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Related Questions
A mass is attached to the end of a spring and set into oscillation on a horizontal frictionless surface by releasing it from a compressed position. The record of time is started when the oscillating mass first passes through the equilibrium position, and the position of the mass at any time is described by
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The question is incomplete. The complete question is :
A mass is attached to the end of a spring and set into oscillation on a horizontal frictionless surface by releasing it from a compressed position. The record of time is started when the oscillating mass first passes through the equilibrium position, and the position of the mass at any time is described by x = (4.7 cm)sin[(7.9 rad/s)πt].
Determine the following:
(a) frequency of the motion
(b) period of the motion
(c) amplitude of the motion
(d) first time after t = 0 that the object reaches the position x = 2.6 cm
Solution :
Given equation : x = (4.7 cm)sin[(7.9 rad/s)πt].
Comparing it with the general equation of simple harmonic motion,
x = A sin (ωt + Φ)
A = 4.7 cm
ω = 7.9 π
a). Therefore, frequency, [tex]$f=\frac{\omega}{2 \pi}$[/tex]
[tex]$=\frac{7.9 \pi}{2 \pi}$[/tex]
= 3.95 Hz
b). The period, [tex]$T=\frac{1}{f}$[/tex]
[tex]$T=\frac{1}{3.95}[/tex]
= 0.253 seconds
c). Amplitude is A = 4.7 cm
d). We have,
x = A sin (ωt + Φ)
[tex]$x_t=4.7 \sin (7.9 \pi t)$[/tex]
[tex]$2.6 = 4.7 \sin (7.9 \pi t)$[/tex]
[tex]$\sin (7.9 \pi t) = \frac{26}{47}$[/tex]
[tex]$7.9 \pi t = \sin^{-1}\left(\frac{26}{47}\right)$[/tex]
Hence, t = 0.0236 seconds.
A grade 12 Physics student shoots a basketball
from the ground at a hoop which is 2.0 m above
her release. The shot was at a velocity of 10 m/s
and at an angle of 80° to the ground.
a. Determine the vertical velocity of the ball
when it is at the level of the net. You
should get two answers.
Please show ALL steps
Answers
Answer:
7.84 m/s
Explanation:
Height, h = 2 m
Initial velocity, u = 10 m/s
Angle, A = 80°
(a) Let the time taken to go to the net is t.
Use second equation of motion
[tex]h = u t + 0.5 at^2\\\\- 2 = - 10 sin 80 t - 4.9 t^2\\\\4.9 t^2 + 9.8 t - 2 = 0 \\\\t= \frac{- 9.8\pm\sqrt{9.8^2 + 4\times 4.9\times 2}}{9.8}\\\\t = \frac{- 9.8 \pm 11.6}{9.8}\\\\t = - 2.2 s , 0.2 s[/tex]
Time cannot be negative.
So, t = 0.2 s
The vertical velocity at t = 0.2 s is
v = u + at
v = 10 sin 80 - 9.8 x0.2
v = 9.8 - 1.96 = 7.84 m/s
A child is outside his home playing with a metal hoop and stick. He uses the stick to keep the hoop of radius 45.0 cm rotating along the road surface. At one point the hoop coasts downhill and picks up speed. (a) If the hoop starts from rest at the top of the hill and reaches a linear speed of 6.35 m/s in 11.0 s, what is the angular acceleration, in rad/s2, of the hoop? rad/s2 (b) If the radius of the hoop were smaller, how would this affect the angular acceleration of the hoop? i. The angular acceleration would decrease. ii. The angular acceleration would increase. iii. There would be no change to the angular acceleration.
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Answer:
a) [tex] \alpha = 1.28 rad/s^{2} [/tex]
b) Option ii. The angular acceleration would increase
Explanation:
a) The angular acceleration is given by:
[tex] \omega_{f} = \omega_{0} + \alpha t [/tex]
Where:
[tex] \omega_{f} [/tex]: is the final angular speed = v/r
v: is the tangential speed = 6.35 m/s
r: is the radius = 45.0 cm = 0.45 m
[tex]\omega_{0}[/tex]: is the initial angular speed = 0 (the hoop starts from rest)
t: is the time = 11.0 s
α: is the angular acceleration
Hence, the angular acceleration is:
[tex] \alpha = \frac{\omega}{t} = \frac{v}{r*t} = \frac{6.35 m/s}{0.45 m*11.0 s} = 1.28 rad/s^{2} [/tex]
b) If the radius were smaller, the angular acceleration would increase since we can see in the equation that the radius is in the denominator ([tex] \alpha = \frac{v}{r*t} [/tex]).
Therefore, the correct option is ii. The angular acceleration would increase.
I hope it helps you!
What has a wind speed of 240 kph or greater?
