Answer:
a)[tex]t_1=\frac{w_1-w_o}{\alpha}=\frac{w_1}{\alpha}sec[/tex]
b)[tex]\theta_1=\frac{w_1^2}{2\alpha}rad[/tex]
c)[tex]t_2=\frac{\alpha t_1}{5\alpha}=\frac{t_1}{5}sec[/tex]
Explanation:
1) Basic concepts
Angular displacement is defined as the angle changed by an object. The units are rad/s.
Angular velocity is defined as the rate of change of angular displacement respect to the change of time, given by this formula:
[tex]w=\frac{\Delat \theta}{\Delta t}[/tex]
Angular acceleration is the rate of change of the angular velocity respect to the time
[tex]\alpha=\frac{dw}{dt}[/tex]
2) Part a
We can define some notation
[tex]w_o=0\frac{rad}{s}[/tex],represent the initial angular velocity of the wheel
[tex]w_1=?\frac{rad}{s}[/tex], represent the final angular velocity of the wheel
[tex]\alpha[/tex], represent the angular acceleration of the flywheel
[tex]t_1[/tex] time taken in order to reach the final angular velocity
So we can apply this formula from kinematics:
[tex]w_1=w_o +\alpha t_1[/tex]
And solving for t1 we got:
[tex]t_1=\frac{w_1-w_o}{\alpha}=\frac{w_1}{\alpha}sec[/tex]
3) Part b
We can use other formula from kinematics in order to find the angular displacement, on this case the following:
[tex]\Delta \theta=wt+\frac{1}{2}\alpha t^2[/tex]
Replacing the values for our case we got:
[tex]\Delta \theta=w_o t+\frac{1}{2}\alpha t_1^2[/tex]
And we can replace [tex]t_1[/tex]from the result for part a, like this:
[tex]\theta_1-\theta_o=w_o t+\frac{1}{2}\alpha (\frac{w_1}{\alpha})^2[/tex]
Since [tex]\theta_o=0[/tex] and [tex]w_o=0[/tex] then we have:
[tex]\theta_1=\frac{1}{2}\alpha \frac{w_1^2}{\alpha^2}[/tex]
And simplifying:
[tex]\theta_1=\frac{w_1^2}{2\alpha}rad[/tex]
4) Part c
For this case we can assume that the angular acceleration in order to stop applied on the wheel is [tex]\alpha_1 =-5\alpha \frac{rad}{s}[/tex]
We have an initial angular velocity [tex]w_1[/tex], and since at the end stops we have that [tex]w_2 =0[/tex]
Assuming that [tex]t_2[/tex] represent the time in order to stop the wheel, we cna use the following formula
[tex]w_2 =w_1 +\alpha_1 t_2[/tex]
Since [tex]w_2=0[/tex] if we solve for [tex]t_2[/tex] we got
[tex]t_2=\frac{0-w_1}{\alpha_1}=\frac{-w_1}{-5\alpha}[/tex]
And from part a) we can see that [tex]w_1=\alpha t_1[/tex], and replacing into the last equation we got:
[tex]t_2=\frac{\alpha t_1}{5\alpha}=\frac{t_1}{5}sec[/tex]
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
The average coefficient of volume expansion for carbon tetrachloride is 5.81 x 10^–4 (°C)–1 . If a 50.0-gal steel container is filled completely with carbon tetrachloride when the temperature is 10.0°C, how much will spill over when the temperature rises to 30.0°C. The coefficient of expansion for steel is11 x 10^-6.
Answer:0.548 gallon
Explanation:
Given
Average coefficient of volume expansion for carbon Tetrachloride [tex]\beta =5.81\times 10^{-4} /^{\circ}C[/tex]
Volume of steel container [tex]V=50 gallon[/tex]
Initial temperature [tex]T_i=10^{\circ}C[/tex]
Final temperature [tex]T_f=30^{\circ}C[/tex]
[tex]\Delta T=20^{\circ}C[/tex]
Coefficient of expansion for steel is [tex]\alpha =11\times 10^{-6}[/tex]
[tex]\beta =3\alpha =3\times 11\times 10^{-6}[/tex]
[tex]\beta =33\times 10^{-6}/^{\circ}C[/tex]
[tex]\Delta V_{spill}=\Delta V_{liquid}-\Delta V_{steel}[/tex]
[tex]\Delta V_{spill}=(\beta _{carbon}-\beta _{steel})V_0(\Delta T)[/tex]
[tex]\Delta V_{spill}=(5.81\times 10^{-4}-33\times 10^{-6})50\times 20[/tex]
[tex]\Delta V_{spill}=0.548\ gallon[/tex]
Approximately 0.04927 gal of carbon tetrachloride will spill over when the temperature rises to 30.0°C inside a 50.0-gal steel container.
