A horizontal force of P=100 N is just sufficient to hold the crate from sliding down the plane, and a horizontal force of P=350 N is required to just push the crate up the plane. Determine the coefficient of static friction between the plane and the crate, and find the mass of the crate.
Answers
"down/up the plane" suggests an inclined plane, but no angle is given so I'll call it θ for the time being.
The free body diagram for the crate in either scenario is the same, except for the direction in which static friction is exerted on the crate. With the P = 100 N force holding up the crate, static friction points up the incline and keeps the crate from sliding downward. When P = 350 N, the crate is pushed upward, so static friction points down. (see attached FBDs)
Using Newton's second law, we set up the following equations.
• p = 100 N
∑ F (parallel) = f + p cos(θ) - mg sin(θ) = 0
∑ F (perpendicular) = n - p sin(θ) - mg cos(θ) = 0
• P = 350 N
∑ F (parallel) = P cos(θ) - F - mg sin(θ) = 0
∑ F (perpendicular) = N - P sin(θ) - mg cos(θ) = 0
(where n and N are the magnitudes of the normal force in the respective scenarios; ditto for f and F which denote static friction, so that f = µn and F = µN, with µ = coefficient of static friction)
Solve for n and N :
n = p sin(θ) + mg cos(θ)
N = P sin(θ) - mg cos(θ)
Substitute these into the corresponding equations containing µ, and solve for µ :
µ = (mg sin(θ) - p cos(θ)) / (mg cos(θ) + p sin(θ))
µ = (P cos(θ) - mg sin(θ)) / (P sin(θ) + mg cos(θ))
Next, you would set these equal and solve for m :
(mg sin(θ) - p cos(θ)) / (mg cos(θ) + p sin(θ)) = (P cos(θ) - mg sin(θ)) / (P sin(θ) + mg cos(θ))
...
Once you find m, you back-substitute and solve for µ, but as you might expect the result will be pretty complicated. If you take a simple angle like θ = 30°, you would end up with
m ≈ 36.5 kg
µ ≈ 0.256
The coefficient of static friction between the plane and the crate is μ = 0.256 and the mass of the crate is m=36.4 kg.
From the given,
The force that opposes the crate by sliding is P = 100N
In X-axis, the sum of forces is zero.
ΣF = 0
Pcosθ - mgsinθ-Ff = 0
Ff = Pcosθ - mgsinθ
In Y-axis
Psinθ - mgcosθ - N = 0
N = Psinθ-mgcosθ
Frictional force, Ff = μN, μ is the coefficient of friction
Ff = μN
Pcos30- mgsin30 + μ( Psin30+mgcos30) = 0
μ = mgsin30-Pcos30/Psin30+mgcos30 ------1
The block is sliding with the horizontal force, F = 350N
X-axis
P₂cosθ - mgsinθ-Ff = 0
Y-axis
P₂sinθ - mgcosθ - N = 0
N = P₂sinθ-mgcosθ
μ = P₂cos30-mgsin30/P₂sin30-mgcos30 -----2
Equate equations 1 and 2
mgsin30-Pcos30/Psin30+mgcos30 =P₂cos30-mgsin30/P₂sin30-mgcos30
4.905m-86.6/50+8.49 = 303.1-4.905m/175+8.49
41.7m² + 123m - 1.516×10⁴ = 0
-41.7m² +2330m -1.516×10⁴(4.905-86.6)(175+8.49) =(303.1-4.905)(50+8.49)
83.4m² - 2207m -3.03×10⁴ = 0
m= 36.4 kg
Hence, the mass of the crate is 36.4 Kg.
Substitute the value of m in equation 1,
μ = 4.905(36.4) - 86.6 / 50 + 8.49
μ = 0.256
Thus, the coefficient of static friction is 0.256.
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Related Questions
When should a line graph be used?
A. When the independent variable is continuous and does not show a relationship to the dependent variable
B. When the independent variable is composed of categories and does not show a relationship
C. When the independent variable is continuous and shows a casual link to the dependent variable
D. When there is no independent variable
Answers
Distillation is the separation of multiple Choose... components based on their different Choose... . As the mixture is heated and the first component Choose... , its Choose... form travels through the distillation set-up and Choose... into a different container.
