Answer:
107 N, option d
Explanation:
Given that
mass of the ball, m = 0.2 kg
initial velocity of the ball, u = 20 m/s
final velocity of the ball, v = -12 m/s
time taken, Δt = 60 ms
Solving this question makes us remember "Impulse Theorem"
It states that, "that the product between the average force applied and the duration of the collision is equal to the change in momentum of the object"
Mathematically, it is represented as
FΔt = m(v - u), where
F = the average force
Δt = time taken
m = mass of the ball
v = final velocity of the ball
u = initial velocity of the ball
From the question we were given, if we substitute the values in it, we have
F = ?
Δt = 60 ms = 0.06s
m = 0.2 kg
v = -12 m/s
u = 20 m/s
F = 0.2(-12 - 20) / 0.06
F = (0.2 * -32) / 0.06
F = -6.4 / 0.06
F = -106.7 N
Thus, the magnitude is 107 N
A solid sphere has a temperature of 556 K. The sphere is melted down and recast into a cube that has the same emissivity and emits the same radiant power as the sphere. What is the cube's temperature in kelvins
Answer:
Cube temperature = 526.83 K
Explanation:
Volume of the cube and sphere will be the same.
Now, volume of cube = a³
And ,volume of sphere = (4/3)πr³
Thus;
a³ = (4/3)πr³
a³ = 4.1187r³
Taking cube root of both sides gives;
a = 1.6119r
Formula for surface area of sphere is;
As = 4πr²
Also,formula for surface area of cube is; Ac = 6a²
Thus, since a = 1.6119r,
Then, Ac = 6(1.6119r)²
Ac = 15.5893r²
The formula for radiant power is;
Q' = eσT⁴A
Where;
e is emissivity
σ is Stefan boltzman constant = 5.67 x 10^(-8) W/m²k
T is temperate in kelvin
A is Area
So, for the cube;
(Qc)' = eσ(Tc)⁴(Ac)
For the sphere;
(Qs)' = eσ(Ts)⁴(As)
We are told (Qc)' = (Qs)'
Thus;
eσ(Tc)⁴(Ac) = eσ(Ts)⁴(As)
eσ will cancel out to give;
(Tc)⁴(Ac) = (Ts)⁴(As)
Since we want to find the cube's temperature Tc,
(Tc)⁴ = [(Ts)⁴(As)]/Ac
Plugging in relevant figures, we have;
(Tc)⁴ = [556⁴ × 4πr²]/15.5893r²
r² will cancel out to give;
(Tc)⁴ = [556⁴ × 4π]/15.5893
Tc = ∜([556⁴ × 4π]/15.5893)
Tc = 526.83 K
A jeep starts from rest with a constant acceleration of 4m/s2.At the same time a car travels with a constant speed of 36km/h overtake and passes the jeep how far beyond the starting point will the jeep overtakes the car?
Answer:
25m
Explanation:
Let's assume the Jeep attains a velocity of 36km/h ; a constant speed same with that of the car.
While the Jeep is accelerating to that speed, the car with that speed passes it.
Now we can calculate the time taken for the Jeep to attain the velocity of 36km/h on her constant acceleration.
This time is t = v/a; from Newton's Law of Motion:
a = V-U / t ; a-acceleration
V is final velocity = 36km/h
U is initial velocity 0 since the body starts from rest.
Hence t = 36000/3600 ÷ 4 = 2.5s
Note conversting from km/h to m/s we multiply by 1000/3600.
But the distance covered by the car while the Jeep just accelerates is
S = U × t = 10× 2.5 = 25m.
Note From Newton's law of Motion, distance for constant speed is defined as: U × t
Hence the Car would be 25m off the starting point just as the Jeep accelerates. It would overtake the Jeep when it just covers 25m from the Jeep starting point.
