Answer:
False
Explanation:
You have a circle so think back to circular motion. Theres 2 directions, centripetal and tangential. The problem tells you there's a constant tangential speed so tangential acceleration is 0. However there is a centripetal acceleration acting on the ball that holds it in its circular motion (i.e. tension, or gravity). Since centripetal is perpendicular to the tangential direction, acceleration and velocity are in different directions.
A wave moves in a rope with a certain wavelength. A second wave is made to move in the same rope with twice the wavelength of the first wave. The frequency of the second wave is _______________ the frequency of the first wave.
Answer:
The frequency of the second wave is half of the frequency of first one.
Explanation:
The wavelength of the second wave is double is the first wave.
As we know that the frequency is inversely proportional to the wavelength of the velocity is same.
velocity = frequency x wavelength
So, the ratio of frequency of second wave to the first wave is
[tex]\frac{f_2}{f_1} =\frac{\lambda _1}{\lambda _2}\\\\\frac{f_2}{f_1} =\frac{\lambda _1}{2\lambda _1}\\\\\frac{f_2}{f_1} =\frac{1}{2}\\\\[/tex]
The frequency of the second wave is half of the frequency of first one.
A body starts from rest and accelerates uniformly at 5m/s. Calculate the time taken by the body to cover a distance of 1km
Answer:
20 seconds
Explanation:
We are given 2 givens in the first statement
v0=0 and a=5
And we are trying to find time needed to cover 1km or 1000m.
So we use
x-x0=v0t+1/2at²
Plug in givens
1000=0+2.5t²
solve for t
t²=400
t=20s
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
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.
A charge of 0.20uC is 30cm from a point charge of 3.0uC in vacuum. what work is required to bring the 0.2uC charge 18cm closer to the 3.0uC charge?
Answer:
The correct answer is "[tex]4.49\times 10^{10} \ joules[/tex]".
Explanation:
According to the question,
The work will be:
⇒ [tex]Work=-\frac{kQq}{R}[/tex]
[tex]=-\frac{1}{4 \pi \varepsilon \times (18-30)\times 3\times 0.2}[/tex]
[tex]=-\frac{1}{4 \pi \varepsilon \times (-12)\times 3\times 0.2}[/tex]
[tex]=\frac{0.3978}{\varepsilon }[/tex]
[tex]=4.49\times 10^{10} \ joules[/tex]
Thus the above is the correct answer.
We have that the workdone is mathematically given as
W=4.49*10e10 J
From the question we are told
A charge of 0.20uC is 30cm from a point charge of 3.0uC in vacuum. what work is required to bring the 0.2uC charge 18cm closer to the 3.0uC charge?WorkdoneGenerally the equation for the workdone is mathematically given as
W=-kQq/R
Therefore
0.3978/ε0 =-1/(4πε0*(18-30)*3*0.2
Hence
W=4.49*10e10 JFor more information on Charge visit
https://brainly.com/question/9383604
Cell phone conversations are transmitted by high-frequency radio waves. Suppose the signal has wavelength 35 cm while traveling through air. What are the
(a) frequency and
(b) wavelength as the signal travels through 3-mm-thick window glass into your room?
Answer:
(a) 8.57 x 10^8 Hz
(b) 23.3 cm
Explanation:
Wavelength = 35 cm = 0.35 m
speed =3 x10^8 m/s
Let the frequency is f.
(a) The relation is
speed = frequency x wavelength
3 x 10^8 = 0.35 x f
f = 8.57 x 10^8 Hz
(b) refractive index of glass is 1.5
The relation for the refractive index and the wavelength is
wavelength in glass= wavelength in air/ refractive index.
Wavelength in glass= 35/1.5 = 23.3 cm
a vessel with mass 10kg intially moving withthe velocicity 12m s along the x axis explodes into three exactly identical pieces Just after the explosion one piece moves with speed 10 m s along the x axis and asecond piece moves with speed 10 m s along the y axis What iis the magnitude of the component of velocity of the third piece along the y axiss
Answer:
Explanation:
Apply law of conservation of momentum along y-axis.
Initially there was no momentum along y-axis. So there will be nil momentum along y-axis again finally.
Let the mass of each piece after breaking be m .
Momentum of piece moving along positive y-axis
= m x 10 = 10m .
Let the component of velocity of third piece along y-axis be v .
Its momentum along the same direction = m v .
Total momentum along y -axis = 10 m + m v
According to law of conservation of momentum
10 m + mv = 0
v = - 10 m/s .
Component of velocity of the third piece along y-axis will be - 10 m/s .
In other words it will be along negative y-axis with speed of 10 m/s.
1.- Que distancia recorrió una carga de 2,5x10-6 coul, generando así un campo eléctrico de 55new/coul.
