Answer:
The charge at the center of the conducting sphere is zero.
Explanation:
A principle of conductor materials is that the electric field inside a conductor in electrostatic state is always zero. The gauss law says that the flux of a electric field in a closed surface is proportional to the charge enclosed by the surface. Then if the Electric field inside of a conductor is zero, imperatively the charge anywhere inside the conductor is zero too, so the charge at the center of the sphere is zero.
In the time interval from 0 s to 15 s, the acceleration of a particle is given by a = (0.1 m/s 3 )t for one-dimensional motion. If a particle starts from rest at the origin, calculate its instantaneous velocity at any time during the interval 4 s to 14 s. 1. v(t) = (0.1 m/s 3 )t 2 2. v(t) = (0.45 m/s 3 )t 2 3. v(t) = (0.05 m/s 3 )t 2 4. v(t) = (0.9 m/s 3 )t 2
Answer:
3. v(t) = (0.05 m/s 3 )t 2
Explanation:
Instantaneous acceleration = 0.1 t
dv /dt = 0.1 t
dv = 0.1 t dt
Integrating on both sides
v(t) = ( 0.1 /2) t²
= .05 t²
Two bicyclist, originally separated by a distance of 20 miles, are each traveling at a uniform speed of 10 miles per hour toward each other down a long straight east-west aligned road. As the bikes travel, a bumble bee flying at a uniform speed of 25 miles per hour travels from the front wheel of one bike to the other, instantaneously reversing its course each time it encounters one of the bicycle wheels. Alas, eventually the two bicycles collide, crushing the poor bumble bee between their front wheels. What distance does the bee travel before his demise?
Answer:
D = 25 miles
Explanation:
To solve this problem, we just need to know how much time it took both bicyclists to collide and that will be the same amount of time that the bee flew at 25miles per hour. With those values we could calculate the distance it traveled.
Since both bicyclists collide, we know that Xa=Xb, so:
Xa = V*t = 10*t and Xb = 20 - V*t = 20 - 10*t
10*t = 20 - 10*t Solving for t:
t = 1 hour Now we can calculate the distance for the bee:
D = Vbee * t = 25 * 1 = 25 miles
On the interstate, the speed limit is 60 mi/h (about 100 km/h) a skilled driver can safely decelerate at about 6.1m/s^2. How long would it take to stop the vehicle? What would the total stopping distance be?
Answer:
4.56 seconds
63.25 m
Explanation:
t = Time taken for the car to stop
u = Initial velocity = 60 km/h = 100 km/h = 100000/3600 = 27.78 m/s
v = Final velocity = 0
s = Displacement
a = Acceleration = -6.1 m/s²
Time taken by the car to stop
[tex]v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-27.78}{-6.1}\\\Rightarrow t=4.56\ s[/tex]
Time taken by the car to stop is 4.56 seconds
[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow s=27.78\times 4.56+\frac{1}{2}\times -6.1\times 4.56^2\\\Rightarrow s=63.25\ m[/tex]
The total stopping distance would be 63.25 m
Calculate the time (in seconds) needed for a car to accelerating from 0 m/s to 10 m/s at 5 m/s^2?
Answer: 2 s
Explanation: In order to solve this problem we have to use the formule of the final speed getting with a constant acceleration, it is given by;
Vfinal=Vo+a*t where Vo is zero.
so then t=Vfinal/a = (10 m/s)/5 m/s^2= 2 s
Final answer:
To calculate the time taken for a car to accelerate from 0 m/s to 10 m/s at 5 m/s², use the given equation and solve for time, resulting in 2 seconds.
Explanation:
The time taken for a car to accelerate from 0 m/s to 10 m/s at 5 m/s² can be calculated using the equation:
Final velocity = Initial velocity (u) + acceleration (a) x time (t)
Plugging in the values:
10 m/s = 0 m/s + 5 m/s² * t
10 m/s = 5t m/s²
Solving for t:
t = 10 m/s / 5 m/s² = 2 seconds
Three forces act on a moving object. One force has a magnitude of 83.7 N and is directed due north. Another has a magnitude of 59.5 N and is directed due west. What must be (a) the magnitude and (b) the direction of the third force, such that the object continues to move with a constant velocity? Express your answer as a positive angle south of east.
Answer:
[tex]|\vec{F}_3| = 102.92 \ N[/tex][tex]\theta = 57 \° 24 ' 48''[/tex]Explanation:
For an object to move with constant velocity, the acceleration of the object must be zero:
[tex]\vec{a} = \vec{0}[/tex].
As the net force equals acceleration multiplied by mass , this must mean:
[tex]\vec{F}_{net} = m \vec{a} = m * \vec{0} = \vec{0}[/tex].
So, the sum of the three forces must be zero:
[tex]\vec{F}_1 + \vec{F}_2 + \vec{F}_3 = \vec{0}[/tex],
this implies:
[tex]\vec{F}_3 = - \vec{F}_1 - \vec{F}_2[/tex].
To obtain this sum, its easier to work in Cartesian representation.
First we need to define an Frame of reference. Lets put the x axis unit vector [tex]\hat{i}[/tex] pointing east, with the y axis unit vector [tex]\hat{j}[/tex] pointing south, so the positive angle is south of east. For this, we got for the first force:
[tex]\vec{F}_1 = 83.7 \ N \ (-\hat{j})[/tex],
as is pointing north, and for the second force:
[tex]\vec{F}_2 = 59.9 \ N \ (-\hat{i})[/tex],
as is pointing west.
