Final answer:
The longer wire will have twice the resistance of the shorter wire.
Explanation:
The resistance of a wire depends on both its length and cross-sectional area. In this case, the longer wire has twice the length and twice the cross-sectional area of the shorter wire. The resistance of a wire is directly proportional to its length and inversely proportional to its cross-sectional area. Therefore, the longer wire will have twice the resistance of the shorter wire.
(a) Neil A. Armstrong was the first person to walk on the moon. The distance between the earth and the moon is 3.85 × 108 m. Find the time it took for his voice to reach the earth via radio waves. (b) Someday a person will walk on Mars, which is 5.60 × 1010 m from the earth at the point of closest approach. Determine the minimum time that will be required for a message from Mars to reach the earth via radio waves.
Answer:
(a):The time what it takes his voice reach the earth via radio waves is t1= 1.28 seconds.
(b): The time what it takes his voice reach from Mars to the earth via radio waves is t2= 186.66 seconds.
Explanation:
d1= 3.85 *10⁸ m
d2= 5.6 *10¹⁰ m
C= 300,000,000 m/s
t1= d1/C
t1= 1.28 s
t2= d2/C
t2= 186.66 s
A solid iron cylinder weighs 600 N. It has a density of 7860 kg/m^3. (a) Calculate the cylinder's volume. (b) How much would the cylinder weigh if it is completely submerged in water?
Answer:
7.66 (litres); 523.66 (N).
Explanation:
all the details are provided in the attached picture. Note, P_i means 'Weight of iron', ρ_i means "Density of iron', ρ_w means 'Density of water', F_f means 'Arhimede force'.
The answers are marked with red colour.
Find the moment of inertia of a hoop (a thin-walled, hollow ring) with mass M = 1kg and radius R = 1m about an axis perpendicular to the hoop’s plane at an edge. (Express your answer in units of kg*m^2).
Answer:
2 kg m^2
Explanation:
M = 1 kg, R = 1 m
The moment of inertia of the hoop about its axis perpendicular to its plane is
I = M R^2
The moment of inertia of the hoop about its edge perpendicular to it splane is given by the use of parallel axis theorem
I' = I + M x (distance between two axes)^2
I' = I + M R^2
I' = M R^2 + M R^2
I' = 2 M R^2
I' = 2 x 1 x 1 x 1 = 2 kg m^2
The moment of inertia of a hoop about an axis perpendicular to its plane at an edge is calculated using the Parallel-Axis Theorem and for a hoop with mass M = 1kg and radius R = 1m, it is 2 kg*m².
Calculation of the Moment of Inertia for a Hoop
To find the moment of inertia of a hoop with mass M = 1kg and radius R = 1m about an axis perpendicular to the hoop's plane at an edge, we use the Parallel-Axis Theorem. The moment of inertia of a hoop about its central axis (through its center, perpendicular to the plane) is MR². According to the Parallel-Axis Theorem, the moment of inertia about an axis parallel to this but passing through the edge of the hoop is given by I = MR₂ + MR₂ (because the distance from the central axis to the outer edge of the hoop is R). Thus, the moment of inertia for the hoop about the edge is 2MR₂ which simplifies to 2 * 1kg * (1m)² = 2 kg*m².
Bill and Janet are pulling on opposite sides of a table. Bill pulls with a force of
450 N to the left, and Janet pulls with a force of 300 N to the right. What is
the net force on the table?
O
A. 450 N to the right
O
B. 450 N to the left
C. 150 N to the left
O
D. 150 N to the right
Answer:
C. 150 N to the left
Explanation:
If we take right to be positive and left to be negative, then:
∑F = -450 N + 300 N
∑F = -150 N
The net force is 150 N to the left.
Answer:
(C) 150 N to the left
Explanation:
It is given that,
Force acting in left side, F = 450 N
Force acting in right side, F' = 300 N
Let left side is taken to be negative while right side is taken to be positive. So,
F = -450 N
F' = +300 N
The net force will act in the direction where the magnitude of force is maximum. Net force is given by :
[tex]F_{net}=-450\ N+300\ N[/tex]
[tex]F_{net}=-150\ N[/tex]
So, the net force on the table is 150 N and it is acting to the left side. Hence, the correct option is (c).
