Answer:decreases by 30.55%
Explanation:
Given
length of wire is increased by 20 % keeping volume constant
Let the length of wire be L and its area of cross section be A
Thus new length=1.2 L
Volume is constant
[tex]AL=1.2 L\times A'[/tex]
A'=0.833 A
and resistance is given by
[tex]R=\frac{\rho L}{A}[/tex]
where [tex]\rho [/tex]=resistivity
New resistance [tex]R'=\frac{\rho\times 1.2L}{0.833A}[/tex]
R'=1.44 R
heat produced for same potential
[tex]H_1=\frac{V^2t}{R}[/tex]
[tex]H_2=\frac{V^2t}{1.44R}=0.694H_1[/tex]
% change in heat
[tex]\frac{H_2-H_1}{H_1}\times 100[/tex]
[tex]=\frac{0.694-1}{1}[/tex]
=30.55 decreases
Answer:
30.55 %
Explanation:
Assumptions:
l = initial length of the wireL = final length of the wirev = initial volume of the wireV = final volume of the wirea = initial cross sectional area of the wireA = final cross sectional area of the wireh = initial heat of generated by the wireH = final heat generated by the wireP = potential difference across the wiret = time for which the potential difference is created across the wirer = initial resistance of the wireR = final resistance of the wire[tex]\Delta H[/tex] = change in heat producedAccording to the question, we have
[tex]L = l + 20\ \% l = \dfrac{120l}{100}\\V=v\\\Rightarrow LA=la\\\Rightarrow A= \dfrac{la}{L}\\\Rightarrow A= \dfrac{la}{\dfrac{120l}{100}}\\\Rightarrow A= \dfrac{100a}{120}[/tex]
Using the formula of resistance of a wire in terms of its length, cross sectional area and the resistivity of the material, we have
[tex]r = \dfrac{\rho l}{a}\\R=\dfrac{\rho L}{A}=\dfrac{\rho\times \dfrac{120l}{100} }{\dfrac{100a}{120}}=(\dfrac{120}{100})^2\dfrac{\rho l}{a}= 1.44r\\[/tex]
Using the formula of heat generated by the wire for potential diofference created across its end for time t, we have
[tex]h = \dfrac{P^2}{r}t\\H = \dfrac{P^2}{R}t= \dfrac{P^2}{1.44r}t\\\therefore \Delta H = h-H\\\Rightarrow \Delta H = \dfrac{P^2}{r}t-\dfrac{P^2}{1.44r}t\\\Rightarrow \Delta H = \dfrac{P^2t}{r}(-\dfrac{1}{1.44})\\\Rightarrow \Delta H = \dfrac{P^2t}{r}(\dfrac{0.44}{1.44})\\\therefore \textrm{Percentage change in the heat produced}= \dfrac{\Delta H}{h}\times 100\ \%= \left (\dfrac{\dfrac{P^2t}{r}(\dfrac{0.44}{1.44})}{\dfrac{P^2}{r}t} \right )\times 100\ \% = 30.55\ \%[/tex]
Hence, the percentage change in the heat produced in the wire is 30.55 %.
Box 1 and box 2 are whirling around a shaft with a constant angular velocity of magnitude ω. Box 1 is at a distance d from the central axis, and box 2 is at a distance 2d from the axis. You may ignore the mass of the strings and neglect the effect of gravity. Express your answer in terms of d, ω, m1 and m2, the masses of box 1 and 2. (a) Calculate TB, the tension in string B (the string connecting box 1 and box 2). (b) Calculate TA, the tension in string A (the string connecting box 1 and the shaft).
Answer:
a) TB = m2 * w^2 * 2*d
b) TA = m1 * w^2 * d + m2 * w^2 * 2*d
Explanation:
The tension on the strings will be equal to the centripetal force acting on the boxes.
The centripetal force is related to the centripetal acceleration:
f = m * a
The centripetal acceleration is related to the radius of rotation and the tangential speed:
a = v^2 / d
f = m * v^2 / d
The tangential speed is:
v = w * d
Then
f = m * w^2 * d
For the string connecting boxes 1 and 2:
TB = m2 * w^2 * 2*d
For the string connecting box 1 to the shaft
TA = m1 * w^2 * d + m2 * w^2 * 2*d
At cruise conditions, air flows into a jet engine at a steady rate of 60 lbm/s. Fuel enters the engine at a steady rate of 0.59 lbm/s. The average velocity of the exhaust gases is 1485 ft/s relative to the engine. If the engine exhaust effective cross-sectional area is 4.0 ft2, estimate the density of the exhaust gases in lbm/ft3.
Answer:
ρ=0.0102lbm/ft^3
Explanation:
To solve this problem we must take into account the equation of continuity, this indicates that the sum of the mass flows that enter a system is equal to the sum of all those that leave.
Therefore, to find the mass flow of exhaust gases we must add the mass flows of air and fuel.
m=0.59+60=60.59lbm/s( mass flow of exhaust gases)
The equation that defines the mass flow (amount of mass that passes through a pipe per unit of time) is as follows
m=ρVA
Where
ρ=density
V=velocity
m=mass flow
A=cross-sectional area
solving for density
ρ=m/VA
ρ=60.59/{(1485)(4)}
ρ=0.0102lbm/ft^3
A 0.10- kg ball is thrown straight up into the air with
aninitial speed of 15m/s. Find the momentum of the ball (a) at
itsmaximum height and (b) halfway to its maximum height.
Answer:(a)0,(b)1.061 kg-m/s
Explanation:
Given
mass of ball is 0.10 kg
Initial speed is 15 m/s
Maximum height reached by ball is h
[tex]v^2-u^2=2as[/tex]
final velocity =0
[tex]-\left ( 15\right )^2=-2\times 9.81\times s[/tex]
[tex]s=\frac{15^2}{2\times 9.81}=11.467 m[/tex]
thus momentum of ball at maximum height is 0 as velocity is zero
For halfway to maximum height
[tex]v^2-u^2=2as_0[/tex]
where [tex]s_0=\frac{11.467}{2}[/tex]
[tex]s_0=5.73 m[/tex]
[tex]v^2=15^2-2\times 9.81\times 5.73[/tex]
v=10.61 m/s
Thus its momentum is
[tex]mv=0.10\times 10.61=1.061 kg-m/s[/tex]
Part A A conducting sphere is charged up such that the potential on its surface is 100 V (relative to infinity). If the sphere's radius were twice as large, but the charge on the sphere were the same, what would be the potential on the surface relative to infinity?
Answer:
[tex]V_{2}=\frac{V_{1}}{2}[/tex]
Explanation:
The potential of a conducting sphere is the same as a punctual charge,
First sphere:
[tex]V_{1}=k*q_{1}/r_{1}[/tex] (1)
Second sphere:
[tex]V_{2}=k*q_{2}/r_{2}[/tex] (2)
But, second sphere's radius is twice first sphere radius, and their charges the same:
[tex]q_{1}=q_{2}[/tex]
[tex]r_{2}=2*r_{1}[/tex]
If we divide the equations (1) and (2), to solve V2:
[tex]V_{2}=V_{1}*r_{1}/r_{2}=V_{1}/2[/tex]
Suppose that a steel bridge, 1000 m long, were built without any expansion joints. Suppose that only one end of the bridge was held fixed. What would the difference in the length of the bridge be between winter and summer, taking a typical winter temperature as 0°C, and a typical summer temperature as 40°C? The coefficient of thermal expansion of steel is 10.5 × 10 -6 K -1. 0.37 cm 0.11 mm 0.42 m 0.42 mm 0.11 m
Answer:
0.42 m
Explanation:
For thermal expansion, the formula used is as follows
L = L₀ ( 1 + α t )
L is length after rise of temperature by t , L₀ is length before rise of temperature , α is coefficient of thermal expansion and t is rise in temperature.
L - L₀ = L₀α t
Difference of length = L₀α t
L₀ = 1000 m , α = 10.5 x 10⁻⁶ , t = 40
Difference of length = 1000 x 10.5 x 10⁻⁶x 40
= 0.42 m .
Two blocks of masses 20 kg and 8.0 kg are connected togetherby
a light string and rest on a frictionless level surface.Attached to
the 8-kg mass is a second light string, which a personuses to pull
both blocks horizontally. If the two blocks systemacdereates at 0.5
m/s2, what is the tension in the second stringattached to the 8-kg
mass?
Answer:
14 N
Explanation:
The tension in the second string is puling both the masses of 20 kg and 8 kg with acceleration of 0.5 m s⁻²
So tension in the second string = total mass x acceleration
= 28 x .5 = 14 N . Ans..
The atomic radii of a divalent cation and a monovalent anion are 0.19 nm and 0.126 nm, respectively. (a) Calculate the force of attraction between these two ions at their equilibrium interionic separation (i.e., when the ions just touch one another). Enter your answer for part (a) in accordance to the question statement N (b) What is the force of repulsion at this same separation distance
Final answer:
The force of attraction between the divalent cation and monovalent anion can be calculated using Coulomb's law. Plugging the values into the formula gives a force of -1.742 N.
Explanation:
The force of attraction between two ions can be calculated using Coulomb's law. The formula is given by F = k * (Q1 * Q2) / r^2, where F is the force, k is the Coulomb constant (9 x 10^9 Nm^2/C^2), Q1 and Q2 are the charges of the ions, and r is the distance between their centers. In this case, since the anion is monovalent and the cation is divalent, the charges would be -1 and +2 respectively.
To calculate the force of attraction, we need to find the equilibrium interionic separation. Given that the atomic radii of the divalent cation and monovalent anion are 0.19 nm and 0.126 nm respectively, their total separation would be the sum of their radii, which is 0.316 nm.
Plugging these values into the formula, we get:
F = (9 x 10^9 Nm^2/C^2) * (-1 C) * (+2 C) / (0.316 x 10^-9)^2 = -1.742 N
A house is 54.0 ft long and 48 ft wide, and has 8.0-ft-high ceilings. What is the volume of the interior of the house in cubic meters and cubic centimeters? Answers needs to be in appropriate significant figures.
Explanation:
Length of house, l = 54 ft = 16.45 m
Breadth of house, b = 48 ft = 14.63 m
Height of house, h = 8 ft = 2.43 m
We need to find the volume of the interior of the house. The house is in the form of cuboid. The volume of cuboid is given by :
[tex]V=l\times b\times h[/tex]
[tex]V=16.45\ m\times 14.63\ m\times 2.43\ m[/tex]
[tex]V=584.81\ m^3[/tex]
Since, [tex]1\ m^3=1000000\ cm^3[/tex]
or the volume of the interior of the house, [tex]V=5.84\times 10^8\ cm^3[/tex]
Hence, this is the required solution.
A galvanometer has an internal resistance of 100 Ω and deflects full-scale at 2.00 mA. What size resistor should be added to the galvanometer to convert it to a milliammeter capable of reading up to 4.00 mA, and how should this resistor be connected to the galvanometer?
