Answer:
t=67.7s
Explanation:
From this question we know that:
Vo = 6m/s
a = 1.8 m/s2
D = 1500m
And we also know that:
[tex]X=V_{o}*t + \frac{a*t^{2}}{2}[/tex] Replacing the known values:
[tex]1500=6t+0.9*t^{2}[/tex] Solving for t we get 2 possible answers:
t1 = -44.3s and t2 = 67.7s Since negative time represents an instant before the beginning of the movement, t1 is discarded. So, the final answer is:
t = 67.7s
A plastic rod has been bent into a circle of radius R = 8.00 cm. It has a charge Q1 = +2.70 pC uniformly distributed along one-quarter of its circumference and a charge Q2 = −6Q1 uniformly distributed along the rest of the circumference. Take V = 0 at infinity. (a) What is the electric potential at the center C of the circle? V (b) What is the electric potential at point P, which is on the central axis of the circle at distance D = 6.71 cm from the center?
The electric potential at the center (C) of the circle formed by the rod is zero due to the symmetry and cancellations of charges. Calculating the potential at point P requires a detailed approach considering the distance D and the charge distribution.
Explanation:The question involves calculating the electric potential at two different points with reference to a specially configured plastic rod. For part (a), the total charge along the circumference is given by the sum of Q1 and Q2, where Q2 = -6Q1. However, due to the uniform distribution and the symmetry of the setup, the electric potential at the center of the circle (point C) would effectively be zero because the contributions from the positively and negatively charged segments cancel each other out.
For part (b), calculating the electric potential at point P, which lies on the axis and is a distance D from the center, would require the application of principles of superposition and the integration of electric potential contributions from each infinitesimal segment of the charged parts of the rod. Considering the arrangement and the charges' nature, this part of the solution involves more intricate calculations and requires knowing the specific distribution of the charges along the rod.
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 projectile is fired into the air at a nonzero angle with the horizontal. When the projectile reaches its maximum height, the speed of the projectile is 21.5% of its original speed. What angle was the projectile originally fired at?
Answer:
Answer:
77.58°
Explanation:
Let the projectile is projected at an angle θ and the velocity of projection is u.
At the maximum height, the projectile has only horizontal component of velocity.
Let the velocity at maximum height is v.
According to the question, the velocity at maximum height is 21.5% of initial velocity.
v = 21.5 % of u
u Cos θ = 21.5 % of u
u Cos θ = 0.215 u
Cos θ = 0.215
θ = 77.58°
Thus, the angle of projection is 77.58°.
Explanation:
A rocket accelerates at 25m/s^2 for 5s before it runs out of fuel and dies. What is the maximum height reached by the rocket?
Answer:
1109m
Explanation:
The distance h₁ traveled by the rocket during acceleration:
[tex]h_1=\frac{1}{2}at^2[/tex]
The velocity v₁ at this height h₁:
[tex]v_1=at[/tex]
When the rocket runs out of fuel, energy is conserved:
[tex]E=\frac{1}{2}mv_1^2+mgh_1=mgh_2[/tex]
Solving for h₂:
[tex]h_2=\frac{v_1^2}{2g}+h_1=\frac{(at)^2}{2g}+\frac{1}{2}at^2[/tex]
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.
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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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A proton accelerates from rest in a uniform electric field of 700 N/C. At one later moment, its speed is 1.10 Mm/s (nonrelativistic because v is much less than the speed of light). (a) Find the acceleration of the proton. m/s2 (b) Over what time interval does the proton reach this speed? s (c) How far does it move in this time interval?
The proton's acceleration is calculated using the electric field strength and the electron charge. The time to reach 1.1 Mm/s is computed using this acceleration and final speed. The displacement of the proton is derived using the time and acceleration.
Explanation:This question involves analyzing the motion of a proton in an electric field, using basic principles from physics, specifically the concept of acceleration and kinematics.
(a): Acceleration of the proton
: The electric force exerted on the proton can be calculated by F=qE where q is the proton's charge (1.6 x 10^-19 C), and E is the supplied electric field (700N/C). Knowing that acceleration a=F/m (F=force, m=mass of proton = 1.67 x 10^-27 kg), we can find the acceleration.
(b): Time taken to reach the speed
: Using the formula v=u+at (u=initial speed=0, a=acceleration from part (a), v=final speed=1.1 x 10^6 m/s) we can solve for t, the time taken.
(c): Displacement of the proton
: Displacement can be calculated by using the formula s= ut + 1/2at²(where u=initial speed, t=time taken from part (b), a=acceleration from part a).
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A neutral solid metal sphere of radius 0.1 m is at the origin, polarized by a point charge of 9 ✕ 10−8 C at location −0.6, 0, 0 m. At location 0, 0.08, 0 m, what is the electric field contributed by the polarization charges on the surface of the metal sphere?
