The question is incomplete. Here is the complete question.
The isotope of Plutonium 238Pu is used to make thermoeletric power sources for spacecraft. Suppose that a space probe was launched in 2012 with 3.5 kg of 238Pu.
(a) If the half-life of 238Pu is 87.7 yr, write a function of the form [tex]Q(t)=Q_{0}e^{-kt}[/tex] to model the quantity Q(t) of 238Pu left after t years. Round ythe value of k to 3 decimal places. Do not round intermediate calculations.
(b) If 1.7kg of 238Pu is required to power the spacecraft's data transmitter, for low long after launch would scientists be able to receive data? Round to the nearest year. Do not round intermediate calculations.
Answer: (a) [tex]Q(t)=3.5e^{-0.0079t}[/tex]
(b) 91 years.
Explanation:
(a) Half-life is time it takes a substance to decrease to half of itself, i.e.:
Q(t) = [tex]0.5Q_{0}[/tex]
[tex]0.5Q_{0}=Q_{0}e^{-87.7k}[/tex]
[tex]0.5=e^{-87.7k}[/tex]
[tex]ln(0.5)=ln(e^{-87.7k})[/tex]
[tex]ln(0.5)=-87.7k[/tex]
[tex]k = \frac{ln(0.5)}{-87.7}[/tex]
k = 0.0079
Knowing k and [tex]Q_{0}[/tex]=3.5kg, function is [tex]Q(t)=3.5e^{-0.0079t}[/tex]
(b) Using function:
[tex]Q(t)=3.5e^{-0.0079t}[/tex]
[tex]1.7=3.5e^{-0.0079t}[/tex]
[tex]e^{-0.0079t}=\frac{1.7}{3.5}[/tex]
[tex]e^{-0.0079t}=0.4857[/tex]
[tex]ln(e^{-0.0079t})=ln(0.4857)[/tex]
[tex]-0.0079t=-0.7221[/tex]
[tex]t = \frac{-0.7221}{-0.0079}[/tex]
t = 91.41
t ≈ 91 years
Scientists will be able to receive data for approximately 91 years.
Matter's resistance to a change in motion is called
Answer:
Inertia! I hope this helps!
Answer:
inertia
Explanation
Inertia.
The linear density rho in a rod 3 m long is 8/ x + 1 kg/m, where x is measured in meters from one end of the rod. Find the average density rhoave of the rod.
Answer:
The average density of the rod is 1.605 kg/m.
Explanation:
The average density of the rod is given by:
[tex] \rho = \frac{m}{l} [/tex]
To find the average density we need to integrate the linear density from x₁ = 0 to x₂ = 3, as follows:
[tex] \int_{0}^{3} \frac{8}{3(x + 1)}dx [/tex]
[tex] \rho = \frac{8}{3} \int_{0}^{3} \frac{1}{(x + 1)}dx [/tex] (1)
Using u = x+1 → du = dx → u₁= x₁+1 = 0+1 = 1 and u₂ = x₂+1 = 3+1 = 4
By entering the values above into (1), we have:
[tex] \rho = \frac{8}{3} \int_{0}^{3} \frac{1}{u}du [/tex]
[tex]\rho = \frac{8}{3}*log(u)|_{1}^{4} = \frac{8}{3}[log(4) - log(1)] = 1.605 kg/m[/tex]
Therefore, the average density of the rod is 1.605 kg/m.
I hope it helps you!
The average density of the rod is [tex]1.605 \;\rm kg/m^{3}[/tex].
Given data:
The length of rod is, L = 3 m.
The linear density of rod is, [tex]\rho=\dfrac{8}{x+1} \;\rm kg/m[/tex].
To find the average density we need to integrate the linear density from x₁ = 0 to x₂ = 3, The expression for the average density is given as,
[tex]\rho' = \int\limits^3_0 { \rho} \, dx\\\\\\\rho' = \int\limits^3_0 { \dfrac{m}{L}} \, dx\\\\\\\rho' = \int\limits^3_0 {\dfrac{8}{3(x+1)}} \, dx[/tex]............................................................(1)
Using u = x+1
du = dx
u₁= x₁+1 = 0+1 = 1
and
u₂ = x₂+1 = 3+1 = 4
By entering the values above into (1), we have:
[tex]\rho' =\dfrac{8}{3} \int\limits^3_0 {\dfrac{1}{u}} \, du\\\\\\\rho' =\dfrac{8}{3} \times [log(u)]^{4}_{1}\\\\\\\rho' =\dfrac{8}{3} \times [log(4)-log(1)]\\\\\\\rho' =1.605 \;\rm kg/m^{3}[/tex]
Thus, we can conclude that the average density of the rod is [tex]1.605 \;\rm kg/m^{3}[/tex].