Now, our third force will be:
[tex]\vec{F}_3 = - 83.7 \ N \ (-\hat{j}) - 59.9 \ N \ (-\hat{i})[/tex]
[tex]\vec{F}_3 = 83.7 \ N \ \hat{j} + 59.9 \ N \ \hat{i}[/tex]
[tex]\vec{F}_3 = (59.9 \ N , 83.7 \ N ) [/tex]
But, we need the magnitude and the direction.
To find the magnitude, we can use the Pythagorean theorem.
[tex]|\vec{R}| = \sqrt{R_x^2 + R_y^2}[/tex]
[tex]|\vec{F}_3| = \sqrt{(59.9 \ N)^2 + (83.7 \ N)^2}[/tex]
[tex]|\vec{F}_3| = 102.92 \ N[/tex]
this is the magnitude.
To find the direction, we can use:
[tex]\theta = arctan(\frac{F_{3_y}}{F_{3_x}})[/tex]
[tex]\theta = arctan(\frac{83.7 \ N }{ 59.9 \ N })[/tex]
[tex]\theta = 57 \° 24 ' 48''[/tex]
and this is the angle south of east.
Unpolarized light with an intensity of 432 W/m^2 passes through three polarizing filters in a row, each of which is rotated 30 degrees from the one before it. Find the intensity of the light that emerges at the end
Answer:
40.5 W/m²
Explanation:
Intensity of unpolarized light = 432 W/m² = I₀
Intensity of light as it passes through first polarizer
I = I₀×0.5
⇒I = 432×0.5
⇒I = 216 W/m²
Intensity of light as it passes through second polarizer. So, Intensity after it passes through first polarizer is the input for the second polarizer
I = I₀×0.5 cos²30
⇒I = 216×0.75
⇒I = 162 W/m²
Intensity of light as it passes through third polarizer.
I = I₀×0.5 cos²30×cos²60
⇒I = 162×0.25
⇒I = 40.5 W/m²
Intensity of light at the end is 40.5 W/m²
A train is travelling towards the station on a straight track. It is a certain distance from the station when the engineer applies the brakes. Initially, the train is moving at 50 m/s and slows down at 2.5 m/s^2. If the train stops at the station, how far before the station did the engineer apply the brake?
Answer:
500 m
Explanation:
t = Time taken
u = Initial velocity = 50 m/s
v = Final velocity = 0
s = Displacement
a = Acceleration = -2.5 m/s²
Equation of motion
[tex]v=u+at\\\Rightarrow 0=50-2.5\times t\\\Rightarrow \frac{-50}{-2.5}=t\\\Rightarrow t=20\ s[/tex]
Time taken by the train to stop is 20 seconds
[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow s=50\times 20+\frac{1}{2}\times -2.5\times 20^2\\\Rightarrow s=500\ m[/tex]
∴ The engineer applied the brakes 500 m from the station
A 1.10 μF capacitor is connected in series with a 1.92 μF capacitor. The 1.10 μF capacitor carries a charge of +10.1 μC on one plate, which is at a potential of 51.5 V. (a) Find the potential on the negative plate of the 1.10 μF capacitor. (b) Find the equivalent capacitance of the two capacitors.
Answer:
(a). The potential on the negative plate is 42.32 V.
(b). The equivalent capacitance of the two capacitors is 0.69 μF.
Explanation:
Given that,
Charge = 10.1 μC
Capacitor C₁ = 1.10 μF
Capacitor C₂ = 1.92 μF
Capacitor C₃ = 1.10 μF
Potential V₁ = 51.5 V
Let V₁ and V₂ be the potentials on the two plates of the capacitor.
(a). We need to calculate the potential on the negative plate of the 1.10 μF capacitor
Using formula of potential difference
[tex]V_{1}=\dfrac{Q}{C_{1}}[/tex]
Put the value into the formula
[tex]V_{1}=\dfrac{10.1 \times10^{-6}}{1.10\times10^{-6}}[/tex]
[tex]V_{1}=9.18\ V[/tex]
The potential on the second plate
[tex]V_{2}=V-V_{1}[/tex]
[tex]V_{2}=51.5 -9.18[/tex]
[tex]V_{2}=42.32\ v[/tex]
(b). We need to calculate the equivalent capacitance of the two capacitors
Using formula of equivalent capacitance
[tex]C=\dfrac{C_{1}\timesC_{2}}{C_{1}+C_{2}}[/tex]
Put the value into the formula
[tex]C=\dfrac{1.10\times10^{-6}\times1.92\times10^{-6}}{(1.10+1.92)\times10^{-6}}[/tex]
[tex]C=6.99\times10^{-7}\ F[/tex]
[tex]C=0.69\ \mu F[/tex]
Hence, (a). The potential on the negative plate is 42.32 V.
(b). The equivalent capacitance of the two capacitors is 0.69 μF.
People hoping to travel to other worlds are faced with huge challenges. One of the biggest is the time required for a journey. The nearest star is 4.1×1016m away. Suppose you had a spacecraft that could accelerate at 1.3 g for half a year, then continue at a constant speed. (This is far beyond what can be achieved with any known technology.) Part A How long would it take you to reach the nearest star to earth?
Answer:
The travel would take 6.7 years.
