Answer:
cobina
Explanation:
me 2
the two ropes are used to vertically lower a 255 kg piano from exactly 4 m form a seocnd sotry window to the ground how much work is done by each of the three forces
Complete Question
The Question diagram is attached below
Answer:
a) [tex]W_{Fg}= 12500 Nm[/tex]
b) [tex]W_{T_1}= - 6339.3Nm[/tex]
c) [tex]W_{T_2}= - 3662.8Nm[/tex]
Explanation:
From the question we are told that:
Mass [tex]m=255kg[/tex]
Distance [tex]d=4m[/tex]
Generally the equation for Work done is mathematically given by
[tex]W=F*d[/tex]
For [tex]F_g[/tex]
[tex]W_{Fg}=2500 x 5.3[/tex]
[tex]W_{Fg}= 12500 Nm[/tex]
For [tex]T_1[/tex]
[tex]W_{T_1}= - {1830 sin(60) x 4}[/tex]
[tex]W_{T_1}= - 6339.3Nm[/tex]
For [tex]T_2[/tex]
[tex]W_{T_2}= - {1295 sin(45) x 4}[/tex]
[tex]W_{T_2}= - 3662.8Nm[/tex]
Q.3. The equivalent resistance across AB is:
(a)1
(c)2
(b)3
(d)4
Answer:
1 ohm
Explanation:
First of all, the equivalent resistance for two resistors (r₁ and r₂) in parallel is given by:
1 / Eq = (1 / r₁) + (1 / r₂)
The equivalent resistance for resistance for two resistors (r₁ and r₂) in series is given by:
Eq = r₁ + r₂
Hence as we can see from the circuit diagram, 2Ω // 2Ω, and 2Ω // 2Ω, hence:
1/E₁ = 1/2 + 1/2
1/E₁ = 1
E₁ = 1Ω
1/E₂ = 1/2 + 1/2
1/E₂ = 1
E₂ = 1Ω
This then leads to E₁ being in series with E₂, hence the equivalent resistance (E₃) of E₁ and E₂ is:
E₃ = E₁ + E₂ = 1 + 1 = 2Ω
The equivalent resistance (Eq) across AB is the parallel combination of E₃ and the 2Ω resistor, therefore:
1/Eq = 1/E₃ + 1/2
1/Eq = 1/2 + 1/2
1/Eq = 1
Eq = 1Ω
You are asked to build an LC circuit that oscillates at 13 kHz. In addition, you are told that the maximum current in the circuit can be 0.14 A and the maximum energy stored in the capacitor must be 1.4x10-5 J. What value of inductor and capacitor should you choose
Answer:
a)[tex]L=0.00142H[/tex]
b) [tex]C=2.65*10^{-12}[/tex]
Explanation:
From the question we are told that:
Frequency[tex]F=13kHz[/tex]
Current [tex]I=0.14A[/tex]
Capacitor[tex]C_e=1.4*10^{-5}J[/tex]
Generally the equation for Energy in the inductor is mathematically given by
Where L is now subject
[tex]L=\frac{2C_e}{I^2}[/tex]
[tex]L=\frac{2*1.4*10^{-5}}{(0.14)^2}[/tex]
[tex]L=0.00142H[/tex]
Generally the equation for Value of Capacitor is mathematically given by
[tex]C=\frac{1}{(2 \pi f)^2} L[/tex]
[tex]C=\frac{1}{(2 3.142 13*10^3Hz)^2} *0.00142[/tex]
[tex]C=2.65*10^{-12}[/tex]
As the speed of a particle approaches the speed of light, the momentum of the particle Group of answer choices approaches zero. decreases. approaches infinity. remains the same. increases.
Answer:
approaches infinity
Explanation:
There are two momentums, the classical momentum which is equal to the product of mass and velocity, and the relativistic momentum, the one we should look at when we work with high speeds, and this happens because massive objects have a speed limit, in this case, we are approaching the speed of light, so we need to work with the relativistic momentum instead of the classical momentum.
The relativistic momentum can be written as:
[tex]p = \frac{1}{\sqrt{1 - \frac{u^2}{c^2} } } *m*u[/tex]
where
u = speed of the object relative to the observer, in this case we have that u tends to c, the speed of light.
m = mass of the object
c = speed of light.