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If you wanted to make your own lenses for a telescope, what features of a lens do you think would affect the images that you can see
Answer:
Therefore the characteristics to be found are:
* the focal length must be large and the focal length of the eyepiece must be small
* The diameter of the objective lens should be as large as possible, to be able to collect small without need from light
* The system must be configured to the far sight tip,
Explanation:
The length of the telescope is
L = f_ocular + f_objetive
the magnification of the telescope is
m = - f_objective / f_ocular
These are the two equations that describe the behavior of the telescope. Therefore the characteristics to be found are:
* the focal length must be large and the focal length of the eyepiece must be small
* The diameter of the objective lens should be as large as possible, to be able to collect small without need from light
* The system must be configured to the far sight tip,
"Determine the magnitude of the net force of gravity acting on the Moon during an eclipse when it is directly between Earth and the Sun."
Answer:
Net force = 2.3686 × 10^(20) N
Explanation:
To solve this, we have to find the force of the earth acting on the moon and the force of the sun acting on the moon and find the difference.
Now, from standards;
Mass of earth;M_e = 5.98 × 10^(24) kg
Mass of moon;M_m = 7.36 × 10^(22) kg
Mass of sun;M_s = 1.99 × 10^(30) kg
Distance between the sun and earth;d_se = 1.5 × 10^(11) m
Distance between moon and earth;d_em = 3.84 × 10^(8) m
Distance between sun and moon;d_sm = (1.5 × 10^(11)) - (3.84 × 10^(8)) = 1496.96 × 10^(8) m
Gravitational constant;G = 6.67 × 10^(-11) Nm²/kg²
Now formula for gravitational force between the earth and the moon is;
F_em = (G × M_e × M_m)/(d_em)²
Plugging in relevant values, we have;
F_em = (6.67 × 10^(-11) × 5.98 × 10^(24) × 7.36 × 10^(22))/(3.84 × 10^(8))²
F_em = 1.9909 × 10^(20) N
Similarly, formula for gravitational force between the sun and moon is;
F_sm = (G × M_s × M_m)/(d_sm)²
Plugging in relevant values, we have;
F_se = (6.67 × 10^(-11) × 1.99 × 10^(30) ×
7.36 × 10^(22))/(1496.96 × 10^(8))²
F_se = 4.3595 × 10^(20) N
Thus, net force = F_se - F_em
Net force = (4.3595 × 10^(20) N) - (1.9909 × 10^(20) N) = 2.3686 × 10^(20) N
When you slosh the water back and forth in a tub at just the right frequency, the water alternately rises and falls at each end, remaining relatively calm at the center. Suppose the frequency to produce such a standing wave in a 55m wide tub is 0.80 Hz.
Required:
What is the speed of the water wave?
Answer:
The speed of the water wave is [tex]v = 88 \ m/s[/tex]
Explanation:
From the question we are told that
The width of the tube is [tex]L = 55 \ m[/tex]
The fundamental frequency is [tex]f = 0.80 \ Hz[/tex]
Generally the fundamental frequency is mathematically represented as
[tex]f = \frac{v}{2 * L }[/tex]
=> [tex]v = f * 2 * L[/tex]
substituting values
[tex]v = 0.8 * 2 * 55[/tex]
[tex]v = 88 \ m/s[/tex]
The speed of the water wave will be 88 m/s.
Given information:
When you slosh the water back and forth in a tub at just the right frequency, the water alternately rises and falls at each end, remaining relatively calm at the center.
The frequency of the standing wave is [tex]f=0.8[/tex] Hz.
The width of the tub is [tex]w=55[/tex] m.
Let v be the speed of the standing wave.
The speed of the wave can be calculated as,
[tex]v=2wf\\v=2\times 55\times 0.8\\v=88\rm\; m/s[/tex]
Therefore, the speed of the water wave will be 88 m/s.
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A plastic dowel has a Young's Modulus of 1.50 ✕ 1010 N/m2. Assume the dowel will break if more than 1.50 ✕ 108 N/m2 is exerted.
(a) What is the maximum force (in kN) that can be applied to the dowel assuming a diameter of 2.40 cm?