Answer
The resistor has to be 100
Explanation:
We will have to use the Current Divider Rule, that rule states:
[tex]Ig=\frac{Req}{Rg}*(It)\\[/tex]
where:
Ig= Galvanometer current
It= Total current
Rg= Galvanometer Resistor
Req= Equivalent circuit resistor
For the case of two resistor in parallel:
[tex]Req=\frac{R1*Rg}{R1+Rg}[/tex]
now:
[tex]Req=\frac{2mA}{4mA}*100\\[/tex]
Req=50Ω
having the Equivalent resistor we can calculate R1 reformulating the Req formula:
[tex]R1=\frac{(Rg*Req)}{Rg-Req}\\[/tex]
R1=100 Ω
So now when a 4mA current flows into the new circuit, 2mA will go through the Galvanometer deflecting the full scale.
To convert the galvanometer to a milliammeter capable of reading up to 4.00 mA, add a 100 Ω shunt resistor in parallel.
To convert a galvanometer with an internal resistance of 100 Ω and a full-scale deflection of 2.00 mA into a milliammeter capable of reading up to 4.00 mA, you need to add a shunt resistor parallel to the galvanometer.
The shunt resistor will divert the excess current, allowing the galvanometer to properly read the increased current range.
Given values:Calculate the voltage corresponding to the galvanometer's full-scale deflection:
V = I * RV = 2.00 mA * 100 Ω V = 0.2 V.Determine the current that will flow through the shunt resistor:
4.00 mA - 2.00 mA = 2.00 mA.Calculate the resistance of the shunt resistor (Rsh) using Ohm's law:
Rsh = V / I = 0.2 V / 2.00 mA = 100 Ω.Therefore, a resistor of 100 Ω should be connected in parallel to the galvanometer to enable it to read up to 4.00 mA.
What must the charge (sign and magnitude) of a particle of mass 1.47 g be for it to remain stationary when placed in a downward-directed electric field of magnitude 680 N/C ? Use 9.81 m/ s 2 for the magnitude of the acceleration due to gravity. View Available Hint(s) nothing C C Submit Part B What is the magnitude of an electric field in which the electric force on a proton is equal in magnitude to its weight? Use 1.67×10−27 kg for the mass of a proton, 1.60×10−19 C for the magnitude of the charge on an electron, and 9.81 m/ s 2 for the magnitude of the acceleration due to gravity. View Available Hint(s) nothing N/C N / C Submit Provide Feedback Next
Answer:
Part a)
[tex]q = -21.2 \times 10^{-6} C[/tex]
Part b)
[tex]E = 1.02 \times 10^{-7} N/C[/tex]
Explanation:
Part a)
As we know that charge and its sign that remains in equilibrium is under gravity must be such that it will balance the gravitational force by electric force
so here we have
[tex]mg = qE[/tex]
[tex]1.47 \times 10^{-3} (9.81) = q(680)[/tex]
[tex]q = 21.2 \times 10^{-6} C[/tex]
and its sign must be negative so that it will have upward electric force
so it is
[tex]q = -21.2 \times 10^{-6} C[/tex]
Part b)
as we know that weight of proton is balanced by electric force
so we will have
[tex]qE = mg[/tex]
[tex](1.6 \times 10^{-19})E = (1.67 \times 10^{-27})(9.81)[/tex]
[tex]E = 1.02 \times 10^{-7} N/C[/tex]
A. The charge (sign and magnitude) of the particle of mass 1.47 g is –2.12×10¯⁵ C
B. The magnitude of the electric field in which the electric force on the proton is equal in magnitude to its weight is 1.02×10¯⁷ N/C
What is electric field strength?This is simply defined as the force per unit charge. Mathematically, it is expressed as:
E = F / Q
Where
E is the electric field strength F is the force Q is the charge A. How to determine the charge Mass (m) = 1.47 g = 1.47 / 1000 = 1.47×10¯³ KgElectric field strength (E) = 680 N/CAcceleration due to gravity (g) = 9.81 m/s²Charge (Q) =?E = F/Q
E = mg / Q
680 = (1.47×10¯³ × 9.81) / Q
Cross multiply
680 × Q = (1.47×10¯³ × 9.81)
Divide both side by 680
Q = (1.47×10¯³ × 9.81) / 680
Q = 2.12×10¯⁵ C
Q = –2.12×10¯⁵ C (since it is stationary)
B. How to determine the electric field strength Mass (m) = 1.67×10¯²⁷ KgAcceleration due to gravity (g) = 9.81 m/s²Charge (Q) = 1.60×10¯¹⁹ CElectric field strength (E) =?E = mg / Q
E = (1.67×10¯²⁷ × 9.81) / 1.60×10¯¹⁹
E = 1.02×10¯⁷ N/C
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A rocket sled for testing equipment under large accelerations starts at rest and accelerates according to the expression a = (3.6 m/s 3 )t + (5.6 m/s 2 ). How far does the sled move in the time interval t = 0 to t = 1.6 s ? Answer in units of m.
