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)
the lamp cord is 85cm long and comprises cupper wire. Calculate the wire‘s resistance?
radius of a wire is 1.8mmm,Use value of resistivity for Cu as 1.75 × 10-8Ωm.
Answer:
R = 0.0015Ω
Explanation:
The formula for calculating the resistivity of a material is expressed as;
ρ = RA/l
R is the resistance
ρ is the resistivity
A is the area of the wire
l is the length of the wire
Given
l = 85cm = 0.85m
A = πr²
A = 3.14*0.0018²
A = 0.0000101736m²
ρ = 1.75 × 10-8Ωm.
Substitute into the formula
1.75 × 10-8 = 0.0000101736R/0.85
1.4875× 10-8 = 0.0000101736R
R = 1.4875× 10-8/0.0000101736
R = 0.0015Ω
A light bulb, attached (by itself) to an ideal 10 V battery, becomes hotter as time goes on. The bulb's filament is made of tungsten, a metal, which becomes more resistive as its temperature increases. Which statement below is true?
a. As time goes on and the bulb grows hotter, the voltage across the bulb increases so the bulb would grow brighter.
b. As time goes on and the bulb grows hotter, the current through the bulb increases so the bulb would grow brighter.
c. As time goes on and the bulb grows hotter, the voltage across the bulb decreases so the bulb would grow dimmer.
d. As time goes on and the bulb grows hotter, the current through the bulb decreases so the bulb would grow dimmer.
Answer:
the correct one is d
Explanation:
The filament of the bulb fulfills the law of ohm
V = i R
indicate that the filament resistance increases with temperature, as the voltage remains constant if the resistance of the filament increases the current should decrease,
When examining the different statements, the correct one is d
The mass flow rate through a centrifugal compressor is 1 kg/s. If air enters at 1 bar and 288k and leaves at 200 kN/m² and 370k, determine the power of the compressor. Take Cp = 1.103 kJ (kg.K), R = 287 kJ (kg.k)
We have that the power of the compressor is
[tex]H_p=24.242hp[/tex]
From the question we are told that:
Flow rate [tex]W=1kg/s[/tex]
inlet Pressure [tex]P_1=1 bar[/tex]
inlet Temperature [tex]T_1= 288k[/tex]
Outlet Temperature [tex]T_2= 370k[/tex]
Outlet Pressure [tex]P_2=200 kN/m^2=2bars[/tex]
[tex]Cp = 1.103 kJ[/tex]
[tex]R = 287 kJ (kg.k)[/tex]
Generally, the equation for Adiabatic head is mathematically given by
[tex]H=\frac{ZRT_1}{Cp-1/K}[\frac{P_2}{P_1}^{(Cp-1)/Cp}-1][/tex]
Where
[tex]Z=Compressibility\ factor[/tex]
[tex]Z=0.99[/tex]
Therefore
[tex]H=\frac{(0.99)(287)(288)}{(1.103)-1/(1.103)}[\frac{(200)}{(1)}^{((1.103)-1)/(1.103)}-1][/tex]
[tex]H=560925.5958 J/kg[/tex]
[tex]H=5.6*10^5J/kg[/tex]
Generally, the equation for centrifugal compressor power is mathematically given by
[tex]H_p=\frac{WH}{E*33000}[/tex]
Where
E is efficiency (adiabatic)
[tex]E=70\%=0.7[/tex]
Therefore
[tex]H_p=\frac{(1)(5.6*10^5)}{0.7*33000}[/tex]
[tex]H_p=24.242hp[/tex]
In conclusion
The power of the compressor is
[tex]H_p=24.242hp[/tex]
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A rubber ball and a steel ball are dropped from the same height onto a concrete floor. They have the same mass, and lets assume that they both rebound to the same height after hitting the ground. How would the average forces experienced by the balls compare. Group of answer choices The steel ball would experience the greater average force The average forces would be the same The rubber ball would experience the greater average force
Answer:
The average forces would be the same
Explanation:
Both have the same velocity on impact as they fell from the same height.