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Answer:
SUPER TYPHOON (STY), a tropical cyclone with maximum wind speed exceeding 220 kph or more than 120 knots.
A cylindrical tank with radius 7 m is being filled with water at a rate of 2 m3/min. How fast is the height of the water increasing (in m/min)?
Answers
Answer:
0.013 m/min
Explanation:
Applying,
dV/dt = (dh/dt)(dV/dh)............. Equation 1
Where
V = πr²h................ Equation 2
Where V = volume of the tank, r = radius, h = height.
dV/dh = πr²............ Equation 3
Substitute equation 3 into equation 1
dV/dt = πr²(dh/dt)
From the question,
Given: dV/dt = 2 m³/min, r = 7 m, π = 3.14
Substitute these values into equation 3
2 = (3.14)(7²)(dh/dt)
dh/dt = 2/(3.14×7²)
dh/dt = 0.013 m/min
A ball has a mass of 4.65kg and approximates a ping pong ball of mass 0.060kg that is at rest by striking it in an elastic collision. The initial velocity of the bowling ball is 5.00m/s, determine the final velocities of both masses after the collision.
Answers
Answer:
Look at work
Explanation:
Elastic Collision: Ki=Kf
M1=4.65kg
M2: 0.060kg
v1=5m/s
v2=0m/s
4.65*5+0.060*0=4.65*v1'+0.060*v2'
23.25+0=4.65v1'+0.060v2'
Also since it is an elastic collision we can use
v1+v1'=v2+v2'
4.65+v1'=v2'
4.65+v1'=v2'
Substitute into the earlier equation
23.25=4.65v1'+0.060(4.65+v1')
Expand
23.25=4.65v1'+0.279+0.06v1'
Solve for v1'
22.971=4.71v1'
v1'=4.88m/s
v2'=4.65+4.88=9.53m/s
prove mathematically :
1. v = u + at
2. s = ut+1*2 at
Answers
Answer:
a.v=u+v/2
a.v=s/t
combining two equation we get,
u+v/2=s/t
(u+v)t/2=s
(u+v)t/2=s
{u+(u+at)}t/2=s
(u+u+at)t/2=s
(2u+at)t/2=s
2ut+at^2/2=s
2ut/2+at^2/2=s
UT +1/2at^2=s
proved
a=v-u/t
at=v-u
u+at=v
crushing chalk into powder is and irreversible change. is this example a physical or chemical change?Why?
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Answer:
It is a example of physical change
1. A message signal m(t) has a bandwidth of 5kHz and a peak magnitude of 2V. Estimate the bandwidth of the signal u(t) obtained when m(t) frequency modulates a carrier with a) kf = 10 Hz/V, b) kf = 100 Hz/V, and c) kf = 1000 Hz/V.
Answers
Answer:
3v at 5.3 herts
Explanation:
What is the strength of the magnetic field a distance 4.4 mm above the center of a circular loop of radius 0.8 mm and current 474.1 A
Answers
Answer:
B = 0.118 T
Explanation:
From Biot-Savart Law:
[tex]B = \frac{\mu_o I}{2\pi r}[/tex]
where,
B = strength of magnetic field = ?
μ₀ = 4π x 10⁻⁷ Tm/A
I = current enclosed = 474.1 A
r = radius = 0.8 mm = 8 x 10⁻⁴ m
Therefore,
[tex]B = \frac{(4\pi\ x\ 10^{-7}\ Tm/A)(474.1\ A)}{2\pi(8\ x\ 10^{-4}\ m)}[/tex]
B = 0.118 T
The outer surface of a spacecraft in space has an emissivity of 0.44 and a solar absorptivity of 0.3. If solar radiation is incident on the spacecraft at a rate of 950 W/m2, determine the surface temperature of the spacecraft when the radiation emitted equals the solar energy absorbed.
Answers
Answer:
[tex]T=326.928K[/tex]
Explanation:
From the question we are told that:
Emissivity [tex]e=0.44[/tex]
Absorptivity [tex]\alpha =0.3[/tex]
Rate of solar Radiation [tex]R=0.3[/tex]
Generally the equation for Surface absorbed energy is mathematically given by
[tex]E=\alpha R[/tex]
[tex]E=0.3*950[/tex]
[tex]E=285W/m^2[/tex]
Generally the equation for Emitted Radiation is mathematically given by
[tex]\mu=e(\sigmaT^4)[/tex]
Where
T=Temperature
[tex]\sigma=5.67*10^8Wm^{-2}K_{-4}[/tex]
Therefore
[tex]\alpha*E=e \sigma T^4[/tex]
[tex]0.3*(950)=0.44(5.67*10^-8)T^4[/tex]
[tex]T=326.928K[/tex]
What is the maximum distance the train can travel if it accelerates from rest until it reaches its cruising speed and then runs at that speed for 15 minutes
Answers
The question is incomplete. The complete question is :
A high-speed bullet train accelerates and decelerates at the rate of 4 ft/s^2. Its maximum cruising speed is 90 mi/h. What is the maximum distance the train can travel if it accelerates from rest until it reaches its cruising speed and then runs at that speed for 15 minutes?