Answers
Answer:
Explanation:
Distillation is the separation of multiple LIQUID components based on their different BOILING POINT. As the mixture is heated and the first component SEPARATES, its PURE form travels through the distillation set-up and GOES into a different container
A spacecraft on its way to Mars has small rocket engines mounted on its hull; one on its left surface and one on its back surface. At a certain time, both engines turn on. The one on the left gives the spacecraft an acceleration component in the x direction of
ax = 5.10 m/s2,
while the one on the back gives an acceleration component in the y direction of
ay = 7.30 m/s2.
The engines turn off after firing for 670 s, at which point the spacecraft has velocity components of
vx = 3670 m/s and vy = 4378 m/s.
What was the magnitude and the direction of the spacecraft's initial velocity before the engines were turned on? Express the magnitude as m/s and the direction as an angle measured counterclockwise from the +x axis.
magnitude m/s
direction ° counterclockwise from the +x-axis
Answers
Answer:
a) v = 517.99 m / s, b) θ = 296.3º
Explanation:
This is an exercise in kinematics, we are going to solve each axis independently
X axis
the acceleration is aₓ = 5.10 1 / S², they are on for t = 670 s and reaches a speed of vₓ= 3670 m / s, let's use the relation
vₓ = v₀ₓ + aₓ t
v₀ₓ = vₓ - aₓ t
v₀ₓ = 3670 - 5.10 670
v₀ₓ = 253 m / s
Y axis
the acceleration is ay = 7.30 m / s², with a velocity of 4378 m / s after
t = 670 s
v_y = v_{oy} + a_y t
v_{oy} = v_y - a_y t
v_oy} = 4378 - 7.30 670
v_{oy} = -513 m / s
to find the velocity modulus we use the Pythagorean theorem
v = [tex]\sqrt{v_o_x^2 + v_o_y^2}[/tex]
v = [tex]\sqrt{253^2 +513^2}[/tex]
v = 517.99 m / s
to find the direction we use trigonometry
tan θ ’= [tex]\frac{v_o_y}{v_o_x}[/tex]
θ'= tan⁻¹ [tex]\frac{voy}{voy}[/tex]
θ'= tan⁻¹ (-513/253)
tea '= -63.7
the negative sign indicates that it is below the ax axis, in the fourth quadrant
to give this angle from the positive side of the axis ax
θ = 360 - θ
θ = 360 - 63.7
θ = 296.3º
0. The temperature of source is 500K with source energy 2003, what is the temperature of sink with sink energy 100 J? a. 500 K b. 300 K c. 250 K d. 125 K
Answers
Answer:
c. 250k
Explanation:
The temperature of the sink is approximately 250 K.
To find the temperature of the sink, we can use the formula for the efficiency of a heat engine:
Efficiency = 1 - (Temperature of Sink / Temperature of Source)
Given that the temperature of the source (T_source) is 500 K and the source energy (Q_source) is 2003 J, and the sink energy (Q_sink) is 100 J, we can rearrange the formula to solve for the temperature of the sink (T_sink):
Efficiency = (Q_source - Q_sink) / Q_source
Efficiency = (2003 J - 100 J) / 2003 J
Efficiency = 1903 J / 2003 J
Efficiency = 0.9497
Now, plug the efficiency back into the first equation to solve for T_sink:
0.9497 = 1 - (T_sink / 500 K)
T_sink / 500 K = 1 - 0.9497
T_sink / 500 K = 0.0503
Now, isolate T_sink:
T_sink = 0.0503 * 500 K
T_sink = 25.15 K
Since the temperature should be in Kelvin, we round down to the nearest whole number, which is 25 K. Thus, the temperature of the sink is approximately 250 K.
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A 1640 kg merry-go-round with a radius of 7.50 m accelerates from rest to a rate of 1.00 revolution per 8.00 s. Estimate the merry-go-round as a solid cylinder and determine the net work needed for this acceleration.