Explanation:
The equation for an object moving in a straight line with acceleration is:
x = x0 + v0 t + 1/2a*t²
where:
x = position at time t
x0 = initial position
v0 = initial velocity
a = acceleration
t = time
In a movement with constant speed, a = 0 and the equation for the position will be:
x = x0 + v t
where v = velocity
Let´s calculate the position from the Earth after half a year moving with an acceleration of 1.3 g = 1.3 * 9.8 m/s² = 12.74 m/s²:
Seconds in half a year:
1/2 year = 1.58 x 10⁷ s
x = 0 m + 0 m/s + 1/2 * 12.74 m/s² * (1.58 x 10⁷ s)² = 1.59 x 10¹⁵ m
Now let´s see how much time it takes the travel to the nearest star after this half year.
The velocity will be the final velocity achived after the half-year travel with an acceleration of 12.74 m/s²
v = v0 + a t
Since the spacecraft starts from rest, v0 = 0
v = 12.74 m/s² * 1.58 x 10⁷ s = 2.01 x 10 ⁸ m/s
Using the equation for position:
x = x0 + v t
4.1 x 10¹⁶ m = 1.59 x 10¹⁵ m + 2.01 x 10 ⁸ m/s * t
(4.1 x 10¹⁶ m - 1.59 x 10¹⁵ m) / 2.01 x 10 ⁸ m/s = t
t = 2.0 x 10⁸ s * 1 year / 3.2 x 10 ⁷ s = 6.2 years.
The travel to the nearest star would take 6.2 years + 0.5 years = 6.7 years.
A ball is batted straight up into the air and reaches a maxium height 65.6 m (a) How long did it take to reach this height? (b) What was the pop-up velocity of the ball?
Answer:
a) 3.65 seconds
b) 35.87 m/s
Explanation:
s = Displacement = 65.6 m
u = Initial velocity
v = Final velocity
t = Time taken
a = Acceleration due to gravity = 9.81 m/s² (downward direction is taken as positive and upward is taken as negative)
b) Equation of motion
[tex]v^2-u^2=2as\\\Rightarrow 0^2-u^2=2\times -9.81\times 65.6\\\Rightarrow u=\sqrt{2\times 9.81\times 65.6}\\\Rightarrow u=35.87\ m/s[/tex]
Initial pop up velocity is 35.87 m/s
a)
[tex]v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-35.87}{-9.81}\\\Rightarrow t=3.65\ s[/tex]
It took 3.65 seconds to reach this height
A kangaroo can jump over an object 2.46 m high. (a) Calculate its vertical speed (in m/s) when it leaves the ground.
m/s
(b) How long (in s) is it in the air?
Answer:
a) 6.95 m/s
b) 1.42 seconds
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration due to gravity = 9.81 m/s²
[tex]v^2-u^2=2as\\\Rightarrow -u^2=2as-v^2\\\Rightarrow -u^2=2\times -9.81\times 2.46-0^2\\\Rightarrow u=\sqrt{2\times 9.81\times 2.46}\\\Rightarrow u=6.95\ m/s[/tex]
a) The vertical speed when it leaves the ground. is 6.95 m/s
[tex]v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-6.95}{-9.81}\\\Rightarrow t=0.71\ s[/tex]
Time taken to reach the maximum height is 0.71 seconds
[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow 2.46=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{2.46\times 2}{9.81}}\\\Rightarrow t=0.71\ s[/tex]
Time taken to reach the ground from the maximum height is 0.71 seconds
b) Time it stayed in the air is 0.71+0.71 = 1.42 seconds
A baseball leaves the bat with a speed of 40 m/s at an angle of 35 degrees. A 12m tall fence is placed 130 m from the point the ball was struck. Assuming the batter hit the ball 1m above ground level, does the ball go over the fence? If not, how does the ball hit the fence? If yes, how far beyond the fence does the ball land?
Answer:
The ball land at 3.00 m.
Explanation:
Given that,
Speed = 40 m/s
Angle = 35°
Height h = 1 m
Height of fence h'= 12 m
We need to calculate the horizontal velocity
Using formula of horizontal velocity
[tex]V_{x}=V_{i}\cos\theta[/tex]
[tex]V_{x}=40\times\cos35[/tex]
[tex]V_{x}=32.76\ m/s[/tex]
We need to calculate the time
Using formula of time
[tex]t = \dfrac{d}{v}[/tex]
[tex]t=\dfrac{130}{32.76}[/tex]
[tex]t=3.96\ sec[/tex]
We need to calculate the vertical velocity
[tex]v_{y}=v_{y}\sin\theta[/tex]
[tex]v_{y}=40\times\sin35[/tex]
[tex]v_{y}=22.94\ m/s[/tex]
We need to calculate the vertical position
Using formula of distance
[tex]y(t)=y_{0}+V_{i}t+\dfrac{1}{2}gt^2[/tex]
Put the value into the formula
[tex]y(3.96)=1+22.94\times3.96+\dfrac{1}{2}\times(-9.8)\times(3.96)^2[/tex]
[tex]y(3.96)=15.00\ m[/tex]
We need to calculate the distance
[tex]s = y-h'[/tex]
[tex]s=15.00-12[/tex]
[tex]s=3.00\ m[/tex]
Hence, The ball land at 3.00 m.
An air-filled parallel plate capacitor has circular plates with radius r=20.0 cm, separated distance 4.00 mm. The capacitor is connected to 70.0 V battery. a) Find the capacitance, Co, and the charge on the plates, Qo. b) Find the electric field strength between the plates. c) While connected to the battery, a dielectric sheet with dielectric constant k is inserted between the plates (it fills the entire space). What are the capacitance, the charge, and the voltage now?