Answer:
9.6 m
Explanation:
This is a case of motion under variable acceleration . So no law of motion formula will be applicable here. We shall have to integrate the given equation .
a = 3.6 t + 5.6
d²x / dt² = 3.6 t + 5.6
Integrating on both sides
dx /dt = 3.6 t² / 2 + 5.6 t + c
where c is a constant.
dx /dt = 1.8 t² + 5.6 t + c
when t = 0 , velocity dx /dt is zero
Putting these values in the equation above
0 = 0 +0 + c
c = 0
dx /dt = 1.8 t² + 5.6 t
Again integrating on both sides
x = 1.8 t³ / 3 + 5.6 x t² /2 + c₁
x = 0.6 t³ + 2.8 t² + c₁
when t =0, x = 0
c₁ = 0
x = 0.6 t³ + 2.8 t²
when t = 1.6
x = .6 x 1.6³ + 2.8 x 1.6²
= 2.4576 + 7.168
= 9.6256
9.6 m
The sled moves approximately 16.896 meters from t=0 s to t=1.6 s when it accelerates according to a = (3.6 m/s 3 )t + (5.6 m/s 2 )
Explanation:The calculation for the distance moved by the sled involves using the equations of motion. For a sled that starts from rest and accelerates according to the function a = (3.6 m/s³) t + (5.6 m/s²), we first find the velocity function by integrating the acceleration function. That is, v(t) = ∫a dt = ∫[(3.6 m/s³)t + (5.6 m/s²)] dt = (1.8 m/s²) t² + (5.6 m/s) t + C. Given that the sled starts from rest, the constant C is zero. Therefore the velocity function is v(t) = (1.8 m/s²) t² + (5.6 m/s) t.
Next, we find the displacement function by integrating the velocity function. That is, x(t) = ∫v dt = ∫[(1.8 m/s²) t² + (5.6 m/s) t] dt = (0.6 m/s) t³ + (2.8 m/s) t² + D. Given the initial condition that the sled starts from rest, the constant D is also zero. So the displacement function is x(t) = (0.6 m/s) t³ + (2.8 m/s) t².
Finally, substituting t = 1.6 s into the displacement function will yield the distance moved by the sled in the time interval from t = 0 to t = 1.6 s, which comes out to be x(1.6 s) = (0.6 m/s) (1.6 s)³ + (2.8 m/s) (1.6 s)² = 9.728 m + 7.168 m = 16.896 m. So, the sled moves approximately 16.896 m.
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Two objects attract each other with a gravitational force
ofmagnitude 1.00 x 10-8 N when separated by 20.0 cm.
Ifthe total mass of the two objects is 5.00 kg, what is the mass
ofeach?
Answer:
mass of bigger object M= 3 kg and mass of smaller object m= 2 kg
Explanation:
force between two objects F= 1.00×10^-8 N
distance between them R= 20 cm = 0.2 m
total mass M+m= 5 kg
We know that
[tex]F= G\frac{Mm}{R^2}[/tex]
putting the values we get
[tex]10^{-8}= 6.67\times10^{-11}\frac{M\times m}{0.2^2}[/tex]
M×m= 6
and M+m= 5
therefore we can calculate that M= 3 kg and m= 2 kg
mass of bigger object M= 3 kg and mass of smaller object m= 2 kg
If AB has a bearing of following 234° 51' 48" and a anti-clockwise angle from AB to C is measured as 80° Calculate the bearing AC. (Enter as numeric value of ddd.mmss e.g. 100° 20' 30" would be entered as 100.2030. Marked out of 5.00 P
Answer:
the bearing of the line AC will be 154° 51' 48"
Explanation:
given,
bearing of the line AB = 234° 51' 48"
an anticlockwise measure of an angle to the point C measured 80°
to calculate the bearing of AC.
As the bearing of line AB is calculated clockwise from North direction.
the angle is moved anticlockwise now the bearing of AC will be calculated by
= bearing of line AB - 80°
= 234° 51' 48" - 80°
= 154° 51' 48".
= 154.5148
so, the bearing of the line AC will be 154° 51' 48"
At some point as you are hiking near a lake, you determine that your campsite is 1.50 km away from you in the direction 30.0° E of N. However, to get back to your campsite, you will need to walk around the lake. You set off due north and walk for 600 m. You then turn in the direction 20.0 W of N and walk an addol1.20 km, before turning and walking directly to your campsite. How far and in what direction was the last leg of your hike?
Answer:
[tex]r_2 = 976.65 m[/tex]
Direction is 19 degree South of East
Explanation:
Let say initial position is our reference
so we will have campsite position given as
[tex]r = 1.50 km[/tex] at 20 degree E of N
now we will have
[tex]r = 1500 sin20\hat i + 1500 cos20\hat j[/tex]
[tex]r = 513 \hat i + 1409.5\hat j[/tex]
now our displacement to walk around is given as
[tex]d_1 = 600 \hat j[/tex]
then we move 20 degree W of N and move 1200 m
so we will have
[tex]d_2 = 1200 sin20(-\hat i) + 1200cos20\hat j[/tex]
so our final position is given as
[tex]r_1 = d_1 + d_2[/tex]
[tex]r_1 = 600\hat j - 410.4 \hat i + 1127.6\hat j[/tex]
[tex]r_1 = -410.4 \hat i + 1727.6\hat j[/tex]
now we know that
[tex]r_1 + r_2 = r[/tex]
so final leg of the displacement is given as
[tex]r_2 = r - r_1[/tex]
[tex]r_2 = (513 \hat i + 1409.5\hat j) - (-410.4 \hat i + 1727.6\hat j)[/tex]
[tex]r_2 = 923.4\hat i - 318.1 \hat i[/tex]
so magnitude is given as
[tex]r_2 = \sqrt{923.4^2 + 318.1^2}[/tex]
[tex]r_2 = 976.65 m[/tex]
direction is given as
[tex]\theta = tan^{-1}\frac{y}{x}[/tex]
[tex]\theta = tan^{-1}\frac{-318.1}{923.4}[/tex]
[tex]\theta = -19 degree[/tex]
so it is 19 degree South of East
Consider the following situations:
A car is moving along a straight road at a constant speed.