______Kn
(b) If a force of this magnitude is applied compressively, by how much (in mm) does the 26.0 cm long dowel shorten? (Enter the magnitude.)
mm
Answer:
a
[tex]F = 67867.2 \ N[/tex]
b
[tex]\Delta L = 2.6 \ mm[/tex]
Explanation:
From the question we are told that
The Young modulus is [tex]Y = 1.50 *10^{10} \ N/m^2[/tex]
The stress is [tex]\sigma = 1.50 *10^{8} \ N/m^2[/tex]
The diameter is [tex]d = 2.40 \ cm = 0.024 \ m[/tex]
The radius is mathematically represented as
[tex]r =\frac{d}{2} = \frac{0.024}{2} = 0.012 \ m[/tex]
The cross-sectional area is mathematically evaluated as
[tex]A = \pi r^2[/tex]
[tex]A = 3.142 * (0.012)^2[/tex]
[tex]A = 0.000452\ m^2[/tex]
Generally the stress is mathematically represented as
[tex]\sigma = \frac{F}{A}[/tex]
=> [tex]F = \sigma * A[/tex]
=> [tex]F = 1.50 *10^{8} * 0.000452[/tex]
=> [tex]F = 67867.2 \ N[/tex]
Considering part b
The length is given as [tex]L = 26.0 \ cm = 0.26 \ m[/tex]
Generally Young modulus is mathematically represented as
[tex]E = \frac{ \sigma}{ strain }[/tex]
Here strain is mathematically represented as
[tex]strain = \frac{ \Delta L }{L}[/tex]
So
[tex]E = \frac{ \sigma}{\frac{\Delta L }{L} }[/tex]
[tex]E = \frac{\sigma }{1} * \frac{ L}{\Delta L }[/tex]
=> [tex]\Delta L = \frac{\sigma * L }{E}[/tex]
substituting values
[tex]\Delta L = \frac{ 1.50*10^{8} * 0.26 }{ 1.50 *10^{10 }}[/tex]
[tex]\Delta L = 0.0026[/tex]
Converting to mm
[tex]\Delta L = 0.0026 *1000[/tex]
[tex]\Delta L = 2.6 \ mm[/tex]
A lab technician uses laser light with a wavelength of 650 nmnm to test a diffraction grating. When the grating is 42.0 cmcm from the screen, the first-order maxima appear 6.09 cmcm from the center of the pattern. How many lines per millimeter does this grating have?
Answer:
221 lines per millimetre
Explanation:
We know that for a diffraction grating, dsinθ =mλ where d = spacing between grating, θ = angle to maximum, m = order of maximum and λ = wavelength of light.
Since the grating is 42.0 cm from the screen and its first order maximum (m = 1) is at 6.09 cm from the center of the pattern,
tanθ = 6.09 cm/42.0 cm = 0.145
From trig ratios, cot²θ + 1 = cosec²θ
cosecθ = √((1/tanθ)² + 1) = √((1/0.145)² + 1) = √48.562 = 6.969
sinθ = 1/cosecθ = 1/6.969 = 0.1435
Also, sinθ = mλ/d at the first-order maximum, m = 1. So
sinθ = (1)λ/d = λ/d
Equating both expressions we have
0.1435 = λ/d
d = λ/0.1435
Now, λ = 650 nm = 650 × 10⁻⁹ m
d = 650 × 10⁻⁹ m/0.1435
d = 4529.62 × 10⁻⁹ m per line
d = 4.52962 × 10⁻⁶ m per line
d = 0.00452962 × 10⁻³ m per line
d = 0.00452962 mm per line
Since d = width of grating/number of lines of grating
Then number of lines per millimetre = 1/grating spacing
= 1/0.00452962
= 220.77 lines per millimetre
≅ 221 lines per millimetre since we can only have a whole number of lines.
We can reasonably model a 75 W incandescent light bulb as a sphere 6.0 cm in diameter. Typically only about 5% of the energy goes to visible light; the rest goes largely to non-visible infrared radiation. (a) What is the visible light intensity at the surface of the bulb
Answer:
Visible light intensity at the surface of the bulb (I) = 331 W/m²
Explanation:
Given:
Energy = 75 W
Radius = 6 /2 = 3 cm = 3 × 10⁻² m
Energy goes to visible light = 5% = 0.05
Find:
Visible light intensity at the surface of the bulb (I)
Computation:
Visible light intensity at the surface of the bulb (I) = P / 4A
Visible light intensity at the surface of the bulb (I) = (0.05)(75) / 4π(3 × 10⁻²)²
Visible light intensity at the surface of the bulb (I) = 3.75 / 4π(9 × 10⁻⁴)
Visible light intensity at the surface of the bulb (I) = 331 W/m²
A charge of 15 is moving with velocity of 6.2 x17 which makes an angle of 48 degrees with respect to the magnetic field. If the force on the particle is 4838 N, find the magnitude of the magnetic field.
a. 06.0T.
b. 08.0T.
c. 07.0T.
d. 05.0 T.