So, as u tends to c, we will have:
[tex]\lim_{u \to c} p = \frac{1}{\sqrt{1 - \frac{u^2}{c^2} } } *m*u[/tex]
Notice that when u tends to c, the denominator on the first term tends to zero, thus, the relativistic momentum of the object will tend to infinity.
Then the correct option is infinity, as the particle speed approaches the speed of light, the relativistic momentum of the particle tends to infinity.
A physics instructor wants to project a spectrum of visible-light colors from 400 nm to 700 nm as part of a classroom demonstration. She shines a beam of white light through a diffraction grating that has 600 lines per mm, projecting a pattern on a screen 2.9 m behind the grating.
Required:
How wide is the spectrum corresponding to m=1?
Answer:
Dr = 263 10⁻⁶ m
Explanation:
The diffraction pattern for constructive interference is described by
a sin θ = m λ
in this it indicates that the order of diffraction is m = 1
Let's use a direct proportion rule to find the separation of two slits. If there are 600 lines in 1 me, what is the distance between 2 slits
a = 2 lines 1/600
a = 2/600
a = 3.33 10⁻³ mm = 3.33 10⁻⁴ cm
let's use trigonometry
tan θ = y / L
as the measured angles are small
tan θ = sin θ / cos θ sin θ
sin θ = y / L
we substitute
a y/L = λ
y = λ L / a
for λ = 400 10-9 m
I = 400 10⁻⁹ 2.9 / 3.33 10⁻³
i = 346.89 10⁻⁶ m
f
or λ = 700 nm
y_f = 700 10⁻⁻⁹ 2.9 / 3.33 10⁻³
y_f = 609.609 10⁻⁶ m
the separation of this spectrum
Δr = v_f - i
Dr = (609.609 - 346) 10 ⁻⁶
Dr = 263 10⁻⁶ m
A rocket explodes into two fragments, one 25 times heavier than the other. The magnitude of the momentum change of the lighter fragment is A) 25 times as great as the momentum change of the heavier fragment. B) The same as the momentum change of the heavier fragment. C) 1/25 as great as the momentum change of the heavier fragment. D) 5 times as great as the momentum change of the heavier fragment. E) 1/4 as great as the momentum change of the heavier fragment.
Answer:
B) The same as the momentum change of the heavier fragment.
Explanation:
Since the initial momentum of the system is zero, we have
0 = p + p' where p = momentum of lighter fragment = mv where m = mass of lighter fragment, v = velocity of lighter fragment, and p' = momentum of heavier fragment = m'v' where m = mass of heavier fragment = 25m and v = velocity of heavier fragment.
0 = p + p'
p = -p'
Since the initial momentum of each fragment is zero, the momentum change of lighter fragment Δp = final momentum - initial momentum = p - 0 = p
The momentum change of heavier fragment Δp' = final momentum - initial momentum = p' - 0 = p' - 0 = p'
Since p = -p' and Δp = p and Δp' = -p = p ⇒ Δp = Δp'
So, the magnitude of the momentum change of the lighter fragment is the same as that of the heavier fragment.
So, option B is the answer
An object of mass 80 kg is released from rest from a boat into the water and allowed to sink. While gravity is pulling the object down, a buoyancy force of 1/50 times the weight of the object is pushing the object up (weight=mg). If we assume that water resistance exerts a force on the abject that is proportional to the velocity of the object, with proportionality constant 10 N-sec/m, find the equation of motion of the object. After how many seconds will the velocity of the object be 40 m/s? Assume that the acceleration due to gravity is 9.81 m/sec^2.
Answer:
a) Fnet = mg - Fb - Fr
b) 8.67 secs
Explanation:
mass of object = 80 kg
Buoyancy force = 1/50 * weight ( 80 * 9.81 ) = 15.696
Proportionality constant = 10 N-sec/m
a) Calculate equation of motion of the object
Force of resistance on object due to water = Fr ∝ V
= Fr = Kv = 10 V
Given that : Fb( due to buoyancy ) , Fr ( Force of resistance ) acts in the positive y-direction on the object while mg ( weight ) acts in the negative y - direction on the object.
