The magnitudes of vectors A and B can be determined by resolving each vector into its x and y components, setting up equations for their sum, and solving the equations simultaneously.
Explanation:To find the magnitudes of vectors A and B when their sum with vector C results in a zero vector (A + B + C = 0), we can break down each vector into its x and y components and then solve the component equations separately.
Since vector C points in the negative y-direction and has a magnitude of 21.0 m, its components are Cx = 0 m and Cy = -21.0 m.
Vector A has a positive y-component and makes an angle of α = 43.4° with the positive x-axis. Thus, the components of A are Ax = A cos(α) and Ay = A sin(α).
Vector B also has a positive y-component and makes an angle of β = 27.7° with the negative x-axis, which is the same as 180° - β with the positive x-axis. So, B's components are Bx = -B cos(180° - β) = -B cos(β) and By = B sin(180° - β) = B sin(β).
The equations for the x and y components of the vector sum being zero are:
Substituting the component expressions in, we get:
These are two equations with two unknowns (the magnitudes A and B), which can be solved simultaneously to find the values of A and B. The solution involves elementary algebra and the use of trigonometric identities.
A toy rocket, launched from the ground, rises vertically with an acceleration of 18 m/s^2 for 18 s until its motor stops. Disregarding any air resistance, what max- imum height above the ground will the rocket achieve? The acceleration of gravity is 9.8 m/s^2. Answer in units of km.
Answer:
8271.92 m
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration
Equation of motion
[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow s=0\times t+\frac{1}{2}\times 18\times 18^2\\\Rightarrow s=2916\ m[/tex]
The height reached at 18 seconds is 2916 m
[tex]v=u+at\\\Rightarrow v=0+18\times 18\\\Rightarrow v=324\ m/s[/tex]
The velocity at 2916 m is 324 m/s
[tex]v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-324^2}{2\times -9.8}\\\Rightarrow s=5355.92\ m[/tex]
Maximum height reached by the toy rocket is 2916+5355.92 = 8271.92 m
The maximum height above the ground that the rocket will achieve is 8,271.92m.
To get the maximum height, we need to calculate the height reached by the rocket.
Using the equation:
[tex]S=ut+\frac{1}{2}at^2\\S =0(t) + \frac{1}{2}\times18\times 18^2\\S=9 \times 18^2\\S=2,916m[/tex]
Get the velocity also using the equation of motion:
[tex]v=u+at\\v=0+18(18)\\v=324m/s[/tex]
Get the maximum height above the ground the rocket will achieve:
[tex]v^2=u^2+2as\\324^2=0^2+2(9.8)s\\104,976=19.6s\\s=\frac{104,976}{19.6}\\s= 5,355.92m[/tex]
The total maximum height reached by rocket is 2,916 + 5,355.92 = 8,271.92m.
Learn more here: https://brainly.com/question/24018491
The driver of a car slams on the brakes when he sees a tree blocking the road. The car slows uniformly with acceleration of −5.25 m/s2 for 4.15 s, making straight skid marks 62.5 m long, all the way to the tree. With what speed (in m/s) does the car then strike the tree? You may want to calculate the initial velocity of the car first. m/s (b) What If? If the car has the same initial velocity, and if the driver slams on the brakes at the same distance from the tree, then what would the acceleration need to be (in m/s2) so that the car narrowly avoids a collision? m/s2
Answer:
Part a)
Final speed of the car is
[tex]v_f = 4.17 m/s[/tex]
Part b)
Acceleration of the car is
[tex]a = -5.39 m/s^2[/tex]
Explanation:
As we know that car makes a skid of 62.5 m
here acceleration of the car is
[tex]a = - 5.25 m/s^2[/tex]
now we have
[tex]d = v_i t + \frac{1}{2}at^2[/tex]
[tex]62.5 = v_i (4.15) + \frac{1}{2}(-5.25)(4.15^2)[/tex]
[tex]v_i = 25.95 m/s[/tex]
Part a)
Speed of the car by which it will hit the tree
[tex]v_f = v_i + at[/tex]
[tex]v_f = 25.95 - (5.25)(4.15)[/tex]
[tex]v_f = 4.17 m/s[/tex]
Part b)
Now if car will stop after travelling same distance which same initial speed
Then we can use kinematics
[tex]v_f^2 - v_i^2 = 2 a d[/tex]
[tex]0 - 25.95^2 = 2(a)(62.5)[/tex]
[tex]a = -5.39 m/s^2[/tex]
Acceleration of a object is the rate of change of velocity of the object per unit time.
a) The speed does the car then strike the tree 4.17 m/s.b) The acceleration need to be, so that the car narrowly avoids a collision -5.39 m/s squared.What is the acceleration of a object?Acceleration of a object is the rate of change of velocity of the object per unit time.
Given information-
The car slows uniformly with acceleration of −5.25 m/s squared for 4.15 s.
The skid made by the car is 62.5 m long.
The initial velocity of the car can be find out using the distance formula of motion as,
[tex]d=ut+\dfrac{1}{2}at^2[/tex]
Put the values as,
[tex]62.5=u\times4.15+\dfrac{1}{2}\times(-5.25)\times(4.15)^2\\u=25.95[/tex]
Thus the value of initial velocity is 25.95 m/s.
(a) The speed does the car then strike the tree-The velocity formula using the equation of motion can be given as,
[tex]v=u+at\\v=25.95+(-5.25)\times4.15\\v=4.17\rm m/s[/tex]
Thus, the speed does the car then strike the tree 4.17 m/s.
(b) The acceleration need to be, so that the car narrowly avoids a collision-To stop the car the final velocity of it must be 0. Thus,
[tex]2ad=v^2-u^2\\2\times a\times62.5=0-25.95^2\\a=-5.93\rm m/s^2[/tex]
The acceleration need to be, so that the car narrowly avoids a collision -5.39 m/s squared.