Explanation:To calculate how much carbon tetrachloride will spill over when the temperature rises, we need to find the change in volume for the steel container and the carbon tetrachloride. The change in volume can be calculated using the formula:
ΔV = V * β * ΔT
where ΔV is the change in volume, V is the initial volume, β is the coefficient of volume expansion, and ΔT is the change in temperature.
Using the given values, the change in volume for the steel container is 0.000605 gal and for the carbon tetrachloride is 0.04927 gal. Therefore, approximately 0.04927 gal of carbon tetrachloride will spill over when the temperature rises to 30.0°C.
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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
5. (Serway 9th ed., 7-3) In 1990, Walter Arfeuille of Belgium lifted a 281.5-kg object through a distance of 17.1 cm using only his teeth. (a) How much work was done on the object by Arfeuille in this lift, assuming the object was lifted at constant speed? (b) What total force was exerted on Arfeuille’s teeth during the lift? (Ans. (a) 472 J; (b) 2.76 kN)
Para resolver este problema es necesario aplicar los conceptos de Fuerza, dados en la segunda Ley de Newton y el concepto de Trabajo, como expresión de la fuerza necesaria para realizar una actividad en una distancia determinada.
El trabajo se define como
W = F*d
Where,
F = Force
d = Distance
At the same time we have that the Force by second's Newton law is equal to
F = mg
Where,
m = mass
g = Gravitational acceleration
PART A) Using our values and replacing we have that
[tex]W = F*d\\W = mg*d\\W=281.5*9.8(17.1*10^{-2}\\W = 471.738 J\approx 472J[/tex]
PART B) Using Newton's Second law we have that,
[tex]F = mg \\F= 281.5*9.8\\F= 2758.7 N \approx 2.76kN[/tex]
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]
Our two intrepid relacar drivers are named Pam and Ned. We use these names to make it easy to remember: measurements made by Pam are primed (x', t') and those made by Ned are not primed (x, t). v is the velocity of Pam (the other frame of reference) as measured by Ned. What is the interpretation of v'?
There are 4 possible choices for this answer:
a. The velocity of Pam as measured by Ned.
b. The velocity of Pam as measured by Pam.
c. The velocity of Ned as measured by Ned.
d. The velocity of Ned as measured by Pam.
The velocity of Ned as measured by Pam is the interpretation of v.
Answer: Option D
Explanation:
According to question, we know that this is an issue depending on the logical and translation of the factors. From the measured information taken what is gathered by the two people is communicated and we have given as:
The Ned reference framework : (x, t)
The Pam reference framework : [tex]\left(x^{\prime}, t^{\prime}\right)[/tex]
From the reference framework, we realize that ν is the speed of Pam (the other reference framework) as estimation by Ned.
At that point, [tex]v^{\prime}[/tex] is the speed of Ned (from the other arrangement of the reference) as estimation by Pam.
In the context of relativity, the velocity denoted by v' represents the velocity of Ned as measured by Pam. Primed quantities typically refer to measurements in the moving reference frame, making option (d) correct.
Explanation:The question is asking about the interpretation of the notation v', which is used in a relativistic physics context. The correct interpretation of v' in the context of relativity would be d. The velocity of Ned as measured by Pam. This is because primed quantities (e.g. x', t', v') typically refer to measurements made in the moving reference frame relative to the unprimed frame. Since Pam's frame is the one moving with velocity v in Ned's frame, v' would be the velocity Ned appears to have when observed from Pam's frame. Accordingly, option (d) is the correct choice.
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A beam of light traveling in air strikes a glass slab at an angle of incidence less than 90°. After entering the glass slab, what does the beam of light do? (There could be more than one correct choice.)
A. I follows the same path as before it struck the glass.
B. It bends closer to the normal at the point of contact.
C. It follows the normal to the glass slab.
D. It bends away from the normal at the point of contact.
E. It slows down.
Answer:
True A and B
Explanation:
Let's propose the solution of the exercise before seeing the affirmations.
