The magnitude of the forces acting at the top are;
[tex]\mathbf{F_{Top, \ x}}[/tex] = 132.95 N
[tex]\mathbf{F_{Top, \ y}}[/tex] = 0
The magnitude of the forces acting at the bottom are;
[tex]\mathbf{F_{Bottom, \ x}}[/tex] = [tex]\mathbf{ F_f}[/tex] = -132.95 N
[tex]\mathbf{F_{Bottom, \ y}}[/tex] = 784.8 N
The known parameters in the question are;
The mass of the person, m₁ = 70.0 kg
The length of the ladder, l = 6.00 m
The mass of the ladder, m₂ = 10.0 kg
The distance of the base of the ladder from the house, d = 2.00 m
The point on the roof the ladder rests = A frictionless plastic rain gutter
The location of the center of mass of the ladder, C.M. = 2 m from the bottom of the ladder
The location of the point the person is standing = 3 meters from the bottom
g = The acceleration due to gravity ≈ 9.81 m/s²
The required parameters are;
The magnitudes of the forces on the ladder at the top and bottom
The strategy to be used;
Find the angle of inclination of the ladder, θ
At equilibrium, the sum of the moments about a point is zero
The angle of inclination of the ladder, θ = arccos(2/6) ≈ 70.53 °C
Taking moment about the point of contact of the ladder with the ground, B gives;
[tex]\sum M_B[/tex] = 0
Therefore;
[tex]\sum M_{BCW}[/tex] = [tex]\sum M_{BCCW}[/tex]
Where;
[tex]\sum M_{BCW}[/tex] = The sum of clockwise moments about B
[tex]\sum M_{BCCW}[/tex] = The sum of counterclockwise moments about B
Therefore, we have;
[tex]\sum M_{BCW}[/tex] = 2 × (2/6) × 10.0 × 9.81 + 3.0 × (2/6) × 70 × 9.81
[tex]\sum M_{BCCW}[/tex] = [tex]F_R[/tex] × √(6² - 2²)
Therefore, we get;
2 × (2/6) × 10.0 × 9.81 + 3.0 × (2/6) × 70 × 9.81 = [tex]F_R[/tex] × √(6² - 2²)
[tex]F_R[/tex] = (2 × (2/6) × 10.0 × 9.81 + 3.0 × (2/6) × 70 × 9.81)/(√(6² - 2²)) ≈ 132.95
The reaction force on the wall, [tex]F_R[/tex] ≈ 132.95 N
We note that the magnitude of the reaction force at the roof, [tex]F_R[/tex] = The magnitude of the frictional force of bottom of the ladder on the floor, [tex]F_f[/tex] but opposite in direction
Therefore;
[tex]F_R[/tex] = [tex]-F_f[/tex]
[tex]F_f[/tex] = - [tex]F_R[/tex] ≈ -132.95 N
Similarly, at equilibrium, we have;
∑Fₓ = [tex]\sum F_y[/tex] = 0
The vertical component of the forces acting on the ladder are, (taking forces acting upward as positive;
[tex]\sum F_y[/tex] = -70.0 × 9.81 - 10 × 9.81 + [tex]F_{By}[/tex]
∴ The upward force acting at the bottom, [tex]F_{By}[/tex] = 784.8 N
Therefore;
The magnitudes of the forces at the ladder top and bottom are;
At the top;
[tex]\mathbf{F_{Top, \ x}}[/tex] = [tex]F_R[/tex] ≈ 132.95 N←
[tex]\mathbf{F_{Top, \ y}}[/tex] = 0 (The surface upon which the ladder rest at the top is frictionless)
At the bottom;
[tex]\mathbf{F_{Bottom, \ x}}[/tex] = [tex]F_f[/tex] ≈ -132.95 N →
[tex]\mathbf{F_{Bottom, \ y}}[/tex] = [tex]F_{By}[/tex] = 784.8 N ↑
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The temperature of a body falls from 30°C to 20°C in 5 minutes. The air
temperature is 13°C. Find the temperature after a further 5 minutes.
Answer:
15.88°C I am not 100% sure this is right but I am 98% sure this IS right
1. Estimate the buoyant force that air exerts on a man. (To do this, you can estimate his volume by knowing his weight and by assuming that his weight density is about equal to that of water. Assume his weight is 940 N.) answer in N
2.On a perfect fall day, you are hovering at low altitude in a hot-air balloon, accelerated neither upward nor downward. The total weight of the balloon, including its load and the hot air in it, is 17000 N.