Hence,
a) The speed does the car then strike the tree 4.17 m/s.b) The acceleration need to be, so that the car narrowly avoids a collision -5.39 m/s squared.Learn more about the acceleration here;
https://brainly.com/question/605631
A particle starts from the origin at t = 0 with an initial velocity having an x component of 29.8 m/s and a y component of −18 m/s. The particle moves in the xy plane with an x component of acceleration only, given by 5.21 m/s 2 . Determine the x component of velocity after 7.22 s. Answer in units of m/s.
Answer:
Vx(7,22s)=67,42m/s
Explanation:
V(t)=Vo+a(t) [tex]Vx(7,22s)=Vxo+5,21\frac{m}{s^{2} } *7,22s
Vx(7,22s)=29,8m/s +37,62m/s=67,42m/s[/tex]
you just consider the x component of your velocity, then you apply the Velocity formula of uniform accelerated rectilinear motion, and you get to the result, by replacing t (time) by the value given 7,22s.
Suppose you take a 50gram ice cube from the freezer at an initial temperature of -20°C. How much energy would it take to completely vaporize the ice cube? (Hint: think of this process as four separate steps and calculate the energy needed for each one.)
Answer:
The amount of energy required is [tex]152.68\times 10^{3}Joules[/tex]
Explanation:
The energy required to convert the ice to steam is the sum of:
1) Energy required to raise the temperature of the ice from -20 to 0 degree Celsius.
2) Latent heat required to convert the ice into water.
3) Energy required to raise the temperature of water from 0 degrees to 100 degrees
4) Latent heat required to convert the water at 100 degrees to steam.
The amount of energy required in each process is as under
1) [tex]Q_1=mass\times S.heat_{ice}\times \Delta T\\\\Q_1=50\times 2.05\times 20=2050Joules[/tex]
where
[tex]'S.heat_{ice}[/tex]' is specific heat of ice =[tex]2.05J/^{o}C\cdot gm[/tex]
2) Amount of heat required in phase 2 equals
[tex]Q_2=L.heat\times mass\\\\\therefore Q_{2}=334\times 50=16700Joules[/tex]
3) The amount of heat required to raise the temperature of water from 0 to 100 degrees centigrade equals
[tex]Q_3=mass\times S.heat\times \Delta T\\\\Q_1=50\times 4.186\times 100=20930Joules[/tex]
where
[tex]'S.heat_{water}[/tex]' is specific heat of water=[tex]4.186J/^{o}C\cdot gm[/tex]
4) Amount of heat required in phase 4 equals
[tex]Q_4=L.heat\times mass\\\\\therefore Q_{4}=2260\times 50=113000Joules\\\\[/tex][tex]\\\\\\\\[/tex]Thus the total heat required equals [tex]Q=Q_{1}+Q_{2}+Q_{3}+Q_{4}\\\\Q=152.68\times 10^{3}Joules[/tex]
It would take 200 kJ of energy to completely vaporize the ice cube.
Explanation:The process of completely vaporizing an ice cube can be broken down into four separate steps, each requiring a certain amount of energy:
Heating the ice cube from -20°C to 0°C: This step requires 4.1 kJ of energy.Melting the ice at 0°C: This step requires 133.6 kJ of energy.Heating the resulting water from 0°C to 37°C: This step requires 61.9 kJ of energy.Vaporizing the water at 37°C: This step requires an additional 0 kJ of energy, since the water is already at its boiling point.In total, it would take 200 kJ of energy to completely vaporize the ice cube.
A rocket carrying a satellite is accelerating straight up from the earth's surface. At 1.15 s after liftoff, the rocket clears the top of its launch platform, 70 m above the ground. After an additional 4.70 s, it is 1.15 km above the ground. Part A
Calculate the magnitude of the average velocity of the rocket for the 4.70 s part of its flight.
Express your answer in meters per second. Part B
Calculate the magnitude of the average velocity of the rocket the first 5.85 s of its flight.
Express your answer in meters per second.
Answer:
a) [tex]v=230 m/s[/tex]
b) [tex]v=196.5 m/s[/tex]
Explanation:
a) The formula for average velocity is
[tex]v=\frac{y_{2}-y_{1} }{t_{2}-t_{1} }[/tex]
For the first Δt=4.7s
[tex]v=\frac{(1150-70)m}{(4.7)s} =230 m/s[/tex]
b) For the secont Δt=5.85s we know that the displacement is 1150m. So, the average velocity is:
[tex]v=\frac{(1150)m}{(5.85)s}=196.5m/s[/tex]
A car and a truck are both traveling at the constant speed of 35 mi/h; the car is 40 ft behind the truck. The driver of the car wants to pass the truck, i.e., he wishes to place his car at B, 40 ft in front of the truck, and then resume the speed of 35 mi/h. The maximum acceleration of the car is 5 ft/s2 and the maximum deceleration obtained by applying the brakes is 20 ft/s2. What is the shortest time in which the driver of the car can complete the passing operation if he does not at any time exceed a speed of 50 mi/h
Final answer:
To find the shortest time for the passing maneuver, we must consider the car's acceleration and deceleration phases and the distance to be traveled. The task involves calculations using the kinematic equations for uniformly accelerated motion.
Explanation:
To calculate the shortest time in which the driver of the car can complete the passing operation, we must consider the car's acceleration, maximum speed, and the distance to be covered. The car must not only reach a point 40 feet in front of the truck but also get back to the speed of 35 mi/h after overtaking, all while not exceeding a speed of 50 mi/h. The entire maneuver consists of accelerating to the maximum speed, maintaining that speed for a portion of the pass, and then decelerating back to 35 mi/h.
First, convert speeds from miles per hour to feet per second:
35 mi/h = 51.33 ft/s (approximately),
50 mi/h = 73.33 ft/s (approximately).
