To calculate average speed, divide the total distance by the total time taken. In this case, the average speed is 30.0 km/h.
Explanation:To calculate average speed, we divide the total distance traveled by the total time taken. In this case, you drove 10.0 km in 20.0 minutes. To convert minutes to hours, we divide by 60. The average speed in kilometers per hour is calculated as:
Average speed = Total distance / Total time = 10.0 km / (20.0 min / 60 min/h) = 30.0 km/h
Two parallel plates 19 cm on a side are given equal and opposite charges of magnitude 2.0 ✕ 10^−9 C. The plates are 1.8 mm apart. What is the electric field (in N/C) at the center of the region between the plates? (Enter the magnitude.)
Answer:
E = 5291.00 N/C
Explanation:
Expression for capacitance is
[tex]C = \frac{\epsilon A}{d}[/tex]
where
A is area of square plate
D = DISTANCE BETWEEN THE PLATE
[tex]C = \frac{\epsilon\times(19\times 10^{-2})^2}{1.5\times 10^{-3}}[/tex]
[tex]C = 24.06 \epsilon[/tex]
[tex]C = 24.06\times 8.854\times 10^{-12} F[/tex]
[tex]C =2.1\times 10^{-10} F[/tex]
We know that capacitrnce and charge is related as
[tex]V = \frac{Q}{C}[/tex]
[tex]= \frac{2\tiimes 10^{-9}}{2.\times 10^{-10}}[/tex]
v = 9.523 V
Electric field is given as
[tex]E = \frac{V}{d}[/tex]
= [tex]\frac{9.52}{1.8*10^{-3}}[/tex]
E = 5291.00 V/m
E = 5291.00 N/C
When a field goal kicker kicks a football as hard as he can at 45° to the horizontal, the ball just clears the 3-m-high crossbar of the goalposts 45.7 m away. (a) What is the maximum speed the kicker can impart to the football? (b) In addition to clearing the crossbar, the football must be high enough in the air early during its flight to clear the reach of the onrushing defensive lineman. If the lineman is 4.6 m away and has a vertical reach of 2.5 m, can he block the 45.7-m field goal attempt? (c) What if the lineman is 1.0 m away?
Answer:
Part a)
[tex]v = 21.9 m/s[/tex]
Part b)
[tex]y = 4.17 m[/tex]
So he will not able to block the goal
Part c)
[tex]y = 0.98 m[/tex]
yes he can stop the goal
Explanation:
As we know by the equation of trajectory of the ball
[tex]y = x tan\theta - \frac{gx^2}{2v^2cos^2\theta}[/tex]
[tex]y = 3 m[/tex]
[tex]x = 45.7 m[/tex]
[tex]\theta = 45 degree[/tex]
now from above equation we have
[tex]y = 45.7 tan 45 - \frac{9.81(45.7)^2}{2v^2cos^245}[/tex]
[tex]3 = 45.7 - \frac{20488}{v^2}[/tex]
[tex]v^2 = 479.81[/tex]
[tex]v = 21.9 m/s[/tex]
Part b)
If lineman is 4.6 m from the football
[tex]y = x tan\theta - \frac{gx^2}{2v^2cos^2\theta}[/tex]
[tex]y = 4.6tan45 - \frac{9.81(4.6^2)}{2(21.9^2)cos^245}[/tex]
[tex]y = 4.17 m[/tex]
So he will not able to block the goal
Part c)
If lineman is 1 m from the football
[tex]y = x tan\theta - \frac{gx^2}{2v^2cos^2\theta}[/tex]
[tex]y = 1tan45 - \frac{9.81(1^2)}{2(21.9^2)cos^245}[/tex]
[tex]y = 0.98 m[/tex]
yes he can stop the goal
A thin soap bubble of index of refraction 1.33 is viewed with light of wavelength 550.0nm and appears very bright. Predict a possible value of the thickness of the soap bubble.
Answer:
Thickness = 103.38 nm
Explanation:
Given that:
The refractive index of the thin soap bubble = 1.33
The wavelength of the light = 550 nm
The minimum thickness that produces a bright fringe can be calculated by using the formula shown below as:
[tex]Thickness=\frac {\lambda}{4\times n}[/tex]
Where, n is the refractive index of the thin soap bubble = 1.33
[tex]{\lambda}[/tex] is the wavelength
So, thickness is:
[tex]Thickness=\frac {550\ nm}{4\times 1.33}[/tex]
Thickness = 103.38 nm
Pions have a half-life of 1.8 x 10^-8 s. A pion beam leaves an accelerator at a speed of 0.8c. What is the expected distance over which half the pions should decay?