Use the Bohr model to address this question. When a hydrogen atom makes a transition from the 5 th energy level to the 2nd, counting the ground level as the first,
A. What is the energy E of the emitted photon in electron volts?、
B. What is the wavelength in nanometers of the emitted photon?
C. What is the radius of the hydrogen atom in nanometers in its initial 5th energy level?
Answer:
A. 2.82 eV
B. 439nm
C. 59.5 angstroms
Explanation:
A. To calculate the energy of the photon emitted you use the following formula:
[tex]E_{n1,n2}=-13.4(\frac{1}{n_2^2}-\frac{1}{n_1^2})[/tex] (1)
n1: final state = 5
n2: initial state = 2
Where the energy is electron volts. You replace the values of n1 and n2 in the equation (1):
[tex]E_{5,2}=-13.6(\frac{1}{5^2}-\frac{1}{2^2})=2.82eV[/tex]
B. The energy of the emitted photon is given by the following formula:
[tex]E=h\frac{c}{\lambda}[/tex] (2)
h: Planck's constant = 6.62*10^{-34} kgm^2/s
c: speed of light = 3*10^8 m/s
λ: wavelength of the photon
You first convert the energy from eV to J:
[tex]2.82eV*\frac{1J}{6.242*10^{18}eV}=4.517*10^{-19}J[/tex]
Next, you use the equation (2) and solve for λ:
[tex]\lambda=\frac{hc}{E}=\frac{(6.62*10^{-34} kg m^2/s)(3*10^8m/s)}{4.517*10^{-19}J}=4.39*10^{-7}m=439*10^{-9}m=439nm[/tex]
C. The radius of the orbit is given by:
[tex]r_n=n^2a_o[/tex] (3)
where ao is the Bohr's radius = 2.380 Angstroms
You use the equation (3) with n=5:
[tex]r_5=5^2(2.380)=59.5[/tex]
hence, the radius of the atom in its 5-th state is 59.5 anstrongs
A) The energy E of the emitted photon in electron volts is; E = 2.856 eV
B) The wavelength in nanometers of the emitted photon is; λ = 434.4nm
C) The radius of the hydrogen atom in nanometers in its initial 5th energy level is; rₙ = 1.323 nm
A) Formula for the energy E of the emitted photons is;
E = -13.6([tex]\frac{1}{n_{2}^2} - \frac{1}{n_{1}^2}[/tex])
We are given;
n₂ = 5
n₁ = 2
Thus;
E = -13.6([tex]\frac{1}{5^2} - \frac{1}{2^2}[/tex])
E = 2.856 eV
B) The formula for the wavelength is;
λ = hc/E
where;
h is Planck's constant = 6.626 × 10⁻³⁴ m².kg/s
c is speed of light = 3 × 10⁸ m/s
E is energy of photon
λ is wavelength of the photon
Earlier we saw that E = 2.856 eV. Converting to Joules gives;
E = 4.5758 × 10⁻¹⁹ J
Thus;
λ = (6.626 × 10⁻³⁴ × 3 × 10⁸)/(4.5758 × 10⁻¹⁹)
λ = 4.344 × 10⁻⁷ m
Converting to nm gives;
λ = 434.4nm
C) Formula for the radius of the hydrogen atom is;
rₙ = n²a₀
where;
a₀ is bohr's radius = 5.292 × 10⁻¹¹ m
n = 5
Thus;
rₙ = 5² × 5.292 × 10⁻¹¹
rₙ = 1.323 × 10⁻⁹
rₙ = 1.323 nm
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A machinist turns the power on to a grinding wheel, which is at rest at time t = 0.00 s. The wheel accelerates uniformly for 10 s and reaches the operating angular velocity of 25 rad/s. The wheel is run at that angular velocity for 37 s and then power is shut off. The wheel decelerates uniformly at 1.5 rad/s2 until the wheel stops. In this situation, the time interval of angular deceleration (slowing down) is closest to: A machinist turns the power on to a grinding wheel, which is at rest at time t = 0.00 s. The wheel accelerates uniformly for 10 s and reaches the operating angular velocity of 25 rad/s. The wheel is run at that angular velocity for 37 s and then power is shut off. The wheel decelerates uniformly at 1.5 rad/s2 until the wheel stops. In this situation, the time interval of angular deceleration (slowing down) is closest to:__________.
a) 19 s
b) 17 s
c) 21 s
d) 23 s
e) 15 s
Starting from rest, the wheel attains an angular velocity of 25 rad/s in a matter of 10 s, which means the angular acceleration [tex]\alpha[/tex] is
[tex]25\dfrac{\rm rad}{\rm s}=\alpha(10\,\mathrm s)\implies\alpha=2.5\dfrac{\rm rad}{\mathrm s^2}[/tex]
For the next 37 s, the wheel maintains a constant angular velocity of 25 rad/s, meaning the angular acceleration is zero for the duration. After this time, the wheel undergoes an angular acceleration of -1.5 rad/s/s until it stops, which would take time [tex]t[/tex],
[tex]0\dfrac{\rm rad}{\rm s}=25\dfrac{\rm rad}{\rm s}+\left(-1.5\dfrac{\rm rad}{\mathrm s^2}\right)t\implies t=16.666\ldots\,\mathrm s[/tex]
which makes B, approximately 17 s, the correct answer.