The car needs to cover an initial 40 ft behind the truck and a further 40 ft to be ahead, for a total of 80 ft. We would calculate the time needed to accelerate to 50 mi/h (maximum passing speed), the time spent traveling at this top speed, and then the time to decelerate back to 35 mi/h, all while covering the required distance. The actual calculations involve using the kinematic equations for uniformly accelerated motion; however, without further details on the constraints such as road length, traffic laws, and exact acceleration and deceleration phases, a precise number cannot be produced.
You are racing your dog next to a football field. You run from the 10 yard line to the 50 yard line, then turn around and run back to the 30 yard line. What is your distance? What is your displacement?
Answer:60 m,20 m
Explanation:
Given
You ran 10 yard line to 50 yard line therefore total distance traveled is 40 yard
Finally you turned around and came back to 30 yard line
Therefore total distance is 40+20=60 m(From 50 yard line to 30 yard line it is 20 yards)
Displacement=30-10=20 yards
An Earth satellite needs to have its orbit changed so the new orbit will be twice as far from the center of Earth as the original orbit. The new orbital period will be twice as long as the original period. O true O false
Answer:
False.
Explanation:
From Kepler's Third Law of plenetary motion, we know that:
"The square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit."
Or, as expressed in mathematical terms:
[tex]\frac{a^3}{T^2}=constant[/tex], where a is the semi-major axis of the orbit (the distance from the center), and T is the orbital period of the satellite.
From this expression we can clearly see that if the orbit's semi-major axis is doubled, orbital period will be [tex]\sqrt{8}[/tex] times longer to compensate the variation.
A squirrel runs along an overhead telephone wire that stretches from the top of one pole to the next. The creature is initially at position xi = 3.65 m, as measured from the center of the wire segment. It then undergoes a displacement of Δx = -6.81 m. What is the squirrel\'s final position xf? xf = _____ m
Answer:
xf = - 3.16 m
Explanation:
the squirrel was initially in the position xi = 3.65 m, then it had a displacement of Δx = -6.81 m.
The negative sign indicates that it moved in the opposite direction, so we must subtract this displacement Δx = -6.81 m to the initial position xi = 3.65 m, to find its final position.
3.65 m - 6.81 m = - 3.16 m
An architect is redesigning a rectangular room on the blueprints of a house. She decides to double the width of the room, increase the length by 50 percent, and increase the height by 20 percent. By what factor has the volume of the room increased?
Answer:3.6
Explanation:
Given
Architect decide to double the width
length increases by 50 percent
height increases by 20 percent
Let L,B,H be length,height and width of rectangular room
L'=1.5L
B'=2B
H'=1.2H
therefore new volume V'=L'\times B'\times H'
[tex]V'=1.5L\times 2B\times 1.2H[/tex]
V'=3.6LBH
new volume is 3.6 times the original
Your FHE brother (mass 144.1 kg) is canoeing on Utah Lake with his sister (mass 81.6 kg). He had the only paddle but managed to drop it in the lake. It has drifted just out of his reach along the long axis of the canoe. His sister has taken Physics 121 and suggests that if they trade places she might be able to reach the paddle. Due to the high "slime" content of the lake and the slick surface of the canoe, the canoe moves in the water without any friction or drag. The passengers sit 2.03 meters apart and the canoe has a mass of 48.7 kg. After the passengers switch places, how much closer to the paddle (which does not move) is the end of the canoe? (Enter a positive number.)
Answer:
The end of the canoe is 46.27 cm closer to the paddle.Explanation:
This problem can be solved using the center of mass of the system.
As there are no external forces in the horizontal direction, and the external forces in the vertical direction, the gravitational force and the buoyancy, cancel each other, we know:
[tex]\vec{F}_{net_{external}} = M \frac{d\vec{P}}{dt} = 0[/tex]
where M is the total mass of the system, and [tex]\vec{P}[/tex] is the linear momentum of the system. This implies:
[tex]\frac{d\vec{P}}{dt} = 0[/tex]
so
[tex]\vec{P} = constant[/tex]
the linear momentum of the system is:
[tex]\vec{P} = M \vec{r}_{cm}[/tex]
where [tex] \vec{r}_{cm}[/tex] is the position of the center of mass. As the mass in this problem is constant, this implies that the position of the center of mass is conserved.
Finding the center of massFor a system formed by i particles, each of mass [tex]m_i[/tex] and position [tex]\vec{r}_i[/tex], the center of mass can be found with the equation:
[tex]\vec{r}_{cm} = \frac{\sum\limits_i m_i \vec{r}_i }{M}[/tex]
where M is the total mass of the system, found with the equation:
[tex]M = \sum\limits_i m_i [/tex]
For this problem, the total mass of the system is :
[tex]M = m_{brother} + m_{sister} + m_{canoe}[/tex]
[tex]M = 144.1 \ kg + 81.6 \ kg + 48.7 \ kg[/tex]
[tex]M = 274.4 \ kg[/tex]
To find the positions, we need an frame of reference. Putting the origin of coordinate at the position of the sister.