Answer: a) 278 * 10^-12 F and 19.4 * 10^-9 C
b) 17.44 * 10^3 N/C and c) C=k*C0 and V=70/k
Explanation: In order to solve this problem we have to use the expression of the capacitor of parallel plates as:
C=A*ε0/d where A is the area of the plates and d the distance between them
C=Π r^2*ε0/d
C=π*0.2^2*8.85*10^-12/0.004=278 * 10^-12F= 278 pF
then
ΔV= Q/C
so Q= ΔV*C=70V*278 pF=19.4 nC
The electric field between the plates is given by:
E= Q/(A*ε0)=19.4 nC/(π*0.2^2*8.85*10^-12)=17.44 *10^3 N/C
If it is introduced a dielectric between the plates, then the new C is increased a factor k while the potential between the plates decreases a factor 1/k.
You're driving down the highway late one night at 25 m/s when a deer steps onto the road 58min front of you. Because you are distracted by your phone, your reaction time before stepping on the brakes is 1.50s, and the maximum deceleration of your car is 10m/s^2. (A) Do you hit the deer? If not, how much distance is between you and the deer when you stop? If you hit the deer, how fast are you going?) (B) What is the maximum initial speed you could have so that you do not hit the deer?
Answer:
(A) Yes, the driver hits the deer with a speed of 14.66 m/s.
(B) 22.21 m/s
Explanation:
Assume:
u = initial velocity of the driverv = final speed of the drivera = acceleration of the drivert = time interval for which the brakes were applieds = displacement of the driver during brakes applicationPart (A):
Before the brakes were applied, the driver moves with a constant velocity of 25 m/s for 1.5 s.
Let us find out distance traveled travel by the driver in this time interval.
[tex]x = 25\times 1.5 = 37.5\ m[/tex]
After the brakes were applied, we have
[tex]u = 25\ m/s\\v = 0\ m/s\\a = -10\ m/s^2\\[/tex]
Using the formula for constant acceleration motion, we have
[tex]v^2=u^2+2as\\\Rightarrow s = \dfrac{v^2-u^2}{2a}\\\Rightarrow s = \dfrac{(0)^2-(25)^2}{2\times (-10)}\\\Rightarrow s = 31.25\ m[/tex]
This means the driver moves a distance of 31.25 m to stop from the time he applies brakes.
So, the total distance traveled by the driver in the entire journey = 37.5 m + 31.25 m = 68.75 m.
Since, the distance traveled by the driver (68.75 m) is greater than the distance of the deer from the driver (58.0 m). So, the driver hits the deer before stopping.
For calculating the speed of the driver with which the driver hits the deer, we have to calculate the actual distance to be traveled by the driver from the instant he applies brake so that the deer does not get hit. This distance is given by:
s = 58 m - 37.5 m = 20.5 m
Now, again using the equation for constant acceleration, we have
[tex]v^2=u^2+2as\\\Rightarrow v^2=(25)^2+2(-10)(20.5)\\\Rightarrow v^2=625-410\\\Rightarrow v^2=215\\\Rightarrow v=\pm \sqrt{215}\\\Rightarrow v=\pm 14.66\ m/s\\[/tex]
Since the driver hits the deer, this means the speed must be positive.
[tex]\therefore v = 14.66 m/s[/tex]
Hence, the driver hits the deer at a speed of 14.66 m/s.
Part (B):
Let the maximum velocity of the driver so that it does not hit the deer be u.
Then,
[tex]x = u\times 1.5 = 1.5u[/tex]
After travelling this distance, the driver applies brakes till it just reaches the position of the deer and does not hit her.
This much distance is S (let).
Using the equation of constant acceleration, we have
[tex]v^2=u^2+2aS\\\Rightarrow (0)^2=u^2+2(-10)(58-x)\\\Rightarrow 0=u^2-1160+30u\\\Rightarrow u^2+30u-1160=0\\[/tex]
On solving the above quadratic equation, we have
[tex]u = 22.21\ m/s\,\,\, or u = -52.21\ m/s[/tex]
But, this speed can not be negative.
So, u = 22.21 m/s.
Hence, the driver could have traveled with a maximum speed of 22.21 m/s so that he does not hit the deer.
Sound 1 has an intensity of 47.0 W/m^2. Sound 2 has an intensity level that is 2.6 dB greater than the intensity level of sound 1. What is the intensity of sound 2? Express your answer using two significant figures.
Answer:
The intensity of Sound 2 up to two significant digits is [tex]77 W/m^{2}[/tex]
Solution:
As per the question:
Intensity of Sound 1, [tex]I_{a} = 47.0 W/m^{2}[/tex]
Intensity of Sound 2, [tex]I_{b} = 2.6 dB + I_{a}(in dB)[/tex]
Now,
The intensity of sound in decibel (dB) is:
[tex]I_{dB} = 10log_{10}\frac{I}{I_{c}}[/tex]
where
[tex]I_{c} = 1\times 10^{- 12} W/m^{2}[/tex] = threshold or critical sound intensity
Now,
Intensity of Sound 1, [tex]I_{a}[/tex] in dB is given by:
[tex]I_{a} = 10log_{10}\frac{47.0}{1\times 10^{- 12}} = 136.72 dB[/tex]
Therefore,
[tex]I_{b} = 2.6 dB + I_{a}(dB) = 2.6 + 136.27 = 138.87 dB[/tex]
Now,
The Intensity, [tex]I_{b}[/tex] in [tex]W/m^{2}[/tex] is given by:
[tex]I_{b}dB = 10log_{10}\frac{I_{b}}{I_{c}}[/tex]
[tex]138.87 = 10log_{10}\frac{I_{b}}{1\times 10^{- 12}}[/tex]
[tex]\frac{138.87}{10} = log_{10}\frac{I_{b}}{1\times 10^{- 12}}[/tex]
[tex]I_{b} = 10^{13.887}\times 1\times 10^{- 12}} = 7.709\times 1\times 10^{- 12}}[/tex]
[tex]I_{b} = 77.09 W/m^{2}[/tex]
The work done to compress a spring with a force constant of 290.0 N/m a total of 12.3 mm is: a) 3.57 J b) 1.78 J c) 0.0219 J d) 23.6 J
Answer:
Work done, W = 0.0219 J
Explanation:
Given that,
Force constant of the spring, k = 290 N/m
Compression in the spring, x = 12.3 mm = 0.0123 m
We need to find the work done to compress a spring. The work done in this way is given by :
[tex]W=\dfrac{1}{2}kx^2[/tex]
[tex]W=\dfrac{1}{2}\times 290\times (0.0123)^2[/tex]
W = 0.0219 J
So, the work done by the spring is 0.0219 joules. Hence, this is the required solution.