A 1.2-kg ball drops vertically onto the floor, hitting with a speed of 25 m/s. Consider the impulse during this collision. Would the magnitude of the impulse be greater: (i) if the ball rebounded with a speed of 10 m/s (the ball was made of rubber), or (ii) if the ball stuck to the floor (the ball was made of clay)? Support your answer with a calculation.
Answer:
3kg
Explanation:
impulse = MV
then
m1v1=m2v2
when the values are subtitude
then
m2=1.2*25/10
m2=30kg//
Two ocean liners, each with a mass of 42,000 metric tons, are moving on parallel courses, 93 m apart. What is the magnitude of the acceleration of one of the liners toward the other due to their mutual gravitational attraction? Treat the ships as particles.
Answer:
0.324×10⁻³ m/s²
Explanation:
G=Gravitational constant=6.67408×10⁻⁸ m³/kg s
m₁=m₂=Mass of the ships=42000×10³ kg
r=Distance between the ships=93 m
The ships are considered particles
From Newtons Law of Universal Gravitation
[tex]F=G\frac {m_1m_2}{r^2}\\Here\ m_1=m_2\ hence\ m_1\times m_2=m_1^2\\\Rightarrow F=G\frac {m_1^2}{r^2}\\\Rightarrow F=\frac{6.67408\times 10^{-8}\times (42000\times 10^3)^2}{93^2}\\\Rightarrow F=13612.0674\ N\\[/tex]
[tex]F=ma\\\Rightarrow 13612.0674=42000\times 10^3a\\\Rightarrow a=0.324\times 10^{-3}\ m/s^2[/tex]
∴Acceleration of one of the liners toward the other due to their mutual gravitational attraction is 0.324×10⁻³
While on a sailboat at anchor, you notice that 15 waves pass its bow every minute. The waves have a speed of 6.0 m/s . Part A What is the distance between two adjacent wave crests
Answer:
Distance between two adjacent wave crests = 24m
Explanation:
Distance= speed × time
Distance traveled by waves in 60 seconds (15 crests)= 15 × distance
15 × distance = 6,0 (meters/second) × 60 seconds
distance = (360 meters) / 15 = 24 meters (between two adyacent waves)
What are the basic primitive solids?
Answer:
A primitive solid is a 'building block' that you can use to work with in 3D. Rather than extruding or revolving an object, AutoCAD has some basic 3D shape commands at your disposal.
Explanation:
The compressor of an air conditioner draws an electric current of 23.7 A when it starts up. If the start-up time is 2.35 s long, then how much electric charge passes through the circuit during this period?
Answer:
Electric charge, Q = 55.69 C
Explanation:
It is given that,
Electric current drawn by the compressor, I = 23.7 A
Time taken, t = 2.35 s
We need to find the electric charge passes through the circuit during this period. The definition of electric current is given by total charge divided by total time taken.
[tex]I=\dfrac{q}{t}[/tex]
Where,
q is the electric charge
[tex]q=I\times t[/tex]
[tex]q=23.7\ A\times 2.35\ s[/tex]
q = 55.69 C
So, the electric charge passes through the circuit during this period is 55.69 C. Hence, this is the required solution.
Water vapor can react reversibly with solid carbon to yield a mixture of hydrogen gas and carbon monoxide. Suppose you continually add more water vapor to the reaction. In what direction does this shift the equilibrium?
Answer:
Product side
Explanation:
When water vapor reacts reversibly with solid carbon to yield a mixture of hydrogen gas and carbon monoxide and we continually add more water vapor to the reaction the equilibrium of the reaction shifts to the product side.
Because gaseous water is reactant that appears in the reaction quotient expression.
[tex]H_{2}O+ C_{s}\leftrightharpoons H_{2}_{g}+ CO[/tex]
When we add more water vapor to the reaction the product formation is increased. The reaction goes in forward direction affecting the equilibrium.
Aircraft sometimes acquire small static charges. Suppose a supersonic jet has a 0.55 μC charge and flies due west at a speed of 685 m/s over the Earth's south magnetic pole, where the 8.00 x 10^-5 T magnetic field points straight up. What is the magnitude of the magnetic force on the plane, in newtons?
