Answer:
The wave equation is [tex]\frac{d^{2}u }{dt^{2} }[/tex] = [tex]c^{2}[/tex] [tex]\frac{d^{2}u }{dx^{2} }[/tex]
a sinusoidal wave can be u = Acos( ax + bt) + B*sin(ax + bt)
where A, a, B and b are real constants. (here you also can add a phase to the arguments of the sin and cosine)
then [tex]\frac{d^{2}u }{dt^{2} }[/tex] = [tex]b^{2}[/tex]*( -Acos(ax + bt) - B*sin(ax + bt))
and [tex]c^{2}[/tex] [tex]\frac{d^{2}u }{dx^{2} }[/tex]= [tex]ac^{2}[/tex]*( -Acos(ax + bt) - B*sin(ax + bt))
then if a*c = b, this is a solution of the wave equation.
A basketball center holds a basketball straight out, 2.0 m above the floor, and releases it. It bounces off the floor and rises to a height of 1.5 m. a) What is the ball's velocity just before it hits the floor? b) What is the ball's velocity just after it leaves the floor? c) If the ball is in contact with the floor for 0.02 seconds, what are the magnitude and direction of the ball's average acceleration while in contact with the floor?
The basketball's velocity just before hitting the floor is approximately -6.26 m/s (downward), while just after it leaves the floor, it reaches around 5.42 m/s (upward). During the brief 0.02 seconds it's in contact with the floor, its average acceleration is enormous, about 584 m/s², and directed upward.
Explanation:This question involves the physics concept of velocity and acceleration related to a bouncing basketball. Let's dig into the details one by one:
(a) The ball's velocity just before hitting the floor:
Using Physics, the velocity of an object just before it hits the ground can be calculated using the equation for motion that involves falling from a height namely v² = u² + 2gh, where u is the initial velocity, g is acceleration due to gravity, v is the final velocity and h is the height. Since the ball was released, the initial velocity (u) is 0, g is approximately 9.8 m/s² (negative because the ball is falling downwards), and h is -2.0 m (negative because it is below the release point). Solving for v, the equation transforms to v = sqrt(u² + 2gh) = sqrt(0 + 2*(-9.8)*(-2)) = sqrt(39.2) which is approximately 6.26 m/s (negative, indicating downward).
(b) The ball's velocity just after it leaves the floor:
Assuming the ball reaches the height of 1.5 m with uniform acceleration, we can use the same equation, but this time treating it as a ground-to-air motion with initial velocity 0, g = 9.8 m/s² (positive because the motion is upward), and h = 1.5 m. Solving for v, we get v = sqrt(u² + 2gh) = sqrt(0 + 2*9.8*1.5) = sqrt(29.4) which is approximately 5.42 m/s (positive, indicating upward).
(c) Average acceleration while the ball is in contact with the floor:
Acceleration can be calculated using the formula a = (v_final - v_initial) / t, where v_final is the final velocity, v_initial is the initial velocity, and t is the time. The change in velocity here is the difference between the velocity just after the ball leaves the floor and the velocity just before it hits the floor, i.e., (5.42 m/s - -6.26 m/s) = 11.68 m/s. Given that the ball is in contact with the floor for 0.02 seconds, the average acceleration is therefore a = (11.68 m/s) / 0.02 s = 584 m/s². This is considerably higher than g because while in contact with the floor, the ball is being rapidly decelerated and then accelerated in the opposite direction due to the impact force. The direction is upward, same as the final velocity.
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A force of 1.4 N is exerted on a 6.6 g rifle bullet. What is the bullet's acceleration?
Answer:
The acceleration of the bullet is 212.12 m/s²
Explanation:
Given that,
Force = 1.4 N
Mass = 6.6 g
We need to calculate the acceleration
Using newton's second law
[tex]F=ma[/tex]
[tex]a=\dfrac{F}{m}[/tex]
Where, F = force
m = mass
Put the value into the formula
[tex]a=\dfrac{1.4 }{6.6\times10^{-3}}[/tex]
[tex]a=212.12\ m/s^2[/tex]
Hence, The acceleration of the bullet is 212.12 m/s²
A +71 nC charge is positioned 1.9 m from a +42 nC charge. What is the magnitude of the electric field at the midpoint of these charges, in units of N/C?
Answer:
The net Electric field at the mid point is 289.19 N/C
Given:
Q = + 71 nC = [tex]71\times 10^{- 9} C[/tex]
Q' = + 42 nC = [tex]42\times 10^{- 9} C[/tex]
Separation distance, d = 1.9 m
Solution:
To find the magnitude of electric field at the mid point,
Electric field at the mid-point due to charge Q is given by:
[tex]\vec{E} = \frac{Q}{4\pi\epsilon_{o}(\frac{d}{2})^{2}}[/tex]
[tex]\vec{E} = \frac{71\times 10^{- 9}}{4\pi\8.85\times 10^{- 12}(\frac{1.9}{2})^{2}}[/tex]
[tex]\vec{E} = 708.03 N/C[/tex]
Now,
Electric field at the mid-point due to charge Q' is given by:
[tex]\vec{E'} = \frac{Q'}{4\pi\epsilon_{o}(\frac{d}{2})^{2}}[/tex]
[tex]\vec{E'} = \frac{42\times 10^{- 9}}{4\pi\8.85\times 10^{- 12}(\frac{1.9}{2})^{2}}[/tex]
[tex]\vec{E'} = 418.84 N/C[/tex]
Now,
The net Electric field is given by:
[tex]\vec{E_{net}} = \vec{E} - \vec{E'}[/tex]
[tex]\vec{E_{net}} = 708.03 - 418.84 = 289.19 N/C[/tex]
A single point charge is placed at the center of an imaginary cube that has 10 cm long edges. The electric flux out of one of the cube's sides is -1 kN·m^2/C. How much charge is at the center?
