Explanation:
v = ωr
v = (550 rev/min) (2π rad/rev) (1 min / 60 s) (3.4 m)
v = 196 m/s
5) What is the weight of a body in earth. if its weight is 5Newton
in moon?
Answer:
8.167
Explanation:
Two cars honking at the same frequency of 800 Hz are moving toward a pedestrian standing at the curb. One car is moving with speed 25 m/s, and the second car is moving at 21 m/s. What is the frequency of the beats heard by the pedestrian? Take the speed of sound as 343 m/s.
Answer:
The frequency of the beats heard by the pedestrian is 9.33 Hz
Explanation:
Frequency of the two cars, F = 800 Hz
speed of sound in air, V = 343 m/s
The sound wave length, λ = ?
λ = V / F
λ = 343 / 800
λ = 0.42875 m
The frequency of the car moving at 25 m/s heard by the pedestrian
f₁ = v₁ / λ
f₁ = 25 / 0.42875
f₁ = 58.31 Hz
The frequency of the car moving at 21 m/s heard by the pedestrian
f₂ = v₂ / λ
f₂ = 21 / 0.42875
f₂ = 48.98 Hz
The beat frequency = f₁ - f₂
= 58.31 - 48.98
= 9.33 Hz
Therefore, the frequency of the beats heard by the pedestrian is 9.33 Hz
The electric field must be zero inside a conductor in electrostatic equilibrium, but not inside an insulator. It turns out that we can still apply Gauss's law to a Gaussian surface that is entirely within an insulator by replacing the right-hand side of Gauss's law, Qin/eo, with Qin/e, where ε is the permittivity of the material. (Technically, Eo is called the vacuum permittivity.) Suppose that a 70 nC point charge is surrounded by a thin, 32-cm-diameter spherical rubber shell and that the electric field strength inside the rubber shell is 2500 N/C.
What is the permittivity of rubber?
Answer:
The permittivity of rubber is [tex]\epsilon = 8.703 *10^{-11}[/tex]
Explanation:
From the question we are told that
The magnitude of the point charge is [tex]q_1 = 70 \ nC = 70 *10^{-9} \ C[/tex]
The diameter of the rubber shell is [tex]d = 32 \ cm = 0.32 \ m[/tex]
The Electric field inside the rubber shell is [tex]E = 2500 \ N/ C[/tex]
The radius of the rubber is mathematically evaluated as
[tex]r = \frac{d}{2} = \frac{0.32}{2} = 0.16 \ m[/tex]
Generally the electric field for a point is in an insulator(rubber) is mathematically represented as
[tex]E = \frac{Q}{ \epsilon } * \frac{1}{4 * \pi r^2}[/tex]
Where [tex]\epsilon[/tex] is the permittivity of rubber
=> [tex]E * \epsilon * 4 * \pi * r^2 = Q[/tex]
=> [tex]\epsilon = \frac{Q}{E * 4 * \pi * r^2}[/tex]
substituting values
[tex]\epsilon = \frac{70 *10^{-9}}{2500 * 4 * 3.142 * (0.16)^2}[/tex]
[tex]\epsilon = 8.703 *10^{-11}[/tex]
Problem 2: A 21-W horizontal beam of light of wavelength 430 nm, travelling at speed c, passes through a rectangular opening of width 0.0048 m and height 0.011 m. The light then strikes a screen at a distance 0.36 m behind the opening.
a) E = 4.62E-19 Joules
b) N = 4.543E+19 # of photons emitted
c) If the beam of light emitted by the source has a constant circular cross section whose radius is twice the height of the opening the beam is approaching, find the flow density of photons as the number of photons passing through a square meter of cross-sectional area per second.
d) Calculate the number of photons that pass through the rectangular opening per second.
e) The quantity NO that you found in part (d) gives the rate at which photons enter the region between the opening and the screen. Assuming no reflection from the screen, find the number of photons in that region at any time.
f) How would the number of photons in the region between the opening and the screen change, if the photons traveled more slowly? Assume no change in any other quantity, including the speed of light.
Answer:
a. E = 4.62 × 10⁻¹⁹J
b. n = 4.54 × 10¹⁹photons
c. 2.99 × 10²²photons/m²
d. 1.58 ×10¹⁸photons/seconds
e. n = 1.896 × 10 ⁹ photons
f. the number of photons will be more if it travels slowly
Explanation:
Carbon dioxide initially at 50 kPa, 400 K, undergoes a process in a closed system until its pressure and temperature are 2 MPa and 800 K, respectively. Assuming an ideal gas behaviour, find the entropy change of the carbon dioxide by assuming that the specific heats are constant. For the gas, take Cp = 0.846 kJ/kg.K and R = 0.1889 kJ/kg.K
Answer:
[tex]\Delta S=-0.11\frac{kJ}{kg*K}[/tex]
Explanation:
Hello,
In this case, we can compute the entropy change by using the following equation containing both pressure and temperature:
[tex]\Delta S=Cp\ ln(\frac{T_2}{T_1} )-R\ ln(\frac{p_2}{p_1} )[/tex]
Thus, we use the given data to obtain (2 MPa = 2000 kPa):
[tex]\Delta S=0.846\frac{kJ}{kg*K} \ ln(\frac{800K}{400K} )-0.1889\frac{kJ}{kg*K} \ ln(\frac{2000kPa}{50kPa} )\\\\\Delta S=-0.11\frac{kJ}{kg*K}[/tex]
Best regards.
