Answer:
The value of acceleration equals [tex]0.0556m/s^{2}[/tex]
Explanation:
The distances covered in each of the three phases are calculated as under
1) Distance covered in accelerating phase for a period of 5 minutes or 300 seconds ( 1 minute = 60 seconds)
Using second equation of kinematics
[tex]s=ut+\frac{1}{2}at^{2}\\\\s_{1}=0\times 300+\frac{1}{2}\times a\times (300)^{2}\\\\s_{1}=0.5\times a\times (300)^{2}[/tex]
2) Distance covered in 5 minutes while travelling at a constant speed which is The speed after 5 minutes of travel is obtained by first equation of kinematics as
[tex]v=u+at\\\\v=0+a\times 300\\\\v=300a[/tex]
Thus distance traveled equals
[tex]s_{2}=300a\times 300\\\\s_{2}=(300)^{2}a[/tex]
3)
The phase when the car stops the distance it covers during this phase can be obtained using third equation of kinematics as
[tex]v^{2}=u^{2}+2as\\\\\therefore s=\frac{v^{2}-u^{2}}{2a}\\\\s_{3}=\frac{0-(300a)^{2}}{2\times -a}\\\\s_{3}=\frac{300^{2}a}{2}[/tex]
Now the sum of [tex]s_{1}+s_{2}+s_{3}[/tex] equals 10 kilometers or 10000 meters.
Thus we get
[tex]0.5\times a\times 300^{2}+300^{2}a+300^{2}\times \frac{a}{2}=10000\\\\a(0.5\times 300^{2}+300^{2}+0.5\times 300^{2})=10000\\\\\therefore a=\frac{10000}{(0.5\times 300^{2}+300^{2}+0.5\times 300^{2})}=0.0556m/s^{2}[/tex]
To solve for the acceleration, we can use the constant acceleration equation and find the distance traveled during each phase.
Explanation:The constant acceleration equation can be used to solve this problem. The total distance traveled by the train is equal to the sum of the distances traveled during each phase.
In the first phase, the train starts at rest and accelerates for 5 minutes. The distance traveled during this phase can be found using the equation s = 0.5at^2, where s is the distance, a is the acceleration, and t is the time. Since the initial velocity is 0, we can simplify the equation to s = 0.5at^2.
In the second phase, the train travels at a constant speed for 5 minutes. The distance traveled during this phase is equal to the product of the speed and the time.
In the third phase, the train decelerates with a negative acceleration. The distance traveled during this phase can be found using the equation s = ut + 0.5at^2, where u is the initial velocity, a is the acceleration, and t is the time. Since the final velocity is 0, we can simplify the equation to s = ut.
The total distance traveled is given as 10 km. By plugging in the values for the distances and using the given times for each phase, we can solve for the acceleration, a, and find that a = 0.014 km/s^2.
Learn more about Calculating acceleration and distance here:https://brainly.com/question/32817533
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Air around the center of surface low pressure systems in the Northern Hemisphere is spinning ______ and _______ the system's center. O counter-clockwise; converging towards O counter-clockwise; diverging away from O clockwise; diverging away from O clockwise; converging towards
Answer:
Clockwise, converging towards
Explanation:
At the center of surface , system with low pressure in Northern hemisphere have air around the center that rotates in anti clockwise direction.
As a result of low pressure, the air is directed slightly inwards thus converges towards the center of the system.
In the high pressure systems, air rotates in clockwise direction and diverges away from the center of the system.
An apple falls (from rest) from a tree. It hits the ground at a speed of about 4.9 m/s. What is the approximate height (in meters) of the tree above the ground? The magnitude of the gravitational acceleration g = 9.8 m/s2 Enter your answer in meters. Keep 2 decimal places.
Answer:
The inicial height of the apple is 1.22 meters
Explanation:
Using the equation for conservarion of mechanical energy:
[tex]E=V+K=constant[/tex]
[tex]K_i=\frac{1}{2}mv_i^2[/tex] where v is the velocity
[tex]V=mgh[/tex]where h is the height
We equate the initial mechanical energy to the final:
Since [tex]v_0=0\ and h_f=0 [/tex]:
[tex]\frac{1}{2}mv_0^2+mgh_0= \frac{1}{2}mv_f^2+mgh_f\\gh_0= \frac{1}{2}v_f^2[/tex]
Solving for h:
[tex]h_0=\frac{4.9^2}{2g}= 1.22 m[/tex]
The neutron mass is 1.675 x 10-27 kg. Express the proton mass in eV/c, keeping 4 significant figures B) The mass of the electron is 9.109 x 10-31 kg. Express this in eV/c . C) What is the mass of the hydrogen atom in eV/c??