Answer:
Explanation:
Given:
Length of each side of the cube, [tex]L=10\ cm[/tex]
The Elecric flux through one of the side of the cube is, [tex]\phi =-1 kNm^2/C.[/tex]
The net flux through a closed surface is defined as the total charge that lie inside the closed surface divided by [tex]\epsilon_0[/tex]
Since Flux is a scalar quantity. It can added to get total flux through the surface.
[tex]\phi_{total}=\dfrac{Q_{in}}{\epsilon_0}\\6\times {-1}\times 10^3=\dfrac{Q_{in}}{\epsilon_0}\\\\Q_{in}=-6}\times 10^3\epsilon_0\\Q_{in}=-6}\times 10^3\times8.85\times 10^{-12}\\Q_{in}=-53.1\times10^{-9}\ C[/tex]
So the the charge at the centre is calculated.
The Bonneville Salt Flats, located in Utah near the border with Nevada, not far from interstate I80, cover an area of over 30000 acres. A race car driver on the Flats first heads north for 6.71 km, then makes a sharp turn and heads southwest for 1.33 km, then makes another turn and heads east for 3.67 km. How far is she from where she started?
Final answer:
The race car driver is approximately 7.644 km from where she started.
Explanation:
To find the distance from where the race car driver started, we can use the Pythagorean theorem. We can consider the north-south movement as one leg of a right triangle, and the east-west movement as the other leg. The distance from the starting point is equal to the square root of the sum of the squares of the two legs. In this case, the north-south leg is 6.71 km and the east-west leg is 3.67 km. Using the Pythagorean theorem, the distance from the starting point is:
d = √((6.71 km)² + (3.67 km)²)
d = √(44.9041 km² + 13.4689 km²)
d = √(58.3730 km²)
d = 7.644 km
Therefore, the race car driver is approximately 7.644 km from where she started.
Answer the following question. Show your work to receive credit. A hypothetical element has an atomic weight of 48.68 amu. It consists of three isotopes having masses of 47.00 amu, 48.00 amu, and 49.00 amu. The lightest-weight isotope has a natural abundance of 10.0%. What is the percent abundance of the heaviest isotope?
Answer:
percent abundance of the heaviest isotope is 78 %
Explanation:
given data
atomic weight = 48.68 amu
mass 1 = 47 amu
mass 2 = 48 amu
mass 3 = 49 amu
natural abundance = 10 %
to find out
percent abundance of the heaviest isotope
solution
we consider here percent abundance of the heaviest isotope is x
so here lightest isotope = 47 amu of 10 % ..........1
and heaviest isotope = 49 amu of x ................2
and middle isotope = 48 amu of 100 - 10 - x ........3
so
average mass = add equation 1 + 2 + 3
average mass = 10% ( 47) + x% ( 49) + (90 - x) % (48)
48.68 = 4.7 + 0.49 x + 43.2 - 0.48 x
x = 0.78
so percent abundance of the heaviest isotope is 78 %
In a thunderstorm, electric charge builds up on the water droplets or ice crystals in a cloud. Thus, the charge can be considered to be distributed uniformly throughout the cloud. The charge builds up until the electric field at the surface of the cloud reaches the value at which the surrounding air "breaks down."
In general, the term "breakdown" refers to the situation when a dielectric (insulator) such as air becomes a conductor. In this case, it means that, because of a very strong electric field, the air becomes highly ionized, enabling it to conduct the charge from the cloud to the ground or another nearby cloud. The ionized air then emits light as the electrons and ionized atoms recombine to form excited molecules that radiate light. The resulting large current heats up the air, causing its rapid expansion. These two phenomena account for the appearance of lightning and the sound of thunder.
The point of this problem is to estimate the maximum amount of charge that a cloud can contain before breakdown occurs. For the purposes of this problem, take the cloud to be a sphere of diameter 1.00 km . Take the breakdown electric field of air to be Eb=3.00106N/C .
question 1:
Estimate the total charge q on the cloud when the breakdown of the surrounding air begins.
Express your answer numerically in coulombs, to three significant figures, using ?0=8.8510?12C2/(N?m2) .
Question 2:
Assuming that the cloud is negatively charged, how many excess electrons are on this cloud?