Both have the same velocity after the bounce because they reach the same height.
Both have the same mass
Both will thus experience the same impulse because both have the same change in momentum.
Therefore both experience the same average force.
Four identical balls are thrown from the top of a cliff, each with the same speed. The
first is thrown straight up, the second is thrown at 30° above the horizontal, the third
at 30° below the horizontal, and the fourth straight down. How do the speeds and
kinetic energies of the balls compare as they strike the ground? Ignore the effects of
air resistance. Explain fully using the concepts from this unit.
The comparison of the speeds and kinetic energy of the identical balls are as follows
The speed and the kinetic energy of the first and fourth ball are equal, while the speed and kinetic energy of the second and third balls are equal
The reason for the above comparison results areas follows;
Known parameters;
First ball is thrown straight up
Second ball is thrown 30° above the horizontal
Third ball it thrown 30° below the horizontal
The fourth ball is thrown straight down
Unknown:
Comparison of the speed and kinetic energy of the four balls
Method:
The kinetic energy, K.E. = (1/2) × m × v²
The velocity of the ball, v = u × sin(θ)
Where;
u = The initial velocity of the ball
θ = The reference angle
For the first ball thrown straight up, we have;
θ = 90°
∴ [tex]v_y[/tex] = u
The final velocity of the ball as it strikes the ground is v₂ = u² + 2gh
Where;
h = The height of the cliff
∴ Kinetic energy of first ball, K.E.₁ = (1/2) × m × (u₁² + 2gh)²
For the second ball thrown 30° to the horizontal, we have;
K.E. = (1/2) × m × ((u×sin30)² + 2·g·h)² = K.E. = (1/2) × m × ((0.5·u)² + 2·g·h)²
Kinetic energy K.E.₂ = (1/2) × m × ((0.5·u₂)² + 2·g·h)²
For the third ball thrown at 30° below the horizontal, we have;K.E. = (1/2) × m × ((u×sin30)² + 2·g·h)² = K.E. = (1/2) × m × ((0.5·u)² + 2·g·h)²
Kinetic energy K.E.₃ = (1/2) × m × ((0.5·u₃)² + 2·g·h)²
For the fourth ball thrown straight down, we have;Kinetic energy K.E.₄ = (1/2) × m × (u₄² + 2gh)²
Therefore, as the ball strike the ground, the speed and the kinetic energy of the first and fourth ball are equal, while the speed and kinetic energy of the second and third balls are equal
u₁ = u₄, K.E₁ = K.E.₄, u₂ = u₃, K.E₂ = K.E.₃
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An object moving with a constant
acceleration changes its velocity from
10ms' to 20 ms' in five seconds. What is the
distance travelled in five seconds
Answer:
Acceleration:
[tex]{ \tt{a = \frac{v - u}{t} }} \\ { \tt{a = \frac{20 - 10}{5} }} \\ { \tt{a = 2 \: m {s}^{ - 2} }}[/tex]
From third equation:
[tex]{ \bf{ {v}^{2} = {u}^{2} + 2as}} \\ { \tt{s = \frac{ {20}^{2} - {10}^{2} }{2 \times 2} }} \\ = { \tt{s = 75 \: m}}[/tex]
Answer:
Formula = m/s
Explanation:
The answer is 10 m / 5 seconds = 2 meters distance
The answer is 20 m / 5 seconds = 4 meters distance
Physics question plz help ASAP
what is the value of x if x-36=5?
Answer:
Therefore, the value of x is 41
Explanation:
x=5+36
x=41
A bicyclist moves along a straight line with an initial velocity vo and slows downs. Which of the following the best describes the signs set for the initial position, initial velocity and the acceleration ?
The sign set after the slowdown of the bicycle will be positive for the position, negative for velocity, and negative for acceleration.