We use the law of refraction
n₁ sin θ₁ = n₂. Sin θ₂
Where n₁ and n₂ are the refractive indices of the two means, θ₁ and θ₂ are the angles of incidence and refraction, respectively
sin θ₂ = (n1 / n2) sin θ₁
Let's apply this equation to the case presented. The index of refraction and airs is 1 (n1 = 1)
Sin θ₂ = (1 / n2) sin θ₁
the angle θ₂ which is the refracted angle is less than the incident angle
Let's analyze the statements time
A. False. We saw that it deviates
B. True Approaches normal (vertical axis)
C. False It deviates, but it is not parallel to normal
D. False It deviates, but approaching the normal not moving away
E. True. Because its refractive index is higher than air,
When a beam of light with an angle of incidence less than 90° enters a denser medium like a glass slab, it bends closer to the normal, bends away from the normal, and slows down, dependent on the refractive indices of the two media. Here options B, D, and E are correct.
When a beam of light travels from air into a denser medium, such as a glass slab, it undergoes refraction. Refraction is the bending of light as it passes from one medium to another with a different optical density.
The angle of incidence, the angle formed between the incident ray and the normal (a line perpendicular to the surface at the point of incidence), plays a crucial role in determining the behavior of the refracted light.
These statements are correct. The degree to which the light bends depends on the refractive indices of the two media. In this case, as light enters the glass slab, it slows down due to the higher refractive index of glass compared to air.
The bending of light towards the normal and slowing down are characteristic behaviors of light when it travels from a less dense to a denser medium. Here options B, D, and E are correct.
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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 1.1 kg block is initially at rest on a horizontal frictionless surface when a horizontal force in the positive direction of an x axis is applied to the block. The force is given by F with arrow(x) = (2.4 − x2)i hat N, where x is in meters and the initial position of the block is x = 0.
(a) What is the kinetic energy of the block as it passes through x = 2.0 m?
(b) What is the maximum kinetic energy of the block between x = 0 and x = 2.0 m?
Answer with Explanation:
Mass of block=1.1 kg
Th force applied on block is given by
F(x)=[tex](2.4-x^2)\hat{i}N[/tex]
Initial position of the block=x=0
Initial velocity of block=[tex]v_i=0[/tex]
a.We have to find the kinetic energy of the block when it passes through x=2.0 m.
Initial kinetic energy=[tex]K_i=\frac{1}{2}mv^2_i=\frac{1}{2}(1.1)(0)=0[/tex]
Work energy theorem:
[tex]K_f-K_i=W[/tex]
Where [tex]K_f=[/tex]Final kinetic energy
[tex]K_i[/tex]=Initial kinetic energy
[tex]W=Total work done[/tex]
Substitute the values then we get
[tex]K_f-0=\int_{0}^{2}F(x)dx[/tex]
Because work done=[tex]Force\times displacement[/tex]
[tex]K_f=\int_{0}^{2}(2.4-x^2)dx[/tex]
[tex]K_f=[2.4x-\frac{x^3}{3}]^{2}_{0}[/tex]
[tex]K_f=2.4(2)-\frac{8}{3}=2.13 J[/tex]
Hence, the kinetic energy of the block as it passes thorough x=2 m=2.13 J
b.Kinetic energy =[tex]K=2.4x-\frac{x^3}{3}[/tex]
When the kinetic energy is maximum then [tex]\frac{dK}{dx}=0[/tex]
[tex]\frac{d(2.4x-\frac{x^3}{3})}{dx}=0[/tex]
[tex]2.4-x^2=0[/tex]
[tex]x^2=2.4[/tex]
[tex]x=\pm\sqrt{2.4}[/tex]
[tex]\frac{d^2K}{dx^2}=-2x[/tex]
Substitute x=[tex]\sqrt{2.4}[/tex]
[tex]\frac{d^2K}{dx^2}=-2\sqrt{2.4}<0[/tex]
Substitute x=[tex]-\sqrt{2.4}[/tex]
[tex]\frac{d^2K}{dx^2}=2\sqrt{2.4}>0[/tex]
Hence, the kinetic energy is maximum at x=[tex]\sqrt{2.4}[/tex]
Again by work energy theorem , the maximum kinetic energy of the block between x=0 and x=2.0 m is given by
[tex]K_f-0=\int_{0}^{\sqrt{2.4}}(2.4-x^2)dx[/tex]
[tex]k_f=[2.4x-\frac{x^3}{3}]^{\sqrt{2.4}}_{0}[/tex]
[tex]K_f=2.4(\sqrt{2.4})-\frac{(\sqrt{2.4})^3}{3}=2.48 J[/tex]
Hence, the maximum energy of the block between x=0 and x=2 m=2.48 J
At 16°C, a rod is exactly 23.59 cm long on a steel ruler. Both the rod and the ruler are placed in an oven at 260°C, where the rod now measures 23.83 cm on the same ruler. What is the coefficient of thermal expansion for the material of which the rod is made? The linear expansion coefficient of steel is 11 x 10-6 /C°.