Answer:
[tex]E_{polarization} = 2.21 \times 10^3(-0.99\hat i -0.132\hat j)[/tex]
Explanation:
As we know that electric field inside any conducting shell is always zero
so we can say that
[tex]E_{charge} + E_{polarization} = 0[/tex]
here we know that
[tex]E_{charge} = \frac{kq}{r^2} \hat r[/tex]
here we know that
[tex]\hat r = \frac{(0 - -0.6)\hat i + (0.08 - 0)\hat j}{\sqrt{0.6^2 + 0.08^2}}[/tex]
[tex]\hat r = 0.99\hat i + 0.132\hat j[/tex]
now we will have
[tex]E_{polarization} = - E_{charge}[/tex]
[tex]E_{polarization} = - \frac{(9\times 10^9)(9 \times 10^{-8})}{0.6^2 + 0.08^2}(0.99\hat i + 0.132\hat j)[/tex]
[tex]E_{polarization} = 2.21 \times 10^3(-0.99\hat i -0.132\hat j)[/tex]
We have that the electric field contributed by the polarization charges on the surface of the metal sphere is
E polarization=2.21*10^3(-0.99i-0.132j))From the question we are told
A neutral solid metal sphere of radius 0.1 m is at the origin, polarized by a point charge of 9 ✕ 10−8 C at location −0.6, 0, 0 m. At location 0, 0.08, 0 m,what is the electric field contributed by the polarization charges on the surface of the metal sphere?E polarizationGenerally the equation for the electric field inside any conducting shell is mathematically given as
E charge+E polarization=0
Therefore
[tex]E charge=\frac{kq}{r^2}r'\\\\E charge=\frac{0--0.6i+0.08-0j}{\sqrt{0.6^2+0.08^2}}\\\\r'=0.99i+0.132j[/tex]
Therefore
E polarization=-E charge
Therefore
E polarization=2.21*10^3(-0.99i-0.132j))For more information on this visit
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The average lifetime of μ-mesons with a speed of 0.95c is measured to be 6 x 10^6 s. Find the average lifetime of μ-mesons in a system in which they are at rest?
Answer:
19.2*10^6 s
Explanation:
The equation for time dilation is:
[tex]t = \frac{t'}{\sqrt{1-\frac{v^2}{c^2}}}[/tex]
Then, if it is observed to have a life of 6*10^6 s, and it travels at 0.95 c:
[tex]t = \frac{6*10^6}{\sqrt{1-\frac{(0.95c)^2}{c^2}}} = 19.2*10^6 s[/tex]
It has a lifetime of 19.2*10^6 s when observed from a frame of reference in which the particle is at rest.
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
Suppose your hair grows at the rate of 1/36 inches per day. Find the rate at which it grows in nanometers per second. Because the distance between atoms in a molecule is on the order of 0.1 nm, you answer suggests how rapidly layers of atoms are assembled in this protein synthesis. Your units should be "atomic layers/sec"
Answer:
hair grows at rate of 8.166 nm/s
Explanation:
given data:
grow rate= (1/36)in/day
we know that
1 day = 86400 seconds
therefore we have [tex]= \frac{1}{36}inc / 86400 sec[/tex]
first change equation to [tex]\frac{1}{36}inc / 86400 sec.[/tex]
Then you find x in/s [tex]= \frac{1}{36}inc / 86400 sec[/tex]
[tex]x/s = \frac{1}{36}inc / 86400 sec[/tex]
[tex](86400s)x = (\frac{1}{36}inc) s[/tex]
[tex]x = \frac{\frac{1}{36}\ inc}{86400} = 3.21 x 10^{-7} in/second[/tex]
Now convert in/sec into meters/second.
we know that 1m = 39.37in
[tex]x/1m = \frac{3.21 x10^{-7} in}{39.37\ in}[/tex]
[tex](39.37in)x = (3.21 *10^{-7} in) m[/tex]
[tex]x = \frac{3.21 * 10^{-7}}{39.37}m[/tex]
x = [tex]8.166 * 10^{-9}[/tex] m/second
Now convert [tex]8.166 * 10^{-9}[/tex] m/second into nm/second
we know that [tex]1nm = 10^{-9}m[/tex]
[tex]x/nm = \frac{8.166x 10^{-9} m/second}{10^{-9} m}[/tex]
x =8.166 nm/second
therefore hair grows at rate of 8.166 nm/s
Final answer:
Hair grows at approximately 8.164 nanometers per second or 81.64 atomic layers per second, based on the conversion from 1/36 inches per day.
Explanation:
Converting Hair Growth Rate to Nanometers per Second
To find the rate at which hair grows in nanometers per second, given that hair grows at the rate of 1/36 inches per day, we first convert inches per day to nanometers per day, then to nanometers per second. Since 1 inch equals 25,400,000 nanometers, the hair growth rate per day in nanometers is:
1/36 inch/day × 25,400,000 nm/inch = 705,555.56 nm/day.
Next, we convert this to nanometers per second. There are 86,400 seconds in a day, so:
705,555.56 nm/day ÷ 86,400 seconds/day = 8.164 nm/second.
Considering that the distance between atoms in a molecule is on the order of 0.1 nm, the rate of hair growth can be seen as:
8.164 nm/second ÷ 0.1 nm = 81.64 atomic layers/second.