The time interval of angular deceleration is 16.667 seconds, whose closest integer is 17 seconds. (B. 17 s.)
Let suppose that the grinding wheel has uniform Acceleration and Deceleration. In this question we need to need to calculate the time taken by the grinding wheel to stop, which is found by means of the following Kinematic formula:
[tex]t = \frac{\omega - \omega_{o}}{\alpha}[/tex] (1)
Where:
[tex]\omega_{o}[/tex] - Initial angular velocity, in radians per second.
[tex]\omega[/tex] - Final angular velocity, in radians per second.
[tex]\alpha[/tex] - Angular acceleration, in radians per square second.
[tex]t[/tex] - Time, in seconds.
If we know that [tex]\omega = 0\,\frac{rad}{s}[/tex], [tex]\omega_{o} = 25\,\frac{rad}{s}[/tex] and [tex]\alpha = -1.5\,\frac{rad}{s^{2}}[/tex], then the time taken by the grinding wheel to stop:
[tex]t = \frac{0\,\frac{rad}{s}-25\,\frac{rad}{s}}{-1.5\,\frac{rad}{s^{2}} }[/tex]
[tex]t = 16.667\,s[/tex]
The time interval of angular deceleration is 16.667 seconds. (Answer: B)
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I really need help with this question someone plz help !
Answer:
The answer is option 2.
Explanation:
Both sides are pulling the rope with equal force where the rope doesn't move. So they have a balanced forces.
Escaping from a tomb raid gone wrong, Lara Croft (m = 62.0 kg) swings across an alligator-infested river from an 8.80-m-long vine. If her speed at the bottom of the swing is 6.30 m/s and she makes it safely across the river, what is the minimum breaking strength of the vine?
Answer:
887.2 N
Explanation:
Given that
Mass of Lara, m = 62 kg
Length of the vine, l = 8.8 m
Speed of the swing, v = 6.3 m/s Then,
We start by calculating her weight.
w = mass * acceleration
w = mg
w = 62 * 9.8
w = 607.6 N
F(c) = mv²/l, on substituting
F(c) = (62 * 6.3²) / 8.8
F(c) = 2460.78 / 8.8
F(c) = 279.6 N
T - 607.6 N = 279.6 N
T = 607.6 N + 279.6 N
T = 887.2 N
Thus, the minimum breaking strength of the vine is 887.2 N
The minimum breaking strength of the vine is about 900N.
TensionTension is a force developed by a rope, string, or cable when stretched under an applied force.
Given:
Mass (m) = 62 kg, length (l) = 8.8 m, velocity (v) = 6.3 m/s, g = 10 m/s².
Weight(W) = m * g = 62 * 10 = 620N
centripetal force = F(c) = mv²/l = 62 * 6.3² / 8.8 = 279.6N
T - W = F(c)
T - 620 = 279.6
T = 900N
The minimum breaking strength of the vine is about 900N.
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An airplane flies 2500 miles east in 245 seconds what is the velocity of the plane?
Speed = (distance) / (time)
Speed = (
Velocity = speed, and its direction
The velocity of the plane is 10.2 miles per second East.
(about 48 times the speed of sound)
A 3 kg mass object is pushed 0.6 m into a spring with spring constant 210 N/m on a frictionless horizontal surface. Upon release, the object moves across the surface until it encounters a rough incline. The object moves UP the incline and stops a height of 1.5 m above the horizontal surface.
(a) How much work must be done to compress the spring initially?
(b) Compute the speed of the mass at the base of the incline.
(c) How much work was done by friction on the incline?
Answer with Explanation:
We are given that
Mass of spring,m=3 kg
Distance moved by object,d=0.6 m
Spring constant,k=210N/m
Height,h=1.5 m
a.Work done to compress the spring initially=[tex]\frac{1}{2}kx^2=\frac{1}{2}(210)(0.6)^2=37.8J[/tex]
b.