What is velocity?The rate at which an object's position changes when observed from a specific point of view and when measured against a specific unit of time is known as its velocity.
According to Que, when a bicyclist moves in a straight line and slows down, then the velocity decrease as displacement is decreasing, and the acceleration also decreases only displacement increases.
Therefore, the sign set for the position is +ve, for velocity it is -ve, and for acceleration also -ve
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A biker slows down after traveling in a long, straight line at initial velocity v0. Which of the following the best \sdescribes the signs set for the initial position, initial velocity and the acceleration? Initial position Initial velocity Acceleration
A. Positive Negative Negative
B. Positive Positive Negative
C. Negative Positive Negative
D. Negative Negative Positive
E. Negative Negative Negative
A 50-turn coil has a diameter of . The coil is placed in a spatially uniform magnetic field of magnitude so that the face of the coil and the magnetic field are perpendicular. Find the magnitude of the emf, , induced in the coil if the magnetic field is reduced to zero unfiformly
Answer:
EMF = 51.01 Volt
Explanation:
A 50-turn coil has a diameter of 15 cm. The coil is placed in a spatially uniform magnetic field of magnitude 0.500~\text{T}0.500 T so that the plane of the coil makes an angle of 30^\circ30 ∘ with the magnetic field. Find the magnitude of the emf induced in the coil if the magnetic field is reduced to zero uniformly in 0.100~\text{s}0.100 s
We have,
Number of turn in the coil, N = 50
The diameter of the coil, d = 15 cm
Radius, r = 7.5 cm = 0.075 m
Initial magnetic field, [tex]B_i=0.5\ T[/tex]
The plane of the coil makes an angle of 30° with the magnetic field.
The magnetic field reduced to zero in 0.1 seconds
We need to find the emf induced in the coil. We know that, emf is equal to the rate of change of magnetic flux. So,
[tex]\epsilon=\dfrac{BNA\cos\theta}{t}\\\\\epsilon=\dfrac{0.5\times 50\times \pi \times 0.075\cos(30)}{0.1}\\\\\epsilon=51.01\ V[/tex]
So, the induced emf in the coil is 51.01 V.
A 10 n force is applied horizontally on a box to move it 10 m across a frictionless surface. How much work was done to move the box?
Answer:
[tex]\boxed {\boxed {\sf 100 \ J}}[/tex]
Explanation:
We are asked to calculate the work done to move a box.
Work is the product of force and distance or displacement.
[tex]W= F*d[/tex]
A 10 Newton force is applied horizontally on the box. Since the surface is frictionless, there is no force of friction, and the net force is 10 Newtons. The force moves the box 10 meters.
F= 10 N d= 10 mSubstitute the values into the formula.
[tex]W= 10 \ N * 10 \ m[/tex]
Multiply.
[tex]W= 100 \ N*m[/tex]
Let's convert the units. 1 Newton meter is equal to 1 Joule, therefore our answer of 100 Newton meters is equal to 100 Joules.
[tex]W= 100 \ J[/tex]
100 Joules of work was done to move the box.
The magnetic flux that passes through one turn of a 8-turn coil of wire changes to 5.0 Wb from 8.0 Wb in a time of 0.098 s. The average induced current in the coil is 140 A. What is the resistance of the wire
Answer:
Resistance is 1.75 ohms
Explanation:
Magnetic flux:
[tex]{ \phi{ = NBA}}[/tex]
N is number of turns, N = 8
B is magnetic flux
A is area of projection.
From faradays law:
[tex]E = - \frac{ \triangle \phi}{t} [/tex]
where E is the Electro motive force.
But E = IR
where I is current and R is resistance:
[tex]IR = \frac{( \phi_{1} - \phi _{2}) }{t} \\ \\ 140 \times R = \frac{8 \times (8 - 5)}{0.098} \\ \\ R = \frac{24}{0.098 \times 140} \\ \\ resistance = 1.75 \: ohms[/tex]
what is meant by specific latent heat of vaporization of water is -2.26mjkg^-1 or -2.26mj/kg?