Therefore, hair grows at approximately 81.64 atomic layers/second.
In a thundercloud there may be an electric charge of 24 C near the top of the cloud and −24 C near the bottom of the cloud. If these charges are separated by about 2 km, what is the magnitude of the electric force between these two sets of charges? The value of the electric force constant is 8.98755 × 109 N · m2 /C 2 .
Answer:
Electric force, [tex]F=1.29\times 10^6\ N[/tex]
Explanation:
Given that,
Charge 1, [tex]q_1=24\ C[/tex]
Charge 2, [tex]q_2=-24\ C[/tex]
Distance between charges, d = 2 km
The electric force is given by :
[tex]F=k\dfrac{q_1q_2}{d^2}[/tex]
k is the electrostatic constant
[tex]F=8.98755\times 10^9\times \dfrac{(24)^2}{(2\times 10^3)^2}[/tex]
F = 1294207.2 N
or
[tex]F=1.29\times 10^6\ N[/tex]
Hence, this is the required solution.
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
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.
The electric field at the point x=5.00cm and y=0 points in the positive x direction with a magnitude of 10.0 N/C . At the point x=10.0cm and y=0 the electric field points in the positive x direction with a magnitude of 17.0 N/C . Assume this electric field is produced by a single point charge. Part AFind the charge's location.Part BFind the magnitude of the charge.
Answer:
(A). The location of the charge is 26.45 cm.
(B). The magnitude of the charge is 51.1 pC.
Explanation:
Given that,
Distance in x axis = 5.00 cm
Electric field = 10.0 N/C
Distance in x axis = 10.0 cm
Electric field = 17.0 N/C
Since, q is the same charge, two formulas can be set equal using the two different electric fields.
(a). We need to calculate the location of the charge
Using formula of force
[tex]F = qE[/tex]....(I)
Using formula of electric force
[tex]F =\dfrac{kq^2}{d^2}[/tex]....(II)
From equation (I) and (II)
[tex]qE=\dfrac{kq^2}{d^2}[/tex]
[tex]E=\dfrac{kq}{d^2}[/tex]
[tex]q=\dfrac{E(x-r)^2}{k}[/tex]...(III)
For both points,
[tex]\dfrac{E(x-r)^2}{k}=\dfrac{E(x-r)^2}{k}[/tex]
Put the value into the formula
[tex]\dfrac{10.0\times(x-5.00)^2}{k}=\dfrac{17.0\times(x-10.0)^2}{k}[/tex]
[tex]10.0\times(x-5.0)^2=17.0\times(x-10.0)^2[/tex]
Take the square root of both sides
[tex]3.162(x-5.0)=4.123(x-10.0)[/tex]
[tex]3.162x-3.162\times5.0=4.123x-4.123\times10.0[/tex]
[tex]3.162x-4.123x=-4.123\times10.0+3.162\times5.0[/tex]
[tex]0.961x=25.42[/tex]
[tex]x=\dfrac{25.42}{0.961}[/tex]
[tex]x=26.45\ cm[/tex]
(B). We need to calculate the charge
Using equation (III)
[tex]q=\dfrac{E(x-r)^2}{k}[/tex]
Put the value into the formula
[tex]q=\dfrac{10.0(26.45\times10^{-2}-5.00\times10^{-2})^2}{9\times10^{9}}[/tex]
[tex]q=5.11\times10^{-11}\ C[/tex]
[tex]q=51.1\ pC[/tex]
Hence, (A). The location of the charge is 26.45 cm.
(B). The magnitude of the charge is 51.1 pC.
The charge is located at x = 2.25 cm along the x-axis and has a magnitude of approximately 8.41 x 10⁻¹² C. These were determined using the relationships of electric field magnitude and distance from the point charge. The analysis involved solving the electric field equations at different points based on Coulomb's law.
The electric field due to a point charge follows the equation:
E = k * |q| / r²
where E is the electric field, k is Coulomb's constant (8.99 x 10⁹ Nm²/C²), q is the charge, and r is the distance from the charge.
Part A: Finding the Charge's Location
At x = 5.00 cm, E1 = 10.0 N/C and at x = 10.0 cm, E2 = 17.0 N/C.
Assume the charge q is located at x = x0.
For x = 5.00 cm, the distance r1 is |x0 - 5.00 cm|.
For x = 10.0 cm, the distance r2 is |x0 - 10.0 cm|.
Using the electric field equations:
10.0 = k * |q| / (|x0 - 5.00|²)
17.0 = k * |q| / (|x0 - 10.0|²)
Divide the second equation by the first to eliminate q:
(17.0 / 10.0) = (|x0 - 5.00|² / |x0 - 10.0|²)
Solving this, we get |x0 - 5.00|² = 1.7 * |x0 - 10.0|².
Let u = x0 - 5.00 and v = x0 - 10.0, hence v = u - 5.00.
Substituting and solving gives the location x0 = 2.25 cm.
Part B: Finding the Magnitude of the Charge
Use the equation E = k * |q| / r² with one of the electric field values from Part A.