Final answer:
The correct answer is b) 1.78 J, which is calculated using the formula W = 1/2kx² for the work done on a spring.
Explanation:
The work done to compress a spring is given by the formula W = 1/2kx², where k is the spring constant and x is the displacement in meters. For a spring with a force constant of 290.0 N/m compressed by 12.3 mm (which should be converted to meters by dividing by 1000, thus 0.0123 m), the work done can be calculated as follows:
W = 1/2 × 290.0 N/m × (0.0123 m)² = 1/2 × 290.0 × 0.00015129 = 1.78 J.
Therefore, the correct answer is b) 1.78 J.
To practice Problem-Solving Strategy 21.1 Coulomb's Law. Three charged particles are placed at each of three corners of an equilateral triangle whose sides are of length 2.3 cm . Two of the particles have a negative charge: q1 = -6.7 nC and q2 = -13.4 nC . The remaining particle has a positive charge, q3 = 8.0 nC . What is the net electric force acting on particle 3 due to particle 1 and particle 2?
Answer:
241.27 N
Explanation:
Both the negative charge will pull the positive charge towards it . Let the forces be F₁ and F₂
F₁ = [tex]\frac{9\times10^9\times8\times6.7\times10^{-18}}{(2.3\times10^{-2})^2}[/tex]
F₁ = 91.19 X 10⁻⁵ N
F₂=[tex]\frac{9\times10^9\times8\times13.4\times10^{-18}}{(2.3\times10^{-2})^2}[/tex]
F₂ = 182.38 X 10⁻⁵ N
F₁ and F₂ act at 60 degree so their resultant will be calculated as follows
R² = (91.19)² +(182.38)² + 2 X 91.19 X 182.38 Cos 60
R² = 58209.30
R = 241.27 N
Two 2.5-cm-diameter disks face each other, 1.5 mm apart. They are charged to ± 14 nC. Part A.
What is the electric field strength between the disks?
Express your answer in newtons per coulomb.
Part B.
A proton is shot from the negative disk toward the positive disk. What launch speed must the proton have to just barely reach the positive disk?
Express your answer in meters per second.
The electric field strength between the disks can be found using the formula E = (k × q₁ × q₂) / (q × r²), where k is the Coulomb constant, q₁ and q₂ are the charges on the disks, and r is the distance between them. To calculate the launch speed of the proton, we can use the conservation of energy principle and equate the potential energy gained by the proton with its kinetic energy.
Explanation:Part A:
To find the electric field strength between the disks, we can use the formula:
Electric field strength (E) = Electric force (F) / Charge (q)
The electric force between the disks can be calculated using the formula:
Electric force (F) = (k × q₁ × q₂) / r₂
where k is the Coulomb constant, q₁ and q₂ are the charges on the disks, and r is the distance between them.
Substituting the given values, we get:
E = (k × q₁ × q₂) / (q × r₂)
Now, we can calculate the electric field strength.
Part B:
To calculate the launch speed of the proton, we can use the conservation of energy principle. The potential energy gained by the proton when moved from negative disk to positive disk can be equated to its kinetic energy.
Potential energy (PE) = Kinetic energy (KE)
PE = qV (where V is the potential difference between the disks)
KE = (1/2)mv2 (where m is the mass of the proton, v is the launch speed)
Equating the two equations, we get:
qV = (1/2)mv2
Now, we can solve for the launch speed.
Learn more about Electric Field here:https://brainly.com/question/8971780
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The electric field strength between the disks is around 79402.08N/C. The proton must be launched at a speed of roughly 2.86*10^6 m/s to just barely reach the positive disk.
Explanation:This problem involves concepts from electrodynamics and quantum mechanics. To answer Part A, we can use the formula E=σ/ε0 where σ is the surface charge density and ε0 is the permittivity of free space. The surface charge density σ=Q/A where Q is charge and A is the area of the disk, A = π*(d/2)^2. Plugging this in, we get E=~79402.08 N/C.
For Part B, the energy that a proton must have to reach the other plate is equal to the work done by the electric field in bringing the proton from the negative to the positive plate, which is W=qE*d, where q is the charge of the proton and d is the distance between the plates. Equating this to kinetic energy,1/2mv^2, we get v = sqrt((2*q*E*d)/m). Substituting appropriate values gives us v=~2.86*10^6 m/s.
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At one point in space, the electric potential energy of a 15 nC charge is 42 μJ . Part A) What is the electric potential at this point?
Express your answer to two significant figures and include the appropriate units.