The total charge on a cloud when breakdown of the surrounding air begins is calculated using Gauss's Law. The breakdown field strength, radius of the cloud and permittivity of free space are needed. The number of excess electrons in the cloud is then found by dividing the total charge by the charge of an electron.
Explanation:The charge q on a spherically symmetric object, such as our cloud, when an electric field E is induced on its surface can be calculated using Gauss's Law, which states that the total charge enclosed by a Gaussian surface is equal to the electric field times the surface area of the Gaussian surface divided by the permittivity of free space (ε0). In our case, E = Eb (the breakdown field strength), the radius of the sphere, r = 0.5km = 500m (half the diameter), and ε0 = 8.85 x 10^-12 C2/N·m2. Therefore, q = (4πr2Eb)ε0.
For question 2: To calculate the number of excess electrons in the cloud, we should recall that the charge of an electron is -1.602 x 10^-19 C. Therefore, the number of excess electrons, N, can be calculated using N = q / Charge of an electron.
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Computethe maximum height that a projectile can
reach if it is launchedwith speed V o at angle thetarelative to the horizontal. If an
object is thrown directly upwardswith a speed of 330m/s, the
typical speed of sound in the air atroom temperature, how high can
it get?
Answer:
A. [tex]H=\frac{v_{0}^{2}Sin^{2}\theta }{2g}[/tex]
B. 5556.1 m
Explanation:
A.
Launch speed, vo
Angle of projection = θ
The value of vertical component of velocity at maximum height is zero. Let the maximum height is H.
Use third equation of motion in vertical direction
[tex]v_{y}^{2}=u_{y}^{2}+2a_{y}H[/tex]
[tex]0^{2}=\left (v_{0}Sin\theta \right )^{2}-2gH[/tex]
[tex]H=\frac{v_{0}^{2}Sin^{2}\theta }{2g}[/tex]
B.
u = 330 m/s
Let it goes upto height H.
V = 0 at maximum height
Use third equation of motion in vertical direction
[tex]v^{2}=u^{2}+2as[/tex]
[tex]0^{2}=330^{2}-2\times 9.8\times H[/tex]
H = 5556.1 m
A solid cylinder of cortical bone has a length of 500mm, diameter of 2cm and a Young’s Modulus of 17.4GPa. Determine the spring constant ‘k’
Answer:
The spring constant is [tex]1.09\times10^{9}\ N/m[/tex]
Explanation:
Given that,
length = 500 mm
Diameter = 2 cm
Young's modulus = 17.4 GPa
We need to calculate the young's modulus
Using formula of young's modulus
[tex]Y=\dfrac{\dfrac{F}{A}}{\dfrac{\Delta l}{l}}[/tex]....(I)
[tex]Y=\dfrac{Fl}{\Delta l A}[/tex]
From hook's law
[tex]F=kx[/tex]
[tex]k=\dfrac{F}{x}[/tex]
[tex]F=k\times\Delta l[/tex]....(II)
Put the value of F in equation
[tex]Y=\dfrac{k\times\Delta l\times l}{\Delta l A}[/tex]
[tex]Y=\dfrac{kl}{A}[/tex]
We need to calculate the spring constant
[tex]k = \dfrac{YA}{l}[/tex]....(II)
We need to calculate the area of cylinder
Using formula of area of cylinder
[tex]A=2\pi\times r\times l[/tex]
Put the value into the formula
[tex]A=2\pi\times 1\times10^{-2}\times500\times10^{-3}[/tex]
[tex]A=0.0314\ m^2[/tex]
Put the value of A in (II)
[tex]k=\dfrac{1.74\times10^{10}\times0.0314}{500\times10^{-3}}[/tex]
[tex]k=1.09\times10^{9}\ N/m[/tex]
Hence, The spring constant is [tex]1.09\times10^{9}\ N/m[/tex]
Three vectors →a, →b, and →c each have a magnitude of 50 m and lie in an xy plane. Their directions relative to the positive direction of the x axis are 30°, 195°, and 315°, respectively. What are (a) the magnitude and (b) the angle of the vector →a+→b+→c and (c) the magnitude and (d) the angle of →a−→b+→c? What are the (e) magnitude and (f) angle of a fourth vector →d such that (→a+→b)−(→c+→d)=0 ?
Answer:
a) 38.27 b) 322.5°
c) 126.99 d) 1.17°
e) 62.27 e) 139.6°
Explanation:
First of all we have to convert the coordinates into rectangular coordinates, so:
a=( 43.3 , 25)
b=( -48.3 , -12.94)
c=( 35.36 , -35.36)
Now we can do the math easier (x coordinate with x coordinate, and y coordinate with y coordinate):
1.) a+b+c=( 30.36 , -23.3) = 38.27 < 322.5°
2.) a-b+c=( 126.96 , 2.6) = 126.99 < 1.17°
3.) (a+b) - (c+d)=0 Solving for d:
d=(a+b) - c = ( -40.36 , 47.42) = 62.27 < 139.6°
. On a safari, a team of naturalists sets out toward a research station located 9.6 km away in a direction 42° north of east. After traveling in a straight line for 3.1 km, they stop and discover that they have been traveling 25° north of east, because their guide misread his compass. What is the direction (relative to due east) of the displacement vector now required to bring the team to the research station?