Answer:
the expected distance is 4.32 m
Explanation:
given data
half life time = 1.8 × [tex]10^{-8}[/tex] s
speed = 0.8 c = 0.8 × 3 × [tex]10^{8}[/tex]
to find out
expected distance over
solution
we know c is speed of light in air is 3 × [tex]10^{8}[/tex] m/s
we calculate expected distance by given formula that is
expected distance = half life time × speed .........1
put here all these value
expected distance = half life time × speed
expected distance = 1.8 × [tex]10^{-8}[/tex] × 0.8 × 3 × [tex]10^{8}[/tex]
expected distance = 4.32
so the expected distance is 4.32 m
A 4.55 nF parallel-plate capacitor contains 27.5 μJ of stored energy. By how many volts would you have to increase this potential difference in order for the capacitor to store 55.0 μJ of potential energy?
Express your answer in volts as an integer.
Answer:
[tex] \Delta V=V_{2}-V_{1}=45.4V[/tex]
Explanation:
The energy, E, from a capacitor, with capacitance, C, and voltage V is:
[tex]E=\frac{1}{2} CV^{2}[/tex]
[tex]V=\sqrt{2E/C}[/tex]
If we increase the Voltage, the Energy increase also:
[tex]V_{1}=\sqrt{2E_{1}/C}[/tex]
[tex]V_{2}=\sqrt{2E_{2}/C}[/tex]
The voltage difference:
[tex]V_{2}-V_{1}=\sqrt{2E_{2}/C}-\sqrt{2E_{1}/C}[/tex]
[tex]V_{2}-V_{1}=\sqrt{2*55*10^{-6}/4.55*10^{-9}}-\sqrt{2*27.5*10^{-6}/4.55*10{-9}}=45.4V[/tex]
A two stage rocket is launched moving vertically with acceleration 5.0 m/s^2. After 10.0 s, the first stage of the rocket is ejected; the second stage is now accelerating at 8.0 m/s^2. What is the distance between the first and second stages 4.0 s after separation?
Answer:264 m
Explanation:
Given
acceleration of rocket([tex]a_1[/tex])= 5 m/s^2[/tex]
velocity after 10 s
v=u+at
[tex]v=0+5\times 10[/tex]
v=50 m/s
after first stage rocket is ejected
acceleration of second stage=[tex]8 m/s^2[/tex]
distance between first and second part after 4 sec
[tex]s=u_1t+\frac{1}{2}at^2 [/tex]
here [tex]u_1=50 m/s[/tex]
[tex]s=50\times 4+\frac{1}{2}\times 8\times 4^2[/tex]
s=200+64=264 m
A fire hose held near the ground shoots water at a speed of 6.5 m/s. At what angle(s) should the nozzle point in order that the water land 2.5 m away (Fig. 3–36)? Why are there two different angles? Sketch the two trajectories.
Answer:
17.72° or 72.28°
Explanation:
u = 6.5 m/s
R = 2.5 m
Let the angle of projection is θ.
Use the formula for the horizontal range
[tex]R=\frac{u^{2}Sin2\theta }{g}[/tex]
[tex]2.5=\frac{6.5^{2}Sin2\theta }{9.8}[/tex]
Sin 2θ = 0.58
2θ = 35.5°
θ = 17.72°
As we know that the range is same for the two angles which are complementary to each other.
So, the other angle is 90° - 17.72° = 72.28°
Thus, the two angles of projection are 17.72° or 72.28°.
To which types of thermodynamic processes will the relation Tds =đQ apply? What is the correct term for the symbol T in this expression, and what exactly does dQ represent? (5) (b) Calculate the change in entropy when tkg of ice is melted at 273.16K Calculate the change in entropy when lkg of water is heated from 273 16K to 300K (15) (c) Why is the change in entropy much larger for one of the calculations in part (b) of this question
Answer:
a) Reversible processes
b) 1222.73 J/K and 392.144 J/K
c) Because of breaking bonds that are a stable state, while accelerating molecules in the other is less difficult.
Explanation:
Since it is equality, it should be applied to reversible processes, and even better to isothermal processes, since the Temperature remains constant. The T is the absolute temperature (measured in °K) and dQ is the heat absorbed by the system, which DEPENDS on the process.
b) The heat absorbed in the fusion process depends on the latent heat of fusion, L, of the water, which is 334 J/g. It says t kg, but I assume it was a mistake in the typing, so the change in entropy is calculated for 1 kg of water melting as follows:
[tex]\Delta S=\frac{\Delta Q}{T}=\frac{mass*L}{T}=\frac{1kg*334J/g*(1000 g/kg)}{273.16 K} \\\Delta S= 1222.73 J/K[/tex]
Now, to proceed with the change of entropy for water heated from 273.16K to 300K we use the specific heat of water, which is 4184 J/kg°K as follows:
[tex]\Delta S=m_{mass}c_{water}*ln(\frac{T_f}{T_i} )=1kg*4184\frac{J}{Kg*K}*ln(\frac{300}{273.16})\\ \Delta S=392.144 J/K[/tex]
c) In the solid state, water molecules have different bonds with other water molecules creating the crystals. In the liquid state, each molecule moves freely with less interaction between molecules. So, it required more energy to break these bonds and alter this ordered state than just accelerating the molecules in the liquid state.