Answer:
3.014 x 10⁻⁸ N
Explanation:
q = magnitude of charge on the supersonic jet = 0.55 μC = 0.55 x 10⁻⁶ C
v = speed of the jet = 685 m/s
B = magnitude of magnetic field in the region = 8 x 10⁻⁵ T
θ = angle between the magnetic field and direction of motion = 90
magnitude of the magnetic force is given as
F = q v B Sinθ
F = (0.55 x 10⁻⁶) (685) (8 x 10⁻⁵) Sin90
F = 3.014 x 10⁻⁸ N
A cylinder is being flattened so that its volume does not change. Find the rate of change of radius when r = 2 inches and h = 5 inches, if the height is decreasing at 0.7 in/sec. Hint: what is the rate of change of volume?
Answer:
[tex]\frac{dr}{dt} = 0.14 in/s[/tex]
Explanation:
As the volume of the cylinder is constant here so we can say that its rate of change in volume must be zero
so here we can say
[tex]\frac{dV}{dt} = 0[/tex]
now we have
[tex]V = \pi r^2 h[/tex]
now find its rate of change in volume with respect to time
[tex]\frac{dV}{dt} = 2\pi rh\frac{dr}{dt} + \pi r^2\frac{dh}{dt}[/tex]
now we know that
[tex]\frac{dV}{dt} = 0 = \pi r(2h \frac{dr}{dt} + r\frac{dh}{dt})[/tex]
given that
h = 5 inch
r = 2 inch
[tex]\frac{dh}{dt} = - 0.7 in/s[/tex]
now we have
[tex]0 = 2(5) \frac{dr}{dt} + 2(-0.7)[/tex]
[tex]\frac{dr}{dt} = 0.14 in/s[/tex]
Final answer:
The change in radius δr/δt of a cylinder with a constant volume is found to be 0.28 inches per second when the height is decreasing at 0.7 inches per second and r = 2 inches, h = 5 inches.
Explanation:
The question involves using calculus to find the rate of change of the radius of a cylinder given a constant volume and a known rate of change in height. The rate of change of volume for a cylinder, which is 0 because the volume doesn't change, can be described as δV/δt = (πr²) (δh/δt) + (2πrh) (δr/δt) = 0. Given the height is decreasing at 0.7 in/sec, we can find the rate of change of the radius δr/δt. Using the known values of r = 2 inches and h = 5 inches, we can solve for δr/δt.
Starting with the equation for the volume of a cylinder V = πr²h, since the volume is constant, we take the derivative with respect to time to obtain 0 = π(2×r×δr/δt×h + r²×δh/δt). Substituting the known values gives 0 = π(2×2×δr/δt×5 + 2²×(-0.7)), which simplifies to 0 = 20πδr/δt - 5.6π. From this we can solve for δr/δt = 5.6π / 20π = 0.28 in/sec.
The rate of change of the radius is 0.28 inches per second when the height is decreasing at 0.7 inches per second, and the radius is 2 inches while the height is 5 inches.
What's the average required power to raise a 150-kg drum to a height of 20 m in a time of 1.0 min? (Answer in kW)
Answer:
Power required, P = 0.49 kW
Explanation:
It is given that,
Mass of drum, m = 150 kg
Height, h = 20 m
Time, t = 1 min = 60 sec
Average power required is given by work done divided by total time taken. It is given by :
[tex]P=\dfrac{W}{t}[/tex]
W = F.d
Here, W = mgh
So, [tex]P=\dfrac{mgh}{t}[/tex]
[tex]P=\dfrac{150\ kg\times 9.8\ m/s^2\times 20\ m}{60\ s}[/tex]
P = 490 watts
or P = 0.49 kW
So, the average power required to raise the drum is 0.49 kilo-watts. Hence, this is the required solution.
Final answer:
The average power required to lift a 150-kg drum to a height of 20 meters in 1 minute is calculated by first finding the work done and then dividing by time. The result is approximately 0.4905 kW.
Explanation:
Calculating Required Power to Lift a Drum
The problem involves finding the average power required to raise a certain weight to a specified height over a defined period of time. The weight in question is a 150-kg drum, the height is 20 meters, and the time is 1 minute (60 seconds).