Please help! Average speed. Show work!
Answer:
3.78 m/s
Explanation:
Recall that the formula for average speed is given by
Speed = Distance ÷ Time taken
Where,
Speed = we are asked to find this
Distance = given as 340m
Time taken = 1.5 min = 1.5 x 60 = 90 seconds
Substituting the values into the equation:
Speed = Distance ÷ Time taken
= 340 meters ÷ 90 seconds
= 3.777777 m/s
= 3.78 m/s (round to nearest hundredth)
Answer:
3.78 m/sOption D is the right option.
solution,
Distance travelled=340 metre
Time taken= 1.5 minutes ( 1 minute=60 seconds)
=1.5* 60
=90 seconds
Now,
Average speed:
[tex] \ \frac{distance \: travelled \: }{time \: taken} \\ = \frac{340 \: m}{90 \: s} \\ = 3.78 \: metre \: per \: second[/tex]
Hope this helps...
Good luck on your assignment..
If an all electric house uses 2400 kWh in a month, what will the amount of the bill for electricity be at a rate of 9.5 cents per kWh?
Answer:
$228Explanation:
Given data
Total electricity consumption = 2400 kWh
Rate of consumption = 9.5 cent per kWh
converting cent to dollar we have. 9.5/100= $0.095
hence the bill is 2400*0.095= $228
Use Kepler's third law to determine how many days it takes a spacecraft to travel in an elliptical orbit from a point 6 590 km from the Earth's center to the Moon, 385 000 km from the Earth's center.
Answer:
1.363×10^15 seconds
Explanation:
The spaceship travels an elliptical orbit from a point of 6590km from the earth center to the moon and 38500km from the earth center.
To calculate the time taken from Kepler's third Law :
T^2 = ( 4π^2/GMe ) r^3
Where Me is the mass of the earth
r is the average distance travel
G is the universal gravitational constant. = 6.67×10-11 m3 kg-1 s-2
π = 3.14
Me = mass of earth = 5.972×10^24kg
r =( r minimum + r maximum)/2 ......1
rmin = 6590km
rmax = 385000km
From equation 1
r = (6590+385000)/2
r = 391590/2
r = 195795km
From T^2 = ( 4π^2/GMe ) r^3
T^2 = (4 × 3.14^2/ 6.67×10-11 × 5.972×10^24) × 195795^3
= ( 4×9.8596/ 3.983×10^14 ) × 7.5059×10^15
= 39.4384/ 3.983×10^14 ) × 7.5059×10^15
= (9.901×10^14) × 7.5059×10^15
T^2 = 7.4321× 10^30
T =√7.4321× 10^30
T = 2.726×10^15 seconds
The time for one way trip from Earth to the moon is :
∆T = T/2
= 2.726×10^15 /2
= 1.363×10^15 secs
If the frequency of a periodic wave is cut in half while the speed remains the same, what happens to the wavelength
Answer:
The wavelength becomes twice the original wavelength
Explanation:
Recall that for regular waves, the relationship between wavelength, velocity (i.e speed) and frequency is given by
v = fλ
where,
v = velocity,
f = frequency
λ = wavelength
Before a change was made to the frequency, we have: v₁ = f₁ λ₁
After a change was made to the frequency, we have: v₂ = f₂ λ₂
We are told that the speed remains the same, so
v₁ = v₂
f₁ λ₁ = f₂ λ₂ (rearranging this)
f₁ / f₂ = λ₂/λ₁ --------(1)
we are given that the frequency is cut in half.
f₂ = (1/2) f₁ (rearranging this)
f₁/f₂ = 2 -------------(2)
if we substitute equation (2) into equation (1):
f₁ / f₂ = λ₂/λ₁
2 = λ₂/λ₁
λ₂ = 2λ₁
Hence we can see that the wavelength after the change becomes twice (i.e doubles) the initial wavelength.
A car moving at a speed of 25 m/s enters a curve that traces a circular quarter turn of radius 129 m. The driver gently applies the brakes, slowing the car with a constant tangential acceleration of magnitude 1.2 m/s2.a) Just before emerging from the turn, what is the magnitudeof the car's acceleration?
b) At that same moment, what is the angle q between the velocity vector and theacceleration vector?
I am having trouble because this problem seems to have bothradial and tangential accleration. I tried finding the velocityusing V^2/R, but then that didnt take into account thedeceleration. Any help would be great.