Given two charges, pick the answers that are true : The two charges act on one another no matter how far apart they are, all the way to infinity
If charge "A" is attracted to "B", then charge "B" is repelled from "A"
Plus and plus repel
If charge "A" is attracted to "B", then charge "B" is equally attracted to "A"
Minus and minus repel
Plus and minus attract
Answer:
All true except the second one.
Explanation:
The two charges act on one another no matter how far apart they are, all the way to infinity . True. Coulomb law doesn't have a distance limitation.
If charge "A" is attracted to "B", then charge "B" is repelled from "A" . False. This violates Newton 3rd Law (Action-Reaction)
Plus and plus repel . True . Charges of same sign repel and opposite sign atract.
If charge "A" is attracted to "B", then charge "B" is equally attracted to "A" . True. This is Newton 3rd Law (Action-Reaction)
Minus and minus repel . True . Charges of same sign repel and opposite sign atract.
Plus and minus attract . True . Charges of same sign repel and opposite sign atract.
A simple pendulum of mass 0.50 kg and length 0.75 m is held still and then released from an angle of 10° at t = 0. At what time does the pendulum first reach its maximum kinetic energy? (a) 0.43 s, (b) 0.53 s, (c) 1.1 s, (d) 1.7 s, (e) 3.4 s. AD string on a guitar is 648 mm long, has a mass of 1.92 g and a fundamental frequency of 147 Hz. How far from the end of the string is the fret associated with the G note, which has a frequency of 110 Hz? (a) 81.5 mm, (b) 163 mm, (c) 245 mm, (d) 326 mm, (e) 408 mm.
Answer:
Explanation:
The motion of the pendulum will be be SHM with time period equal to
T = [tex]2\pi\sqrt{\frac{l}{g} }[/tex]
l = .75 m , g = 9.8
T = [tex]2\pi\sqrt{\frac{.75}{9.8} }[/tex]
T = 1.73 s .
Time to reach the point of maximum velocity or maximum kinetic energy
= T /4
= 1.73 /4
= 0.43 s
For notes on Guitar , The formula is
n = [tex]\frac{1}{2l} \sqrt{\frac{T}{m} }[/tex]
n is fequency of the note , T is tension of string , m is mass per unit length of string , l is length of string.
For fundamental note , l = .648 m and f = 147 Hz
147 = [tex]\frac{1}{2\times.648} \sqrt{\frac{T}{m} }[/tex]
For G note
110 = [tex]\frac{1}{2\times l} \sqrt{\frac{T}{m} }[/tex]
[tex]\frac{147}{110} =\frac{l}{.648}[/tex]
l = 866 mm
Distance from the end of string
866 - 648 = 218 mm or 245 mm
Option c ) is correct .
You throw a ball vertically from an apartment balcony to the ground 16.0 m below. Find the ball's initial velocity if it hits the ground 1.02 s after you release it.
Answer:
u(intial velocty)= 10.688 m/s apprx
Explanation:
given data:
height of apartment = 16 m
time of hiitng = 1.02 s
Using equation of motion we have
h = ut + 0.5gt2 putting all value to get inital velocity value
16 = u(1.02)+ 0.5(9.8)(1.02)^2
SOLVING FOR u
16 - 5.09 = U1.02
u(intial velocty)= 10.688 m/s apprx
Direction will be towards ground
To find the ball's initial velocity, use the equation of motion for free fall and solve for v0. The initial velocity of the ball is approximately 10.78 m/s.
Explanation:To find the ball's initial velocity, we can use the equation of motion for free fall: d = v0t + 0.5gt2, where d is the distance, v0 is the initial velocity, t is the time, and g is the acceleration due to gravity. In this case, d = 16.0 m, t = 1.02 s, and g = 9.8 m/s2. Plugging in these values, we can solve for v0.
16.0 = v0(1.02) + 0.5(9.8)(1.02)2
Simplifying the equation gives us:
16.0 = 1.02v0 + 5.01
Subtracting 5.01 from both sides gives:
10.99 = 1.02v0
Finally, dividing both sides by 1.02 gives:
v0 ≈ 10.78 m/s
Two blocks of masses 20 kg and 8.0 kg are connected togetherby
a light string and rest on a frictionless level surface.Attached to
the 8- kg mass is a second light string, which a personuses to pull
both blocks horiaontally. If the two- block systemaccelerates at
0.5 m/s ^2, what is the tension in the second stringattached to the
8-kg mass?
Answer:
The tension is 14 N
Explanation:
For this problem we have to use newton's law, so:
[tex]F=m*a[/tex]
The second string is connected to the mass of 8 kg, but the mass of 8 kg is connected to the mass of 20kg, so we can say that the second string is handling the two masses. so:
[tex]F=28kg*0.5\frac{m}{s^2}=14N[/tex]
Who among the following educators did not contribute to the study or play?
