Answer:
(a) 0.9 s
(b) 7.78 m/s
Explanation:
height, h = 4 m
Horizontal distance, d = 7 m
Let it takes time t to reach the ground and u be the initial velocity of the jet.
(a) Use second equation of motion in vertical direction
[tex]s = ut + \frac{1}{2}at^{2}[/tex]
In vertical direction, uy = 0 m/s, a = g = - 9.8 m/s^2, h = - 4 m
By substituting the values, we get
[tex]-4 = 0 - \frac{1}{2}\times 9.8\times t^{2}[/tex]
t = 0.9 second
Thus, the time taken by water jet in air is 0.9 second.
(b) Use
Horizontal distance = horizontal velocity x time
d = u t
7 = u x 0.9
u = 7.78 m/s
Thus, the initial velocity of water jet is 7.78 m/s.
The duration the water is in the air is found using the formula for the motion under gravity, which depends on the vertical distance and the acceleration due to gravity. After finding the time, the initial velocity of the water is calculated using the horizontal distance and the fraction of time the water was in motion.
Explanation:To determine how long the water is in the air (a), and the initial velocity of the water (b) when a water gun is fired horizontally from a hill, we can use the principles of projectile motion. The time a projectile is in the air is solely determined by its vertical motion. Since the water gun is fired horizontally, it has an initial vertical velocity of 0 m/s.
For part (a), the time (t) it takes for the water to reach the ground can be calculated using the formula for the motion under gravity, which is y = 0.5 * g * t2, where y is the vertical distance (4 meters in this case) and g is the acceleration due to gravity (approximated to 9.81 m/s2). Solving for t gives us the time the water is in the air.
For part (b), once we have the time, we can use the horizontal distance (7 meters) to find the initial velocity (v0) using the formula x = v0 * t. This provides the initial horizontal velocity of the water gun's jet. The overall process involves solving for time first and then using that time to find the initial velocity.
A rocket sled used in old Air Force test could accelerate a volunteer at rates of 90m/s^2. With this acceleration being constant, how long would it take for one of these rocket sleds to travel the length of a one kilometer test track if it starts from rest? A) 11.1sec B) 22.2sec C)4.71sec D) 32.2sec
Answer:
option (c) 4.71 seconds
Explanation:
Given:
Distance to be covered = 1 km = 1000 m
Acceleration, a = 90 m/s²
From Newton's equation of motion
[tex]s=ut+\frac{1}{2}at^2[/tex]
where,
s is the distance
u is the initial speed = 0 m/s ( since it starts from rest )
a is the acceleration
t is the time
on substituting the respective values, we get
[tex]1000=0\times t+\frac{1}{2}\times90\times t^2[/tex]
or
45t² = 1000
or
t² = 22.22
or
t = 4.71 seconds
Hence, the correct answer is option (c) 4.71
An infinitely long line charge of uniform linear charge density λ = -3.00 µC/m lies parallel to the y axis at x = -3.00 m. A point charge of 3.00 µC is located at x = 1.00 m, y = 2.00 m. Find the electric field at x = 2.00 m, y = 1.50 m.
Answer:
E=[8.1X-9.63Y]*10^{3}N/m
Explanation:
Field in the point is the sum of the point charge electric field and the field of the infinite line.
First, we calculate the point charge field:
[tex]E_{Charge}=\frac{1}{4\pi \epsilon_0} *\frac{Q}{||r_p -r||^2} *Unitary vector\\||r_p -r||^2=(x_p-x)^2+(y_p -y)^2=1.25 m^2\\Unitary Vector=\frac{(r_p -r)}{||r_p -r||}=\frac{(x_p-x)X+(y_p -y)Y}{||r_p -r||}\\=\frac{2\sqrt{5} }{5}X- \frac{\sqrt{5} }{5}Y \\E_{Charge}=K*\frac{3\mu C}{1.25m^2}*(\frac{2\sqrt{5} }{5}X- \frac{\sqrt{5} }{5}Y)[/tex]
It is vectorial, where X and Y represent unitary vectors in X and Y. we recall the Coulomb constant k=[tex]\frac{1}{4\pi \epsilon_0}[/tex] and not replace it yet. Now we compute the line field as follows:
[tex]E_{Line}=\frac{\lambda}{2\pi \epsilon_0 distance} *Unitary Vector\\Unitary Vector=X[/tex] (The field is only in the perpendicular direction to the wire, which is X)
[tex]E_{Line}=\frac{-3\mu C/m*2}{2*2\pi \epsilon_0 5*m}X=K*\frac{6\mu C/m}{ 5*m}(-X)[/tex]
We multiplied by 2/2 in order to obtain Coulomb constant and express it that way. Finally, we proceed to sum the fields.
[tex]E=K*\frac{3\mu C}{1.25m^2}*(\frac{2\sqrt{5} }{5}X- \frac{\sqrt{5} }{5}Y)+K*\frac{6\mu C}{ 5*m^2}(-X)\\E=K*[2.15-1.2]X-K*[1.07]Y \mu N/m\\E=K*[0.9X-1.07Y] \mu C/m^2\\E=[8.1X-9.63Y]*10^{3}N/m[/tex]
N women, each of mass m, stand on a railway flatcar of mass M. They jump off one end of the flatcar with velocity u relative to the car. The car rolls in the opposite direction without friction. (a) What is the final velocity of the flatcar if all the women jump at the same time? (b) What is the final velocity of the flatcar if the women jump off one at a time? (c) Which answer, either (a) or (b), is greater? Try to make physical and/or intuitive sense of this answer and comment on your thoughts.