To calculate power, you need to first calculate the work done, which is the product of force and displacement. Since we are lifting the object vertically, the force is the weight of the object, which is mass (m) times the acceleration due to gravity (g), and the displacement is the height (h). The formula for work (W) is:
W = m × g × h
The average power (P), then, is the work done over the time (t) it takes to do it:
P = W / t
Using the values given:
mass (m) = 150 kg
acceleration due to gravity (g) = 9.81 m/s²
height (h) = 20 m
time (t) = 60 s
First, calculate the work done:
W = 150 kg × 9.81 m/s² × 20 m = 29430 J
Now calculate the power required:
P = 29430 J / 60 s = 490.5 W
To convert this into kilowatts (kW), divide by 1000:
P = 0.4905 kW
Therefore, the average power required to lift the drum is approximately 0.4905 kW.
A 1400 kg wrecking ball hangs from a 20-m-long cable. The ball is pulled back until the cable makes an angle of 23.0 ∘ with the vertical. Part A By how much has the gravitational potential energy of the ball changed?
Answer:
21813.46 J
Explanation:
The Potential energy stored at the time it is hanging freely is mgl.
the potential energy stored at the time when it makes a angle 23 degree from the vertical is mglCos 23
Change in gravitational potential energy
U = mgl = mgl cos 23
U = mgl (1 - cos23)
U = 1400 x 9.8 x 20 ( 1 - Cos 23)
U = 21813.46 J
A 3.0 kg ball tied to the end of a 50 cm long string being swung in a circle in a horizontal plane at constant speed. If the tension of the string is 3.8 N, what is the speed of the ball?
Question 31 options:
A)
1.2 m/s
B)
1.0 m/s
C)
1.4 m/s
D)
0.8 m/s
Answer:
The speed of the ball is 0.8 m/s.
(D) is correct option.
Explanation:
Given that,
mass of ball = 3.0 kg
length = 50 cm
We need to calculate the speed of the ball
Using relation between the centripetal force and tension
[tex]\dfrac{mv^2}{r}=T[/tex]
[tex]v^2=\dfrac{T\times r}{m}[/tex]
Where, m = mass
T = tension
r = radius
Put the value into the formula
[tex]v=\sqrt{\dfrac{3.8\times50\times10^{-2}}{3.0}}[/tex]
[tex]v=0.8\ m/s[/tex]
Hence, The speed of the ball is 0.8 m/s.
You are driving along a highway at 35.0 m/s when you hear the siren of a police car approaching you from behind and you perceive the frequency as 1310 Hz. You are relieved that he is in pursuit of a different speed when he continues past you, but now you perceive the frequency as 1240 Hz. What is the frequency of the sirenin the police car
Answer:
1270.44 Hz
Explanation:
[tex]v_{L}[/tex] = velocity of the our car = 35.0 m/s
[tex]v_{P}[/tex] = velocity of the police car = ?
[tex]v_{S}[/tex] = velocity of the sound = 343 m/s
[tex]f_{app}[/tex] = frequency observed as police car approach = 1310 Hz
[tex]f_{rec}[/tex] = frequency observed as police car go away = 1240 Hz
[tex]f[/tex] = actual frequency of police siren
Frequency observed as police car approach is given as
[tex]f_{app}= \frac{(v_{s}-v_{L})f}{v_{s} -v_{P} }[/tex]
inserting the values
[tex]1310 = \frac{(343 - 35)f}{343 -v_{P} }[/tex] eq-1
Frequency observed as police car goes away is given as
[tex]f_{rec}= \frac{(v_{s} + v_{L})f}{v_{s} + v_{P} }[/tex]
inserting the values
[tex]1240 = \frac{(343 + 35)f}{343 + v_{P} }[/tex] eq-2
Dividing eq-1 by eq-2
[tex]\frac{1310}{1240} = \left ( \frac{343 - 35}{343 - v_{P} } \right )\frac{(343 + v_{P})}{343 + 35 }\\[/tex]
[tex]v_{P}[/tex] = 44.3 m/s
Using eq-1
[tex]1310 = \frac{(343 - 35)f}{343 - 44.3 }[/tex]
f = 1270.44 Hz
A cosmic ray electron moves at 6.5x 10^6 m/s perpendicular to the Earth's magnetic field at an altitude where the field strength is 10x 10^-5 T. What is the radius, in meters, of the circular path the electron follows?