Solution :
Given :
Speed of the bullet train, v = 90 mi/h
= [tex]$90 \times \frac{5280}{3600}$[/tex]
= 132 ft/s
Time = 15 minutes
= 15 x 60
= 900 s
Acceleration from rest,
[tex]$a(t) = 4 \ ft/s^2$[/tex]
[tex]$v(t) = 4t + C$[/tex]
Since, v(0) = 0, then C = 0, so velocity is
v(t) = 4t ft/s
Then find the position function,
[tex]$s(t) = \frac{4}{2}t^2 + C$[/tex]
[tex]$=2t^2+C$[/tex]
It is at position 0 when t = 0, so C = 0, and the final position function for only the time it is accelerating is :
[tex]$s(t) = 2t^2$[/tex]
Time to get maximum cruising speed is :
4t = 132
t = 33 s
Distance travelled (at cruising speed) by speed to get the remaining distance travelled.
[tex]$900 \ s \times 132 \ \frac{ft}{s} = 118800 \ ft$[/tex]
Total distance travelled, converting back to miles,
[tex]$2178 + 118800 = 120978\ ft . \ \frac{mi}{5280 \ ft}$[/tex]
= 22.9125 mi
Therefore, the distance travelled is 22.9125 miles
A rope, under a tension of 221 N and fixed at both ends, oscillates in a second-harmonic standing wave pattern. The displacement of the rope is given by y = (0.10 m)(sin πx/2) sin 12πt, where x = 0 at one end of the rope, x is in meters, and t is in seconds.
What are:
a. the length of the rope.
b. the speed of the waves on the rope
c. the mass of the rope
d. If the rope oscillates in a third-harmonic standing wave pattern, what will be the period of oscillation.
Answers
Answer:
sup qwertyasdfghjk
Explanation:
The force an ideal spring exerts on an object is given by , where measures the displacement of the object from its equilibrium position. If , how much work is done by this force as the object moves from to
Answers
Answer:
The correct answer is "1.2 J".
Explanation:
Seems that the given question is incomplete. Find the attachment of the complete query.
According to the question,
x₁ = -0.20 mx₂ = 0 mk = 60 N/mNow,
⇒ [tex]W=\int_{x_1}^{x_2}F \ dx[/tex]
⇒ [tex]=\int_{x_1}^{x_2}-kx \ dx[/tex]
⇒ [tex]=-k \int_{-0.20}^{0}x \ dx[/tex]
By putting the values, we get
⇒ [tex]=-(60)[\frac{x^2}{2} ]^0_{-0.20}[/tex]
⇒ [tex]=-60[\frac{0}{2}-\frac{0.04}{2} ][/tex]
⇒ [tex]=1.2 \ J[/tex]
Find the volume of cuboid of side 4cm. Convert it in SI form
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Answer:
0.000064 cubic meters.
Explanation:
Given the following data;
Length of side = 4 centimeters
Conversion:
100 centimeters = 1 meters
4 cm = 4/100 = 0.04 meters
To find the volume of cuboid;
Mathematically, the volume of a cuboid is given by the formula;
Volume of cuboid = length * width * height
However, when all the sides are equal the formula is;
Volume of cuboid = L³
Volume of cuboid = 0.04³
Volume of cuboid = 0.000064 cubic meters.
When an apple falls towards the earth,the earth moves up to meet the apple. Is this true?If yes, why is the earth's motion not noticeable?
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Answer:
because the mass of the apple is very less compared to the mass of earth. Due to less mass the apple cannot produce noticable acceleration in the earth but the earth which has more mass produces noticable acceleration in the apple. thus we can see apple falling on towards the earth but we cannot see earth moving towards the apple.
In the figure, particle A moves along the line y = 31 m with a constant velocity v with arrow of magnitude 2.8 m/s and parallel to the x axis. At the instant particle A passes the y axis, particle B leaves the origin with zero initial speed and constant acceleration a with arrow of magnitude 0.35 m/s2. What angle between a with arrow and the positive direction of the y axis would result in a collision?
Answers
Answer:
59.26°
Explanation:
Since a is the acceleration of the particle B, the horizontal component of acceleration is a" = asinθ and the vertical component is a' = acosθ where θ angle between a with arrow and the positive direction of the y axis.
Now, for particle B to collide with particle A, it must move vertically the distance between A and B which is y = 31 m in time, t.