Answer:
r = 20.22 m
Explanation:
Given that,
Charge,[tex]q=2.5\times 10^{-6}\ C[/tex]
Electric field, [tex]E=55\ N/C[/tex]
We need to find the distance. We know that, the electric field a distance r is as follows :
[tex]E=\dfrac{kq}{r^2}\\\\r=\sqrt{\dfrac{kq}{E}}\\\\r=\sqrt{\dfrac{9\times 10^9\times 2.5\times 10^{-6}}{55}}\\\\r=20.22\ m[/tex]
So, the required distance is 20.22 m.
PLZ help asap :-/
............................
Explanation:
[16][tex]\underline{\boxed{\large{\bf{Option \; A!! }}}} [/tex]
Here,
[tex]\rm { R_1} [/tex] = 2Ω[tex]\rm { R_2} [/tex] = 2Ω[tex]\rm { R_3} [/tex] = 2Ω[tex]\rm { R_4} [/tex] = 2ΩWe have to find the equivalent resistance of the circuit.
Here, [tex]\rm { R_1} [/tex] and [tex]\rm { R_2} [/tex] are connected in series, so their combined resistance will be given by,
[tex]\longrightarrow \rm { R_{(1,2)} = R_1 + R_2} \\ [/tex]
[tex]\longrightarrow \rm { R_{(1,2)} = (2 + 2) \; Omega} \\ [/tex]
[tex]\longrightarrow \rm { R_{(1,2)} = 4 \; Omega} \\ [/tex]
Now, the combined resistance of [tex]\rm { R_1} [/tex] and [tex]\rm { R_2} [/tex] is connected in parallel combination with [tex]\rm { R_3} [/tex], so their combined resistance will be given by,
[tex]\longrightarrow \rm {\dfrac{1}{ R_{(1,2,3)}} = \dfrac{1}{R_{(1,2)}} + \dfrac{1}{R_3} } \\ [/tex]
[tex]\longrightarrow \rm {\dfrac{1}{ R_{(1,2,3)}} = \Bigg ( \dfrac{1}{4} + \dfrac{1}{2} \Bigg ) \;\Omega} \\ [/tex]
[tex]\longrightarrow \rm {\dfrac{1}{ R_{(1,2,3)}} = \Bigg ( \dfrac{1 + 2}{4} \Bigg ) \;\Omega} \\ [/tex]
[tex]\longrightarrow \rm {\dfrac{1}{ R_{(1,2,3)}} = \Bigg ( \dfrac{3}{4} \Bigg ) \;\Omega} \\ [/tex]
Reciprocating both sides,
[tex]\longrightarrow \rm {R_{(1,2,3)}= \dfrac{4}{3} \;\Omega} \\ [/tex]
Now, the combined resistance of [tex]\rm { R_1} [/tex], [tex]\rm { R_2} [/tex] and [tex]\rm { R_3} [/tex] is connected in series combination with [tex]\rm { R_4} [/tex]. So, equivalent resistance will be given by,
[tex]\longrightarrow \rm {R_{(1,2,3,4)}= R_{(1,2,3)} + R_4} \\ [/tex]
[tex]\longrightarrow \rm {R_{(1,2,3,4)}= \Bigg ( \dfrac{4}{3} + 2 \Bigg ) \; \Omega} \\ [/tex]
[tex]\longrightarrow \rm {R_{(1,2,3,4)}= \Bigg ( \dfrac{4 + 6}{3} \Bigg ) \; \Omega} \\ [/tex]
[tex]\longrightarrow \rm {R_{(1,2,3,4)}= \Bigg ( \dfrac{10}{3} \Bigg ) \; \Omega} \\ [/tex]
[tex]\longrightarrow \bf {R_{(1,2,3,4)}= 3.33 \; \Omega} \\ [/tex]
Henceforth, Option A is correct.
_________________________________[17][tex]\underline{\boxed{\large{\bf{Option \; B!! }}}} [/tex]
Here, we have to find the amount of flow of current in the circuit. By using ohm's law,
[tex] \longrightarrow [/tex] V = IR
[tex] \longrightarrow [/tex] 3 = I × 3.33
[tex] \longrightarrow [/tex] 3 ÷ 3.33 = I
[tex] \longrightarrow [/tex] 0.90 Ampere = I
Henceforth, Option B is correct.
____________________________[tex] \tt \purple{Hope \; it \; helps \; you, Army! \heartsuit } \\ [/tex]
A 100-W light bulb is left on for 20.0 hours. Over this period of time, how much energy did the bulb use?
Answer:
Power = Energy/time
Energy = Power xtime.
Time= 20hrs
Power = 100Watt =0.1Kw
Energy = 0.1 x 20 = 2Kwhr.
This Answer is in Kilowatt-hour ...
If the one given to you is in Joules
You'd have to Change your time to seconds
Then Multiply it by the power of 100Watts.
~~~~~NEED HELP ASAP~~~~~
A point on a rotating wheel (thin loop) having a constant angular velocityy of 300 rev/min, the wheel has a radius of 1.5m and a mass of 30kg. (I = mr^2)
a.) Determine the linear regression
b.) At this given angular velocity, what is the rotational kinetic energy?
