The bridge was approximately 11.01 meters high. The swimmers were moving at a speed of around 14.72 m/s when they hit the water. If the bridge were twice as high, the drop time would be about 2.12 seconds.
Explanation:The subject in question involves principles of physics, specifically gravitational acceleration. When an object is dropped, it falls under the influence of gravity. This is usually around 9.81 m/s2 on Earth. Using the formula for motion d=1/2gt^2, where d is the distance or height, g is the acceleration due to gravity, and t is the time it takes for the object to fall.
a.) For calculation of the bridge's height, plug in the values of time (t=1.5s) and gravity (g=9.81m/s^2) into the equation, we get d=1/2*(9.81)*(1.5)^2 = 11.01 meters. The height of the bridge from which they jumped is approximately 11 meters.
b.) To calculate how fast the swimmers were moving when they hit the water, we can use the equation v=gt, where v is the velocity or speed. Plugging in the values, we get v=(9.81)*(1.5) = 14.72 m/s. So, the swimmers were moving at about 14.7 meters per second when they hit the water.
c.) If the bridge were twice as high, the time of drop would be found using the equation t= sqrt(2d/g), where d is now twice the original distance = 2*11.01 = 22.02 meters. Substituting the given values, we get t= sqrt(2(22.02)/9.81) = 2.12 seconds. The swimmer's drop time would be approximately 2.12 seconds.
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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
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
In quantum mechanics, the fundamental constant called Planck's constant, h, has dimensions of [ML^2T^-1 ]. Construct a quantity with the dimensions of length using h, a mass m, and c, the speed of light.
Answer:
h/(m*c)
Explanation:
Hi!
Lets denote the units of X as [X]
Since the dimentions of h are:
[tex][h] = \frac{ML^{2}}{T}[/tex]
If we divide [h] by the units of mass, we get:
[tex]\frac{[h]}{[m]} = \frac{L^{2}}{T}}[/tex]
Also we know that:
[tex][c] = \frac{L}{T}[/tex]
So:
[tex][\frac{h}{mc}] = \frac{L^{2}}{T}}*\frac{T}{L}=L[/tex]
Therefore
h/(mc) has dimentiosn of length
To construct a quantity with the dimensions of length using Planck's constant (h), mass (m), and the speed of light (c), the formula L = h/(mc) can be used. This length scale is relevant in the realms of quantum mechanics and high-energy physics.
Explanation:The student has asked how to construct a quantity with the dimensions of length using Planck's constant (h), a mass (m), and the speed of light (c). To achieve this, we can use the formula for the Planck length given by:
Lp = √hG/c³
Where G is Newton's gravitational constant. However, since we need to construct a length using just h, m, and c without G, we can rearrange the Planck length equation to:
L = h/(mc)
This gives a length L which is dependent on the mass m in addition to Planck's constant h and the speed of light c. This length scale is significant in quantum mechanics and high-energy physics, where extremely small distances are relevant.
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
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 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.
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Tectonic plates are large segments of the earth's crust that move slowly. Suppose one such plate has an average speed of 6.0 cm per year. (a) What distance does it move in 71 seconds at this speed?
m
(b) What is its speed in miles per million years?
mi/My
Answer:
1.35×10⁻⁷ m
37.278 mi/My
Explanation:
Speed of the tectonic plate= 6 cm/yr
Converting to seconds
[tex]6=\frac{6}{365.25\times 24\times 60\times 60}[/tex]
So in one second it will move
[tex]\frac{6}{365.25\times 24\times 60\times 60}[/tex]
In 71 seconds
[tex]71\times \frac{6}{365.25\times 24\times 60\times 60}=1.35\times 10^{-5}\ cm[/tex]
The tectonic plate will move 1.35×10⁻⁵ cm or 1.35×10⁻⁷ m
Convert to mi/My
1 cm = 6.213×10⁻⁶ mi
1 M = 10⁶ years
[tex]6\times 6.213\times 10^{-6}\times 10^6=37.278\ mi/My[/tex]
Speed of the tectonic plate is 37.278 mi/My
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.
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.
A concave mirror produces a real image that is three times as large as the object. If the object is 20 cm in front of the mirror, what is the image distance? What is the focal length of this mirror?
The distance of the image is 60 cm and the focal length of this concave mirror is 30 cm.