Answer:
[tex]5.28\times 10^{-5}\ /^{\circ}C[/tex]
Explanation:
[tex]L_0[/tex] = Original length of rod
[tex]\alpha[/tex] = Coefficient of linear expansion = [tex]1.62\times 10^{-5}\ /^{\circ}C[/tex]
Initial temperature = 16°C
Final temperature = 260°C
Change in length of a Steel is given by
[tex]\Delta L=\alpha L_0\Delta T\\\Rightarrow \Delta L=11\times 10^{-6}\times 23.83\times (260-16)\\\Rightarrow \Delta L=0.06395972\ cm[/tex]
Change in material rod length will be
[tex]23.83-23.59+0.0639572=0.3039572\ cm[/tex]
The coefficient of thermal expansion is given by
[tex]\alpha=\frac{\Delta L}{L_0\Delta T}\\\Rightarrow \alpha=\frac{0.3039572}{23.59\times (260-16)}\\\Rightarrow \alpha=5.28\times 10^{-5}\ /^{\circ}C[/tex]
The coefficient of thermal expansion for the material is [tex]5.28\times 10^{-5}\ /^{\circ}C[/tex]
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
The temperature of the Earth's surface is maintained by radiation from the Sun. By making the approximation that the Sun is a black body, but now assuming that the Earth is a grey body with albedo A (this means that it reflects a fraction A of the incident energy), show that the ratio of the Earth's temperature to that of the Sun is given by T_Earth = T_Sun (1 - A)^1/4 Squareroot R_Sun/2d, where R_Sun is the radius of the Sun and the Earth-Sun separation is D.
Answer:
T_t = Ts (1-A[tex])^{1/4}[/tex] √ (Rs/D)
Explanation:
The black body radiation power is given by Stefan's law
P = σ A e T⁴
This power is distributed over a spherical surface, so the intensity of the radiation is
I = P / A
Let's apply these formulas to our case. Let's start by calculating the power emitted by the Sun, which has an emissivity of one (e = 1) black body
P_s = σ A_s 1 T_s⁴
This power is distributed in a given area, the intensity that reaches the earth is
I = P_s / A
A = 4π R²
The distance from the Sun Earth is R = D
I₁ = Ps / 4π D²
I₁ = σ (π R_s²) T_s⁴ / 4π D²
I₁ = σ T_s⁴ R_s² / 4D²
Now let's calculate the power emitted by the earth
P_t = σ A_t (e) T_t⁴
I₂ = P_t / A_t
I₂ = P_t / 4π R_t²2
I₂ = σ (π R_t²) T_t⁴ / 4π R_t²2
I₂ = σ T_t⁴ / 4
The thermal equilibrium occurs when the emission of the earth is equal to the absorbed energy, the radiation affects less the reflected one is equal to the emitted radiation
I₁ - A I₁ = I₂
I₁ (1 - A) = I₂
Let's replace
σ T_s⁴ R_s²/4D² (1-A) = σ T_t⁴ / 4
T_s⁴ R_s² /D² (1-A) = T_t⁴
T_t⁴ = T_s⁴ (1-A) (Rs / D) 2
T_t = Ts (1-A[tex])^{1/4}[/tex] √ (Rs/D)
I examine the same second hand on the clock. Again, there are two points called A and B on the clock, with A farther from the center than B. Which of the following is true?
a. Point A has a higher angular acceleration about the center than Point B.
b. Point A has a lower angular acceleration about the center than Point B.
c. The angular acceleration for both points is 0.
d. None of the above.
I examine the same second hand on the clock. Again, there are two points called A and B on the clock, with A farther from the center than B. "The angular acceleration for both points is 0" is true.
Answer: Option C
Explanation:
The clock is in rotator motion. All the three hands of clock, move in same direction, but different speeds. And hence, we count hours, minutes and seconds. And, when we take each hand, they move related to the centre of the clock, where all the three are attached.
So, there is a centripetal acceleration which depends upon the velocity. But, the motion is uniform everywhere in the circle. The hands have no tangential acceleration. And hence, there is no angular acceleration, which is derived from tangential one. So, at any point, the angular acceleration is zero.