Let's use E1 = 10.0 N/C and r1 = |2.25 cm - 5.00 cm| = 2.75 cm = 0.0275 m:
10.0 = (8.99 x 10⁹ N*m²/C²) * |q| / (0.0275 m)².
Solving for q, we get q ≈ 8.41 x 10⁻¹² C.
Thus, the charge is located at 2.25 cm along the x-axis with a magnitude of 8.41 x 10⁻¹² C.
What is the frequency of a wave whose wavelength is 0.7m and whose velocity is 120m/s?
Answer:
f=171.43Hz
Explanation:
Wave frequency is the number of waves that pass a fixed point in a given amount of time.
The frequency formula is: f=v÷λ, where v is the velocity and λ is the wavelength.
Then replacing with the data of the problem,
f=[tex]\frac{120\frac{m}{s} }{0.7m}[/tex]
f=171.43[tex]\frac{1}{s}[/tex]
f=171.43 Hz (because [tex]\frac{1}{s} = Hz[/tex], 1 hertz equals 1 wave passing a fixed point in 1 second).
A curved road has a radius of 120 m and a cant angle of 48 degrees. What is the maximum speed to stay on the curve in the absence of friction? a. 16 m/s
b. 28 m/s
c. 24 m/s
d. 36 m/s
Answer:
Maximum speed, v = 36 m/s
Explanation:
Given that,
The radius of the curved road, r = 120 m
Road is at an angle of 48 degrees. We need to find the maximum speed of stay on the curve in the absence of friction. On a banked curve, the angle at which it is cant is given by :
[tex]tan\theta=\dfrac{v^2}{rg}[/tex]
g is the acceleration due to gravity
[tex]v=\sqrt{rg\ tan\theta}[/tex]
[tex]v=\sqrt{120\times 9.8\times \ tan(48)}[/tex]
v = 36.13 m/s
or
v = 36 m/s
So, the maximum speed to stay on the curve in the absence of friction is 36 m/s. Hence, this is the required solution.
A 20 kg mass is moving down a ramp and there is friction. Describe all the forces acting on it using terms such as weight, spring force, gravitational force, etc.
Answer:
Explanation:
mass of block, m = 20 kg
let teh angle of inclination of the ramp with the horizontal is θ.
Weight of the block = mg
There are two components of the weight
The component of weight of block along the plane = mg Sinθ
The component of weight of block normal to the plane = mg Cosθ
Friction force is acting opposite to the motion, i.e., along the plane and upwards, f = μN = μ mg Cosθ
Look at the diagram to find the directions of force.
A bicyclist is finishing his repair of a flat tire when a friend rides by with a constant speed of 3.0 m/s . Two seconds later the bicyclist hops on his bike and accelerates at 2.0 m/s^2 until he catches his friend. A.) How much time does it take until he catches his friend (after his friend passes him)? B.) How far has he traveled in this time? C.) What is his speed when he catches up?
Answer:
A.) t₁= 4.37 s
B.) d₁= 19.09 m
C.) [tex]v_{f1} = 8,74 \frac{m}{s}[/tex]
Explanation:
Bicyclist kinematics :
Bicyclist moves with uniformly accelerated movement:
d₁ = v₀₁*t₁ + (1/2)a₁*t₁² equation (1)
d₁ : distance traveled by the cyclist
v₀₁: initial speed (m/s)
a₁: acceleration (m/s²)
t₁: time it takes the cyclist to catch his friend (s)
Friend kinematics:
Friend moves with constant speed:
d ₂= v₂*t₂ Equation (2)
d₂ : distance traveled by the friend (m)
v₂: friend speed (m/s)
t₂ : time that has elapsed since the cyclist meets the friend until the cyclist catches him.
Data
v₀₁=0
a₁ = 2.0 m/s²
v₂ = 3.0 m/s
Problem development
A.) When the bicyclist catches his friend, d₁ = d₂=d and t₂=t₁+2
in the equation (1) :
d = 0 + (1/2)2*t₁² = t₁²
d = t₁² equation (3)
in the equation (2) :
d = 3*(t₁+2) ₂ = 3*t₁+6
d = 3t₁+6 equation (4)
Equation (3) = Equation (4)
t₁² = 3t₁+6
t₁² - 3t₁ - 6 = 0 ,we solve the quadratic equation
t₁= 4.37 s : time it takes the cyclist to catch his friend
B.) d₁ : distance traveled by the cyclist
In the Equation (2) d₁= (1/2)2* 4.37² = 19.09 m
C.) speed of the cyclist when he catches his friend
[tex]v_{f1} = v_{o1} + a*t[/tex]
[tex]v_{f1} =0 + 2* 4.37 [/tex]
[tex]v_{f1} = 8.74 \frac{m}{s}[/tex]
A jet lands on an aircraft carrier at 140 mi/h. What is its acceleration if it stops in 2.0 seconds?