Answer:[tex]\theta =49.76^{\circ}[/tex] North of east
Explanation:
Given
Research station is 9.6 km away in [tex]42^{\circ}[/tex]North of east
after travelling 3.1 km [tex]25^{\circ}[/tex] north of east
Position vector of safari after 3.1 km is
[tex]r_2=3.1cos25\hat{i}+3.1sin25\hat{j}[/tex]
Position vector if had traveled correctly is
[tex]r_0=9.6cos42\hat{i}+9.6sin42\hat{j}[/tex]
Now applying triangle law of vector addition we can get the required vector[tex](r_1)[/tex]
[tex]r_1+r_2=r_0[/tex]
[tex]r_1=(9.6cos42-3.1cos25)\hat{i}+(9.6sin42-3.1sin25)\hat{j}[/tex]
[tex]r_1=4.325\hat{i}+5.112\hat{j}[/tex]
Direction is given by
[tex]tan\theta =\frac{y}{x}=\frac{5.112}{4.325}[/tex]
[tex]\theta =49.76^{\circ}[/tex]
What is the energy stored between 2 Carbon nuclei that are 1.00 nm apart from each other? HINT: Carbon nuclei have 6 protons and 1.00 nm = 1.00x10^-9m
A. 8.29x10^-18J
B. 2.30x10^-19J
C. 8.29x10^-10J
D. 8.29x10^-9J
E. 0 J
Answer:
[tex]A. 8.29\times 10^{-18}\ J[/tex]
Explanation:
Given that:
p = magnitude of charge on a proton = [tex]1.6\times 10^{-19}\ C[/tex]
k = Boltzmann constant = [tex]9\times 10^{9}\ Nm^2/C^2[/tex]
r = distance between the two carbon nuclei = 1.00 nm = [tex]1.00\times 10^{-9}\ m[/tex]
Since a carbon nucleus contains 6 protons.
So, charge on a carbon nucleus is [tex]q = 6p=6\times 1.6\times 10^{-19}\ C=9.6\times 10^{-19}\ C[/tex]
We know that the electric potential energy between two charges q and Q separated by a distance r is given by:
[tex]U = \dfrac{kQq}{r}[/tex]
So, the potential energy between the two nuclei of carbon is as below:
[tex]U= \dfrac{kqq}{r}\\\Rightarrow U = \dfrac{kq^2}{r}\\\Rightarrow U = \dfrac{9\times10^9\times (9.6\times 10^{-19})^2}{1.0\times 10^{-9}}\\\Rightarrow U =8.29\times 10^{-18}\ J[/tex]
Hence, the energy stored between two nuclei of carbon is [tex]8.29\times 10^{-18}\ J[/tex].
The driver of a car traveling on the highway suddenly slams on the brakes because of a slowdown in traffic ahead. A) If the car’s speed decreases at a constant rate from 74 mi/h to 50 mi/h in 3.0 s, what is the magnitude of its acceleration, assuming that it continues to move in a straight line?Answer is in mi/h^2B) What distance does the car travel during the braking period?Answer is in ft
To determine the magnitude of acceleration, use the formula (final velocity - initial velocity) / time. To calculate the distance traveled during the braking period, use the formula (initial velocity + final velocity) / 2 * time.
Explanation:The magnitude of acceleration can be calculated using the formula:
acceleration = (final velocity - initial velocity) / time
Convert the speeds to feet per second by multiplying by 1.46667 (1 mile = 5280 feet, 1 hour = 3600 seconds).Calculate the acceleration by plugging in the values:acceleration = (50 mi/h * 1.46667 ft/s - 74 mi/h * 1.46667 ft/s) / 3 s
The distance traveled during the braking period can be calculated using the formula:
distance = (initial velocity + final velocity) / 2 * time
Calculate the distance by plugging in the values:distance = (74 mi/h * 1.46667 ft/s + 50 mi/h * 1.46667 ft/s) / 2 * 3 s
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In 1865, Jules Verne proposed sending men to the Moon by firing a space capsule from a 220-m-long cannon with final speed of 10.97 km/s. What would have been the unrealistically large acceleration experienced by the space travelers during their launch? (A human can stand an acceleration of 15g for a short time.) Compare your answer with the free-fall acceleration, 9.80 m/s^2.
The acceleration experienced by the space travelers during their launch would be considered unrealistically large. It would exceed both the limits of human endurance and the acceleration experienced during free fall.
Explanation:To calculate the acceleration experienced by the space travelers during their launch, we can use the formula:
Acceleration = (Final Velocity - Initial Velocity) / Time
In this case, the final velocity is given as 10.97 km/s, and the initial velocity is 0 m/s (since the spaceship starts from rest). The time is not provided, so we cannot calculate the exact acceleration. However, we can compare it to the acceleration that a human can withstand, which is 15g for a short time. One g is equivalent to the acceleration due to gravity, which is approximately 9.8 m/s².
So, 15g is equal to 15 * 9.8 m/s² = 147 m/s². Therefore, any acceleration larger than 147 m/s² would be unrealistically large for the space travelers during their launch.