Answer:
Explanation:
Given
Velocity of women relative to the car is u
Let v be the velocity of car after women jump off
Therefore women velocity relative to the ground is v-u
Conserving momentum
Mv+Nm(v-u)=0
Mv+Nmv=Nmu
[tex]v=\frac{Nm}{M+Nm}[/tex]
(b)Let [tex]v_r[/tex] be the velocity of car after r women has jumped and [tex]v_N[/tex] be the final velocity of the car after N women have jumped.
After r women jumped and N-r on cart momentum is given by
[tex]P_r=\left [ M+\left ( N-r\right )m\right ]v_r[/tex]
After next woman jumps car has velocity of [tex]v_{r+1}[/tex] while the woman has velocity is [tex]v_{r+1}-u[/tex](relative to the ground)
Total momentum
[tex]P_{r+1}=\left [ M+\left ( N-r-1\right )m\right ]v_{r+1}+m\left ( v_{r+1}-u\right )[/tex]
Since total momentum in horizontal direction is conserved then
[tex]P_{r+1}=P_r[/tex]
[tex]v_{r+1}-v_r=\frac{mu}{\left ( M+\left ( N-r\right )m\right )}[/tex]
Summing the above expression from r=0 to r=N-1 we get
[tex]\sum_{j=1}^{j=N}\frac{mu}{\left ( M+jm\right )}[/tex]
(c)Answer in part b is greater because j\leq N[/tex]
Thus Velocity in part b is greater.
Intuitively if you jump after an another person you will impart extra momentum to the cart compared to when all persons jumped off simultaneously.
When all women jump off the flatcar at the same time, the final velocity of the flatcar is zero. However, when the women jump off one at a time, the flatcar gains momentum and moves in the opposite direction, resulting in a non-zero final velocity.
Explanation:To answer the given question, we can apply the principle of conservation of momentum. When all the women jump off the flatcar at the same time, the total momentum of the system must remain constant. Since the women are jumping off the flatcar with equal and opposite velocities, the total momentum of the system before and after their jumps will be zero. Therefore, the final velocity of the flatcar will also be zero.
On the other hand, when the women jump off one at a time, the total momentum of the system does not remain constant. Initially, the flatcar and the first woman have a total momentum of zero. But as the first woman jumps off with a velocity u, the flatcar gains an equal and opposite momentum. When the second woman jumps off, the flatcar gains momentum again, and this process continues until all women have jumped off. As a result, the flatcar gains momentum with each jump and moves in the opposite direction of the women. Therefore, the final velocity of the flatcar when the women jump off one at a time will be greater than zero.
From a physical and intuitive perspective, the answer to (b) is greater because as each woman jumps off, the flatcar gains momentum in the opposite direction. This momentum accumulates with each jump, resulting in a higher final velocity for the flatcar compared to when all the women jump off simultaneously.
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If an astronaut has a weight of 450 N on the earth, what would be her weight on the moon? A. 101 Ib В. 46 N C. 77 N D. 7.8 N
Explanation:
Given that,
Weight of the astronaut on the surface of Earth, W = 450 N
We know that the acceleration due to gravity on the earth is, [tex]g=9.8\ m/s^2[/tex]
Weight of an object is given by, W = mg
[tex]m=\dfrac{W}{g}[/tex]
[tex]m=\dfrac{450\ N}{9.8\ m/s^2}[/tex]
m = 45.91 kg
Also, the acceleration due to gravity on the surface of moon is one -sixth of the acceleration due to gravity on the surface of Earth, [tex]g'=1.62\ m/s^2[/tex]
As mass remains constant. So, the weight on the moon is :
W' = mg'
[tex]W'=45.91\times 1.62[/tex]
W' = 74.37 N
Hence, this is the required solution.
The car is traveling at a constant speed v0 = 80 km/hr on the level portion of the road. When the 4-percent (tan q = 4/100) incline is encountered, the driver does not change the throttle setting and consequently the car decelerates at the constant rate g sin q. Determine the speed of the car: (a) 28 seconds after passing point A ,and (b) when s = 620 m.
Answer:
11.27 m /s
2.98 m / s.
Explanation:
80 km / h = 22.22 m /s
Tanq = 4 / 100
Sinq = .0399
Deceleration acting on inclined plane = g sinq
= 9.8 x .0399
= .3910
Initial speed u = 22.22 m/s
acceleration = - .3910 ms⁻²
v = u - a t
= 22.22 - .3910 x 28
= 22.22 - 10.95
= 11.27 m /s
b ) v² = u² - 2 a s
v² = ( 22.22) ² - 2 x .3910 x 620
= 493.7284 - 484.84
= 8.8884
v = 2.98 m / s.
The Joule (J) is a unit of : kinetic energy
electric energy
potential energy
work all of the above
Answer: work all of the above
Explanation: kinetic, electric and potential are energies so their units must be energy and the Joule is.
Answer:
work all of the above
Explanation:
A charge of -3.35 nC is placed at the origin of an xy-coordinate system, and a charge of 1.75 nC is placed on the y axis at y = 3.90 cm . A. If a third charge, of 5.00 nC , is now placed at the point x = 2.70 cm , y = 3.90 cm find the x and y components of the total force exerted on this charge by the other two charges.
B. Find the magnitude of this force.
C. Find the direction of this force. ( ° below the +x axis )
Answer:
Explanation:
Force due to charges 1.75 and 5 nC is given below
F =K Q₁Q₂ / d²
F₁ = [tex]\frac{9\times10^9\times1.75\times10^{-9}\times5\times 10^{-9}}{(2.7\times10^{-2})^2}[/tex]
F₁ = 10.8 X 10⁻⁵ N . It will at in x direction.