Part B) If a 20 nC charge were placed at this point, what would its electric potential energy be?
Express your answer to two significant figures and include the appropriate units.
Answer:
Part A:
[tex]\rm 2.8\times 10^3\ Volts.[/tex]
Part B:
[tex]\rm 5.6\times 10^{-5}\ J.[/tex]
Explanation:
Part A:
Potential energy of charge at the given point, [tex]\rm U=42\ \mu J=42\times 10^{-6}\ J.[/tex]Charge, [tex]\rm q=15\ nC = 15\times 10^{-9}\ C.[/tex]The potential energy at a point due to a charge is defined as
[tex]\rm U=qV[/tex].
where,
V = electric potential at that point.
Therefore,
[tex]\rm V=\dfrac{U}{q}=\dfrac{42\times 10^{-6}}{15\times 10^{-9}}=2.8\times 10^3\ Volts.[/tex]
Part B:
Now, if the charge at that point is replaced with [tex]\rm q_1 = 20\ nC = 20\times 10^{-9}\ C.[/tex], then the electric potential energy at that point is given by
[tex]\rm U=q_1V = 20\times 10^{-9}\times 2.8\times 10^3=5.6\times 10^{-5}\ J.[/tex]
Electric potential energy is the energy which is required to move a unit charge from a point to another point in the electric field.
a) The electric potential at this point when 15 nC charge is placed at one point in space is [tex]2.8\times10^3[/tex] V.b) The electric potential energy, when 20 nC charge is placed at one point in space is [tex]5.6\times10^{-5}[/tex] J.What is electric potential energy?Electric potential energy is the energy which is required to move a unit charge from a point to another point in the electric field.
It can be given as,
[tex]U=qV[/tex]
Here, [tex]q[/tex] is the charge and [tex]V[/tex] is the electric potential.
Thus the electric potential can be given as,
[tex]V=\dfrac{U}{q}[/tex]
Given information-
The value of electric potential energy is 42 μJ.
a) The electric potential at this point when 15nC charge is placed at one point in space.As the value of electric potential energy is 42 μJ and charge is 15 nC. Then using the above formula, the electric potential can be given as,
[tex]V=\dfrac{42\times10^{-6}}{15\times10^{-9}}\\V=2.8\times10^3\rm volts[/tex]
Thus the value of electric potential is [tex]2.8\times10^3[/tex] V.
b) The electric potential energy, when 20 nC charge is placed at one point in space.As the value of electric potential is [tex]2.8\times10^3[/tex] V and charge is 20 nC. Then using the above formula, the electric potential can be given as,
[tex]U=({2.8\times10^{3})}\times({20\times10^{-9}})\\V=5.6\times10^{-5}\rm J[/tex]
Thus the value of electric potential energy is [tex]5.6\times10^{-5}[/tex] J.
Hence,
a) The electric potential at this point when 15 nC charge is placed at one point in space is [tex]2.8\times10^3[/tex] V.b) The electric potential energy, when 20 nC charge is placed at one point in space is [tex]5.6\times10^{-5}[/tex] J.Learn more about the electric potential energy here;
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A woman throws a javelin 35 mph at an angle 30 degrees from the ground. Neglecting wind resistance or the height the javelin thrower, how fast will the javelin strike the ground?
Answer:
35 mph
Explanation:
The key of this problem lies in understanding the way that projectile motion works as we are told to neglect the height of the javelin thrower and wind resistance.
When the javelin is thown, its velocity will have two components: a x component and a y component. The only acceleration that will interact with the javelin after it was thown will be the gravety, which has a -y direction. This means that the x component of the velocity will remain constant, and only the y component will be affected, and can be described with the constant acceleration motion properties.
When an object that moves in constant acceleration motion, the time neccesary for it to desaccelerate from a velocity v to 0, will be the same to accelerate the object from 0 to v. And the distance that the object will travel in both desaceleration and acceleration will be exactly the same.
So, when the javelin its thrown, it willgo up until its velocity in the y component reaches 0. Then it will go down, and it will reach reach the ground in the same amount of time it took to go up and, therefore, with the same velocity.
What is the minimum amount of energy required for an 80-
kgclimber carrying a 20- kg pack to climb Mt. Everest, 8850
mhigh?
Answer:
minimum amount of energy required is 8.673 MJ
Explanation:
given data
mass m1 = 80 kg
additional mass m2 = 20 kg
height = 8850 m
to find out
minimum amount of energy required
solution
we know here total mass is m = m1 + m2
m = 80 + 20
m = 100 kg
so now apply energy equation for climb high that is
energy = m×g×h ....................1
here m is mass and g is 9.8 and h is height
put here all value in equation 1
energy = m×g×h
energy = 100 ×9.8×8850
energy = 8673000 J
so minimum amount of energy required is 8.673 MJ
Final answer:
To climb Mt. Everest at 8850 m high, an 80-kg climber carrying a 20-kg pack requires a minimum of 867,300 Joules (867.3 kJ) of energy, based on the gravitational potential energy calculation.
Explanation:
The question relates to calculating the minimum amount of energy required for an 80-kg climber carrying a 20-kg pack to climb Mt. Everest, 8850 m high. To calculate the energy, we use the formula for gravitational potential energy, which is Potential Energy (PE) = mass (m) × gravity (g) × height (h). The mass of the climber and the pack combined is 100 kg (80 + 20), the acceleration due to gravity (g) is approximately 9.8 m/s2, and the height (h) is 8850 m. Therefore, the required energy can be calculated as PE = 100 kg × 9.8 m/s2 × 8850 m.