Comparing this with the free-fall acceleration, which is approximately 9.8 m/s², we can see that the acceleration proposed by Jules Verne would be much larger than both the limits of human endurance and the acceleration experienced during free fall.
What phase difference between two otherwise identical harmonic waves, moving in the same direction along a stretched string, will result in the combined wave having an amplitude 0.6 times that of the amplitude of either of the combining waves? Express your answer in degrees.
Answer:
[tex]\theta=145[/tex]
Explanation:
The amplitude of he combined wave is:
[tex]B=2Acos(\theta/2)\\[/tex]
A, is the amplitude from the identical harmonic waves
B, is the amplitude of the resultant wave
θ, is the phase, between the waves
The amplitude of the combined wave must be 0.6A:
[tex]0.6A=2Acos(\theta/2)\\ cos(\theta/2)=0.3\\\theta/2=72.5\\\theta=145[/tex]
A ball at the end of a string of 2.2m length rotates at
aconstant speed in a horizontal circle. It makes 5.8 rev/s. What
isthe frequency of motion( ans. in Hz).?
Answer:
The frequency of motion is 5.8 Hz.
Explanation:
frequency of motion of any object is defined as the number of times the object repeats it's motion in 1 second.
mathematically frequency equals [tex]f=\frac{1}{T}[/tex]
where,
'T' is the time it takes for the object to complete one revolution. Since it is given that the ball completes 5.8 revolutions in 1 seconds thus the time it takes for 1 revolution equals [tex]\frac{1}{5.8}s[/tex]
Hence[tex]T=\frac{1}{5.8}s[/tex]
thus the frequency equals
[tex]\frac{1}{\frac{1}{5.8}}s^{-1}\\\\f=5.8s^{-1}=5.8Hz[/tex]
Answer:
5.8 Hz.
Explanation:
Given:
The length of the string to which the ball is attached, L = 2.2 m. The number of revolutions, the ball is making in unit time = 5.8 rev/s.The frequency of the motion of a rotating object is defined the total number of revolutions the object makes in unit time. It is measured in units of Hertz (Hz) or per second.
It is given that the ball is making 5.8 revolutions in one second, therefore, its frequency of motion would be the same, i.e., 5.8 Hz.
Given a particle that has the velocity v(t) = 3 cos(mt) = 3 cos (0.5t) meters, a. Find the acceleration at 3 seconds. b. Find the displacement at 2 seconds.
Answer:
Explanation:
a ) V = 3 cos(0.5t)
differentiating with respect to t
dv /dt = -3 x .5 sin0.5t
= -1.5 sin0.5t.
acceleration = - 1.5 sin 0.5t
when t = 3 s
acceleration = - 1.5 sin 1.5
= - 1.496 ms⁻²
v = 3 cos.5t
b ) dx/dt = 3 cos 0.5 t
dx = 3 cos 0.5 t dt
integrating on both sides
x = 3 sin .5t / .5
x = 6 sin0.5t
At t = 2 s
x = 6 sin 1
x = 5.05 m
A toy car runs off the edge of a table that is 1.807 m high. The car lands 0.3012 m from the base of the table. How long does it take for the car to fall? The acceleration due to gravity is 9.8 m/s^2. Answer in units of s. What is the horizontal velocity of the car? Answer in units of m/s.
Final answer:
The time it takes for the car to fall is 0.606 s and the horizontal velocity of the car is 0.497 m/s.
Explanation:
To find the time it takes for the toy car to fall, we can use the kinematic equation:
h = (1/2)gt^2
Where h is the height, g is the acceleration due to gravity, and t is the time. Rearranging the equation, we can solve for t:
t = sqrt(2h / g)
Plugging in the values, we have:
t = sqrt(2 * 1.807 / 9.8) = 0.606 s
To find the horizontal velocity of the car, we can use the equation:
v = d / t
Where v is the velocity, d is the horizontal distance, and t is the time. Plugging in the values, we have:
v = 0.3012 / 0.606 = 0.497 m/s
An object starts at Xi = -4m with an initial velocity of 4m/s. It experiences an acceleration of -2 m/s^2 for 2 seconds, followed by an acceleration of -6 m/s^2 for 4 seconds. 1. After 2 seconds, what is the objects velocity? 2. After 6 seconds, what is the objects velocity? 3. After 6 seconds, what is the objects total displacement?