What is focal length of the lens?The focal length of the lens is length of the distance between the middle of the lens to the focal point.
It can be find out using the following formula as,
[tex]\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}[/tex]
Here, (v)is the distance of the image, (u) is the distance of the object, and (f) is the focal length of the lens.
Here, the concave mirror produces a real image that is three times as large as the object. The object is 20 cm in front of the mirror, and the concave mirror produces a real image that is three times as large as the object.
Hence, the value of magnification of the mirror is 3.
The object distance is 20 cm thus the image distance, using the magnification formula, can be given as,
[tex]m=\dfrac{v}{u}\\3=\dfrac{v}{20}\\v=60\rm cm[/tex]
Put the values in the lens formula as,
[tex]\dfrac{1}{60}+\dfrac{1}{20}=\dfrac{1}{f}\\f=30 \rm \; cm[/tex]
Hence, the distance of the image is 60 cm and the focal length of this concave mirror is 30 cm.
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Final answer:
Using the mirror equation and magnification formula, the image distance is found to be -60 cm and the focal length of the concave mirror is calculated to be 30 cm.
Explanation:
The question involves determining the image distance and the focal length of a concave mirror that produces a real image three times larger than the object. The object is placed 20 cm in front of the mirror. To find these values, we can use the mirror equation which is 1/f = 1/do + 1/di, where f is the focal length, do is the object distance, and di is the image distance.
Given that the magnification (m) is -3 (negative sign since the image is real and inverted), and magnification is also equal to -di/do, we can express di as -3do. We are told that do is 20 cm, so di is -3(20 cm) = -60 cm (negative because the image is on the same side as the object).
Now, we can plug these values into the mirror equation to find the focal length (f):
1/f = 1/do + 1/di1/f = 1/20 + 1/(-60)1/f = 1/20 - 1/601/f = (3-1)/601/f = 2/60f = 60/2f = 30 cmTherefore, the image distance is -60 cm and the focal length of the concave mirror is 30 cm.
A car is driven east for a distance of 47 km, then north for 28 km, and then in a direction 35° east of north for 27 km. Determine (a) the magnitude of the car's total displacement from its starting point and (b) the angle (from east) of the car's total displacement measured from its starting direction.
Answer:
(a) 82 Km
(b) 32°
Explanation:
First you should draw the vectors in the cartesian plane (please see the picture below).
As the car is driven east for a distance of 47 Km, your first vector should be drawn from the origin (0,0) and on the x axis
Then the car is driven north for a distance of 28 Km, so your second vector should be drawn from the origin and on the y axis.
And the car finally goes east of north for 27Km, so the third vector should be drawn from the origin east of north forming an 35° angle with x axis.
Then you should find the components of the vector in x and y:
For Vector 1 ([tex]V_{1}[/tex])
[tex]V_{1x}=47Km[/tex]
[tex]V_{1y}=0[/tex]
For Vector 2 ([tex]V_{2}[/tex])
[tex]V_{2x}=0[/tex]
[tex]V_{2y}=28Km[/tex]
For Vector 3 ([tex]V_{3}[/tex])
[tex]V_{3x}=27cos37^{o}[/tex]
[tex]V_{3x}=21.56[/tex]
[tex]V_{3y}=27sin37^{o}[/tex]
[tex]V_{3y}=16.25[/tex]
To find the magnitud of the car´s total displacement, (R) you should add up all the x and y components.
For the x component:
[tex]R_{x}=V_{1x}+V_{2x}+V_{3x}[/tex]
[tex]R_{x}=47+0+21.56[/tex]
[tex]R_{x}=68.56Km[/tex]
For the x component:
[tex]R_{y}=V_{1y}+V_{2y}+V_{3y}[/tex]
[tex]R_{y}=0+28+16.25[/tex]
[tex]R_{y}=44.25Km[/tex]
Now please see the second picture that is showing the components x and y as the sides of a right triangle, and we are going to use the Pythagorean theorem to find the resultant, R.
[tex]R=\sqrt{R_{x}^{2}+R_{y}^{2}[/tex]
[tex]R=\sqrt{68.56^{2}+44.25^{2}}[/tex]
[tex]R=82Km[/tex]
And to find the angle of the car´s total displacement (α), we use the same right triangle with the relationship between its legs.