Answer:
The specific latent heat of a substance is the amount of energy required to change the state of one kilo of the substance without change in it temperature.The latent heat of vaporization or evaporation is the heat given to some mass to convert if from the liquid to the vapor phase.
Hey guys,I hope u r gonna answer this question fast,SI system is extended from of MKS system.Why? I will be waiting for the answer. Good luck thank u
Answer:
Because SI system has fundamental units of MKS System
Answer:
Explanation: the unit of length ,mass , and time are same in both the system , thus, the SI system is the extended from of MKS system.
Consider two closely spaced and oppositely charged parallel metal plates. The plates are square with sides of length L and carry charges Q and -Q on their facing surfaces. What is the magnitude of the electric field in the region between the plates
Answer:
E_ {total} = [tex]\frac{Q }{L^2 \epsilon_o}[/tex]
Explanation:
In this exercise you are asked to calculate the electric field between two plates, the electric field is a vector
E_ {total} = E₁ + E₂
E_ {total} = 2 E
where E₁ and E₂ are the fields of each plate, we have used that for the positively charged plate the field is outgoing and for the negatively charged plate the field is incoming, therefore in the space between the plates for a test charge the two fields point in the same direction
to calculate the field created by a plate let's use Gauss's law
Ф = ∫ E . dA = q_{int} /ε₀
As a Gaussian surface we use a cylinder with the base parallel to the plate, therefore the direction of the electric field and the normal to the surface are parallel, therefore the scalar product is reduced to the algebraic product.
E 2A = q_{int} / ε₀
where the 2 is due to the surface has two faces
indicate that the surface has a uniform charge for which we can define a surface density
σ = q_{int} / A
q_{int} = σ A
we substitute
E 2A = σ A /ε₀
E = σ / 2ε₀
therefore the total field is
E_ {total} = σ /ε₀
let's substitute the density for the charge of the whole plate
σ= Q / L²
E_ {total} = [tex]\frac{Q }{L^2 \epsilon_o}[/tex]
A cycle track is 500 metres long. A cyclist completes 10 laps (that is, he rides completely round the track 10 times).
a) How many kilometres has the cyclist travelled?
b) On average it took the cyclist 50 second to complete one lap (that it, to ride round just one).
(i) What was the average speed of the cyclist?
(ii) How long in minutes and seconds dit it take the cyclist to complete the 10 laps?
c) Near the end of the run the cyclist put on a spurt. During this spurt it took the cyclist 2 seconds to increase speed from 8 m/s to 12 m/s. What was the cyclist's acceleration during this spurt?
Explanation:
a) D_t = 500m*10laps
D_t = 5000m or 5km
b)
I) v = d/t
v = 500m/50s
v = 10m/s
ii) 10m/s = 5000m/t
t = 5000m/10m/s
t = (500s)*(1min/60s)
t = 8'20" or 8 mins and 20 sec
c) v_f = v_0 + a*t
v_f-v_0 = a*t
a = (v_f-v_0)/t
a = (12m/s-8m/s)/2s
a = (4m/s)/2s
a = 2m/s²
Part(a),
The distance travelled by the cyclist is 5 km.
Part(b),
(i) The average speed is 10 m/s
(ii) The time taken to cover 10 laps is 8 minutes and 20 seconds.
