac{4.91^3}{4}\\\\=29.59[/tex]
So,
[tex]\dfrac{M}{M'}=\dfrac{85750}{29.59}\\\\=2897.93\approx 2898[/tex]
So, the mass of human is 2898 times of the mass of cat.
Match each term with the best description.
a. Tightly woven fabric used to smother and extinguish a fire.
b. Consists of absorbent material that can be ringed around a chemical spill until the spill can be neutralized.
c. Device used to control small fires in an emergency situation
d. Provides chemical. physical. Health, and safety information regarding chemical reagents and supplies
1. Spill containment kit
2. Safety Data sheet
3. Fume hood
4. Fire extinguisher
5. Fire blanket
Answer:
A - 5
B - 1
C - 4
D -2
Explanation:
I don't have one i just know...
The fire blanket is a tightly woven fabric. The spill containment kit consists of absorbent material. Fire extinguishers control small fires and the safety data sheet provides chemical, health, and safety information.
(a) The fire blanket is a blanket, which may be quickly thrown over a fire to snuff out the flames, and comprises fire-resistant materials.
Hence, option (a) matches with option (5)
(b) In order to contain a chemical spill, absorbent items like pads, socks, or booms are frequently included in spill containment kits.
Hence, option (b) correctly matches with option (1).
(c) A fire extinguisher is a tool used to put out small fires during emergencies.
Hence, option (c) correctly matches with option (4).
(d) A Safety Data Sheet (SDS) gives in-depth details regarding a specific chemical or chemical mixture. It provides information about the physical characteristics of the chemical, any potential risks, safe handling and storage practices, emergency response strategies, and more.
Hence, option (d) correctly matches option (2).
To learn more about Fire extinguishers, here:
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An ice chest at a beach party contains 12 cans of soda at 3.78 °C. Each can of soda has a mass of 0.35 kg and a specific heat capacity of 3800 J/(kg C°). Someone adds a 6.48-kg watermelon at 29.4 °C to the chest. The specific heat capacity of watermelon is nearly the same as that of water. Ignore the specific heat capacity of the chest and determine the final temperature T of the soda and watermelon in degrees Celsius.
Answer:
T = 13.25°C
Explanation:
From the law of conservation of energy:
Heat Lost by Watermelon = Heat Gained by Cans
[tex]m_wC_w\Delta T_w = m_cC_c\Delta T_c[/tex]
where,
[tex]m_w[/tex] = mass of watermelon = 6.48 kg
[tex]m_c[/tex] = mass of cans = (12)(0.35 kg) = 4.2 kg
[tex]C_w[/tex] = specific heat capacity of watermelon = 3800 J/kg.°C
[tex]C_c[/tex] = specific heat capacity of cans = 4200 J/kg.°C
[tex]\Delta T_w[/tex] = Change in Temprature of watermelon = 29.4°C - T
[tex]\Delta T_c[/tex] = Change in Temperature of cans = T - 3.78°C
T = final temperature = ?
Therefore,
[tex](4.2\ kg)(3800\ J/kg.^oC)(29.4^oC-T)=(6.48\ kg)(4200\ J/kg^oC)(T-3.78^oC)\\469224\ J-(15960\ J/^oC)T = (27216\ J/^oC)T-102876.48\ J\\469224\ J + 102876.48\ J = (27216\ J/^oC)T+(15960\ J/^oC)T\\\\T = \frac{572100.48\ J}{43176\ J/^oC}[/tex]
T = 13.25°C
Imagine you were given a converging lens and a meter stick and sent outside on a sunny day. In a few sentences, describe a method to measure, as accurately as possible, the focal length of the lens using only the lens, a meter stick, and your outside surroundings. Explain your reasoning
Answer:
the Sun is the object that is at a very great distance and the focus point of the sun's shape is the distance to the magnesia, this image is equal to the seal distance
Explanation:
The method for measuring the focal length of a lens is based on the use of the constructor's equation
[tex]\frac{1}{f } = \frac{1}{p} + \frac{1}{q}[/tex]
where q and q are the distance to the object and the image respectively, f is the focal length.