[tex]tan(\alpha)=\frac{44.25}{68.56}[/tex]
[tex]tan(\alpha)=0.64[/tex]
[tex]\alpha=tan^{-1}(0.64)[/tex]
[tex]\alpha =32^{o}[/tex]
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).
A small block with mass 0.0475 kg slides in a vertical circle of radius 0.425 m on the inside of a circular track. During one of the revolutions of the block, when the block is at the bottom of its path, point A, the magnitude of the normal force exerted on the block by the track has magnitude 3.95 N . In this same revolution, when the block reaches the top of its path, point B, the magnitude of the normal force exerted on the block has magnitude 0.670 N. How much work was done on the block by friction during the motion of the block from point A to point B?
Answer:0.10283 J
Explanation:
Given
mass of block(m)=0.0475 kg
radius of track (r)=0.425 m
when the Block is at Bottom Normal has a magnitude of 3.95 N
Force acting on block at bottom
[tex]N-mg=\frac{mu^2}{r}[/tex]
[tex]N=mg+\frac{mu^2}{r}[/tex]
[tex]3.95=0.0475\timess 9.81+\frac{0.0475\times u^2}{0.425}[/tex]
[tex]u^2=31.172[/tex]
[tex]u=\sqrt{31.172}[/tex]
u=5.583 m/s
At top point
[tex]N+mg=\frac{mv^2}{r}[/tex]
[tex]0.670+0.0475\timess 9.81=\frac{mv^2}{r}[/tex]
[tex]v^2=10.1639[/tex]
v=3.188 m/s
Using Energy conservation
[tex]\frac{mu^2}{2}=\frac{mv^2}{2}+mg\left ( 2r\right )+W_f[/tex]
[tex]W_f=0.4989-0.39607[/tex]
[tex]W_f=0.10283 J[/tex]
i.e. 0.10283 J of energy is wasted while moving up.
The work done on the block by friction during the motion of the block from point A to point B is 0.102 J.
Velocity of the block at top circle
The velocity of the block at the top of the vertical circle is calculated as follows;
[tex]W + mg = \frac{mv^2}{r} \\\\0.67 + 0.0475(9.8)= \frac{0.0475v^2}{0.425} \\\\1.136 = 0.112 v^2\\\\v^2 = 10.14\\\\v = 3.18 \ m/s[/tex]
Velocity of the block at bottom circleThe velocity of the block at the bottom of the vertical circle is calculated as follows;
[tex]W - mg = \frac{mv^2}{r} \\\\3.95 - 0.0475(9.8)= \frac{0.0475v^2}{0.425} \\\\3.485 = 0.112 v^2\\\\v^2 = 31.12\\\\v = 5.58 \ m/s[/tex]
Work done by frictionThe work done by friction is the change in the kinetic energy of the block.
[tex]W _f =P.E_f - \Delta K.E \\\\W_f = mgh - \frac{1}{2} m(v_f^2 - v_i^2)\\\\Wf = 0.0475\times 9.8(2 \times 0.425) - \frac{1}{2} \times 0.0475(5.58^2 - 3.18^2)\\\\W_f = 0.396 - 0.498\\\\W_f = -0.102 \ J[/tex]
Thus, the work done on the block by friction during the motion of the block from point A to point B is 0.102 J.
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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].
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.
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.
The electric field at 4 cm from the center of a long copper rod of radius 2 cm has a magnitude of 4 N/C and is directed outward from the axis of the rod. (a) How much charge per unit length (in C/m) exists on the copper rod?
(b) What would be the electric flux (in N · m^2/C) through a cube of side 3 cm situated such that the rod passes through opposite sides of the cube perpendicularly?
To find the charge per unit length on the copper rod, use the formula E = k * λ / r. Plug in the given values to find the charge per unit length. The electric flux through the cube can be calculated by multiplying the electric field by the area of one face of the cube.
Explanation:(a) To find the charge per unit length on the copper rod, we can use the formula for an electric field (E = k * λ / r), where E is the electric field, k is the electrostatic constant, λ is the charge per unit length, and r is the distance from the center of the rod. Rearranging the formula, we can solve for λ: (λ = E * r / k). Plugging in the given values, we have: λ = (4 N/C) * (0.04 m) / (9 x 10^9 Nm^2/C^2) = 1.78 x 10^-10 C/m.