Using y = ut + 1/2a't² where u = initial velocity of particle B = 0 m/s, t = time taken for collision, a' = vertical component of particle B's acceleration = acosθ.
So, y = ut + 1/2a't²
y = 0 × t + 1/2(acosθ)t²
y = 0 + 1/2(acosθ)t²
y = 1/2(acosθ)t² (1)
Also, both particles must move the same horizontal distance to collide in time, t.
Let x be the horizontal distance,
x = vt (2)where v = velocity of particle A = 2.8 m/s and t = time for collision
Also, using x = ut + 1/2a"t² where u = initial velocity of particle B = 0 m/s, t = time taken for collision, a" = horizontal component of particle B's acceleration = asinθ.
So, x = ut + 1/2a"t²
x = 0 × t + 1/2(ainsθ)t²
x = 0 + 1/2(asinθ)t²
x = 1/2(asinθ)t² (3)
Equating (2) and (3), we have
vt = 1/2(asinθ)t² (4)
From (1) t = √[2y/(acosθ)]
Substituting t into (4), we have
v√[2y/(acosθ)] = 1/2(asinθ)(√[2y/(acosθ)])²
v√[2y/(acosθ)] = 1/2(asinθ)(2y/(acosθ)
v√[2y/(acosθ)] = ytanθ
√[2y/(acosθ)] = ytanθ/v
squaring both sides, we have
(√[2y/(acosθ)])² = (ytanθ/v)²
2y/acosθ = (ytanθ/v)²
2y/acosθ = y²tan²θ/v²
2/acosθ = ytan²θ/v²
1/cosθ = aytan²θ/2v²
Since 1/cosθ = secθ = √(1 + tan²θ) ⇒ sec²θ = 1 + tan²θ ⇒ tan²θ = sec²θ - 1
secθ = ay(sec²θ - 1)/2v²
2v²secθ = aysec²θ - ay
aysec²θ - 2v²secθ - ay = 0
Let secθ = p
ayp² - 2v²p - ay = 0
Substituting the values of a = 0.35 m/s, y = 31 m and v = 2.8 m/s into the equation, we have
ayp² - 2v²p - ay = 0
0.35 × 31p² - 2 × 2.8²p - 0.35 × 31 = 0
10.85p² - 15.68p - 10.85 = 0
dividing through by 10.85, we have
p² - 1.445p - 1 = 0
Using the quadratic formula to find p,
[tex]p = \frac{-(-1.445) +/- \sqrt{(-1.445)^{2} - 4 X 1 X (-1)}}{2 X 1} \\p = \frac{1.445 +/- \sqrt{2.088 + 4}}{2} \\p = \frac{1.445 +/- \sqrt{6.088}}{2} \\p = \frac{1.445 +/- 2.4675}{2} \\p = \frac{1.445 + 2.4675}{2} or p = \frac{1.445 - 2.4675}{2} \\p = \frac{3.9125}{2} or p = \frac{-1.0225}{2} \\p = 1.95625 or -0.51125[/tex]
Since p = secθ
secθ = 1.95625 or secθ = -0.51125
cosθ = 1/1.95625 or cosθ = 1/-0.51125
cosθ = 0.5112 or cosθ = -1.9956
Since -1 ≤ cosθ ≤ 1 we ignore the second value since it is less than -1.
So, cosθ = 0.5112
θ = cos⁻¹(0.5112)
θ = 59.26°
So, the angle between a with arrow and the positive direction of the y axis would result in a collision is 59.26°.
The cannon on a battleship can fire a shell a maximum distance of 33.0 km.
(a) Calculate the initial velocity of the shell.
Answers
Answer:
v = 804.23 m/s
Explanation:
Given that,
The maximum distance covered by a cannon, d = 33 km = 33000 m
We need to find the initial velocity of the shell. Let it is v. It can be calculated using the conservation of energy such that,
[tex]v=\sqrt{2gh} \\\\v=\sqrt{2\times 9.8\times 33000} \\\\v=804.23\ m/s[/tex]
So, the initial velocity of the shell is 804.23 m/s.
The 1 kg box is sliding along a frictionless surface. It collides with and sticks to the 2 kg box. Afterward, the speed of the two boxes is:__________.
A) 0 m/s
B) 1 m/s
C) 2 m/s
D) 3 m/s
E) Not enough info
Answers
Answer:
The correct option is (E).
Explanation:
Given that,
Mass of object 1, m₁ = 1 kg
Mass of object 2, m₂ = 2 kg
They collides after the collision. We need to find the speed of the two boxes after the collision.
The initial speeds of both boxes is not given. So, we can't put the values of their speeds in the momentum conservation equation.
So, the information is not enough.