Answer:
8.7 m/s^2
82.15°
Explanation:
Given:-
- The initial speed of the car, vi = 25 m/s
- The radius of track, r = 129 m
- Car makes a circular " quarter turn "
- The constant tangential acceleration, at = 1.2 m/s^2
Solution:-
- We will solve the problem using rotational kinematics. Determine the initial angular velocity of car ( wi ) as follows:
[tex]w_i = \frac{v_i}{r} \\\\w_i = \frac{25}{129}\\\\w_i = 0.19379 \frac{rad}{s}[/tex]
- Now use the constant tangential acceleration ( at ) and determine the constant angular acceleration ( α ) for the rotational motion as follows:
at = r*α
α = ( 1.2 / 129 )
α = 0.00930 rad/s^2
- We know that the angular displacement from the initial entry to the exit of the turn is quarter of a turn. The angular displacement would be ( θ = π/2 ).
- Now we will use the third rotational kinematic equation of motion to determine the angular velocity at the exit of the turn (wf) as follows:
[tex]w_f^2 = w_i^2 + 2\alpha*theta\\\\w_f = \sqrt{0.19379^2 + 0.00930\pi } \\\\w_f = 0.25840 \frac{rad}{s}[/tex]
- We will use the evaluated final velocity ( wf ) and determine the corresponding velocity ( vf ) as follows:
[tex]v_f = r*w_f\\\\v_f = 129*0.2584\\\\v_f = 33.33380 \frac{x}{y}[/tex]
- Now use the formulation to determine the centripetal acceleration ( ac ) at this point as follows:
[tex]a_c = \frac{v_f^2}{r} \\\\a_c = \frac{33.3338^2}{129} \\\\a_c = 8.6135 \frac{m}{s^2}[/tex]
- To determine the magnitude of acceleration we will use find the resultant of the constant tangential acceleration ( at ) and the calculated centripetal acceleration at the exit of turn ( ac ) as follows:
[tex]|a| = \sqrt{a^2_t + a_c^2} \\\\|a| = \sqrt{1.2^2 + 8.6135^2} \\\\|a| = 8.7 \frac{m}{s^2}[/tex]
- To determine the angle between the velocity vector and the acceleration vector. We need to recall that the velocity vector only has one component and always tangential to the curved path. Hence, the velocity vector is parallel to the tangential acceleration vector ( at ). We can use the tangential acceleration ( at ) component of acceleration ( a ) and the centripetal acceleration ( ac ) component of the acceleration and apply trigonometric ratio as follows:
[tex]q = arctan \frac{a_c}{a_t} = arctan \frac{8.7}{1.2} \\\\q = 82.15 ^.[/tex]
Answer: The angle ( q ) between acceleration vector ( a ) and the velocity vector ( v ) at the exit of the turn is 82.15° .
In a uniform electric field, the magnitude of torque is given by:-
Answer:
Electric dipole
Explanation:
the axis of a dipole makes an angle with the electric field. depending on the direction (clockwise/anticlockwise) we can get the torque (positive/negative).
hope this helps :D
A 100-m long transmission cable is suspended between two towers. If the mass density is 2.01 kg/m and the tension in the cable is 3.00 x 104 N, what is the speed of transverse waves on the cable
Answer:
The speed is [tex]v =122.2 \ m/s[/tex]
Explanation:
From the question we are told that
The length of the wire is [tex]L = 100 \ m[/tex]
The mass density is [tex]\mu = 2.01 \ kg/m[/tex]
The tension is [tex]T = 3.00 *10^{4} \ N[/tex]
Generally the speed of the transverse cable is mathematically represented as
[tex]v = \sqrt{\frac{T}{\mu} }[/tex]
substituting values
[tex]v = \sqrt{\frac{3.0 *10^{4}}{2.01} }[/tex]
[tex]v =122.2 \ m/s[/tex]
Lightbulbs are typically rated by their power dissipation when operated at a given voltage. Which of the following lightbulbs has the largest resistance when operated at the voltage for which it's rated?
A. 0.8 W, 1.5 V
B. 6 W 3 V
C. 4 W, 4.5 V
D. 8 W, 6 V
Answer:
The arrangement with the greatest resistance is the light bulb of option C. 4 W, 4.5 V
Explanation:
The equation for electric power is
power P = IV
also, I = V/R,
substituting into the equation, we have
[tex]P = \frac{V^{2} }{R}[/tex]
[tex]R = \frac{V^{2} }{P}[/tex]
a) [tex]R = \frac{1.5^{2} }{0.8}[/tex] = 2.8 Ω
b) [tex]R = \frac{3^{2} }{6}[/tex] = 1.5 Ω
c) [tex]R = \frac{4.5^{2} }{4}[/tex] 5.06 Ω
d) [tex]R = \frac{6^{2} }{8}[/tex] = 4.5 Ω
from the calculations, one can see that the lightbulb with te greates resistance is
C. 4 W, 4.5 V
"In order to eliminate glare from a horizontal surface such as water, what is the orientation of the polarizing axes in polarizing sunglasses?"