The calculation gives us PE = 867,300 Joules or 867.3 kJ. This is the minimum amount of energy required for the climber and his pack to reach the summit of Mt. Everest, assuming no energy is lost to friction or other forces.
A stone is thrown straight up with an initial speed of 12.0m/s, off the edge of a 75m tall building. A. How long does it take the stone to reach the ground?
B. How high above the ground does the stone go?
C. At what time is the stone 50.0m from the bottom of the building?
D. How fast is the stone moving when it is 50m from the bottom of the building?
Answer:
A) 3.11 seconds
B) 82.33 m
C) 5.32 seconds
D) 25.2 m/s
Explanation:
t = Time taken
u = Initial velocity = 12 m/s
v = Final velocity
s = Displacement
a = Acceleration due to gravity = 9.81 m/s²
[tex]v=u+at\\\Rightarrow 0=12-9.8\times t\\\Rightarrow \frac{-12}{-9.81}=t\\\Rightarrow t=1.22 \s[/tex]
Time taken to reach maximum height is 1.22 seconds
[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow s=12\times 1.22+\frac{1}{2}\times -9.81\times 1.22^2\\\Rightarrow s=7.33\ m[/tex]
So, the stone would travel 7.33 m up
B) So, total height stone would fall is 7.33+75 = 82.33 m
[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow 82.33=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{82.33\times 2}{9.81}}\\\Rightarrow t=4.1\ s[/tex]
A) Total time taken by the stone to reach the ground from the time it left the person's hand is 4.1+1.22 = 5.32 seconds.
C) When s = 82.33-50 = 33.33 m
[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow 32.33=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{32.33\times 2}{9.81}}\\\Rightarrow t=2.57\ s[/tex]
The time it takes from the person's hand throwing the ball to the distance of 50 m from the ground is 2.57+1.22 = 3.777 seconds
D)
[tex]v=u+at\\\Rightarrow v=0+9.81\times 2.57\\\Rightarrow v=25.2\ m/s[/tex]
The stone is moving at 25.2 m/s when it is 50m from the bottom of the building
Determine the speed of ocean currents (m/s) moving water near Africa at 30 W passing Puerto Rico (70W) in one year.
Answer:
The speed of ocean currents approximately is 0.4667 m/s
Solution:
As per the question:
The ocean currents water moves at 30 W near Africa and 70 W near Puerto Rico
Time taken, t = 1 year = [tex]365\times 24\times 3600 = 3.15\times 10^{7} s[/tex]
The distance between Puerto Rico and Africa, x = [tex]1.46\times 10^{7} m[/tex]
Thus
The speed of the ocean currents there:
speed, [tex]v = \frac{d}{t} = \frac{1.47\times 10^{7}}{3.15\times 10^{7}}[/tex]
[tex]s = 0.4667 m/s[/tex]
A tungsten target is struck by electrons that have been accelerated from rest through a 27.3-kV potential difference. Find the shortest wavelength of the radiation emitted.
Answer:
4.555 x 10^-11 m
Explanation:
Potential difference, V = 27.3 kV
Let the wavelength is λ.
The energy associated with the potential difference, E = 27.3 keV
E = 27.3 x 1000 x 1.6 x 10^-19 J = 4.368 x 10^-15 J
The energy associated with the wavelength is given by
[tex]E=\frac{hc}{\lambda }[/tex]
Where, h is Plank's constant = 6.63 x 10^-34 Js
c is velocity of light 3 x 10^8 m/ s
By substituting the values, we get
[tex]4.368\times10^{-15}=\frac{6.634\times10^{-34}\times 3\times 10^{8}}{\lambda }[/tex]
λ = 4.555 x 10^-11 m
An iron vat is 11 m long at room temperature (20°C). How much longer is it when it contains boiling water at 1 atm pressure?
The iron vat will be 0.01056 meters (or 10.56 millimeters) longer when it contains boiling water at 1 atm pressure.
Given data:
To calculate the change in length of the iron vat when it is heated from room temperature (20°C) to boiling water temperature, we can use the linear thermal expansion formula:
ΔL = α * L * ΔT
Where:
ΔL = Change in length
α = Coefficient of linear expansion of iron (given)
L = Original length of the vat
ΔT = Change in temperature
Given:
Original length, L = 11 m
Coefficient of linear expansion of iron, α = 12 x 10^-6 /°C (approximate value for iron)
Change in temperature, ΔT = Boiling point of water (100°C) - Room temperature (20°C) = 80°C
Now, plug in the values and calculate:
[tex]\Delta L = (12 * 10^{-6} ) * (11 m) * (80 ^\circ C)[/tex]
ΔL ≈ 0.01056 m
Hence, the iron vat will be 0.01056 meters longer.
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The iron vat will be approximately 1.056 cm longer when it contains boiling water at 1 atm pressure compared to at room temperature due to linear thermal expansion. This is calculated using the relevant formula and the coefficients for iron.
Explanation:The subject of this question is linear thermal expansion, related to physics. When the iron vat is heated by boiling water, it expands. The length of the expansion can be calculated using the formula ΔL = αLΔT, where ΔL is the change in length, α is the coefficient of linear expansion for iron (12 ×10^-6/°C), L is the original length, and ΔT is the change in temperature. Considering the original length L as 11 m and ΔT as the difference between 100°C (boiling point of water at 1 atm pressure) and 20°C (room temperature), which is 80°C. So, plug in these values we have ΔL = 12 ×10^-6/°C * 11m * 80°C = 0.01056 m or about 1.056 cm. Therefore, the iron vat is about 1.056 cm longer when it contains boiling water at 1 atm pressure compared to at room temperature.