A car is moving along a straight road while slowing down.
A car is moving along a straight road while speeding up.
A hockey puck slides along a horizontal icy (frictionless) surface.
A hockey puck slides along a rough concrete surface.
A cockroach is speeding up from rest.
A rock is thrown horizontally; air resistance is negligible.
A rock is thrown horizontally; air resistance is substantial.
A rock is dropped vertically; air resistance is negligible.
A rock is dropped vertically; air resistance is substantial.
Part H
Which of these situations describe the motion shown in the motion diagram at point A?
Type the letters corresponding to all the right answers in alphabetical order. Do not use commas. For instance, if you think that only situations C and D are correct, type CD.
Answer:
AD
Explanation:
A motion diagram is used to represent the motion of an object. Each point represents the location of the object after equal intervals of time, and the velocity of the object is represented using arrows.
The motion diagram for this problem has been attached here. The question is asking us to analyze the motion at point A. We can make the following observations for the motion diagram at point A:
- The direction of the arrows does not change --> this means that the object is moving on a straight line
- The length of the arrows remains constant at A --> this means that the object is moving at constant speed
Of all the situations listed in the question, only two of them are compatible with these observations:
A) A car is moving along a straight road at a constant speed.
D) A hockey puck slides along a horizontal icy (frictionless) surface. (Frictionless surface means there is no friction acting on the puck, so the acceleration is zero and therefore the speed is constant).
All the other options involves a non-zero acceleration, so that the velocity is not constant. Therefore, only A and D are the correct options.
Write down the equation which describes the simple harmonic motion of a 0.4 kg mass on a spring with a spring constant k = 100 N/m, that starts its motion (at t = 0 s) at its maximum positive displacement of +0.5 m.
Answer:
The equation which describes the simple harmonic motion is [tex]x=0.5\cos(15.81t)[/tex]
Explanation:
Given that,
Mass = 0.4 kg
Spring constant = 100 N/m
Maximum displacement = 0.5 m
We need to calculate the angular frequency
[tex]\omega=\sqrt{\dfrac{k}{m}}[/tex]
[tex]\omega=\sqrt{\dfrac{100}{0.4}}[/tex]
We need to find the equation which describes the simple harmonic motion
Using equation of simple harmonic motion
[tex]x = A\cos\omega t[/tex]
Where, A = amplitude
[tex]\omega [/tex] = angular frequency
Put the value of angular frequency
[tex]x=A\cos\sqrt{\dfrac{k}{m}t}[/tex]
Put the value in the equation
[tex]x=0.5\cos\sqrt{\dfrac{100}{0.4}t}[/tex]
[tex]x=0.5\cos(15.81t)[/tex]
Hence, The equation which describes the simple harmonic motion is [tex]x=0.5\cos(15.81t)[/tex]
The alternator in a car consists of a rectangular coil, with 250 turns of wire and area 0.01 m^2, rotating in a 0.1 T magnetic field. (The field is produced by a direct-current electromagnet.) If the rotation rate is 10^3 rpm, what is the peak output voltage?
Answer:
26.17 V
Explanation:
In a alternator , an ac voltage is produced whose magnitude of voltage varies sinusoidally.
Maximum voltage induced = nBAω
where n is no of turns , B is magnetic field , A is area of the coil and ω is angular velocity .
Rate of rotation n = 1000rpm = 1000/60 rps = 16.67 rps
angular velocity ω = 2π n = 2π x 16.67 = 104.68 rad / s
Putting all values in the given equation above
Max voltage = 250 x .1 x .01 x 104.68
= 26.17 V
A light ray is incident on a plane surface separating two transparent media of refractive indices of 1.65 and 1.35. If the incident angle is 35º and the light ray originates in the medium of higher refractive index. (a) Compute the angle of refraction. (b) Repeat part (a), assuming that the light ray originates in the medium of lower refractive index.
Explanation:
Given
Refractive Index [tex](n_1)=1.65[/tex]
[tex]n_2=1.35[/tex]
angle of incident[tex]=35 ^{\circ}[/tex]
Light originates in the medium of higher refractive index
Let [tex]\theta _1[/tex] and [tex]\theta _2 [/tex]be the incident and refraction angles respectively thus according to snell's law
[tex]n_1sin\theta _1=n_2sin\theta _2[/tex]
[tex]1.65\times sin35=1.35\times sin\theta _2[/tex]
[tex]sin\theta _2=0.701[/tex]
[tex]\theta _2=44.507^{\circ}[/tex]
(b)Ray originates from lighter medium
[tex]1.35\times sin35=1.65\times sin\theta _2'[/tex]
[tex]sin\theta _2'=0.469[/tex]
[tex]\theta _2'=27.96\approx 28^{\circ}[/tex]
A 77.0 kg ice hockey goalie, originally at rest, catches a 0.150 kg hockey puck slapped at him at a velocity of 22.0 m/s. Suppose the goalie and the ice puck have an elastic collision and the puck is reflected back in the direction from which it came. What would their final velocities (in m/s) be in this case? (Assume the original direction of the ice puck toward the goalie is in the positive direction. Indicate the direction with the sign of your answer.)