Complete question:
A charge of 15C is moving with velocity of 6.2 x 10³ m/s which makes an angle of 48 degrees with respect to the magnetic field. If the force on the particle is 4838 N, find the magnitude of the magnetic field.
a. 0.06 T
b. 0.08 T
c. 0.07 T
d. 0.05 T
Answer:
The magnitude of the magnetic field is 0.07 T.
Explanation:
Given;
magnitude of the charge, q = 15C
velocity of the charge, v = 6.2 x 10³ m/s
angle between the charge and the magnetic field, θ = 48°
the force on the particle, F = 4838 N
The magnitude of the magnetic field can be calculated by applying Lorentz force formula;
F = qvBsinθ
where;
B is the magnitude of the magnetic field
B = F / vqsinθ
B = (4838) / (6.2 x 10³ x 15 x sin48)
B = 0.07 T
Therefore, the magnitude of the magnetic field is 0.07 T.
A laboratory electromagnet produces a magnetic field of magnitude 1.38 T. A proton moves through this field with a speed of 5.86 times 10^6 m/s.
a. Find the magnitude of the maximum magnetic force that could be exerted on the proton.
b. What is the magnitude of the maximum acceleration of the proton?
c. Would the field exert the same magnetic force on an electron moving through the field with the same speed? (Assume that the electron is moving in the direction as the proton.)
1. Yes
2. No
.Answer;
Using Fmax=qVB
F=(1.6*10^-19 C)(5.860*10^6 m/s)(1.38 T)
ANS=1.29*10^-12 N
2. Using Amax=Fmax/ m
Amax =(1.29*10^-12 N) / (1.67*10^-27 kg)
ANS=1.93*10^15 m/s^2*
3. No, the acceleration wouldn't be the same. Since The magnitude of the electron is equal to that of the proton, but the direction would be in the opposite direction and also Since an electron has a smaller mass than a proton
Does the moon light originate from the moon only
Answer:
No
Explanation:
Moon has no light of its own. It just shines because its surface reflects light from the sun and that's what we see.
:-)
PLEASE HELP FAST The object distance for a convex lens is 15.0 cm, and the image distance is 5.0 cm. The height of the object is 9.0 cm. What is the height of the image?
Answer:
The image height is 3.0 cm
Explanation:
Given;
object distance, [tex]d_o[/tex] = 15.0 cm
image distance, [tex]d_i[/tex] = 5.0 cm
height of the object, [tex]h_o[/tex] = 9.0 cm
height of the image, [tex]h_i[/tex] = ?
Apply lens equation;
[tex]\frac{h_i}{h_o} = -\frac{d_i}{d_o}\\\\ h_i = h_o(-\frac{d_i}{d_o})\\\\h_i = -9(\frac{5}{15} )\\\\h_i = -3 \ cm[/tex]
Therefore, the image height is 3.0 cm. The negative values for image height indicate that the image is an inverted image.
The magnetic field of a plane-polarized electromagnetic wave moving in the z-direction is given by in SI units. What is the frequency of the wave
Complete Question
The magnetic field of a plane-polarized electromagnetic wave moving in the z-direction is given by
[tex]B=1.2* 10^{-6} sin [2\pi[(\frac{z}{240} ) - ( \frac{t * 10^7}{8} ) ] ][/tex] in SI units.
Answer:
The value is [tex]f = 1.98918*10^{5}\ Hz[/tex]
Explanation:
From the question we are told that
The magnetic field is [tex]B=1.2* 10^{-6} sin [2\pi[(\frac{z}{240} ) - ( \frac{t * 10^7}{8} ) ] ][/tex]
This above equation can be modeled as
[tex]B=1.2* 10^{-6} sin [2\pi[(\frac{z}{240} ) - ( \frac{t * 10^7}{8} ) ] ] \equiv A sin ( kz -wt )[/tex]
So
[tex]w = \frac{10^7}{8}[/tex]
Generally the frequency is mathematically represented as
[tex]f = \frac{w}{2 \pi}[/tex]
=> [tex]f = \frac{ \frac{10^7}{8} }{2 \pi}[/tex]
=> [tex]f = 1.98918*10^{5}\ Hz[/tex]
A battery establishes a voltage V on a parallel-plate capacitor. After the battery is disconnected, the distance between the plates is doubled without loss of charge. Accordingly:_____.
a. stay same
b. increases
c. decreases
d. the capacitance decreases and the voltage between the plates increases.