By conservation law of energy
Initial energy of spring=Kinetic energy of object
[tex]37.8=\frac{1}{2}(3)v^2[/tex]
[tex]v^2=\frac{37.8\times 2}{3}[/tex]
[tex]v=\sqrt{\frac{37.8\times 2}{3}}[/tex]
v=5.02 m/s
c.Work done by friction on the incline,[tex]w_{friction}=P.E-spring \;energy[/tex]
[tex]W_{friction}=3\times 9.8\times 1.5-37.8=6.3 J[/tex]
When you "crack" a knuckle, you suddenly widen the knuckle cavity, allowing more volume for the synovial fluid inside it and causing a gas bubble suddenly to appear in the fluid. The sudden production of the bubble, called "cavitation", produces a sound pulse---the cracking sound. Assume that the sound is transmitted uniformly in all directions and that it fully passes from the knuckle interior to the outside, at a distance of 0.29 m from your ear. If the pulse has a sound level of 61 dB at your ear, what is the rate at which energy is produced by the cavitation
Answer:If a wave y(x, t) = (6.0 mm) sin(kx + (600 rad/s)t + Φ) travels along a string, how much time does any given point on the string take to move between displacements y = +2.0 mm and y = -2.0 mm?
Explanation:
The index of refraction of Sophia's cornea is 1.387 and that of the aqueous fluid behind the cornea is 1.36. Light is incident from air onto her cornea at an angle of 17.5° from the normal to the surface. At what angle to the normal is the light traveling in the aqueous fluid?
Answer:
17.85°
Explanation:
To find the angle to the normal in which the light travels in the aqueous fluid you use the Snell's law:
[tex]n_1sin\theta_1=n_2sin\theta_2[/tex]
n1: index of refraction of Sophia's cornea = 1.387
n2: index of refraction of aqueous fluid = 1.36
θ1: angle to normal in the first medium = 17.5°
θ2: angle to normal in the second medium
You solve the equation (1) for θ2, next, you replace the values of the rest of the variables:
[tex]\theta_2=sin^{-1}(\frac{n_1sin\theta_1}{n_2})\\\\\theta_2=sin^{-1}(\frac{(1.387)(sin17.5\°)}{1.36})=17.85\°[/tex]
hence, the angle to normal in the aqueous medium is 17.85°
Parallel light rays with a wavelength of 610nm fall on a single slit. On a screen 3.10m away, the distance between the first dark fringes on either side of the central maximum is 4.00mm.
What is the width of the slit?
Answer:
The width of the slit will be ".946 mm".
Explanation:
The given values are:
Wavelength = 610 × 10⁻⁹
Length, L = 3 m
As we know,
⇒ [tex]\frac{y}{L} = \frac{m(wavelength)}{a}[/tex]
On putting the estimated values, we get
⇒ [tex]\frac{2\times 10^{-3}}{3.1} = \frac{(1)(610 X 10^{-9})}{a}[/tex]
On applying cross-multiplication, we get
⇒ [tex]a=9.46\times 10^{-4}[/tex]
⇒ [tex]a = .946 mm[/tex]
A string is stretched between fixed supports separated by 72.0 cm. It is observed to have resonant frequencies of 370 and 555 Hz, and no other resonant frequencies between these two.(a) What is the lowest resonant frequency for this string?(b) What is the wave speed for this string?