Fnet = mg - Fb - Fr
∴ Equation of motion of the object ( Ma = mg - Fb - Fr )
b) Calculate how long before velocity of the object hits 40 m/s
Ma = mg - Fb - Fr
a = 9.81 - 0.1962 - 0.125 V = 9.6138 - 0.125 V
V = u + at ---- ( 1 )
u = 0
V = 40 m/s
a = 9.6138 - 0.125 V
back to equation 1
40 = 0 + ( 9.6138 - 0.125 (40) ) t
40 = 4.6138 t
∴ t = 40 / 4.6138 = 8.67 secs
A cylindrical swimming pool has a radius 2m and depth 1.3m .it is completely filled with salt water of specific gravity 1.03.The atmospheric preassure is 1.013 x 10^5 Pa.
a.calculate the density of salt water.
Answer:
the density of the salt water is 1030 kg/m³
Explanation:
Given;
radius of the cylindrical pool, r = 2 m
depth of the pool, h = 1.3 m
specific gravity of the salt water, γ = 1.03
The atmospheric pressure, P₀ = 1.013 x 10⁵ Pa
Density of fresh water, [tex]\rho _w[/tex] = 1000 kg/m³
The density of the salt water is calculated as;
[tex]Specific \ gravity \ of \ salt\ water \ (\gamma _s_w) = \frac{density \ of \ salt \ water \ (\rho_{sw})}{density \ of \ fresh \ water \ (\rho_{w})} \\\\1.03 = \frac{\rho_{sw}}{1000 \ kg/m^3}\\\\\rho_{sw} = 1.03 \times 1000 \ kg/m^3\\\\\rho_{sw} = 1030 \ kg/m^3[/tex]
Therefore, the density of the salt water is 1030 kg/m³
System A consists of a mass m attached to a spring with a force constant k; system B has a mass 2m attached to a spring with a force constant k; system C has a mass 3m attached to a spring with a force constant 6k; and system D has a mass m attached to a spring with a force constant 4k. Rank these systems in order of decreasing period of oscillation.
Answer:
T₂ > T₁ > T₃ >T₄
Explanation:
In a simple harmonic motion the angular velocity is
w = [tex]\sqrt{\frac{k}{m} }[/tex]
angular velocity and period are related
w = 2π / T
we substitute
T = [tex]2 \pi \ \sqrt{\frac{m}{k} }[/tex]
let's find the period for each case
a) mass m
spring constant k
T₁ = 2π [tex]\sqrt{\frac{m}{k} }[/tex]
b) mass 2m
spring constant k
T₂ = 2π [tex]\sqrt{\frac{2m}{k} }[/tex]
T₂ = T₁ √2
T₂ = T₁ 1.41
c) masses 3m
spring constant 6k
T₃ = 2π [tex]\sqrt{\frac{3m}{6k} }[/tex]
T₃ = 2π [tex]\sqrt{\frac{m}{k} } \ \sqrt{0.5}[/tex]
T₃ = T₁ 0.707
d) mass m
spring constant 4k
T₄ = 2π [tex]\sqrt{ \frac{m}{4k} }[/tex]
T₄ = 2π [tex]\sqrt{\frac{m}{k} } \ \sqrt{0.25}[/tex]
T₄ = T₁ 0.5
now let's order the periods in decreasing order
T₂ > T₁ > T₃ >T₄
Light of frequency f falls on a metal surface and ejects electrons of maximum kinetic energy K by the photoelectric effect. If the frequency of this light is doubled, the maximum kinetic energy of the emitted electrons will be
The question is incomplete, the complete question is;
Light of frequency f falls on a metal surface and ejects electrons of maximum kinetic energy K by the photoelectric effect.
Part A If the frequency of this light is doubled, the maximum kinetic energy of the emitted electrons will be If the frequency of this light is doubled, the maximum kinetic energy of the emitted electrons will be
K/2.
K.
2K.
greater than 2K.
Answer:
2K
Explanation:
Given that the kinetic energy of photo electrons is given by;
K= E -Wo
Where;
K = kinetic energy
E= energy of incident photon
Wo = work function
But;
E= hf
Wo = fo
h= Plank's constant
f= frequency of incident photon
fo= Threshold frequency
So:
K= hf - hfo
Where the frequency of incident light is doubled;
K= 2hf - hfo
Hence, maximum kinetic energy of the emitted electrons in this case will be 2K
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%
A planet of mass m = 4.25 x 1024 kg orbits a star of mass M = 6.75 x 1029 kg in a circular path. The radius of the orbits R = 8.85 x 107 km. What is the orbital period Tplanet of the planet in Earth days?