Answers
Solution :
Given data :
Mass of the merry-go-round, m= 1640 kg
Radius of the merry-go-round, r = 7.50 m
Angular speed, [tex]$\omega = \frac{1}{8}$[/tex] rev/sec
[tex]$=\frac{2 \pi \times 7.5}{8}$[/tex] rad/sec
= 5.89 rad/sec
Therefore, force required,
[tex]$F=m.\omega^2.r$[/tex]
[tex]$$=1640 \times (5.89)^2 \times 7.5[/tex]
= 427126.9 N
Thus, the net work done for the acceleration is given by :
W = F x r
= 427126.9 x 7.5
= 3,203,451.75 J
Wood is an example of
A. Metalloid
B. Insulator
C. Nonmetal
D. Conductor
Answers
Many types of decorative lights are connected in parallel. If a set of lights is connected to a 110 V source and the filament of each bulb has a hot resistance of what is the currentthrough each bulb
Answers
Answer:
i₀ = V / R_i
Explanation:
For this exercise we use Ohm's law
V = i R
i = V / R
the equivalent resistance for
[tex]\frac{1}{R_{eq}}[/tex] = ∑ [tex]\frac{1}{R_i}[/tex]
if all the bulbs have the same resistance, there are N bulbs
[tex]\frac{1}{ R_{eq}} = \frac{N}{R_i}[/tex]
R_{eq} = R_i / N
we substitute
i = N V / Ri
where i is the total current that passes through the parallel, the current in a branch is
i₀ = i / N
i₀ = V / R_i
A satellite of mass m, originally on the surface of the Earth, is placed into Earth orbit at an altitude h. (a) Assuming a circular orbit, how long does the satellite take to complete one orbit
Answers
Answer:
T = 5.45 10⁻¹⁰ [tex]\sqrt{(R_e + h)^3}[/tex]
Explanation:
Let's use Newton's second law
F = ma
force is the universal force of attraction and acceleration is centripetal
G m M / r² = m v² / r
G M / r = v²
as the orbit is circular, the speed of the satellite is constant, so we can use the kinematic relations of uniform motion
v = d / T
the length of a circle is
d = 2π r
we substitute
G M / r = 4π² r² / T²
T² = [tex]\frac{4\pi ^2 }{GM} \ r^3[/tex]
the distance r is measured from the center of the Earth (Re), therefore
r = Re + h
where h is the height from the planet's surface
let's calculate
T² = [tex]\frac{4\pi ^2}{ 6.67 \ 10^{-11} \ 1.991 \ 10^{30}}[/tex] (Re + h) ³
T = [tex]\sqrt{29.72779 \ 10^{-20}} \ \sqrt[2]{R_e+h)^3}[/tex]
T = 5.45 10⁻¹⁰ [tex]\sqrt{(R_e + h)^3}[/tex]
A person jumps out of an airplane above the surface of the Earth, and falls a distance h before opening their parachute. Once the prachute is open the person coasts to the ground a distance d at constant velocity.
a. The work done on the person by the Earth is:
b. The change in gravitational potential energy of the person + Earch system is:
Answers
Answer:
a) W_total = mg (2h + d) , b) E_total = - mg (h + d)
Explanation:
a) We must solve this problem in two parts, the first for the accelerated movement and the second for the movement with constant speed
Let's look for work for the part that is in free fall
y = y₀ + v₀ t - ½ g t²
when he jumps out of a plane his vertical speed is zero
y =y₀ - ½ g t²
dy = 0 - ½ g 2t dt
the work in this first part is
W₁ = ∫ F dy
W₁ = mg ∫ g t dt
W₁ = m g² t² / 2
the time it takes to travel the distance y₀-y = h is
y₀-y = ½ g t²
t =[tex]\sqrt{2h/g}[/tex]
we substitute
W₁ = m g² 2h / g
W₁ = m g 2h
now we look for the work for the part with constant speed
since the velocity is constant let's use the uniform motion ratio
W₂ = F d
W₂ = mg d
the total work is
W_total = W₁ + W₂
W_total = 2mgh + m gd
W_total = mg (2h + d)
b) The change in gravitational potential energy
U = mg Δy
in the part with accelerated movement
U₁ = mg h
in the part with uniform movement
U₂ = mg d
the total potential energy is
E_total = U₁ + U₂
E_total = - mg (h + d)
The atoms in your body are mostly empty space . And so are the atoms in any wall. Why then is your body unable to pass through walls ?