Answer:
Acceleration, [tex]a=-31.29\ m/s^2[/tex]
Explanation:
It is given that,
Initial speed of the aircraft, u = 140 mi/h = 62.58 m/s
Finally, it stops, v = 0
Time taken, t = 2 s
Let a is the acceleration of the aircraft. We know that the rate of change of velocity is called acceleration of the object. It is given by :
[tex]a=\dfrac{v-u}{t}[/tex]
[tex]a=\dfrac{0-62.58}{2}[/tex]
[tex]a=-31.29\ m/s^2[/tex]
So, the acceleration of the aircraft is [tex]31.29\ m/s^2[/tex] and the car is decelerating. Hence, this is the required solution.
Three balls are kicked from the ground level at some angles above horizontal with different initial speeds. All three balls reached in their paths the samemaximum height. Ball A covered the total horizontal range D; ball B covered the total horizontal range 1.5D and ball C covered total horizontal range 2D when they hit the ground. Which of these three balls spent the greatest time in flight? (section 4.3)
Answer:
They all hit at the same time
Explanation:
Let the time of flight is T.
The maximum height is H and the horizontal range is R.
The formula for the time of flight is
[tex]T=\frac{2uSin\theta }{g}[/tex] ..... (1)
Te formula for the maximum height is
[tex]H=\frac{u^{2}Sin^{2}\theta }{2g}[/tex] .... (2)
From equation (1) and (2), we get
[tex]\frac{T^{2}}{H}=\frac{\frac{4u^{2}Sin^{2}\theta }{g^{2}}}{\frac{u^{2}Sin^{2}\theta }{2g}}[/tex]
[tex]\frac{T^{2}}{H}=\frac{8}{g}[/tex]
[tex]T=\sqrt{\frac{8H}{g}}[/tex]
here, we observe that the time of flight depends on the maximum height and according to the question, the maximum height for all the three balls is same so the time of flight of all the three balls is also same.
Find the volume of a rectangular cube of sides 2 mm, 3 mm, and 5 mm.
Answer:
Volume=30mm^3
Explanation:
The equation to find the volume of a cube is to multiply all its sides.
V = a.b.c
where a, b, c correspond to the sides of the cube
a=2mm
b=3mm
c=5mm
V=(2mm)(3mm)(5mm)=30mm^3
the volume of the rectangular cube is 30mm^3
A person jumps off a diving board 5.0 m above the water's surface into a deep pool. The person's downward motion stops 2.2 m below the surface of the water. Calculate the speed of the diver just before striking the water, in m/s (ignore air friction).
Answer:
9.9 m/s
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 v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 5+0^2}\\\Rightarrow v=9.9\ m/s[/tex]
If the body has started from rest then the initial velocity is 0. In order to find the velocity just before hitting the water then the distance at which the downward motion stops is irrelevant.
Hence, the speed of the diver just before striking the water is 9.9 m/s
The speed of the diver just before striking the water is 11.88 m/s.
Explanation:To calculate the speed of the diver just before striking the water, we can use the concept of conservation of energy. At the top of the dive, the diver has potential energy, which is converted to kinetic energy as the diver falls. When the diver reaches the surface of the water, all of the potential energy has been converted to kinetic energy. We can use the equations for potential and kinetic energy to solve for the speed of the diver.
Given the height of the diving board (5.0 m) and the depth the diver stops (2.2 m below the surface), we can calculate the total distance the diver falls (5.0 m + 2.2 m = 7.2 m). Using the equations for potential energy (PE = mgh) and kinetic energy (KE = 0.5mv^2), we can set the two equal to each other and solve for the speed (v) of the diver:
PE = KE
mgh = 0.5mv^2
mg(7.2 m) = 0.5mv^2
gh = 0.5v^2
v^2 = 2gh
v = sqrt(2gh)
Substituting the values for g (acceleration due to gravity) and h (height of fall), we can calculate the speed of the diver:
v = sqrt(2 * 9.8 m/s^2 * 7.2 m) = sqrt(141.12) m/s = 11.88 m/s
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.
An effective treatment for some cancerous tumors involves irradiation with "fast" neutrons. The neutrons from one treatment source have an average velocity of 3.1 3 107 m/s. If the velocities of individual neutrons are known to within 2% of this value, what is the uncertainty in the position of one of them?
Answer:
[tex]5.07\cdot 10^{-14} m[/tex]
Explanation:
The velocity of the neutrons is
[tex]v=3.13\cdot 10^7 m/s[/tex]
The mass of a neutron is
[tex]m=1.66\cdot 10^{-27} kg[/tex]
So their momentum is
[tex]p=mv=(1.66\cdot 10^{-27})(3.13\cdot 10^7)=5.20\cdot 10^{-20}kg m/s[/tex]
The relative uncertainty on the velocity is 2 %. Assuming that the mass of the neutron is known with negligible uncertainty, then the relative uncertainty on the momentum of the neutron is equal to the relative uncertainty on the velocity, so 2%. Therefore, the absolute uncertainty on the momentum is
[tex]\sigma_p = 0.02 p =0.02(5.20\cdot 10^{-20})=1.04\cdot 10^{-21} kg m/s[/tex]
Heisenber's uncertainty principle states that
[tex]\sigma_x \sigma_p \geq \frac{h}{4\pi}[/tex]
where
[tex]\sigma_x[/tex] is the uncertainty on the position
h is the Planck constant
Solving for [tex]\sigma_x[/tex], we find the minimum uncertainty on the position:
[tex]\sigma_x \geq \frac{h}{4\pi \sigma_p}=\frac{6.63\cdot 10^{-34}}{4\pi(1.04\cdot 10^{-21})}=5.07\cdot 10^{-14} m[/tex]
The question involves the application of the Heisenberg Uncertainty Principle in Quantum Mechanics to calculate the uncertainty in the position of a particle (in this case, a neutron) given the uncertainty in its velocity.