Answer:
Explanation:
For elestic collision
v₁ = [tex]\frac{(m_1-m_2)u_1}{m_1+m_2} +\frac{2m_2u_2}{m_1+m_2}[/tex]
[tex]v_2 = [tex]\frac{(m_2-m_1)u_2}{m_1+m_2} +\frac{2m_1u_1}{m_1+m_2}[/tex][/tex]
Here u₁ = 0 , u₂ = 22 m/s , m₁ = 77 kg , m₂ = .15 kg , v₁ and v₂ are velocity of goalie and puck after the collision.
v₁ = 0 + ( 2 x .15 x22 )/ 77.15
= .085 m / s
Velocity of goalie will be .085 m/s in the direction of original velocity of ball before collision.
v₂ = (.15 - 77)x 22 / 77.15 +0
= - 21.91 m /s
=Velocity of puck will be - 21.91 m /s in the direction opposite to original velocity of ball before collision.
A student aims a cork gun at a small dot on the wall. The wall is 10m from the students, and 8m above the level of the gun. The gun can launch corks with a speed of 12m/s. How high must the gun be aimed to hit the spot?
Answer:[tex]\theta =57.99\approx 58 ^{\circ}[/tex]
Explanation:
Given
wall is 10 m from student
8 m high
Therefore student should launch at an angle \theta such that its maximum height is 8 m and its range is 20 m
For maximum height(h)[tex]=\frac{u^2sin^2\theta }{2g}[/tex]
Range (R)[tex]=\frac{u^2sin2\theta }{g}[/tex]
[tex]8=\frac{12^2sin^2\theta }{2g}-----1[/tex]
[tex]20=\frac{12^2sin2\theta }{g}----2[/tex]
divide 1 & 2
[tex]\frac{20}{8}=\frac{2sin2\theta }{sin^2\theta }[/tex]
[tex]\frac{5}{2}=\frac{2\times 2\times sin\theat \cdot cos\theta }{sin^2\theta }[/tex]
[tex]tan\theta =\frac{8}{5}[/tex]
[tex]\theta =57.99\approx 58 ^{\circ}[/tex]
A charge Q= 2 C is distributed uniformly through out a bar of length L=2.5 m. The bar is placed horizontally in free space. A second charge q = 10−9C is placed along the line of the bar a distance d= 2m away in space, measured from the right end of the bar. What force is exerted on charge q by the charged bar?
Answer:
F = 2 N
Explanation:
Let the rod is made up of large number of point charges
so here force due to one small part which is at distance "x" from the end of the rod on the given charge is
[tex]dF = \frac{kdq q}{(d + x)^2}[/tex]
here we know that
[tex]dq = \frac{Q}{L} dx[/tex]
so we have
[tex]F = \int \frac{k Q q dx}{L(d + x)^2}[/tex]
[tex]F = \int_0^L \frac{kQq dx}{L(d + x)^2}[/tex]
[tex]F = \frac{kQq}{L} ( \frac{1}{d} - \frac{1}{d + L})[/tex]
now plug in all data
[tex]F = \frac{(9\times 10^9)(2C)(10^{-9})}{2.5} (\frac{1}{2} - \frac{1}{(2 + 2.5)})[/tex]
[tex]F = 2N[/tex]
An elevator ascends from the ground with uniform speed. A time Tį later, a boy drops a marble through a hole in the floor. A time T2 after that (i.e. Ti +T2 after start) the marble hits the ground. Find an expression for the height of the elevator at time Ti. (Local gravity is g.) What checks can you make?
Answer:
The height of the elevator at [tex]T_{i}[/tex] is [tex]\frac{gT_{2}^{2}}{2(1 + \frac{T_{2}}{T_{i}})}[/tex]
Solution:
As per the question:
Let us assume:
The velocity with which the elevator ascends be u'
The height attained by the elevator at time, [tex]T_{i}[/tex] be h
Thus
[tex]u' = \frac{h}{T_{i}}[/tex] (1)_
Now, with the help of eqn (2) of motion, we can write:
[tex]h = - u'T_{2} + \frac{1}{2}gT_{2}^{2}[/tex]
Using eqn (1):
[tex]h = - \frac{h}{T_{i}}T_{2} + \frac{1}{2}gT_{2}^{2}[/tex]
[tex]h + \frac{h}{T_{i}}T_{2} = \frac{1}{2}gT_{2}^{2}[/tex]
[tex]h(1 + \frac{h}{T_{i}}T_{2}) = \frac{1}{2}gT_{2}^{2}[/tex]
[tex]h = \frac{gT_{2}^{2}}{2(1 + \frac{T_{2}}{T_{1}})}[/tex]
A hawk flying at 38 m/s emits a cry whose frequency is 440 Hz. A wren is moving in the same direction as the hawk at 17 m/s. (Assume the speed of sound is 343 m/s.) (a) What frequency does the wren hear (in Hz) as the hawk approaches the wren? Hz (b) What frequency does the wren hear (in Hz) after the hawk passes the wren?
Answer:
frequency wren hear when approaches is 470.29 Hz
frequency wren hear after hawk pass is 415.75 Hz
Explanation:
given data
hawk velocity Vs= 38 m/s
frequency f = 440 Hz
wren velocity Vo= 17 m/s
speed of sound s = 343 m/s
to find out
What frequency wren hear and What frequency wren hear after hawk pass
solution
we apply here frequency formula that is
[tex]f1=f ( \frac{s+V_o}{s+V_s} )[/tex] .........1
here f1 is frequency hear by observer
put here all value as Vo and Vs negative because it approaches
[tex]f1=f ( \frac{s-V_o}{s-V_s} )[/tex]
[tex]f1=440 ( \frac{343-17}{343-38} )[/tex]
f1 = 470.29 Hz
so frequency wren hear when approaches is 470.29 Hz
and
after passing from equation 1 we take both Vo and Vs as positive
[tex]f1=f ( \frac{s+V_o}{s+V_s} )[/tex]
[tex]f1=440 ( \frac{343+17}{343+38} )[/tex]
f1 = 415.75 Hz
so frequency wren hear after hawk pass is 415.75 Hz
Determine (a) how many seconds there are in 1.00 year? (b) Using your answer from part (a), how many years are there in 1.00 second?
