Answer:
The speed of ocean currents approximately is 0.4667 m/s
Solution:
As per the question:
The ocean currents water moves at 30 W near Africa and 70 W near Puerto Rico
Time taken, t = 1 year = [tex]365\times 24\times 3600 = 3.15\times 10^{7} s[/tex]
The distance between Puerto Rico and Africa, x = [tex]1.46\times 10^{7} m[/tex]
Thus
The speed of the ocean currents there:
speed, [tex]v = \frac{d}{t} = \frac{1.47\times 10^{7}}{3.15\times 10^{7}}[/tex]
[tex]s = 0.4667 m/s[/tex]
Suppose you are sitting on a rotating stool holding a 2 kgmass
in each outstretched hand. If you suddenly drop the masses,will
your angular velocity increase, decrease, or stay the same?Please
Explain.
Answer:Increase
Explanation:
Given
You are holding 2 kg mass in each outstreched hand
If the masses are dropped then Moment of inertia will decease by [tex]2mr^2[/tex]
Where m=2 kg
r=length of stretched arm
Since angular momentum is conserved therefore decrease in Moment of inertia will result in increase of angular velocity
as I[tex]\omega [/tex]=constant
I=Moment of inertia
[tex]\omega [/tex]=angular velocity
An inflatable raft (unoccupied) floats down a river at an approximately constant speed of 5.6 m/s. A child on a bridge, 71 m above the river, sees the raft in the river below and attempts to drop a small stone onto the raft. The child releases the stone from rest. In order for the stone to hit the raft, what must be the horizontal distance between the raft and the bridge when the child releases the stone?
Answer:
21.28 m
Explanation:
height, h = 71 m
velocity of raft, v = 5.6 m/s
let the time taken by the stone to reach to raft is t.
use second equation of motion for stone
[tex]h = ut + \frac{1}{2}at^{2}[/tex]
u = 0 m/s, h = 71 m, g = 9.8 m/s^2
71 = 0 + 0.5 x 9.8 x t^2
t = 3.8 s
Horizontal distance traveled by the raft in time t
d = v x t = 5.6 x 3.8 = 21.28 m
A nonconducting container filled with 25 kg of water at 23°C is fitted with a stirrer, which is made to turn by gravity acting on a weight of mass 32 kg. The weight falls slowly through a distance of 5 m in driving the stirrer. Assume that all work done on the weight is transferred to the water and that the local acceleration of gravity is 9.8 m·s−2, determine:
(a) The amount of work done on the water.
(b) The internal-energy change of the water.
(c) The final temperature of the water, for which Cp =4.18 kJ/kgC.
(d) The amount of heat that must be removed from the water to return it to it initial temperature.
Explanation:
Given that,
Weight of water = 25 kg
Temperature = 23°C
Weight of mass = 32 kg
Distance = 5 m
(a). We need to calculate the amount of work done on the water
Using formula of work done
[tex]W=mgh[/tex]
[tex]W=32\times9.8\times5[/tex]
[tex]W=1568\ J[/tex]
The amount of work done on the water is 1568 J.
(b). We need to calculate the internal-energy change of the water
Using formula of internal energy
The change in internal energy of the water equal to the amount of the work done on the water.
[tex]\Delta U=W[/tex]
[tex]\Delta U=1568\ J[/tex]
The change in internal energy is 1568 J.
(c). We need to calculate the final temperature of the water
Using formula of the change internal energy
[tex]\Delta U=mc_{p}\Delta T[/tex]
[tex]\Delta U=mc_{p}(T_{2}-T_{1})[/tex]
[tex]T_{2}=T_{1}+\dfrac{\Delta U}{mc_{p}}[/tex]
[tex]T_{2}=23+\dfrac{1568}{25\times4.18\times10^{3}}[/tex]
[tex]T_{2}=23.01^{\circ}\ C[/tex]
The final temperature of the water is 23.01°C.
(d). The amount of heat removed from the water to return it to it initial temperature is the change in internal energy.
The amount of heat is 1568 J.
Hence, This is the required solution.
The work done on the water is 1568 Joules, which is also the internal-energy change of the water. The final temperature of the water is 23.015°C and to return the water to its initial temperature, 1568 Joules of heat must be removed.
Explanation:(a) The amount of work done on the water is calculated using the formula for gravitational potential energy which depends on the weight's height, mass and acceleration due to gravity. Therefore, work done= mass × gravity × height = 32 kg × 9.8 m·s−2 × 5 m = 1568 Joules.
(b) As per the Law of Conservation of Energy, the work done on the water is converted completely into the internal energy of the water, so the internal-energy change of the water is 1568 Joules.
(c) The final temperature of the water can be calculated using the formula q = m × c × Δt, where 'q' is heat-transfer, 'm' is mass, 'c' is specific heat capacity and 'Δt' is change in temperature. Rearranging, we find Δt = q /(m × c). Substituting the known values gives Δt = 1568 J /(25 kg × 4.18 kJ/kgC) = 0.015 °C. Adding this to the initial temperature, we find the final temperature of the water is 23.015°C.
(d) To return the water to its initial temperature, the heat equal to the increase in internal energy must be removed. Hence, the amount of heat to be removed from the water = 1568 Joules.