Part(c),
The acceleration is 2 m/s²
What is speed?Speed is defined as the ratio of the time distance travelled by the body to the time taken by the body to cover the distance. Speed is the ratio of the distance travelled by time. The unit of speed in miles per hour.
a) The distance will be calculated as
D = 500m*10laps
D = 5000m or 5km
b) The average speed is calculated as,
I) v = d/t
v = 500m/the 50s
v = 10m/s
The time will be calculated as,
ii) 10m/s = 5000m/t
t = 5000m/10m/s
t = (500s)*(1min/60s)
t = 8'20" or 8 mins and 20 sec
The acceleration is calculated as,
c) vf = v0 + at
vf-v0 = at
a = (vf-v0)/t
a = (12m/s-8m/s)/2s
a = (4m/s)/2s
a = 2m/s²
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A 2000 kg truck has put its front bumper against the rear bumper of a 2500 kg SUV to give it a push. With the engine at full power and good tires on good pavement, the maximum forward force on the truck is 18,000 N.
What is the maximum possible acceleration the truck can give the SUV?
At this acceleration, what is the force of the SUV's bumper on the truck's bumper?
Answer:
The net magnitude of the force of the SUV's bumper on the truck's bumper is 9120 N.
Explanation:
Concepts and reason
The concept required to solve this problem is Newton’s second law of motion.
Initially, write an expression for the force according to the Newton’s second law of motion. Later, rearrange the expression for the acceleration. Finally, substitute the value of the acceleration obtained to find the new force.
Fundamentals
According to the Newton’s second law of motion, the net force is equal to the product of the mass and the acceleration of an object. The expression for the Newton’s second law of motion is as follows:
F = maF=ma
Here, m is mass and a is the acceleration.
(a)
Rearrange the equation F = maF=ma for a.
a = \frac{F}{m}a=
m
F
Substitute 18,000 N for F and \left( {2300{\rm{ kg + 2400 kg}}} \right)(2300kg+2400kg) for m in the equation a = \frac{F}{m}a=
m
F
.
\begin{array}{c}\\a = \frac{{18,000{\rm{ N}}}}{{\left( {2300{\rm{ kg + 2400 kg}}} \right)}}\\\\ = \frac{{18,000{\rm{ N}}}}{{\left( {4700{\rm{ kg}}} \right)}}\\\\ = 3.83{\rm{ m/}}{{\rm{s}}^2}\\\end{array}
a=
(2300kg+2400kg)
18,000N
=
(4700kg)
18,000N
=3.83m/s
2
(b)
Substitute 3.83{\rm{ m/}}{{\rm{s}}^2}3.83m/s
2
for a and 2400 kg for m in the equation F = maF=ma .
\begin{array}{c}\\F = \left( {2400{\rm{ kg}}} \right)\left( {3.83{\rm{ m/}}{{\rm{s}}^2}} \right)\\\\ = 9120{\rm{ N}}\\\end{array}
F=(2400kg)(3.83m/s
2
)
=9120N
Ans: Part a
The maximum possible acceleration the truck can give the SUV is 3.83{\rm{ m/}}{{\rm{s}}^2}3.83m/s
2
.
Part b
The net magnitude of the force of the SUV's bumper on the truck's bumper is 9120 N.
The maximum possible acceleration the truck can give the SUV is equal to 4 m/s².
The force of the SUV's bumper on the truck's bumper is 10000N
What is acceleration?Acceleration of an object can be described as as the change in the velocity of an object w.r.t. time. The acceleration is a vector quantity, contains both magnitude and direction. Acceleration is the second derivative of position w.r.t. time and the first derivative of velocity w.r.t. time.
According to Newton's second law of motion, the force is equal to the product of acceleration and mass of an object.
F = ma
And, a = F/m
Given, the mass of the ruck , m = 2000 Kg
The mass of the SUV, M = 2500 Kg
The total mass of the both = 2000 + 2500 = 4500 Kg
The maximum force on the trick , F = 18000 N
The maximum acceleration of the truck can give the SUV:
[tex]a_{max} = \frac{F_{max}}{m+M}[/tex]
a = 18000/4500
a = 4 m/s²
The force of the SUV's bumper on the truck's bumper will be:
[tex]F_{max} -f= ma_{max}[/tex]
[tex]f= 18000-2000\times 4[/tex]
[tex]f =10000N[/tex]
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A parallel plate capacitor creates a uniform electric field of and its plates are separated by . A proton is placed at rest next to the positive plate and then released and moves toward the negative plate. When the proton arrives at the negative plate, what is its speed
Complete Question
A parallel plate capacitor creates a uniform electric field of 5 x 10^4 N/C and its plates are separated by 2 x 10^{-3}'m. A proton is placed at rest next to the positive plate and then released and moves toward the negative plate. When the proton arrives at the negative plate, what is its speed?