Force due to other charge placed at origin
F₂ = [tex]\frac{9\times10^9\times1.75\times10^{-9}\times5\times 10^{-9}}{22.5\times10^{-4}}[/tex]
F₂ = 3.5 x 10⁻⁵ N.
Its x component
= F₂ Cos θ
= 3.5 x 10⁻⁵ x 3.9/ 4.74
= 2.88 x 10⁻⁵ N
Its y component
F₂ sin θ
= 3.5 x 10⁻⁵ x 2.7/4.743
= 1.99 x 10⁻⁵ N
Total x component
= 10.8 X 10⁻⁵ +2.88 x 10⁻⁵
= 13.68 x 10⁻⁵ N.
Magnitude of total force F
F² = (13.68 x 10⁻⁵)² + (1.99 x 10⁻⁵ )²
F = 13.82 X 10⁻⁵ N
Direction θ with x axis .
Tanθ = 1.99/ 13.68
θ = 8 °
A +1.0μC charge sits at the origin, another +1.0μC sits at x = +0.01m, and another -2.0μC sits at x = +0.02m. Calculate the magnitude and direction of the electric field at the point x = +0.2m. (1 μC = 1.0x10^-6C). Draw a picture first.
Answer:
A)Ep=-81.3N/C :Electric field at the point x = +0.2
Ep Magnitude =81.3N/C
Direction of the electric field ( Ep): -x
B)Graphic attached
Explanation:
Conceptual analysis
The electric field at a point P due to a point charge is calculated as follows:
E = k*q/d²
E: Electric field in N/C
q: charge in Newtons (N)
k: electric constant in N*m²/C²
d: distance from charge q to point P in meters (m
Graphic attached
The attached graph shows the field due to the charges:
The electric field at a point P due to several point charges is the vector sum of the electric field due to individual charges.
E₁: Electric Field at point Xp=0.2 m due to charge q₁. As the charge q1 is positive (q₁+) ,the field leaves the charge.
E₁: Electric Field at point Xp=0.2 m due to charge q₂. As the charge q1 is positive (q₂+) ,the field leaves the charge
E₃: Electric Field at point Xp=0.2 m due to charge q₃. As the charge q₃ is negative (q₃-), the field enters the charge.
Equivalence
+1.0μC=1* 10⁶C
Data
q₁=+1.0μC=1* 10⁶C
q₂=+1.0μC=1* 10⁶C
q₃=-2.0μC=-2* 10⁶C
Xp=0.2m
x₁=0
x₂=0.01 m
x₃=0.02m
Calculation of the distances of the charges to the point P
d= Xp-x
d₁=Xp-x₁= 0.2-0= 0.2m
d₂=Xp-x₂=0.2-0.01= 0.19m
d₃=Xp-x₃=0.2-002= 0.18m
Calculation of electric fields due to charges q1, q2 and q3 at point P
E₁=k*q₁/d₁²=9*10⁹*1*10⁻⁶/0.2²=225*10³N/C
E₂=k*q₂/d₂²=9*10⁹*1*10⁻⁶/0.19²=249.3*10³N/C
E₃=-k*q₃/d₃²=9*10⁹*2*10⁻⁶/0.18²=-555.6*10³N/C
Calculation of electric field at point P due to charges q₁, q₂ and q₃
To calculate Ep, the electric fields E₁,E₂ and E₃ are added algebraically:
Ep=E₁+E₂ +E₃
Ep=(225*10³+249.3*10³ -555.6*10³)N/C
Ep=-81.3N/C
Ep Magnitude =81.3N/C in -x direction
What is the relationship between the slope of the position graph of an object and its velocity?
Answer:
Slope of position time is velocity
Explanation:
The position time graph means a relation between the position of the object and the time which is represent on a graph.
The graph line shows that how the position of an object changes with respect to time.
The slope of the position time graph shows the rate of change of position of the object with respect to time.
The rate of change of position with respect to time is called velocity.
thus, the slope of position time graph gives the velocity of the object.
What is the beat frequency between a note at 350 Hz and a note at 353 Hz?
Answer:
[tex]f_{beat}=f_2-f_1=353Hz -350Hz=3Hz[/tex]
Explanation:
When there are two waves with similar frequency, the superposition of these waves create a new wave with a particular beat. The beat frequency from this constructive superposition is equal to the value of the difference in frequency of the two waves.
[tex]f_{beat}=f_2-f_1=353Hz -350Hz=3Hz[/tex]
Two identical small insulating balls are suspended by separate 0.26-m threads that are attached to a common point on the ceiling. Each ball has a mass of 7.75 10^-4 kg. Initially the balls are uncharged and hang straight down. They are then given identical positive charges and, as a result, spread apart with an angle of 35° between the threads. (a) Determine the charge on each ball. (answer in C)
The charge on each ball is approximately [tex]\( 8.08 \times 10^{-9} \) C[/tex].
Step 1
To determine the charge on each ball, let's analyze the forces acting on one of the charged balls when they are in equilibrium. The forces are:
1. Gravitational force [tex](\( F_g \))[/tex] downward.
2. Tension [tex](\( T \))[/tex] in the thread.
3. Electrostatic force [tex](\( F_e \))[/tex] between the two charges.
Step 2
Given:
Length of the thread [tex](\( L \)) = 0.26 m[/tex]
Mass of each ball [tex](\( m \)) = \( 7.75 \times 10^{-4} \) kg[/tex]
Angle between the threads = 35°, so each thread makes an angle of 17.5° with the vertical.