(a) What is the weight of the displaced air?
answer in N
(b) What is the volume of the displaced air?
answer in m^3
Solution :
1. We know that : Buoyant force = weight of the liquid displace
= volume displaced x density of the fluid
Now volume of the man = [tex]$\frac{\text{mass}}{\text{density}}$[/tex]
Mass = weight / g
[tex]$=\frac{940}{9.8}$[/tex]
= 95.92 kg
And density = 1000 [tex]kg/m^3[/tex]
Therefore,
[tex]$\text{volume} = \frac{\text{mass}}{\text{density}}$[/tex]
[tex]$=\frac{95.92}{1000}$[/tex]
= 0.0959 [tex]m^3[/tex]
We know density of air = 1.225 [tex]kg/m^3[/tex]
∴ Mass of air displaced = 0.0959 x 1.225
= 0.1175 kg
Weight of the air displaced = 1.1515 N
Therefore, the buoyant force = 1.1515 N
2). As the balloon is not accelerated, the net force acting on it is zero.
Thus the weight that acts downwards = buoyant force upwards
So, the weight of the air displaced = weight of the balloon
= 17000 N
Therefore, the mass of the air displaced = volume of the air displaced (volume of the balloon) x density of air
[tex]$\frac{17000}{9.8} = \text{volume of air} \times 1.225$[/tex]
[tex]$\text{Volume of air displaced} = \frac{1700}{9.8 \times 1.225}$[/tex]
= 1416.0766 [tex]m^3[/tex]
A 40 kg boy is standing on the edge of a stationary 30-kg platform that is free to rotate without friction. The boy tries to walk around the platform in a counterclockwise direction. As he does:
a. the platform doesn't rotate.
b. the platform rotates in a clockwise direction just fast enough so that the boy remains stationary relative to the ground.
c. the platform rotates in a clockwise direction while the boy goes around in a counterclockwise direction relative to the ground.
d. both go around with equal angular velocities but in opposite directions
Answer:
the correct one is C
Explanation:
To find the answer, let's propose the solution of the problem
We create a system formed by the child and the platform so that all the forces have been internal and the angular momentum is conserved.
Initial instant. Before starting to walk
L₀ = 0
Final moment. After the child is walking
L_f = I₁ w₁ + m r v₂
where index 1 is used for the platform and index 2 for the child
linear and angular velocity are related
v₂ = w₂ r
angular momentum is conserved
0 = I₁ w₁ + m r (w₂ r)
w₁ = [tex]- \frac{m r^2}{I1} \ w_2[/tex]
the moment of inertia of the platform bringing it closer to a disk or cylinder
I₁ = [tex]\frac{1}{2}[/tex] M r²
sustitute
w₁ = [tex]- \frac{2 m }{M} \ w_2[/tex]
W₁ = - [tex]- \frac{2 40}{30} \ w_2 = - \frac{8}{3} \ w_2[/tex]
from here we can see that the platform and the child rotate in the opposite direction and with different angular speeds
when examining the answers the correct one is C
Answer:
Option C (the platform rotates in a clockwise direction while the boy goes around in a counterclockwise direction relative to the ground)Explanation:
relative to the ground the boy moves in a counter clockwise motion , now the boy and the wheel are one system
so by conservation of angular momentum their net sum of angular momentum relative to a point outside the system(say ground) should be zero
so the wheel moves in a clockwise direction , their angular velocity may or may not be same depending on I. so option D is wrong
option B is wrong because relative to ground their angular momentum should be equal and opposite
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A 55kg bungee jumper has fallen far enough that her bungee cord is beginning to stretch and resist her downward motion . Find the ( magnitude and direction ) exerted on her by the bungee cord at an instant when her downward acceleration has a magnitude of 7.1m/s2
Answer:
148.5 N
Explanation:
Given that,
The mass of a bungee jumper, m = 55 kg
The downward acceleration, a = 7.1 m/s²
We need to find the net force acting on the jumper. As it is moving in downward direction, net force is :
T = m(g-a)
Put all the values,
T = 55(9.8 - 7.1)
= 148.5 N
So, the force exerted on the bungee cord is 148.5 N.
Answer:
The downward force is 148.5 N.