Answers
First of all, both are not a single sheet of atom. There are many layers of atoms, so the empty part gets beside each other, so there are less empty part. Secondly, there are so many atoms that the probability that they will have empty space at the same place necessary, is negligible.
This was something from logic.
The reason I was taught in my class was that only a limited number of electrons can be in a given orbit, so atoms cannot overlap each other.
A coin and feather are dropped in a moon. what will fall earlier on ground.give reasons.if they are dropped in the earth,which one will fall faster?
Answers
Answer:
When an object is dropped, the "principal" force that acts on that object is the gravitational force.
Thus, in the absence of air resistance and such, the acceleration of the object will be equal to the gravitational acceleration:
g = 9.8m/s^2
So, when we drop objects in the moon (where there is no air) the acceleration of every object will be exactly the same. (so there is no dependence in the mass or shape of the object)
Thus, if we drop a coin and a feather in the moon, both objects will fall with the same acceleration, and then both objects will hit the ground at the same time.
But if we are in Earth, we can not ignore the air resistance (a force that acts in the opposite direction than the movement of the object)
And this force depends on the shape and mass of the object (for example, something with a really larger surface and really thin, like a sheet of paper will be more affected by this force than a small rock)
Then here, when the air resistance applies, we should expect that the heavier and smaller object (the coin) to be less affected by this force, then the resistance that the coin experiences is smaller, then the coin falls "faster" than the feather.
Which one of the following statements concerning resistors in "parallel" is true? Question 7 options: The voltage across each resistor is the same. The current through each resistor is the same. The total current through the resistors is the sum of the current through each resistor. The power dissipated by each resistor is the same.
Answers
Answer: The correct statement is:
--> The voltage across each resistor is the same.
Explanation:
RESISTORS are defined as the components of an electric circuit which are capable of creating resistance to the file of electric current in the circuit. They work by converting electrical energy into heat, which is dissipated into the air. These resistors can be divided into two according to their arrangements in the electric cell. It include:
--> Resistors in parallel and
--> Resistors in series
RESISTORS are said to be in parallel when two or more resistance or conductors are connected to common terminals so that the potential difference ( voltage) across each conductor IS THE SAME but with different current flow through each of them. Also, Individual resistances diminish to equal a smaller total resistance rather than add to make the total.
A heavy truck moving with 20 km/hr hits a car at rest. A physics student argued that
the maximum velocity the car suddenly gains is 40 km/hr. Do you agree with it?
Explain with necessary theory
Answers
Answer:
Yes
Explanation:
speed of truck = 20 km/h
Initially the car at rest.
maximum velocity of car = 40 km/h
When the truck and the car collide, the momentum of the truck transferred to car.
So, the car can attain the speed of 40 km/h.
11. An object moves in circular path with constant speed
a. Is the object's velocity constant? Explain.
b. Is its acceleration constant? Explain.
Answers
Answer:
B. Is its acceleration constant
Explanation:
Uniform circular motion can be described as the motion of an object in a circle at a constant speed. As an object moves in a circle, it is constantly changing its direction. ... An object undergoing uniform circular motion is moving with a constant speed. Nonetheless, it is accelerating due to its change in direction.
A ball on a frictionless plane is swung around in a circle at constant speed. The acceleration points in the same direction as the velocity vector.
a. True
b. False
Answers
Answer:
False
Explanation:
You have a circle so think back to circular motion. Theres 2 directions, centripetal and tangential. The problem tells you there's a constant tangential speed so tangential acceleration is 0. However there is a centripetal acceleration acting on the ball that holds it in its circular motion (i.e. tension, or gravity). Since centripetal is perpendicular to the tangential direction, acceleration and velocity are in different directions.
А pressure gauge with a measurement range of 0-10 bar has a quoted inaccuracy of £1.0% f.s. (+1% of full-scale reading). (a) What is the maximum measurement error expected for this instrument? (b) What is the likely measurement error expressed as a percentage of the or reading if this pressure gauge is measuring a pressure of 1 bar?
Answers
Answer:
I am not able to answer this question please don't mind...Explanation:
please marks me as brainliests...A horse gallops a distance of 10 kilometers in a time of 30 minutes its average speed is?