Answer:
a) 0 m/s
b) - 24 m/s
c) - 68 m
Explanation:
Given:
Initial distance = - 4 m
Initial velocity, u = 4 m/s
1) acceleration, a = - 2 m/s² for time, t = 2 seconds
thus,
velocity after 2 seconds will be
from Newton's equation of motion
v = u + at
v = 4 + (-2) × 2
v = 0 m/s
2) Velocity after 2 second is the initial velocity for this case
given acceleration = - 6 m/s² for 4 seconds
thus,
final velocity, v = 0 + ( - 6 ) × 4 = - 24 m/s
here the negative sign depicts the velocity in opposite direction to the initial direction of motion
thus, velocity after 6 seconds = - 24 m/s
3) Now,
Total displacement in 6 seconds
= Displacement in 2 seconds + Displacement in 4 seconds
From Newton's equation of motion
[tex]s=ut+\frac{1}{2}at^2[/tex]
where,
s is the distance
u is the initial speed
a is the acceleration
t is the time
thus,
= [tex]0\times2+\frac{1}{2}\times(-2)\times2^2[/tex] + [tex]0\times4+\frac{1}{2}\times-6\times4^2[/tex]
= - 16 - 48
= - 64 m
Hence, the final displacement = - 64 - 4 = - 68 m
A particle with a charge of -60.0 nC is placed at the center of a nonconducting spherical shell of inner radius 20.0 cm and outer radius 33.0 cm. The spherical shell carries charge with a uniform density of-1.30 μC/m^3. A proton moves in a circular orbit just outside the shcal shell. Calculate the speed of the proton.
Answer:
Explanation:
Total volume of the shell on which charge resides
= 4/3 π ( R₁³ - R₂³ )
= 4/3 X 3.14 ( 33³ - 20³) X 10⁻⁶ m³
= 117 x 10⁻³ m³
Charge inside the shell
-117 x 10⁻³ x 1.3 x 10⁻⁶
= -152.1 x 10⁻⁹ C
Charge at the center
= - 60 x 10⁻⁹ C
Total charge inside the shell
= - (152 .1 + 60 ) x 10⁻⁹ C
212.1 X 10⁻⁹C
Force between - ve charge and proton
F = k qQ / R²
k = 9 x 10⁹ .
q = 1.6 x 10⁻¹⁹ ( charge on proton )
Q = 212.1 X 10⁻⁹ ( charge on shell )
R = 33 X 10⁻² m ( outer radius )
F = [tex]\frac{9\times10^9\times1.6\times10^{-19}\times212.1\times 10^{-9}}{(33\times10^{-2})^2}[/tex]
F = 2.8 X 10⁻¹⁵ N
This force provides centripetal force for rotating proton
mv² / R = 2.8 X 10⁻¹⁵
V² = R X 2.8 X 10⁻¹⁵ / m
= 33 x 10⁻² x 2.8 x 10⁻¹⁵ /( 1.67 x 10⁻²⁷ )
[ mass of proton = 1.67 x 10⁻²⁷ kg)
= 55.33 x 10¹⁰
V = 7.44 X 10⁵ m/s
The dimensionless parameter is used frequently in relativity. As y becomes larger and larger than 1, it means relativistic effects are becoming more and more important. What is y if v = 0.932c? A. 0.546 B. 0.362 C. 2.76 D. 3.83 E. 7.61
The relativistic factor y (gamma) quantifies the relativistic effects based on the speed of an object. In your given case with v = 0.932c, gamma would be approximately 2.61, indicating significant relativistic effects. The provided options do not include this value, suggesting some error.
Explanation:The dimensionless parameter used in relativity is denoted as y (gamma), and it is referred to as the relativistic factor. It is associated with the relative speed of an object and is defined by the equation y = 1/√(1 − (v²/c²)), where v is the object's velocity and c is the speed of light.
For your case where v = 0.932c, we'll substitute this into the formula and solve for y. This results in y being approximately equal to 2.61. This isn't available in the provided options, which would indicate an error in the question or provided choices.
Without a doubt, as y approaches and surpasses 1, the relativistic effects become more noticeable in an object's behavior. Significant relativistic effects mean that the classical interpretation, where we neglect these effects, becomes increasingly inaccurate.
Learn more about Relativistic Factor here:https://brainly.com/question/31975532
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A friend tells you that a scientific law cannot be changed. State whether or not your friend is correct and then briefly explain your answer.
Answer with Explanation:
The person stating that the scientific law cannot be changed is not correct in his claim.
A scientific law can be defined as a statement that is deemed to explain or predict a natural process and is validated by repeated experiments.
In the light of above statement we may conclude that if the law is validated by experiments it should not be subjected to change but this is not the case as an experiment may be done in the future that invalidates the law.
A classical example of this case is the whole of Newtonian law of gravitation. Classically if we analyse this law it is perfect to describe the motion of solar system , planets ,satellites found in nature hence we should find it absolutely correct and since rockets are designed on the basis of this law we can safely assume that it is correct but it is not the case as the gravity is described radically differently by Einstein by his general theory of relativity thus invalidating the newton's law of gravity.
But we still use the Newton's law of gravity as the error's that are involved in the results of the newton's law are not significant to influence any radical departure from the design philosophy of objects such as rocket's or satellites. It always boils down to the accuracy that we need but theoretically we can say that newton's law of gravity was invalidated by Einstein's general theoty of relativity.
Calculate the velocity of a car (in m/s) that starts from rest and accelerates at 5 m/s^2 for 6 seconds.
Answer:
The final velocity of the car is 30 m/s.