A. Howard Gardner
B. Jean Jacques Rousseau
C. Johann Amos Comenius
D. Friedrich Froebel
Answer:
A. Howard Gardner
Explanation:
Howard Gardner was the educator who did not contribute to study or play.
- His parents fled Nazi Germany with their first son who died in accident before Howard was born.
-He was not allowed to play sports.
- He was an excellent pianist.
-He attended Harvard university.
-Influenced by Erickson
Among Howard Gardner, Jean Jacques Rousseau, Johann Amos Comenius, and Friedrich Froebel, Friedrich Froebel is the one most directly associated with the study of play for establishing the kindergarten system.
Explanation:The question asks who among the listed educators did not contribute to the study or theory of play. Let's examine the contributions of each figure to determine the answer:
Howard Gardner is known for his theory of multiple intelligences, which includes the idea of bodily-kinesthetic intelligence, relating to the way we control our body motions and handle objects, suggesting a recognition of play's importance.Jean Jacques Rousseau emphasized naturalism and the importance of allowing children to interact with their environment, which is in line with the principles of learning through play.Johann Amos Comenius advocated for educational reform that included playful learning, which he saw as a natural way for children to learn.Friedrich Froebel is recognized as the creator of the concept of kindergarten, where play is a fundamental educational approach.Considering the contributions of these educators, Friedrich Froebel is the one who stands out the most as being directly linked to the concept of play in education by virtue of establishing the kindergarten system.
A flea walking along a ruler moves from the 45 cm mark to the 27 cm mark. It does this in 3 seconds. What is the speed? What is the velocity?
(Define increasing numbers to be the positive direction and decreasing numbers to be the negative direction.)
Answer:
Speed= 6cm/s and velocity= 6cm/s in the negative direction
Explanation:
the change in position is from 45cm to 27 cm (moving towards the negative x direction)
[tex] \Delta x = 45 cm - 27 cm = 18 cm[/tex]
And the change in time:
[tex] \Delta t= 3 s[/tex]
Now we must define the difference between speed and velocity:
Speed is a scalar quantity, which means that it is a number. Velocity is also a number but you must also indicate the direction of the movement.
Thus, the speed is:
[tex] speed= \Delta x/ \Delta t = 18cm/3s=6cm/s[/tex]
An the velocity is:
6cm/s in the negative direction
An object with a charge of -2.9 μC and a mass of 1.0×10^−2 kg experiences an upward electric force, due to a uniform electric field, equal in magnitude to its weight. Find the magnitude of the electric field?
If the electric charge on the object is doubled while its mass remains the same, find the direction and magnitude of its acceleration.?
Answer:
[tex]E=3.38*10^{4}N/C[/tex]The new acceleration will have a value equal to gravity (9.81m/s^2) but pointing upwards.Explanation:
The electric Force and gravitational Force are the same:
[tex]m*g=E*q[/tex]
[tex]E=m*g/q=1.0*10^{-2}*9.81/(2.9*10^{-6})=3.38*10^{4}N/C[/tex]
If the Electric field is doubled, the Electric Force is doubled, so its new value is twice the weight of the object. If we add this new electric force (upwards), with the weight (downwards), we have a resulting force upwards = one time the weight. in conclusion the acceleration will have a value equal to gravity but pointing upwards.
Machmer Hall is 400 m North and 180 m West of Witless.
Astudent is late for her appointment in Machmer and wants to
runthere directly from Witless.
What is the distance from Witless to Machmer?
Answer:
The distance from Witless to Machmer is 438.63 m.
Explanation:
Given that,
Machmer Hall is 400 m North and 180 m West of Witless.
We need to calculate the distance
Using Pythagorean theorem
[tex]D = \sqrt{(d_{m})^2+(d_{w})^2}[/tex]
Where, [tex]d_{m}[/tex] =distance of Machmer Hall
[tex]d_{w}[/tex] =distance of Witless
Put the value into the formula
[tex]D = \sqrt{(400)^2+(180)^2}[/tex]
[tex]D=438.63\ m[/tex]
Hence, The distance from Witless to Machmer is 438.63 m.
An airplane is heading due south at a speed of 430 km/h . A wind begins blowing from the northwest at a speed of 85.0 km/h (average).What should the plane's course shift be so that it will fly due south?
Answer:
It should fly 8° to west of south at 430km/h
Explanation:
According to the diagram. X components for both velocities must have the same magnitude in order to get the resultant velocity due south.
[tex]V_{w}*cos(45) = V_{A}*sin(\alpha )[/tex] Solving for α:
α = 8.03°
A placekicker must kick a football from a point 36.0 m (about 40 yards) from the goal. Half the crowd hopes the ball will clear the crossbar, which is 3.05 m high. When kicked, the ball leaves the ground with a speed of 22.8 m/s at an angle of 51.0° to the horizontal. (a) By how much does the ball clear or fall short of clearing the crossbar? (Enter a negative answer if it falls short.)
