Answer:
4.99 mg of vitamin C are in the beaker.
Explanation:
Given that,
Weight of vitamin = 0.0499 g
Molar mass = 176.124 g/mol
Weight of water = 100.0 ml
We need to calculate the mg of vitamin C in the beaker
We dissolve 0.0499 g vitamin C in water to from 100.0 ml solution.
100 ml solution contain 49.9 mg vitamin C
Now, we take 10 ml of this vitamin C solution in breaker
Since, 100 ml solution =49.9 mg vitamin C
Therefore,
[tex]10\ ml\ solution =\dfrac{10\times49.9}{100}[/tex]
[tex]10\ ml\ solution =4.99\ mg[/tex]
Hence, 4.99 mg of vitamin C are in the beaker.
A wave pulse travels down a slinky. The mass of the slinky is m = 0.89 kg and is initially stretched to a length L = 6.2 m. The wave pulse has an amplitude of A = 0.27 m and takes t = 0.408 s to travel down the stretched length of the slinky. The frequency of the wave pulse is f = 0.44 Hz.1) What is the speed of the wave pulse? 17.035 m/s
2) What is the tension in the slinky? 33.92 N
3) What is the average speed of a piece for the slinky as a complete wave pulse passes? .4704 m/s
4) What is the wavelength of the wave pulse? 34.765 m
1) The wave speed is 15.2 m/s
2) The tension in the slinky is 33.2 N
3) The average speed of a piece of the slinky during one pulse is 0.475 m/s
4) The wavelength is 34.5 m
Explanation:
1)
The motion of a wave pulse along the slinky is a uniform motion, therefore its speed is given by the equation for uniform motion:
[tex]v=\frac{L}{t}[/tex]
where
L is the length covered
t is the time elapsed
For the wave in this problem, we have:
L = 6.2 m is the length of the slinky
t = 0.408 s is the time taken for a pulse to travel across the length os the slinky
Substituting, we find the wave speed
[tex]v=\frac{6.2}{0.408}=15.2 m/s[/tex]
2)
The speed of a wave on a slinky can be found with the same expression for the wave speed along a string:
[tex]v=\sqrt{\frac{T}{m/L}}[/tex]
where
T is the tension in the slinky
m is the mass of the slinky
v is the wave speed
L is the length
In this problem, we have:
m = 0.89 kg is the mass of the slinky
L = 6.2 m is the length
Therefore, we can re-arrange the equation to find the tension in the slinky, T:
[tex]T=v^2 (\frac{m}{L})=(15.2)^2 (\frac{0.89}{6.2})=33.2 N[/tex]
3)
The average speed of a piece of the slinky as a complete wave pulse passes is the total displacement done by a piece of slinky during one period, which is 4 times the amplitude, divided by the time taken for one complete oscillation, the period:
[tex]v_{avg} = \frac{4A}{T}[/tex]
where
A is the amplitude
T is the period
Here we have:
A = 0.27 m is the amplitude of the wave
The period is the reciprocal of the frequency, therefore
[tex]T=\frac{1}{f}[/tex]
where f = 0.44 Hz is the frequency of this wave. Substituting and solving, we find
[tex]v_{avg} = \frac{4A}{1/f}=4Af=4(0.27)(0.44)=0.475 m/s[/tex]
4)
The wavelength of the wave pulse can be found by using the wave equation:
[tex]v=f\lambda[/tex]
where
v is the wave speed
f is the frequency
[tex]\lambda[/tex] is the wavelength
For the pulse in this problem, we have
v = 15.2 m/s
f = 0.44 Hz
Substituting, we find the wavelength:
[tex]\lambda=\frac{v}{f}=\frac{15.2}{0.44}=34.5 m[/tex]
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The speed of the wave pulse is 17.035 m/s, the tension in the slinky is 33.92 N, the average speed of a piece is 17.035 m/s, and the wavelength of the wave pulse is 34.765 m.
Explanation:1) The speed of the wave pulse can be calculated using the formula v = λf, where v is the speed, λ is the wavelength, and f is the frequency. In this case, the frequency is given as 0.44 Hz and the wavelength can be calculated as λ = v/f = 17.035/0.44 = 38.716 m.
The speed of the wave pulse is therefore 17.035 m/s.
2) The tension in the slinky can be determined using the formula T = 2μv², where T is the tension, μ is the mass per unit length, and v is the speed of the wave pulse. The mass per unit length can be calculated as μ = m/L = 0.89/6.2 = 0.143 kg/m.
The tension is then T = 2 * 0.143 * (17.035)² = 33.92 N.
3) The average speed of a piece of the slinky can be calculated as v_avg = λ/T, where λ is the wavelength and T is the period of the wave pulse. The period can be calculated as T = 1/f = 1/0.44 = 2.2727 s.
The average speed is then v_avg = 38.716/2.2727 = 17.035 m/s.
4) The wavelength of the wave pulse is given as λ = v/f = 17.035/0.44 = 34.765 m.
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A lidless shoebox is made of five rectangular pieces of cardboard forming its base and four sides. Its dimensions are: base length L = 48 cm, base width W = 25 cm, and side height H = 11 cm. The base lies in quadrant I of the xy-plane, with its length along the x-axis and its width along the y-axis. The box’s sides extend in the positive z-direction. The cardboard is thin and has an area mass density of s = 0.075 g/cm2.(a) What is the x-coordinate, in centimeters, of the center of mass of the shoebox's base?
The x-coordinate of the center of mass of the lidless shoebox's base, which is a rectangle, will be at half its length. Therefore, the x-coordinate is 24 cm.