Explanation:
Given that,
Initial speed of the car, u = 0
Acceleration of the car, [tex]a=5\ m/s^2[/tex]
Time taken, t = 6 s
Let v is the final velocity of the car. It can be calculated using first equation of kinematics as :
[tex]v=u+at[/tex]
[tex]v=at[/tex]
[tex]v=5\ m/s^2\times 6\ s[/tex]
v = 30 m/s
So, the final velocity of the car is 30 m/s. Hence, this is the required solution.
The age of the universe is thought to be about 14 billion years. Assuming two significant figures, (a) write this in exponential notation in units of years, and (b) use the method shown in class to convert this to seconds. Give your answer in exponential notation.
Answer:
a) 14×10⁹ years
b) 4.4×10¹⁷ seconds
Explanation:
1 billion years = 1000000000 years
1000000000 years = 10⁹ years
14 billion years
= 14×1000000000 years
= 14000000000 years
= 14×10⁹ years
∴ 14 billion years = 14×10⁹ years
b) 1 year = 365.25×24×60×60 seconds
14×10⁹ years = 14×10⁹×365.25×24×60×60
= 441806400×10⁹ seconds
Rounding off, we get
= 4.4×10⁸×10⁹
= 4.4×10¹⁷ seconds
∴ 14 billion years = 4.4×10¹⁷ seconds
Starting from the front door of your ranch house, you walk 50.0 m due east to your windmill, and then you turn around and slowly walk 40.0 m west to a bench where you sit and watch the sunrise. It takes you 28.0 s to walk from your house to the windmill and then 42.0 s to walk from the windmill to the bench.
(a) For the entire trip from your front door to the bench, what is your average velocity?
(b) For the entire trip from your front door to the bench, what is your average speed?
Answer:
Average velocity
[tex]v=\frac{d}{t}\\ v=\frac{10m}{70s}\\v=.1428 \frac{m}{s}[/tex]
Average speed,
[tex]S=\frac{D}{t}\\ S=\frac{90}{70}\\ S=1.29\frac{m}{s}[/tex]
Explanation:
(a)Average velocity
We have to find the average velocity. We know that velocity is defined as the rate of change of displacement with respect to time.
To find the average velocity we have to find the total displacement.
since displacement along east direction is 50m
and displacement along west=40m
so total displacement,
[tex]d=50m-40m\\d=10m[/tex]
total time,
[tex]t=28 s+42 s\\t=70 s[/tex]
therefore, average velocity
[tex]v=\frac{d}{t}\\ v=\frac{10m}{70s}\\v=.1428 \frac{m}{s}[/tex]
(b)Average Speed:
Average speed is defined as the ratio of total distance to the total time
it means
Average speed= total distance/total time
here total distance,
[tex]D= 50m+40m\\D=90m[/tex]
and total time,
[tex]t= 28s+40s\\t=70s[/tex]
therefore,
Average speed,
[tex]S=\frac{D}{t}\\ S=\frac{90}{70}\\ S=1.29\frac{m}{s}[/tex]
What happens to the electric potential energy of a negatively charged ion as it moves through the water from the negative probe to the positive probe?
Answer:
Decreases.
Explanation:
Electric potential energy is the potential energy which is associated with the configuration of points charge in a system and it is the result of conservative coulomb force.
When the negatively charge ion is at the position of the negative probe than its potential energy is positive when it is move towards the positive probe it's potential energy becomes negative due to the negative ion.
Therefore, potential energy is decreases when negative charge ion moves through the water from negative probe to positive probe.
The driver of a car slams on the brakes when he sees a tree blocking the road. The car slows uniformly with acceleration of -5.75 m/s^2 for 4.40 s, making straight skid marks 60.0 m long, all the way to the tree. With what speed does the car then strike the tree? m/s
The car strikes the tree at approximately 10.7 m/s after slowing down with a uniform acceleration of -5.75 m/s^2 for 60.0 m.
To calculate the speed with which the car strikes the tree, we'll use the kinematic equations for uniformly accelerated motion. Given the uniform acceleration of -5.75 m/s2 and the time of 4.40 s, we can use the following equation to find the initial velocity (vi) before the car started braking:
v = vi + at
Where:
By rearranging the equation to solve for the initial velocity (vi), we get:
vi = v - at
Substituting the known values:
vi = 0 - (-5.75 m/s2 * 4.40 s)
vi = 25.3 m/s
This is the speed with which the car was travelling before it hit the brakes. However, to find the speed at which the car strikes the tree, we must consider the distance of the skid marks. Using the kinematic equation for distance (d), where d equals the initial velocity times time plus half the acceleration times time squared:
d = vit + rac{1}{2}at2
Since the distance to the tree is 60.0 m and we're looking for the speed at the end of this distance, we rearrange the equation to solve for final velocity (v). But first, we need to calculate the time it takes to stop over this distance. We can use the formula:
d = rac{vi2 - v2}{2a}
By solving for v we find:
v = [tex]\sqrt{x}[/tex]{vi2 - 2ad}
v = [tex]\sqrt{x}[/tex]{(25.3 m/s)2 - 2(-5.75 m/s2)(60.0 m)}
v = [tex]\sqrt{x}[/tex]{(640.09) - (-690)}
v = 10.7 m/s
Therefore, the car strikes the tree at approximately 10.7 m/s.