Answer:
a) 31557600 seconds b) 3.168x10⁻⁸ years
Explanation:
1 hour is equivalent to 3600 seconds, that multiplied by 24 hours (1 day) gives 86.400 seconds
a) how many seconds there are in 1.00 year?
[tex]\frac{86400 s}{1 day} x 365,25 days[/tex] ⇒ 31557600 s
So in 1 year there are a total of 31557600 seconds.
b) how many years are there in 1.00 second?
Using the quantity from part a) it is get:
[tex]1 s x \frac{1 year}{31557600 s}[/tex] ⇒ 3.168x10⁻⁸ years
So in 1 second there are a total of 3.168x10⁻⁸ years.
A sphere of radius r = 5cm carries a uniform volume charge density rho = 400 nC/m^3. Q. What is the total charge Q of the sphere?
Answer:
The total charge Q of the sphere is [tex]2.094\times10^{-10}\ C[/tex].
Explanation:
Given that,
Radius = 5 cm
Charge density [tex]J= 400\ nC/m^3[/tex]
We need to calculate the total charge Q of the sphere
Using formula of charge
[tex]q=\rho V[/tex]
Where, [tex]\rho[/tex] = charge density
V = volume
Put the value into the formula
[tex]q=\rho\times(\dfrac{4}{3}\pi r^3)[/tex]
Put the value into the formula
[tex]q=\dfrac{4}{3}\times\pi\times400\times10^{-9}\times(5\times10^{-2})^3[/tex]
[tex]q=2.094\times10^{-10}\ C[/tex]
Hence, The total charge Q of the sphere is [tex]2.094\times10^{-10}\ C[/tex].
A student witnesses a flash of lightning and then t=2.5s later the student hears assiciated clap of thunder. (please show work) (Part A) Sound travels at 343 m?s in the air, What distance from the student is the lightning strikes, in meters? (Part B) Light travels at 3.0x108 m/s in the air. How long, t1, in seconds did it take the light to reach the student's eyes after the flash?
Answer:
857.5 m
2.8583×10⁻⁶ seconds
Explanation:
Time taken by the sound of the thunder to reach the student = 2.5 s
Speed of sound in air is 343 m/s
Speed of light is 3×10⁸ m/s
Distance travelled by the sound = Time taken by the sound × Speed of sound in air
⇒Distance travelled by the sound = 2.5×343 = 857.5 m
⇒Distance travelled by the sound = 857.5 m
Time taken by light = Distance the light travelled / Speed of light
[tex]\text{Time taken by light}=\frac{857.5}{3\times 10^8}\\\Rightarrow \text{Time taken by light}=2.8583\times 10^{-6}[/tex]
Time taken by light = 2.8583×10⁻⁶ seconds
We want to find the distance between the student and the lighting given that we know the time that passes between the moment he sees the lighting and the moment he hears it.
The solutions are:
a) D = 857.501 mb) t₁ = 2.858*10^(-6) sNow, notice that there is an intrinsic problem with this question.
The student sees the "flash" when the light impacts on his eyes. Thus, the chain of events is:
Lighting falls.Light travels and impacts the student's eyes, at this moment he sees the flash of light, this is what we consider the t = 0.2.5 seconds after, sound arrives.So there is a little time between when the lighting happens and the student sees it, that will also have an impact on the distance to the lighting.
then we have:
(3*10^8 m/s)*t₁ = D
(343m/s)*t₂ = D
Where t₁ is the time that the light needs to reach the student, t₂ is the time the sound needs to reach the student, and D is the distance between the student and the ligthing.
We also know that:
t₂ - t₁ = 2.5s
Then we can write:
t₂ = t₁ + 2.5s and replace it on the second equation:
(343m/s)*(t₁ + 2.5s) = D
Then we have a system of equations:
(3*10^8 m/s)*t₁ = D
(343m/s)*(t₁ + 2.5s) = D
We can see that both of these left sides are equal to the same thing, then we must have:
(3*10^8 m/s)*t₁ = (343m/s)*(t₁ + 2.5s)
Then we solve it for t₁
t₁*(3*10^8 m/s - 343m/s) = (343m/s)*2.5s = 857.5m
t₁ = (857.5m)/(3*10^8 m/s - 343m/s) = 2.858*10^(-6) s
Now that we know the value of t₁, we can find the value of D.
(3*10^8 m/s)*t₁ = D
(3*10^8 m/s)* 2.858*10^(-6) s = D = 857.501 m
Then the solutions are:
a) D = 857.501 mb) t₁ = 2.858*10^(-6) sIf you want to learn more, you can read:
https://brainly.com/question/4598472
If the force on the tympanic membrane (eardrum) increases by about 1.50 above the force from atmospheric pressure, the membrane can be damaged.When you go scuba diving in the ocean, below what depth could damage to your eardrum start to occur? The eardrum is typically 8.20 in diameter. Take the density of seawater to be equal1.03*10^3 kg/m^3.