Answer
Polaroid sunglasses with transmission axes that are oriented vertically so the glasses absorb the horizontally polarised light and as the glare is composed of that orientation of light they therefore reduce the glare.
a sandbag is released from a rising air balloon and hits the ground 7 seconds later. from what height was the sandbag dropped from if at teh moment of release the balloon was traveling upward at 3m/s
Answer:
0.46
Explanation:
We can solve this problem using the suvat equations of motions.
the height, s is what we are trying to find out
initial velocity, u = 3 m/s from the question
final velocity, v = 0 m/s since it hits the ground it will stop moving downwards
acceleration, a = -9.81, due to gravity
time, t =7, from the question.
v² = u² + 2as ⇒ v²-u² = 2as
⇒ s = [tex]\frac{v^{2} -u^{2} }{2a}[/tex]
⇒ s = [tex]\frac{0^{2} -3^{2} }{-19.62}[/tex]
⇒ s = 9/19.62
⇒ s ≈ 0.46
The frequency of a physical pendulum comprising a nonuniform rod of mass 1.15 kg pivoted at one end is observed to be 0.658 Hz. The center of mass of the rod is 42.5 cm below the pivot point. What is the rotational inertia of the pendulum around its pivot point
Answer:
The rotational inertia of the pendulum around its pivot point is [tex]0.280\,kg\cdot m^{2}[/tex].
Explanation:
The angular frequency of a physical pendulum is measured by the following expression:
[tex]\omega = \sqrt{\frac{m\cdot g \cdot d}{I_{o}} }[/tex]
Where:
[tex]\omega[/tex] - Angular frequency, measured in radians per second.
[tex]m[/tex] - Mass of the physical pendulum, measured in kilograms.
[tex]g[/tex] - Gravitational constant, measured in meters per square second.
[tex]d[/tex] - Straight line distance between the center of mass and the pivot point of the pendulum, measured in meters.
[tex]I_{O}[/tex] - Moment of inertia with respect to pivot point, measured in [tex]kg\cdot m^{2}[/tex].
In addition, frequency and angular frequency are both related by the following formula:
[tex]\omega =2\pi\cdot f[/tex]
Where:
[tex]f[/tex] - Frequency, measured in hertz.
If [tex]f = 0.658\,hz[/tex], then angular frequency of the physical pendulum is:
[tex]\omega = 2\pi \cdot (0.658\,hz)[/tex]
[tex]\omega = 4.134\,\frac{rad}{s}[/tex]
From the formula for the physical pendulum's angular frequency, the moment of inertia is therefore cleared:
[tex]\omega^{2} = \frac{m\cdot g \cdot d}{I_{o}}[/tex]
[tex]I_{o} = \frac{m\cdot g \cdot d}{\omega^{2}}[/tex]
Given that [tex]m = 1.15\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]d = 0.425\,m[/tex] and [tex]\omega = 4.134\,\frac{rad}{s}[/tex], the moment of inertia associated with the physical pendulum is:
[tex]I_{o} = \frac{(1.15\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (0.425\,m)}{\left(4.134\,\frac{rad}{s} \right)^{2}}[/tex]
[tex]I_{o} = 0.280\,kg\cdot m^{2}[/tex]
The rotational inertia of the pendulum around its pivot point is [tex]0.280\,kg\cdot m^{2}[/tex].
A stretch spring has an elastic potential energy of 35 J when it is stretched 0.54m. What is the spring constant of the spring?
Answer:
240.1 N/m
Explanation:
Applying the formula for the potential energy stored in a stretched spring,
E = 1/2ke².................... Equation 1
Where E = potential energy, k = spring constant, e = extension.
make k the subject of the equation,
k = 2E/e².................. Equation 2
Given: E = 35 J, e = 0.54 m
Substitute into equation 2
k = 2(35)/0.54²
k = 70/0.2916
k = 240.1 N/m.
Hence the spring constant of the spring is 240.1 N/m
Answer:
240 N/m
Explanation:
edge2o2o
A hollow sphere of radius 1.66 m is in a region where the electric field is radial and directed toward the center of the sphere. If the magnitude of the field at the surface of the sphere is 21.5 N/C, what is the net electric flux through the spherical surface?
Answer:
The value of flux will be "744.1 N.m²/C".
Explanation:
The given values are:
Magnitude,
E = 21.5 N/C
Radius,
R = 1.66 m
As we know,
⇒ [tex]Flux = Area\times E[/tex]
On putting the estimated values, we get
⇒ [tex]=-21.5\tines (4\times \pi\times 1.66^2 )[/tex]
⇒ [tex]=744.1 \ N.m^2/C[/tex]
A spring attached to the ceiling is stretched one foot by a four pound weight. The mass is set in motion by pulling down 2 feet and then released in a medium with a damping force numerically equal to the velocity.
a. Find the Hook's law spring constant k.
b. Find its natural frequency ω.
c. Form a differential equation for the position x(t) of the mass.
d. Determine the solution for the position (in alternate form).
e. Find the times at which the mass passes the equilibrium second time heading up.
Given that,
Weight = 4 pound
[tex]W=4\ lb[/tex]
Stretch = 2 feet
Let the force be F.