Answer:
Centripetal Acceleration 18.75 m/s^2, Rotational Kinetic Energy 843.75 J
Explanation:
a Linear acceleration (we cant find tangential acceleration with the givens so we will find centripetal)
a= ω^2*r
ω= 300rev/min
convert into rev/s
300/60= 5rev/s
a= 18.75m/s^2
b) use Krot= 1/2 Iω^2
plug in gives
1/2(30*2.25)(25)= 843.75 J
what is time taken by radio wave to go and return back from communication satellite to earth??
Answer:
Radio waves are used to carry satellite signals. These waves travel at 300,000 km/s (the speed of light). This means that a signal sent to a satellite 38,000 km away takes 0.13 s to reach the satellite and another 0.13 s for the return signal to be received back on Earth.
Explanation:
hope it help
d. On the afternoon of January 15, 1919, an unusually warm day in Boston, a 17.7-m-high, 27.4-m-diameter cylindrical metal tank used for storing molasses ruptured. Molasses flooded into the streets in a 5-m-deep stream, killing pedestrians and horses and knocking down buildings. The molasses had a density of 1600 kg>m3 . If the tank was full before the accident, what was the total outward force the molasses exerted on its sides
Answer:
F = 1.638 x 10⁸ N = 163.8 MN
Explanation:
The total force exerted by the molasses is given as:
F = PA
where,
F = Force exerted by the molasses = ?
P = Pressure = ρgh
ρ = density of molasses = 1600 kg/m³
g = acceleration due to gravity = 9.81 m/s²
h = height of tank = 17.7 m
A = cross-sectional area of tank = πr²
r = radius of tank = 27.4 m/2 = 13.7 m
Therefore,
[tex]F = \rho ghA = \rho gh(\pi r^2)\\\\F = (1600\ kg/m^3)(9.81\ m/s^2)(17.7\ m)(\pi)(13.7\ m)^2[/tex]
F = 1.638 x 10⁸ N = 163.8 MN
A 31 kg block is initially at rest on a horizontal surface. A horizontal force of 83 N is required to set the block in motion. After it is in motion, a horizontal force of 55 N i required to keep it moving with constant speed. From this information, find the coefficients of static and kinetic friction
Answer:
The static and kinetic coefficients of friction are 0.273 and 0.181, respectively.
Explanation:
By Newton's Laws of Motion and definition of maximum friction force, we derive the following two formulas for the static and kinetic coefficients of friction:
[tex]\mu_{s} = \frac{f_{s}}{m\cdot g}[/tex] (1)
[tex]\mu_{k} = \frac{f_{k}}{m\cdot g}[/tex] (2)
Where:
[tex]\mu_{s}[/tex] - Static coefficient of friction, no unit.
[tex]\mu_{k}[/tex] - Kinetic coefficient of friction, no unit.
[tex]f_{s}[/tex] - Static friction force, in newtons.
[tex]f_{k}[/tex] - Kinetic friction force, in newtons.
[tex]m[/tex] - Mass, in kilograms.
[tex]g[/tex] - Gravitational constant, in meters per square second.
If we know that [tex]f_{s} = 83\,N[/tex], [tex]f_{k} = 55\,N[/tex], [tex]m = 31\,kg[/tex] and [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], then the coefficients of friction are, respectively:
[tex]\mu_{s} = \frac{83\,N}{(31\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}[/tex]
[tex]\mu_{s} = 0.273[/tex]
[tex]\mu_{k} = \frac{55\,N}{(31\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}[/tex]
[tex]\mu_{k} = 0.181[/tex]
The static and kinetic coefficients of friction are 0.273 and 0.181, respectively.
The relation of mass m, angular velocity o and radius of the circular path r of an object with the centripetal force is-
a. F = m²wr
b. F = mwr²
c. F = mw²r
d. F = mwr.
Answer:
Correct option not indicated
Explanation:
There are few mistakes in the question. The angular velocity ought to have been denoted with "ω" and not "o" (as also suggested in the options).
The formula to calculate a centripetal force (F) is
F = mv²/r
Where m is mass, v is velocity and r is radius
where
While the formula to calculate a centrifugal force (F) is
F = mω²r
where m is mass, ω is angular velocity and r is radius of the circular path.
From the above, it can be denoted that the relationship been referred to in the question is that of a centrifugal force and not centripetal force, thus the correct option should be C.
NOTE: Centripetal force is the force required to keep an object moving in a circular path/motion and acts inward towards the centre of rotation while centrifugal force is the force felt by an object in circular motion which acts outward away from the centre of rotation.
ACCORDING TO NEWTON'S THIRD LAW EVERY ACTION HAS EQUAL AND OPPOSITE REACTION BUT THEN WHY DON'T WE FLY WHEN WE FART??
Answer:
Your fart only has so much force, not nearly enough to launch you into oblivion. Your fart and you still exert a force onto each other, so I guess, hypothetically, you could fly if you really, really try hard enough. Just make sure you don't try too hard and prolapse as a result :)