Explanation:This is a question regarding the application of the Heisenberg Uncertainty Principle, a fundamental concept in Quantum Mechanics. The uncertainty in position (Δx) can be calculated using this principle which states that the product of the uncertainties in position and momentum of a particle is always greater than or equal to half of Planck's constant.
The uncertainty in the momentum (Δp) is given by the product of the uncertainty (0.02) and the average velocity (3.1 × 107 m/s) of the neutron. Thus, Δp = 0.02 × neutron mass (1.675 × 10-27 kg) × average velocity.
Then from the Uncertainty Principle (Δx × Δp ≥ 0.5 × h), we get: Δx ≥ 0.5 × Planck's constant / Δp. By substituting the appropriate values, we can calculate the uncertainity in the position.
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In a flying ski jump, the skier acquires a speed of 110 km/h by racing down a steep hill and then lifts off into the air from a horizontal ramp. Beyond this ramp, the ground slopes downward at an angle of 45◦. (a) Assuming that the skier is in a free-fall motion after he leaves the ramp, at what distance down the slope will he land? What is his displacement vector from the point of ‘lift off’?
Answer:
Approximately [tex]\displaystyle\rm \left[ \begin{array}{c}\rm191\; m\\\rm-191\; m\end{array}\right][/tex].
Explanation:
Consider this [tex]45^{\circ}[/tex] slope and the trajectory of the skier in a cartesian plane. Since the problem is asking for the displacement vector relative to the point of "lift off", let that particular point be the origin [tex](0, 0)[/tex].
Assume that the skier is running in the positive x-direction. The line that represents the slope shall point downwards at [tex]45^{\circ}[/tex] to the x-axis. Since this slope is connected to the ramp, it should also go through the origin. Based on these conditions, this line should be represented as [tex]y = -x[/tex].
Convert the initial speed of this diver to SI units:
[tex]\displaystyle v = \rm 110\; km\cdot h^{-1} = 110 \times \frac{1}{3.6} = 30.556\; m\cdot s^{-1}[/tex].
The question assumes that the skier is in a free-fall motion. In other words, the skier travels with a constant horizontal velocity and accelerates downwards at [tex]g[/tex] ([tex]g \approx \rm -9.81\; m\cdot s^{-2}[/tex] near the surface of the earth.) At [tex]t[/tex] seconds after the skier goes beyond the edge of the ramp, the position of the skier will be:
[tex]x[/tex]-coordinate: [tex]30.556t[/tex] meters (constant velocity;)[tex]y[/tex]-coordinate: [tex]\displaystyle -\frac{1}{2}g\cdot t^{2} = -\frac{9.81}{2}\cdot t^{2}[/tex] meters (constant acceleration with an initial vertical velocity of zero.)To eliminate [tex]t[/tex] from this expression, solve the equation between [tex]t[/tex] and [tex]x[/tex] for [tex]t[/tex]. That is: express [tex]t[/tex] as a function of [tex]x[/tex].
[tex]x = 30.556\;t\implies \displaystyle t = \frac{x}{30.556}[/tex].
Replace the [tex]t[/tex] in the equation of [tex]y[/tex] with this expression:
[tex]\begin{aligned} y = &-\frac{9.81}{2}\cdot t^{2}\\ &= -\frac{9.81}{2} \cdot \left(\frac{x}{30.556}\right)^{2}\\&= -0.0052535\;x^{2}\end{aligned}[/tex].
Plot the two functions:
[tex]y = -x[/tex], [tex]\displaystyle y= -0.0052535\;x^{2}[/tex],and look for their intersection. Refer to the diagram attached.
Alternatively, equate the two expressions of [tex]y[/tex] (right-hand side of the equation, the part where [tex]y[/tex] is expressed as a function of [tex]x[/tex].)
[tex]-0.0052535\;x^{2} = -x[/tex],
[tex]\implies x = 190.35[/tex].
The value of [tex]y[/tex] can be found by evaluating either equation at this particular [tex]x[/tex]-value: [tex]x = 190.35[/tex].
[tex]y = -190.35[/tex].
The position vector of a point [tex](x, y)[/tex] on a cartesian plane is [tex]\displaystyle \left[\begin{array}{l}x \\ y\end{array}\right][/tex]. The coordinates of this skier is approximately [tex](190.35, -190.35)[/tex]. The position vector of this skier will be [tex]\displaystyle\rm \left[ \begin{array}{c}\rm191\\\rm-191\end{array}\right][/tex]. Keep in mind that both numbers in this vectors are in meters.