Moon does not have atmosphere as we have on Earth. On the earth, you can see the ground in someone’s shadow; on the moon, you can’t—the shadow is deep black. Explain the scientific reason behind this difference
One of the scientific characteristics that can make the difference between the Earth and the moon is the so-called Rayleigh dispersion effect. This concept is identified as the dispersion of visible light or any other electromagnetic radiation by particles whose size is much smaller than the wavelength of the dispersed photons. Our atmosphere allows even our 'shadows' to be clear.
On the moon under the absence of the atmosphere or any other mechanism that allows absorbing or failing to re-irradiate sunlight towards the area in its shadows, which makes the shadows on the moon look darker.
A spherical balloon is made from a material whose mass is 3.30 kg. The thickness of the material is negligible compared to the 1.25 m radius of the balloon. The balloon is filled with helium (He) at a temperature of 345 K and just floats in air, neither rising nor falling. The density of the surrounding air is 1.19 kg/m³ and the molar mass of helium is 4.0026×10-3 kg/mol. Find the absolute pressure of the helium gas.
Answer:
563712.04903 Pa
Explanation:
m = Mass of material = 3.3 kg
r = Radius of sphere = 1.25 m
v = Volume of balloon = [tex]\frac{4}{3}\pi r^3[/tex]
M = Molar mass of helium = [tex]4.0026\times 10^{-3}\ kg/mol[/tex]
[tex]\rho[/tex] = Density of surrounding air = [tex]1.19\ kg/m^3[/tex]
R = Gas constant = 8.314 J/mol K
T = Temperature = 345 K
Weight of balloon + Weight of helium = Weight of air displaced
[tex]mg+m_{He}g=\rho vg\\\Rightarrow m_{He}=\rho vg-m\\\Rightarrow m_{He}=1.19\times \frac{4}{3}\pi 1.25^3-3.3\\\Rightarrow m_{He}=6.4356\ kg[/tex]
Mass of helium is 6.4356 kg
Moles of helium
[tex]n=\frac{m}{M}\\\Rightarrow n=\frac{6.4356}{4.0026\times 10^{-3}}\\\Rightarrow n=1607.85489[/tex]
Ideal gas law
[tex]P=\frac{nRT}{v}\\\Rightarrow P=\frac{1607.85489\times 8.314\times 345}{\frac{4}{3}\pi 1.25^3}\\\Rightarrow P=563712.04903\ Pa[/tex]
The absolute pressure of the Helium gas is 563712.04903 Pa
Final answer:
To find the absolute pressure of helium gas in the balloon, we can use the ideal gas law. Since the thickness of the material is negligible compared to the radius of the balloon, we can consider the balloon as a sphere. Once we have the number of moles, we can substitute the values into the ideal gas law and solve for P, the absolute pressure.
Explanation:
To find the absolute pressure of helium gas in the balloon, we can use the ideal gas law: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature. In this case, we need to solve for P. Since the thickness of the material is negligible compared to the radius of the balloon, we can consider the balloon as a sphere and use the formula for the volume of a sphere: V = (4/3)πr^3, where r is the radius. Given that the radius is 1.25 m and the volume is known, we can calculate the number of moles of helium using the ideal gas law and the molar mass of helium.
Once we have the number of moles, we can substitute the values into the ideal gas law and solve for P, the absolute pressure. Remember to convert the temperature from Celsius to Kelvin by adding 273.15 to the given temperature.
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]
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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Two circular plates of radius 9cm are separated in air by 2.0mm, forming a parallel plate capacitor. A battery is connected across the plates. At a particular time, t1, the rate at which the charge is flowing through the battery from one plate to the other is 5A. (a)What is the time rate of change of the electric field between the plates at t1? (b)Compute the displacement current between the plates at t1, and show it is equal to 5A.