Answers
Answer:
20 km/hr
Explanation:
Distance = 10km
Time = 30 minutes = 1/2 hour
Average Speed = Total distance / Total Time Taken
= 10 ÷ 1/2
= 10 x 2
= 20 km/hr
Average speed = (distance covered) / (time to cover the distance)
Average speed = (10 km) / (30 minutes)
Average speed = 1/3 km/min
Most people would probably want to see it in a more convenient, more familiar unit, such as km/hour or m/second.
(10 km / 30 min) x (60 min / hour) = (10 x 60 / 30) (km-min / min-hour)
Average speed = 20 km/hour
AvgSpd = (10 km / 30 min) x (1,000 m / km) x (min / 60 sec)
AvgSpd = (10x1,000 / 30x60) (km-m-min / min-km-sec)
Averge Speed = 5.56 m/s
Electrical charges are of two types. True False
Answers
Answer:
Electrical charges r of 2 types its true.they are positive and negative.hope it helps.stay safe healthy and happy..Answer: Think its true
In order to keep a leaking ship from sinking, it is necessary to pump 12.0 lb of water each second from below deck up a height of 2.00 m and over the side. What is the minimum horse-
power motor that can be used to save the ship?
Answers
Answer:
P = 0.14 hp
Explanation:
The power required by the ship is given as:
[tex]P = \frac{Work}{Time} = \frac{Potential\ Eenrgy}{t}\\\\P = \frac{mgh}{t}[/tex]
where,
P = Power = ?
m = mass to pump = (12 lb)(1 kg/2.20 lb) = 5.44 kg
g = acceleration due to gravity = 9.81 m/s²
h = height = 2 m
t = time = 1 s
Therefore,
[tex]P = \frac{(5.44\ kg)(9.81\ m/s^2)(2\ m)}{1\ s}\\\\P = 106.8\ W[/tex]
Converting to horsepower (hp):
[tex]P = (106.8\ W)(\frac{1\ hp}{746\ W})[/tex]
P = 0.14 hp
Drag the titles to the correct boxes to complete the pairs.
Answers
A simple pendulum takes 2.00 s to make one compete swing. If we now triple the length, how long will it take for one complete swing
Answers
Answer:
3.464 seconds.
Explanation:
We know that we can write the period (the time for a complete swing) of a pendulum as:
[tex]T = 2*\pi*\sqrt{\frac{L}{g} }[/tex]
Where:
[tex]\pi = 3.14[/tex]
L is the length of the pendulum
g is the gravitational acceleration:
g = 9.8m/s^2
We know that the original period is of 2.00 s, then:
T = 2.00s
We can solve that for L, the original length:
[tex]2.00s = 2*3.14*\sqrt{\frac{L}{9.8m/s^2} }\\\\\frac{2s}{2*3.14} = \sqrt{\frac{L}{9.8m/s^2}}\\\\(\frac{2s}{2*3.14})^2*9.8m/s^2 = L = 0.994m[/tex]
So if we triple the length of the pendulum, we will have:
L' = 3*0.994m = 2.982m
The new period will be:
[tex]T = 2*3.14*\sqrt{\frac{2.982m}{9.8 m/s^2} } = 3.464s[/tex]
The new period will be 3.464 seconds.
why is the water drawn from the bottom of the dam rather than the top?
Answers
Answer:
because minerals can be gotten from the bottom
Explanation:
it's self explanatory
A 2090-kg test rocket is launched vertically from the launch pad. Its fuel (of negligible mass) provides a thrust force so that its vertical velocity as a function of time is given by v(t) =At+Bt^2 , where A and B are constants and time is measured from the instant the fuel is ignited. The rocket has an upward acceleration of 1.50m/s 2 at the instant of ignition and, 1.00 s later, an upward velocity of 2.00 m/s. (a) Determine A and B , including their SI units. (b) At 4.00 s after fuel ignition, what is the acceleration of the rocket, and (c) what thrust force does the burning fuel exert on it, assuming no air resistance? Express the thrust in newtons and as a multiple of the rocket’s weight. (d) What was the initial thrust due to the fuel?