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An airplane in the process of taking off travels with a speed of 80 m/s at an angle of 15° above the horizontal. What is the ground speed of the airplane? O 80 m/s O 21 m/s O 77 m/s O 2.6 m/s
Answer:
Option C
Explanation:
given,
velocity of airplane = 80 m/s
angle with the horizontal = 15°
speed of the ground= ?
when the plane is taking off the horizontal component of the velocity is v cosθ
so,
ground speed of the airplane is = [tex]v\times cos\theta[/tex]
= [tex]80 \times cos 15^0[/tex]
v = 77.27 m/s
horizontal velocity of the air plane comes out to be 77.27 m/s ≅ 77 m/s
so, the correct option is Option C
Ethyl alcohol has a boiling point of 78.0°C, a freezing point of -114°C, a heat of vaporization of 879 kJ/kg, a heat of fusion of 109 kJ/kg, and a specific heat of 2.43 kJ/kg.K. How much energy must be removed from 0.651 kg of ethyl alcohol that is initially a gas at 78.0°C so that it becomes a solid at -114°C?
Answer:
946.92 kJ
Explanation:
This process has 3 parts:
1. The first part, where the temperature of Ethyl alcohol remains constant and it changes from gas to liquid.
2. The second part, where the temperature drops from 78°C to -114°C
3. The third parts, where the temperature remains constant and it changes from liquid to solid.
The energy lost in a phase change is:
Q = m*cl
The energy lost because of the drop in temperature is:
[tex]Q = m c(T_2-T_1)[/tex]
cl is the heat of vaporization or heat of fusion, depending on the type of phase change. c is the specific heat.
So, the energy lost in each part is:
1. [tex]Q_1 = 0.651kg*879 kJ/kg = 572.23 kJ[/tex]
2. [tex]Q_2 = 0.651kg*2.43 kJ/kgK(78.0^oC - (-114^oC)) = 303.73 kJ[/tex]
3. [tex]Q_3 = 0.651kg*109kJ/kg = 70.96 kJ[/tex]
Then, the total energy removed should be:
Q = Q1 + Q2 + Q3 = 572.23 kJ + 303.73kJ + 70.96kJ = 946.92 kJ
A car is going 7 m/s when it begins to accelerate. Sixty meters further down the road, the car is going 24 m/s. a) What was the acceleration of the car? b) How much time did the change from 7 m/s to 24 m/s take?
Answer:
acceleration = 4.4 m/s²
time is 3.86 s
Explanation:
given data
initial speed = 7 m/s
final speed = 24 m/s
distance = 60 m
to find out
acceleration and time when change speed change
solution
we will apply here equation of motion for acceleration
v²-u² = 2×a×s .................1
here v is final speed and u is initial speed and s is distance and a is acceleration
put here all these value
24²-7² = 2×a×60
so
a = 4.4
acceleration = 4.4 m/s²
and
now find time by equation of motion
v = u +at
put her value
24 = 7 + 4.4 (t)
t = 3.86
so time is 3.86 s
Your hair grows at the rate of 0.0330 mm/hr. What is the rate of your hair growth in ft/year?
Answer: 0.95 ft/year
Explanation: In order to explain this question we have to convert the units so
if we have a rate equal to 0.033 mm/hr then 0.033 mm *24*365 hr/year
then 1 m=3.28 feet
8760*0.033 *10^-3m* 3.28 feet/m=1.08*10^-4 feet*8760=0.95 feet/year
A highway curve forms a section of a circle. A car goes around the curve. Its dashboard compass shows that the car is initially heading due east. After it travels 830. m, it is heading 15.0° south of east. Find the radius of curvature of its path. (Use the correct number of significant figures.)
Answer:
R = 3170.36m or R = 186.5m
Explanation:
For this problem, we have either trajectory (a), assuming that the car was going south-east, or trajectory (b), assuming the car was going north-east.
In both cases, we know that S = 830m = θ * R. Finding θ, will lead us to the value of R.
For option a:
θ = 15° = 0.2618 rad
[tex]R = \frac{S}{\theta} = 3170.36m[/tex]
For option b:
θ = 270° - 15° = 4.45 rad
[tex]R = \frac{S}{\theta} = 186.5m[/tex]
Three objects are dropped from the top of a building. The first is thrown straight down with a velocity v, the second is thrown straight up with a velocity 2v, and the third is simply dropped. Which one has the highest speed when it hits the ground?
Answer:
Second ball
Explanation:
When a ball is thrown up with a certain velocity when the object reaches the same point from where it was thrown the velocity of the object becomes equal to the velocity with which the ball was thrown.
First ball
[tex]v_g_1^2-u^2=2as\\\Rightarrow v_g_1=\sqrt{2as+u^2}\\\Rightarrow v_g_1=\sqrt{2as+v^2}[/tex]
Second ball
[tex]v_g_2^2-u^2=2as\\\Rightarrow v_g_2=\sqrt{2as+u^2}\\\Rightarrow v_g_2=\sqrt{2as+4v^2}[/tex]
Third ball
[tex]v_g_3^2-u^2=2as\\\Rightarrow v_g_3=\sqrt{2as+0^2}\\\Rightarrow v_g_3=\sqrt{2as}[/tex]
From the equations above it can be seen that the second ball will have the highest velocity when it hits the ground.
So, [tex]v_g_3<v_g_1<v_g_2[/tex]
A startled armadillo leaps upward, rising 0.540 m in the first 0.216 s. (a) What is its initial speed as it leaves the ground? (b)What is its speed at the height of 0.540 m? (c) How much higher does it go? Use g=9.81 m/s^2.