In a flying ski jump, the skier acquires a speed of 110 km/h by racing down a steep hill and then lifts off into the air from a horizontal ramp. Assuming that the skier is in a free-fall motion, the distance down the slope that he will land is 190.3 m and the total displacement vector from the point of lift-off is 269.15 m
From the given information;
the speed of the skier V = 110 km/hrTo convert the speed into m/s, we have:
[tex]\mathbf{= 110 \ \times (\dfrac{1000 \ m}{3600 \ s})}[/tex]
= 30.556 m/s
Since the speed race down the steep hill before flying horizontally into the air, then we can assert that the;
[tex]\mathbf{V_{horizontal} = 30.556 \ m/s}[/tex][tex]\mathbf{V_{vertical } = 0 \ m/s}[/tex]Now, for a motion under free-fall, to determine the distance(S) of the slope using the second equation of motion, we have:
The distance along the vertical axis to be:
[tex]\mathbf{S_v= ut + \dfrac{1}{2} gt^2}[/tex]
where the initial velocity = 0 m/s
[tex]\mathbf{S_v= (0)t + \dfrac{1}{2} gt^2}[/tex]
[tex]\mathbf{S_v= \dfrac{1}{2} gt^2}[/tex]
[tex]\mathbf{S_v= \dfrac{1}{2} (9.81)t^2}[/tex]
[tex]\mathbf{S_v=4.905t^2}[/tex]
The horizontal distance [tex]\mathbf{S_x = ut }[/tex]
[tex]\mathbf{S_x = 30.556t }[/tex]From the angle at which the ground begins to slope, taking the tangent of the angle with respect to the distance, we have:
[tex]\mathbf{tan \ \theta = \dfrac{opposite}{adjacent}}[/tex]
where;
θ = 45°[tex]\mathbf{tan \ 45^0 = \dfrac{S_v}{S_x}}[/tex]
[tex]\mathbf{tan \ 45^0 = \dfrac{4.905t^2}{30.556t}}[/tex]
1 × 30.556t = 4.905t²
Divide both sides by t, we have:
30.556 = 4.905t
[tex]\mathbf{t = \dfrac{30.556}{4.905}}[/tex]
t = 6.229 sec
However, since the time at which the slope will land is known, we can easily determine the value of the vertical and horizontal distances.
i.e.
[tex]\mathbf{S_v=4.905(6.229)^2}[/tex][tex]\mathbf{S_v=4.905\times 38.800}[/tex][tex]\mathbf{S_v=190.314 \ m}[/tex]The horizontal distance [tex]\mathbf{S_x = 30.556t }[/tex]
[tex]\mathbf{S_x = 30.556(6.229) }[/tex][tex]\mathbf{S_x = 190.33\ m}[/tex]The total displacement vector from the point of lift-off is:
= [tex]\mathbf{190.314 \sqrt{2}\ m}[/tex]
= 269.15 m
Therefore, we can conclude that the distance down the slope that he will land is 190.3 m and the displacement down the slope that the skier will land is 269.15 m
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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.
Suppose that you observe light emitted from a distant star to be at a wavelength of 525 nm. The wavelength of light to an observer on the distant star is 950 nm. What is the velocity of the star relative to you (in units of the speed of light c), and is it moving towards or away from you?
Answer:
The velocity of the star is 0.532 c.
Explanation:
Given that,
Wavelength of observer = 525 nm
Wave length of source = 950 nm
We need to calculate the velocity
If the direction is from observer to star.
From Doppler effect
[tex]\lambda_{0}=\sqrt{\dfrac{c+v}{c-v}}\times\lambda_{s}[/tex]
Put the value into the formula
[tex]525=\sqrt{\dfrac{c+v}{c-v}}\times950[/tex]
[tex]\dfrac{c+v}{c-v}=(\dfrac{525}{950})^2[/tex]
[tex]\dfrac{c+v}{c-v}=0.305[/tex]
[tex]c+v=0.305\times(c-v)[/tex]
[tex]v(1+0.305)=c(0.305-1)[/tex]
[tex]v=\dfrac{0.305-1}{1+0.305}c[/tex]
[tex]v=−0.532c[/tex]
Negative sign shows the star is moving toward the observer.
Hence, The velocity of the star is 0.532 c.
Assume the Earth is a ball of perimeter 40, 000 kilometers. There is a building 20 meters tall at point a. A robot with a camera placed at 1.75 m. above the surface of the Earth starts walking away from the building. At what distance from a, computed on the surface, does the robot cease to see the (top of the) building?
Answer:
Approximately 21 km.
Explanation:
Refer to the not-to-scale diagram attached. The circle is the cross-section of the sphere that goes through the center C. Draw a line that connects the top of the building (point B) and the camera on the robot (point D.) Consider: at how many points might the line intersects the outer rim of this circle? There are three possible cases:
No intersection: There's nothing that blocks the camera's view of the top of the building.Two intersections: The planet blocks the camera's view of the top of the building.One intersection: The point at which the top of the building appears or disappears.There's only one such line that goes through the top of the building and intersects the outer rim of the circle only once. That line is a tangent to this circle. In other words, it is perpendicular to the radius of the circle at the point A where it touches the circle.