Explanation:The center of mass is the coordinate that averages the distribution of mass in a physical object. To calculate the center of mass, one must consider the dimensions of the object. For a rectangular object like the shoebox's base with length, L = 48 cm, and width, W = 25 cm, the x-coordinate of the center of mass will be half the length. Therefore, the x-coordinate of the center of mass of the shoebox's base is L/2 = 48 cm / 2 = 24 cm. This is because in this rectangular coordinate system, the center of mass is located at the middle along each dimension (L, W).
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The x-coordinate of the center of mass of the shoebox's base is half its length which is 24 cm.
Explanation:
The center of mass of an object is basically the average position of all its parts. Given that the shoebox's base lies on the xy-plane, the x-coordinate for its center of mass will simply be half the length of the base. Hence, using the length L = 48 cm, the x-coordinate for the center of mass is: L/2 = 48/2 = 24 cm. Therefore, the center of mass of the shoebox's base along the x-axis lies at 24 cm.
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The radius of Earth is 6370 km in the Earth reference frame. The cosmic ray is moving at 0.880Co relative to Earth.
a. In the reference frame of a cosmic ray how wide does Earth seem along the flight direction?
b. In the reference frame of a cosmic ray how wide does Earth seem perpendicular to the flight direction?
Express your answer with the appropriate units.
Answer:
6052114.67492 m
[tex]12.742\times 10^{6}\ m[/tex]
Explanation:
v = Velocity of cosmic ray = 0.88c
c = Speed of light = [tex]3\times 10^8\ m/s[/tex]
d = Width of Earth = Diameter of Earth = [tex]12.742\times 10^{6}\ m[/tex]
When the cosmic ray is moving towards Earth then in the frame of the cosmic ray the width of the Earth appears smaller than the original
This happens due to length contraction
Length contraction is given by
[tex]d_e=d\sqrt{1-\frac{v^2}{c^2}}\\\Rightarrow d_e=12.742\times 10^{6}\sqrt{1-\frac{0.88^2c^2}{c^2}}\\\Rightarrow d_e=6052114.67492\ m[/tex]
The Earth's width is 6052114.67492 m
Contraction only occurs in the cosmic ray's frame of reference in the direction of the ray. But in perpendicular direction the width remains unchanged.
Hence, the width is [tex]12.742\times 10^{6}\ m[/tex]
When a 6.35-g sample of magnesium metal is burned, it produces enough heat to raise the temperature of 1,910 g of water from 24.00°C to 33.10°C. How much heat did the magnesium release when it burned?
Answer:
the heat released by the magnesium is 72 kJ
Explanation:
the heat exchanged will be
Q = m*c*(T final - T initial)
where Q= heat released, c= specific heat capacity, T initial= initial temperature of water, T final = final temperature of water
Assuming the specific heat capacity of water as c= 1 cal/g°C=4.186 J/g°C
replacing values
Q = m * c* (T final - T initial) = 1910 g * 4.186 J/g°C*(33.10 °C - 24°C) = 72 kJ
Consider a rod of length L rotated about one of its ends instead of about its center of mass. If the mass of the rod is 5 kg, and the length is 2 meters, calculate the magnitude of the moment of inertia (I).
Answer:
[tex]I_{edge} = 6.67 kg.m^2[/tex]
Explanation:
given,
mass of rod = 5 Kg
Length of rod = 2 m
R = 1 m
moment of inertial from one edge of the rod = ?
moment of inertia of rod through center of mass
[tex]I_{CM}= \dfrac{1}{12}M(L)^2[/tex]
using parallel axis theorem
[tex]I_{edge} = I_{CM} + MR^2[/tex]
[tex]I_{edge} =\dfrac{1}{12}ML^2+ M(\dfrac{L}{2})^2[/tex]
[tex]I_{edge} =\dfrac{1}{12}ML^2+ \dfrac{ML^2}{4}[/tex]
[tex]I_{edge} =\dfrac{1}{3}ML^2[/tex]
now, inserting all the given values
[tex]I_{edge} =\dfrac{1}{3}\times 5 \times 2^2[/tex]
[tex]I_{edge} = 6.67 kg.m^2[/tex]
An electron moves at 2.40×106 m/s through a region in which there is a magnetic field of unspecified direction and magnitude 7.10×10−2 T .
a. What is the largest possible magnitude of the force on the electron due to the magnetic field? Express your answer in newtons to two significant figures. Fmax = nothing N.
(b) If the actual acceleration of theelectron is one-fourth of the largest magnitude in part (a), whatis the angle between the electron velocity and the magneticfield?
Answer:
a) F = 2.7 10⁻¹⁴ N , b) a = 2.97 10¹⁶ m / s² c) θ = 14º
Explanation:
The magnetic force on the electron is given by the expression
F = q v x B
Which can be written in the form of magnitude and the angle found by the rule of the right hand
F = q v B sin θ
where θ is the angle between the velocity and the magnetic field
a) the maximum magnitude of the force occurs when the velocity and the field are perpendicular, therefore, without 90 = 1
F = e v B
F = 1.6 10⁻¹⁹ 2.40 10⁶ 7.10 10⁻²
F = 2.73 10⁻¹⁴ N
F = 2.7 10⁻¹⁴ N
b) Let's use Newton's second law
F = m a
a = F / m
a = 2.7 10⁻¹⁴ / 9.1 10⁻³¹
a = 2.97 10¹⁶ m / s²
The actual acceleration (a1) is a quarter of this maximum
a1 = ¼ a
a1 = 7.4 10¹⁵ m / s²
With this acceleration I calculate the force that is executed on the electron
F = ma
e v b sin θ= ma
sin θ = ma / (e v B)
sin θ = 9.1 10⁻³¹ 7.4 10¹⁵ / (1.6 10⁻¹⁹ 2.40 10⁶ 7.10 10⁻²)
sin θ = 6.734 10⁻¹⁵ / 27.26 10⁻¹⁵
sin θ = 0.2470
θ = 14.3º
The aorta carries blood away from the heart at a speed of about 42 cm/s and has a radius of approximately 1.1 cm. The aorta branches eventually into a large number of tiny capillaries that distribute the blood to the various body organs. In a capillary, the blood speed is approximately 0.064 cm/s, and the radius is about 5.5 x 10-4 cm. Treat the blood as an incompressible fluid, and use these data to determine the approximate number of capillaries in the human body.