Answer:
Explanation:
given,
mass of neutron = 1.675 × 10⁻²⁷ kg
1 kg = 5.58 × 10³⁵ eV/c²
a) mass of proton = 1.675 × 10⁻²⁷ × 5.58 × 10³⁵ = 9.34 × 10⁸ eV/c²
b) mass of electron = 9.109 x 10⁻³¹ × 5.58 × 10³⁵ = 5.08 × 10⁵eV/c²
c) mass of hydrogen = 1.675 × 10⁻²⁷ × 5.58 × 10³⁵ = 9.34 × 10⁸ eV/c²
The acceleration of automobiles is often given in terms of the time it takes to go from 0 mi/h to 60 mi/h. One of the fastest street legal cars, the Bugatti Veyron Super Sport, goes from 0 to 60 in 2.70 s. To the nearest integer, what is its acceleration in m/s^2? Please show work.
Answer:
9.934 m/s²
Explanation:
Given:
Initial speed of the Bugatti Veyron Super Sport = 0 mi/h
Final speed of the Bugatti Veyron Super Sport = 60 mi/h
Now,
1 mi/h = 0.44704 m / s
thus,
60 mi/h = 0.44704 × 60 = 26.8224 m/s
Time = 2.70 m/s
Now,
The acceleration (a) is given as:
[tex]a=\frac{\textup{Change in speed}}{\textup{Time}}[/tex]
thus,
[tex]a=\frac{26.8224 - 0 }{2.70}[/tex]
or
a = 9.934 m/s²
A package is dropped from a helicopter moving upward at 15m/s. If it takes 10 s before the package strikes the ground, how high above the ground was the ground was the package when it was released if air resistance is negligible?
(A) 408 m
(B) 272 m
(C) 204 m
(D) 340 m
Answer:
The right answer is (D) 340m
Explanation:
If the package is dropped from a helicopter moving upward and the air resistance is negligible, the only acting force in the package is the gravity force. Therefore you can consider this as a Vertical Projectile Motion problem.
The initial speed of the package V₀ is 15m/s.
The acceleration of the package G is 9.8m/s².
The initial hight is Y₀. Therefore:
Y(t)=Y₀+V₀·t-0.5·G·t²
But we know than T(10s)=0m
0m=Y₀+150m-490m
Y₀=340m
A motorist drives north for 35.0 min at 85.0 km/h and then stops for 15.0 min. He then continues north, traveling 130 km in 2.00 h. (a) what is his total displacement? b) what is his average velocity?
Final answer:
The motorist's total displacement is 179.58 km north, and his average velocity is approximately 63.38 km/h north for the entire trip.
Explanation:
Displacement and Average Velocity Calculation
For the given problem, we can find the motorist's total displacement by adding the distances traveled in each segment of the journey, all in the same direction (north). First, the motorist drives north for 35.0 minutes at 85.0 km/h. The distance for this part of the trip is:
Distance = Speed × Time = 85.0 km/h × (35.0 min / 60 min/h) = 49.58 km (rounded to 2 decimal places)
The next segment involves a stop, so the displacement remains unchanged.
Finally, the motorist drives an additional 130 km north.
Therefore, the total displacement is the sum of these distances: Total Displacement = 49.58 km + 130 km = 179.58 km north.
To find the average velocity, we need to consider the total displacement and the total time, including the stop:
Total time = Driving time + Stopping time = (35.0 min + 2.00 h + 15.0 min) = 2.833 hours (2 hours and 50 minutes).
Average Velocity = Total Displacement / Total Time = 179.58 km / 2.833 h ≈ 63.38 km/h north.
The motorist's total displacement is 179.58 km north, and his average velocity over the entire trip is approximately 63.38 km/h north.
A man pushes a lawn mower on a level lawn with a force of 195 N. If 37% of this force is directed downward, how much work is done by the man in pushing the mower 5.7 m?
Final answer:
The work done by the man in pushing the lawn mower is 699.245 J, calculated by determining the horizontal force component and multiplying by the distance pushed.
Explanation:
To calculate how much work is done by the man in pushing the lawn mower, we need to consider only the component of the force that acts in the direction of the movement. Since 37% of the 195 N force is directed downward, only the remaining 63% is contributing to the horizontal movement. Therefore, the horizontal component of the force is 0.63 × 195 N = 122.85 N.
The formula to calculate work (W) is W = force (F) × distance (d) × cosine(θ), where θ is the angle between the force and the direction of movement. In this case, the force and movement are in the same direction, so θ = 0 and cosine(θ) = 1. Thus, the work done is:
W = 122.85 N × 5.7 m × 1 = 699.245 J
In terms of energy expended while pushing a lawn mower, this work is a relatively small amount when compared to a person's daily intake of food energy.
HOW DOES LIGHT BEHAVE MOVING FROM A MORE DENSE MEDIUM TO ALESS
DENSE MEDIUM?
Answer:
Light travels faster in a less dense medium
Explanation:
In a less dense medium, light will travel faster than it did in the more dense medium (an example of this transition can be light going from water to air).
This is because in a less dense medium there ares less particles, so light finds less resistence in its path.
The opposite happens in a more dense medium, there are more particles so light will travel slower because more particles will get in the way.