The camera needs to be on this tangent line when the building starts to disappear. To find the length of the arc that the robot has travelled, start by finding the angle [tex]\angle \mathrm{B\hat{C}D}[/tex] which corresponds to this minor arc.
This angle comes can be split into two parts:
[tex]\angle \mathrm{B\hat{C}D} = \angle \mathrm{B\hat{C}A} + \angle \mathrm{A\hat{C}D}[/tex].
Also,
[tex]\angle \mathrm{B\hat{A}C} = \angle \mathrm{D\hat{A}C} = 90^{\circ}[/tex].
The radius of this circle is:
[tex]\displaystyle r = \frac{c}{2\pi} = \rm \frac{4\times 10^{7}\; m}{2\pi}[/tex].
The lengths of segment DC, AC, BC can all be found:
[tex]\rm DC = \rm \left(1.75 \displaystyle + \frac{4\times 10^{7}\; m}{2\pi}\right)\; m[/tex];[tex]\rm AC = \rm \displaystyle \frac{4\times 10^{7}}{2\pi}\; m[/tex];[tex]\rm BC = \rm \left(20\; m\displaystyle +\frac{4\times 10^{7}}{2\pi} \right)\; m[/tex].In the two right triangles [tex]\triangle\mathrm{DAC}[/tex] and [tex]\triangle \rm BAC[/tex], the value of [tex]\angle \mathrm{B\hat{C}A}[/tex] and [tex]\angle \mathrm{A\hat{C}D}[/tex] can be found using the inverse cosine function:
[tex]\displaystyle \angle \mathrm{B\hat{C}A} = \cos^{-1}{\rm \frac{AC}{BC}} [/tex]
[tex]\displaystyle \angle \mathrm{D\hat{C}A} = \cos^{-1}{\rm \frac{AC}{DC}} [/tex]
[tex]\displaystyle \angle \mathrm{B\hat{C}D} = \cos^{-1}{\rm \frac{AC}{BC}} + \cos^{-1}{\rm \frac{AC}{DC}}[/tex].
The length of the minor arc will be:
[tex]\displaystyle r \theta = \frac{4\times 10^{7}\; \rm m}{2\pi} \cdot (\cos^{-1}{\rm \frac{AC}{BC}} + \cos^{-1}{\rm \frac{AC}{DC}}) \approx 20667 \; m \approx 21 \; km[/tex].
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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What is the net charge on a sphere that has the following? a) 5.29 x 10^6 electrons and 7.07 x 10^6 protons
b) 225 electrons and 115 protons
Answer:
Explanation:
Electrons will create negative and protons will create positive charge . 5.29 x 10⁶ electrons will neutralize same no of protons . So
( 7.07 - 5.29 ) x 10⁶ protons will create net positive charge
magnitude of this charge can be calculated as follows
no of protons x charge on one proton
(7.07 - 5.29)x 10⁶ x 1.6 x 10⁻¹⁹ C
2.848 X 10⁻¹³ C is the net charge on the sphere. Ans
b )
Protons and electrons have equal but opposite charges .
115 protons will neutralize the charge of 115 electrons .
225-115 = 110 electrons will remain un-neutralized .
Charge on these un-neutralized electrons will be calculated as follows .
no of electrons x charge on a single electrons
110 x 1.6 x 10⁻¹⁹ C
176 X 10⁻¹⁹C is the net charge on the sphere. Ans
The net charge on a sphere is derived from the difference between its protons and electrons, multiplied by the elemental charge. For case (a), the charge is 2.848 x 10⁻¹³ C, and for case (b), it is -1.76 x 10⁻¹⁷ C.
he net charge on a sphere can be calculated by considering the difference between the number of protons and electrons, then multiplying by the fundamental charge of an electron or proton, which is approximately 1.6 x 10⁻¹⁹ Coulombs (C).
Case (a):
Electrons: 5.29 x 10⁶Protons: 7.07 x 10⁶The net charge is determined by:Net Charge = (Number of Protons - Number of Electrons) x 1.6 x 10⁻¹⁹ CNet Charge = (7.07 x 10⁶ - 5.29 x 10⁶) x 1.6 x 10⁻¹⁹ CNet Charge = 1.78 x 10⁶ x 1.6 x 10⁻¹⁹CNet Charge = 2.848 x 10⁻¹³ CCase (b):
Electrons: 225Protons: 115Net Charge = (Number of Protons - Number of Electrons) x 1.6 x 10⁻¹⁹ CNet Charge = (115 - 225) x 1.6 x 10⁻¹⁹ CNet Charge = -110 x 1.6 x 10⁻¹⁹ CNet Charge = -1.76 x 10⁻¹⁷ CTherefore, the net charge on each sphere is 2.848 x 10⁻¹³ C for (a) and -1.76 x 10⁻¹⁷ C for (b).