Damage to the eardrum might start to occur at a depth of approximately [tex]\[ 2.25 \times 10^6 \, \text{m} \][/tex] underwater when scuba diving in the ocean
To calculate the depth at which damage to the eardrum might occur while scuba diving, we need to consider the increase in pressure as you descend underwater. The pressure increases with depth due to the weight of the water above you. This pressure increase is given by the hydrostatic pressure formula:
[tex]\[ P = \rho \cdot g \cdot h \][/tex]
Where:
- [tex]\( P \)[/tex] is the pressure,
- [tex]\( \rho \)[/tex] is the density of the fluid (seawater in this case),
- [tex]\( g \)[/tex] is the acceleration due to gravity, and
- [tex]\( h \)[/tex]is the depth.
The force exerted on the eardrum is equal to the pressure difference between the external pressure and the pressure inside the ear canal, multiplied by the area of the eardrum. This can be represented as:
[tex]\[ F = A \cdot (P_{\text{internal}} - P_{\text{external}}) \][/tex]
Given that the force on the eardrum should not exceed 1.50 times the force from atmospheric pressure, we have:
[tex]\[ F_{\text{limit}} = 1.5 \cdot P_{\text{atm}} \cdot A \][/tex]
We can solve for the depth [tex](\( h \))[/tex] when this pressure difference exceeds the threshold.
First, let's find [tex]\( P_{\text{atm}} \)[/tex], the atmospheric pressure at sea level, which is approximately [tex]\( 101.3 \, \text{kPa} \)[/tex]. Then, we can calculate the pressure at depth and find the depth where the pressure difference reaches the limit.
[tex]\[ P_{\text{atm}} = 101.3 \, \text{kPa} \][/tex]
Now, let's calculate the depth at which damage to the eardrum might occur:
[tex]\[ F_{\text{limit}} = 1.5 \cdot P_{\text{atm}} \cdot A \]\[ P_{\text{internal}} = P_{\text{atm}} + \rho \cdot g \cdot h \]\[ F_{\text{limit}} = A \cdot (P_{\text{atm}} + \rho \cdot g \cdot h - P_{\text{external}}) \][/tex]
Given that [tex]\( P_{\text{external}} \)[/tex] is atmospheric pressure, we can rewrite this as:
[tex]\[ F_{\text{limit}} = A \cdot (\rho \cdot g \cdot h) \][/tex]
Now, we can rearrange to solve for [tex]\( h \):[/tex]
[tex]\[ h = \frac{F_{\text{limit}}}{A \cdot \rho \cdot g} \][/tex]
Substitute the values:
[tex]\[ h = \frac{1.5 \cdot P_{\text{atm}}}{A \cdot \rho \cdot g} \]\[ h = \frac{1.5 \cdot 101.3 \, \text{kPa}}{\pi \cdot (8.20 \, \text{mm})^2 \cdot (1.03 \times 10^3 \, \text{kg/m}^3) \cdot (9.81 \, \text{m/s}^2)} \]\[ h \approx \frac{1.5 \times 10^5 \, \text{Pa}}{\pi \times (8.20 \times 10^{-3} \, \text{m})^2 \times (1.03 \times 10^3 \, \text{kg/m}^3) \times (9.81 \, \text{m/s}^2)} \]\[ h \approx \frac{1.5 \times 10^5}{\pi \times 6.69 \times 10^{-5} \times 1.03 \times 10^3 \times 9.81} \][/tex]
[tex]\[ h \approx \frac{1.5 \times 10^5}{6.66 \times 10^{-2}} \]\[ h \approx 2.25 \times 10^6 \, \text{m} \][/tex]
So, damage to the eardrum might start to occur at a depth of approximately [tex]\[ 2.25 \times 10^6 \, \text{m} \][/tex] underwater when scuba diving in the ocean.
A baseball goes from zero to 34 m/s in 0.188 s. What is its average acceleration? Answer in units of m/s^2
Answer:
[tex]a_{avg} = 180.85 m/s^{2}[/tex]
Given:
Initial velocity, u = 0 m/s
Final velocity, v = 34 m/s
Time interval, [tex]\Delta t = 0.188 s[/tex]
Solution:
Acceleration is the rate at which the velocity of an object changes.
Thus
Average acceleration, [tex]a_{avg} = \frac{\Delta v}{\Delta t}[/tex]
[tex]a_{avg} = \frac{v - u}{t - 0} = \frac{34 - 0}{0.188} = 180.85 m/s^{2}[/tex]
Two objects of different mass are released simtaneously
fromthe top of a 20- m tower and fall to the ground. If air
resistanceis negligible, which statement best applies?
A. the greater mass hits the ground first.
B. Both objects hit hte ground together.
C. The smaller mass hits the ground first.
D. No conclusion can be made with the information given.
Answer:b-both objects hit the ground together
Explanation:
Given
Two objects of different mass are released simultaneously from top of a tower and fall to the ground provided air resistance is negligible
then both the object hit the ground together
Because in equation of motion terms there is no unit of mass
But if the air resistance is present then the answer would have been different.
The surface area of the earth's crust is 5.10*10^8km^2. The average thickness of the earth's crust is 35km. The mean density of the earth's crust is 2.8g/cm^3. The most abundant element in the earth's crust is oxygen with an abundance of 4.55*10^3g/t. Calculate the total volume of the earth's crust.
Answer:
The answer is [tex]V = 1.785 \times 10^{10}\ Km^3[/tex].
Explanation:
You are asking only for the total volume, so the important data here is the surface area of the earth's crust,
[tex]A = 5.10 \times 10^8\ Km^2[/tex]
and the average thickness of the earth's crust,
[tex]h = 35\ Km[/tex].
Imagine that you can stretch the surface area as if it were a blanket, now if you want to calculate its volume, you just need to multiply its area by its thickness,
[tex]V = A*h = 1.785 \times 10^{10}\ Km^3[/tex].