Answer:
Radius, r = 0.36 meters
Explanation:
It is given that,
Speed of cosmic ray electron, [tex]v=6.5\times 10^6\ m/s[/tex]
Magnetic field strength, [tex]B=10\times 10^{-5}\ T=10^{-4}\ T[/tex]
We need to find the radius of circular path the electron follows. It is given by :
[tex]qvB=\dfrac{mv^2}{r}[/tex]
[tex]r=\dfrac{mv}{qB}[/tex]
[tex]r=\dfrac{9.1\times 10^{-31}\ kg\times 6.5\times 10^6\ m/s}{1.6\times 10^{-19}\times 10^{-4}\ T}[/tex]
r = 0.36 meters
So, the radius of circular path is 0.36 meters. Hence, this is the required solution.
beam of light, traveling in air(the index of refraction for air is 1), strikes the surface of mineral oil at an ngle of 30° with the normal to the surface.( the index of refraction for this mineral oil is 1.38) What is the ngle of refraction? What is the speed of the light traveling in mineral oil?
Answer:
2.17 x 10^8 m/s
Explanation:
Angle of incidence, i = 30 degree, refractive index of mineral oil, n = 1.38
Let r be the angle of refraction.
By use of Snell's law
n = Sin i / Sin r
Sin r = Sin i / n
Sin r = Sin 30 / 1.38
Sin r = 0.3623
r = 21.25 degree
Let the speed of light in oil be v.
By the definition of refractive index
n = c / v
Where c be the speed of light
v = c / n
v = ( 3 x 10^8) / 1.38
v = 2.17 x 10^8 m/s
A gas expands from an initial volume of 0.040 m^3 and an initial pressure of 210 kPa to a final volume of 0.065 m^3 while its temperature is kept constant. How much work is done by the system?
Answer:
286
Explanation:
p1v1/T1=p2v2/T2
then subtitude your values
T2=760*0.65*273/210*0.040
T2=135/8.4=16+273=289
The pressure-volume work done by the system during the expansion is 5.25 x 10³J.
What is done by pressure-volume work?The work that is done when a fluid is compressed or expanded is known as pressure-volume work. Pressure-volume work occurs whenever there is a change in volume but the outside pressure stays the same.
Here,
The pressure of the system of gas, P = 210 kPa
Initial volume of the system of gas, V₁ = 0.04 m³
Final volume of the system of gas, V₂ = 0.065 m³
The expression for the pressure-volume work done by a system of gas is given by,
Work done,
W = PΔV
W = P(V₂ - V₁)
Applying the values of P, V₁ and V₂,
W = 210 x 10³(0.065 - 0.04)
W = 210 x 10³x 0.025
W = 5.25 x 10³J
Hence,
The pressure-volume work done by the system during the expansion is 5.25 x 10³J.
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Two ideal gases have the same mass density and the same absolute pressure. One of the gases is helium (He), and its temperature is 175 K The other gas is neon (Ne). What is the temperature of the neon?
In this situation, the neon gas should also have a temperature of 175 K, the same as the helium gas, given all the conditions and the principle of the ideal gas law.
Explanation:The problem you're working on involves understanding the ideal gas law, which is PV = nRT. This equates the pressure (P), volume (V) of the gas, and temperature (T). There are two important points to consider. Firstly, the number of molecules or moles (n) and the ideal gas constant (R) aren't changing in this situation. Secondly, the volume and pressure are the same for both gases.
Given that the masses, volume, and pressure are the same for both helium and neon, we conclude that the number of moles (n) for helium and neon are equal (because the mass density is the same and for ideal gas mass density (ρ) = PM/RT, where M is molar mass). The temperature ratio should be the same as the ratio of Kelvin temperatures of two gases.
So, we can safely say that since the gases obey the ideal gas law, and all conditions are held constant aside from the identity of the gas and the temperature, the temperature of the neon gas must also be 175 Kelvin like the helium gas.
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A 24-g rifle bullet traveling 280 m/s buries itself in a 3.7-kg pendulum hanging on a 2.8-m-long string, which makes the pendulum swing upward in an arc. Part A Determine the vertical and horizontal components of the pendulum's maximum displacement.