Step 3
First, calculate the gravitational force:
[tex]\[ F_g = mg = 7.75 \times 10^{-4} \, \text{kg} \times 9.8 \, \text{m/s}^2 = 7.595 \times 10^{-3} \, \text{N} \][/tex]
The electrostatic force must balance the horizontal component of the tension:
[tex]\[ F_e = T \sin(17.5^\circ) \][/tex]
The vertical component of the tension balances the gravitational force:
[tex]\[ T \cos(17.5^\circ) = F_g \][/tex]
[tex]Solving for \( T \):[/tex]
[tex]\[ T = \frac{F_g}{\cos(17.5^\circ)} = \frac{7.595 \times 10^{-3} \, \text{N}}{\cos(17.5^\circ)} \approx 7.99 \times 10^{-3} \, \text{N} \][/tex]
Step 4
Now calculate the horizontal component of the tension:
[tex]\[ T \sin(17.5^\circ) = 7.99 \times 10^{-3} \, \text{N} \times \sin(17.5^\circ) \approx 2.41 \times 10^{-3} \, \text{N} \][/tex]
Step 5
This is the electrostatic force:
[tex]\[ F_e = \frac{k q^2}{r^2} \][/tex]
where [tex]\( r \)[/tex] is the distance between the balls.
Step 6
To find [tex]\( r \)[/tex]:
[tex]\[ r = 2L \sin(17.5^\circ) = 2 \times 0.26 \, \text{m} \times \sin(17.5^\circ) \approx 0.156 \, \text{m} \][/tex]
Using Coulomb's law:
[tex]\[ 2.41 \times 10^{-3} \, \text{N} = \frac{(8.99 \times 10^9 \, \text{N m}^2/\text{C}^2) q^2}{(0.156 \, \text{m})^2} \][/tex]
Step 7
Solving for [tex]\( q \)[/tex]:
[tex]\[ q^2 = \frac{2.41 \times 10^{-3} \, \text{N} \times (0.156 \, \text{m})^2}{8.99 \times 10^9 \, \text{N m}^2/\text{C}^2} \approx 6.53 \times 10^{-17} \, \text{C}^2 \][/tex]
[tex]\[ q \approx \sqrt{6.53 \times 10^{-17}} \approx 8.08 \times 10^{-9} \, \text{C} \][/tex]
If the retina is 1.7 cm from the lens in the eye, how large is the image on the retina of a person of height 1.5 m standing 9.0 m away?
Answer:
The image height is 0.00283 m or 0.283 cm and is inverted due to the negative sign
Explanation:
u = Object distance = 9 m
v = Image distance = 1.7 cm (as the image is forming on the retina)
[tex]h_u[/tex]= Object height = 1.5 m
Magnification
[tex]m=-\frac{v}{u}=\frac{h_v}{h_u}\\\Rightarrow -\frac{v}{u}=\frac{h_v}{h_u}\\\Rightarrow -\frac{0.017}{9}=\frac{h_v}{1.5}\\\Rightarrow h_v=-0.00283\ m[/tex]
The image height is 0.00283 m or 0.283 cm and is inverted due to the negative sign
The image formed on the retina is 1.8 cm behind the lens.
Explanation:The image formed on the retina can be calculated using the lens formula: 1/f = 1/do + 1/di, where f is the focal length of the lens, do is the object distance from the lens, and di is the image distance.
In this case, the lens-to-retina distance is given as 2.00 cm, so di = -2.00 cm. The object distance do can be calculated as the difference between the person's height and the distance at which they are standing: do = 9.0 m - 1.5 m = 7.5 m. Substituting the values into the lens formula, we get 1/2.00 cm = 1/7.5 m + 1/di. Solving for di, we find that the image is formed 1.8 cm behind the lens.
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In a 5000 m race, the athletes run 12 1/2 laps; each lap is 400 m.Kara runs the race at a constant pace and finishes in 17.9 min.Hannah runs the race in a blistering 15.3 min, so fast that she actually passes Kara during the race. How many laps has Hannah run when she passes Kara?
Answer:
No. of laps of Hannah are 7 (approx).
Solution:
According to the question:
The total distance to be covered, D = 5000 m
The distance for each lap, x = 400 m
Time taken by Kara, [tex]t_{K} = 17.9 min = 17.9\times 60 = 1074 s[/tex]
Time taken by Hannah, [tex]t_{H} = 15.3 min = 15.3\times 60 = 918 s[/tex]
Now, the speed of Kara and Hannah can be calculated respectively as:
[tex]v_{K} = \frac{D}{t_{K}} = \frac{5000}{1074} = 4.65 m/s[/tex]
[tex]v_{H} = \frac{D}{t_{H}} = \frac{5000}{918} = 5.45 m/s[/tex]
Time taken in each lap is given by:
[tex](v_{H} - v_{K})t = x[/tex]
[tex](5.45 - 4.65)\times t = 400[/tex]
[tex]t = \frac{400}{0.8}[/tex]
t = 500 s
So, Distance covered by Hannah in 't' sec is given by:
[tex]d_{H} = v_{H}\times t[/tex]
[tex]d_{H} = 5.45\times 500 = 2725 m[/tex]
No. of laps taken by Hannah when she passes Kara:
[tex]n_{H} = \frac{d_{H}}{x}[/tex]
[tex]n_{H} = \frac{2725}{400} = 6.8[/tex] ≈ 7 laps
The number of laps that Hannah has run when she passes Kara is 7 laps.