Answer:
The number of capillaries is
[tex]N=2.625x10^9[/tex] Capillaries
Explanation:
[tex]v_{aorta}=42cm/s[/tex], [tex]r_{aorta}=1.1 cm[/tex], [tex]v_{cap}=0.064cm/s[/tex], [tex]r_{cap}=5.5x10^{4}cm[/tex],
To find the number of capillaries in the human body use the equation:
[tex]N_{cap}=\frac{v_{aorta}*\pi*r_{aorta}^2}{v_{cap}*\pi*r_{cap}^2}[/tex]
So replacing numeric
[tex]N_{cap}=\frac{42cm/s*\pi*(1.1cm)^2}{0.064cm/s*\pi*(5.5x10^{-4}cm)^2}[/tex]
Now we can find the number of capillaries
[tex]N=26250000000[/tex]
[tex]N=2.625x10^9[/tex] Capillaries
In the SI system, the unit of current, the ampere, is defined by this relationship using an apparatus called an Ampere balance. What would be the force per unit length of two infinitely long wires, separated by a distance 1 m, if 1 A of current were flowing through each of them? Express your answer numerically in newtons per meter. F/L = N/m
Answer:
[tex]\dfrac{F}{l}=2\times 10^{-7}\ N/m[/tex]
Explanation:
It is given that,
Distance between two infinitely long wires, d = 1 m
Current flowing in both of the wires, I = 1 A
The magnetic field in a wire is given by :
[tex]B=\dfrac{\mu_oI}{2\pi d}[/tex]
The force per unit length acting on the two infinitely long wires is given by :
[tex]\dfrac{F}{l}=\dfrac{\mu_o I_1I_2}{2\pi d}[/tex]
[tex]\dfrac{F}{l}=\dfrac{4\pi \times 10^{-7}\times 1^2}{2\pi \times 1}[/tex]
[tex]\dfrac{F}{l}=2\times 10^{-7}\ N/m[/tex]
So, the force acting on the parallel wires is [tex]2\times 10^{-7}\ N/m[/tex]. Hence, this is the required solution.
The force per unit length between the two wires is [tex]\( 2 \times 10^{-7} \)[/tex] newtons per meter.
The force per unit length (F/L) between two infinitely long wires carrying current can be calculated using Ampere force law, which is given by:
[tex]\[ \frac{F}{L} = \frac{\mu_0 I_1 I_2}{2 \pi r} \][/tex]
Plugging in the values, we get:
[tex]\[ \frac{F}{L} = \frac{(4\pi \times 10^{-7} \text{ N/A}^2) \times (1 \text{ A}) \times (1 \text{ A})}{2 \pi \times (1 \text{ m})} \][/tex]
Simplifying the expression by canceling out [tex]\( \pi \)[/tex] and multiplying the numerical values, we have:
[tex]\[ \frac{F}{L} = \frac{(4 \times 10^{-7} \text{ N/A}^2) \times (1 \text{ A})^2}{2 \times (1 \text{ m})} \][/tex]
[tex]\[ \frac{F}{L} = \frac{4 \times 10^{-7} \text{ N}}{2 \text{ m}} \][/tex]
[tex]\[ \frac{F}{L} = 2 \times 10^{-7} \text{ N/m} \][/tex]
Therefore, the force per unit length between the two wires is [tex]\( 2 \times 10^{-7} \)[/tex] newtons per meter.
The answer is: [tex]2 \times 10^{-7} \text{ N/m}[/tex]
The gauge pressure inside an alveolus with a 200 µm radius is 25 mmHg, while the blood pressure outside is only 10 mmHg. Assuming the alveolus acts like a spherical bubble, what is the surface tension of the fluid and membrane around the outside of the alveolus? How does this compare to the surface tension of water?
Answer:
The surface tension is 0.0318 N/m and is sufficiently less than the surface tension of the water.
Solution:
As per the question:
Radius of an alveolus, R = [tex]200\mu m = 200\times 10^{- 6}\ m[/tex]
Gauge Pressure inside, [tex]P_{in} = 25\ mmHg[/tex]
Blood Pressure outside, [tex]P_{o} = 10\ mmHg[/tex]
Now,
Change in pressure, [tex]\Delta P = 25 - 10 = 15\ mmHg = 1.99\times 10^{3}\ Pa[/tex]
Since the alveolus is considered to be a spherical shell
The surface tension can be calculated as:
[tex]\Delta P = \frac{4\pi T}{R}[/tex]
[tex]T = \frac{1.99\times 10^{3}\times 200\times 10^{- 6}}{4\pi} = 0.0318\ N/m = 0.318\ mN/m[/tex]
And we know that the surface tension of water is 72.8 mN/m
Thus the surface tension of the alveolus is much lesser as compared to the surface tension of water.
If you slide down a rope, it’s possible to create enough thermal energy to burn your hands or your legs where they grip the rope. Suppose a 40 kg child slides down a rope at a playground, descending 2.0 m at a constant speed. How much thermal energy is created as she slides down the rope?