Explanation:
mass, m = 55 kg
downwards acceleration, a = 7.1 m/s^2
Let the force is F.
According to the newton's second law
m g - F = m a
F = m (g - a)
F = 55 (9.8 - 7.1)
F = 148.5 N
In Young's double slit experiment, 402 nm light gives a fourth-order bright fringe at a certain location on a flat screen. What is the longest wavelength of visible light that would produce a dark fringe at the same location? Assume that the range of visible wavelengths extends from 380 to 750 nm.
Answer:
λ₂ = 357.3 nm
Explanation:
The expression for double-slit interference is
d sin θ = m λ constructive interference
d sin θ = (m + ½) λ destructive interference.
The initial data corresponds to a constructive interference, they indicate that we are in the fourth order (m = 4), let's look for the separation of the slits
d sin θ = m λ₁
now ask for destructive interference for m = 4
d sin θ = (m + ½) λ₂
we match these two expressions
m λ₁ = (m + ½) λ₂
λ₂ = ( m / m + ½) λλ₁
let's calculate
λ₂ =[tex]\frac{4}{(4.000 +0.5) \ 401}[/tex]
λ₂ = 357.3 nm
A 30-year-old astronaut goes off on a long-term mission in a spacecraft that travels at speeds close to that of light. The mission lasts exactly 20 years as measured on Earth. Biologically speaking, at the end of the mission, the astronaut's age would be:_______.
a) exactly 50 years.
b) exactly 25 years.
c) exactly 30 years.
d) less than 50 years.
e) more than 50 years.
Answer:
I think D) less than 50 years
Biologically speaking, at the end of the mission, the astronaut's age would be less than 50 years. The correct option is d.
Who is an astronaut?An astronaut observes and performs the experiments based on the universe.
A 30-year-old astronaut goes off on a long-term mission in a spacecraft that travels at speeds close to that of light. The mission lasts exactly 20 years as measured on Earth.
Due to special relativity, between space and Earth, both moving with different speeds.
The total age will be less than 30 +20 =50 years. In space, he is moving with speed of light. So, time will move slowly. As measured with respect to Earth, exact time spent in space 20 years will be less on Earth.
So, biologically speaking, at the end of the mission, the astronaut's age would be less than 50 years.
Thus, the correct option is d.
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Two circular coils are concentric and lie in the same plane.The inner coil contains 120 turns of wire, has a radius of 0.012m,and carries a current of 6.0A. The outer coil contains 150turns and has a radius of 0.017 m. What must be the magnitudeand direction (relative to the current in the inner coil) ofthe current in the outer coil, such that the net magnetic field atthe common center of the two coils is zero?
Answer:
[tex]I_2=6.8A[/tex]
Explanation:
From the question we are told that:
Turns of inner coil [tex]N_1=120[/tex]
Radius of inner coil [tex]r_1=0.012m[/tex]
Current of inner coil [tex]I_1=6.0A[/tex]
Turns of Outer coil [tex]N_2=150[/tex]
Radius of Outer coil [tex]r_2=0.017m[/tex]
Generally the equation for Magnetic Field is mathematically given by
[tex]B =\frac{ \mu N I}{2R}[/tex]
Therefore
Condition for the net Magnetic field to be zero
[tex]\frac{N_1* I_1}{( 2 * r_1 )}=\frac{N_2 * I_2}{2 * r_2}[/tex]
[tex]I_2=\frac{(N_1* I_1)*(( 2 * r_2)}{( 2 * r_1)*N_2}[/tex]
[tex]I_2=\frac{(120*6.0)*(( 2 * 0.017)}{( 2 * 0.012)*150}[/tex]
[tex]I_2=6.8A[/tex]
What is the de Broglie wavelength of a red blood cell with a mass of 1.00 * 10-11 g that is moving with a speed of 0.400 cm/s? Do we need to be concerned with the wave nature of the blood cells when we describe the flow of blood in the body?
Answer:
The wavelength is "[tex]=16.5675\times 10^{-18} \ m[/tex]".