Answer:
(a) f = 185 Hz
(b) v = 266.4 m/s
Explanation:
(a) The lowest frequency can be calculated by using the following formula for the calculation of the modes (resonant frequencies) in a string:
[tex]f_n=\frac{nv}{2L}[/tex]
[tex]f_n=nf[/tex]
n: order of the mode
v: velocity of the waves in the string
L: length of the string = 72.0cm = 0.72m
fn: frequency of the n-th mode
With the information about two consecutive modes you can find the lowest resonant frequency. First you find the resonant mode n:
[tex]f_n=nf\\\\f_{n-1}=(n-1)f\\\\\frac{f_n}{f_{n-1}}=\frac{n}{n-1}[/tex]
you solve the previous equation for n:
[tex](n-1)f_n=nf_{n-1}\\\\555n-555=370n\\\\n=3[/tex]
With this information you can calculate the lowest resonant frequency:
[tex]f_n=nf\\\\f=\frac{f_n}{n}=\frac{555}{3}=185Hz[/tex]
b) You have information about two consecutive modes fn, fn-1. Then, you can calculate the velocity of the waves:
[tex]f_{n}-f_{n-1}=n\frac{v}{2L}-(n-1)\frac{v}{2L}\\\\f_n-f_{n-1}=\frac{v}{2L}\\\\v=2L(f_n-f_{n-1})[/tex]
fn = 555 Hz
fn-1: 370 Hz
[tex]v=2(0.72m)(555-370)Hz=266.4\frac{m}{s}[/tex]´
hence, the velocityof the waves in the string is 266.4 m/s
A rifle fires a 2.05 x 10-2-kg pellet straight upward, because the pellet rests on a compressed spring that is released when the trigger is pulled. The spring has a negligible mass and is compressed by 8.01 x 10-2 m from its unstrained length. The pellet rises to a maximum height of 4.46 m above its position on the compressed spring. Ignoring air resistance, determine the spring constant.
Answer:
Spring Constant = 279.58 N/m
Explanation:
We are given;
Mass; m = 2.05 x 10^(-2) kg = 0.0205 kg
Distance of compression; x = 8.01 × 10^(-2) m = 0.0801 m
Maximum height; h = 4.46 m
The formula for the energy in the spring is given by;
E = ½kx²
where:
k is the spring constant
x is the distance the spring is compressed.
Now, this energy of the spring will be equal to the energy of the pellet at its highest point. Energy of pallet = mgh So;
½kx² = mgh
Plugging in the relevant values, we have;
½ * k * 0.0801² = 0.0205 * 9.81 * 4.46
0.003208005k = 0.8969
k = 0.8969/0.003208005
k = 279.58 N/m
A water slide is constructed so that swimmers, starting from rest at the top of the slide, leave the end of the slide traveling horizontally. One person hits the water 5.00 m from the end of the slide in a time of 0.504 s after leaving the slide. Ignore friction and air resistance. Find the height H.
Answer:
4.93 m
Explanation:
According to the question, the computation of the height is shown below:
But before that first we need to find out the speed which is shown below:
As we know that
[tex]Speed = \frac{Distance}{Time}[/tex]
[tex]Speed = \frac{5}{0.504}[/tex]
= 9.92 m/s
Now
[tex]v^2 - u^2 = 2\times g\times h[/tex]
[tex]9.92^2 = 2\times 9.98 \times h[/tex]
98.4064 = 19.96 × height
So, the height is 4.93 m
We simply applied the above formulas so that the height i.e H could arrive
The starships of the Solar Federation are marked with the symbol of the Federation, a circle, whereas starships of the Denebian Empire are marked with the Empire's symbol, an ellipse whose major axis is n times its minor axis (a=nb in the figure ).
How fast, relative to an observer, does an Empire ship have to travel for its markings to be confused with those of a Federation ship? Use c for the speed of light in a vacuum.
Express your answer in terms of n and c.
Complete question
The complete question is shown on the first uploaded image
Answer:
The velocity is [tex]v = c* \sqrt{1 - \frac{1}{n^2} }[/tex]
Explanation:
From the question we are told that
a = nb
The length of the minor axis of the symbol of the Federation, a circle, seen by the observer at velocity v must be equal to the minor axis(b) of the Empire's symbol, (an ellipse)
Now this length seen by the observer can be mathematically represented as
[tex]h = t \sqrt{1 - \frac{v^2}{c^2} }[/tex]
Here t is the actual length of the major axis of of the Empire's symbol, (an ellipse)
So t = a = nb
and b is the length of the minor axis of the symbol of the Federation, (a circle) when seen by an observer at velocity v which from the question must be the length of the minor axis of the of the Empire's symbol, (an ellipse)
i.e h = b
So
[tex]b = nb [\sqrt{1 - \frac{v^2}{c^2} } ][/tex]
[tex][\frac{1}{n} ]^2 = 1 - \frac{v^2}{c^2}[/tex]
[tex]v^2 =c^2 [1- \frac{1}{n^2} ][/tex]
[tex]v^2 =c^2 [\frac{n^2 -1}{n^2} ][/tex]
[tex]v = c* \sqrt{1 - \frac{1}{n^2} }[/tex]
During last year’s diving competition, the divers always pull their limbs in and curl up their bodies when they do flips. Just before entering the water, they fully extend their limbs to enter straight down as shown. Explain the effect of both actions on their angular velocities and kinetic energy (support your answer with working). Also explain the effect on their angular momentum.