The elongation of the spring after the mass attached is
[tex]x=2-1=1\ feet[/tex]
(a). We need to calculate the value of spring constant
Using Hooke's law
[tex]F=kx[/tex]
[tex]k=\dfrac{F}{x}[/tex]
Where, F = force
k = spring constant
x = elongation
Put the value into the formula
[tex]k=\dfrac{4}{1}[/tex]
[tex]k=4[/tex]
(b). We need to calculate the mass
Using the formula
[tex]F=mg[/tex]
[tex]m=\dfrac{F}{g}[/tex]
Where, F = force
g = acceleration due to gravity
Put the value into the formula
[tex]m=\dfrac{4}{32}[/tex]
[tex]m=\dfrac{1}{8}\ lb[/tex]
We need to calculate the natural frequency
Using formula of natural frequency
[tex]\omega=\sqrt{\dfrac{k}{m}}[/tex]
Where, k = spring constant
m = mass
Put the value into the formula
[tex]\omega=\sqrt{\dfrac{4}{\dfrac{1}{8}}}[/tex]
[tex]\omega=\sqrt{32}[/tex]
[tex]\omega=4\sqrt{2}[/tex]
(c). We need to write the differential equation
Using differential equation
[tex]m\dfrac{d^2x}{dt^2}+kx=0[/tex]
Put the value in the equation
[tex]\dfrac{1}{8}\dfrac{d^2x}{dt^2}+4x=0[/tex]
[tex]\dfrac{d^2x}{dt^2}+32x=0[/tex]
(d). We need to find the solution for the position
Using auxiliary equation
[tex]m^2+32=0[/tex]
[tex]m=\pm i\sqrt{32}[/tex]
We know that,
The general equation is
[tex]x(t)=A\cos(\sqrt{32t})+B\sin(\sqrt{32t})[/tex]
Using initial conditions
(I). [tex]x(0)=2[/tex]
Then, [tex]x(0)=A\cos(\sqrt{32\times0})+B\sin(\sqrt{32\times0})[/tex]
Put the value in equation
[tex]2=A+0[/tex]
[tex]A=2[/tex].....(I)
Now, on differentiating of general equation
[tex]x'(t)=-\sqrt{32}A\sin(\sqrt{32t})+\sqrt{32}B\cos(\sqrt{32t})[/tex]
Using condition
(II). [tex]x'(0)=0[/tex]
Then, [tex]x'(0)=-\sqrt{32}A\sin(\sqrt{32\times0})+\sqrt{32}B\cos(\sqrt{32\times0})[/tex]
Put the value in the equation
[tex]0=0+\sqrt{32}B[/tex]
So, B = 0
Now, put the value in general equation from equation (I) and (II)
So, The general solution is
[tex] x(t)=2\cos\sqrt{32t}[/tex]
(e). We need to calculate the time
Using formula of time
[tex]T=\dfrac{2\pi}{\omega}[/tex]
Put the value into the formula
[tex]T=\dfrac{2\pi}{4\sqrt{2}}[/tex]
[tex]T=1.11\ sec[/tex]
Hence, (a). The value of spring constant is 4.
(b). The natural frequency is 4√2.
(c). The differential equation is [tex]\dfrac{d^2x}{dt^2}+32x=0[/tex]
(d). The solution for the position is [tex] x(t)=2\cos\sqrt{32t}[/tex]
(e). The time period is 1.11 sec.
A soft tennis ball is dropped onto a hard floor from a height of 1.50 m and rebounds to a height of 1.10 m. (a) Calculate its velocity just before it strikes the floor. (b) Calculate its velocity just after it leaves the floor on its way back up. (c) Calculate its acceleration during contact with the floor if that contact lasts 3.50 ms (3.50×10−3s). (d) How much did the ball compress during its collision with the floor, assuming the floor is absolutely rigid
Answer:
(a) v = 5.42m/s
(b) vo = 4.64m/s
(c) a = 2874.28m/s^2
(d) Δy = 5.11*10^-3m
Explanation:
(a) The velocity of the ball before it hits the floor is given by:
[tex]v=\sqrt{2gh}[/tex] (1)
g: gravitational acceleration = 9.8m/s^2
h: height where the ball falls down = 1.50m
[tex]v=\sqrt{2(9.8m/s^2)(1.50m)}=5.42\frac{m}{s}[/tex]
The speed of the ball is 5.42m/s
(b) To calculate the velocity of the ball, after it leaves the floor, you use the information of the maximum height reached by the ball after it leaves the floor.
You use the following formula:
[tex]h_{max}=\frac{v_o^2}{2g}[/tex] (2)
vo: velocity of the ball where it starts its motion upward
You solve for vo and replace the values of the parameters:
[tex]v_o=\sqrt{2gh_{max}}=\sqrt{2(9.8m/s^2)(1.10m)}=4.64\frac{m}{s}[/tex]
The velocity of the ball is 4.64m/s
(c) The acceleration is given by:
[tex]a=\frac{\Delta v}{\Delta t}=\frac{v_o-v}{3.50*10^{-3}s}=\frac{4.64m/s-(-5.42m)/s}{3.50*10^{-3}s}=2874.285\frac{m}{s^2}[/tex]
[tex]a=\frac{\Delta v}{\Delta t}=\frac{v_o-v}{3.50*10^{-3}s}=\frac{4.64m/s-5.42m/s}{3.50*10^{-3}s}=-222.85\frac{m}{s^2}[/tex]
The acceleration of the ball is 2874.28/s^2
(d) The compression of the ball is:
[tex]\Deta y=\frac{v^2}{2(a)}=\frac{(5.42m/s)^2}{2(2874.28m/s^2)}=5.11*10^{-3}m[/tex]
THe compression of the ball when it strikes the floor is 5.11*10^-3m
In a double-slit interference experiment you are asked to use laser light of different wavelengths and determine the separation between adjacent maxima. You observe that this separation is greatest when you illuminate the double slit with In a double-slit interference experiment you are asked to use laser light of different wavelengths and determine the separation between adjacent maxima. You observe that this separation is greatest when you illuminate the double slit with:_________.