I need help with this problem can anybody help me please , it’s physics 2 course
Answers
Answer:
ਹੈਲੋ, ਇੰਡੀਆ ਦਾ ਆਪਣਾ ਵੀਡੀਓ ਐਪ - ਰੋਪੋਸੋ ਤੇ Manjeet Warval ਦਾ ਵੀਡੀਓ ਦੇਖੋ | ਨਾਲ ਹੀ PM ਮੋਦੀ ਦੇ 'ਵੋਕਲ ਫ਼ਾਰ ਲੋਕਲ' ਮੋਹਿਮ ਨੂੰ ਸਫ਼ਲ ਬਣਾਉਣ ਲਈ ਰੋਪੋਸੋ ਤੇ 5 ਕਰੋੜ ਤੋਂ ਜ਼ਿਆਦਾ ਭਾਰਤੀਆਂ ਦੇ ਨਾਲ ਜੋੜੋ| ਹੁਣੇ ਰੋਪੋਸੋ ਐਪ ਡਾਊਨਲੋਡ ਕਰੋ ਅਤੇ 100 ਕੋਇਨਜ਼ ਪਾਣ ਲਈ 24 ਘੰਟੇ ਦੇ ਅੰਦਰ ਸਾਇਨ ਅੱਪ ਕਰੋ|ਹੈਲੋ, ਇੰਡੀਆ ਦਾ ਆਪਣਾ ਵੀਡੀਓ ਐਪ - ਰੋਪੋਸੋ ਤੇ Manjeet Warval ਦਾ ਵੀਡੀਓ ਦੇਖੋ | ਨਾਲ ਹੀ PM ਮੋਦੀ ਦੇ 'ਵੋਕਲ ਫ਼ਾਰ ਲੋਕਲ' ਮੋਹਿਮ ਨੂੰ ਸਫ਼ਲ ਬਣਾਉਣ ਲਈ ਰੋਪੋਸੋ ਤੇ 5 ਕਰੋੜ ਤੋਂ ਜ਼ਿਆਦਾ ਭਾਰਤੀਆਂ ਦੇ ਨਾਲ ਜੋੜੋ| ਹੁਣੇ ਰੋਪੋਸੋ ਐਪ ਡਾਊਨਲੋਡ ਕਰੋ ਅਤੇ 100 ਕੋਇਨਜ਼ ਪਾਣ ਲਈ 24 ਘੰਟੇ ਦੇ ਅੰਦਰ ਸਾਇਨ ਅੱਪ ਕਰੋ|ਹੈਲੋ, ਇੰਡੀਆ ਦਾ ਆਪਣਾ ਵੀਡੀਓ ਐਪ - ਰੋਪੋਸੋ ਤੇ Manjeet Warval ਦਾ ਵੀਡੀਓ ਦੇਖੋ | ਨਾਲ ਹੀ PM ਮੋਦੀ ਦੇ 'ਵੋਕਲ ਫ਼ਾਰ ਲੋਕਲ' ਮੋਹਿਮ ਨੂੰ ਸਫ਼ਲ ਬਣਾਉਣ ਲਈ ਰੋਪੋਸੋ ਤੇ 5 ਕਰੋੜ ਤੋਂ ਜ਼ਿਆਦਾ ਭਾਰਤੀਆਂ ਦੇ ਨਾਲ ਜੋੜੋ| ਹੁਣੇ ਰੋਪੋਸੋ ਐਪ ਡਾਊਨਲੋਡ ਕਰੋ ਅਤੇ 100 ਕੋਇਨਜ਼ ਪਾਣ ਲਈ 24 ਘੰਟੇ ਦੇ ਅੰਦਰ ਸਾਇਨ ਅੱਪ ਕਰੋ|ਹੈਲੋ, ਇੰਡੀਆ ਦਾ ਆਪਣਾ ਵੀਡੀਓ ਐਪ - ਰੋਪੋਸੋ ਤੇ Manjeet Warval ਦਾ ਵੀਡੀਓ ਦੇਖੋ | ਨਾਲ ਹੀ PM ਮੋਦੀ ਦੇ 'ਵੋਕਲ ਫ਼ਾਰ ਲੋਕਲ' ਮੋਹਿਮ ਨੂੰ ਸਫ਼ਲ ਬਣਾਉਣ ਲਈ ਰੋਪੋਸੋ ਤੇ 5 ਕਰੋੜ ਤੋਂ ਜ਼ਿਆਦਾ ਭਾਰਤੀਆਂ ਦੇ ਨਾਲ ਜੋੜੋ| ਹੁਣੇ ਰੋਪੋਸੋ ਐਪ ਡਾਊਨਲੋਡ ਕਰੋ ਅਤੇ 100 ਕੋਇਨਜ਼ ਪਾਣ ਲਈ 24 ਘੰਟੇ ਦੇ ਅੰਦਰ ਸਾਇਨ ਅੱਪ ਕਰੋ|ਹੈਲੋ, ਇੰਡੀਆ ਦਾ ਆਪਣਾ ਵੀਡੀਓ ਐਪ - ਰੋਪੋਸੋ ਤੇ Manjeet Warval ਦਾ ਵੀਡੀਓ ਦੇਖੋ | ਨਾਲ ਹੀ PM ਮੋਦੀ ਦੇ 'ਵੋਕਲ ਫ਼ਾਰ ਲੋਕਲ' ਮੋਹਿਮ ਨੂੰ ਸਫ਼ਲ ਬਣਾਉਣ ਲਈ ਰੋਪੋਸੋ ਤੇ 5 ਕਰੋੜ ਤੋਂ ਜ਼ਿਆਦਾ ਭਾਰਤੀਆਂ ਦੇ ਨਾਲ ਜੋੜੋ| ਹੁਣੇ ਰੋਪੋਸੋ ਐਪ ਡਾਊਨਲੋਡ ਕਰੋ ਅਤੇ 100 ਕੋਇਨਜ਼ ਪਾਣ ਲਈ 24 ਘੰਟੇ ਦੇ ਅੰਦਰ ਸਾਇਨ ਅੱਪ ਕਰੋ|ਹੈਲੋ, ਇੰਡੀਆ ਦਾ ਆਪਣਾ ਵੀਡੀਓ ਐਪ - ਰੋਪੋਸੋ ਤੇ Manjeet Warval ਦਾ ਵੀਡੀਓ ਦੇਖੋ | ਨਾਲ ਹੀ PM ਮੋਦੀ ਦੇ 'ਵੋਕਲ ਫ਼ਾਰ ਲੋਕਲ' ਮੋਹਿਮ ਨੂੰ ਸਫ਼ਲ ਬਣਾਉਣ ਲਈ ਰੋਪੋਸੋ ਤੇ 5 ਕਰੋੜ ਤੋਂ ਜ਼ਿਆਦਾ ਭਾਰਤੀਆਂ ਦੇ ਨਾਲ ਜੋੜੋ| ਹੁਣੇ ਰੋਪੋਸੋ ਐਪ ਡਾਊਨਲੋਡ ਕਰੋ ਅਤੇ 100 ਕੋਇਨਜ਼ ਪਾਣ ਲਈ 24 ਘੰਟੇ ਦੇ ਅੰਦਰ ਸਾਇਨ ਅੱਪ ਕਰੋ|ਹੈਲੋ, ਇੰਡੀਆ ਦਾ ਆਪਣਾ ਵੀਡੀਓ ਐਪ - ਰੋਪੋਸੋ ਤੇ Manjeet Warval ਦਾ ਵੀਡੀਓ ਦੇਖੋ | ਨਾਲ ਹੀ PM ਮੋਦੀ ਦੇ 'ਵੋਕਲ ਫ਼ਾਰ ਲੋਕਲ' ਮੋਹਿਮ ਨੂੰ ਸਫ਼ਲ ਬਣਾਉਣ ਲਈ ਰੋਪੋਸੋ ਤੇ 5 ਕਰੋੜ ਤੋਂ ਜ਼ਿਆਦਾ ਭਾਰਤੀਆਂ ਦੇ ਨਾਲ ਜੋੜੋ| ਹੁਣੇ ਰੋਪੋਸੋ ਐਪ ਡਾਊਨਲੋਡ ਕਰੋ ਅਤੇ 100 ਕੋਇਨਜ਼ ਪਾਣ ਲਈ 24 ਘੰਟੇ ਦੇ ਅੰਦਰ ਸਾਇਨ ਅੱਪ ਕਰੋ|ਹੈਲੋ, ਇੰਡੀਆ ਦਾ ਆਪਣਾ ਵੀਡੀਓ ਐਪ - ਰੋਪੋਸੋ ਤੇ Manjeet Warval ਦਾ ਵੀਡੀਓ ਦੇਖੋ | ਨਾਲ ਹੀ PM ਮੋਦੀ ਦੇ 'ਵੋਕਲ ਫ਼ਾਰ ਲੋਕਲ' ਮੋਹਿਮ ਨੂੰ ਸਫ਼ਲ ਬਣਾਉਣ ਲਈ ਰੋਪੋਸੋ ਤੇ 5 ਕਰੋੜ ਤੋਂ ਜ਼ਿਆਦਾ ਭਾਰਤੀਆਂ ਦੇ ਨਾਲ ਜੋੜੋ| ਹੁਣੇ ਰੋਪੋਸੋ ਐਪ ਡਾਊਨਲੋਡ ਕਰੋ ਅਤੇ 100 ਕੋਇਨਜ਼ ਪਾਣ ਲਈ 24 ਘੰਟੇ ਦੇ ਅੰਦਰ ਸਾਇਨ ਅੱਪ ਕਰੋ|
Explanation:
The 2 capacitors in the middle are connected in parallel so simply add their capacitance together:
[tex]5.0\:\mu\text{F} + 8.0\:\mu\text{F} = 13.0\:\mu \text{F}[/tex]
Now we have 3 capacitors connected in series so their equivalent capacitance [tex]C_{eq}[/tex] is
[tex]\dfrac{1}{C_{eq}} = \dfrac{1}{10.0\:\mu \text{F}} + \dfrac{1}{13.0\:\mu \text{F}} + \dfrac{1}{9.0\:\ mu \text{F}} [/tex]
or
[tex]C_{eq} = 3.5\:\mu \text{F}[/tex]
A system gains 1500J of heat and 2200J of work is done by the system on its surroundings. Determine the change in internal energy of the system
Answers
Answer:
-700
formula is heat gained - work done
The change in internal energy if A system gains 1500J of heat and 2200J of work is done by the system on its surroundings, is 700 joules.