Answer:
Thermal energy will be equal to 784 J
Explanation:
We have given that mass of the child m = 40 kg
Height h = 2 m
Acceleration due to gravity [tex]g=9.8m/sec^2[/tex]
We have to find the thermal energy '
The thermal energy will be equal to potential energy
And we know that potential energy is given by
[tex]W=mgh=40\times 9.8\times 2=784J[/tex]
So the thermal energy will be equal to 784 J
A cube has a density of 1800 kg/m3 while at rest in the laboratory. What is the cube's density as measured by an experimenter in the laboratory as the cube moves through the laboratory at 91.0 % of the speed of light in a direction perpendicular to one of its faces? You may want to review (Pages 1040 - 1043) .
Answer:
4341.44763 kg/m³
Explanation:
[tex]\rho'[/tex] = Actual density of cube = 1800 kg/m³
[tex]\rho[/tex] = Density change due to motion
v = Velocity of cube = 0.91c
c = Speed of light = [tex]3\times 10^8\ m/s[/tex]
Relativistic density is given by
[tex]\rho=\frac{\rho'}{\sqrt{1-\frac{v^2}{c^2}}}\\\Rightarrow \rho=\frac{1800}{\sqrt{1-\frac{0.91^2c^2}{c^2}}}\\\Rightarrow \rho=\frac{1800}{\sqrt{1-0.91^2}}\\\Rightarrow \rho=4341.44763\ kg/m^3[/tex]
The cube's density as measured by an experimenter in the laboratory is 4341.44763 kg/m³
10%) Problem 7: Water flows through a water hose at a rate of Q1 = 620 cm3/s, the diameter of the hose is d1 = 1.99 cm. A nozzle is attached to the water hose. The water leaves the nozzle at a velocity of v2 = 10.4 m/s. show answer No Attempt 17% Part (a) Enter an expression for the cross-sectional area of the hose, A1, in terms of its diameter, d1.
Answer:
a) [tex]A_1 = \frac{\pi d_1^2}{4}[/tex]
Explanation:
a) the cross-sectional area of the hose would be the square of radius times pi. And since the sectional radius is half of its diameter d. We can express the cross-sectional area A1 in term of diameter d1
[tex]A_1 = \pi r_1^2 = \pi (d_1/2)^2 = \frac{\pi d_1^2}{4}[/tex]
Two astronauts on opposite ends of a spaceship are comparing lunches. One has an apple, the other has an orange. They decide to trade. Astronaut 1 tosses the 0.132 kg apple toward astronaut 2 with a speed of 1.25 m/s. The 0.143 kg orange is tossed from astronaut 2 to astronaut 1 with a speed of 1.14 m/s. Unfortunately, fruits collide, sending the orange off with a speed 1.03 m/s and an angle of 43.0° with respect to its original direction of motion. Using conservation of linear momentum, find the final speed and direction of the apple. Assume an elastic collision occurs. Give the apple’s direction relative to its original direction of motion.
Answer:
Explanation:
We shall consider direction towards left as positive Let the required velocity be v and let v makes an angle φ
Applying law of conservation of momentum along direction of original motion
m₁ v₁ - m₂ v₂ = m₂v₃ - m₁ v₄
0.132 x 1.25 - .143 x 1.14 = 1.03 cos43 x .143 - v cos θ
v cos θ = .8
Applying law of conservation of momentum along direction perpendicular to direction of original motion
1.03 sin 43 x .143 = .132 x v sinθ
v sinθ = .76
squaring and adding
v² = .76 ² + .8²
v = 1.1 m /s
Tan θ = .76 / .8
θ = 44°
Kristen is spinning on the ice at 40 rad/s about her longitudinal axis when she abducts her arms and doubles her radius of gyration about her longitudinal axis from 32 cm to 64 cm. If her angular momentum is conserved, what is her angular velocity about her longitudinal axis after she increases her radius of gyration (in rad/s)
Answer: I = k^2m.. equa1
I = moment of inertia
M = mass of skater
K = radius of gyration.
When her angular momentum is conserved we have
Iw = I1W1... equ 2
Where I = with extended arm, w = angular momentum =40rads/s, I1 = inertia when hands her tucked in, w1 = angular momentum when hands are tucked in.
Substituting equation 1 into equ2 and simplifying to give
W = (k/k1)^2W..equation 3
Where'd k= 64cm, k1 = 32cm, w = angular momentum when hands is tucked in= 40rad/s
Substituting figures into equation 3
W1 = 10rad/s
Explanation:
Assuming a centroidal axis of the skater gives equation 1
Final answer:
When the moment of inertia is doubled, the angular velocity will decrease by half due to the conservation of angular momentum.
Explanation:
When the moment of inertia is doubled, the angular velocity will decrease by half due to the conservation of angular momentum. In this case, Kristen's initial angular velocity is 40 rad/s, and her initial moment of inertia is 32 cm. After doubling her radius of gyration to 64 cm, her final moment of inertia is 128 cm. Using the conservation of angular momentum equation, we can calculate her final angular velocity:
Initial Angular Momentum = Final Angular Momentum
Initial Angular Velocity * Initial Moment of Inertia = Final Angular Velocity * Final Moment of Inertia
Substituting the values: 40 rad/s * 32 cm = Final Angular Velocity * 128 cm
Simplifying the equation: Final Angular Velocity = 10 rad/s
Therefore, Kristen's angular velocity about her longitudinal axis after increasing her radius of gyration is 10 rad/s.
A ball on a string travels once around a circle with a circumference of 2.0 m. The tension in the string is 5.0 N. how much work is done by tension?
Answer:0
Explanation:
Given
circumference of circle is 2 m
Tension in the string [tex]T=5 N[/tex]
[tex]2\pi r=2[/tex]
[tex]r=\frac{2}{2\pi }=\frac{1}{\pi }=0.318 m[/tex]
In this case Force applied i.e. Tension is Perpendicular to the Displacement therefore angle between Tension and displacement is [tex]90^{\circ}[/tex]
[tex]W=\int\vec{F}\cdot \vec{r}[/tex]
[tex]W=\int Fdr\cos 90 [/tex]
[tex]W=0[/tex]
The work done by the ball as it travels once around the circular string is 0.