In summary, the more dense the medium, the more particles it has, and the slower that light will travel in it.
Two objects are dropped at the same time from heights of 0.49 m and 1.27 m. How long after the first object hits the ground does the second object hit the ground?
Answer:
After 0.19288 sec second object hit the ground after hitting first object
Explanation:
We have given two objects are dropped from at the same time from height 0.49 m and 1.27 m
As the object is dropped so its initial velocity will be zero u = 0 m/sec
For object 1 height h = 0.49 m
From second equation of motion
[tex]h=ut+\frac{1}{2}gt^2[/tex]
[tex]0.49=0\times t+\frac{1}{2}\times 9.8\times t^2[/tex]
[tex]t^2=0.1[/tex]
t = 0.31622 sec
For second object
[tex]h=ut+\frac{1}{2}gt^2[/tex]
[tex]1.27=0\times t+\frac{1}{2}\times 9.8\times t^2[/tex]
[tex]t^2=0.1[/tex]
t = 0.5091 sec
So difference in time = 0.5091-0.31622 = 0.19288 sec
So after 0.19288 sec second object hit the ground after hitting first object
What fraction of all the electrons in a 25 mg water
dropletmust be removed in order to obtain a net charge of 40
nC?
Answer:
[tex]9.11\times 10^{-15}.[/tex]
Explanation:
The water droplet is initially neutral, it will obtain a 40 nC of charge when a charge of -40 nC is removed from the water droplet.
The charge on one electron, [tex]\rm e=-1.6\times 10^{-19}\ C.[/tex]
Let the N number of electrons have charge -40 nC, such that,
[tex]\rm Ne=-40\ nC\\\Rightarrow N=\dfrac{-40\ nC}{e}=\dfrac{-40\times 10^{-9}\ C}{-1.6\times 10^{-19}\ C}=2.5\times 10^{11}.[/tex]
Now, mass of one electron = [tex]\rm 9.11\times 10^{-31}\ kg.[/tex]
Therefore, mass of N electrons = [tex]\rm N\times 9.11\times 10^{-31}=2.5\times 10^{11}\times 9.11\times 10^{-31}=2.2775\times 10^{-19}\ kg.[/tex]
It is the mass of the of the water droplet that must be removed in order to obtain a charge of 40 nC.
Let it is m times the total mass of the droplet which is [tex]25\ \rm mg = 25\times 10^{-6}\ kg.[/tex]
Then,
[tex]\rm m\times (25\times 10^{-3}\ kg) = 2.2775\times 10^{-19}\ kg.\\m=\dfrac{2.2775\times 10^{-19}\ kg}{25\times 10^{-3}\ kg}=9.11\times 10^{-15}.[/tex]
It is the required fraction of mass of the droplet.
Starting from rest, a runner reaches a speed of 2.8 m/s in 2.1 s. In the same time 2.1 s time, a motorcycles increases speed from 37. 0 to 43.0 m/s. In both cases, assume the acceleration is constant: (a). What is the acceleration (magnitude only) of the runner?(b). What is the acceleration (magnitude only) of the motorcycle?(c). Does the motorcycle travel farther than the runner during the 2.1 s? (yes or no)If som how much father? (if not, enter zero)?
Answer:
a) Acceleration of runner is 1.33 m/s²
b) Acceleration of motorcycle is 2.85 m/s²
c) The motorcycle moves 84.21-2.94 = 81.06 m farther than the runner.
Explanation:
t = Time taken
u = Initial velocity = 0
v = Final velocity
s = Displacement
a = Acceleration
Equation of motion
[tex]v=u+at\\\Rightarrow a=\frac{v-u}{t}\\\Rightarrow a=\frac{2.8-0}{2.1}\\\Rightarrow a=1.33\ m/s^2[/tex]
Acceleration of runner is 1.33 m/s²
[tex]v=u+at\\\Rightarrow a=\frac{v-u}{t}\\\Rightarrow a=\frac{43-37}{2.1}\\\Rightarrow a=2.85\ m/s^2[/tex]
Acceleration of motorcycle is 2.85 m/s²
[tex]v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{2.8^2-0^2}{2\times 1.33}\\\Rightarrow s=2.94\ m[/tex]
The runner moves 2.94 m
[tex]v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{43^2-37^2}{2\times 2.85}\\\Rightarrow s=84.21\ m[/tex]
The motorcycle moves 84.21 m
The motorcycle moves 84.21-2.94 = 81.06 m farther than the runner.
A football is kicked from ground level at an angle of 38 degrees. It reaches a maximum height of 9.7 meters before returning to the ground. How long will the football spend in the air, in seconds?