Answer:
a) 3.6 m/s
b) 1.53 m/s
c) 0.12 m
Explanation:
t = Time taken = 0.216 s
u = Initial velocity
v = Final velocity
s = Displacement = 0.54 m
a = Acceleration due to gravity = 9.81 m/s² (negative upward)
[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow 0.54=u\times 0.216+\frac{1}{2}\times -9.81\times 0.216^2\\\Rightarrow u=\frac{0.54+\frac{1}{2}\times 9.81\times 0.216^2}{0.216}\\\Rightarrow u=3.6\ m/s[/tex]
Initial speed as it leaves the ground is 3.6 m/s
[tex]v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times -9.81\times 0.54+3.6^2}\\\Rightarrow v=1.53\ m/s[/tex]
Speed at the height of 0.540 m is 1.53 m/s
[tex]v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-3.6^2}{2\times -9.81}\\\Rightarrow s=0.66\ m[/tex]
The total height the armadillo leaps is 0.66 m
So, the additional height is 0.66-0.54 = 0.12 m
Your throw a ball straight upward at an initial speed of 5 m/s. How many times does the ball pass a point 2 m above the point you launched it from? Draw an x-t and a v-t diagram for the motion of this ball.
Answer:
The ball never passes 2m high, Hmax=1.27m
Explanation:
we assume the ball doesn't bounce when it hits the ground.
We calculate the maximum height, Vf = 0.
[tex]v_{o}^{2}=2gH_{max}\\H_{max}=v_{o}^{2}/(2g)=5^{2}/(2*9.81)=1.27m[/tex]
So, the ball never passes 2m high.
Kinematics equations:
[tex]x(t)=v_{o}t-1/2*g*t^{2}\\v(t)=v_{o}-gt[/tex]
Find annexed the graphics of x(t) and v(t)
A bullet has a mass of 8 grams and a muzzle velocity of 340m/sec. A baseball has a mass of 0.2kg and is thrown by the pitcher at 40m/sec. What is the momentum of the baseball? What is the momentum of the bullet?
Answer:
Momentum of bullet
[tex]P = 2.72 kg m/s[/tex]
momentum of baseball
[tex]P = 8 kg m/s[/tex]
Explanation:
As we know that momentum is defined as the product of mass and velocity
here we know that
mass of the bullet = 8 gram
velocity of bullet = 340 m/s
momentum of the bullet is given as
[tex]P = mv[/tex]
[tex]P = (\frac{8}{1000})(340)[/tex]
[tex]P = 2.72 kg m/s[/tex]
Now we have
mass of baseball = 0.2 kg
velocity of baseball = 40 m/s[/tex]
momentum of baseball is given as
[tex]P = (0.2)(40)[/tex]
[tex]P = 8 kg m/s[/tex]
A man strikes one end of a thin rod with a hammer. The speed of sound in the rod is 15 times the speed of sound in air. A woman, at the other end with her ear close to the rod, hears the sound of the blow twice with a 0.12 s interval between; one sound comes through the rod and the other comes through the air alongside the rod. If the speed of sound in air is 343 m/s, what is the length of the rod?
Answer:
44.1 m
Explanation:
Given:
[tex]V_a[/tex] = speed of sound in air = 343 m/s[tex]V_r[/tex] = speed of sound in the rod = [tex]15V_a[/tex][tex]\Delta t[/tex] = times interval between the hearing the sound twice = 0.12 sAssumptions:
[tex]l[/tex] = length of the rod[tex]t[/tex] = time taken by the sound to travel through the rod [tex]T[/tex] = time taken by the sound to travel to through air to the same point = [tex]t+\Delta t = t+0.12\ s[/tex]We know that the distance traveled by the sound in a particular medium is equal to the product of the speed of sound in that medium and the time taken.
For traveling sound through the rod, we have
[tex]l=V_r t\\\Rightarrow t = \dfrac{l}{V_r}[/tex]..........eqn(1)
For traveling sound through the air to the women ear for traveling the same distance, we have
[tex]l=V_aT\\\Rightarrow l=V_a(t+0.12)\\\Rightarrow l=V_a(\dfrac{l}{V_r}+0.12)\,\,\,\,\,\,(\textrm{From eqn (1)})\\\Rightarrow l=V_a(\dfrac{l}{15V_a}+0.12)\\\Rightarrow l=\dfrac{l}{15}+0.12V_a\\\Rightarrow l-\dfrac{l}{15}=0.12V_a\\\Rightarrow \dfrac{14l}{15}=0.12V_a\\\Rightarrow l = \dfrac{15}{14}\times 0.12V_a\\\Rightarrow l = \dfrac{15}{14}\times 0.12\times 343\\\Rightarrow l = \dfrac{15}{14}\times 0.12\times 343\\\Rightarrow l = 44.1\ m[/tex]
Hence, the length of the rod is 44.1 m.
The length of the rod can be calculated using the difference in hearing times and the speed of sound in the rod and air. Using the formula for distance (speed x time), and given that the speed of sound in the rod is 15 times the speed of sound in air, the length of the rod is found to be approximately 41.16 meters.
Explanation:In this problem, we know that the speed of sound in the rod is 15 times the speed of sound in the air, and that the woman hears the sound of the strike twice with a 0.12 second gap. The first sound is transmitted through the rod and the second, through the air. Therefore, we can use this information to conclude that the difference in time is the amount of time it takes for the sound to travel the length of the rod in air after it already traveled through the rod.
The speed of sound in the rod is 15 times the speed of sound in air, which is given as 343 m/s. So, the speed of sound in the rod is 15 * 343 = 5145 m/s.
We are looking for the distance travelled, which is the length of the rod. We can find the distance by using the formula distance = speed x time. In this case we are calculating distance as time taken for sound to travel through air minus the time taken to travel through the rod. Therefore, the length of the rod can be calculated to be 343 m/s * 0.12 s = 41.16 meters.