Answer:
d.
Explanation:
Since, the capacitance( decreases )
therefore voltage between the plates(increases ).
Hence, option d is correct.
C =εA/d.
d is doubled, therefore C decrease ( inverse relation).
D) The capacitance decreases and the voltage between the plates increases.
BatteryA battery establishes a voltage V on a parallel-plate capacitor. After the battery is disconnected, the distance between the plates is doubled without loss of charge. Accordingly, the capacitance decreases and the voltage between the plates increases.
The capacitance - (decreases)
The voltage between the plates- (increases ).
Thus, the correct answer is D.
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A rectangular coil having N turns and measuring 15 cm by 25 cm is rotating in a uniform 1.6-T magnetic field with a frequency of 75 Hz. The rotation axis is perpendicular to the direction of the field. If the coil develops a sinusoidal emf of maximum value 56.9 V, what is the value of N?
A) 2
B) 4
C) 6
D) 8
E) 10
Answer:
A) 2
Explanation:
Given;
magnetic field of the coil, B = 1.6 T
frequency of the coil, f = 75 Hz
maximum emf developed in the coil, E = 56.9 V
area of the coil, A = 0.15 m x 0.25 m = 0.0375 m²
The maximum emf in the coil is given by;
E = NBAω
Where;
N is the number of turns
ω is the angular velocity = 2πf = 2 x 3.142 x 75 = 471.3 rad/s
N = E / BAω
N = 56.9 / (1.6 x 0.0375 x 471.3)
N = 2 turns
Therefore, the value of N is 2
A) 2
Two waves are traveling in the same direction along a stretched string. The waves are 45.0° out of phase. Each wave has an amplitude of 7.00 cm. Find the amplitude of the resultant wave.
Answer:
The amplitude of the resultant wave is 12.93 cm.
Explanation:
The amplitude of resultant of two waves, y₁ and y₂, is given as;
Y = y₁ + y₂
Let y₁ = A sin(kx - ωt)
Since the wave is out phase by φ, y₂ is given as;
y₂ = A sin(kx - ωt + φ)
Y = y₁ + y₂ = 2A Cos (φ / 2)sin(kx - ωt + φ/2 )
Given;
phase difference, φ = 45°
Amplitude, A = 7.00 cm
Y = 2(7) Cos (45 /2) sin(kx - ωt + 22.5° )
Y = 12.93 cm
Therefore, the amplitude of the resultant wave is 12.93 cm.
The block moves up an incline with constant speed. What is the total work WtotalWtotalW_total done on the block by all forces as the block moves a distance LLL
Answer:
External force W₁ = F L
Friction force W₂ = - fr L
weight component W₃ = - mg sin θ L
Y Axis Force W=0
Explanation:
When the block rises up the plane with constant velocity, it implies that the sum of the forces is zero.
For these exercises it is indicated to create a reference system with the x axis parallel to the plane and the y axis perpendicular
let's write the equations of translational equilibrium in given exercise
X axis
F - fr -Wₓ = 0
F = fr + Wₓ
the components of the weight can be found using trigonometry
Wₓ = W sin θ
[tex]W_{y}[/tex] = W cos θ
let's look for the work of these three forces
W = F x cos θ
External force
W₁ = F L
since the displacement and the force have the same direction
Friction force
W₂ = - fr L
since the friction force is in the opposite direction to the displacement
For the weight component
W₃ = - mg sin θ L
because the weight component is contrary to displacement
Y Axis
N- Wy = 0
in this case the forces are perpendicular to the displacement, the angle is 90º and the cosine 90 = 0
therefore work is worth zero
A cylindrical container with a cross-sectional area of 66.2 cm2 holds a fluid of density 856 kg/m3 . At the bottom of the container the pressure is 119 kPa . Assume Pat = 101 kPa
A) What is the depth of the fuild?
B) Find the pressure at the bottom of the container after an additional 2.35×10−3 m3 of this fluid is added to the container. Assume that no fluid spills out of the container.