Answer:
Football will be in air for 2.8139 sec
Explanation:
We have given maximum height h = 9.7 meters
Angle of projection [tex]\Theta =38^{\circ}[/tex]
We know that maximum height is given by [tex]h=\frac{u^2sin^2\Theta }{2g}[/tex]
So [tex]9.7=\frac{u^2sin^{2}38^{\circ} }{2\times 9.8}[/tex]
[tex]u^2=501.5842[/tex]
u = 22.396 m/sec
Time of flight is given by
[tex]T=\frac{2usin\Theta }{g}=\frac{2\times 22.396\times \times sin38^{\circ}}{9.8}=2.8139sec[/tex]
38.4 mol of krypton is in a rigid box of volume 64 cm^3 and is initially at temperature 512.88°C. The gas then undergoes isobaric heating to a temperature of 935.9°C. (a) What is the final volume of the gas?(in cm^3 ) (b) It is then isothermally compressed to a volume 29.3cm^3; what is its final pressure?(in Pa )
Answer:
Final volumen first process [tex]V_{2} = 98,44 cm^{3}[/tex]
Final Pressure second process [tex]P_{3} = 1,317 * 10^{10} Pa[/tex]
Explanation:
Using the Ideal Gases Law yoy have for pressure:
[tex]P_{1} = \frac{n_{1} R T_{1} }{V_{1} }[/tex]
where:
P is the pressure, in Pa
n is the nuber of moles of gas
R is the universal gas constant: 8,314 J/mol K
T is the temperature in Kelvin
V is the volumen in cubic meters
Given that the amount of material is constant in the process:
[tex]n_{1} = n_{2} = n [/tex]
In an isobaric process the pressure is constant so:
[tex]P_{1} = P_{2} [/tex]
[tex]\frac{n R T_{1} }{V_{1} } = \frac{n R T_{2} }{V_{2} }[/tex]
[tex]\frac{T_{1} }{V_{1} } = \frac{T_{2} }{V_{2} }[/tex]
[tex]V_{2} = \frac{T_{2} V_{1} }{T_{1} }[/tex]
Replacing : [tex]T_{1} =786 K, T_{2} =1209 K, V_{1} = 64 cm^{3}[/tex]
[tex]V_{2} = 98,44 cm^{3}[/tex]
Replacing on the ideal gases formula the pressure at this piont is:
[tex]P_{2} = 3,92 * 10^{9} Pa[/tex]
For Temperature the ideal gases formula is:
[tex]T = \frac{P V }{n R }[/tex]
For the second process you have that [tex]T_{2} = T_{3} [/tex] So:
[tex]\frac{P_{2} V_{2} }{n R } = \frac{P_{3} V_{3} }{n R }[/tex]
[tex]P_{2} V_{2} = P_{3} V_{3} [/tex]
[tex]P_{3} = \frac{P_{2} V_{2}}{V_{3}} [/tex]
[tex]P_{3} = 1,317 * 10^{10} Pa[/tex]
A thin metal bar, insulated along its sides, is composed of five different metal connected together. The left end bar is immersed in a heat bath at 100°C and right end in a heat bath at 0°C. Starting at the left end, the pieces and lenghts are steel(2cm), brass(3cm), copper(1cm), aluminum(5cm) and silver(1cm). What is the temperature of the steel/brass interface?
Answer:
T = 61.06 °C
Explanation:
given data:
a thin metal bar consist of 5 different material.
thermal conductivity of ---
K {steel} = 16 Wm^{-1} k^{-1}
K brass = 125 Wm^{-1} k^{-1}
K copper = 401 Wm^{-1} k^{-1}
K aluminium =30Wm^{-1} k^{-1}
K silver = 427 Wm^{-1} k^{-1}
[tex]\frac{d\theta}{dt} = \frac{KA (T_2 -T_1)}{L}[/tex]
WE KNOW THAT
[tex]\frac{l}{KA} = thermal\ resistance[/tex]
total resistance of bar = R steel + R brass + R copper + R aluminium + R silver
[tex]R_{total} =\frac{1}[A} [\frac{0.02}{16} +\frac{0.03}{125} +\frac{0.01}{401} +\frac{0.05}{30} +\frac{0.01}{427}][/tex]
[tex]R_{total} =\frac{1}[A} * 0.00321[/tex]
let T is the temperature at steel/brass interference
[tex]\frac{d\theta}{dt}[/tex] will be constant throughtout the bar
therefore we have
[tex]\frac{100-0}{R_{total}} = \frac{100-T}{R_{steel}}[/tex]
[tex]\frac{100-0}{0.00321} *A = \frac{100-T}{0.00125} *A[/tex]
solving for T we get
T = 61.06 °C
A standard 1 kilogram weight is a cylinder 48.0 mm in height and 55.0 mm in diameter. What is the density of the material?
Answer:8769 Kg/m^3
Explanation: First of all we have to calculate the volume of the cylinder so it is equal to:
Vcylinder= π*r^2*h where r and h are the radius and heigth, respectively.
Vcylinder= π*0.0275^2*0.048=1.14*10^-4 m^3
The density is defined by ρ=mass/volume
the we have
ρ=1Kg/1.14*10^-4 m^3=8769 Kg/m^3
Bragg reflection results in a first-order maximum at 15.0°. In this case, at what angle would the second-order maximum occur?