The given parameters;
circumference of the circle, P = 2 mtension in the string, T = 5 NThe work-done by a body is the dot product of applied force and displacement.
For one complete rotation around the circumference of the circle, the displacement of the object is zero.
The work-done by the ball when its makes a complete rotation around the circle is calculated as;
Work-done = F x r
where;
r is the displacement of the ball, = 0Work-done = 5 x 0 = 0
Thus, the work done by the ball as it travels once around the circular string is 0.
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A household refrigerator with a COP of 1.2 removes heat from the refrigerated space at a rate of 40 kJ/min. Determine the rate of heat transfer to the kitchen air in kilojoules per minute to three significant digits. Pay attention to the units asked for in the answer !!!
Answer:
73.3 kJ / min
Explanation:
COP or coefficient of performance of a refrigerator is defined as ratio of heat extracted from the refrigerator to electrical imput to the refrigerator
If Q₁ be the heat extracted out and Q₂ be the heat given out to the surrounding
Imput energy = Q₂ - Q₁
so COP = Q₁ / Q₂ - Q₁
Given
COP = 1.2
Q₁ = 40kJ
Substituting the values
1.2 = 40 / (Q₂ - 40)
1.2 (Q₂ - 40) = 40
1.2 Q₂ = 2.2 X 40
Q₂ = 73.3 kJ / min
Four equal masses m are so small they can be treated as points, and they are equally spaced along a long, stiff wire of neglible mass. The distance between any two adjacent masses is ℓ . m m m m ℓ ℓ ℓ What is the rotational inertia Icm of this system about its center of mass?
A.) (1/2)ml^2
B.) ml^2
C.) 5ml^2
D.) 6ml^2
E.) 2ml^2
F.) 3ml^2
G.) 7ml^2
H.) 4ml^2.
Answer:C
Explanation:
Given
Four masses are attached to the wire such that distance between two mass is L
therefore the Length of wire is 4 L
and the center of mass is at 2L
moment of inertia is distribution of mass from its rotational axis
thus moment of Inertia I is given by
[tex]I=m\times (\frac{3L}{2})^2+m\times (\frac{L}{2})^2+m\times (\frac{3L}{2})^2+m\times (\frac{L}{2})^2[/tex]
[tex]I=2\times m\times (\frac{L}{2})^2+2\times m\times (\frac{3L}{2})^2[/tex]
[tex]I=\frac{2mL^2}{4}+\frac{18mL^2}{4}[/tex]
[tex]I=5mL^2[/tex]
The rotational inertia of the four-point mass system about its center of mass, with equal spacing between adjacent masses, would be 2mℓ^2.
Explanation:The rotational inertia of a system around any particular axis is given by sum of the product of the mass of each particle and square of perpendicluar distance from the axis of rotation. In the case of our four-point mass system, the masses are arranged such that they are all equidistant from the center. Therefore, the total inertia, Icm, is the sum of the inertias of each individual mass.
Assuming the center of rotation is halfway between the second and third mass, there will be two masses at distance ℓ/2 and two at 3ℓ/2. Therefore, Icm= 2* m*((ℓ/2)^2) + 2* m*((3ℓ/2)^2) = 2mℓ^2. Hence, the correct answer is E.) 2mℓ^2.
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A fish looks up toward the surface of a pond and sees the entire panorama of clouds, sky, birds, and so on, contained in a narrow cone of light, beyond which there is darkness. What is going on here to produce this vision? How large (in degrees) is the opening angle of the cone of light received by the fish?
Answer:
Total internal reflection is going on. The refractive index of water is about 1.3, so sin 90/sin r=1/sinr=1.3. So the fish can only see objects outside the water within about 50 degrees of the vertical
A child whirls a ball in a vertical circle. Assuming the speed of the ball is constant (an approximation), when would the tension in the cord connected to the ball be greatest?
a. A little after the bottom of the circle when the ball is climbing.
b. A little before the bottom of the circle when the ball is descending quickly.
c. At the bottom of the circle.
d. Nowhere; the cord is stretched the same amount at all points.
e. At the top of the circle.
Answer:
C. At the bottom of the circle.
Explanation:
Lets take
Radius of the circle = r
Mass = m
Tension = T
Angular speed = ω
The radial acceleration towards = a
a= ω² r
Weight due to gravity = mg
At the bottom conditionT - m g = m a
T = m ω² r + m g
At the top conditionT + m g = m a
T= m ω² r -m g
From above equation we can say that tension is grater when ball at bottom of the vertical circle.
Therefore the answer is C.
C. At the bottom of the circle.
Final answer:
The tension in the cord connected to a ball whirled in a vertical circle is greatest at the bottom of the circle because the tension must overcome gravity and provide the centripetal force to keep the ball in motion.
Explanation:
The question asks, "A child whirls a ball in a vertical circle. Assuming the speed of the ball is constant (an approximation), when would the tension in the cord connected to the ball be greatest?" The correct answer is c. At the bottom of the circle.
When the ball is at the bottom of the circle, the tension in the string is highest because it must counteract both the gravitational force pulling downwards and provide enough centripetal force to keep the ball moving in a circular path. At this point, the sum of the gravitational force and the centripetal force dictates the necessary tension, making it greater than at any other point in the ball's circular trajectory. This is in contrast to the top of the circle, where the tension is lowest since gravity assists in providing the centripetal force needed, sometimes reducing the tension in the string to nearly zero if the ball moves at the minimum speed required to continue in circular motion.
A solid cylinder of radius 10.0 cm rolls down an incline with slipping. The angle of the incline is 30°. The coefficient of kinetic friction on the surface is 0.400. What is the angular acceleration of the solid cylinder? What is the linear acceleration?