1. yellow light.
2. red light.
3. blue light.
4. green light.
5. The separation is the same for all wavelengths.
Answer:
Red light
Explanation:
This because All interference or diffraction patterns depend upon the wavelength of the light (or whatever wave) involved. Red light has the longest wavelength (about 700 nm)
what are the rays that come of the sun called? A. Ultraviolent rays B. Gamma rays C. soundwaves D. sonic rays
Answer:
The answer is ultraviolet rays...
Explanation:
...because the ozone layer protects us from the UV rays of the sun.
A crate of mass 9.2 kg is pulled up a rough incline with an initial speed of 1.58 m/s. The pulling force is 110 N parallel to the incline, which makes an angle of 20.2° with the horizontal. The coefficient of kinetic friction is 0.400, and the crate is pulled 5.10 m.A) How much work is done by the gravitational force on thecrate?
B) Determine the increase in internal energy of the crate-inclinesystem owing to friction.
C) How much work is done by the 100N force on the crate?
D) What is the change in kinetic energy of the crate?
E) What is the speed of the crate after being pulled 5.00m?
Given that,
Mass = 9.2 kg
Force = 110 N
Angle = 20.2°
Distance = 5.10 m
Speed = 1.58 m/s
(A). We need to calculate the work done by the gravitational force
Using formula of work done
[tex]W_{g}=mgd\sin\theta[/tex]
Where, w = work
m = mass
g = acceleration due to gravity
d = distance
Put the value into the formula
[tex]W_{g}=9.2\times(-9.8)\times5.10\sin20.2[/tex]
[tex]W_{g}=-158.8\ J[/tex]
(B). We need to calculate the increase in internal energy of the crate-incline system owing to friction
Using formula of potential energy
[tex]\Delta U=-W[/tex]
Put the value into the formula
[tex]\Delta U=-(-158.8)\ J[/tex]
[tex]\Delta U=158.8\ J[/tex]
(C). We need to calculate the work done by 100 N force on the crate
Using formula of work done
[tex]W=F\times d[/tex]
Put the value into the formula
[tex]W=100\times5.10[/tex]
[tex]W=510\ J[/tex]
We need to calculate the work done by frictional force
Using formula of work done
[tex]W=-f\times d[/tex]
[tex]W=-\mu mg\cos\theta\times d[/tex]
Put the value into the formula
[tex]W=-0.4\times9.2\times9.8\cos20.2\times5.10[/tex]
[tex]W=-172.5\ J[/tex]
We need to calculate the change in kinetic energy of the crate
Using formula for change in kinetic energy
[tex]\Delta k=W_{g}+W_{f}+W_{F}[/tex]
Put the value into the formula
[tex]\Delta k=-158.8-172.5+510[/tex]
[tex]\Delta k=178.7\ J[/tex]
(E). We need to calculate the speed of the crate after being pulled 5.00m
Using formula of change in kinetic energy
[tex]\Delta k=\dfrac{1}{2}m(v_{2}^2-v_{1}^{2})[/tex]
[tex]v_{2}^2=\dfrac{2\times\Delta k}{m}+v_{1}^2[/tex]
Put the value into the formula
[tex]v_{2}^2=\dfrac{2\times178.7}{9.2}+1.58[/tex]
[tex]v_{2}=\sqrt{\dfrac{2\times178.7}{9.2}+1.58}[/tex]
[tex]v_{2}=6.35\ m/s[/tex]
Hence, (A). The work done by the gravitational force is -158.8 J.
(B). The increase in internal energy of the crate-incline system owing to friction is 158.8 J.
(C). The work done by 100 N force on the crate is 510 J.
(D). The change in kinetic energy of the crate is 178.7 J.
(E). The speed of the crate after being pulled 5.00m is 6.35 m/s
A parallel-plate capacitor with circular plates of radius R is being discharged. The displacement current through a central circular area, parallel to the plates and with radius R/2, is 9.2 A. What is the discharging current?