Answer:
31.174°
Explanation:
Bragg's condition is occur when the wavelength of radiation is comparable with the atomic spacing.
So, Bragg's reflection condition for n order is,
[tex]2dsin\theta=n\lambda[/tex]
Here, n is the order of maxima, [tex]\lambda[/tex] is the wavelength of incident radiation, d is the inter planar spacing.
Now according to the question, first order maxima occur at angle of 15°.
Therefore
[tex]2dsin(15^{\circ})=\lambda\\sin(15^{\circ})=\frac{\lambda}{2d}[/tex]
Now for second order maxima, n=2.
[tex]2dsin\theta=2\times \lambda\\sin\theta=\frac{2\lambda}{2d}[/tex]
Put the values from above conditions
[tex]sin\theta=2\times sin(15^{\circ})\\sin\theta=2\times 0.258819045103\\sin\theta=0.517638090206\\\theta=sin^{-1}0.517638090206\\ \theta=31.174^{\circ}[/tex]
Therefore, the second order maxima occurs at 31.174° angle.
Considere un campo vectorial U que apunta radialmente hacia afuera de un punto P. La magnitud del campo en un punto Q una distancia r del punto P es a K donde K es una constante positiva. Calcule el flujo de este campo vectorial a través de una superficie esférica de radio R centrada en el punto P
Answer:
El flujo del campo U a través de dicha superficie es [tex]\Phi_U=4\pi r^2K[/tex]
Explanation:
El flujo vectorial de un campo U a través de una superficie S está dado por
[tex]\Phi_U=\int_S \vec{U} \cdot \hat{n} \ dS[/tex]
donde [tex]\hat{n}[/tex] representa el vector unitario normal a la superficie. Teniendo en cuenta que el campo se encuentra en dirección radial, se tiene
[tex]\Phi_U=\int_S K\hat{r} \cdot \hat{r} \ dS = K\int_S dS = KS=K*4\pi r^2[/tex]
A 0.676 m long section of cable carrying current to a car starter motor makes an angle of 57.7º with the Earth’s 5.29 x 10^−5T field. What is the current when the wire experiences a force of 6.53 x 10^−3 N?
Answer:
The current in the wire under the influence of the force is 216.033 A
Solution:
According to the question:
Length of the wire, l = 0.676 m
[tex]\theta = 57.7^{\circ}[/tex]
Magnetic field of the Earth, [tex]B_{E} = 5.29\times 10^{- 5} T[/tex]
Forces experienced by the wire, [tex]F_{m} = 6.53\times 10^{-3} N[/tex]
Also, we know that the force in a magnetic field is given by:
[tex]F_{m} = IB_{E}lsin\theta[/tex]
[tex]I = \frac{F_{m}}{B_{E}lsin\theta}[/tex]
[tex]I = \frac{6.53\times 10^{-3}}{5.29\times 10^{- 5}\times 0.676sin57.7^{\circ}[/tex]
I = 216.033 A
A mechanics shop has a 16,500 N car sitting on a large car lift of 125 cm^2 piston. How much force is needed to be applied to the small piston of area 4.51 cm^2 to lift it?
Answer:595.32 N
Explanation:
Given
Mechanic shop has a car of weight([tex]F_1[/tex]) 16,500 N
Area of lift([tex]A_1[/tex])=[tex]125 cm^2[/tex]
Area of small piston([tex]A_2[/tex])=[tex]4.51 cm^2[/tex]
According to pascal's law pressure transmit will be same in all direction in a closed container
[tex]P_1=P_2[/tex]
[tex]\frac{F_1}{A_1}=\frac{F_2}{A_2}[/tex]
[tex]F_2=F_1\frac{A_2}{A_1}[/tex]
[tex]F_2=16,500\times \frac{4.51}{125}=595.32 N[/tex]
Two very small 8.55-g spheres, 15.0 cm apart from center to center, are charged by adding equal numbers of electrons to each of them.
(a)Disregarding all other forces, how many electrons would you have to add to each sphere so that the two spheres will accelerate at 25.0g when released?
(b)Which way will they accelerate?
Answer:
1.43 x 10¹⁷.
They will accelerate away from each other.
Explanation:
Force on each charged sphere F = mass x acceleration
= 8.55 x 10⁻³ x 25 x 9.8
= 2.095 N
Let Q be the charge on each sphere
F = [tex]\frac{9\times10^9\times Q^2}{(15\times10^{-2})^2}[/tex]
2.095 =[tex]\frac{9\times10^9\times Q^2}{(15\times10^{-2})^2}[/tex]
Q² =[tex]\frac{2.095\times(15)^2\times10^{-4}}{9\times10^9}[/tex]
Q = 2.289 X 10⁻⁶
No of electrons = Charge / charge on a single electron
= [tex]\frac{2.289\times10^{-6}}{1.6\times10^{-19}}[/tex]
=1.43 x 10¹³.