Answer:
A. h = 2.15 m
B. Pb' = 122 KPa
Explanation:
The computation is shown below:
a) Let us assume the depth be h
As we know that
[tex]Pb - Pat = d \times g \times h \\\\ ( 119 - 101) \times 10^3 = 856 \times 9.8 \times h[/tex]
After solving this,
h = 2.15 m
Therefore the depth of the fluid is 2.15 m
b)
Given that
height of the extra fluid is
[tex]h' = \frac{2.35 \times 10^{-3}}{ area} \\\\ h' = \frac{2.35 \times 10^{-3}} { 66.2 \times 10^{-4}}[/tex]
h' = 0.355 m
Now let us assume the pressure at the bottom is Pb'
so, the equation would be
[tex]Pb' - Pat = d \times g \times (h + h')\\\\Pb' = 856 \times 9.8 \times ( 2.15 + 0.355) + 101000[/tex]
Pb' = 122 KPa
(A) The depth of the fluid is 2.14 m.
(B) The new pressure at the bottom of container is 121972 Pa.
Given data:
The cross-sectional area of the container is, [tex]A =66.2 \;\rm cm^{2}=66.2 \times 10^{-4} \;\rm m^{2}[/tex].
The density of fluid is, [tex]\rho = 856 \;\rm kg/m^{3}[/tex].
The container pressure at bottom is, [tex]P=119 \;\rm kPa=119 \times 10^{3} \;\rm Pa[/tex].
The atmospheric pressure is, [tex]P_{at}=101 \;\rm kPa=101 \times 10^{3}\;\rm Pa[/tex].
(A)
The given problem is based on the net pressure on the container, which is equal to the difference between the pressure at the bottom and the atmospheric pressure. Then the expression is,
[tex]P_{net} = P-P_{at}\\\\\rho \times g \times h= P-P_{at}[/tex]
Here, h is the depth of fluid.
Solving as,
[tex]856\times 9.8 \times h= (119-101) \times 10^{3}\\\\h=\dfrac{ (119-101) \times 10^{3}}{856\times 9.8}\\\\h= 2.14 \;\rm m[/tex]
Thus, the depth of the fluid is 2.14 m.
(B)
For an additional volume of [tex]2.35 \times 10^{-3} \;\rm m^{3}[/tex] to the liquid, the new depth is,
[tex]V=A \times h'\\\\h'=\dfrac{2.35 \times 10^{-3}}{66.2 \times 10^{-4}}\\\\h'=0.36 \;\rm m[/tex]
Now, calculate the new pressure at the bottom of the container as,
[tex]P'-P_{at}= \rho \times g \times (h+h')\\\\\P'-(101 \times 10^{3})= 856 \times 9.8 \times (2.14+0.36)\\\\P'=121972 \;\rm Pa[/tex]
Thus, we can conclude that the new pressure at the bottom of container is 121972 Pa.
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A 1.25-kg ball begins rolling from rest with constant angular acceleration down a hill. If it takes 3.60 s for it to make the first complete revolution, how long will it take to make the next complete revolution?
Answer:
The time taken is [tex]\Delta t = 1.5 \ s[/tex]
Explanation:
From the question we are told that
The mass of the ball is [tex]m = 1.25 \ kg[/tex]
The time taken to make the first complete revolution is t= 3.60 s
The displacement of the first complete revolution is [tex]\theta = 1 rev = 2 \pi \ radian[/tex]
Generally the displacement for one complete revolution is mathematically represented as
[tex]\theta = w_i t + \frac{1}{2} * \alpha * t^2[/tex]
Now given that the stone started from rest [tex]w_i = 0 \ rad / s[/tex]
[tex]2 \pi =0 + 0.5* \alpha *(3.60)^2[/tex]
[tex]\alpha = 0.9698 \ s[/tex]
Now the displacement for two complete revolution is
[tex]\theta_2 = 2 * 2\pi[/tex]
[tex]\theta_2 = 4\pi[/tex]
Generally the displacement for two complete revolution is mathematically represented as
[tex]4 \pi = 0 + 0.5 * 0.9698 * t^2[/tex]
=> [tex]t^2 = 25.9187[/tex]
=> [tex]t= 5.1 \ s[/tex]
So
The time taken to complete the next oscillation is mathematically evaluated as
[tex]\Delta t = t_2 - t[/tex]
substituting values
[tex]\Delta t = 5.1 - 3.60[/tex]
[tex]\Delta t = 1.5 \ s[/tex]
The time for the ball to complete the next revolution is 1.5 s.