Answer:
The discharging current is [tex]I_d = 36.8 \ A[/tex]
Explanation:
From the question we are told that
The radius of each circular plates is R
The displacement current is [tex]I = 9.2 \ A[/tex]
The radius of the central circular area is [tex]\frac{R}{2}[/tex]
The discharging current is mathematically represented as
[tex]I_d = \frac{A}{k} * I[/tex]
where A is the area of each plate which is mathematically represented as
[tex]A = \pi R ^2[/tex]
and k is central circular area which is mathematically represented as
[tex]k = \pi [\frac{R}{2} ]^2[/tex]
So
[tex]I_d = \frac{\pi R^2 }{\pi * [ \frac{R}{2}]^2 } * I[/tex]
[tex]I_d = \frac{\pi R^2 }{\pi * \frac{R^2}{4} } * I[/tex]
[tex]I_d = 4 * I[/tex]
substituting values
[tex]I_d = 4 * 9.2[/tex]
[tex]I_d = 36.8 \ A[/tex]
A cylinder initially contains 1.5 kg of air at 100 kPa and 17 C. The air is compressed polytropically until the volume is reduced by one-half. The polytropic exponent is 1.3. Determine the work done (absolute value) and heat transfer for this process.
Answer:
Work done = 96.15 KJ
Heat transferred = 24 KJ
Explanation:
We are given;
Mass of air;m = 1.5 kg
Initial pressure;P1 = 100 KPa
Initial temperature;T1 = 17°C = 273 + 17 K = 290 K
Final volume;V2 = 0.5V1
Since the polytropic exponent is 1.3,thus it means;
P2 = P1[V1/V2]^(1.3)
So,P2 = 100(V1/0.5V1)^(1.3)
P2 = 100(2)^(1.3)
P2 = 246.2 KPa
To find the final temperature, we will make use of combined gas law.
So,
(P1×V1)/T1 = (P2×V2)/T2
T2 = (P2×V2×T1)/(P1×V1)
Plugging in the known values;
T2 = (246.2×0.5V1×290)/(100×V1)
V1 cancels out to give;
T2 = (246.2×0.5×290)/100
T2 = 356.99 K
The boundary work for this polytropic process is given by;
W_b = - ∫P. dv between the initial boundary and final boundary.
Thus,
W_b = - (P2.V2 - P1.V1)/(1 - n) = -mR(T2 - T1)/(1 - n)
R is gas constant for air = 0.28705 KPa.m³/Kg.K
n is the polytropic exponent which is 1.3
Thus;
W_b = -1.5 × 0.28705(356.99 - 290)/(1 - 1.3)
W_b = 96.15 KJ
The formula for the heat transfer is given as;
Q_out = W_b - m.c_v(T2 - T1)
c_v for air = 0.718 KJ/Kg.k
Q_out = 96.15 - (1.5×0.718(356.99 - 290))
Q_out = 24 KJ
A circular coil of wire of 200 turns and diameter 2.0 cm carries a current of 4.0 A. It is placed in a magnetic field of 0.70 T with the plane of the coil making an angle of 30° with the magnetic field. What is the magnetic torque on the coil?
Answer:
0.087976 Nm
Explanation:
The magnetic torque (τ) on a current-carrying loop in a magnetic field is given by;
τ = NIAB sinθ --------- (i)
Where;
N = number of turns of the loop
I = current in the loop
A = area of each of the turns
B = magnetic field
θ = angle the loop makes with the magnetic field
From the question;
N = 200
I = 4.0A
B = 0.70T
θ = 30°
A = π d² / 4 [d = diameter of the coil = 2.0cm = 0.02m]
A = π x 0.02² / 4 = 0.0003142m² [taking π = 3.142]
Substitute these values into equation (i) as follows;
τ = 200 x 4.0 x 0.0003142 x 0.70 sin30°
τ = 200 x 4.0 x 0.0003142 x 0.70 x 0.5
τ = 200 x 4.0 x 0.0003142 x 0.70
τ = 0.087976 Nm
Therefore, the torque on the coil is 0.087976 Nm
The average density of the body of a fish is 1080kg/m^3 . To keep from sinking, the fish increases its volume by inflating an internal air bladder, known as a swim bladder, with air.
By what percent must the fish increase its volume to be neutrally buoyant in fresh water? Use 1.28kg/m^3 for the density of air at 20 degrees Celsius. (change in V/V)
Answer:
Increase of volume (F) = 8.01%
Explanation:
Given:
Density of fish = 1,080 kg/m³
Density of water = 1,000 kg/m³
density of air = 1.28 kg/m³
Find:
Increase of volume (F)
Computation:
1,080 kg/m³ + [F × 1.28 kg/m³ ] = (1+F) × 1,000 kg/m³
1,080 + 1.28 F =1,000 F + 1,000
80 = 998.72 F
F = 0.0801 (Approx)
F = 8.01% (Approx)
A Ferris wheel starts at rest and builds up to a final angular speed of 0.70 rad/s while rotating through an angular displacement of 4.9 rad. What is its average angular acceleration
Answer:
The average angular acceleration is 0.05 radians per square second.
Explanation:
Let suppose that Ferris wheel accelerates at constant rate, the angular acceleration as a function of change in angular position and the squared final and initial angular velocities can be clear from the following expression:
[tex]\omega^{2} = \omega_{o}^{2} + 2 \cdot \alpha\cdot (\theta-\theta_{o})[/tex]
Where:
[tex]\omega_{o}[/tex], [tex]\omega[/tex] - Initial and final angular velocities, measured in radians per second.
