Answer:
(i)8m/s²(ii)40m/s
Explanation:
according to the formula
½at²=s.
then substituting the data
½a•5²=100
a=8m/s²
v=at=8•5=40m/s
Answer:
(I)
[tex]{ \bf{s = ut + \frac{1}{2} a {t}^{2} }} \\ 100 = (0 \times 5) + \frac{1}{2} \times a \times {5}^{2} \\ 200 = 25a \\ { \tt{acceleration = 8 \: m {s}^{ -2} }}[/tex]
(ii)
[tex]{ \bf{v = u + at}} \\ v = 0 + (8 \times 5) \\ { \tt{final \: velocity = 40 \: m {s}^{ - 1} }}[/tex]
A tire is filled with air at 22oC to a gauge pressure of 240 kPa. After driving for some time, if the temperature of air inside the tire is 45oC, what fraction of the original volume of air must be removed to maintain the pressure at 240 kPa?
Answer:
7.8% of the original volume.
Explanation:
From the given information:
Temperature [tex]T_1[/tex] = 22° C = 273 + 22 = 295° C
Pressure [tex]P_1[/tex] = 240 kPa
Temperature [tex]T_2[/tex] = 45° C
At initial temperature and pressure:
Using the ideal gas equation:
[tex]P_1V_1 =nRT_1[/tex]
making V_1 (initial volume) the subject:
[tex]V_1 = \dfrac{nRT_1}{P_1}[/tex]
[tex]V_1 = \dfrac{nR*295}{240}[/tex]
Provided the pressure maintained its rate at 240 kPa, when the temperature reached 45° C, then:
the final volume [tex]V_2[/tex] can be computed as:
[tex]V_2 = \dfrac{nR*318}{240}[/tex]
Now, the change in the volume ΔV = V₂ - V₁
[tex]\Delta V = \dfrac{nR*318}{240}- \dfrac{nR*295}{240}[/tex]
[tex]\Delta V = \dfrac{23nR}{240}[/tex]
∴
The required fraction of the volume of air to keep up the pressure at (240) kPa can be computed as:
[tex]= \dfrac{\dfrac{23nR}{240}}{ \dfrac{295nR}{240}}[/tex]
[tex]= {\dfrac{23nR}{240}} \times { \dfrac{240}{295nR}}[/tex]
[tex]= 0.078[/tex]
= 7.8% of the original volume.
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.
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]
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
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 particle of mass 1.2 mg is projected vertically upward from the ground with a velocity of 1.62 x 10 cm/h. Use the above information to answer the following four questions: 7. The kinetic energy of the particle at time t = 0 s is A. 1.215 x 10-3 J B. 2.430 J C. 1215 J D. 9.72 x 106 J E. OJ (2)
Answer:
K = 0 J
Explanation:
Given that,
The mass of the particle, m = 1.2 mg
The speed of the particle, [tex]v=1.62\times 10\ cm/h[/tex]
We need to find the kinetic energy of the particle at time t = 0 s.
At t = 0 s, the particle is at rest, v = 0
So,
[tex]K=\dfrac{1}{2}mv^2[/tex]
If v = 0,
[tex]K=0\ J[/tex]
So, the kinetic energy of the particle at time t = 0 s is 0 J.
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.
When a rigid body rotates about a fixed axis, all the points in the body have the same Group of answer choices linear displacement. angular acceleration. centripetal acceleration. tangential speed. tangential acceleration.
Answer:
angular acceleration.
Explanation:
Newton's law of universal gravitation states that the force of attraction (gravity) acting between the Earth and all physical objects is directly proportional to the Earth's mass, directly proportional to the physical object's mass and inversely proportional to the square of the distance separating the Earth's center and that physical object.
Generally, when a rigid body is made to rotate about a fixed axis, all the points in the body would typically have the same angular acceleration, angular displacement, and angular speed.
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.
The gravitational field strength due to its planet is 5N/kg What does it mean?
Answer:
The weight of an object is the force on it caused by the gravity due to the planet. The weight of an object and the gravitational field strength are directly proportional. For a given mass, the greater the gravitational field strength of the planet, the greater its weight.
Weight can be calculated using the equation:
weight = mass × gravitational field strength
This is when:
weight (W) is measured in newtons (N)
mass (m) is measured in kilograms (kg)
gravitational field strength (g) is measured in newtons per kilogram (N/kg)
There are two beakers of water on the table. We can compare the average kinetic energy of the water molecules in the two beakers by measuring their
A temperatures.
B volumes.
C densities.
D masses.
Answer: masses
Explanation:
Trust me
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
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 :)
A student claimed that thermometers are useless because a
thermometer always registers its own temperature. How would you
respond?
[
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
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.