Explanation:
Given:
Mass,
m = [tex]1\times 10^{-11} \ g[/tex]
Speed,
V = [tex]0.400 \ cm/s[/tex]
or,
= [tex]0.4\times 10^{-2}[/tex]
According to De Broglie,
The wavelength will be:
⇒ [tex]\lambda = \frac{h}{mV}[/tex]
[tex]=\frac{6.627\times 10^{-34}}{1\times 10^{-11}\times 10^{-3}\times 0.4\times 10^{-2}}[/tex]
[tex]=16.5675\times 10^{-18} \ m[/tex]
So, blood cells move these wavelength.
list at least types of motion
Answer:
These four are rotary, oscillating, linear and reciprocating. Each one moves in a slightly different way and each type of achieved using different mechanical means that help us understand linear motion and motion control.
(I got this off the web so credits to the rightful owner and I hope you have good day :)
Both of these questions are the same but their answers in the answer key are different. Why?
A regulation soccer field for international play is a rectangle with a length between 100 m and a width between 64 m and 75 m. What are the smallest and largest areas that the field could be?
Answer:
The smallest and largest areas could be 6400 m and 7500 m, respectively.
Explanation:
The area of a rectangle is given by:
[tex] A = l*w [/tex]
Where:
l: is the length = 100 m
w: is the width
We can calculate the smallest area with the lower value of the width.
[tex] A_{s} = 100 m*64 m = 6400 m^{2} [/tex]
And the largest area is:
[tex] A_{l} = 100 m*75 m = 7500 m^{2} [/tex]
Therefore, the smallest and largest areas could be 6400 m and 7500 m, respectively.
I hope it helps you!
Answer:
the largest areas that the field could be is [tex]A_l[/tex]=7587.75 m
the smallest areas that the field could be is [tex]A_s[/tex]=6318.25 m
Explanation:
to the find the largest and the smallest area of the field measurement error is to be considered.
we have to find the greatest possible error, since the measurement was made nearest whole mile, the greatest possible error is half of 1 mile and that is 0.5m.
therefore to find the largest possible area we add the error in the mix of the formular for finding the perimeter with the largest width as shown below:
[tex]A_l[/tex]= (L+0.5)(W+0.5)
(100+0.5)(75+0.5) = (100.5)(75.5) = 7587.75 m
To find the smallest length we will have to subtract instead of adding the error factor value of 0.5 as shown below:
[tex]A_s[/tex]= (L-0.5)(W-0.5)
(100-0.5)(64-0.5) = (99.5)(63.5) = 6318.25 m
difference between wavefront and wavelets
Answer:
A wavefront is the locus of all the particles which are in phase. A wavelet is an oscilation that starts from zero, then the amplitude increases and later decreases to zero
A runner has a temperature of 40°c and is giving off heat at the rate of 50cal/s (a) What is the rate of heat loss in watts? (b) How long will it take for this person's temperature to return to 37°c if his mass is 90kg.
Answer:
(a) 209 Watt
(b) 4482.8 seconds
Explanation:
(a) P = 50×4.18
Where P = rate of heat loss in watt
P = 209 Watt
Applying,
Q = cm(t₁-t₂)................ Equation 1
Where Q = amount of heat given off, c = specific heat capacity capacity of human, m = mass of the person, t₁ and t₂ = initial and final temperature.
From the question,
Given: m = 90 kg, t₁ = 40°C, t₂ = 37°C
Constant: c = 3470 J/kg.K
Substtut these values into equation 1
Q = 90×3470(40-37)
Q = 936900 J
But,
P = Q/t.............. Equation 2
Where t = time
t = Q/P............ Equation 3
Given: P = 209 Watt, Q = 936900
Substitute into equation 3
t = 936900/209
t = 4482.8 seconds
A projectile is launched straight upwards at 75 m/s. Three seconds later, its velocity is...?
Answer:
V = V0 + a t
V = 75 - 9.8 * 3 = 45.6 m/s
The final velocity of the projectile after 3 seconds is equal to 45.6 m/s.
What is the equation of motion?The equations of motion can be defined as the relation of the motion of a physical system as the function of time and set up the relationship between the displacement (s), acceleration, velocity (v & u), and time of a moving system.
Given, the initial velocity of the projectile, u = 75 m/s
The time taken by the projectile, t = 3 sec
The acceleration due to gravity upward, g = - 9.8 m/s²
From the first equation of motion we can calculate the final velocity of the projectile:
v = u + at
v = u - gt
v = 75 - 9.8 ×(3)
v = 75 - 29.4
v = 45.6 m/s
Therefore, the final velocity of the projectile after three seconds is 45.6 m/s.