To solve this problem it is necessary to apply the expressions related to the calculation of angular acceleration in cylinders as well as the calculation of linear acceleration in these bodies.
By definition we know that the angular acceleration in a cylinder is given by
[tex]\alpha = \frac{2\mu_k g cos\theta}{r}[/tex]
Where,
[tex]\mu_k[/tex] = Coefficient of kinetic friction
g = Gravitational acceleration
r= Radius
[tex]\theta[/tex]= Angle of inclination
While the tangential or linear acceleration is given by,
[tex]a = g(sin\theta-\mu_k cos\theta)[/tex]
ANGULAR ACCELERATION, replacing the values that we have
[tex]\alpha = \frac{2\mu_k g cos\theta}{r}[/tex]
[tex]\alpha = \frac{2(0.4)(9.8) cos(30)}{10*10^{-2}}[/tex]
[tex]\alpha = 67.9rad/s[/tex]
LINEAR ACCELERATION, replacing the values that we have,
[tex]a = (9.8)(sin30-(0.4)cos(30))[/tex]
[tex]a = 1.5m/s^2[/tex]
Therefore the linear acceleration of the solid cylinder is [tex]1.5 m/s^2[/tex]
A particle moves along a line so that its velocity at time t is v(t) = t2 − t − 20 (measured in meters per second).
(a) Find the displacement of the particle during 3 ≤ t ≤ 9.
(b) Find the distance traveled during this time period. SOLUTION (a) By this equation, the displacement is s(9) − s(3) = 9 v(t) dt 3 = 9 (t2 − t − 20) dt 3
Answer:
(a) [tex]\displaystyle s(t)= \frac{t^3}{3}-\frac{t^2}{2}-20t+C\ \ \ \ \forall\ \ 3\leqslant t\leqslant 9[/tex]
(b) 78 m
Explanation:
Physics' cinematics as rates of change.
Velocity is defined as the rate of change of the displacement. Acceleration is the rate of change of the velocity.
[tex]\displaystyle v=\frac{ds}{dt}[/tex]
Knowing that
[tex]\displaystyle v(t)= t^2 - t - 20[/tex]
(a) To find the displacement we need to integrate the velocity
[tex]\displaystyle \frac{ds}{dt}=t^2 - t - 20[/tex]
[tex]\displaystyle ds=(t^2 - t - 20)dt[/tex]
[tex]\displaystyle s(t)= \int(t^2 - t - 20)dt=\frac{t^3}{3}-\frac{t^2}{2}-20t+C\ \ \ \ \forall \ \ \ 3\leqslant t\leqslant 9[/tex]
(b) The displacement can be found by evaluating the integral
[tex]\displaystyle d=\int_{3}^{9} (t^2 - t - 20)dt[/tex]
[tex]\displaystyle d=\left | \frac{t^3}{3}-\frac{t^2}{2}-20t \right |_3^9=\frac{45}{2}+\frac{111}{2}=78\ m[/tex]
To find displacement, we integrate the velocity function from 3 to 9. The displacement is the integral of the velocity function. For distance, we integrate the absolute value of the velocity function from 3 to 9 because distance is a scalar quantity and includes total path length.
Explanation:To solve the problem, we need to find the displacement and distance traveled by the particle. For part (a), the displacement for the given time period is obtained by integrating the velocity function from 3 to 9. The displacement is the integral of the velocity function v(t) = t2 - t - 20 from 3 to 9, which gives us the value s(9) - s(3).
For part (b), to calculate the distance traveled, we need to integrate the absolute value of the velocity function because the distance is always positive. Negative values would represent backward motion, but the distance traveled includes total path length and does not care about direction.
So, the distance traveled from time 3 to 9 would be ∫ |t2 - t - 20| dt from 3 to 9. The calculation of this integral will give the distance traveled.
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At the Indianapolis 500, you can measure the speed of cars just by listening to the difference in pitch of the engine noise between approaching and receding cars. Suppose the sound of a certain car drops by a factor of 2.40 as it goes by on the straightaway. How fast is it going? (Take the speed of sound to be 343 m/s.)
To develop this problem it is necessary to apply the concepts related to the Dopler effect.
The equation is defined by
[tex]f_i = f_0 \frac{c}{c+v}[/tex]
Where
[tex]f_h[/tex]= Approaching velocities
[tex]f_i[/tex]= Receding velocities
c = Speed of sound
v = Emitter speed
And
[tex]f_h = f_0 \frac{c}{c+v}[/tex]
Therefore using the values given we can find the velocity through,
[tex]\frac{f_h}{f_0}=\frac{c-v}{c+v}[/tex]
[tex]v = c(\frac{f_h-f_i}{f_h+f_i})[/tex]
Assuming the ratio above, we can use any f_h and f_i with the ratio 2.4 to 1
[tex]v = 353(\frac{2.4-1}{2.4+1})[/tex]
[tex]v = 145.35m/s[/tex]
Therefore the cars goes to 145.3m/s
You pour 250 g of tea into a Styrofoam cup, initially at 80?C and stir in a little sugar using a 100-g aluminum 20?C spoon and leave the spoon in the cup. Assume the specific heat of tea is 4180 J/kg??C and the specific heat of aluminum is 900 J/kg??C.
What is the highest possible temperature of the spoon when you finally take it out of the cup?
Answer: 75ºC
Explanation:
Assuming that the Styrofoam is perfectly adiabatic, and neglecting the effect of the sugar on the system, the heat lost by the tea, can only be transferred to the spoon, reaching all the system to a final equilibrium temperature.