[tex]\vec{r}_{sister_i} = 0 \ \hat{i}[/tex]
The brother is 2.03 meters apart in the positive direction:
[tex]\vec{r}_{brother_i} = 2.03 m \ \hat{i}[/tex]
and the center of mass of the canoe must be at half this distance:
[tex]\vec{r}_{canoe_i} = 1.015 m \ \hat{i}[/tex]
taking all this, the center of mass of the system is at :
[tex]\vec{r}_{cm} = \frac{ 81.6 \ kg * 0 \ \hat{i} + 144.1 \ kg \ 2.03 m \ \hat{i} + 48.7 \ kg \ 1.015 m \ \hat{i} }{274.4 \ kg}[/tex]
[tex]\vec{r}_{cm} = 1.2462 \ m\ \hat{i} [/tex]
Finding the displacementAfter the switch, we found that the center of mass has to be displaced. If its displaced a distance d, we find that the new positions must be:
[tex]\vec{r}_{sister_f} = 2.03 m \ \hat{i} + d \ \hat{i}[/tex]
[tex]\vec{r}_{brother_f} = 0 \ \hat{i} + d \ \hat{i}[/tex]
[tex]\vec{r}_{canoe_f} = 1.015 m \ \hat{i} + d \ \hat{i}[/tex]
where the brother and the sister had switched the original position. Now, the center of mass will be located at:
[tex]\vec{r}_{cm} = 1.2462 \ m\ \hat{i} [/tex]
And the equation is :
[tex]\vec{r}_{cm} = \frac{ 81.6 \ kg * (2.03 m \ \hat{i} + d \ \hat{i} ) + 144.1 \ kg \ d\ \hat{i} + 48.7 \ kg \ ( 1.015 m \ \hat{i} + d \ \hat{i}) }{274.4 \ kg}[/tex]
[tex]\vec{r}_{cm} = 0.7838 \ m \ \hat{i} + \frac{ 81.6 \ kg d \ \hat{i} ) + 144.1 \ kg \ d\ \hat{i} + 48.7 \ kg \ d \ \hat{i}) }{274.4 \ kg}[/tex]
[tex]\vec{r}_{cm} = 0.7838 \ m \ \hat{i} + \frac{ 81.6 \ kg + 144.1 \ kg + 48.7 \ kg \ }{274.4 \ kg} \ d \ \hat{i}[/tex]
[tex]\vec{r}_{cm} = 0.7838 \ m \ \hat{i} + d \ \hat{i}[/tex]
[tex]1.2462 \ m\ \hat{i} = 0.7838 \ m \ \hat{i} + d \ \hat{i}[/tex]
[tex]d \hat{i} = 0.4627 \ m \ \hat{i} [/tex]
And this is how much closer the paddle is to the end of the canoe.
After the passengers switch places, the distance to the paddle (which does not move) is the end of the canoe is 46.27 m.
What is center of mass?The center of mass of a object or for a system of object is the point, where the center of distribution of mass in space.
FHE brother (mass 144.1 kg) is canoeing on Utah Lake with his sister (mass 81.6 kg) and canoe has a mass of 48.7 kg.
Let the position of the sister at the origin, his brother is at 2.03 meters. The position of the center of the mass of the canoe is at half distance (1.015 m). Thus, the center of mass of the system is,
[tex]r_{COM}=\dfrac{144.1(2.03)+81.6(0)+48.7(1.015)}{144.1+81.6+48.7}\\r_{COM}=1.2462[/tex]
For the displacement, the equation can be given as,
[tex]r_{COM}=\dfrac{144.1(d)+81.6(2.03+d)+48.7(1.015+d)}{144.1+81.6+48.7}\\1.2462=\dfrac{144.1(d)+81.6(2.03+d)+48.7(1.015+d)}{144.1+81.6+48.7}\\d=0.4627\rm \; m[/tex]
Thus, after the passengers switch places, the distance to the paddle (which does not move) is the end of the canoe is 46.27 m.
Learn more about the center of mass here;
https://brainly.com/question/13499822
A solid conducting sphere of radius 2.00 cm has a charge of 6.88 μC. A conducting spherical shell of inner radius 4.00 cm and outer radius 5.00 cm is concentric with the solid sphere and has a charge of −2.96 μC. Find the electric field at the following radii from the center of this charge configuration. (a) r = 1.00 cm magnitude:_________ N/C
direction:____________
(b) r = 3.00 cm magnitude:____________ N/C
direction:_________________
(c) r = 4.50 cm magnitude:_____________ N/C direction:______________
(d) r = 7.00 cm magnitude:_____________ N/C
direction:______________
Explanation:
Given that,
Radius R= 2.00
Charge = 6.88 μC
Inner radius = 4.00 cm
Outer radius = 5.00 cm
Charge = -2.96 μC
We need to calculate the electric field
Using formula of electric field
[tex]E=\dfrac{kq}{r^2}[/tex]
(a). For, r = 1.00 cm
Here, r<R
So, E = 0
The electric field does not exist inside the sphere.
(b). For, r = 3.00 cm
Here, r >R
The electric field is
[tex]E=\dfrac{kq}{r^2}[/tex]
Put the value into the formula
[tex]E=\dfrac{9\times10^{9}\times6.88\times10^{-6}}{(3.00\times10^{-2})^2}[/tex]
[tex]E=6.88\times10^{7}\ N/C[/tex]
The electric field outside the solid conducting sphere and the direction is towards sphere.
(c). For, r = 4.50 cm
Here, r lies between R₁ and R₂.
So, E = 0
The electric field does not exist inside the conducting material
(d). For, r = 7.00 cm
The electric field is
[tex]E=\dfrac{kq}{r^2}[/tex]
Put the value into the formula
[tex]E=\dfrac{9\times10^{9}\times(-2.96\times10^{-6})}{(7.00\times10^{-2})^2}[/tex]
[tex]E=5.43\times10^{6}\ N/C[/tex]
The electric field outside the solid conducting sphere and direction is away of solid sphere.
Hence, This is the required solution.
While John is traveling along a straight interstate highway, he notices that the mile marker reads 245 km. John travels until he reaches the 135 km marker and then retraces his path to the 176 km marker. What is John’s resultant displacement from the 245 km marker? Answer in units of km.
Answer:Displacement =69 km
Explanation:
Given
First John notices 245 km marker i.e. he is 245 away from destination.
John travels until he reach 135 km marker i.e. he is 135 km away from destination.
After that he retraces his path to reach 176 km marker
so his resultant displacement is i.e. shortest distance between initial and final point is 245-176 =69 km
A bird flying at a height of 12 m doubles its speed as it descends to a height of 6.0 m. The kinetic energy has changed by a factor of : a) 2 b) 4 c) 1 d) 0.25
Answer:
Option d)
Explanation:
As the Kinetic energy of the body ids due to the speed of the body and depends on it, it does not depend on the height of the body.