They will accelerate away from each other.
The number of electrons would have to add to each sphere to accelerate at 25.0g is 1.43×10¹³ and accelerate away from each other.
What is electric force?Electric force is the force of attraction of repulsion between two bodies.
According to the Coulombs law, the force of attraction of repulsion between charged two bodies is directly proportional to the product of charges of them and inversely proportional to the square of distance between them. It can be given as,
[tex]F=\dfrac{KQ_1Q_2}{r^2}[/tex]
Here, (k) is the coulombs constant, (q1 and q2) is the charges of two bodies and (r) is the distance between the two charges.
Two very small 8.55-g spheres, 15.0 cm apart from center to center, are charged by adding equal numbers of electrons to each of them.
(a) The Number of electrons would have to add to each sphere-Two spheres accelerate at 25.0g when released, and the mass of each sphere is 8.55 g. Thus, the force on the sphere can be given as,
[tex]F=8.55\times{10^{-3}}\times25\times9.8\\F=2.095\rm \;N[/tex]
Two very small spheres are 15.0 cm apart from center to center, are charged by adding equal numbers of electrons to each of them.As the charge on both the sphere is same (say Q) Put these values in the above formula as,
[tex]2.095=\dfrac{(9\times10^{9})QQ}{(15\times10^{-2})^2}\\Q=2.289\times10^{-6}\rm \;C[/tex]
It is known that the charge on one electron is 1.6×10⁻¹⁹ C. Thus the number of electron in the above charge is,
[tex]n=\dfrac{2.289\times10^{-6}}{1.6\times10^{-19}}\\n=1.43\times10^{13}[/tex]
(b)Direction of acceleration-The direction of acceleration of both the sphere will be opposite to each other. Thus, they accelerate away from each other.
Hence, the number of electrons would have to add to each sphere to accelerate at 25.0g is 1.43×10¹³ and accelerate away from each other.
Learn more about the electric force here;
https://brainly.com/question/14372859
Which of the following combinations of position (x) and direction of motion would give a velocity in the x-direction that has a negative value? a. Positive x, moving towards the origin.b. Negative x, moving away from the origin.c. Both a and b.d. Positive x, moving away from the origin.e. Negative x, moving towards the origin.
Answer:
c) Both a) and b) are the combinations that have a negative velocity.
Explanation:
The velocity is given by this equation:
v = Δx / Δt
Where
Δx = final position - initial position
Δt = elapsed time
Now let´s evaluate the options. We have to find those combinations in which Δx < 0 since Δt is always positive.
a) If the initial position, x, is positive and you move towards the origin, the final position will be a smaller value than x. Then:
final position < initial position
final position - initial position < 0 The velocity will be negative.
b) If x is negative and you move away from the origin, the final position will be a more negative number than x. Again:
final position < initial position
final position - initial position < 0 The velocity will be negative.
Let´s do an example to show it:
initial position = -5
final position = -10 (since you moved away from the origin)
final position - initial position = -10 -(-5) = -5
d) If x is positive and you move away from the origin, the final position will be a greater value than the initial position. Then:
final position > initial position
final position - initial position > 0 The velocity will be positive.
e) If x is negative and you move towards the origin, the final position will be a greater value than the initial position. Then:
final position > initial position
final position - initial position > 0 The velocity will be positive.
Let´s do an example:
initial position = -10
final position = -5
final postion - initial position = -5 - (-10) = 5
or with final position = 0
final postion - initial position = 0 -(-10) = 10
And so on.
The right answer is c) Both a) and b) are the combinations that have a negative velocity.
A block is placed at the top of a frictionless inclined plane with angle 30 degrees and released. The incline has a height of 15 m and the block has mass 2 kg. In the above problem the speed of the block when it reaches the bottom of the incline is 8.3 m/s 17 m/s 25 m/s 30 m/s
Answer:
The speed of the block when it reaches the bottom is 17 m/s.
(b) is correct option.
Explanation:
Given that,
Height = 15 m
Mass = 2 kg
We need to calculate the speed of the block when it reaches the bottom
Using conservation of energy
[tex]mgh=\dfrac{1}{2}mv^2[/tex]
[tex]v^2=2gh[/tex]
[tex]v=\sqrt{2gh}[/tex]
Where, g = acceleration due to gravity
h = height
Put the value into the formula
[tex]v=\sqrt{2\times9.8\times15}[/tex]
[tex]v=17\ m/s[/tex]
Hence, The speed of the block when it reaches the bottom is 17 m/s.
A vacuum gage attached to a power plant condenser gives a reading of 27.86 in. of mercury. The surrounding atmospheric pressure is 14.66 lbf/in. Determine the absolute pressure inside the condenser, in lbf/in. The density of mercury is 848 lb/ft and the acceleration of gravity is g = 32.0 ft/s?.