The car strikes the tree at a speed of 0.98 m/s.
To determine the speed at which the car strikes the tree, we can utilize the kinematic equations of motion. We are given the following data:
Initial velocity (Vi): UnknownFinal velocity (Vf): ? (what we need to find)Acceleration (a): -5.75 m/s² (negative as it is a deceleration)Time (t): 4.40 sDistance (d): 60.0 mFirst, we calculate the initial velocity using the equation:
Distance d = Vi * t + 0.5 * a * t²Substituting the given values:
60.0 m = Vi * 4.40 s + 0.5 * (-5.75 m/s²) * (4.40 s)²Solving this:
60.0 m = Vi * 4.40 s - 55.66 mVi * 4.40 s = 115.66 mVi = 115.66 m / 4.40 s = 26.28 m/sNow, to find the final velocity when the car strikes the tree, we use the kinematic equation:
Final velocity (Vf) =Vi + a * tSubstituting the values:
Vf= 26.28 m/s + (-5.75 m/s²) * 4.40 sVf = 26.28 m/s - 25.30 m/sVf = 0.98 m/sSo, the car strikes the tree at a speed of 0.98 m/s.
Suppose that a submarine inadvertently sinks to the bottom of the ocean at a depth of 1000m. It is proposed to lower a diving bell to the submarine and attempt to enter the conning tower. What must the minimum air pressure be in the diving bell at the level of the submarine to prevent water from entering into the bell when the opening valve at the bottom is cracked open slightly? Give your answer in absolute kilopascal. Assume that seawater has a constant density of 1.024 g/cm3.
Answer:
1.004 × 10⁴ kPa
Explanation:
Given data
Depth (h): 1000 mDensity of seawater (ρ): 1.024 × 10³ kg/m³[tex]\frac{1.024g}{cm^{3}}.\frac{1kg}{10^{3}g} .\frac{10^{6}cm^{3}}{1m^{3} } =1.024 \times 10^{3} kg/m^{3}[/tex]
Gravity (g): 9.806 m/s²In order to prevent water from entering, the air pressure must be equal to the pressure exerted by the seawater at the bottom. We can find that pressure (P) using the following expression.
P = ρ × g × h
P = (1.024 × 10³ kg/m³) × (9.806 m/s²) × 1000 m
P = 1.004 × 10⁷ Pa
P = 1.004 × 10⁷ Pa × (1 kPa/ 10³ Pa)
P = 1.004 × 10⁴ kPa
During World War I, the Germans had a gun called Big Bertha that was used to shell Paris. The shell had an initial speed of 2.61 km/s at an initial inclination of 81.9° to the horizontal. The acceleration of gravity is 9.8 m/s^2. How far away did the shell hit? Answer in units of km How long was it in the air? Answer in units of s.
Answer:
The shell hit at a distance of 1.9 x 10² km
The time of flight of the shell was 5.3 x 10² s
Explanation:
The position of the shell is given by the vector "r":
r = (x0 + v0 * t * cos α ; y0 + v0 * t * sin α + 1/2 g t²)
where:
x0 = initial horizontal position
v0 = magnitude of the initial velocity
t = time
α = launching angle
y0 = initial vertical position
g = acceleration of gravity
When the shell hit, the vertical component (ry) of the vector position r is 0. See figure.
Then:
ry = 0 = y0 + v0 * t * sin α + 1/2 g t²
Since the gun is at the center of our system of reference, y0 and x0 = 0
0 = t (v0 sin α + 1/2 g t)
t= 0 is discarded as solution
v0 sin α + 1/2 g t = 0
t = -2v0 sin α / g
t = (-2 * 2610 m/s * sin 81.9°)/ (-9.8 m/s²) = 5.3 x 10² s. This is the time of flight of the shell until it hit.
Then, the distance at which the shell hit is:
Distance = Module of r = ( x0 + v0 * t * cos α; 0) = x0 + v0 * t * cos α
Distance = 2.61 km/s * 5.3 x 10² s * cos 81.9 = 1.9 x 10² km
(This is a non-relativistic warm-up problem, to get you to think about reference frames.) A girl throws a baseball upwards at time t=0. She catches it at exactly t=2.0 seconds. A) Calculate the position of the ball as a function of time (0
Answer:
X(t) = 9.8 *t - 4.9 * t^2
Explanation:
We set a frame of reference with origin at the hand of the girl the moment she releases the ball. We assume her hand will be in the same position when she catches it again. The positive X axis point upwards.The ball will be subject to a constant gravitational acceleration of -9.81 m/s^2.
We use the equation for position under constant acceleration:
X(t) = X0 + V0 * t + 1/2 * a *t^2
X0 = 0 because it is at the origin of the coordinate system.
We know that at t = 2, the position will be zero.
X(2) = 0 = V0 * 2 + 1/2 * -9.81 * 2^2
0 = 2 * V0 - 4.9 * 4
2 * V0 = 19.6
V0 = 9.8 m/s
Then the position of the ball as a function of time is:
X(t) = 9.8 *t - 4.9 * t^2