If we place the object very far away (infinity) the equation remains
[tex]\frac{1}{f} = \frac{1}{q}[/tex]
Therefore with this we can devise a means for measuring the Sun is the object that is at a very great distance and the focus point of the sun's shape is the distance to the magnesia, this image is equal to the seal distance
A metre rule is used to measure the length of a piece of string in a certain experiment. It is found to be 20 cm long to the nearest millimeter. How should thisresult be recorded in a table of results? a. 0.2000m b. 0.200m c. 0.20m d. 0.2m
Answer:
C
Explanation:
20 cm = 0.2m
since uncertainty is 0.1 cm (0.001 m), should be recorded to same number of decimal place as uncertainty
therefore it's 0.200m
A 1200 kg car traveling east at 4.5 m/s crashes into the side of a 2100 kg truck that is not moving. During the collision, the vehicles get stuck together. What is their velocity after the collision? A. 2.9 m/s east B. 1.6 m/s east m C. 2.6 m/s east D. 1.8 m/s east
Answer:
Explanation:
This is a simple Law of Momentum Conservation problem of the inelastic type. The equation for this is
[tex][m_1v_1+m_2v_2]_b=[(m_1+m_2)v]_a[/tex] Filling in:
[tex][1200(4.5)+2100(0)]=[(1200+2100)v][/tex] which simplifies to
5400 + 0 = 3300v
so v = 1.6 m/s to the east, choice B
This diagram shows the magnetic field lines near the ends of two magnets. There is an error in the diagram.
Two bar magnet with the north pole of one near the south pole of the second. field lines are leaving the north pole and bent away from the south pole of the other. Field lines are leaving the south pole of one and bending away from the north pole of the other.
Which change will correct the error in the diagram?
a)changing the N to S
b)reversing the arrows on the left to point toward the N
c)changing the S to N
d)reversing the arrows on the right to point toward the S
Answer:
changing the N to S. that's how the error will be corrected
Answer:
C is the correct answer
Explanation:
i took the test
Driving on asphalt roads entails very little rolling resistance, so most of the energy of the engine goes to overcoming air resistance. But driving slowly in dry sand is another story. If a 1500 kg car is driven in sand at 4.9 m/s , the coefficient of rolling friction is 0.060. In this case, nearly all of the energy that the car uses to move goes to overcoming rolling friction, so you can ignore air drag in this problem.
Required:
a. What propulsion force is needed to keep the car moving forward at a constant speed?
b. What power is required for propulsion at 5.0 m/s?
c. If the car gets 15 mpg when driving on sand, what is the car's efficiency? One gasoline contains 1.4×10 ^8 J of chemical energy.
Answer:
a) [tex]F_p=882N[/tex]
b) [tex]P=4410W[/tex]
c) [tex]V_p'=24135[/tex] ,[tex]n=15.2\%[/tex]
Explanation:
From the question we are told that:
Mass [tex]M=1500kg[/tex]
Velocity [tex]v=4.9m/s[/tex]
Coefficient of Rolling Friction [tex]\mu=0.06[/tex]
a)
Generally the equation for The Propulsion Force is mathematically given by
[tex]F_p=\mu*mg[/tex]
[tex]F_p=0.06*1500*9.81[/tex]
[tex]F_p=882N[/tex]
b)
Therefore Power Required at
[tex]V_p=5.0m/s[/tex]
[tex]P=F_p*V_p[/tex]
[tex]P=882*5[/tex]
[tex]P=4410W[/tex]
c)
[tex]V_p' =15mpg[/tex]
[tex]V_p'=15*\frac{1609}[/tex]
[tex]V_p'=24135[/tex]
Generally the equation for Work-done is mathematically given by
[tex]W=F_p*V_p'[/tex]
[tex]W=882*15*1609[/tex]
[tex]W=2.13*10^7[/tex]
Therefore
Efficiency
[tex]n=\frac{W}{E}*100\%[/tex]
Since
Energy in one gallon of gas is
[tex]E=1.4*10^8J[/tex]
Therefore
[tex]n=\frac{2.1*10^7}{1.4*10^8}*100\%[/tex]
[tex]n=15.2\%[/tex]
A charge Q exerts a 1.2 N force on another charge q. If the distance between the charges is doubled, what is the magnitude of the force exerted on Q by q
Answer:
0.3 N
Explanation:
Electromagnetic force is F= Kq1q2/r^2, where r is the distance between charges. If r is doubled then the force will be 1/4F which is 0.3 N.