(b) Does the ball approach the crossbar while still rising or while falling?
Answer:
a) The ball clears the crossbar by 10.6 m
b) The ball approaches the crossbar while falling
Explanation:
The position of the ball is described by the vector position r (see attached figure):
r = (x0 + v0 t cos α ; y0 + v0 t sin α + 1/2 g t²)
where:
x0 = initial horizontal position
v0 = magnitude of the initial velocity
t = time
α = launching angle
y0 = initial vertical position
g = acceleration due to gravity (-9.8 m/s²)
The vector r is composed by rx and ry (see figure):
r = (rx ; ry)
a) Let´s find the time at which the ball flies a distance of 36.0 m. If at that time the vertical component of the vector r, ry, is equal or greater than 3.05 m, then the ball will clear the crossbar.
rx = x0 + v0 t cos α = 36.0 m
Since the origin of the system of reference is located where the kicker is, x0 = 0.
36.0 m = v0 t cos α
36.0 m /(v0 cos α) = t
36.0 m / (22.8 m/s * cos 51.0°) = t
t = 2.51 s
Now let´s calculate the height of the ball at that time:
ry = y0 + v0 t sin α + 1/2 g t²
Since the kicker is on the ground, y0 = 0
ry = 22.8 m/s * 2.51 s * sin 51.0° - 1/2 * 9.8 m/s² * (2.51 s)² = 13.6 m
Since the crossbar is 3.05 m high, the ball clears it by (13.6 m - 3.05 m) 10.6 m
b) Please see the figure to figure this out ;)
If the ball approaches the crossbar while still rising, the vertical component (vy) of the velocity vector will be positive. In change, if the ball approaches the crossbar while falling the vertical component of the velocity will be negative. See the figure.
The velocity vector is given by this equation:
v = (vx ; vy)
v = ( v0 cos α ; v0 sin α + g t)
Let´s see the vertical component at time t = 2.51
vy = v0 sin α + g t
vy = 22.8 m/s * sin 51.0° - 9.8 m/s² * 2.51 s
vy = -6.88 m/s
Then, the ball approaches the crossbar while falling.
An object moves in one dimensional motion with constant acceleration a = 5 m/s^2. At time t = 0 s, the object is at x0 = 2.7 m and has an initial velocity of v0 = 4.3 m/s. How far will the object move before it achieves a velocity of v = 6.4 m/s?
Your answer should be accurate to the nearest 0.1 m.
The object will move approximately 10.1 m before it achieves a velocity of 6.4 m/s with a constant acceleration of 5 m/s^2.
Explanation:To find the distance the object will move before it achieves a velocity of 6.4 m/s, we can use the equation v^2 = v0^2 + 2ax, where v is the final velocity, v0 is the initial velocity, a is the acceleration, and x is the distance. Rearranging the equation, we get x = (v^2 - v0^2) / (2a). Plugging in the values, we have x = (6.4^2 - 4.3^2) / (2 * 5) = 10.12 m. Therefore, the object will move approximately 10.1 m before it achieves a velocity of 6.4 m/s.
Final answer:
To find how far an object with constant acceleration a = 5 [tex]m/s^2[/tex] and initial velocity of 4.3 m/s moves before reaching a velocity of 6.4 m/s, use the kinematic equation [tex]v^2[/tex] = [tex]v0^2[/tex] + 2a(x - x0). After calculation, the object will move 5.0 m before reaching 6.4 m/s.
Explanation:
We need to calculate how far the object moves before it achieves a velocity of v = 6.4 m/s given constant acceleration a = 5 [tex]m/s^2[/tex], initial velocity v0 = 4.3 m/s, and starting position x0 = 2.7 m. We can use the kinematic equation which relates final velocity, initial velocity, acceleration, and displacement:
[tex]v^2[/tex] = [tex]v0^2[/tex] + 2a(x - x0)
Rearranging for displacement (x - x0):
(x - x0) = ([tex]v^2[/tex] - [tex]v0^2[/tex]) / (2a)
Substituting the given values:
(x - 2.7 m) = [tex](6.4 m/s)^2[/tex] - [tex](4.3 m/s)^2[/tex] / [tex](2 * 5 m/s^2)[/tex]
(x - 2.7 m) = (40.96 [tex]m^2/s^2[/tex] - 18.49 [tex]m^2/s^2[/tex]) / (10 [tex]m/s^2[/tex])
(x - 2.7 m) = 22.47 [tex]m^2/s^2[/tex] / (10 [tex]m/s^2[/tex])
(x - 2.7 m) = 2.247 m
x = 2.7 m + 2.247 m
x = 4.947 m
To the nearest 0.1 m, the object will move 5.0 m before it achieves a velocity of 6.4 m/s.