Answer:
(a) [tex]2.26\times 10^{13}\ N/C.s[/tex]
(b) 5 A
Solution:
As per the question:
Radius of the circular plate, R = 9 cm = 0.09 m
Distance, d = 2.0 mm = [tex]2.0\times 10^{- 3}\ m[/tex]
At [tex]t_{1}[/tex], current, I = 5 A
Now,
Area, A = [tex]\pi R^{2} = \pi 0.09^{2} = 0.025[/tex]
We know that the capacitance of the parallel plate capacitor, C = [tex]\frac{\epsilon_{o} A}{d}[/tex]
Also,
[tex]q = CV[/tex]
[tex]q = \frac{\epsilon A}{d}V[/tex]
Also,
[tex]V = \frac{E}{d}[/tex]
Now,
(a) The rate of change of electric field:
[tex]\frac{dE}{dt} = \frac{dq}{dt}(\frac{1}{A\epsilon_{o}})[/tex]
where
[tex]I = \frac{dq}{dt} = 5\ A[/tex]
[tex]\frac{dE}{dt} = 5\times (\frac{1}{0.025\times 8.85\times 10^{- 12}}) = 2.26\times 10^{13}\ N/C.s[/tex]
(b) To calculate the displacement current:
[tex]I_{D} = epsilon_{o}\times \frac{d\phi}{dt}[/tex]
where
[tex]\frac{d\phi}{dt}[/tex] = Rate of change of flux
[tex]I_{D} = Aepsilon_{o}\times \frac{dE}{dt}[/tex]
[tex]I_{D} = 0.025\times 8.85\times 10^{- 12}\times 2.26\times 10^{13} = 5\ A[/tex]
The rear wheels of a truck support 57.0 % of the weight of the truck, while the front wheels support 43.0 % of the weight. The center of gravity of the truck is 1.68 m in front of the rear wheels.
What is the wheelbase of the truck (the distance between the front and rear wheels)? Express your answer with the appropriate units.
To solve the problem it is necessary to apply the concepts related to torque, as well as the concepts where the Force is defined as a function of mass and acceleration, which in this case is gravity.
Considering the system in equilibrium, we perform sum of moments in the rear wheel (R2)
[tex]\sum M = 0[/tex]
[tex]F_g*1.68-R_1*d = 0[/tex]
[tex]mg*1.68-R_1*d = 0[/tex]
Another of the parameters given in the problem is that the front wheel supports 43% of the weight, that is
[tex]R1=0.43F_g[/tex]
[tex]R1=0.43mg[/tex]
Replacing, we have to
[tex]mg*1.68 -R_1*d = 0[/tex]
[tex]mg*1.68 -0.43mg*d = 0[/tex]
[tex]mg*1.68 =0.43mg*d[/tex]
[tex]1.68 = 0.43*d[/tex]
[tex]d =3.9m[/tex]
Therefore the wheelbase of the truck is 3.9m between the front and the rear.
With the use of physics principles, the truck's wheelbase, given that the rear wheels support 57.0% of the weight, and this weight acts at the center of gravity 1.68 meters in front of the rear wheels, is found to be 3.90 meters.
Explanation:The question asks for the wheelbase of the truck, which can be solved using physics principles such as torque and equilibrium. The center of gravity is the point where all the weight can be considered to be concentrated for the purpose of calculations. Here, we know that the rear wheels support 57.0% of the weight of the truck, and this weight acts at the center of gravity, which is 1.68 meters in front of the rear wheels. Given that the truck is in equilibrium (i.e., not tipping over), the torques about any point caused by the weight must cancel out. Hence, the total distance from the rear wheels (where the 57% weight acts) to the front wheels (where the 43% weight acts) is (1.68 m * 57.0 / 43.0) m = 2.22 meters. Therefore, the wheelbase of the truck is 1.68 m + 2.22 m = 3.90 meters.
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Babe Ruth steps to the plate and casually points to left center field to indicate the location of his next home run. The mighty Babe holds his bat across his shoulder, with one hand holding the small end of the bat.
The bat is horizontal, and the distance from the small end of the bat to the shoulder is 23.5 cm.
The bat has a mass of 1.30 kg and has a center of mass that is 70.0 cm from the small end of the bat.(a) Find the magnitude of the force exerted by the hand.(b) Find the direction of the force exerted by the hand.(c) Find the magnitude of the force exerted by the shoulder.(d) Find the direction of the force exerted by the shoulder.
Answer:
a) Fh = 25.23 N
b) The direction of the force exerted by the hand is pointed to the downward (negative) direction.
c) Fs = 37.97 N
d) The direction of the force exerted by the shoulder is pointed to the upward (positive) direction.
Explanation:
Given data
Distance from the small end of the bat to the shoulder: d = 23.5 cm
Distance from the small end of the bat to the center of mass: r = 70.0 cm
Mass of the bat: m = 1.30 Kg
This situation can be seen in the pic.
a) We can apply
∑τA = 0 ⇒ + (23.5cm)*Fh - (70 - 23.5)cm*m*g = 0
⇒ + (23.5cm)*Fh - (70 - 23.5)cm*(1.30 Kg)*(9.8 m/s²) = 0
⇒ Fh = 25.23 N (↓)
b) The direction of the force exerted by the hand is pointed to the downward (negative) direction.
c) We apply
∑Fy = 0 (↑)
⇒ - Fh + Fs - m*g = 0
⇒ Fs = Fh + m*g
⇒ Fs = 25.23 N + (1.30 Kg)*(9.8 m/s²)
⇒ Fs = 37.97 N (↑)
d) The direction of the force exerted by the shoulder is pointed to the upward (positive) direction.