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A transparent. dielectric coating is applied to glass (εr = 4.μr=1, σ= 0) to eliminate the reflection of red light (wavelength in air of 750 nm).
a. What is the required dielectric constant and minimum thickness of the coating?
b. If violet light (wavelength in air of 420 nm) is shone onto this glass coating (6-0). what percentage of the incident power will be reflected?
Answer:
a) Dielectric constant ( λ ) = 750 * 10^-9 m
minimum thickness of coating ( d ) = 187.5 nm
b) 3.6%
Explanation:
Given data:
wavelength of red light in air = 750 nm
εr = 4
μr = 1, σ = 0
a) Determine the required dielectric constant and min thickness of coating used
Refractive index of coating ( n ) = √εr * μr = √4*1 = 2
the refractive index of glass( ng) = 1.5 which is < 2
λ = 750 * 10^-9 m
Dielectric constant ( λ ) = 750 * 10^-9 m
To determine the minimum thickness we will apply the formula below
d = m λ/2n ; where m = 1
∴ d = 750 nm / 2 ( 2 )
= 187.5 nm
minimum thickness of coating ( d ) = 187.5 nm
b) Determine the percentage of the incident power that will be reflected
R = [ ( n-1 / n + 1 ) - ( n - ng / ng + n ) ]^2
= [ ( 2 - 1 / 2 + 1 ) - ( 2 - 1.5 / 1.5 + 2 ) ]^2
= 0.03628 = 3.6%
An electric fan is running on HIGH. After fan has been running for 1.3 minutes, the LOW button is pushed. The fan slows down to 109 rad/sec in 1.87 seconds. The blades of the fan have a diameter of 6.7 meters and their deceleration rate is 4.7 rad/sec2.
What was the initial angular speed of the fan in rev/sec?
ωi =
Answer:
wo = 18.75 rev / s
Explanation:
This is an exercise in endowment kinematics, it indicates that the final angular velocity is w_f = 109 rad / s, the time to reach this velocity is t = 1.87 s and the deceleration a = 4.7 rad / s²
w_f = w₀ - a t
w₀ = w_f + a t
w₀ = 109 + 4.7 1.87
w₀ = 117.8 rad / s
let's reduce to revolutions / s
w₀ = 117.8 rad / s (1 rev / 2pi rad)
w₀ = 18.75 rev / s
A parallal capacitor consists of two Squere plates each of Side 25cm, 3. Omm apart. If a potential difference of 2000volts is applied, calculate the change in the plate with
1.air
2. paper of relative permittity 2.5, fully the space between them E=8.9×10^-12
Answer:poop
Explanation:
poop
signment
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While an elevator of mass 984 kg moves downward, the tension in the supporting cable is a constant 7730 N. Between t= 0 and t=
4.00 s, the elevator's displacement is 5.00 m downward. What is the elevator's speed at t= 4.00 s?
m/s
Answer:
v = 5.15 m/s
Explanation:
At constant velocity, the cable tension will equal the car weight of 984(9.81) = 9,653 N
As the cable tension is less than this value, the car must be accelerating downward.
7730 = 984(9.81 - a)
a = 1.95 m/s²
kinematic equations s = ut + ½at² and v = u + at
-5.00 = u(4.00) + ½(-1.95)4.00²
u = 2.65 m/s the car's initial velocity was upward at 2.65 m/s
v = 2.65 + (-1.95)(4.00)
v = -5.15 m/s
A scientist who studies the tiny microorganisms of the environment .
geologist
meteorologist
microbiologist
entomologist
Question:- A scientist who studies the tiny microorganisms of the environment
Answer:- MicrobiologistExplanation:-
Microbiologist means a person who studies micro sized living organisms
Microbiologist word is combination of two words Micro and biologist
Micro stands for objects which cannot be seen with the naked eyes and are very small in sizeBiologist a person who studies living forms.Answer:
microbiologist i think
hope it helps
please mark Brainliest if you think the answer is correct
If a negatively charged particle is placed inside a uniform electric field the electric force that will act on that particle points in what direction in reference to the electric field lines?
Answer:
opposite direction
Explanation:
An electric field is defined as a physical field which surrounds the electrically charged particles that exerts force on the other particles on the field.