If the heat transfer process is due only to conduction, we can use this empirical relationship for both objects:
Qt = ct . mt . (tfn – ti)
Qs = cs . ms . (ti – tfn)
If the cup is perfectly adiabatic, it must be Qt = Qs
Using the information provided, and solving for tfinal, we get:
tfinal = (83,600 + 1,800) / (90 + 1045) ºC
tfinal = 75º C
A food department is kept at 2128C by a refrigerator in an environment at 308C. The total heat gain to the food department is estimated to be 3300 kJ/h and the heat rejection in the condenser is 4800 kJ/h. Determine the power input to the compressor, in kW and the COP of the refrigerator.
Answer:
2.2
Explanation:
[tex]Q_u[/tex] = Heat rejection in the condenser = 3300 kJ/h
[tex]Q_L[/tex] = Heat gain to the food department = 4800 kJ/h
Power output is given by
[tex]W=Q_u-Q_L\\\Rightarrow W=4800-3300\\\Rightarrow W=1500\ kJ/h[/tex]
COP of a refrigerator is given by
[tex]COP=\frac{Desired\ effect}{Work}\\\Rightarrow COP=\frac{Q_L}{W}\\\Rightarrow COP=\frac{3300}{1500}\\\Rightarrow COP=2.2[/tex]
The COP of the refrigerator is 2.2
A rubber ball filled with air has a diameter of 24.2 cm and a mass of 0.459 kg. What force is required to hold the ball in equilibrium immediately below the surface of water in a swimming pool? (Assume that the volume of the ball does not change. Indicate the direction with the sign of your answer.)
To solve this problem, it is necessary to apply the concepts related to Newton's second Law as well as to the expression of mass as a function of Volume and Density.
From Newton's second law we know that
F= ma
Where,
m = mass
a = acceleration
At the same time we know that the density is given by,
[tex]\rho = \frac{m}{V} \rightarrow m = \rho V[/tex]
Our values are given as,
[tex]g = 9.8m/s^2[/tex]
[tex]m =0.459 kg[/tex]
D=0.242 m
Therefore the Force by Weight is
[tex]F_w = mg[/tex]
[tex]F_w = 0.459kg * 9.8m/s^2 = 4.498N[/tex]
Now the buoyant force acting on the ball is
[tex]F_B=\rho V g[/tex]
The value of the Volume of a Sphere can be calculated as,
[tex]V = \frac{4}{3} \pi r^3[/tex]
[tex]V = \frac{4}{3} \pi (0.242/2)^3[/tex]
[tex]V = 0.007420m^3[/tex]
[tex]\rho_w = 1000kg/m^3 \rightarrow[/tex] Normal conditions
Then,
[tex]F_B=0.007420*(1000)*(9.8) = 72.716 N[/tex]
Therefore the Force net is,
[tex]F_{net} = F_B -F_w[/tex]
[tex]F_{net} = 72.716N - 4.498N =68.218 N[/tex]
Therefore the required Force is 68.218N
3) A 0.060-kg tennis ball, initially moving at a speed of 12 m/s, is struck by a racket causing it to rebound in the opposite direction at a speed of 18 m/s. What is the change in momentum of the ball?
Answer:
Change in momentum will be 1.8 kgm/sec
Explanation:
We have given mass of the ball m = 0.060 kg
Initial speed = 12 m /sec
And final speed = 18 m/sec
We have to find the change in momentum
Change in momentum is given by [tex]\Delta P=m(v_f-v_i)[/tex]
So [tex]\Delta P=0.060\times (18-(-12))=0.060\times 30=1.8kgm/sec[/tex] ( negative sign is due to finally opposite direction of ball )
So change in momentum will be 1.8 kgm/sec
The change of momentum of the tennis ball after being hit by the racket, reversing its direction, is calculated to be -1.8 kg*m/s.
Explanation:To determine the change in momentum of a tennis ball following a bounce, one must calculate the initial and final momentum and find the difference between the two. Momentum (p) is the product of mass (m) and velocity (v), hence p = mv. The tennis ball in question has a mass of 0.06 kg. The initial momentum is 0.06 kg * 12 m/s = 0.72 kg*m/s. After it is struck by the racket, it rebounds in the opposite direction at a speed of 18 m/s giving it a momentum of 0.06 kg * -18 m/s = -1.08 kg*m/s (negative because direction changed). The change in momentum is the final momentum minus the initial momentum, hence -1.08 kg*m/s - 0.72 kg*m/s = -1.8 kg*m/s.
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tarzan plans to cross a gorge by swinging in an arc from a hanging vine. if his arms are capable of exerting a force of 1400N on nthe vine, what is the maximum speed he can tolerate at the lowest point of his swing? His mass is 80kg, and the vine is 5.5m long
To solve this problem it is necessary to use three key concepts. The first of these is the Centripetal Force which is in the upward direction and would be described as
[tex]F_c = \frac{mv^2}{r}[/tex]
Where,
m= mass
v= Velocity
r = Radius
The second is the voltage generated which is given by 1400N. Finally the third is the force generated by the weight and that would be described by Newton's second law as
F =mg
Where,
m = Mass
g = Acceleration gravity
F = 80*9.8
F = 784N
For balance to exist, the sum of Force must be equal to the Centripetal Force, therefore
[tex]\sum F = F_c[/tex]
[tex]T - mg = \frac{mv^2}{r}[/tex]
Replacing we have
[tex]1400 - 784 = \frac{(80kg)v^2}{5,5}[/tex]
[tex]v^2 = \frac{ 616*5.5}{80}[/tex]
[tex]v = \sqrt{42.35}[/tex]
[tex]v=6.5m/s[/tex]
Therefore the maximum speed he can tolerate at the lowest point of his swing is 6.5m/s
A 1.80-m -long uniform bar that weighs 531 N is suspended in a horizontal position by two vertical wires that are attached to the ceiling. One wire is aluminum and the other is copper. The aluminum wire is attached to the left-hand end of the bar, and the copper wire is attached 0.40 m to the left of the right-hand end. Each wire has length 0.600 m and a circular cross section with radius 0.250 mm .a. What is the fundamental frequency of transverse standing waves for aluminium wire? b. What is the fundamental frequency of transverse standing waves for copper wire?