(b) The electric flux through a surface is given by the formula (Φ = E * A), where Φ is the electric flux, E is the electric field, and A is the area of the surface. In this case, the electric flux through the cube can be calculated by multiplying the electric field (4 N/C) by the area of one face of the cube (3 cm)^2 = (0.03 m)^2 = 9 x 10^-4 m^2. Therefore, the electric flux is Φ = (4 N/C) * (9 x 10^-4 m^2) = 3.6 x 10^-3 Nm^2/C.
(a) The charge per unit length on the copper rod is [tex]\( {8.90 \times 10^{-10} \, \text{C/m}} \).[/tex]
(b) The electric flux through the cube is [tex]\( {0.0036 \, \text{N} \cdot \text{m}^2/\text{C}} \).[/tex]
(a): Charge per Unit Length on the Copper Rod
The electric field ( E ) at a distance ( r ) from the axis of a long charged rod is given by:
[tex]\[E = \frac{2k\lambda}{r}\][/tex]
From the given data:
[tex]\[4 = \frac{2 \cdot 8.99 \times 10^9 \cdot \lambda}{0.04}\][/tex]
Solving for [tex]\( \lambda \):[/tex]
[tex]\[\lambda = \frac{4 \cdot 0.04}{2 \cdot 8.99 \times 10^9}\][/tex]
[tex]\[\lambda = \frac{0.16}{1.798 \times 10^9}\][/tex]
[tex]\[\lambda \approx 8.90 \times 10^{-10} \, \text{C/m}\][/tex]
(b): Electric Flux through a Cube
To find the electric flux [tex]\( \Phi_E \)[/tex] through the cube, we use Gauss's law, which states:
[tex]\[\Phi_E = \oint_S \mathbf{E} \cdot d\mathbf{A} = \frac{q_{\text{enclosed}}}{\epsilon_0}\][/tex]
Since the rod is long and the cube is oriented such that the rod passes through opposite sides perpendicularly, the electric flux [tex]\( \Phi_E \)[/tex] through the cube is:
[tex]\[\Phi_E = E \cdot A\][/tex]
Calculate ( A ):
[tex]\[A = a^2 = (0.03)^2 = 0.0009 \, \text{m}^2\][/tex]
Now, calculate [tex]\( \Phi_E \):[/tex]
[tex]\[\Phi_E = 4 \times 0.0009\][/tex]
[tex]\[\Phi_E = 0.0036 \, \text{N} \cdot \text{m}^2/\text{C}\][/tex]
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]
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
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.
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.
Suppose that a constant force is applied to an object with a mass of 12kg, it’s creates an acceleration of 5m/s^2. The acceleration of another object produced by the same force is 4m/s^2, what is the mass of this object?
Answer:
Mass of second object will be 15 kg
Explanation:
We have given mass of first object = 12 kg
Acceleration [tex]a=5m/sec^2[/tex]
According to second law of motion we know that force F = MA
So force [tex]F=12\times 5=60N[/tex]
As the same force is applied to the second object of acceleration [tex]a=4m/Sec^2[/tex]
So force = ma
[tex]60=m\times 4[/tex]
m = 15 kg
So mass of second object will be 15 kg
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
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'.
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
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 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 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
A rescue helicopter is hovering over a person whose boat has sunk. One of the rescuers throws a life preserver straight down to the victim with an initial velocity of 1.50 m/s and observes that it takes 1.2 s to reach the water. How high above the water was the preserver released? Note that the downdraft of the helicopter reduces the effects of air resistance on the falling life preserver, so that an acceleration equal to that of gravity is reasonable. Round your answer to the nearest whole number.
Answer:
The answer is 9 m.
Explanation:
Using the kinematic equation for an object in free fall:
[tex]y = y_o - v_o-\frac{1}{2}gt^{2}[/tex]
In this case:
[tex]v_o = \textrm{Initial velocity} = 1.5[m/s]\\t = \textrm{air time} = 1.2 [s][/tex][tex]y_o = 0[/tex]
[tex]g = \textrm{gravity} = 9.8 [m/s^{2} ][/tex]
Plugging those values into the previous equation:
[tex]y = 0 - 1.5*1.2-\frac{1}{2}*9.8*1.2^{2} \\y = -8.85 [m] \approx -9 [m][/tex]
The negative sign is because the reference taken. If I see everything from the rescuer point of view.
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.