What is Energy?Energy is the ability to perform work in physics. It could exist in several different forms, such as potential, kinetic, thermal, electrical, chemical, radioactive, etc.
Additionally, there is heat and work, which is energy being transferred from one body to another. Energy is always assigned based on its nature once it has been transmitted. Thus, heat transmitted may manifest as thermal energy while work performed may result in mechanical energy.
Given:
A system gains 1500J of heat and 2200J of work is done by the system on its surroundings,
Calculate the change in internal energy as shown below,
The change in internal energy = heat gained - work done
The change in internal energy = 1500 - 2200
The change in internal energy = -700 J
Thus, the change in internal energy is 700 joules.
To know more about Energy:
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What is the energy equivalent of an object with a mass of 2.5 kg? 5.5 × 108 J 7.5 × 108 J 3.6 × 1016 J 2.25 × 1017 J
Answers
Answer:
E = m c^2 = 2.5 * (3 * 10E8)^2 = 2.25 * 10E17 Joules
Answer:
The answer is D. 2.25 × 1017 J
Explanation:
got it right on edge 2021
What is the escape speed on a spherical asteroid whose radius is 517 km and whose gravitational acceleration at the surface is 0.636 m/s2
Answers
Answer:
810.94 m/s
Explanation:
Applying,
v = √(2gR)............. Equation 1
Where v = escape velocity of the spherical asteroid, g = acceleration due to gravity, R = radius of the earth
From the question,
Given: g = 0.636 m/s², R = 517 km = 517000 m
Substitute these values into equation 1
v = √(2×0.636×517000)
v = √(657624)
v = 810.94 m/s
Hence, the escape velocity is 810.94 m/s
g a horizontal wheel of radius is rotating about a vertical axis. What is the magnitude of the resultant acceleration of a bug that is hanging tightl on the rim of the wheel
Answers
Answer:
a = w² r
Explanation:
In this exercise, indicate that the wheel has angular velocity w, the worm experiences the same angular velocity if it does not move, and has an acceleration towards the center of the circle, according to Newton's second law, called the centripetal acceleration.
a = v² / r
angular and linear variables are related
v = w r
we substitute
a = w² r
where r is the radius of the wheel
A beam of light has a wavelength of 549nm in a material of refractive index 1.50. In a different material of refractive index 1.07, its wavelength will be:_________.
Answers
Explanation:
someone to check if the answer is correct
Two spheres are rolling without slipping on a horizontal floor. They are made of different materials, but each has mass 5.00 kg and radius 0.120 m. For each the translational speed of the center of mass is 4.00 m/s. Sphere A is a uniform solid sphere and sphere B is a thin-walled, hollow sphere. Part B How much work, in joules, must be done on the solid sphere to bring it to rest? Express your answer in joules. VO AE4D ? J WA Request Answer Submit Part C How much work, in joules, must be done on the hollow sphere to bring it to rest? Express your answer in joules. Wa Request
Answers
Answer:
Explanation:
Moment of inertia of solid sphere = 2/5 m R²
m is mass and R is radius of sphere.
Putting the values
Moment of inertia of solid sphere I₁
Moment of inertia of hollow sphere I₂
Kinetic energy of solid sphere ( both linear and rotational )
= 1/2 ( m v² + I₁ ω²) [ ω is angular velocity of rotation ]
= 1/2 ( m v² + 2/5 m R² ω²)
= 1/2 ( m v² + 2/5 m v²)
=1/2 x 7 / 5 m v²
= 0.7 x 5 x 4² = 56 J .
This will be equal to work to be done to stop it.