The kinetic energy is given by:
KE = [tex]\frac{1}{2}mv^{2}[/tex]
where
m = mass of the body
v = kinetic energy of the body
Now, as per the question:
Initial velocity is u
and final velocity, v = 2u
Thus
Initial KE = [tex]\frac{1}{2}mu^{2}[/tex] (1)
FInal KE = [tex]\frac{1}{2}mv^{2}[/tex] = [tex]\frac{1}{2}m(2u)^{2}[/tex]
FInal KE = 4[tex]\frac{1}{2}mu^{2}[/tex] (2)
Thus from eqn (1) and (2):
Final KE = 4(Initial KE)
The Kinetic Energy has changed by a factor of 4
Two point charges are fixed on the y axis: a negative point charge q1 = -27 µC at y1 = +0.21 m and a positive point charge q2 at y2 = +0.35 m. A third point charge q = +9.0 µC is fixed at the origin. The net electrostatic force exerted on the charge q by the other two charges has a magnitude of 23 N and points in the +y direction. Determine the magnitude of q2.
The magnitude of [tex]\( q_2 \)[/tex] is [tex]\( 674.99 \, \mu \text{C} \)[/tex].
To find the magnitude of [tex]\( q_2 \)[/tex], we can use Coulomb's Law and the principle of superposition. Coulomb's Law states that the force between two point charges is given by:
[tex]\[ F = \frac{k \cdot |q_1 \cdot q|}{r_1^2} + \frac{k \cdot |q_2 \cdot q|}{r_2^2} \][/tex]
where ( k ) is Coulomb's constant [tex](\( 8.99 \times 10^9 \, \text{N}\cdot\text{m}^2/\text{C}^2 \))[/tex], [tex]\( q_1 \)[/tex] and [tex]\( q_2 \)[/tex] are the charges, [tex]\( q \)[/tex] is the test charge, and [tex]\( r_1 \)[/tex] and [tex]\( r_2 \)[/tex] are the distances from [tex]\( q \)[/tex] to the charges [tex]\( q_1 \)[/tex] and [tex]\( q_2 \)[/tex], respectively.
Given the problem, we have [tex]\( F = 23 \, \text{N} \)[/tex], [tex]\( q_1 = -27 \times 10^{-6} \, \text{C} \)[/tex], [tex]\( q = 9 \times 10^{-6} \, \text{C} \)[/tex], [tex]\( r_1 = 0.21 \, \text{m} \)[/tex], [tex]\( r_2 = 0.35 \, \text{m} \)[/tex], and we need to find [tex]\( q_2 \)[/tex].
First, let's calculate the force due to [tex]\( q_1 \)[/tex]:
[tex]\[ F_1 = \frac{k \cdot |q_1 \cdot q|}{r_1^2} = \frac{8.99 \times 10^9 \cdot |-27 \times 10^{-6} \cdot 9 \times 10^{-6}|}{(0.21)^2} \][/tex]
[tex]\[ F_1 = \frac{8.99 \times 10^9 \cdot 243 \times 10^{-12}}{0.0441} = \frac{2185.57}{0.0441} = 49536.73 \, \text{N} \][/tex]
Now, the force due to [tex]\( q_2 \)[/tex]:
[tex]\[ F_2 = \frac{k \cdot |q_2 \cdot q|}{r_2^2} \][/tex]
We know the net force is in the positive y-direction, so [tex]\( F_2 \)[/tex] must be positive.
Now, sum the forces and equate to the given net force:
[tex]\[ F = F_1 + F_2 \][/tex]
[tex]\[ 23 = 49536.73 + F_2 \][/tex]
[tex]\[ F_2 = 23 - 49536.73 = -49513.73 \, \text{N} \][/tex]
Now, plug in [tex]\( F_2 \)[/tex] and solve for [tex]\( q_2 \)[/tex]:
[tex]\[ -49513.73 = \frac{8.99 \times 10^9 \cdot |q_2 \cdot 9 \times 10^{-6}|}{(0.35)^2} \][/tex]
[tex]\[ -49513.73 = \frac{8.09 \times 10^9 \cdot |q_2|}{0.1225} \][/tex]
[tex]\[ |q_2| = \frac{-49513.73 \cdot 0.1225}{8.99 \times 10^9} \][/tex]
[tex]\[ |q_2| = -674.99 \times 10^{-6} \][/tex]
Since [tex]\( q_2 \)[/tex] is positive, [tex]\( q_2 = 674.99 \times 10^{-6} \, \text{C} \)[/tex].
Thus, the magnitude of [tex]\( q_2 \)[/tex] is [tex]\( 674.99 \, \mu \text{C} \)[/tex].
A policeman, parked at arn intersection, is passed by a speeder traveling at 120 km/hr. Two seconds later, the policeman starts and uniformly accelerates at 4.0 m/s^2. How far will the policeman travel before he catches the speeder?
Answer:
682.32 m
Explanation:
Speed of the passing speeder = 120 km/hr = 120 × 0.2777 = 33.33 m/s
time after which police man starts = 2 seconds
Acceleration of the policeman = 4.0 m/s²
let the time taken to catch be 't' seconds
Now,
The total distance to be covered by the policeman
= Distance covered by the speeder in the 2 seconds + Distance further traveled by the speeder in time t
thus,
From Newton's equation of motion
[tex]s=ut+\frac{1}{2}at^2[/tex]
where,
s is the total distance traveled by the police man
u is the initial speed = 0
a is the acceleration
t is the time
thus,
[tex]0\times t+\frac{1}{2}\times4t^2[/tex]= 33.33 × 2 + 33.33 × t
or
2t² = 66.66 + 33.33t
or
2t² - 33.33t - 66.66 = 0
on solving the above equation, we get
t = 18.47 seconds (negative value is ignored as time cannot be negative)
therefore,
the total distance covered = 33.33 × 2 + 33.33 × 18.47 = 682.32 m
A rocket is fired straight upward, starting from rest with an acceleration of 25.0 m/s^2. It runs out of fuel at the end of 4.00 s and continues to coast upward, reaching a maximum height before falling back to earth. (a) Find the rocket height and velocity when it runs out of fuel. (b) Find the maximum height the rocket reaches. (c) Find the velocity the instant before the rocket crashes into the ground. (d) Find the total elapsed time from launch to ground impact.