Answer:
the angular speed of the person increases, being able to make more turns and faster.
K₂ = K₁ I₁ / I₂
Explanation:
When the divers are turning the system is isolated, so all the forces are internal and therefore also the torque, therefore the angular momentum is conserved
initial, joint when starting to turn
L₀ = I₁ w₁
final. When you shrink your arms and legs
Lf = I₂ w₂
L₀ = Lf
I₁ w₁ = I₂ w₂
when you shrink your arms and legs the distance to the turning point decreases and since the moment of inertia depends on the distance squared, the moment of inertia also decreases
I₂ <I₁
w₂ = I₁ / I₂ w₁
therefore the angular speed of the person increases, being able to make more turns and faster.
When it goes into the water it straightens the arm and leg, so the moment of inertia increases
I₁> I₂
w₁ = I₂ / I₁ w₂
therefore we see that the angular velocity decreases, therefore the person trains the water like a stone and can go deeper faster.
In both cases the kinetic energy is
K = ½ I w²
the initial kinetic energy is
K₁ = ½ I₁ w₁²
the final kinetic energy is
K₂ = ½ I₂ w₂²
we substitute
K₂ = ½ I₂ (I₁ / I₂ w1² 2
K₂ = ½ I₁² / I₂ w₁² = (½ I₁ w₁²) I₁ / I₂
K₂ = K₁ I₁ / I₂
therefore we see that the kinetic energy increases by factor I₁/I₂
PLEASE HELP !
Complete the following sentence. Choose the right answer from the given ones. The internal energy of the body can be changed A / B / C. A. only when the body is warmed or cooled B. when work is done on the body or heat flow C. only when the body does work
B
HOPE IT HELPS LET ME KNOW IF U NEED EXPLANATION
Consider a system of a cliff diver and the earth. The gravitational potential energy of the system decreases by 24,500 J as the diver drops to the water from a height of 44.0 m. Determine her weight in newtons.
Before she jumped from the cliff, her gravitational potential energy was
GPE = (her weight) x (height of the cliff) .
That's exactly the GPE that she lost on the way down to the water. So we can write
24,500 J = (her weight) x (44.0 m)
Divide each side by 44.0 m:
Her weight = 24,500 J / 44 m
Her weight = 556.8 Newtons
(about 125 pounds)
Davina accelerates a box across a smooth frictionless horizontal surface over a displacement of 18.0 m with a constant 25.0 N force angled at 23.0° below the horizontal. How much work does she do on the box? A. 176 J B. 414 J C. 450 J D. 511 J Group of answer choices
Answer:
W = 414 J, correct is B
Explanation:
Work is defined by
W = ∫ F .dx
where F is the force, x is the displacement and the point represents the dot product
this expression can also be written with the explicit scalar product
W = ∫ F dx cos θ
where is the angle between force and displacement
for this case as the force is constant
W = F x cos θ
calculate
W = 25.0 18.0 cos (-23)
W = 414 J
the correct answer is B
Which is the best description of the scientific theory
Explanation:
a scientific theory is a well substantiated explanation of some aspect of the nature world, based on a body of facts that have been repeatedly confirmed through observation and experiment. search fact-supported theories are not "guesses" but reliable account of the real world .
Two students are pushing their stalled car down the street. If the net force exerted on
the car by the students is 1000 N at an angle of 20° below horizontal, the horizontal
component of the force is:
(a) greater than 1000 N.
(b) less than 1000 N.
(c) sum of the pushing force and the weight of the students.
(d) (a) and (b)
(e) (a) and (c)
Answer:
B
Explanation:
Because the force has 2 components (horizontal and vertical), the horizontal component must be smaller than the total force. The Pythagorean theorem only adds positive values (because they're squared) so it makes sense. Using trigonometry, 100*cos(-20) yields a horizontal force of around 939.7N, which is less than 1000N.