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
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
One way to attack a satellite in Earth orbit is to launch a swarm of pellets in the same orbit as the satellite but in the opposite direction. Suppose a satellite in a circular orbit 810 km above Earth's surface collides with a pellet having mass 3.7 g. (a) What is the kinetic energy of the pellet in the reference frame of the satellite just before the collision? (b) What is the ratio of this kinetic energy to the kinetic energy of a 3.7 g bullet from a modern army rifle with a muzzle speed of 1200 m/s?
Answer:
411087.52089 J
[tex]\frac{K_r}{K_b}=154.31213[/tex]
Explanation:
R = Radius of Earth = 6370000 m
h = Altitude of satellite = 810 km
r = R+h = 63700000+810000 m
m = Mass of bullet = 3.7 g
Velocity of bullet = 1200 m/s
The relative velocity between the pellets and satellite is 2v
Now, the square of velocity is proportional to the kinetic energy
[tex]K\propto v^2[/tex]
[tex]\\\Rightarrow 4K\propto (2v)^2\\\Rightarrow 4K\propto 4v^2[/tex]
Kinetic energy in terms of orbital mechanics is
[tex]K=\frac{GMm}{2r}[/tex]
In this case relative kinetic energy is
[tex]K_r=4\frac{GMm}{2r}\\\Rightarrow K_r=2\frac{6.67\times 10^{-11}\times 5.98\times 10^{24}\times 3.7\times 10^{-3}}{(6370+810)\times 10^3}\\\Rightarrow K_r=411087.52089\ J[/tex]
The relative kinetic energy is 411087.52089 J
The ratio of kinetic energies is given by
[tex]\frac{K_r}{K_b}=\frac{411087.52089}{\frac{1}{2}\times 3.7\times 10^{-3}\times 1200^2}\\\Rightarrow \frac{K_r}{K_b}=154.31213[/tex]
The ratio is [tex]\frac{K_r}{K_b}=154.31213[/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
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
Two objects, X and Y, are held at rest on a horizontal frictionless surface and a spring is compressed between them. The mass of X is 2/5 times the mass of Y. Immediately after the spring is released, X has a kinetic energy of 50 J and Y has a kinetic energy of:
Answer:
20J
Explanation:
Using conservation law of momentum;
since the bodies were at rest, their initial momentum is zero
0 = M1Vx + M2Vy
- M1Vx = M2Vy where Vx is the final velocity of x after the spring has been release and Vy is final velocity of y and M1 and M2 are the masses of x and y
also M1 = 2/5 M2
substitute M1 into the the equation above
-2/5 M2Vx = M2Vy
cancel M2 on both side
-2/5Vx = Vy
comparing the kinetic energy of both x and y
for x K.E = 1/2 M1 Vx²
and y K.E = 1/2M2 Vy²
substitute for M1 = 2/5 M2
K.Ex = 1/2 × 2/5 M2 Vx²
divide K.Ex / K.Ey = (1/2 × 2/5 M2 Vx²) / 1/2 M2 Vy²
cancel the common terms
K.Ex / K.Ey = (2/5 Vx²) / Vy²
substitute -2/5Vx for Vy
(2/5 Vx²) / ( -2/5 Vx)² = (2/5 Vx²) / ( 4/25 Vx²)
cancel Vx²
(2/5) / (4/25) = 2/5 ÷ 4/25 = 2/5 × 25/4 = 5/2
the ratio of x and y kinetic energy is 5:2
since the kinetic energy of x is 50
50 : 20 = 5 : 2 if 10 is used to divide both sides
the kinetic energy of y = 20 J
Final answer:
The kinetic energy of object Y is 312.5 J.
Explanation:
The kinetic energy of an object is given by the equation KE = 1/2 mv^2, where m is the mass of the object and v is its velocity. Since object X has a kinetic energy of 50 J, we can use this equation to find the velocity of X. Rearranging the equation, we have v = sqrt(2KE/m). Plugging in the values, we get v = sqrt(2*50 / (2/5)m) = sqrt(500/m).