There are four charges, each with a magnitude of 1.96 µC. Two are positive and two are negative. The charges are fixed to the corners of a 0.47-m square, one to a corner, in such a way that the net force on any charge is directed toward the center of the square. Find the magnitude of the net electrostatic force experienced by any charge.
Answer:
Magnitude of the resultant force (Fn₁) on q₁
Fn₁ = 0.142N (directed toward the center of the square)
Explanation:
Theory of electrical forces
Because the particle q₁ is close to three other electrically charged particles, it will experience three electrical forces and the solution of the problem is of a vector nature.
Graphic attached
The directions of the individual forces exerted by q₂, q₃ and q₄ on q₁ are shown in the attached figure.
The force (F₁₄) of q₄ on q₁ is repulsive because the charges have equal signs and the forces (F₁₂) and (F₁₃) of q₂ and q₃ on q₁ are attractive because the charges have opposite signs.
Calculation of the forces exerted on the charge q₁
To calculate the magnitudes of the forces exerted by the charges q₂, q₃, and q₄ on the charge q₁ we apply Coulomb's law:
[tex]F_{12} = \frac{k*q_1*q_2}{r_{12}^2}[/tex]: Magnitude of the electrical force of q₂ over q₁. Equation((1)
[tex]F_{13} = \frac{k*q_1*q_3}{r_{13}^2}[/tex]: Magnitude of the electrical force of q₃ over q₁. Equation (2)
[tex]F_{14} = \frac{k*q_1*q_4}{r_{14}^2}[/tex]: Magnitude of the electrical force of q₄ over q₁. Equation (3)
Equivalences
1µC= 10⁻⁶ C
Known data
q₁=q₄= 1.96 µC = 1.96*10⁻⁶C
q₂=q₃= -1.96 µC = -1.96*10⁻⁶C
r₁₂= r₁₃ = 0.47m: distance between q₁ and q₂ and q₁ and q₄
[tex]r_{14} = \sqrt{0.47^2+ 0.47^2}=0.664m[/tex]
k=8.99x10⁹N*m²/C² : Coulomb constant
F₁₂ calculation
We replace data in the equation (1):
[tex]F_{12} = \frac{8.99*10^9*(1.96*10^{-6})^2}{0.47^2}[/tex]
F₁₂ = 0.156 N Direction of the positive x axis (+x)
F₁₃ calculation
We replace data in the equation (2):
[tex]F_{13} = \frac{8.99*10^9*(1.96*10^{-6})^2}{0.47^2}[/tex]
F₁₃ = 0.156 N Direction of the negative y axis (-y)
Magnitude of the net electrostatic force between F₁₃ and F₁₂
[tex]F_{n23}= \sqrt{0.156^2+0.156^2} = 0.22N[/tex] (directed toward the center of the square)
F₁₄ calculation
We replace F₁₄ data in the equation (3):
[tex]F_{14} = \frac{8.99*10^9*(1.96*10^{-6})^2}{0.664^2}[/tex]
F₁₄ = 0.078 N (In the opposite direction to Fn₂₃)
Calculation of the resulting force on q₁: Fn₁
Fn₁ = Fn₂₃ - F₁₄ = 0.22 - 0.078 = 0.142 N
Final answer:
The net electrostatic force on a charge can be found using Coulomb's law, considering the effects of both attraction and repulsion due to the charges' arrangement at the corners of a square. Calculations account for diagonal opposites and adjacent corners, taking advantage of the square's symmetry.
Explanation:
To calculate the magnitude of the net electrostatic force on a charge in a configuration where two positive charges of 1.96 µC and two negative charges of the same magnitude are placed at the corners of a 0.47-m square, we can use Coulomb's law. Coulomb's law states that the electrostatic force (F) between two point charges is proportional to the product of the charges (q1 and q2) and inversely proportional to the square of the distance (r) between them. It is given by F = k * |q1*q2| / r², where k is Coulomb's constant (k ≈ 8.99 x 10⁹ N·m²/C²).
Since the charges are at the corners of a square, the opposing charges will attract, and like charges will repel. Each charge experiences forces due to its interaction with the other three charges. The net force on any charge is the vector sum of the forces exerted by the other three charges. This sum can be simplified as the square has symmetry so that the forces along one diagonal cancel and the forces along the sides of the square add together to pull the charge towards the center.
The diagonal distance (r) is √2 times the side of the square (a), which is r = √2 * 0.47 m. For simplicity, let's calculate the force between one charge and its diagonal opposite, and then double it for symmetry reasons (each charge has two diagonally opposite charges).
The force due to one diagonal is: F_diagonal = (k * (1.96 x 10⁻⁶)^2) / ((√2 * 0.47)²)
Now we need to calculate this force and then double it to account for both diagonals.
The final step will be to add the forces due to charges on adjacent corners, which are of the same polarity and hence exert repulsion. Their vector sum will be directed towards the square's center due to symmetry.