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Two tiny conducting sphere are identical and carry charges of -20 μC and +50 μC. They are separated by a distance of 2.50 cm. What is the magnitude of the force that each sphere experiences, and is the force attractive or repulsive?
Answer:
Force between two spheres will be 14400 N
And as the both charges of opposite nature so force will be attractive
Explanation:
We have given two conducting spheres of charges [tex]q_1=-20\mu C=-20\times 10^{-6}C\ and\ q_2=50\mu C=50\times 10^{-6}C[/tex]
Distance between the spheres = 2.5 cm =0.025 m
According to coulombs law we know that force between two charges is given by [tex]F=\frac{1}{4\pi \varepsilon _0}\frac{q_1q_2}{r^2}=\frac{Kq_1q_2}{r^2}=\frac{9\times 10^9\times 20\times 10^{-6}\times 50\times 10^{-6}}{0.025^2}=14400N[/tex]
As the both charges of opposite nature so force will be attractive
A speed skater moving across frictionless ice at 8.4 m/s hits a 5.7 m -wide patch of rough ice. She slows steadily, then continues on at 6.5 m/s. What is her acceleration on the rough ice?
Answer:
Acceleration, [tex]a=-2.48\ m/s^2[/tex]
Explanation:
Initial speed of the skater, u = 8.4 m/s
Final speed of the skater, v = 6.5 m/s
It hits a 5.7 m wide patch of rough ice, s = 5.7 m
We need to find the acceleration on the rough ice. The third equation of motion gives the relationship between the speed and the distance covered. Mathematically, it is given by :
[tex]v^2-u^2=2as[/tex]
[tex]a=\dfrac{v^2-u^2}{2s}[/tex]
[tex]a=\dfrac{(6.5)^2-(8.4)^2}{2\times 5.7}[/tex]
[tex]a=-2.48\ m/s^2[/tex]
So, the acceleration on the rough ice [tex]-2.48\ m/s^2[/tex] and negative sign shows deceleration.
Calculate the individual positive plate capacity in motive power cell that has 15 plates and a copa of 595 Ah A. 110 Ah B. 75 Ah C. 90 Ah D. 85 Ah
Answer:
The individual positive plate capacity is 85 Ah.
(D) is correct option.
Explanation:
Given that,
Number of plates = 15
Capacity = 595 Ah
We need to calculate the individual positive plate capacity in motive power cell
We have,
15 plates means 7 will make pair of positive and negative.
So, there are 7 positive cells individually.
The capacity will be
[tex]capacity =\dfrac{power}{number\ of\ cells}[/tex]
Put the value into the formula
[tex]capacity =\dfrac{595}{7}[/tex]
[tex]capacity =85\ Ah[/tex]
Hence, The individual positive plate capacity is 85 Ah.
Answer:
SDFGHJKL
Explanation:
If the wavelength of an electron is 4.63 x 10^−7 m, how fast is it moving?
Answer:
it move with velocity 1571 m/s
Explanation:
given data
wavelength λ = 4.63 × [tex]10^{-7}[/tex] m
to find out
how fast is it moving
solution
we will use here de Broglie wavelength equation
that is
wavelength λ = [tex]\frac{h}{mv}[/tex] ..........1
here h is planck constant = 6.626068 × [tex]10^{-34}[/tex]
and m is mass of electron i.e = 9.10938188 × [tex]10^{-31}[/tex]
and v is velocity
put all value we find velocity in equation 1
wavelength λ = [tex]\frac{h}{mv}[/tex]
v = [tex]\frac{6.626068*10^{-34}}{9.10938188*10^{-31}*4.63*10^{-7}}[/tex]
v = 1571.035464
so it move with velocity 1571 m/s
Answer:
[tex]v=1.57*10^{3}\frac{m}{s}[/tex]
Explanation:
As DeBroglie equation proved by Davisson-Germer experiment says, the wavelength of an electron is related with its velocity with the equation:
λ = [tex]\frac{h}{mv}[/tex]
where m is the mass of the electron [tex]m=9.11*10^{-31}kg[/tex], h is the Planck´s constant [tex]h=6.626*10^{-34}J.s[/tex] and v its velocity.
Solving the equation for the velocity of the electron, we have:
v = h/mλ
And replacing the values:
[tex]v=\frac{6.626*10^{-34}J.s}{(9.11*10^{-31}Kg)*(4.63*10^{-7}m)}[/tex]
[tex]v=1570.9\frac{m}{s}[/tex]
[tex]v=1.57*10^{3}\frac{m}{s}[/tex]
A grasshopper makes four jumps. The displacement vectors are (1) 31.0 cm, due west; (2) 26.0 cm, 44.0 ° south of west; (3) 22.0 cm, 56.0 ° south of east; and (4) 23.0 cm, 75.0 ° north of east. Find (a) the magnitude and (b) direction of the resultant displacement. Express the direction as a positive angle with respect to due west.
Answer:
(a) 34.47 cm
(b) [tex]24.09^\circ[/tex] south of west
Explanation:
Let us draw a figure representing the individual displacement vectors in the four jumps as shown in the figure attached with this solution.
Now, let us try to write the four displacement vectors in in terms of unit vectors along the horizontal and the vertical axis.