Speed of each athlete
The speed of each athlete is calculated as follows;
Kara = (5000) / (17.9 x 60) = 4.66 m/s
Hannah = (5000) / (15.3 x 60) = 5.47 m/s
Time taken in each lapThe time taken in each lap if Hannah passes kara is calculated as follow;
(5.47 - 4.66)t = 400
0.81t = 400
t = 493.83 s
Distance covered by Hannah when she passes kara;
d = 493.83 x 5.47 = 2,701.25 m
Number of lapsn = 2,701.25/400
n = 6.8 ≈ 7 laps.
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A long, nonconducting cylinder (radius = 6.0 mm) has a nonuniform volume charge density given by αr2, where α = 6.2 mC/m5 and r is the distance from the axis of the cylinder. What is the magnitude of the electric field at a point 2.0 mm from the axis?
Answer: 2.80 N/C
Explanation: In order to calculate the electric firld inside the solid cylinder
non conductor we have to use the Gaussian law,
∫E.ds=Q inside/ε0
E*2πrL=ρ Volume of the Gaussian surface/ε0
E*2πrL= a*r^2 π* r^2* L/ε0
E=a*r^3/(2*ε0)
E=6.2 * (0.002)^3/ (2*8.85*10^-12)= 2.80 N/C
A 3.50 m -tall, 50.0 cm -diameter concrete column supports a 5.00 x 10^5 kg load. Part A : By how much is the column compressed?
Answer:
The column compressed is 57.142 Pa.
Explanation:
Given that,
Height = 3.50 m
Diameter = 50.0 cm
Radius = 25.0 cm
Load [tex]F=5.00\times10^{5}\ kg[/tex]
We need to calculate the column compressed
Using formula of compressed
[tex]P = \dfrac{F}{A}[/tex]
[tex]P=\dfrac{F}{l\times r}[/tex]
Where, F = load
l = length
r = radius
Put the value into the formula
[tex]P=\dfrac{5.00\times10^{5}}{3.50\times25\times10^{2}}[/tex]
[tex]P=57.142\ Pa[/tex]
Hence, The column compressed is 57.142 Pa.
A step index fiber has a numerical aperture of NA = 0.1. The refractive index of its cladding is 1.465. What is the largest core diameter for which the fiber remains single-moded at the wavelength of 1.3 μm?
Answer:
diameter = 9.951 × [tex]10^{-6}[/tex] m
Explanation:
given data
NA = 0.1
refractive index = 1.465
wavelength = 1.3 μm
to find out
What is the largest core diameter for which the fiber remains single-mode
solution
we know that for single mode v number is
V ≤ 2.405
and v = [tex]\frac{2*\pi *r}{ wavelength} NA[/tex]
here r is radius
so we can say
[tex]\frac{2*\pi *r}{ wavelength} NA[/tex] = 2.405
put here value
[tex]\frac{2*\pi *r}{1.3*10^{-6}} 0.1[/tex] = 2.405
solve it we get r
r = 4.975979 × [tex]x^{-6}[/tex] m
so diameter is = 2 × 4.975979 × [tex]10^{-6}[/tex] m
diameter = 9.951 × [tex]10^{-6}[/tex] m
A straight wire of length L has a positive charge Q distributed along its length. Find the magnitude of the electric field due to the wire at a point located a distance d from one end of the wire along the line extending from the wire. Imagine that distance d is much greater than the length of the wire. Intuitively, what should the magnitude of the electric field at point P be in this case
Answer:
Electric Field at a distance d from one end of the wire is [tex]E=\dfrac{Q}{4\pi \epsilon_0(L+d)d}[/tex]
Electric Field when d is much grater than length of the wire =[tex]\dfrac{Q}{4\pi \epsilon_0\ d^2}[/tex]
Explanation:
Given:
Total charge over the length of the wire = QLength of the wire = LDistance from one end of wire at which electric field is needed to find=dLet dE be the Electric Field due to the small elemental charge on the wire at a distance x from the one end of the wire and let [tex]\lambda[/tex] be the charge density of the wire
[tex]E=\dfrac{dq}{4\pi \epsilon_0x^2}[/tex]
Now integrating it over the entire length varying x from x=d to x=d+L we have and replacing [tex]\lambda=\dfrac{Q}{L}[/tex] we have
[tex]E=\int\dfrac{\lambda dx}{4\pi \epsilon_0x^2}\\E=\dfrac{Q}{4\pi \epsilon_0 (L+d)(d)}[/tex]
When d is much greater than the length of the wire then we have
1+\dfrac{L]{d}≈1
So the Magnitude of the Electric Field at point P = [tex]\dfrac{Q}{4\pi \epsilon_0\ d^2}[/tex]
The vertical velocity of a projectile launch at 35 degree : O continuously decreases O continuously increases O sometimes decreases and sometimes increases O is zero O stays constant.
Answer:Sometime decreases and sometime increases.
Explanation:
Given
Projectile is launched at [tex]35 ^{\circ}[/tex]
so horizontal velocity=ucos35
vertical velocity=usin35
since gravity acts in the vertical direction, therefore, it affects only the vertical component of the projectile.
So Vertical velocity first decrease up to maximum height and then starts to increase.
Two small spherical insulators separated by 2.5 cm, which is much greater than either of their diameters. Both carry positive charge, one +60.0 microCoulombs and the other +6.66 microCoulombs. A third positive charge remains at rest between the two spheres and along the line joining them. What is the position of this charged sphere?
Answer:
1.875 cm from 60 microcoulomb charge.
Explanation:
Let the third charge be Q. Let it be put at x distance from 60 micro coulomb charge for balance.