Answers
Answer:
a) A = 1.50 m / s², B = 1.33 m/s³, b) a = 12.1667 m / s²,
c) I = M (1.5 t + 1.333 t²) , d) ΔI = M 2.833 N
Explanation:
In this exercise give the expression for the speed of the rocket
v (t) = A t + B t²
and the initial conditions
a = 1.50 m / s² for t = 0 s
v = 2.00 m / s for t = 1.00 s
a) it is asked to determine the constants.
Let's look for acceleration with its definition
a = [tex]\frac{dv}{dt}[/tex]
a = A + 2B t
we apply the first condition t = 0 s
a = A
A = 1.50 m / s²
we apply the second condition t = 1.00 s
v = 1.5 1 + B 1²
2 = 1.5 + B
B = 2 / 1.5
B = 1.33 m/s³
the equation remains
v = 1.50 t + 1.333 t²
b) the acceleration for t = 4.00 s
a = 1.50 + 1.333 2t
a = 1.50 + 2.666 4
a = 12.1667 m / s²
c) The thrust
I = ∫ F dt = p_f - p₀
Newton's second law
F = M a
F = M (1.5 + 2 1.333 t) dt
we replace and integrate
I = M ∫ (1.5 + 2.666 t) dt
I = 1.5 t + 2.666 t²/2
I = M (1.5 t + 1.333 t²) + cte
in general the initial rockets with velocity v = 0 for t = 0, where we can calculate the constant
cte = 0
I = M (1.5 t + 1.333 t²)
d) the initial push
For this we must assume some small time interval, for example between
t = 0 s and t = 1 s
ΔI = I_f - I₀
ΔI = M (1.5 1 + 1.333 1²)
ΔI = M 2.833 N
When air expands adiabatically (without gaining or losing heat), its pressure P and volume V are related by the equation PV1.4=C where C is a constant. Suppose that at a certain instant the volume is 420 cubic centimeters and the pressure is 99 kPa and is decreasing at a rate of 7 kPa/minute. At what rate in cubic centimeters per minute is the volume increasing at this instant?
Answers
Answer:
[tex]\frac{dV}{dt}=21.21cm^3/min[/tex]
Explanation:
We are given that
[tex]PV^{1.4}=C[/tex]
Where C=Constant
[tex]\frac{dP}{dt}=-7KPa/minute[/tex]
V=420 cubic cm and P=99KPa
We have to find the rate at which the volume increasing at this instant.
Differentiate w.r.t t
[tex]V^{1.4}\frac{dP}{dt}+1.4V^{0.4}P\frac{dV}{dt}=0[/tex]
Substitute the values
[tex](420)^{1.4}\times (-7)+1.4(420)^{0.4}(99)\frac{dV}{dt}=0[/tex]
[tex]1.4(420)^{0.4}(99)\frac{dV}{dt}=(420)^{1.4}\times (7)[/tex]
[tex]\frac{dV}{dt}=\frac{(420)^{1.4}\times (7)}{1.4(420)^{0.4}(99)}[/tex]
[tex]\frac{dV}{dt}=21.21cm^3/min[/tex]
Answer:
[tex]\dot V=2786.52~cm^3/min[/tex]
Explanation:
Given:
initial pressure during adiabatic expansion of air, [tex]P_1=99~kPa[/tex]
initial volume during the process, [tex]V_1=420~cm^3[/tex]
The adiabatic process is governed by the relation [tex]PV^{1.4}=C[/tex] ; where C is a constant.
Rate of decrease in pressure, [tex]\dot P=7~kPa/min[/tex]
Then the rate of change in volume, [tex]\dot V[/tex] can be determined as:
[tex]P_1.V_1^{1.4}=\dot P.\dot V^{1.4}[/tex]
[tex]99\times 420^{1.4}=7\times V^{1.4}[/tex]
[tex]\dot V=2786.52~cm^3/min[/tex]
[tex]\because P\propto\frac{1}{V}[/tex]
[tex]\therefore[/tex] The rate of change in volume will be increasing.
SCALCET8 3.9.018.MI. A spotlight on the ground shines on a wall 12 m away. If a man 2 m tall walks from the spotlight toward the building at a speed of 1.7 m/s, how fast is the length of his shadow on the building decreasing when he is 4 m from the building
Answers
Answer:
The length of his shadow is decreasing at a rate of 1.13 m/s
Explanation:
The ray of light hitting the ground forms a right angled triangle of height H, which is the height of the building and width, D which is the distance of the tip of the shadow from the building.