[tex]\vec{d}_1= 31\ cm\ west = -31\ cm\ \hat{i}\\\vec{d}_2= 26\ cm\ south\ of\ west = -26\cos 44^\circ\ \hat{i} -26 \sin 44^\circ\ \hat{j}=(-18.72\ \hat{i}-18.06\ \hat{i})\ cm\\\vec{d}_3= 22\ cm\ south\ of\ east = 22\cos 56^\circ\ \hat{i} -22 \sin 56^\circ\ \hat{j}=(12.30\ \hat{i}-18.23\ \hat{i})\ cm\\\vec{d}_4= 23\ cm\ north\ of\ east = 23\cos 75^\circ\ \hat{i} +23\sin \sin 75^\circ\ \hat{j}=(5.95\ \hat{i}+22.22\ \hat{i})\ cm\\[/tex]
Now, the vector sum of all these vector will give the resultant displacement vector.
[tex]\vec{D} = \vec{d}_1+\vec{d}_2+\vec{d}_3+\vec{d}_4\\\Rightarrow \vec{D} = -31\ cm\ \hat{i}+(-18.72\ \hat{i}-18.06\ \hat{i})\ cm+(12.30\ \hat{i}-18.23\ \hat{i})\ cm+(5.95\ \hat{i}+22.22\ \hat{i})\ cm\\\Rightarrow \vec{D} =(-31.47\ \hat{i}-14.07\ \hat{i})\ cm[/tex]
Part (a):
The magnitude of the resultant displacement vector is given by:
[tex]D=\sqrt{(-31.47)^2+(-14.07)^2}\ m = 34.47\ m[/tex]
Part (b):
Since the resultant displacement vector indicates that the final position of the vector lies in the third quadrant, the vector will make some positive angle in the direction south of west given by:
[tex]\theta = \tan^{-1}(\dfrac{14.07}{31.47})= 24.09^\circ[/tex]
To find the resultant displacement of the grasshopper, we can break down the vectors into their x and y components, and then sum up the components separately. After performing the calculations, we find that the magnitude of the resultant displacement is approximately 39.4 cm and the direction is approximately 38.3° south of west.
Explanation:To find the resultant displacement of the grasshopper, we need to add the individual displacement vectors. We can do this by breaking down each vector into its x and y components.
For vector (1) with a magnitude of 31.0 cm due west, the x component is -31.0 cm and the y component is 0.
Similarly, for the other vectors, the x and y components are:
(2): x = -26.0*cos(44.0) cm, y = -26.0*sin(44.0) cm(3): x = 22.0*cos(56.0) cm, y = -22.0*sin(56.0) cm(4): x = 23.0*cos(75.0) cm, y = 23.0*sin(75.0) cmNow, we can sum up the x components and y components separately to find the resultant displacement.
The magnitude of the resultant displacement can be found using the formula:
resultant magnitude = sqrt((sum of x components)^2 + (sum of y components)^2)
The direction of the resultant displacement can be found using the formula:
resultant direction = atan2((sum of y components), (sum of x components))
Plugging in the values and performing the calculations, we find that the magnitude of the resultant displacement is approximately 39.4 cm and the direction of the resultant displacement is approximately 38.3° south of west.
ou drag your feet on a carpeted floor on a dry day and the carpet acquires a net positive charge. a. Will you have an electron deficiency or an excess of electrons?
b. If the load acquired has a magnitude of 2.15 nC, how many elecrtrons were transferred?
Answer:
1) We will have excess of electrons
2) The number of electrons transferred equals [tex]1.343\times 10^{10}[/tex]
Explanation:
Part a)
Since we know that the charge transfer occurs by the transfer of electrons only as it is given that the carpet has acquired a positive charge it means that it has lost some of the electron's since electrons are negatively charged and if a neutral body looses negative charge it will become positively charged. The electron's that are lost by the carpet will be acquired by the feet of the human thus making us negatively charged.Hence we will gain electrons making us excess in electrons.
Part b)
From charge quantinization principle we have
[tex]Q=ne[/tex]
where
Q = charge of body
n = no of electrons
e = fundamental charge
Applying values in the above equation we get
[tex]2.15\times 10^{-9}C=n\times 1.6\times 10^{-19}C\\\\\therefore n=\frac{2.15\times 10^{-9}C}{1.6\times 10^{-19}C}=1.343\times 10^{10}[/tex]
There is a naturally occurring vertical electric field near the Earth’s surface that points toward the ground. In fair weather conditions, in an open field, the strength of this electric field is 95.0 N/C . A spherical pollen grain with a radius of 12.0 μm is released from its parent plant by a light breeze, giving it a net charge of −0.700 fC (where 1 fC=1×10−15 C ). What is the ratio of the magnitudes of the electric force to the gravitational force, ????electric/????grav , acting on the pollen? Pollen is primarily water, so assume that its volume mass density is 1000 kg/m3 , identical to the volume mass density of water.
Answer:
[tex]\frac{F}{W} = 9.37 \times 10^{-4}[/tex]
Explanation:
Radius of the pollen is given as
[tex]r = 12.0 \mu m[/tex]
Volume of the pollen is given as
[tex]V = \frac{4}{3}\pi r^3[/tex]
[tex]V = \frac{4}{3}\pi (12\mu m)^3[/tex]
[tex]V = 7.24 \times 10^{-15} m^3[/tex]
mass of the pollen is given as
[tex]m = \rho V[/tex]
[tex]m = 7.24 \times 10^{-12}[/tex]
so weight of the pollen is given as
[tex]W = mg[/tex]
[tex]W = (7.24 \times 10^{-12})(9.81)[/tex]
[tex]W = 7.1 \times 10^{-11}[/tex]
Now electric force on the pollen is given
[tex]F = qE[/tex]
[tex]F = (-0.700\times 10^{-15})(95)[/tex]
[tex]F = 6.65 \times 10^{-14} N[/tex]
now ratio of electric force and weight is given as
[tex]\frac{F}{W} = \frac{6.65 \times 10^{-14}}{7.1 \times 10^{-11}}[/tex]
[tex]\frac{F}{W} = 9.37 \times 10^{-4}[/tex]
A spring stretches 0.2 cm per newton of applied force. An object is suspended from the spring and a deflection of 3 cm is observed. If g = 9.81 m/s?, what is the mass of the object, in kg?