510 g squirrel with a surface area of 935 cm2 falls from a 4.8-m tree to the ground. Estimate its terminal velocity. (Use the drag coefficient for a horizontal skydiver. Assume that the squirrel can be approximated as a rectanglar prism with cross-sectional area of width 11.6 cm and length 23.2 cm. Note, the squirrel may not reach terminal velocity by the time it hits the gr
Answer:
The terminal velocity is [tex]v_t =17.5 \ m/s[/tex]
Explanation:
From the question we are told that
The mass of the squirrel is [tex]m_s = 50\ g = \frac{50}{1000} = 0.05 \ kg[/tex]
The surface area is [tex]A_s = 935 cm^2 = \frac{935}{10000} = 0.0935 \ m^2[/tex]
The height of fall is h =4.8 m
The length of the prism is [tex]l = 23.2 = 0.232 \ m[/tex]
The width of the prism is [tex]w = 11.6 = 0.116 \ m[/tex]
The terminal velocity is mathematically represented as
[tex]v_t = \sqrt{\frac{2 * m_s * g }{\dho_s * C * A } }[/tex]
Where [tex]\rho[/tex] is the density of a rectangular prism with a constant values of [tex]\rho = 1.21 \ kg/m^3[/tex]
[tex]C[/tex] is the drag coefficient for a horizontal skydiver with a value = 1
A is the area of the prism the squirrel is assumed to be which is mathematically represented as
[tex]A = 0.116 * 0.232[/tex]
[tex]A = 0.026912 \ m^2[/tex]
substituting values
[tex]v_t = \sqrt{\frac{2 * 0.510 * 9.8 }{1.21 * 1 * 0.026912 } }[/tex]
[tex]v_t =17.5 \ m/s[/tex]
On April 13, 2029 (Friday the 13th!), the asteroid 99942 mi Apophis will pass within 18600 mi of the earth-about 1/13 the distance to the moon! It has a density of 2600 kg/m^3, can be modeled as a sphere 320 m in diameter, and will be traveling at 12.6 km/s.
1)If, due to a small disturbance in its orbit, the asteroid were to hit the earth, how much kinetic energy would it deliver?
2)The largest nuclear bomb ever tested by the United States was the "Castle/Bravo" bomb, having a yield of 15 megatons of TNT. (A megaton of TNT releases 4.184x10^15 J of energy.) How many Castle/Bravo bombs would be equivalent to the energy of Apophis?
Answer:
Explanation:
Volume of asteroid = 4/3 x π x 160³
= 17.15 x 10⁶
mass = volume x density
= 17.15 x 10⁶ x 2600
= 445.9 x 10⁸ kg
kinetic energy
= 1/2 x 445.9 x 10⁸ x( 12.6 )² x 10⁶
= 35.4 x 10¹⁷ J .
2 )
energy of 15 megaton
= 4.184 x 10¹⁵ x 15 J
= 62.76 x 10¹⁵ J
No of bombs required
= 35.4 x 10¹⁷ / 62.76 x 10¹⁵
= 56.4 Bombs .
Q) A particle in simple harmonic motion starts its motion from its mean position. If T be the time period, calculate the ratio of kinetic energy and potential energy of the particle at the instant when t = T/12.
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Explanation:
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In a Venn diagram, the separate circles contain characteristics unique to each compared and the intersection contains characteristics that are common to both items being compared. This Venn diagram compares the inner and outer planets. What belongs in the center section?
a. -Revolve around the Sun
-Rotate on an axis
-Generally have rings
b. -Revolve around the Sun
-Rotate on axis
-Generally have moons
c. -Rotate around the Sun
-Revolve on an axis
-Generally have moons
d. -Rotate around the Sun
-Revolve on an axis
-Generally have rings
Answer:
B. revolve around the sun
rotate on an axis
generally have moons
Explanation:
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Volume of an block is 5 cm3. If the density of the block is 250 g/cm3, what is the mass of the block ?
Answer:
The mass of the block is 1250g.
Explanation:
Given that the formula for density is ρ = mass/volume. Then you have to substitute the values into the formula :
[tex]ρ = \frac{mass}{volume} [/tex]
Let density = 250,
Let volume = 5,
[tex]250 = \frac{m}{5} [/tex]
[tex]m = 250 \times 5[/tex]
[tex]m = 1250g[/tex]
A person sits on a freely spinning lab stool that has no friction in its axle. When this person extends her arms, A. her moment of inertia decreases and her angular speed decreases B. her moment of inertia decreases and her angular speed increases C. her moment of inertia increases and her angular speed decreases D. her moment of inertia increases and her angular speed decreases E. her moment of inertia increases and her angular speed remains the same.