Answer:
a) 200m, 100m/s
b) 710.20m
c) -117.98 m/s
d) 26.24 s
Explanation:
To solve this we have to use the formulas corresponding to a uniformly accelerated motion problem:
[tex]V=Vo+a*t[/tex] (1)
[tex]X=Xo+Vo*t+\frac{1}{2}*a*t^2\\[/tex] (2)
[tex]V^2=Vo^2+2*a*X[/tex] (3)
where:
Vo is initial velocity
Xo=intial position
V=final velocity
X=displacement
a)
[tex]X=0+0*4+\frac{1}{2}*25*4^2[/tex]
the intial position is zero because is lauched from the ground and the intial velocitiy is zero because it started from rest.
[tex]X=200m[/tex]
[tex]V=0+25*4[/tex]
[tex]V=100m/s[/tex]
b)
The intial velocity is 100m/s we know that because question (a) the acceleration is -9.8[tex]\frac{m}{s^2}[/tex] because it is going downward.
[tex]0=100^2+2*(-9.8)*X\\X=510.20\\totalheight=200+510.20=710.20m[/tex]
c)
In order to find the velocity when it crashes, we can use the formula (3).
the initial velocity is 0 because in that moment is starting to fall.
[tex]V^2=0^2+2*(-9.8)*(710.20)\\V=-117.98 m/s[/tex]
the minus sign means that the object is going down.
d)
We can find the total amount of time adding the first 4 second and the time it takes to going down.
to calculate the time we can use the formula (2) setting the reference at 200m:
[tex]-200=0+100*t+\frac{1}{2}*(-9.8)*t^2[/tex]
solving this we have: time taken= 22.24 seconds
total time is:
total=22.24+4=26.24 seconds.
How fast does a rocket ship have to go for its length to be contracted to 99% of its rest length?
Answer:
4.2×10⁷ m/s
Explanation:
L = length observed by an observer
L₀ = Proper length
v = Velocity of the rocket ship
Length contraction formula
[tex]L=L_{0}{\sqrt{1-\frac{v^{2}}{c^{2}}}}\\\Rightarrow 0.99=1{\sqrt{1-\frac{v^{2}}{c^{2}}}}\\\Rightarrow \frac{0.99}{1}={\sqrt{1-\frac{v^{2}}{c^{2}}}}\\\Rightarrow 0.99^2=1-\frac{v^{2}}{c^{2}}\\\Rightarrow 0.99^2-1=-\frac{v^2}{c^2}\\\Rightarrow 0.0199\times c^2=v^2\\\Rightarrow v=\sqrt{0.0199}c=v\\\Rightarrow v =\sqrt{0.0199}\times 3\times 10^8\\\Rightarrow v=4.2\times 10^7\ m/s[/tex]
The speed of the ship would be 4.2×10⁷ m/s
What is the de Broglie wavelength for a proton (m = 1.67× 10^−27 kg) moving at a speed of 9.50 × 10^6 m/s? (h = 6.63 × 10^−34 J⋅s)
Answer:
De broglie wavelength = 4.17 pm
Explanation:
We have given mass of proton [tex]m=1.67\times 10^{-27}kg[/tex]
Speed of proton v [tex]v=9.50\times 10^{6}m/sec[/tex]
Plank's constant [tex]h=6.63\times 10^{-34}J-s[/tex]
We have to find de broglie wavelength
De broglie wavelength is given by [tex]\lambda =\frac{h}{mv}=\frac{6.63\times 10^{-34}}{1.67\times 10^{-27}\times 9.5\times 10^6}=4.17\times 10^{-12}m=4.17pm[/tex]
A charge of 9 µC is on the y axis at 1 cm, and a second charge of −9 µC is on the y axis at −1 cm. Find the force on a charge of 7 µC on the x axis at x = 6 cm. The value of the Coulomb constant is 8.98755 × 10^9 N*m^2/C^2. Answer in n units of N.