Now when an electron or a negatively charged particle enters a uniform electric field, the electric forces acts on the negatively charged particles and it forces the particle to move in the direction which is opposite to the direction of the field. In an uniform electric field, the field lines are parallel.
Answer:
Explanation:
negatively charged particle is placed inside uniform electric field
The force on the charge due to the electric field is
F = q E
when the charge is negative so the force on the charge is opposite to the direction of electric field.
The electric field is opposite to the force.
2. What is the average speed of an athlete who runs 1500 m in 4 minutes?
Answer:
375 is the answer.
Explanation:
Speed : Distance / Time taken
S: m/ s
s: 1500/4
375 m / s answer
Answer:
375m per minute
Explanation:
if you are looking for a diffrent unit just multiply your answer by however many minutes are in that time frame
Determine the magnitude as well as direction of the electric field at point A, shown in the above figure. Given the value of k = 8.99 × 1012N/C. where, d= 11 cm Q= 12.5 C
Answer:
The electric field is 9.3 x 10^12 N/C and the direction is away from the charge.
Explanation:
charge, Q = 12.5 C
distance, d = 11 cm = 0.11 m
Let the electric field is E.
[tex]E =\frac{K Q}{d^2}\\\\E = \frac{9\times 10^9\times 12.5}{0.11\times 0.11}\\\\E = 9.3\times 10^{12} N/C[/tex]
The direction of electric filed is away from the charge.
Where is the center of mass of homogeneous body which has a regular
Following the definition of the center of mass, "In physics, the center of mass of a distribution of mass in space is the unique point where the weighted relative position of the distributed mass sums to zero."
(see explanation below)
A bicycle tire with a volume of 0.00210 m^3 is filled to its recommended absolute pressure of 495 kPa on a cold winter day when the tire's temperature is -14°C. The cyclist then brings his bicycle into a hot laundry room at 32°C.
a. If the tire warms up while its volume remains constant, will the pressure increase be greater than, less than, or equal to the manufacturer's stated 10% overpressure limit?
b. Find the absolute pressure in the tire when it warms to 32 degrees Celcius at constant volume.
(A) The pressure will be greater than 10% overpressure limit.
(B) The final pressure will be "582.915 kPa".
Given:
Volume,
[tex]V = 0.0021 \ m^3[/tex]Initial pressure,
[tex]P_o= 495 \ kPa[/tex]Initial temperature,
[tex]T_o = -14^{\circ} C[/tex][tex]= 259 \ K[/tex]
Final temperature,
[tex]T = 32^{\circ} C[/tex](B)
Number of moles,
→ [tex]n = (\frac{P_o V}{RT_o} )[/tex]
then,
The final absolute pressure,
→ [tex]P = \frac{nRT}{V}[/tex]
[tex]= (\frac{P_o V}{RT_o} )(\frac{RT}{V} )[/tex]
[tex]=(\frac{T}{T_o} )P_o[/tex]
[tex]= (\frac{305}{259} )\times 495[/tex]
[tex]= 582.915 \ kPa[/tex]
Thus the above approach is correct.
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An ideal spring is hung vertically from the ceiling. When a 2.0-kg mass hangs at rest from it the spring is extended 6.0 cm from its relaxed length. A downward external force is now applied to the mass to extend the spring an additional 10 cm. While the spring is being extended by the force, the work done by the spring is:
a. -3.6 J
b. -3.3 F
c. -3.4 times 10^-5 J
d. 3.3 J
e. 3.6 J
Answer:
b) - 3.3 J
Explanation:
Given;
mass, m = 2 kg
initial extension of the spring, x = 6 cm = 0.06 m
The weight of the mass on the spring;
W = mg
where;
g is acceleration due to gravity = 9.81 m/s²
W = 2 x 9.81
W = 19.62 N
The spring constant is calculated as;
W = kx
k = W/x
k = 19.62 / 0.06
k = 327 N/m
The work done by the spring when it is extended to an additional 10 cm;
work done = force x distance
distance = extension, x = 10 cm = 0.1 m
The work done by the spring opposes the applied force by acting in opposite direction to the force.
W = - Fx
W = - (kx) x
W = - kx²
W = - (327) x (0.1)²
W = - 3.27 J
W ≅ - 3.3 J
Therefore, the work done by the spring by opposing the applied force is -3.3 J
Simple Pendulum: A 34-kg child on an 18-kg swing set swings back and forth through small angles. If the length of the very light supporting cables for the swing is 4.9 m, how long does it take for each complete back-and-forth swing
Answer:
The correct answer is "4.443 sec".