Also, the height of the man, h which is parallel to H forms a right-angled triangle of width, L which is the length of the shadow.
By similar triangles,
H/D = h/L
L = hD/H
Also, when the man is 4 m from the building, the length of his shadow is L = D - 4
So, D - 4 = hD/H
H(D - 4) = hD
H = hD/(D - 4)
Since h = 2 m and D = 12 m,
H = 2 m × 12 m/(12 m - 4 m)
H = 24 m²/8 m
H = 3 m
Since L = hD/H
and h and H are constant, differentiating L with respect to time, we have
dL/dt = d(hD/H)/dt
dL/dt = h(dD/dt)/H
Now dD/dt = velocity(speed) of man = -1.7 m/s ( negative since he is moving towards the building in the negative x - direction)
Since h = 2 m and H = 3 m,
dL/dt = h(dD/dt)/H
dL/dt = 2 m(-1.7 m/s)/3 m
dL/dt = -3.4/3 m/s
dL/dt = -1.13 m/s
So, the length of his shadow is decreasing at a rate of 1.13 m/s
The sound level measured in a room by a person watching a movie on a home theater system varies from 40 dB during a quiet part to 80 dB during a loud part. Approximately how many times louder is the latter sound
Answers
Answer:
[tex]\alpha=-3.01dB[/tex]
Explanation:
From the question we are told that:
Sound level intensity
[tex]\triangle I=40dB-80dB[/tex]
Generally the equation for intensity level is mathematically given by
[tex]\alpha=10log_{10}(I/I_x)dB[/tex]
Where
I= Intensity measured
[tex]I_x=Threshold\ of\ audibility[/tex]
[tex]I_x= 10-12 W / m2[/tex]
[tex]\alpha= 10 log10 \frac{I_1}{I_x} - 10 log10 \frac{}I_2{I_x}[/tex]
[tex]\alpha= 10 log10 \frac{I_1}{I_2}[/tex]
[tex]\alpha=10 log10\frac{40}{80}[/tex]
[tex]\alpha=-3.01dB[/tex]
Harmonics a.are components of a complex waveform. b.have frequencies that are integer multiples of the frequency of the complex waveform. c.are pure tones. d.have sinusoidal waveforms. e.all of the above
Answers
Answer:
b.have frequencies that are integer multiples of the frequency of the complex waveform
Explanation:
Please correct me if I am wrong
8. If a moving object triples its speed, how much kinetic energy will it have? A. six times as much as before B. three times as much as before C. one third as much as before D. nine times as much as before
Answers
D
Explanation:
KE: 0.5mv²
when v is tripled v² is 9 times its original value
Newspapers often talk about an energy crisis-about running out of certain energy sources in the not-so-distant future. About which kind of energy sources are they talking
Answers
Answer:
Nonrenewable energy
Explanation:
Renewable energy is also known as clean energy and it can be defined as a type of energy that are generated through natural sources or technology-based processes that are replenished constantly. Some examples of these natural sources are water (hydropower), wind (wind energy), sun (solar power), geothermal, biomass, waves etc.
Basically, a renewable energy source is sustainable and as such can not be exhausted.
On the other hand, a non-renewable energy refers to an energy source such as fossil fuels that takes a very long time to be created or their creation happened long ago and isn't likely to happen again e.g uranium.
For example, fossil fuels such as coal, oil, and natural gas, come from deep inside the Earth where they formed over millions of years ago.
In this scenario, the kind of energy the newspaper sources are talking about is a nonrenewable energy source because they are capable of being exhausted in the not-so-distant future.
A closely wound, circular coil with radius 2.70 cm has 800 turns. What must the current in the coil be if the magnetic field at the center of the coil is 0.0750 T
Answers
Answer:
Approximately 4.029 A
Explanation:
We can use the formula that the B field of a few loops all with current in same direction is permeability of free space (mu)* current * Number or loops divided by 2*radius. You are given B field, radius(convert into meters), number of loops and mu is 4pi * 10^-7. Solve for current and you get 4.029 Amperes.