Answer:
x = 0.95 m
y = 0.166 m
Explanation:
m = mass of the bullet = 24 g = 0.024 kg
v = speed of bullet before collision = 280 m/s
M = mass of the pendulum = 3.7 kg
L = length of the string = 2.8 m
h = height gained by the pendulum after collision
V = speed of the bullet and pendulum combination
Using conservation of momentum
m v = (m + M) V
(0.024) (280) = (0.024 + 3.7) V
V = 1.805 m/s
using conservation of energy
Potential energy gained by bullet and pendulum combination = Kinetic energy of bullet and pendulum combination
(m + M) g h = (0.5) (m + M) V²
(9.8) h = (0.5) (1.805)²
h = 0.166 m
y = vertical displacement = h = 0.166 m
x = horizontal displacement
horizontal displacement is given as
x = sqrt(L² - (L - h)²)
x = sqrt(2.8² - (2.8 - 0.166)²)
x = 0.95 m
A hawk flies in a horizontal arc of radius 12.0 m at constant speed 4.00 m/s. (a) Find its centripetal acceleration. (b) It continues to fly along the same horizontal arc, but increases its speed at the rate of 1.20 m/s2. Find the acceleration (magnitude and direction) in this situation at the moment the hawk’s speed is 4.00 m/s.
Answer:
a) [tex]a_c= 1.33 m/s^2 [/tex]
b) a= 1.79 m/s²
θ = 41.98⁰
Explanation:
arc radius = 12 m
constant speed = 4.00 m/s
(a) centripetal acceleration
[tex]a_c=\frac{v^2}{R}[/tex]
[tex]a_c=\frac{4^2}{12} [/tex]
= 1.33 m/s²
(b) now we have given
[tex]a_t= \ 1.20 m/s^2 [/tex]
now,
[tex]a=\sqrt{a^2_c+ a^2_t}[/tex]
[tex]a=\sqrt{1.33^2+ 1.20^2}[/tex]
a= 1.79 m/s²
direction
[tex]\theta = tan^{-1}(\frac{a_t}{a_r} )[/tex]
[tex]\theta = tan^{-1}(\frac{1.2}{1.33} )[/tex]
θ = 41.98⁰
The centripetal acceleration of the hawk is 1.33 m/s².
The resultant acceleration of the hawk at the given moment is 1.79 m/s².
The direction resultant acceleration of the hawk is 48⁰.
The given parameters;
radius of the arc, r = 12 mspeed of the hawk, u = 4 m/sacceleration of the hawk, a = 1.2 m/s²The centripetal acceleration of the hawk is calculated as follows;
[tex]a_c = \frac{v^2}{r} \\\\a_c = \frac{(4)^2}{12} \\\\a_c = 1.33 \ m/s^2[/tex]
The resultant acceleration is calculated as;
[tex]a = \sqrt{a_c^2 + a_t} \\\\a = \sqrt{(1.33)^2 + (1.2)^2} \\\\a = 1.79 \ m/s^2[/tex]
The direction of the acceleration is calculated as follows;
[tex]tan(\theta) = \frac{a_c}{a_t} \\\\\theta = tan^{-1} ( \frac{a_c}{a_t} )\\\\\theta = tan^{-1} ( \frac{1.33}{1.2} )\\\\\theta = 48^0[/tex]
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An object is attached to a spring, the spring is stretched by 23.7 cm in the negative direction, and the object is released, oscillating in simple harmonic motion. After 0.317 s, it is again 23.7 cm from the equilibrium position, having passed through the equilibrium position once in those 0.317 s. Determine the speed of the object after 1.39 seconds. Number cm/s
Answer:83.17 cm/s
Explanation:
Let positive x be negative direction and negative x be positive direction in this question
General equation of motion of SHM is
x=[tex]Asin\left ( \omega_{n}t\right )[/tex] --------1
where [tex]\omega [/tex]is natural frequency of motion given by
[tex]\omega_n[/tex]=[tex]\sqrt{\frac{k}{m}}[/tex]
Where K is spring constant
here A=23.7cm
And it is given it is again at 23.7 from equilibrium position having passed through the equilibrium once.