Answer:
(a) 498.4 Hz
(b) 442 Hz
Solution:
As per the question:
Length of the wire, L = 1.80 m
Weight of the bar, W = 531 N
The position of the copper wire from the left to the right hand end, x = 0.40 m
Length of each wire, l = 0.600 m
Radius of the circular cross-section, R = 0.250 mm = [tex].250\times 10^{- 3}\ m[/tex]
Now,
Applying the equilibrium condition at the left end for torque:
[tex]T_{Al}.0 + T_{C}(L - x) = W\frac{L}{2}[/tex]
[tex]T_{C}(1.80 - 0.40) = 531\times \frac{1.80}{2}[/tex]
[tex]T_{C} = 341.357\ Nm[/tex]
The weight of the wire balances the tension in both the wires collectively:
[tex]W = T_{Al} + T_{C}[/tex]
[tex]531 = T_{Al} + 341.357[/tex]
[tex]T_{Al} = 189.643\ Nm[/tex]
Now,
The fundamental frequency is given by:
[tex]f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}[/tex]
where
[tex]\mu = A\rho = \pi R^{2}\rho[/tex]
(a) For the fundamental frequency of Aluminium:
[tex]f = \frac{1}{2L}\sqrt{\frac{T_{Al}}{\mu}}[/tex]
[tex]f = \frac{1}{2L}\sqrt{\frac{T_{Al}}{\pi R^{2}\rho_{Al}}}[/tex]
where
[tex]\rho_{l} = 2.70\times 10^{3}\ kg/m^{3}[/tex]
[tex]f = \frac{1}{2\times 0.600}\sqrt{\frac{189.643}{\pi 0.250\times 10^{- 3}^{2}\times 2.70\times 10^{3}}} = 498.4\ Hz[/tex]
(b) For the fundamental frequency of Copper:
[tex]f = \frac{1}{2L}\sqrt{\frac{T_{C}}{\mu}}[/tex]
[tex]f = \frac{1}{2L}\sqrt{\frac{T_{C}}{\pi R^{2}\rho_{C}}}[/tex]
where
[tex]\rho_{C} = 8.90\times 10^{3}\ kg/m^{3}[/tex]
[tex]f = \frac{1}{2\times 0.600}\sqrt{\frac{341.357}{\pi 0.250\times 10^{- 3}^{2}\times 2.70\times 10^{3}}} = 442\ Hz[/tex]
A copper ring and a wooden ring of the same dimensions are placed in magnetic fields so that there is the same change in magnetic flux through them. Compare the induced EMFs and currents in the rings.
Answer:
emf induces in both iron and copper is same
Explanation:
The induced emf in both the rings would be same.
The induced emf in any material is given by
[tex]\epsilon= -N\frac{d\phi}{dt}[/tex]
N= number of turns in the coil
dΦ= change in magnetic flux
dt= change in time
clearly, induced emf is independent of the material of the object,it only depends upon rate of change of flux.
therefore, emf induces in both iron and copper is same.
The induced emf and currents in the copper ring and the wooden ring are compared based on their conductivity.
Explanation:When a magnetic field changes, it induces an electric field and hence an emf (electromotive force) in a conducting loop. The magnitude of the induced emf depends on the rate of change of the magnetic field. In this case, both the copper ring and the wooden ring experience the same change in magnetic flux, which means they have the same induced electric fields. However, the copper ring, being a good conductor, has a much higher induced emf compared to the wooden ring, which has a lower conductivity.
Wooden Ring:
Conductivity: Wood is a poor conductor of electricity. Its conductivity is significantly lower compared to metals like copper.
Induced Emf and Currents: Due to its poor conductivity, a wooden ring will exhibit a much weaker induced emf and induce much smaller currents when subjected to a changing magnetic field.
In summary, the difference in conductivity between copper and wood directly influences the induced emf and currents in the respective rings. The high conductivity of copper facilitates efficient response to changes in magnetic fields, while the low conductivity of wood results in a much weaker and less efficient induction process.
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A man stands on a merry-go-round that is rotating at 1.58 rad/s. If the coefficient of static friction between the man's shoes and the merry-go-round is μs = 0.45, how far from the axis of rotation can he stand without sliding? (Enter the maximum distance in meters.) m
Answer:
the maximum distance of rotation can he stand without sliding is 1.77 m
Explanation:
given information:
angular velocity , ω = 1.58 rad/s
static friction, μs = 0.45
now we calculate the vertical force
N - W = 0, N is normal force and W is weight
N = W
= m g
next, for the horizontal force we only have frictional force, thus
F(friction) = m a
μs N = m a
μs m g = m a
a = μs g,
now we have to find the acceleration which is both translation and cantripetal.
a = [tex]\sqrt{a_{t} ^{2}+a_{c} ^{2} }[/tex]
[tex]a_{t} ^{2}[/tex] is the acceleration for translation
[tex]a_{t} ^{2}[/tex] = 0
[tex]a_{c} ^{2}[/tex] is centripetal acceleration
[tex]a_{c} ^{2}[/tex] = ω^2r
therefore,
a = [tex]\sqrt{a_{c} ^{2} }[/tex]
= [tex]a_{c} ^{2}[/tex]
= ω^2r
Now, to find the radius, substitute the equation into the following formula
a = μs g
ω^2r = μs g
r = μs g / ω^2
= (0.45 x 9.8) / (1.58)
= 1.77 m