The given parameters;
mass of the ball, m = 1.25 kgtime of motion, t = 3.6 sone complete revolution, θ = 2πThe constant angular acceleration of the ball is calculated as follows;
[tex]\theta = \omega t \ + \ \frac{1}{2} \alpha t^2\\\\2\pi = 0 \ + \ 0.5(3.6)^2 \alpha\\\\2\pi = 6.48 \alpha \\\\\alpha = \frac{2 \pi }{6.48} \\\\\alpha = 0.97 \ rad/s^2[/tex]
The time to complete the next revolution is calculated as follows;
[tex]4\pi = 0 + \frac{1}{2} (0.97)t^2\\\\8\pi = 0.97t^2\\\\t^2 = \frac{8\pi }{0.97} \\\\t^2 = 25.91\\\\t = \sqrt{ 25.91} \\\\t = 5.1 \ s[/tex]
[tex]\Delta t = 5.1 \ s \ - \ 3.6 \ s \\\\\Delta t = 1.5 \ s[/tex]
Thus, the time for the ball to complete the next revolution is 1.5 s.
Learn more here:https://brainly.com/question/20738528
A stereo speaker produces a pure "G" tone, with a frequency of 392 Hz. What is the period T of the sound wave produced by the speaker?
Answer:
The period is [tex]T = 0.00255 \ s[/tex]
Explanation:
From the question we are told that
The frequency is [tex]f = 392 \ Hz[/tex]
Generally the period is mathematically represented as
[tex]T = \frac{1}{f}[/tex]
=> [tex]T = \frac{1}{ 392}[/tex]
=> [tex]T = 0.00255 \ s[/tex]
An archer practicing with an arrow bow shoots an arrow straight up two times. The first time the initial speed is vi and second
time he increases the initial sped to 4v. How would you compare the maximum height in the second trial to that in the first trial?
Answer:
The maximum height reached in the second trial is 16times the maximum height reached in the first trial.
Explanation:
The following data were obtained from the question:
First trial
Initial speed (u) = v
Final speed (v) = 0
Second trial
Initial speed (u) = 4v
Final speed (v) = 0
Next, we shall obtain the expression for the maximum height reached in each case.
This is illustrated below:
First trial:
Initial speed (u) = v
Final speed (v) = 0
Acceleration due to gravity (g) = 9.8 m/s²
Height (h₁) =.?
v² = u² – 2gh₁ (going against gravity)
0 = (v)² – 2 × 9.8 × h₁
0 = v² – 19.6 × h₁
Rearrange
19.6 × h₁ = v²
Divide both side by 19.6
h₁ = v²/19.6
Second trial
Initial speed (u) = 4v
Final speed (v) = 0
Acceleration due to gravity (g) = 9.8 m/s²
Height (h₂) =.?
v² = u² – 2gh₂ (going against gravity)
0 = (4v)² – 2 × 9.8 × h₂
0 = 16v² – 19.6 × h₂
Rearrange
19.6 × h₂ =16v²
Divide both side by 19.6
h₂ = 16v²/19.6
Now, we shall determine the ratio of the maximum height reached in the second trial to that of the first trial.
This is illustrated below:
Second trial:
h₂ = 16v²/19.6
First trial:
h₁ = v²/19.6
Second trial : First trial
h₂ : h₁
h₂ / h₁ = 16v²/19.6 ÷ v²/19.6
h₂ / h₁ = 16v²/19.6 × 19.6/v²
h₂ / h₁ = 16
h₂ = 16 × h₁
From the above illustrations, we can see that the maximum height reached in the second trial is 16times the maximum height reached in the first trial.
Thomas and Lilian are walking down the street to get to the corner store. They walk 5 blocks up the street and turn right by the stop sign. Once they turn at the stop sign they continue walking for 8 more blocks. They make a left, walk 2 blocks and cross the street to arrive at the corner store. While there they purchase a few snacks, sit at the curb, and then walk back home where they originally started. Thomas and Lilian are discussing their walk in reference to their overall displacement and distance. They seem to be in disagreement about their journey. Thomas says their overall displacement and distance are both zero, because they are back where they started. Lilian thinks their total distance and displacement are greater than zero.
Which person do you most agree with?
You are not expected to actually calculate in order to solve this problem.
Answer:
Thomas is correct that the zero displacements
Lilian is right that the distance is greater than zero.
Explanation:
In this problem we have to be clear about the difference between displacement and distance.
The displacement is a vector, that is, it has a modulation and direction, in this case we can draw a vector for the outward trip and another vector for the return trip, both will have the same magnitude, but their directions are opposite, so the resulting vector is zero.
The distance is a scalar and its value coincides with the modulus of the distance vector, in our case the distance is d for the outward journey and d for the return journey, so the total distance is 2d, which is different from zero.