285.3 days
Explanation:
The centripetal force [tex]F_c[/tex] experienced by the planet is the same as the gravitational force [tex]F_G[/tex] so we can write
[tex]F_c = F_G[/tex]
or
[tex]m\dfrac{v^2}{R} = G\dfrac{mM}{R^2}[/tex]
where M is the mass of the star and R is the orbital radius around the star. We know that
[tex]v = \dfrac{C}{T} = \dfrac{2\pi R}{T}[/tex]
where C is the orbital circumference and T is orbital period. We can then write
[tex]\dfrac{4\pi^2R}{T^2} = G\dfrac{M}{R^2}[/tex]
Isolating [tex]T^2[/tex], we get
[tex]T^2 = \dfrac{4\pi^2R^3}{GM}[/tex]
Taking the square root of the expression above, we get
[tex]T = 2\pi \sqrt{\dfrac{R^3}{GM}}[/tex]
which turns out to be [tex]T = 2.47×10^7\:\text{s}[/tex]. We can convert this into earth days as
[tex]T = 2.47×10^7\:\text{s}×\dfrac{1\:\text{hr}}{3600\:\text{s}}×\dfrac{1\:\text{day}}{24\:\text{hr}}[/tex]
[tex]\:\:\:\:\:= 285.3\:\text{days}[/tex]
Physics question plz help ASAP
Two loudspeakers placed 6.00 m apart are driven in phase by an audio oscillator having a frequency range from 1908 Hz to 2471 Hz. A point P is located 4.70 m from one loudspeaker and 3.60 m from the other speaker. At what frequency of the oscillator does the sound reaching point P interfere constructively
Answer:
2164 Hz
Explanation:
Since point P is 4.70 m away from one speaker and 3.60 m away from the other speaker, the path length difference ΔL = 4.70 m - 3.60 m = 1.1 m.
The path length difference ΔL = nλ for a constructive interference where n is an integer and λ = wavelength of sound from oscillator = v/f where v = speed of sound in air = 340 m/s and f = frequency of sound from oscillator.
So, ΔL = nλ = nv/f
So, the frequency from the oscillator is f = nv/ΔL
Substituting the values of v and ΔL into the equation, we have
f = nv/ΔL
f = n340 m/s/1.1 m
f = n309.09 /s
f = 309.09n Hz
We now insert values of n that will gives us a frequency in the range 1908 Hz to 2471 Hz.
The value of n that will give us a frequency in the range is n = 7
So, when n = 7,
f = 309.09n Hz
f = 309.091 × 7 Hz
f = 2163.64 Hz
f ≅ 2164 Hz
So, the frequency of the oscillator that will produce a constructive interference at P is 2164 Hz.
friction always opposes the _____
Answer:
Friction always opposes the motion
I HOPE ITS RIGHT IF NOT THEN SORRYHAVE A GREAT DAY :)
Which of the following is not an example of approximate simple harmonic motion
Answer:
where are the options
it's not full question
If a car drives 10 mph South, this is an example of a:
A. Displacement
B. Velocity
C. Speed
D. Distance
Answer:
杰杰伊杜杜杜伊格富尔杰迪耶赫分离福音
Explanation:
莱德利 · 赫耶尔伊 3uritievrirjrirhruebwkwieheoo2hfjcbvi3hd
Answer:
B velocity
Explanation:
An observer on Earth sees spaceship 1 fly by at 0.80c. 6.0 years later, the observer on Earth sees spaceship 2 fly by at 0.80c, traveling in the same direction as the first. Both spaceships continue to travel with constant velocities. An observer in spaceship 1 observes Earth to pass spaceship 2 ____ years after Earth passed spaceship 1.
Answer:
[tex]t_2=10[/tex]
Explanation:
From the question we are told that:
Velocity of both spaceships [tex]V=0.8c[/tex]
Time [tex]t_1=6[/tex]
Generally the equation for time of spaceship 2 through earth is mathematically given by
[tex]t_2=\frac{t_1}{\sqrt{1-v^2}}[/tex]
[tex]t_2=\frac{6}{\sqrt{1-0.8^2}}[/tex]
[tex]t_2=10[/tex]
which characteristic of nuclear fission makes it hazardous?
Answer:The radioactive waste
Explanation:Fission is the splitting of a heavy unstable nucleus into two Lighter nuclei
A uniform electric field of strength E points to the right. An electron is fired with a velocity v0 to the right and travels a distance d before coming to a stop. An second electron is then fired upwards through the same field at a velocity of v0. After the electron moving vertical has traveled vertically upwards a distance d, how far will it have moved horizontally?