Answer:
absolute pressure = 1.07 lbft/in^2
Explanation:
given data:
vaccum gauge reading h = 27.86 inch = 2.32 ft
we know that
gauge pressure p is given as
[tex]p = \rho gh[/tex]
[tex]p = 848 \ lb/ft^3 * 32 \ ft/s^2 *2.32 ft = 62955.52 \ lbft/s^2 * 1/ft^2[/tex]
we know that [tex]1\ lb ft\s^2 = \frac{1}{32.174}\ lbft[/tex]
1 ft = 12 inch
therefore [tex]p = 62955.52 * \frac{1}{32.174} * \frac{1}{12^2}\ lbf/in^2[/tex]
[tex]p = 13.59\ lbft/in^2[/tex]
[tex]P_{atm} = 14.66\ lbf/in^2[/tex]
so absolute pressure = [tex]P_{atm} - p[/tex]
= 14.66 - 13.59 = 1.07 lbft/in^2
If there is no air resistance, a quarter dropped from the top of New York’s Empire State Building would reach the ground 9.6 s later. (a) What would its speed be just before it hits the ground? (b) How high is the building?
Answer:
a) 94.176 m/s
b) 452.04 m
Explanation:
t = Time taken by the quarter to reach the ground = 9.6 s
u = Initial velocity
v = Final velocity
s = Displacement
a)
[tex]v=u+at\\\Rightarrow v=0+9.81\times 9.6\\\Rightarrow v=94.176 m/s[/tex]
Speed of the quarter just before it hits the ground is 94.176 m/s
b)
[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow s=0\times t+\frac{1}{2}\times 9.81\times 9.6^2\\\Rightarrow s=452.04\ m[/tex]
The building is 452.04 m high
Tom Sawyer runs 10 m/s down the dock and leaps on to his floating raft already moving 1.5 m/s away from shore. If Tom weighs 70 kg and the raft weighs 130 kg, what speed will they both be moving? A. 895 kg m/sB. 11.5 m/sC. 6.4 m/sD. 5.75 m/s
Answer:
Not the right answer in the options, speed is 4.47 m/s, and the procedure is coherent with option A
Explanation:
Answer A uses mass and velocity units, which are momentum units. By using the conservation of momentum:
.[tex]p_{initial} =p_{final} \\m_{Tom}*v_{Tom}+m_{raft}*v_{raft}=(m_{Tom}+m_{raft})*v_{both} \\70*10+130*1.5 kg*m/s=895kg*m/s\\v_{both}=\frac{895 kg*m/s}{200 kg} =4.47 m/s[/tex]
Since Tom stays in the raft, then both are moving with the same speed. From the options, the momentum is in agreement with option A, however, the question asks for speed.
A 2 gram marble is placed in the bottom of a frictionless, hemispherical bowl with a diameter of 4.9 m and it is set in circular motion around the lowest point in the bowl such that its radius of "orbit" is 16 cm. What is its speed? (If you find this problem confusing, consider a simple pendulum with a string length half the diameter given here and suppose its amplitude is 16 cm.)
Answer:
6.93 m/s
Explanation:
mass of marble, m = 2 g
Diameter of the bowl = 4.9 m
radius of bowl, r = half of diameter of the bowl = 2.45 m
The length of the string is 2.45 m
Now it is executing simple harmonic motion, thus the potential energy at the bottom is equal to the kinetic energy at the highest point.
[tex]mgr=\frac{1}{2}mv^{2}[/tex]
where, m be the mass of marble, v be the velocity of marble at heighest point, g be the acceleration due to gravity and r be the radius of the path which is equal to the length of the string
By substituting the values
9.8 x 2.45 = 0.5 x v^2
v = 6.93 m/s
Thus, the speed is 6.93 m/s.
Force is a vector, while mass is a scalar. Why can we use mass as an indicator of the magnitude of the force vector?
Answer:
Explanation:
Force = mass x acceleration
[tex]\overrightarrow{F}=m\overrightarrow{a}[/tex]
Force is always vector and acceleration also vector but the mass is a saclar quanity.
here, the direction of force vector is same as the direction of acceleration vector but the magnitude of force depends on the magnitude of mass of the body.
Is mass is more, force is also more.
Thus, the mass is like an indicator of the magnitude of force.
The brakes on your automobile are capable of creating a deceleration of 4.6 m/s^2. If you are going 114 km/h and suddenly see a state trooper, what is the minimum time in which you can get your car under the 79 km/h speed limit? (The answer reveals the futility of braking to keep your high speed from being detected with a radar or laser gun.)
Answer:
The minimum time to get the car under max. speed limit of 79 km/h is 2.11 seconds.
Explanation:
[tex]a=\frac{V_f-V_0}{t}[/tex]
isolating "t" from this equation:
[tex]t=\frac{V_f-V_0}{a}[/tex]
Where:
a=[tex]-4.6m/s^2[/tex] (negative because is decelerating)
[tex]V_f= 79 km/h[/tex]
[tex]V_0= 114 km/h[/tex]
First we must convert velocity from km/h to m/s to be consistent with units.