The magnitude of the force exerted on Q by q when the distance between them is doubled is 0.3 N
Coulomb's law equationF = Kq₁q₂ / r²
Where
F is the force of attraction K is the electrical constant q₁ and q₂ are two point charges r is the distance apart Data obtained from the question Initial distance apart (r₁) = rInitial force (F₁) = 1.2 NFinal distance apart (r₂) = 2rFinal force (F₂) =? How to determine the final forceFrom Coulomb's law,
F = Kq₁q₂ / r²
Cross multiply
Fr² = Kq₁q₂
Kq₁q₂ = constant
F₁r₁² = F₂r₂²
With the above formula, we can obtain the final force as follow:
F₁r₁² = F₂r₂²
1.2 × r² = F₂ × (2r)²
1.2r² = F₂ × 4r²
Divide both side by 4r²
F₂ = 1.2r² / 4r²
F₂ = 0.3 N
Learn more about Coulomb's law:
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g How much buoyancy force, in N, a person with a mass of 70 kg experiences by just standing in air
Answer:
686.7N
Explanation:
Given data
Mass= 70kg
We know that the buoyant force experienced by the person is equal to the weight of the person
Hence the weight is
Weight = mass* Acceleration
Weight= 70*9.81
Weight= 686.7N
Therefore the weight is 686.7N
In a double-slit experiment, the slit spacing is 0.120 mm and the screen is 2.00 m from the slits. Find the wavelength (in nm) if the distance between the central bright region and the third bright fringe on a screen is 2.75 cm.
Answer:
[tex]\lambda=550nm[/tex]
Explanation:
From the question we are told that:
The slit spacing [tex]d_s=0.120mm=>0.120*10^{-3}[/tex]
Screen distance [tex]d_{sc}=2.0m[/tex]
Third Distance [tex]X=2.75cm=>2.75*10^{-2}[/tex]
Generally the equation for Wavelength is mathematically given by
[tex]\lambda=\frac{Xd_s}{n*d_{sc}}[/tex]
Where
n=number of screens
[tex]n=3[/tex]
Therefore
[tex]\lambda=\frac{2.75*10^{-2}*0.120*10^{-3}}{3*2}[/tex]
[tex]\lambda=0.055*10^{-5}[/tex]
[tex]\lambda=550nm[/tex]
what is Friction
short note on friction
Answer:
Explanation:
Friction can be defined as a force that resists the relative motion of two objects when there surface comes in contact. Thus, it prevents two surface from easily sliding over or slipping across one another. Also, friction usually reduces the efficiency and mechanical advantage of machines but can be reduced through lubrication.
Generally, there are four (4) main types of friction and these includes;
I. Static friction.
II. Rolling friction.
III. Sliding friction.
IV. Fluid friction.
A 5 kg object is moving in one dimension along the x-axis with a speed of 2 m/s. An external impulse acts on the force causing the speed of the object to increase to 5 m/s. The impulse lasted for 3 s. What is the average net force (in N) exerted on the object
Answer:
The correct answer is "15 Kg.m/s".
Explanation:
Given values are:
Mass,
m = 5 Kg
Initial velocity,
u = 2 m/s
Final velocity,
v = 5 m/s
Now,
The magnitude of change in linear momentum will be:
= [tex]m\times (v - u)[/tex]
By substituting the values, we get
= [tex]5\times (5 - 2)[/tex]
= [tex]5\times 3[/tex]
= [tex]15 \ Kg.m/s[/tex]
helppp!!! what's the answer to this??
when an ideal capacitor is connected across an ac voltage supply of variable frequency, the current flowing
a) is in phase with voltage at all frequencies
b) leads the voltage with a phase independent of frequency
c) leads the voltage with a phase which depends on frequency
d) lags the voltage with a phase independent of frequency
what would be the correct option?
Answer:
(b)
Explanation:
The voltage always lags the current by 90°, regardless of the frequency.
A 12.5-m fire truck ladder is leaning against a wall. Find the distance d the ladder goes up the wall (above the fire truck) if the ladder makes an angle of with the horizontal
Complete Question
A 12.5-m fire truck ladder is leaning against a wall. Find the distance d the ladder goes up the wall (above the fire truck) if the ladder makes an angle of
40° 16' with the horizontal.
Answer:
[tex]d=8.01m[/tex]
Explanation:
From the question we are told that:
Length of ladder [tex]l=12.5m[/tex]
Angle [tex]\theta=40° 16'=20.26 \textdegree[/tex]
Generally the Trigonometric equation for distance d it goes up the wall is mathematically given by
[tex]d=l sin \theta[/tex]
[tex]d=12.5 sin 40.26[/tex]
[tex]d=8.01m[/tex]