i.e. it covers this distance in [tex]\frac{T}{2}[/tex] sec
where T is the time period of oscillation i.e. returning to same place after T sec
therefore T=0.634 sec
differentiating equation 1 we get
v=[tex]A\omega_n[/tex]cos[tex]\left ( \omega_{n}t\right )[/tex]
and [tex]T\times \omega_n[/tex]=[tex]2\pi[/tex]
[tex]\omega_n[/tex]=9.911rad/s
[tex]v[/tex]=[tex]23.7\times 9.911cos\left (789.261\degree\right)[/tex]
v=83.17 cm/s
A burnt paper on the road has a picture, which shows a speed boat runs fast on the lake and produces V-like water waves. This remind you of Moessbauer Effect and Cherenkov Radiation. What are these?
Answer:
Moessbauer Effect = eggy eggs
Explanation:
An industrial machine requires a solid, round piston connecting rod 200 mm long, between pin ends that is subjected to a maximum compression force of 80,000 N. Using a factor of safety of 2.5, what diameter is required if aluminum is used with properties Sy = 496 MPa and E = 71 GPa?
Answer:
diameter is 13.46 mm
Explanation:
length of rod = 200 mm = 0.2 m
compression force = 80,000 N
factor of safety = 2.5
Sy = 496 MPa
E = 71 GPa
to find out
diameter
solution
first we calculate the allowable stress i.e. = Sy/factor of safety
allowable stress = 496/ 2.5= 198.4 MPa 198.4 × [tex]10^{6}[/tex] Pa
now we calculate the diameter d by the Euler's equation i.e.
critical load = [tex]\pi ^{2}[/tex] E × moment of inertia / ( K × length )² ..........1
now we calculate the critical load i.e. allowable stress × area
here area = [tex]\pi[/tex] /4 × d²
so critical load = 198.4 × [tex]\pi[/tex] /4 × d²
and K = 1 for pin ends
and moment of inertia is = [tex]\pi[/tex] / 64 × [tex]d^{4}[/tex]
put all value in equation 1 and we get d
198.4 ×[tex]10^{6}[/tex] × [tex]\pi[/tex] /4 × d² = [tex]\pi ^{2}[/tex] 71 × [tex]10^{9}[/tex] × [tex]\pi[/tex] / 64 × [tex]d^{4}[/tex] / ( 1 × 0.2 )²
155.8229× [tex]10^{6}[/tex] × d² = 700.741912× [tex]10^{9}[/tex]× 0.049087× [tex]d^{4}[/tex] / 0.04
d=0.01346118 m
d = 13.46 mm
diameter is 13.46 mm
Ignoring the mass of the spring, a 5 kg mass hanging from a coiled spring having a constant k= 50 N/m will have a period of oscillation of about: (A) 10 sec., (B) 5 sec., (C) 2 sec., (D) 0.1 secC., (E) 1 min.
Answer:
Period of oscillation, T = 2 sec
Explanation:
It is given that,
Mass of the object, m = 5 kg
Spring constant of the spring, k = 50 N/m
This object is hanging from a coiled spring. We need to find the period of oscillation of the spring. The time period of oscillation of the spring is given by :
[tex]T=2\pi\sqrt{\dfrac{m}{k}}[/tex]
[tex]T=2\pi\sqrt{\dfrac{5\ kg}{50\ N/m}}[/tex]
T = 1.98 sec
or
T = 2 sec
So, the period of oscillation is about 2 seconds. Hence, this is the required solution.
An object with a mass of 7.8 kg is pulled on a horizontal surface by a horizontal pull of 50 N to the right. The friction force on this object is 30 N to the left. What is the acceleration of the object? 0.16 m/s^2
10 m/s^2
2.6 m/s^2
6.4 m/s^2
Answer:
2.6 m/s²
Explanation:
m = mass of the object = 7.8 kg
Consider the right direction as positive and left direction as negative
F = horizontal applied force towards right = 50 N
f = frictional force acting towards left = - 30 N
a = acceleration
Force equation for the motion of the object is given as
F + f = ma
50 - 30 = (7.8) a
a = 2.6 m/s²
What is the hydrostatic pressure at 20,000 leagues under the sea? (a league is the distance a person can walk in one hour) ?) 40 kPa b) 100 ps? c) 1300 Pad) 2000 psi e) none of these answers
Answer:
alternative E- none of these answers
Explanation:
Hydrostatic pressure is the pressure exerted by a fluid at equilibrium at a given point within the fluid, due to the force of gravity. Hydrostatic pressure increases in proportion to depth measured from the surface because of the increasing weight of fluid exerting downward force from above.