Force on this charge due to first charge
= [tex]\frac{k\times60\times10^{-6}Q}{x^2}[/tex]
Force on this charge due to second charge
= [tex]\frac{k\times6.66\times10^{-6}Q}{(2.5-x)^2}[/tex]
Since both these forces are equal [tex]\frac{k\times60\times10^{-6}Q}{x^2}=\frac{k\times6.66\times10^{-6}Q}{(2.5-x)^2}[/tex]
[tex]\frac{60}{6.66} = \frac{x^2}{(2.5-x)^2}[/tex][tex]\frac{x}{2.5 -x} = \frac{3}{1}[/tex]
x = 1.875
1.875 cm from 60 microcoulomb charge.
A sports car accelerates uniformly from rest to a speed of 87 mi/hr in 8s. Determine: a.The acceleration of the car
b.The distance the car travels in the first 8 s
c.The velocity of the car after the first 10 s
Answer:
Part a)
[tex]a = 4.86 m/s^2[/tex]
Part b)
[tex]d = 155.52 m[/tex]
Part c)
[tex]v_f = 48.6 m/s[/tex]
Explanation:
As we know that car start from rest and reach to final speed of 87 mph
so we have
[tex]v_f = 87 mph = 38.88 m/s[/tex]
now we have
Part a)
acceleration is rate of change in velocity
[tex]a = \frac{v_f - v_i}{t}[/tex]
[tex]a = \frac{38.88 - 0}{8}[/tex]
[tex]a = 4.86 m/s^2[/tex]
Part b)
distance moved by car with uniform acceleration is given as
[tex]d = \frac{v_f + v_i}{2} t[/tex]
[tex]d = \frac{38.88 + 0}{2} 8[/tex]
[tex]d = 155.52 m[/tex]
Part c)
As we know that the car start from rest
so final speed after t = 10 s
[tex]v_f = v_i + at[/tex]
[tex]v_f = 0 + (4.86)10[/tex]
[tex]v_f = 48.6 m/s[/tex]
Assume that an oil slick consists of a single layer of molecules and that each molecule occupies a cube 1.4 nm on a side. Determine the area of an oil slick formed by 1.1 m^3 of oil. Answer in units of m^2.
Answer:
The area of oil slick is calculate as [tex]785.7 m^{2}[/tex]
Solution:
Volume of oil slick, [tex]V_{o} = 1.1 m^{3}[/tex]
The thickness of one molecule on a side, w = 1.4 mm = [tex]1.4\times 10^{- 3}[/tex]
Now, in order to determine the area of oil slick, [tex]A_{o}[/tex]:
Volume, V = [tex]Area\times thickness[/tex]
Thus
[tex]Area,\ A_{o} = \frac{V_{o}}{w}[/tex]
[tex]Area,\ A_{o} = \frac{1.1}{1.4\times 10^{- 3}} = 785.7 m^{2}[/tex]
A charge Q is to be divided into two parts, labeled 'q' and 'Q-q? What is the relationship of q to Q if, at any given distance, the Coulomb force between the two is to be maximized?
Answer:
Explanation:
The two charges are q and Q - q. Let the distance between them is r
Use the formula for coulomb's law for the force between the two charges
[tex]F = \frac{Kq_{1}q_{2}}{r^{2}}[/tex]
So, the force between the charges q and Q - q is given by
[tex]F = \frac{K\left ( Q-q \right )q}}{r^{2}}[/tex]
For maxima and minima, differentiate the force with respect to q.
[tex]\frac{dF}{dq} = \frac{k}{r^{2}}\times \left ( Q - 2q \right )[/tex]
For maxima and minima, the value of dF/dq = 0
So, we get
q = Q /2
Now [tex]\frac{d^{2}F}{dq^{2}} = \frac{-2k}{r^{2}}[/tex]
the double derivate is negative, so the force is maxima when q = Q / 2 .
A baseball is hit with a speed of 47.24 m/s from a height of 0.42 meters. If the ball is in the air 5.73 seconds and lands 130 meters from the batters feet, (a) at what angle did the ball leave the bat? (b) with what velocity will the baseball hit the ground?
Answer:
a)the ball will leave the bat at an angle of 61.3° .
b) the velocity at which it will hit the ground will be v = 27.1 m/s
Explanation:
Given,
v = 47.24 m
h = 0.42 m
t = 5.73 s
R = 130 m
a)We know that
R = v cosθ × t
cosθ = [tex]\dfrac{R}{v t } = \dfrac{130}{47.24\times 5.73 } =0.4803[/tex]
θ = 61.3°
the ball will leave the bat at an angle of 61.3° .
b)Vx = v cos(θ) = 47.24 x cos 61.3 = 22.7 m/s
v = u + at
Vy = 47.24 x sin 61.3 - 9.81 x 5.73
= -14.8 m/s
v = [tex]\sqrt{v_x^2 + v_y^2)}[/tex]
v = [tex]\sqrt{22.7^2 + -14.8^2}[/tex]
v = 27.1 m/s
the velocity at which it will hit the ground will be v = 27.1 m/s
In this first example of constant accelerated motion, we will simply consider a car that is initially traveling along a straight stretch of highway at 15 m/s. At t=0 the car begins to accelerate at 2.0 m/s2 in order to pass a truck. What is the velocity of the car after 5.0 s have elapsed?