Answer:
C. her moment of inertia increases and her angular speed decreases
D. her moment of inertia increases and her angular speed decreases
Explanation:
The moment of inertia of a body is the sum of the products of an increment of mass and the square of its distance from the center of rotation. When a spinning person extends her arms, part of her mass increases its distance from the center of rotation, so increases the moment of inertia.
The kinetic energy of a spinning body is jointly proportional to the moment of inertia and the square of the angular speed. Hence an increase in moment of inertia will result in a decrease in angular speed unless there is a change in the rotational kinetic energy.
This effect is used by figure skaters to increase their spin rate by drawing their arms and legs closer to the axis of rotation. Similarly, they can slow the spin by extending arms and legs.
When the person extends her arms, her moment of inertia increases and her angular speed decreases.
_____
Note to those looking for a letter answer
Both choices C and D have identical (correct) wording the way the problem is presented here. You will need to check carefully the wording in any problem you may think is similar.
Crystalline germanium (Z=32, rho=5.323 g/cm3) has a band gap of 0.66 eV. Assume the Fermi energy is half way between the valence and conduction bands. Estimate the ratio of electrons in the conduction band to those in the valence band at T = 300 K. (See eq. 10-11) Assume the width of the valence band is ΔΕV ~ 10 eV.
Answer:
= 8.2*10⁻¹²
Explanation:
Probability of finding an electron to occupy a state of energy, can be expressed by using Boltzmann distribution function
[tex]f(E) = exp(-\frac{E-E_f}{K_BT} )[/tex]
From the given data, fermi energy lies half way between valence and conduction bands, that is half of band gap energy
[tex]E_f = \frac{E_g}{2}[/tex]
Therefore,
[tex]f(E) = exp(-\frac{E-\frac{E_g}{2} }{K_BT} )[/tex]
Using boltzman distribution function to calculate the ratio of number of electrons in the conduction bands of those electrons in the valence bond is
[tex]\frac{n_{con}}{n_{val}} =\frac{exp(-\frac{[E_c-E_g/2]}{K_BT} )}{exp(-\frac{[E_v-E_fg/2}{K_BT} )}[/tex]
[tex]= exp(\frac{-(E_c-E_v}{K_BT} )\\\\=exp(\frac{-(0.66eV)}{(8.617\times10^-^5eV/K)(300K)} )\\\\=8.166\times10^-^1^2\approx8.2\times10^{-12}[/tex]
If you were to drop a rock from a tall building, assuming that it had not yet hit the ground, and neglecting air resistance. What is its vertical displacement (in m) after 4 s? (g = 10 m/s2)
Answer:
d = 80 m
its vertical displacement (in m) after 4 s is 80 m
Explanation:
From the equation of motion;
d = vt + 0.5at^2 ......1
Where;
d = displacement
v = initial velocity = 0 (dropped with no initial speed)
t = time of flight = 4s
a = g = acceleration due to gravity = 10 m/s^2
Substituting the given values into equation 1;
d = 0(4) + 0.5(10 × 4^2)
d = 0.5(10×16)
d = 80 m
its vertical displacement (in m) after 4 s is 80 m
We say that the displacement of a particle is a vector quantity. Our best justification for this assertion is: A. a displacement is obviously not a scalar. B. displacement can be specified by a magnitude and a direction. C. operating with displacements according to the rules for manipulating vectors leads to results in agreement with experiments. D. displacement can be specified by three numbers. E. displacement is associated by motion.
Answer:
Option B - displacement can be specified by a magnitude and a direction.
Explanation:
A Vector quantity is defined as a physical quantity characterized by the presence of both magnitude as well as direction. Examples include displacement, force, torque, momentum, acceleration, velocity e.t.c
Whereas a scalar quantity is defined as a physical quantity which is specified with the magnitude or size alone. Examples include length, speed, work, mass, density, etc.
Displacement is the difference between the initial position and the final position of a body. Displacement is a vector quantity and not a scalar quantity because it can be described by using both magnitude as well as direction.
Looking at the options, the only one that truly justifies this definition is option B.