A student places her 490 g physics book on a frictionless table. She pushes the book against a spring, compressing the spring by 7.10 cm , then releases the book. What is the book's speed as it slides away? The spring constant is 1550 N/m .
Answer:
book speed is 3.99 m/s
Explanation:
given data
mass m = 490 g = 0.490 kg
compressing x = 7.10 cm = 0.0710 m
spring constant k = 1550 N/m
to find out
book speed
solution
we know energy is conserve so
we can say
loss in spring energy is equal to gain in kinetic energy
so
[tex]\frac{1}{2}*k*x^2 = \frac{1}{2}*m*v^2[/tex] ..................1
put here value
[tex]\frac{1}{2}*1550*0.071^2 = \frac{1}{2}*0.490*v^2[/tex]
v = 3.99 m/s
so book speed is 3.99 m/s
The cubit is an ancient unit of length based on the distance between the elbow and the tip of the middle finger of the measurer. Assume that the distance ranged from 43 to 53 cm, and suppose that ancient drawings indicate that a cylindrical pillar was to have a length of 5.0 cubits and a diameter of 3.0 cubits. For the stated range, what are the lower values for (a) the cylinder's length in meters, (b) the cylinder's length in millimeters, and (c) the cylinder's volume in cubic meters? What are the upper values for (d) the cylinder's length in meters, (e) the cylinder's length in millimeters, and (f) the cylinder's volume in cubic meters?
Answer:
a) 215 cm
b) 1290 mm
c) 2.81 m³
d) 265 cm
e) 1590 mm
f) 5.26 m³
Explanation:
Lower value
1 cubit = 43 cm
Length of cylinder = 5 cubits
So,
a) Length of cylinder = 5×43 = 215 cm
Diameter of cylinder = 3 cubit
1 cubit = 43 cm = 430 mm
b) Diameter of cylinder = 3×430 = 1290 mm
Radius of cylinder = 129/2 = 64.5 cm
Volume of cylinder
[tex]v=\pi r^2h\\\Rightarrow v=\pi 0.645^2\times 2.15\\\Rightarrow v=2.81 m^3[/tex]
c) Volume of cylinder = 2.81 m³
Upper value = 53 cm
Length of cylinder = 5 cubits
So,
d) Length of cylinder = 5×53 = 265 cm
Diameter of cylinder = 3 cubit
1 cubit = 53 cm = 530 mm
e) Diameter of cylinder = 3×530 = 1590 mm
Radius of cylinder = 159/2 = 79.5 cm
Volume of cylinder
[tex]v=\pi r^2h\\\Rightarrow v=\pi 0.795^2\times 2.65\\\Rightarrow v=5.26 m^3[/tex]
f) Volume of cylinder = 5.26 m³
A 2000 kg car rounds a circular turn of radius 20 m. If
theroad is flat and the coefficient of friction between tires and
roadis 0.70, how fast can the car go without skidding?
Answer:
The velocity of car is 11.71 m/s.
Explanation:
Given that,
Mass of car = 2000 kg
Radius = 20 m
Coefficient of friction = 0.70
We need to calculate the velocity of car
Using relation centripetal force and frictional force
[tex]F= \dfrac{mv^2}{r}[/tex]...(I)
[tex]F=\mu mg[/tex]...(II)
From equation (I) and (II)
[tex]\dfrac{mv^2}{r}=\mu mg[/tex]
[tex]v=\sqrt{\mu\times r\times g}[/tex]
Put the value into the formula
[tex]v=\sqrt{0.70\times20\times9.8}[/tex]
[tex]v=11.71\ m/s[/tex]
Hence, The velocity of car is 11.71 m/s.
Answer:
car can move at 11 m/s without skidding.
Explanation:
given,
mass of car = 2000 kg
radius of turn = 20 m
μ = 0.7
using centripetal force
F = [tex]\dfrac{mV^2}{R}[/tex].....................(1)
and we know
F = μ N = μ m g........................(2)
equating both the equation (1) and (2)
μ m g = [tex]\dfrac{mV^2}{R}[/tex]
v = [tex]\sqrt{\mu R g}[/tex]
v = [tex]\sqrt{0.7 \times 20 \times 9.81}[/tex]
v = 11.71 m/s
hence, car can move at 11 m/s without skidding.
Find the volume of a slab whose dimensions are L=45.24 cm, W=23.981cm, H=2.33002 cm in proper significant digits
Answer:
The volume of the slab is 2527.84 cm
Explanation:
Only the first two decimals are significant digits because so are the data that the problem gives us.
You mount a ball launcher to a car. It points toward the back of the car 39 degrees above the horizontal and launches balloons at 9.7 m/s. When you drive the car straight ahead at 2.2 m/s, what is the initial angle in degrees of the ball's trajectory as seen by someone standing on the ground?