[tex]\alpha[/tex] - Angular acceleration, measured in radians per square second.
[tex]\theta_{o}[/tex], [tex]\theta[/tex] - Initial and final angular position, measured in radians.
Then,
[tex]\alpha = \frac{\omega^{2}-\omega_{o}^{2}}{2\cdot (\theta-\theta_{o})}[/tex]
Given that [tex]\omega_{o} = 0\,\frac{rad}{s}[/tex], [tex]\omega = 0.70\,\frac{rad}{s}[/tex] and [tex]\theta-\theta_{o} = 4.9\,rad[/tex], the angular acceleration is:
[tex]\alpha = \frac{\left(0.70\,\frac{rad}{s} \right)^{2}-\left(0\,\frac{rad}{s} \right)^{2}}{2\cdot \left(4.9\,rad\right)}[/tex]
[tex]\alpha = 0.05\,\frac{rad}{s^{2}}[/tex]
Now, the time needed to accelerate the Ferris wheel uniformly is described by this kinematic equation:
[tex]\omega = \omega_{o} + \alpha \cdot t[/tex]
Where [tex]t[/tex] is the time measured in seconds.
The time is cleared and obtain after replacing every value:
[tex]t = \frac{\omega-\omega_{o}}{\alpha}[/tex]
If [tex]\omega_{o} = 0\,\frac{rad}{s}[/tex], [tex]\omega = 0.70\,\frac{rad}{s}[/tex] and [tex]\alpha = 0.05\,\frac{rad}{s^{2}}[/tex], the required time is:
[tex]t = \frac{0.70\,\frac{rad}{s} - 0\,\frac{rad}{s} }{0.05\,\frac{rad}{s^{2}} }[/tex]
[tex]t = 14\,s[/tex]
Average angular acceleration is obtained by dividing the difference between final and initial angular velocities by the time found in the previous step. That is:
[tex]\bar \alpha = \frac{\omega-\omega_{o}}{t}[/tex]
If [tex]\omega_{o} = 0\,\frac{rad}{s}[/tex], [tex]\omega = 0.70\,\frac{rad}{s}[/tex] and [tex]t = 14\,s[/tex], the average angular acceleration is:
[tex]\bar \alpha = \frac{0.70\,\frac{rad}{s} - 0\,\frac{rad}{s} }{14\,s}[/tex]
[tex]\bar \alpha = 0.05\,\frac{rad}{s^{2}}[/tex]
The average angular acceleration is 0.05 radians per square second.
Three sleds (30kg sled connected by tension rope B to 20kg sled connected by tension rope A to 10kg sled) are being pulled horizontally on frictionless horizontal ice using horizontal ropes. The pull is horizontal and of magnitude 143N . Required:a. Find the acceleration of the system. b. Find the tension in rope A. c. Find the tension in rope B.
Answer:
a) a = 2.383 m / s², b) T₂ = 120,617 N , c) T₃ = 72,957 N
Explanation:
This is an exercise of Newton's second law let's fix a horizontal frame of reference
in this case the mass of the sleds is 30, 20 10 kg from the last to the first, in the first the horizontal force is applied.
a) request the acceleration of the system
we can take the sledges together and write Newton's second law
T = (m₁ + m₂ + m₃) a
a = T / (m₁ + m₂ + m₃)
a = 143 / (10 +20 +30)
a = 2.383 m / s²
b) the tension of the cables we think through cable A between the sledges of 1 and 20 kg
on the sled of m₁ = 10 kg
T - T₂ = m₁ a
in this case T₂ is the cable tension
T₂ = T - m₁ a
T₂ = 143 - 10 2,383
T₂ = 120,617 N
c) The cable tension between the masses of 20 and 30 kg
T₂ - T₃ = m₂ a
T₃ = T₂ -m₂ a
T₃ = 120,617 - 20 2,383
T₃ = 72,957 N
The ability to distinguish between acceleration and velocity will be critical to your understanding of many other concepts in this course. Some of the most prevalent issues arise in interpreting the sign of both the velocity and acceleration of an object. I would recommend reading through the section "The Sign of the Acceleration" carefully. An object moves with a positive acceleration. Could the object be moving with increasing speed, decreasing speed or constant speed?
Answer:
Object could only be moving with increasing speed.
Explanation:
Let us consider the general formula of acceleration:
a = (Vf - Vi)/t
Vf = Vi + at -------- equation 1
where,
Vf = Final Velocity
Vi = Initial Velocity
a = acceleration
t = time
FOR POSITIVE ACCELERATION:
Vf = Vi + at
since, both acceleration and time are positive quantities. Hence, it means that the final velocity of the object shall be greater than the initial velocity of the object.
Vf > Vi
It clearly shows that if an object moves with positive acceleration. It could only be moving with increasing speed.
Solving the same equation for negative acceleration shows that the final velocity will be less than initial velocity and object will be moving with decreasing speed.
And for the constant velocity final and initial velocities are equal and thus, acceleration will be zero.