Answer:
[tex]v_{f} =25m/s[/tex]
Explanation:
Kinematics equation for constant acceleration:
[tex]v_{f} =v_{o} + at=15+2*5=25m/s[/tex]
Kathy tests her new sports car by racing with Stan, an experienced racer. Both start from rest, but Kathy leaves the starting line 1.00 s after Stan does. Stan moves with a constant acceleration of 3.9 m/s^2 while Kathy maintains an acceleration of 4.73 m/s^2. (a) Find the time at which Kathy overtakes (b) Find the distance she travels before she catches him (c) Find the speeds of both cars at the instant she overtakes him.
Answer:
(a) 10 s
(b) 236.5 m
(c) Kathy's speed = 47.3 m/s
Stan's speed = 42.9 m/s
Explanation:
Given:
[tex]u_k[/tex] = initial speed of Kathy = 0 m/s[tex]u_s[/tex] = initial speed of Stan = 0 m/s[tex]a_k[/tex] = acceleration of Kathy = [tex]4.73\ m/s^2[/tex][tex]a_s[/tex] = acceleration of Stan = [tex]3.9\ m/s^2[/tex]Assumptions:
[tex]v_k[/tex] = final speed of Kathy when see catches Stan[tex]v_s[/tex] = final speed of Stan when Kathy catches him[tex]s_k[/tex] = distance traveled by Kathy to catch Stan[tex]s_s[/tex] = distance traveled by Stan when Kathy catches him[tex]t_k[/tex] = time taken by Kathy to catch Stan = [tex]t[/tex][tex]t_s[/tex] = time interval in which Kathy catches Stan = [tex]t+1[/tex]Part (a):
Kathy will catch Stan only if the distances traveled by each of them are equal at the same instant.
[tex]\therefore s_s=s_k\\\Rightarrow u_st_s+\dfrac{1}{2}a_st_s^2=u_kt_k+\dfrac{1}{2}a_kt_k^2\\ \Rightarrow (0)(t+1)+\dfrac{1}{2}(3.9)(t+1)^2=(0)(t)+\dfrac{1}{2}(4.73)t^2\\ \Rightarrow \dfrac{1}{2}(3.9)(t+1)^2=\dfrac{1}{2}(4.73)t^2\\\Rightarrow (3.9)(t+1)^2=(4.73)t^2\\\Rightarrow \dfrac{(t+1)^2}{t^2}=\dfrac{4.73}{3.9}\\\textrm{Taking square root in both sides}\\\dfrac{t+1}{t}= 1.1\\\Rightarrow t+1=1.1t\\\Rightarrow 0.1t = 1\\\Rightarrow t = 10\\[/tex]
Hence, Kathy catches Stan after 11 s from the Stan's starting times.
Part (b):
Distance traveled by Kathy to catch Stan will be distance the distance traveled by her in 10 s.
[tex]s_s = u_kt_k+\dfrac{1}{2}a_kt_k^2\\\Rightarrow s_s= (0)(t)+\dfrac{1}{2}(4.73)t^2\\\Rightarrow s_s= \dfrac{1}{2}(4.73)(10)^2\\\Rightarrow s_s= 236.5[/tex]
Hence, Kathy traveled a distance of 236.5 m to overtake Stan.
Part (c):
[tex]v_k = u_k+a_kt_k\\\Rightarrow v_k = 0+(4.73)(t)\\\Rightarrow v_k = (4.73)(10)\\\Rightarrow v_k =47.3[/tex]
The speed of Kathy at the instant she catches Stan is 47.3 m/s.
[tex]v_s = u_s+a_st_s\\\Rightarrow v_s = 0+(3.9)(t+1)\\\Rightarrow v_s = (3.9)(10+1)\\\Rightarrow v_s =42.9[/tex]
The speed of Stan at the instant Kathy catches him is 42.9 m/s.
The mass of a rocket decreases as it burns through its fuel. If the rocket engine produces constant force (thrust), how does the acceleration of the rocket change over time? Answers:- it does not chage- it increases- it decreases
Answer:
it increases-
Explanation:
When the mass of a rocket decreases as it burns through its fuel and the force ( thrust) is constant then by newtons second law of motion
F= ma here F is constant this means that ma= constant
⇒ m= F /a this implies that mass is inversely proportional to acceleration.
its means when the mass decreases the acceleration must increase. hence the acceleration increases
Astone is thrown directly upward with an initial speed of 9.6 m/s from a height of 12.8 m. After what time interval (in s) does the stone strike the ground? Use g 9.8 m/s^2 Enter a number with 2 digits behind the decimal point.
Answer:
1.89 seconds
Explanation:
t = Time taken
u = Initial velocity = 9.6 m/s
v = Final velocity
s = Displacement
a = Acceleration due to gravity = 9.81 m/s²
[tex]v=u+at\\\Rightarrow 0=9.6-9.8\times t\\\Rightarrow \frac{-9.6}{-9.8}=t\\\Rightarrow t=0.97 \s[/tex]
Time taken to reach maximum height is 0.97 seconds
[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow s=9.6\times 0.97+\frac{1}{2}\times -9.8\times 0.97^2\\\Rightarrow s=4.7\ m[/tex]
So, the stone would travel 4.7 m up
So, total height ball would fall is 4.7+12.8 = 17.5 m
[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow 17.5=0t+\frac{1}{2}\times 9.8\times t^2\\\Rightarrow t=\sqrt{\frac{17.5\times 2}{9.8}}\\\Rightarrow t=1.89\ s[/tex]
Time taken by the stone to travel 17.5 m is 1.89 seconds
A car with an initial speed of 6.64 m/s accelerates at a uniform rate of 0.85 m/s^2 for 3.7s. The final speed of the car is 9.8 m/s. What is the cars displacement after that time? answer in km.