Answer:
The net force on X is Fx=75.4N
The net force on Y is FY=25.17N
Explanation:
This is an electrostatic problem, we can calculate de force applying the formula:
[tex]F=k*\frac{Q*Q'}{r^2}\\where:\\k=coulomb constant\\r=distance\\Q=charge[/tex]
the force because of charge at 1cm on the X axis, will only have an X component of force, so:
[tex]Fx1=8.98755*10^9*\frac{(9\µC)(7\µC)}{(5*10^{-2}m)^2}\\Fx1=226.48N[/tex]
For the force because of the charge of the Y axis we have to find the distances usign pitagoras, and the angle:
[tex]r=\sqrt{(-1*10^{-2}m)^2+(6*10^{-2}m)^2} \\r=6.08cm=0.0608m[/tex]
we can find the angle with:
[tex]\alpha = arctg(\frac{1cm}{6cm})=9.46^o[/tex]
We now can calculate the force of the X axis because of the second charge:
[tex]Fx2=8.98755*10^9*\frac{(-9\µC)(7\µC)}{(6.08*10^{-2}m)^2}*cos(9.46 )\\Fx2=-151.08N[/tex]
and for the Force on Y axis:
[tex]Fy2=8.98755*10^9*\frac{(-9\µC)(7\µC)}{(6.08*10^{-2}m)^2}*sin(9.46 )\\Fy2=-25.17N[/tex]
The net force on X axis is:
Fx=226.48N-151.08N=75.4N
Fy=-25.17N
Answer:
The magnitude of the resultant force is equal to 0.0216 N
The direction is along the negative y axis
Explanation:
According to the exercise data:
q1 = 9x10^-6 C (0,1)
q2 = -9x10^-6 C (0,-1)
q = 7x10^6 C (6,0)
the distance between load q1 and load q will be equal to:
[tex]r_{1} = \sqrt{(0-1)^{2}+(6-0)^{2} }= 6.08 m[/tex]
The same way to calculate the distance between q2 and q:
[tex]r_{2} =\sqrt{(0-(-1))^{2}+(6-0)^{2} } =6.08 m[/tex]
The force on q due to the load q1, is calculated with the following equation:
[tex]F_{1}=\frac{K*q_{1}*q }{r1^{2} } *(cos45i-sin45j)=\frac{8.987559x10^{9}*9x10^{-6} *7x10^{-6} }{6.08^{2} }*(cos45i-sin45j)=0.0153*(0.707i-0.707j)=0.0108N(i-j)[/tex]
The same way to force F2:
F2 = (8.987559x10^9*-9x10^-6*7x10^-6))/(6.08^2)*(cos45i-sin45j) = 0.0108 N(i-j)
The resultant force:
F = F1 + F2 = 0.0108 N *(-2j) = - 0.0216 N j
The magnitude is equal to 0.0216 N
The direction is along the negative y axis
The magnitude of the acceleration of gravity, in the International System of Measurements, in an object that falls vertically is: a. 32.2 ft / s^2
b. 9.81 m / s
c. 32.2 m / s^2
d. 9.81 m / s^2
Answer:
option D
Explanation:
the correct answer is option D
acceleration of a body under free fall due to the effect of gravity is the called acceleration due to gravity.
Acceleration due to gravity is a constant value.
According to the International System of Measurements the acceleration due to gravity is 9.81 m/s² when an object falls vertically.
acceleration due to gravity is denoted by 'g'.
What is the strength of the electric field between two parallel conducting plates separated by 6 cm and having a potential difference (voltage) between them of 4.18 ×10^4 V ? Give answer in terms of 10^6 V/m.
Answer:
[tex]E=0.697*10^{6}V/m\\[/tex]
Explanation:
If the voltage is constant, the relation between the electric field E and the voltage V , for two plates separated by a distance d, is:
[tex]V=E*d\\[/tex]
d=6cm=0.06m
V=4.18×10^4 V
We solve to find E:
[tex]E=V/d=4.18*10^{4}/0.06=6.97*10^{5}=0.697*10^{6}V/m\\[/tex]
you walk from the park to your friend's house, then back to your house. (a) What is the distance traveled? (b) What is your displacement? If you walk from your house to the library, then back to your house, repeat (a) and (b).
Explanation:
The total distance in a path is called distance.
The shortest distance between two points is called displacement.
a) Here, the distance travelled between the park to your friend's house and back is
Distance between park to friends house + Distance from friend's house to your house.
b) Displacement would be the shortest distance between the park and your house.
a) Distance walked between your house to library and back is
Distance between your house and library + Distance between your house and library
b) Displacement would be zero (0) as the distance between you initial point and final point is zero. Here, the initial and final points are the same
Final answer:
The distance traveled is the total length of the path taken, while displacement is the distance between the initial and final positions in a straight line. Displacement is a vector quantity because it has both magnitude and direction.
Explanation:
a. The distance traveled is the total length of the path taken. In this case, you traveled from the park to your friend's house, and then back to your own house. So, the distance traveled is the sum of these two distances.
b. Displacement is the distance between the initial and final positions, considering only the straight line between them. In this case, since you ended up back at your own house, the displacement is zero because you did not change your position.
c. Displacement is a vector quantity because it has both magnitude (the distance between the initial and final positions) and direction. In this case, the direction is the straight line between the initial and final positions.
What is the acceleration of an electron that has moved between -1.5V and 3.0V? Assume is began at rest.
Answer:
Insufficient data.
Explanation:
Hi!
The data you have is only potential diference, so you know that the variation in potential energy of the electron when moving from -1.5V to 3.0V, is 4.5 V.
But you cannot know the acceleration. For that you need to know the electric field, so you can calculate force.
Calculate work done by the electrostatic force to move the charge with the magnitude of 1nC between two points 2cm spaced, along the equipotential line, corresponding to the potential of 1V
Answer:
work done is = 0
Explanation:
given data
distance = 2 cm
potential = 1 V
charge with magnitude = 1 nC
to find out
work done by the electrostatic force
solution
we know that at equipotential surface is that surface which have equal potential at each every point that we say
work done will be
work done = ∫dw
∫dw = [tex]\int\limits^v1_v2 {q} \, dv[/tex]
here q is charge
so
net work done = q ( v2 - v1 )
and
so v2 = v1 = 0
so
work done is = 0
No work is done in moving a charge along an equipotential line because there is no difference in electric potential along that line, making the work done by the electrostatic force zero.
Explanation:The question asks to calculate the work done by the electrostatic force when moving a charge along an equipotential line. By definition, the potential difference along an equipotential line is zero. Therefore, the work done W by the electric force to move a charge q in an electric potential V along an equipotential line is given by the equation W = qΔV, where ΔV is the change in electric potential. Since ΔV is zero along an equipotential line, the work done is also zero. No work is required to move a charge along an equipotential line because there is no change in electric potential energy.
A can contains 375 mL of soda. How much is left after 308 mL is removed?
Answer:
Volume left, v = 67 mL
Explanation:
Given that,
Volume of soda contained in a can, V = 375 mL
We need to find the volume of soda left after 308 mL of soda is removed, V' = 308 mL
Let v is the left volume of soda. It can be calculated using simple calculations as :
v = V - V'
v = 375 mL - 308 mL
v = 67 mL
So, the left volume in the can is 67 mL. Hence, this is the required solution.
Which of the following must always be the same: a. Time - a second for example b. Distance - a meter for example The speed of Light - 3E8 m/s d. Weight - how much does an apple weight.