Final answer:
The mass of the object is 1.53 kg.
Explanation:
To find the mass of the object, we need to use Hooke's Law which states that the force exerted by a spring is directly proportional to its extension. In this case, the spring stretches 0.2 cm per newton of applied force. The deflection of 3 cm corresponds to an applied force of 15 newtons (0.2 cm per newton * 3 cm).
Using the equation F = mg, where F is the force, m is the mass, and g is the acceleration due to gravity (9.81 m/s^2), we can find the mass:
15 newtons = m * 9.81 m/s^2
m = 15 newtons / 9.81 m/s^2 = 1.53 kg
Therefore, the mass of the object is approximately 1.53 kg.
The mass of the object is approximately 1.53 kg.
Given that the spring stretches 0.2 cm per newton of applied force. Thus, we can say that
for F = - kx.
1 N = - k (0.2 cm)
or, k = spring constant of the given spring = [tex]\frac{F}{x}[/tex] = 5 N/cm
Now, for an object producing deflection of 3 cm, we can say that:
F = - k x = 5 N/cm × 3 cm
or, F = 15 N
This concludes that the weight of the object is 15 N.
Now, W = F = mg
hence, [tex]m = \frac{F}{g}[/tex]
or, m = [tex]\frac{15 \hspace{0.6mm} N}{9.8 \hspace{0.5mm} m/s^2}[/tex]
or, m ≈ 1.53 kg
Suppose you have two identical capacitors. You connect the first capacitor to a battery that has a voltage of 21.2 volts, and you connect the second capacitor to a battery that has a voltage of 12.8 volts. What is the ratio of the energies stored in the capacitors?
Answer:
r=2.743
Explanation:
The energy stored on a capacitor is of type potencial, therfore depends on the capacity to "store" energy. Inthe case of the capacitor, it stores charge (Q), and the equations you use to calculate it are:
[tex]E_p=\frac{Q^2}{2C}=\frac{QV}{2}=\frac{CV^2}{2}[/tex]
In this case we know V and C, therefore we use the last expression:
[tex]E_{p1}=\frac{CV_1^2}{2}[/tex]
[tex]E_{p2}=\frac{CV_2^2}{2}[/tex]
[tex]\frac{E_{p1}}{E_{p2}}=r=\frac{\frac{CV_1^2}{2}}{\frac{CV_2^2}{2}} \\r=(\frac{V_1}{V_2})^2\\r=(\frac{21.2}{12.8})^2[/tex]
r=2.743
What does it mean if a conductor is in "electrostatic equilibrium"? a) The conductor is at rest.
b) The charges in the conductor are not moving.
c) The charges in the conductor are distributed uniformly throughout the conductor.
d) The charges in the conductor are moving in response to an electric field.
e) None of the above.
Answer:
25
Explanation:
A 0.010 kg ball is shot from theplunger of a pinball machine.
Because of a centripetal force of0.025 N, the ball follows a
circulararc whose radius is 0.29 m. What isthe speed of the
ball?
Answer:
v = 0.85 m/s
Explanation:
Given that,
Mass of the ball, m = 0.01 kg
Centripetal force on the ball, F = 0.025 N
Radius of the circular path, r = 0.29 m
Let v is the speed of the ball. The centripetal force of the ball is given by :
[tex]F=\dfrac{mv^2}{r}[/tex]
[tex]v=\sqrt{\dfrac{Fr}{m}}[/tex]
[tex]v=\sqrt{\dfrac{0.025\times 0.29}{0.01}}[/tex]
v = 0.85 m/s
So, the speed of the ball is 0.85 m/s. Hence, this is the required solution.
In an experiment, a rectangular block with height h is allowed to float in two separate liquids. In the first liquid, which is water, it floats fully submerged. In the second liquid it floats with height h/7 above the liquid surface. What is the relative density (the density relative to that of water) of the second liquid?
Answer:
The relative density of the second liquid is 7.
Explanation:
From archimede's principle we know that the force that a liquid exerts on a object equals to the weight of the liquid that the object displaces.
Let us assume that the volume of the object is 'V'
Thus for the liquid in which the block is completely submerged
The buoyant force should be equal to weight of liquid
Mathematically
[tex]F_{buoyant}=Weight\\\\\rho _{1}\times V\times g=m\times g\\\\\therefore \rho _{1}=\frac{m}{V}...............(i)[/tex]
Thus for the liquid in which the block is 1/7 submerged
The buoyant force should be equal to weight of liquid
Mathematically
[tex]F'_{buoyant}=Weight\\\\\rho _{2}\times \frac{V}{7}\times g=m\times g\\\\\therefore \rho _{2}=\frac{7m}{V}.................(ii)[/tex]
Comparing equation 'i' and 'ii' we see that
[tex]\rho_{2}=7\times \rho _{1}[/tex]
Since the first liquid is water thus [tex]\rho _{1}=1gm/cm^3[/tex]
Thus the relative density of the second liquid is 7.