Since object Y has a mass that is 2/5 times the mass of X, its mass is (2/5)m. Therefore, its velocity can be calculated as v = sqrt(500 / (2/5)m) = sqrt(1250/m).
To find the kinetic energy of Y, we use the formula KE = 1/2 mv^2. Plugging in the mass of Y and its velocity, we get KE = 1/2 ((2/5)m) (sqrt(1250/m))^2 = 1/2 (5/10) m (1250/m) = 1/2 * (5/10) * 1250 = 312.5 J.
A cell composed of a platinum indicator electrode and a silver-silver chloride reference electrode in a solution containing both Fe 2 + and Fe 3 + has a cell potential of 0.693 V. If the silver-silver chloride electrode is replaced with a saturated calomel electrode (SCE), what is the new cell potential?
Answer:
0.639 V
Explanation:
The volatge of the cell containing both Ag/AgCl reference electrode and
[tex]Fe^{2+}/Fe^{3+}[/tex] electrode = 0.693 V
Thus,
[tex]E_{cathod}-E_{anode} =0.693 V[/tex]
E_{anode}=0.197 V
Note: potential of the silver-silver chloride reference electrode (0.197 V)
⇒E_{Cthode}= 0.693+0.197 = 0.890V
To calculate the voltage of the cell containing both the calomel reference
electrode and [tex]Fe^{2+}/Fe^{3+}[/tex] electrode as follows
Voltage of the cell = [tex]E_{cathod}-E_{anode}[/tex]
E_{anode}= calomel electrode= 0.241 V
Voltage of the cell = 0.890-0.241 = 0.639 V
Therefore, the new volatge is = 0.639 V
A string of length L, fixed at both ends, is capable of vibrating at 309 Hz in its first harmonic. However, when a finger is placed at a distance ℓ from one end, the remaining length L − ℓ of the string vibrates in its first harmonic with a frequency of 463 Hz. What is the distance ℓ? Express your answer as a ratio of the length L.
Answer:
i = 0.3326 L
Explanation:
A fixed string at both ends presents a phenomenon of standing waves, two waves with the same frequency that are added together. The expression to describe these waves is
2 L = n λ n = 1, 2, 3…
The first harmonic or leather for n = 1
Wave speed is related to wavelength and frequency
v = λ f
λ = v / f
Let's replace in the first equation
2 L = 1 (v / f₁)
For the shortest length L = L-l
2 (L- l) = 1 (v / f₂)
These two equations form our equation system, let's eliminate v
v = 2L f₁
v = 2 (L-l) f₂
2L f₁ = 2 (L-l) f₂
L- l = L f₁ / f₂
l = L - L f₁ / f₂
l = L (1- f₁ / f₂)
.
Let's calculate
l / L = (1- 309/463)
i / L = 0.3326
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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A discus thrower accelerates a discus from rest to a speed of 25.3 m/s by whirling it through 1.29 rev. Assume the discus moves on the arc of a circle 0.96 m in radius. A discus thrower moving in a circle as he prepares to throw the discus. (a) Calculate the final angular speed of the discus. rad/s
Answer:
ω = 26.35 rad/s
Explanation:
given,
speed of discus thrower = 25.3 m/s
whirling through = 1.29 rev
radius of circular arc = 0.96 m
final angular speed of the discus = ?
using formula
v = ω r
v is the velocity of disk
ω is the angular speed of the discus
r is the radius of arc
[tex]\omega = \dfrac{v}{r}[/tex]
[tex]\omega = \dfrac{25.3}{0.96}[/tex]
ω = 26.35 rad/s
Final angular speed of the discus is equal to ω = 26.35 rad/s
The final angular speed is approximately 26.354 rad/s. To find the final angular speed, we use the relation [tex]\omega = v / r[/tex]. With v = 25.3 m/s and r = 0.96 m.
Calculate the final angular speed (ω) using the linear speed provided. Rotational motion relies on the relationship between linear speed (v) and angular speed (ω), which is expressed as [tex]v = r\cdot \omega[/tex], where r is the circular path radius. Rearranging this equation yields [tex]\omega = v / r[/tex], demonstrating the angular speed's direct and inverse relationship.
[tex]\omega = v / r[/tex]Here, v = 25.3 m/s (final linear speed) and r = 0.96 m (radius of the circle):
[tex]\omega = 25.3\left m/s / 0.96\left m \approx 26.354\left rad/s[/tex]Conclusion:
Thus, the final angular speed of the discus is approximately 26.354 rad/s.