Answer:[tex]\theta =32.08 ^{\circ}[/tex]
Explanation:
Given
Ball launcher is mounted on car at angle of [tex]39^{\circ}[/tex]
launching velocity is u=9.7 m/s
Speed of car =2.2 m/s
So in horizontal component of balloon speed of car will be added
Thus [tex]u_x=9.7cos39+2.2=7.53+2.2=9.73 m/s[/tex]
[tex]u_y=9.7sin39=6.10 m/s[/tex]
therefore Appeared trajectory angle is
[tex]tan\theta =\frac{6.10}{9.73}=0.627[/tex]
[tex]\theta =32.08 ^{\circ}[/tex]
Two children on opposite sides of a 13 kg door (I = 1/3 mL 2 ) apply a force to the door. The first child pushes with force F at an angle of 80° relative to the door at a position r 1 from the door’s hinges. The second child pushes with a force of 15 N perpendicular to the door at a position 0.3 m from the door’s hinges. The width of the door is 0.4 m. What is the door’s angular acceleration?
Final answer:
The force that the second child must exert to keep the door from moving is 10.5 N.
Explanation:
To find the force that the second child must exert to keep the door from moving, we need to consider the torque applied by each child.
First, we calculate the torque applied by the first child:
T1 = F1 * r1 * sin(θ)
T1 = (17.5 N) * (0.600 m) * sin(90°) = 10.5 N·m
The torque applied by the second child is:
T2 = F2 * r2
T2 = F2 * (0.450 m)
Since the door is not moving, the net torque must be zero. Therefore:
T1 + T2 = 0
Substituting the values:
10.5 N·m + F2 * (0.450 m) = 0
Simplifying, we find that F2 = -10.5 N * (0.450 m) / (0.450 m) = -10.5 N
Therefore, the force that the second child must exert to keep the door from moving is 10.5 N.
If my mass is 196 lbm and I tackle one of my teammates - while decelerating from a velocity of 6.7 m/s to 0 m/s in 0.5 s, how much force (kN) do I impart to the teammate's body?
Answer:
the force acting on the team mate is 1.19 kN.
Explanation:
given,
mass = 196 lbm
while tackling, the deceleration is from velocity 6.7 m/s to 0 m/s
time taken for deceleration = 0.5 sec
F = mass × acceleration
acceleration = [tex]\dfrac{0-6.7}{0.5}[/tex]
= -13.4 m/s²
1 lbs = 0.453 kg
196 lbs = 196 × 0.453 = 88.79 kg
F = 88.79 × 13.4
F = 1189.786 N = 1.19 kN
hence, the force acting on the team mate is 1.19 kN.
Two balls are kicked with the same initial speeds. Ball A was kicked at the angle 20° above horizontal and ball B was kicked at the angle 75° above horizontal. What ball will have bigger speed at the highest point of their trajectory? O Ball A O Ball B O Their will have equal speeds O Impossible to answer without knowing their actual initial speeds.
Answer:
Ball A
Explanation:
Let the initial speed of the balls be u .
Angle of projection for ball A = 20°
Angle of projection for ball B = 75°
As we know that at highest point, the ball has only horizontal speed which always remains constant throughout the motion because the acceleration in horizontal direction is zero.
Speed of ball A at highest point = u Cos 20° = 0.94 u
Speed of ball B at highest point = u Cos 75° = 0.26 u
So, the ball A has bigger speed than B.
Which of the following has the greatest kinetic energy? A. a mass of 4m at velocity v. B. a mass of 3m at velocity 2v. C. a mass of 2m at velocity 3v. D. a mass of m with a velocity of 4v.
Answer:
Option C
Explanation:
Kinetic energy is the energy that the body possesses by virtue of its motion.
The formula for Kinetic energy is given by [tex]\frac{1}{2} mv^2[/tex]
Using this formula let us find kinetic energy for the bodies given and find out which is the greatest
A) KE = [tex]\frac{1}{2} (4m)(v^2) = 2mv^2[/tex]
B) KE =[tex]\frac{1}{2} (3m)(2v)^2 = 6mv^2[/tex]
C) KE = [tex]\frac{1}{2} (2m)(3v)^2 = 9mv^2[/tex]
D) KE = [tex]\frac{1}{2} (3)(4v)^2 = 8mv^2[/tex]
Comparing these we find that 9mv^2 is the highest.
Hence option C is the answer.
Your code returns a number of 99.123456789 +0.00455679 for your calculation. How should you report it in your lab write-up?
Answer: Your code returns a number of 99.123456789 +0.00455679
Ok, you must see where the error starts to affect your number.
In this case, is in the third decimal.
So you will write 99.123 +- 0.004 da da da.
But you must round your results. In the number you can see that after the 3 comes a 4, so the 3 stays as it is.
in the error, after the 4 comes a 5, so it rounds up.
So the final presentation will be 99.123 +- 0.005
you are discarding all the other decimals because the error "domains" them.