Answer:
[tex]V=1.4*10^5m/s[/tex]
Explanation:
From the question we are told that:
Electric field [tex]B=1.5*10N/C[/tex]
Distance [tex]d=2 x 10^{-3}[/tex]
At negative plate
Generally the equation for Velocity is mathematically given by
[tex]V^2=2as[/tex]
Therefore
[tex]V^2=\frac{2*e_0E*d}{m}[/tex]
[tex]V^2=\frac{2*1.6*10^{-19}(5*10^4)*2 * 10^{-3}}{1.67*10^{-28}}[/tex]
[tex]V=\sqrt{19.2*10^9}[/tex]
[tex]V=1.4*10^5m/s[/tex]
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
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:
what do we mean by thrust?
Answer:
the answer is push example: she thrust her hand into her pocket
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
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Two students try to move a heavy box. One pushes with the force of the 20N while the other pulls with a force of 30N in the same direction. What is the work done by each boy after 10 seconds if the box can\t be moved? Show your equation.
Answer:
Explanation:
Time is not part of the Work equation. That's the only conclusion that you can come to.
Work = Force * distance.
Not enough information is given to go any further. You don't have enough information to calculate the distance.
We don't know if the box can be moved or not. It says heavy. 50 N is really not very much.
I would guess that you are intended to answer that the box didn't move, but it's really hard to tell.
Megan accelerates her skateboard from 0 m/s to 8 m/s in 2 seconds. What is the magnitude of the acceleration of the skateboard?
O 8 m/s^2
O 16 m/s^2
O 2 m/s^2
O 4 m/s^2
Answer:
chk picture for eqn
Explanation:
how much is need to lift a load of 100n placed at a distance of 29 cm from fulcrum if effort is applied at 60cm from the fulcrum on opposite side of the load? calculate mechanical advantage and velocity ratio of the lever
Answer:
206.8965517 n
Explanation:
First, we need to see that 60:29 is 2.078965517:1. Then we need to multiply the energy put 29 cm from the fulcrum by 2.078965517, giving us the end result of our answer.
what additional load will be required to cause the extension of 2.0cm when an elastic wire extend by 1.0cm when a load of 20g range from it
Answer:
The additional load is 20g
A train moving west with an initial velocity of 20 m/s accelerates at 4 m/s2 for 10 seconds. During this tim
moves a distance of
meters.
Answer:
400m
Explanation:
[tex]x = v_{0}t + \frac{at^{2} }{2} \\x = 20*10+\frac{4*10x^{2} }{2} = 400m[/tex]a coach is travelling east wards at 12.6 m/s after 12 second its velocity is 9.5 m/s in the same direction. what is the acceleration and direction of its acceleration?
pls do it with the formula
thx mates :)
[tex]\\ \rm\longmapsto a=\dfrac{v-u}{t}[/tex]
[tex]\\ \rm\longmapsto a=\dfrac{12.6-9.5}{12}[/tex]
[tex]\\ \rm\longmapsto a=\dfrac{3.1}{12}[/tex]
[tex]\\ \rm\longmapsto \overrightarrow{a}=0.25m/s^2[/tex]
Differences between LED and CFL bulb..
Explanation:
CFL bulbs were made to take the place of incandescent bulbs, which generate light as a result of heat. ... LED (light-emitting diode) is a type of bulb that produces light using a narrow band of wavelengths. LED lighting is more energy efficient than CFL bulb.