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A pipe carrying 20°C water has a diameter of 3.8 cm. Estimate the maximum flow speed (cm/s) if the flow must be streamline. (The density of water is 1.00 x 10^3 kg/m^3.)
Answer:
maximum flow speed is 5.278 cm/s
Explanation:
given data
temperature = 20°C
diameter d = 3.8 cm
density of water ρ = 1 × 10³ kg/m³
to find out
maximum flow speed
solution
we have given steam line flow so it is laminar flow
and for laminar flow we know reynolds number is Re= 2000
so
Re = [tex]\frac{\rho*v*d}{\mu}[/tex] ............1
here μ is dynamic viscosity = 0.001003 for 20°C for water
put all value in equation 1
Re = [tex]\frac{\rho*v*d}{\mu}[/tex]
2000 = [tex]\frac{1000*v*3.8*10^{-2}}{0.001003}[/tex]
solve it we get v
v = 0.0527 m/s
so maximum flow speed is 5.278 cm/s
Look online for the growth of the trunk of a tree. Estimate how much time does it take for a water oak to grow one inch in diameter. Estimate the growth rate over a year.
Answer:
8 months needed to grow one inch in diameter by water oak and it grows 24 inches of height in a year.
Explanation:
In 2 different ways the trunk of a tree grows first the height than second in diameter usually in diameter tree grows one ring per year. We can say on counting the number of rings we can determine the age of a tree both of the growth does not occur in the same rate.
Tree grows more higher as compared to the diameter growth and mature trees usually grows 1 inch in diameter every year.
Water oak gets 24 inches height growth every year and 1.5 inch growth in diameter annually means if we divide 1.5 inches by 12 months we gets 0.125 inches growth monthly.so we can say 8 months are needed to grow the diameter of water oak 1 inch.
A circular hole in an aluminum plate is 2.739 cm in diameter at 0.000°C. What is the change in its diameter when the temperature of the plate is raised to 108.0°C? ________ cm
Answer:
[tex]L = 2.746 cm[/tex]
Explanation:
As we know that thermal expansion coefficient of aluminium is given as
[tex]\alpha = 24 \times 10^{-6} per ^oC[/tex]
now we also know that after thermal expansion the final length is given as
[tex]L = L_o(1 + \alpha \Delta T)[/tex]
here we know that
[tex]L_o = 2.739 cm[/tex]
[tex]\alpha = 24 \times 10^{-6}[/tex]
[tex]\Delta T = 108 - 0= 108^oC[/tex]
now we will have
[tex]L = 2.739(1 + 24 \times 10^{-6} (108))[/tex]
[tex]L = 2.746 cm[/tex]
The x component of a velocity vector that has an angle of 37° (to the +x) has a magnitude of 4.8 m/s. What is the magnitude of the velocity? What is the y component of the velocity?
Answer:
Magnitude of velocity V = 6.015 m/sec
Y component of velocity = 3.61 m/sec
Explanation:
We have given magnitude of velocity in x direction, that is horizontal component [tex]V_X=3.8m/sec[/tex]
Angle with x axis = 37°
We know that horizontal component [tex]V_X=Vcos\Theta[/tex]
So [tex]4.8=Vcos37^{\circ}[/tex]
[tex]V=6.015m/sec[/tex]
Y component of velocity is given by [tex]V_Y=Vsin\Theta =6.015sin37^{\circ}=3.61m/sec[/tex]
A proton is at the origin and an electron is at the point x = 0.36 nm , y = 0.39 nm . Find the electric force on the proton.Express your answer using two significant figures. Enter your answers numerically separated by a comma.
Answer:
The electric force on the proton is 8.2x10^-10 N
Explanation:
We use the formula to calculate the distance between two points, as follows:
r = ((x2-x1)^2 + (y2-y1)^2)^1/2, where x1 and x2 are the x coordinate, y2, y1 are the y coordinate. replacing values:
r = ((0.36-0)^2 + (0.39-0)^2)^1/2 = 0.53 nm = 5.3x10^-10 m
We will use the following expression to calculate the electrostatic force:
F = (q1*q2)/(4*pi*eo*r^2)
Here we have:
q1 = q2 = 1.6x10^-19 C, 1/4*pi*eo = 9x10^9 Nm^2C^-2
Replacing values:
F = (1.6x10^-19*1.6x10^-9*9x10^9)/((5.3x10^-10)^2) = 8.2x10^-10 N
The electric force on a proton can be calculated using Coulomb's Law and the charges and distance between the proton and electron. The charge of both particles is ±1.602 × 10^{-19} C, and the distance is calculated from their coordinates on the x and y-axis.
Explanation:To find the electric force on the proton, we can use Coulomb's Law. Coulomb's Law states that the force between two charges is proportional to the product of the charges and inversely proportional to the square of the distance between them.
The formula for Coulomb's Law is:
F = k * |q1 * q2| / r2,
where F is the force, k is Coulomb's constant (approximately 8.9875 × 109 N m2/C2), q1 and q2 are the charges of the particles, and r is the distance between them.
The charge of a proton (q1) and an electron (q2) is approximately ±1.602 × 10−16 C.
The distance r can be found using the Pythagorean theorem, since we have the x and y coordinates:
r = √(x2 + y2) = √(0.362 + 0.392).
After calculating the distance r, we can plug all the values into the Coulomb's Law formula to get the magnitude of the force.