Answer:
The speed of light remains the same.
Explanation:
(a) Time : second, hour minutes are example of time.
(b) Distance : meter, kilometers etc are some example of distance.
(c) The speed of light : It always remains constant. It is equal to [tex]3\times 10^8\ m/s[/tex].
(d) The weight of an object is given by the product of mass of an object and the acceleration due to gravity. As the value of g is not same everywhere, its weight varies.
So, the speed of light is always remains the same. Hence, the correct option is (c).
Answer each of the following questions:
(a) When did the Big Bang occur and what was the result?
(b) Is "string theory" a proven scientific theory? Why or why not?
(c) Where does the Strong Nuclear Force come into play?
(d) What is Cosmology?
Answer:
Part a) Big bang occurred about 13.8 billion years ago. This time is arrived after see the shift in the cosmic background radiation that fills the universe. The result of the big bang was the creation of universe itself. Everything we know space, matter, time was created by the event thus giving birth to the universe and stars, galaxies,e.t.c.
Part b) No string theory is currently not proven as of yet. The basic problem in formulating the theory of everything is to mix 2 completely different theories known as quantum theory of matter and theory of relativity. But this has not been achieved as of yet. There are different theories that try to explain the both quantum and relativistic nature of the matter and string theory is one of the theories that has been proposed to explain the same.The theory is not completely developed as of yet as there are numerous inconsistencies with it such as it proposes 34 dimension's of nature which have not been observed as yet.Also there is an inherent nature of the theory that it deals with quantities with [tex]10^{34}m[/tex] scales of length making it impossible to verify the theory which still is mathematically incomplete.
Part c) Strong Nuclear forces come into action in the nucleus of an atom and holds the nucleon's together. As we know that a pair of protons will repel due to nature of their charge the strong nuclear force holds them together.
Part d) Cosmology is the branch of astronomy that deals with the origin and evolution of the universe. It basically is the study about the universe was formed and how it evolved during different stages.
A transverse sinusoidal wave is moving along a string in the positive direction of an x axis with a speed of 86 m/s. At t = 0, the string particle at x = 0 has a transverse displacement of 4.0 cm from its equilibrium position and is not moving. The maximum transverse speed of the string particle at x = 0 is 18 m/s. (a) What is the frequency of the wave? (b) What is the wavelength of the wave? If the wave equation is of the form y(x, t) = ym sin(kx ± ωt + φ), what are (c) ym, (d) k, (e) ω, (f) φ, and (g) the correct choice of sign in front of ω?
Final answer:
The frequency of the wave is 1075 Hz, and the wavelength is 0.08 m. The wave equation of the form y(x, t) = ym sin(kx ± ωt + φ) represents a sinusoidal wave with specific parameters. The amplitude is 0.04 m, the wave number is 25π m⁻¹, the angular frequency is 6755π rad/s, and the phase angle can be either +π or -π. The choice of sign in front of ω determines the direction of wave propagation.
Explanation:
To find the frequency of the wave, we can use the formula f = v/λ, where f is the frequency, v is the speed of the wave, and λ is the wavelength. In this case, the speed of the wave is given as 86 m/s. Since the wave is sinusoidal, the wavelength is equal to twice the amplitude of the transverse displacement of the string particle at x = 0. So, the wavelength is equal to 2 times 4.0 cm, which is 8.0 cm or 0.08 m.
Substituting these values into the formula, we get f = 86 m/s / 0.08 m = 1075 Hz. Therefore, the frequency of the wave is 1075 Hz.
To find the wavelength of the wave, we can use the formula λ = v/f, where λ is the wavelength, v is the speed of the wave, and f is the frequency. Substituting the given values, we get λ = 86 m/s / 1075 Hz = 0.08 m.
Therefore, the wavelength of the wave is 0.08 m.
The wave equation of the form y(x, t) = ym sin(kx ± ωt + φ) represents a sinusoidal wave. In this equation, ym is the amplitude of the wave, k is the wave number, ω is the angular frequency, and φ is the phase angle. The choice of the sign in front of ω determines the direction of wave propagation. If it is positive, the wave propagates in the positive x-direction, and if it is negative, the wave propagates in the negative x-direction.
In this case, the amplitude of the wave is given as 4.0 cm or 0.04 m. The wave number can be calculated using the formula k = 2π/λ, where k is the wave number and λ is the wavelength. Substituting the given wavelength of 0.08 m, we get k = 2π/0.08 m = 25π m⁻¹.
The angular frequency can be calculated using the formula ω = 2πf, where ω is the angular frequency and f is the frequency. Substituting the given frequency of 1075 Hz, we get ω = 2π × 1075 Hz = 6755π rad/s.
The phase angle is given as ± π. The choice between +π and -π determines the phase of the wave at x = 0 and t = 0.
Therefore, (c) ym = 0.04 m, (d) k = 25π m⁻¹, (e) ω = 6755π rad/s, (f) φ = ±π, and (g) the correct choice of sign in front of ω depends on the desired direction of wave propagation.
A 1500kg space probe is moving at constant velocity of 4200m/sec. In a course change maneuver the rocket engines fire for 5 minutes increasing the probe's velocity to 5000m/sec. What is the probe's change in kinetic energy? A. 3.68x10^6 J. B. 5.52x10^9 J C. 1.1x10^10 J D. not enough info.
Answer:
B)5.52x10^9 J
Explanation:
The equation for calculating kinetic energy is as follows
E=[tex]\frac{1}{2} mv^{2}[/tex]
so what we have to do is calculate the kinetic energy of the space probe in each state and calculate the difference
E=[tex]\frac{1}{2} m.V2^{2}-\frac{1}{2} m.V1^{2}=E[/tex]
E=[tex]\frac{1}{2} m(V2^{2} -V1^{2})=E[/tex]
E=(1500)(5000^2-4200^2)/2
E=5.52x10^9 J