The formula is :
P= d x g x h
p: hydrostatic pressure (N/m²)
d: density (kg/m³) density of seawater is 1,030 kg/m³
g: gravity (m/s²) ≅ 9.8m/s²
h: height (m)
The hydrostatic pressure at 20,000 leagues under the sea is approximately 1,002,500,000,000 Pa.
Explanation:The hydrostatic pressure at 20,000 leagues under the sea can be calculated using the equation for pressure in a fluid, which is given by P = ρgh, where P is the pressure, ρ is the density of the fluid, g is the acceleration due to gravity, and h is the depth below the surface.
Since a league is the distance a person can walk in one hour, we need to convert it to meters. Assuming an average walking speed of 5 km/h, a league is equal to 5 km. Therefore, 20,000 leagues is equal to 100,000 km.
The pressure at this depth can be calculated using the known values: density of seawater is about 1025 kg/m³ and acceleration due to gravity is 9.8 m/s². Plugging in these values, we get P = (1025 kg/m³)(9.8 m/s²)(100,000,000 m) = 1,002,500,000,000 Pa.
Therefore, the correct answer is none of the provided options. The hydrostatic pressure at 20,000 leagues under the sea is approximately 1,002,500,000,000 Pa.
A flat coil of wire consisting of 20 turns, each with an area of 50 cm2, is positioned perpendicularly to a uniform magnetic field that increases its magnitude at a constant rate from 2.0 T to 6.0 T in 2.0 s. If the coil has a total resistance of 0.40 ? What is the magnitude of the induced current?
Answer:
0.5 A
Explanation:
N = 20, A = 50 cm^2 = 50 x 10^-4 m^2, dB = 6 - 2 = 4 T, dt = 2 s, R = 0.4 ohm
The induced emf is given by
e = - N dФ/dt
Where, dФ/dt is the rate of change of magnetic flux.
Ф = B A
dФ/dt = A dB/dt
so,
e = 20 x 50 x 10^-4 x 4 / 2 = 0.2 V
negative sign shows the direction of magnetic field.
induced current, i = induced emf / resistance = 0.2 / 0.4 = 0.5 A
Molecules like DNA may be stretched and are well modeled as springs. An optical trap can pull with a maximum force of 11\; fN11fN (femto-Newtons) and can stretch a DNA molecule by 0.06\; \mu m0.06μm . What is the spring constant of the molecule?
Answer:
1.8 x 10⁻⁷ N/m
Explanation:
[tex]F_{max}[/tex] = maximum force with which the optical trap can pull = 11 x 10⁻¹⁵ N
x = stretch caused in DNA molecule due to the force = 0.06 x 10⁻⁶ m
k = spring constant of the spring
Maximum force is given as
[tex]F_{max}= k x[/tex]
[tex]11\times 10^{-15}= k (0.06\times 10^{-6})[/tex]
k = 1.8 x 10⁻⁷ N/m
The spring constant of the DNA molecule is calculated using Hooke's Law with the given force of 11 fN and stretch distance of 0.06 μm, resulting in a spring constant of approximately 183.33 N/m.
Explanation:To calculate the spring constant (k) of a DNA molecule modeled as a spring, we can use Hooke's Law, which states that the force (F) applied to stretch or compress a spring is directly proportional to the displacement (x) it causes, as represented by the equation F = kx. The optical trap pulls with a maximum force of 11 fN and stretches the DNA molecule by 0.06 μm. Using the given values, the spring constant (k) can be calculated as:
k = F / x
Therefore, k = 11 fN / 0.06 μm, and to ensure the units are consistent, we convert 0.06 μm to meters (0.06 μm = 0.06 x 10-6 m).
k = 11 x 10-15 N / 0.06 x 10-6 m
k = (11 / 0.06) x 10-9 N/m
k ≈ 183.33 N/m
The spring constant of the DNA molecule is therefore approximately 183.33 N/m.