Answer:
[tex]D_l=d[/tex]
Explanation:
From the question we are told that:
The Electric field of strength direction =Right
The Velocity of The First Electron=V_0
The Velocity of The Second Electron=V_0
Therefore
[tex]V_{e1}=V_{e2}[/tex]
Generally, the equation for the Horizontal Displacement of electron is mathematically given by
[tex]D=\frac{at^2}{2}[/tex]
Where
Acceleration is given as
[tex]a=\frac{V_o}{2d}[/tex]
And
Time
[tex]T=\frac{d}{v_0}[/tex]
Therefore horizontal displacement towards the left is
[tex]D_l=\frac{(\frac{V_o}{2d})(\frac{d}{v_0})^2}{2}[/tex]
[tex]D_l=d[/tex]
List what sources of uncertainty go into calculating the wavelength of the laser (no explanation necessary here). (b) Accurately report the uncertainties for these quantities. (c) Explain which of these contributes the most to the final uncertainty on the laser wavelength
Answer:
thanks for da 5points hoi
Explanation: thanks dawg
There can be uncertainty in calculating the wavelength of a laser light due to experimental errors
All measurements have an uncertainty, in the case of direct measurements the uncertainty is equal to the precision of the given instrument.
What are uncertainity in measuring ?Uncertainty means the range of possible values within which the true value of the measurement lies.
What are errors?
The deviation in the value of the measured quantity from the actual quantity or true value is called an error
(a) For the calculation of wavelength of laser light , the sources which can lead to uncertainty are
1. least count of measuring instruments like spectrometer or interferometer
2. Parallax error in the measurement
3. Error in identifying the order of fringes
4.. unable to identify the accurate reading of Vernier or circular scales present in the measuring instruments.
5. Propagating errors
What is least count?
The least count of a measuring instrument is the smallest and accurate value in the measured quantity that can be measured by instrument.
What is propagating error?When you have derived variables, that is, when measurements are made with different instruments, each with a different uncertainty, the way to find the uncertainty or error is that all the errors add up. which increases the uncertainty
b. The uncertainty in measurement due to least count depends on the instrument used for measurement f wavelength. A Michelson's
interferometer has the least count of .0001mm. whereas spectrometer has a least count of 0.5⁰. Hence uncertainty in the measurement by Michelson's interferometer is very less as compared to any other instrument.
C. The maximum uncertainty arises due to the least count , as all other errors can be minimized by taking an average value of many observations but the least count of an instrument do not change so uncertainty within the least count arises.
To know more on errors and uncertainty in measurements here
https://brainly.com/question/24652157
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You have a 1 W light bulb in your lab. It puts out light of only 1 frequency. The wavelength of this light is 500nm. you set up a detector with a surface area of 1 square centimeter facing the light source at a distance of 100m.
Required:
a. Find the number of photons hitting the detector every second.
b. What is the maximum E field of the E M wave hitting the detector?
c. What is the maximum value of the B field of this E M wave?
d. How far away would you have to place the detector to only receive 1 photon per second from the light bulb?
Answer:
a) # _photon = 2.5 10¹⁸ photons / s, b) E = 10⁻² N / C, c) B = 3 10⁻¹¹ T
d) r= 2 10⁹ m
Explanation:
a) Let's solve this exercise in part, let's start by finding the energy of each photon using the Planck relation
E₀ = h f
c = λ f
E₀ = h c /λ
E₀ = 6.63 10⁻³³⁴ 3 10⁸/500 10⁻⁹
E₀ = 3.978 10⁻⁻¹⁹ J
Let's use a direct ratio rule to find the number of photons
#_foton = E / Eo
#_fototn = 1 / 3.978 10⁻¹⁹
# _photon = 2.5 10¹⁸ photons / s
b) The intensity received by the detector is related to the electric field
I = E²
Let's look for the intensity that the detector receives, suppose that the emission is shapeless throughout the space
I = P / A
P = I A
Let's use index 1 for the point on the bulb and index 2 for the point on the detector.