Explanation:
Given:
Mass of child,
= 34 kg
Mass of swing,
= 18 kg
Length,
= 4.9 m
The time period of pendulum will be:
T = [tex]2 \pi \sqrt{4g}[/tex]
= [tex]2 \pi \sqrt{\frac{4.9}{9.8} }[/tex]
= [tex]4.443 \ sec[/tex]
Answer:
The time taken to back and forth is 4.4 s .
Explanation:
Length, L = 4.9 m
let the time period is T.
Acceleration due to gravity, g = 9.8 m/s^2
Use the formula of time period
[tex]T = 2 \pi\sqrt{L}{g}\\\\T = 2 \times 3.14\sqrt{4.9}{9.8}\\\\T = 4.4 s[/tex]
Determine the magnitude as well as direction of the electric field at point A, shown in the above figure. Given the value of k = 8.99 × 1012N/C.
Answer:
Electric field at A = 9.28 x 10¹² N/C
Explanation:
Given:
K = 8.99 x 10¹² N/C
Missing information:
Length = 11 cm = 11 x 10⁻² m
q = 12.5 C
Find:
Electric field at A
Computation:
Electric field = Kq / r²
Electric field at A = [(8.99 x 10¹²)(12.5)] / [11 x 10⁻²]²
Electric field at A = 9.28 x 10¹² N/C
A motor is designed to operate on 117 V and draws a current of 17.7 A when it first starts up. At its normal operating speed, the motor draws a current of 2.78 A. Obtain (a) the resistance of the armature coil, (b) the back emf developed at normal speed, and (c) the current drawn by the motor at one-third normal speed.
Answer:
Resistance of the armature coil = 6.61 ohms
Back emf developed at normal speed = 98.62 V (Approx.)
Current drawn by the motor at one-third normal speed = 12.73 A
Explanation:
Given:
Potential difference V = 117 V
Current = 17.7 A
Motor drawn current = 2.78 A
Find:
Resistance of the armature coil
Back emf developed at normal speed
Current drawn by the motor at one-third normal speed
Computation:
A] Resistance of the armature coil R = V/ I
Resistance of the armature coil = 117 / 17.7
Resistance of the armature coil = 6.61 ohms
B] Back emf developed at normal speed = V- IR
Back emf developed at normal speed = 117 V - (2.78 A)(6.61 ohms)
Back emf developed at normal speed = 117 V - 18.37
Back emf developed at normal speed = 98.62 V (Approx.)
C] Current drawn by the motor at one-third normal speed = 17.7 A - (98.62/3)/(6.61 ohms)
Current drawn by the motor at one-third normal speed = 17.7 - 4.97
Current drawn by the motor at one-third normal speed = 12.73 A
A 771.0-kg copper bar is melted in a smelter. The initial temperature of the copper is 300.0 K. How much heat must the smelter produce to completely melt the copper bar? For solid copper, the specific heat is 386 J/kg • K, the heat of fusion is 205 kJ/kg, and the melting point is 1357 K.
Answer:
4.73 × 10^5
Explanation:
A block with a mass of 0.26 kg is attached to a horizontal spring. The block is pulled back from its equilibrium position until the spring exerts a force of 1.2 N on the block. When the block is released, it oscillates with a frequency of 1.4 Hz. How far was the block pulled back before being released?
Answer:
2
Explanation:
pulling force because of it force
Answer:
5.9 cm
Explanation:
f: frequency of oscillation
frequency of oscillationk: spring constant
frequency of oscillationk: spring constantm: the mass
[tex]f = \frac{1}{2\pi} \sqrt{ \frac{k}{m} } [/tex]
in this problem we know,
F= 1.4 Hz
m= 0.26 kg
By re-arranging the formula we get
[tex]k = {(2\pi \: f )}^{2} m = {(2\pi(1.4hz))}^{2} 0.26kg = 20.1 \frac{n}{m} [/tex]
The restoring force of the spring is:
F= kx
where
F= 1.2 N
k= 20.1 N/m
x: the displacement of the block
[tex]x = \frac{f}{k} = \frac{1.2 \: n}{20.1 \frac{n}{m} } = 0.059m \: = 5.9 \: cm[/tex]