Answer:
7
Explanation:
Let the density of second liquid is d.
Density of water = 1 g/cm^3
In case of equilibrium, according to the principle of flotation, the weight of the body is balanced by the buoyant force acting on the body.
Let A be the area of cross section of block and D be the density of material of block and h be the height.
For first liquid (water):
Weight of block = m x g = A x h x D x g .... (1)
Buoyant force in water = A x h x 1 x g ..... (2)
Equating (1) and (2) we get
A x h x D x g = A x h x 1 x g
D = 1 g/cm^3
For second liquid:
Weight of block = m x g = A x h x D x g .... (1)
Buoyant force in second liquid = A x h/7 x d x g ..... (2)
Equating (1) and (2) we get
A x h x D x g = A x h/7 x d x g
D = d/7
d = 7 D = 7 x 1 = 7 g/cm^3
Thus, the relative density of second liquid is 7.
Two identical conducting spheres, fixed in place, attract each other with an electrostatic force of 0.116 N when their center-to-center separation is 65.4 cm. The spheres are then connected by a thin conducting wire. When the wire is removed, the spheres repel each other with an electrostatic force of 0.0273 N. Of the initial charges on the spheres, with a positive net charge, what was (a) the negative charge on one of them and (b) the positive charge on the other? (Assume the negative charge has smaller magnitude.)
Answer:
Part a)
[tex]q_1 = -1.47 \times 10^{-6} C[/tex]
Part b)
[tex]q_2 = 3.75 \times 10^{-6} C[/tex]
Explanation:
Let the charge on two spheres is q1 and q2
now the force between two charges are
[tex]F = \frac{kq_1q_2}{r^2}[/tex]
[tex]0.116 = \frac{(9\times 10^9)(q_1)(q_2)}{0.654^2}[/tex]
[tex]q_1 q_2 = 5.51 \times 10^{-12}[/tex]
now when we connect then with conducting wire then both sphere will equally divide the charge
so we will have
[tex]q = \frac{q_1-q_2}{2}[/tex]
now we have
[tex]0.0273 = \frac{(9\times 10^9)(\frac{q_1- q_2}{2})^2}{0.654^2}[/tex]
[tex]q_1 - q_2 = 2.28\times 10^{-6} C[/tex]
now we will have
Now we can solve above two equations
Part a)
negative charge on the sphere is
[tex]q_1 = -1.47 \times 10^{-6} C[/tex]
Part b)
positive charge on the sphere is
[tex]q_2 = 3.75 \times 10^{-6} C[/tex]
An object is originally moving at a constant velocity of 8 m/s in the -x direction. It moves at this constant velocity for 3 seconds. Then, a constant acceleration of 7 m/s2 is applied to it in the +x direction for 9 seconds. What is the total distance covered by this object in meters? Please give a detailed explanation.
Answer:
244.64m
Explanation:
First, we find the distance traveled with constant velocity. It's simply multiplying velocity time the time that elapsed:
[tex]x = V*t = -8\frac{m}{s} *3s = -24m[/tex]
After this, the ball will start traveling with a constant acceleration motion. Due to the fact that the acceleration is the opposite direction to the initial velocity, this motion will have 2 phases:
1. The velocity will start to decrease untill it reaches 0m/s.
2. Then, the velocity will start to increase at the rate of the acceleration.
The distance that the ball travels in the first phase can be found with the following expression:
[tex]v^2 = v_0^2 + 2a*d[/tex]
Where v is the final velocity (0m/s), v_0 is the initial velocity (-8m/s) and a is the acceleration (+9m/s^2). We solve for d:
[tex]d = \frac{v^2 - v_0^2}{2a} = \frac{(0m/s)^2 - (-8m/s)^2}{2*7m/s^2}= -4.57m[/tex]
Now, before finding the distance traveled in the second phase, we need to find the time that took for the velocity to reach 0:
[tex]t_1 = \frac{v}{a} = \frac{8m/s}{7m/s^2} = 1.143 s[/tex]
Then, the time of the second phase will be:
[tex]t_2 = 9s - t_1 = 9s - 1.143s = 7.857s[/tex]
Using this, we using the equations for constant acceleration motion in order to calculate the distance traveled in the second phase:
[tex]x = \frac{1}{2}a*t^2 + v_0*t + x_0[/tex]
V_0, the initial velocity of the second phase, will be 0 as previously mentioned. X_0, the initial position, will be 0, for simplicity:
[tex]x = \frac{1}{2}*7\frac{m}{s^2}*t^2 + 0m/s*t + 0m = 216.07m[/tex]
So, the total distance covered by this object in meters will be the sum of all the distances we found:
[tex]x_total = 24m + 4.57m + 216.07m = 244.64m[/tex]
Consider two displacements, one of magnitude 15 m and another of magnitude 20 m. What angle between the directions of this two displacements give a resultant displacement of magnitude (a) 35 m, (b) 5 m, and (c) 25 m.
Answer:
a) 0°
b) 180°
c) 90°
Explanation:
Hello!