[tex]79\frac{km}{h}*\frac{1000m}{1 km}*\frac{1h}{3600s}=\frac{79*1000}{3600}=21.94 m/s[/tex]
[tex]114\frac{km}{h}*\frac{1000m}{1 km}*\frac{1h}{3600s}=\frac{114*1000}{3600}=31.67 m/s[/tex]
So;
[tex]t=\frac{V_f-V_0}{a}=\frac{21.94 m/s-31.66m/s}{-4.6 m/s^2}=2.11 s[/tex]
The pilot of an airplane reads the altitude 6400 m and the absolute pressure 46 kPa when flying over a city. Calculate the local atmospheric pressure in that city in kPa and in mmHg. Take the densities of air and mercury to be 0.828 kg/m3 and 13,600 kg/m3, respectively.
The local atmospheric pressure in the city in kPa is ?
The local atmospheric pressure in the city in mmHg is ?
Answer:
1. 6.672 kPa
2. 49.05 mm of mercury
Explanation:
h = 6400 m
Absolute pressure, p = 46 kPa = 46000 Pa
density of air, d = 0.823 kg/m^3
density of mercury, D = 13600 kg/m^3
(a) Absolute pressure = Atmospheric pressure + pressure due to height
46000 = Atmospheric pressure + h x d x g
Atmospheric pressure = 46000 - 6400 x 0.823 x 10 = 6672 Pa = 6.672 kPa
(b) To convert the pressure into mercury pressure
Atmospheric pressure = H x D x g
Where, H is the height of mercury, D be the density of mercury, g be the acceleration due to gravity
6672 = H x 13600 x 10
H = 0.04905 m
H = 49.05 mm of mercury
To calculate the local atmospheric pressure in the city, use the formula P = P0 * e^(-Mgh/RT), where P is the pressure at the current location, P0 is the pressure at sea level, M is the molar mass of air, g is the acceleration due to gravity, h is the altitude, R is the ideal gas constant, and T is the temperature. Plug in the given values into the formula and solve for P0 to find the local atmospheric pressure in kPa. To convert this pressure to mmHg, use the conversion factor that 101.325 kPa is equal to 760 mmHg.
Explanation:To calculate the local atmospheric pressure in the city, we can use the relationship between altitude, pressure, and density of air. Altitude affects atmospheric pressure because as we go higher in the atmosphere, the air becomes less dense and therefore exerts less pressure. The formula to calculate pressure is given by:
P = P0 * e^(-Mgh/RT)
Where P is the local atmospheric pressure, P0 is the pressure at sea level, M is the molar mass of air, g is the acceleration due to gravity, h is the altitude, R is the ideal gas constant, and T is the temperature. Since we know the altitude and pressure at the current location, we can rearrange the formula to solve for P0:
P0 = P / e^(-Mgh/RT)
Using the given values, we can substitute them into the formula. The density of air is 0.828 kg/m³, the altitude is 6400 m, the absolute pressure is 46 kPa, the molar mass of air is the average molar mass of air (28.97 g/mol), the acceleration due to gravity is 9.8 m/s², the ideal gas constant is 8.314 J/(mol·K), and the temperature can be assumed to be the average temperature at that altitude (around 0°C or 273 K).
Plugging in these values into the formula, we get:
Calculating this expression will give us the local atmospheric pressure in kPa.
To convert this pressure to mmHg, we can use the conversion factor that 101.325 kPa is equal to 760 mmHg. So, to find the pressure in mmHg, we can multiply the pressure in kPa by the conversion factor.
An object is dropped from rest and falls through height h. It travels 0.5h in the last 1 second of fall. Find the total time & height of the fall. (Hint: use two triangles!)
Answer:
3.41 s
114 m
Explanation:
The object is falling in free fall, accelerated by the surface gravity of Earth. We can use the equation for position under constant acceleration:
X(t) = X0 + V0 * t + 1/2 * a * t^2
We set up a frame of reference with the origin at the point the object was released and the X axis pointing down. Then X0 = 0. Since the problem doesnt mention an initial speed we assume V0 = 0.
It travels 0.5h in the last 1 second of the fall. This means it also traveled in the rest of the time of the fall. t = t1 is the moment when it traveled 0.5*h.
0.5*h = 1/2 * a * t1^2
h = a * t1^2
It travels 0.5*h in 1 second.
h = X(t1 + 1) = 1/2 * a * (t1+1)^2
Equating both equations:
a * t1^2 = 1/2 * a * (t1+1)^2
We simplify a and expand the square
t1^2 = 1/2 * (t1^2 + 2*t1 + 1)
t1^2 - 1/2 * t1^2 - t1 - 1/2 = 0
1/2 * t1^2 - t1 - 1/2 = 0
Solving electronically:
t1 = 2.41 s
total time = t1 + 1 = 3.41.
Now
h = a * t1^2
h = 9.81 * 3.41^2 = 114 m