Answer:
So the car displacement after 3.7 sec is 0.030 km
Explanation:
We have given initial velocity u = 6.64 m/sec
Acceleration [tex]a=0.85m/sec^2[/tex]
Time t = 3.7 sec
Final velocity v = 9.8 m/sec
We have to find the displacement after that time
From second equation of motion we know that [tex]s=ut+\frac{1}{2}at^2[/tex], here s is displacement, u is initial velocity, t is time , and a is acceleration
So displacement [tex]s=ut+\frac{1}{2}at^2=6.64\times 3.7+\frac{1}{2}\times 0.85\times 3.7^2=30.386m[/tex]
We know that 1 km = 1000 m
So 30.386 m = 0.030 km
Answer:
The displacement of car after that time is 30.56 m.
Explanation:
Given that,
Initial velocity = 6.64 m/s
Acceleration = 0.85 m/s²
Time = 3.7 s
Final velocity = 9.8 m/s
We need to calculate the displacement
Using equation of motion
[tex]v^2=u^2+2as[/tex]
[tex]s=\dfrac{v^2-u^2}{2a}[/tex]
Put the value into the formula
[tex]s=\dfrac{9.8^2-6.64^2}{2\times0.85}[/tex]
[tex]s =30.56\ m[/tex]
Hence, The displacement of car after that time is 30.56 m.
A differential manometer is used to measure the drop in pressure across a filter at a water (rho = 1.00 g/cm3) processing plant. The density of the manometer fluid is 8.75 g/cm3. The height of the manometer fluid on side 1 is 275 mm. The height of the manometer fluid on side 2 is 293 mm. What is the pressure difference between points 1 and 2 in psi?
Answer:
0.224 psi
Explanation:
The pressure using a differential manometer is calculated with the delta H.
Delta H = 293 - 275 = 18 mm
The formula for the pressure is:
P = rho * g * h,
where rho : density of the fluid inside the manometer
g : gravitational acceleration
h : delta H inside the manometer.
It is importar the use of units.
8.75 g/cm3 = 8750 kg/m3
g = 9.8 m/s2
h = 18 mm = 0.018 m
P = 1543,5 Pa ; 1 psi = 6894.8 Pa
P = 1543,5/6894,8 = 0.224 psi
An air hockey game has a puck of mass 30 grams and a diameter of 100 mm. The air film under the puck is 0.1 mm thick. Calculate the time required after impact for a puck to lose 10% of its initial speed. Assume air is at 15o C and has a dynamic viscosity of 1.75´10-5 N×s/m2 .
Answer:
time required after impact for a puck is 2.18 seconds
Explanation:
given data
mass = 30 g = 0.03 kg
diameter = 100 mm = 0.1 m
thick = 0.1 mm = 1 ×[tex]10^{-4}[/tex] m
dynamic viscosity = 1.75 ×[tex]10^{-5}[/tex] Ns/m²
air temperature = 15°C
to find out
time required after impact for a puck to lose 10%
solution
we know velocity varies here 0 to v
we consider here initial velocity = v
so final velocity = 0.9v
so change in velocity is du = v
and clearance dy = h
and shear stress acting on surface is here express as
= µ [tex]\frac{du}{dy}[/tex]
so
= µ [tex]\frac{v}{h}[/tex] ............1
put here value
= 1.75×[tex]10^{-5}[/tex] × [tex]\frac{v}{10^{-4}}[/tex]
= 0.175 v
and
area between air and puck is given by
Area = [tex]\frac{\pi }{4} d^{2}[/tex]
area = [tex]\frac{\pi }{4} 0.1^{2}[/tex]
area = 7.85 × [tex]\frac{v}{10^{-3}}[/tex] m²
so
force on puck is express as
Force = × area
force = 0.175 v × 7.85 × [tex]10^{-3}[/tex]
force = 1.374 × [tex]10^{-3}[/tex] v
and now apply newton second law
force = mass × acceleration
- force = [tex]mass \frac{dv}{dt}[/tex]
- 1.374 × [tex]10^{-3}[/tex] v = [tex]0.03 \frac{0.9v - v }{t}[/tex]
t = [tex] \frac{0.1 v * 0.03}{1.37*10^{-3} v}[/tex]
time = 2.18
so time required after impact for a puck is 2.18 seconds
To calculate the time required for the puck to lose 10% of its initial speed, you can use the equation for deceleration. First, find the final velocity of the puck using the given 10% decrease. Then, calculate the acceleration of the puck using the equation for acceleration. Finally, substitute the acceleration back into the equation for time to find the answer.
Explanation:To calculate the time required for the puck to lose 10% of its initial speed, we need to find the deceleration of the puck. We can use the equation for deceleration, which is a = (v_f - v_i) / t, where a is the acceleration, v_f is the final velocity, v_i is the initial velocity, and t is the time. In this case, since the puck is losing speed, we can use the negative value of the acceleration.
Given that the initial speed of the puck is v_i = (2 * distance) / t, we can calculate the final velocity v_f = 0.9 * v_i, where 0.9 represents the 10% decrease. Substituting these values into the deceleration equation, we can solve for t as follows:
t = (v_f - v_i) / a = (0.9 * v_i - v_i) / (-a) = 0.1 * v_i / a.
Now we can find the acceleration by using the equation a = (6 * π * η * r) / (m * v_i), where η is the dynamic viscosity of air, r is the radius of the puck (which is half the diameter), m is the mass of the puck, and v_i is the initial velocity. Substituting the given values, we can calculate the acceleration. Finally, substituting the acceleration back into the equation for t, we can calculate the time required for the puck to lose 10% of its initial speed.