To solve this question let a be the vector whose length is 15 m and b the vector of length 20 m
So:
|a | = 15
|b | = 20
Since we are looking for the angle between the vectors we need to calculate the length of the sum of the two vectors, this is:
[tex]|a+b|^{2} = |a|^{2} + |b|^{2} + 2 |a||b|cos(\theta)[/tex]
Now we replace the value of the lengths:
[tex]|a+b|^{2} = 15^{2} + 20^{2} + 2*15*20*cos(\theta)[/tex]
[tex]|a+b|^{2} = 625 + 600*cos(\theta)[/tex] --- (1)
Now, if:
a) |a+b| = 35
First we can see that 20 + 15 = 35, so the angle must be 0, lets check this:
[tex]35^{2} = 625 + 600*cos(\theta)[/tex]
[tex]1225 = 625 + 600*cos(\theta)[/tex]
[tex]600 = 600*cos(\theta)[/tex]
[tex]1= cos(\theta)[/tex]
and :
[tex]\theta = arccos(1)[/tex]
θ = 0
b) |a+b|=5
From eq 1 we got:
[tex]\theta = arccos(\frac{|a+b|^{2}-625}{600})[/tex] --- (2)
[tex]\theta = arccos(\frac{|a+b|^{2}-625}{600})[/tex]
[tex]\theta = arccos(-1)[/tex]
θ = π or θ = 180°
c) |a+b|=25
[tex]\theta = arccos(\frac{|25|^{2}-625}{600})[/tex]
[tex]\theta = arccos(-1)[/tex]
θ = π/2 or θ = 90°
Final answer:
In vector addition, an angle of 0° between two vectors gives a resultant of 35 m, an angle of 180° gives a resultant of 5 m, and the angle for a resultant of 25 m can be found using the Law of Cosines.
Explanation:
The question involves the concept of vector addition and the use of trigonometry to determine the resultant displacement when two vectors are combined at various angles. The displacement vectors have magnitudes of 15 m and 20 m, and we are interested in finding the angles that would result in resultant displacements of 35 m, 5 m, and 25 m, respectively.
For (a) a resultant displacement of 35 m, the two vectors must be added in the same direction. This implies that the angle between them is 0°.
For (b) a resultant displacement of 5 m, the two vectors must be in exactly opposite directions. Since the difference in magnitudes is 5 m, this means that the larger vector (20 m) partially cancels out the smaller vector (15 m). Hence, the angle between them is 180°.
For (c) a resultant displacement of 25 m, we can use the Law of Cosines to determine the angle:
c2 = a2 + b2 - 2ab cos(θ)
Where a = 15 m, b = 20 m, and c = 25 m. Solving this equation will give us the value of θ.
Time dilation: A missile moves with speed 6.5-10 m/s with respect to an observer on the ground. How long will it take the missile's clock to fall behind the ground observer's clock by 1 millisecond? Hint: use the binomial formula:(1+x)a1+ ax.
Answer:
The time taken by missile's clock is [tex]4.6\times 10^{6} s[/tex]
Solution:
As per the question:
Speed of the missile, [tex]v_{m = 6.5\times 10^{3}} m/s[/tex]
Now,
If 'T' be the time of the frame at rest then the dilated time as per the question is given as:
T' = T + 1
Now, using the time dilation eqn:
[tex]T' = \frac{T}{\sqrt{1 + (\frac{v_{m}}{c})^{2}}}[/tex]
[tex]\frac{T'}{T} = \frac{1}{\sqrt{1 + (\frac{v_{m}}{c})^{2}}}[/tex]
[tex]\frac{T + 1}{T} = \frac{1}{\sqrt{1 + (\frac{v_{m}}{c})^{2}}}[/tex]
[tex]1 + \frac{1}{T} = \frac{1}{\sqrt{1 + (\frac{v_{m}}{c})^{2}}}[/tex]
[tex]1 + \frac{1}{T} = (1 + (\frac{v_{m}}{c})^{2})^{- \frac{1}{2}}[/tex] (1)
Using binomial theorem in the above eqn:
We know that:
[tex](1 + x)^{a} = 1 + ax[/tex]
Thus eqn (1) becomes:
[tex]1 + \frac{1}{T} = 1 - \frac{- 1}{2}.\frac{v_{m}^{2}}{c^{2}}[/tex]
[tex]T = \frac{2c^{2}}{v_{m}^{2}}[/tex]
Now, putting appropriate values in the above eqn:
[tex]T = \frac{2(3\times 10^{8})^{2}}{(6.5\times 10^{3})^{2}}[/tex]
[tex]T = 4.6\times 10^{6} s[/tex]
A test charge of +4 µC is placed halfway between a charge of +6 µC and another of +2 µC separated by 20 cm. (a) What is the magnitude of the force (in N) on the test charge?
(b) What is the direction of this force (away from or toward the +6 µC charge)?
a. away from the +6 µC charge or
b. toward the +6 µC charge
Answer:
(a) Magnitude: 14.4 N
(b) Away from the +6 µC charge
Explanation:
As the test charge has the same sign, the force that the other charges exert on it will be a repulsive force. The magnitude of each of the forces will be:
[tex]F_e = K\frac{qq_{test}}{r^2}[/tex]
K is the Coulomb constant equal to 9*10^9 N*m^2/C^2, q and qtest is the charge of the particles, and r is the distance between the particles.
Let's say that a force that goes toward the +6 µC charge is positive, then:
[tex]F_e_1 = K\frac{q_1q_{test}}{r^2}=-9*10^9 \frac{Nm^2}{C^2} \frac{6*10^{-6}C*4*10^{-6}C}{(0.1m)^2} =-21.6 N[/tex]
[tex]F_e_2 = K\frac{q_2q_{test}}{r^2}=9*10^9 \frac{Nm^2}{C^2} \frac{2*10^{-6}C*4*10^{-6}C}{(0.1m)^2} =7.2 N[/tex]
The magnitude will be:
[tex]F_e = -21.6 + 7.2 = -14.4 N[/tex], away from the +6 µC charge