Kinetic energy of hollow sphere ( both linear and rotational )
= 1/2 ( m v² + I₂ ω²) [ ω is angular velocity of rotation ]
= 1/2 ( m v² + 2/3 m R² ω²)
= 1/2 ( m v² + 2/3 m v²)
=1/2 x 5 / 3 m v²
= 0.833 x 5 x 4² = 66.64 J .
This will be equal to work to be done to stop it.
two resistors with resistance values 4.5 ohms and 2.3 ohms are connected in series or parallel across a potential difference of 30V to a light bulb find the current flowing through the light bulb in both cases
Answers
Answer:
Look at work
Explanation:
Series:
I is the same for all resistors so just find the value of Req. In series Req= R1+R2+...+Rn. So here it will be 4.5+2.3=6.8ohms. Ieq=Veq/Req=4.41A. And since current is the same across all resistors the current to the lightbulb is 4.41A.
Parallel:
V is the same for all resistors so start of by finding Req. In parallel, Ieq=I1+I2+...+In. So I1= 30/4.5= 6.67A and I2= 13.04A. Ieq= 6.67+13.04= 19.71A.
The working substance of a certain Carnot engine is 1.50 mol of an ideal monatomic gas. During the isothermal expansion portion of this engine's cycle, the volume of the gas doubles, while during the adiabatic expansion the volume increases by a factor of 5.7. The work output of the engine is 940 J in each cycle. Compute the temperatures of the two reservoirs between which this engine
operates.
Answers
Answer:
The hot temperature is 157.5 K
The cold temperature is 48.8 K
Explanation:
Step 1: Data given
The working substance of a certain Carnot engine is 1.50 mol of an ideal monatomic gas.
The volume increases by a factor of 5.7
The work output of the engine is 940 J in each cycle.
During the isothermal expansion portion of this engine's cycle, the volume of the gas doubles. This means V2 = 2*V1 (and V4 = 2*V3)
Step 2:For a carnot engine:
V2/V1 = V4/V3
Work = nR((T1)ln(V2/V1) - (T2)ln(V4/V3))
⇒with Work = the work done in the cycle = 940J
⇒with n = the number of moles = 1.50 moles
⇒with R = the gas constant = 8.314 J/mol*K
⇒with T1 = the hot temperature
⇒With T2⇒ the cold temperature
where R = 8.31 J/mol K Gas Constant
940J = 1.5moles * 8.314 J/mol*K * (T1*ln(2) - T2*ln(2)))
940 = 1.5 * 8.314 ln(2) * (T1-T2)
(T1-T2) = 940 / (1.5*8.314*ln(2))
(T1-T2) = 108.7K
For the reversible adiabatic expansion: T2 = T1*(V1/V2)^(R/Cv). Where V2/V1 = 5.7 (Because during the adiabatic expansion the volume increases by a factor of 5.7)
For a monatomic ideal gas, Cv = 3/2R
When we combine both, we'll have:
T2 = T1*(1/5.7)^(R/3/2R)
T2 = T1*(1/5.7)^(2/3)
T2= T1 * 0.31
Since we know that (T1-T2) = 108.7K
we have:
T1 - 0.31T1= 108.7K
0.69T1 = 108.7K
T1 = 157.5K
T2 = 157.5*0.31 = 48.8K
The block in the drawing has dimensions L0×2L0×3L0,where L0 =0.2 m. The block has a thermal conductivity of 150 J/(s·m·C˚). In drawings A, B, and C, heat is conducted through the block in three different directions; in each case the temperature of the warmer surface is 35 ˚C and that of the cooler surface is 16 ˚C Determine the heat that flows in 6 s for each case.
Answers
Answer:
1140 J, 6840 J, 10260 J
Explanation:
Lo x 2 Lo x 3 Lo, Lo = 0.2 m, K = 150 J/(s · m · C˚) , T = 35 ˚C, T' = 16 ˚C,
time, t = 6 s
The heat conducted is
[tex]H = \frac{K A (T - T') t}{d}\\\\H = \frac{150\times 3\times 0.2\times 0.2\times (35-16) \times 6}{3\times 0.2}\\\\H = 1140 J[/tex]
The heat conducted is
[tex]H = \frac{K A (T - T') t}{d}\\\\H = \frac{150\times 3\times 0.2\times 2\times0.2\times (35-16) \times 6}{3\times 0.2}\\\\H = 6840 J[/tex]
The heat conducted is
[tex]H = \frac{K A (T - T') t}{d}\\\\H = \frac{150\times 3\times 0.2\times 2\times0.2\times (35-16) \times 6}{2\times 0.2}\\\\H = 10260 J[/tex]
You need to calculate the volume of berm that has a starting cross-sectional area of 118 SF, and an ending cross-sectional area of 245 SF. The berm is 300 ft long and is assumed to taper evenly between the two cross-sectional areas, what is the calculated volume of the berm in cubic feet