The area of a sphere is
A = 4π r²
P = I₁ A₁ = I₂ A₂
I₁ r₁² = I₂ r₂²
I₂ = I₁ r₁²/r₂²
I₂ = I₁ 1 / 100²
I₂ = I₁ 10⁻⁴
we must know the intensity at the output of the bulb suppose that I₁ = 1 J
I₂ = 10⁻⁴ J
let's look for the electric field
E =√I
E = √10⁻⁴
E = 10⁻² N / C
c) for the calculation of the magnetic field we use that the field is in phase
E / B = c
B = E / c
B = 10⁻² / 3 10⁸
B = 3 10⁻¹¹ T
d) Let's use a direct proportions rule if we fear 2.5 10¹⁸ photons in an area A = 4π R² where R = 100 m how many photons are there in the area of the detector r = 1 cm, A’= 10⁻⁴ m²
#_photons = 2.5 10¹⁸ A_detector / A_sphere
#_photons = 2.5 1018 10-4 / 4π 10⁴
#_photons = 2 10⁹ photons in the detector area
for the number of photons to decrease to 1, the radius of the sphere must be 2 10⁹ m
a stone is thrown vertically upwards with a velocity of 20 m per second what will be its velocity when it reaches a height of 10.2 m
Answer:
Explanation:
Here's the info we have:
initial velocity is 20 m/s;
final velocity is our unknown;
displacement is -10.2 m; and
acceleration due to gravity is -9.8 m/s/s. Using the one-dimensional equation
v² = v₀² + 2aΔx and filling in accordingly to solve for v:
[tex]v=\sqrt{(20)^2+2(-9.8)(-10.2)}[/tex] Rounding to the correct number of sig fig's to simplify:
[tex]v=\sqrt{400+2.0*10^2}[/tex] to get
v = [tex]\sqrt{600}=20\frac{m}{s}[/tex] If you don't round like that, the velocity could be 24, or it could also be 24.5 depending on how your class is paying attention to sig figs or if you are at all.
So either 20 m/s or 24 m/s
What do you understand by moment of inertia and torque?
Word limit 50-60
Please don't copy from any sources. You can rewrite. Plagiarism will be check. Thank you.
Answer:
Moment of inertia, in physics, quantitative measure of the rotational inertia of a body—i.e., the opposition that the body exhibits to having its speed of rotation about an axis altered by the application of a torque (turning force). The axis may be internal or external and may or may not be fixed.
1. A 2.7-kg copper block is given an initial speed of 4.0 m/s on a rough horizontal surface. Because of friction, the block finally comes to rest. (a) If the block absorbs 85% of its initial kinetic energy as internal energy, calculate its increase in temperature.
Answer:
ΔT = 0.017 °C
Explanation:
According to the given condition, the change in internal energy of the block must be equal to 85% of its kinetic energy:
Change in Internal Energy = (0.85)(Kinetic Energy)
[tex]mC\Delta T = (0.85)\frac{1}{2}mv^2\\\\C\Delta T = (0.425)v^2\\\\\Delta T = \frac{0.425v^2}{C}[/tex]
where,
ΔT = increase in temperature = ?
v = speed of block = 4 m/s
C = specific heat capacity of copper = 389 J/kg.°C
Therefore,
[tex]\Delta T = \frac{(0.425)(4\ m/s)^2}{389}\\\\[/tex]
ΔT = 0.017 °C
17- How much work is needed for a climber in order to climb 45 m height, where his weight is 70 kg. also, calculate the power required to climb the height in 30 minutes ? g= 9,8 m.sec
Answer:
Work Done= 3150J
Power= 1.75W
Explanation:
Work Done= Force x the distance travelled in the direction of the force (W= f x d)
Weight is a force, i think the qn. stated it wrongly, it should be 70N not 70kg.
Work Done= 70 x 45
=3150J
Power= Work Done/Time
=3150/(30x60)
*convert minutes to seconds since the S.I. unit of Power is joules/seconds(J/s) or watts(W)
=1.75W
Why is oiling done time and again in a sewing machine?
Answer:
to prevent friction on the surfaces
Answer:
Explanation:
Oiling reduces friction between parts with relative motion between them.
Repeated oiling is needed as the film of oil reducing the friction becomes thinner with time as some of the oil gets pushed off of the areas of motion where it can no longer be useful.
Oil also becomes oxidized which reduces the oil's ability to decrease friction.
Oil can also be fouled by dirt, lint or other material. This added material becomes coated in oil and typically gets sequestered away from the moving parts reducing the oil available for lubrication purposes.
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.
calculate the force on an object with mass of 50kg and gravity of 10
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]
Learn more about acceleration, here:
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