The two students have some reason, but neither complete,
The displacement is zero because it is a vector and
the distance is different from zero (2d) because it is a scalar
Thomas is correct that the zero displacements
Lilian is right that the distance is greater than zero.
Therefore I agree with both, because each one has a 50% of the reason
The advantage of a hydraulic lever is A : it transforms a small force acting over a large distance into a large force acting over a small distance. B : it transforms a small force acting over a small distance into a large force acting over a large distance. C : it allows you to exert a larger force with less work. D : it transforms a large force acting over a large distance into a small force acting over a small distance. E : it transforms a large force acting over a small distance into a small force acting over a large distance.
Answer:
A) it transforms a small force acting over a large distance into a large force acting over a small distance.
Explanation:
The hydraulic lever works based on Pascal's law of transmission of pressure through a fluid. In the hydraulic lever, the pressure transmitted is the same.
Pressure transmitted P = F/A
where F is the force applied
and A is the area over which the force is applied.
This pressure can be manipulated on the input end as a small force applied over a small area, and then be transmitted to the output end as a large force over a large area.
F/A = f/a
where the left side of the equation is for the output, and the right side is for the input.
The volume of the displaced fluid will be the same on both ends of the hydraulic lever. Since we know that
volume V = (area A) x (distance d)
this means that the the piston on the input smaller area of the hydraulic lever will travel a greater distance, while the piston on the larger output area of the lever will travel a small distance.
From all these, we can see that the advantage of a hydraulic lever is that it transforms a small force acting over a large distance into a large force acting over a small distance.
How much heat is required to convert 5.0 kg of ice from a temperature of - 20 0C to water at a temperature of 205 0F
Answer:
Explanation:
To convert from °C to °F , the formula is
( F-32 ) / 9 = C / 5
F is reading fahrenheit scale and C is in centigrade scale .
F = 205 , C = ?
(205 - 32) / 9 = C / 5
C = 96°C approx .
Let us calculate the heat required .
Total heat required = heat required to heat up the ice at - 20 °C to 0°C + heat required to melt the ice + heat required to heat up the water at 0°C to
96°C.
= 5 x 2.04 x (20-0) + 5 x 336 + 5 x ( 96-0 ) x 4.2 kJ .
= 204 + 1680 + 2016
= 3900 kJ .
What is temperature?
O A. The force exerted on an area
B. A measure of mass per unit volume
O C. The net energy transferred between two objects
OD. A measure of the movement of atoms or molecules within an
object
Answer:
The net energy transferred between two objects
Explanation:
The physical property of matter that expresses hot or cold is called temperature. It demonstrates the thermal energy. A thermometer is used to measure temperature. It defines the rate to which the chemical reaction occurs. It tells about the thermal radiation emitted from an object.
The correct option that defines temperature is option C.
Answer:
A measure of the movement of atoms or molecules within an object
Explanation:
Process of elimination
A roller coaster starts from rest at its highest point and then descends on its (frictionless) track. Its speed is 26 m/s when it reaches ground level. What was its speed when its height was half that of its starting point?
Answer:
The velocity is [tex]v_h = 19.2 \ m/s[/tex]
Explanation:
From the question we are told that
The speed of the roller coaster at ground level is [tex]v = 26 \ m/s[/tex]
Generally we can define the roller coaster speed at ground level using the an equation of motion as
[tex]v^2 = u^2 + 2 g s[/tex]
u is zero given that the roller coaster started from rest
So
[tex]26^2 = 0 + 2 * g * s[/tex]
So
[tex]s = \frac{26^2}{ 2 * g }[/tex]
=> [tex]s = 37.6 \ m[/tex]
Now the displacement half way is mathematically represented as
[tex]s_{h} = \frac{37.6}{2}[/tex]
[tex]s_{h} = 18.8 \ m[/tex]
So
[tex]v_h ^2 = u^2 + 2 * g * s_h[/tex]
Where [tex]v_h[/tex] is the velocity at the half way point
=> [tex]v_h = \sqrt{ 0 + 2 * 9.8 * 18.8 }[/tex]
=> [tex]v_h = 19.2 \ m/s[/tex]
If one could transport a simple pendulum of constant length from the Earth's surface to the Moon's, where acceleration due to gravity is one-sixth (1/6) that on the Earth, by what factor would be the pendulum frequency be changed
Answer:
The frequency will change by a factor of 0.4
Explanation:
T = 2(pi)*sqrt(L/g)
Since g(moon) = (1/6)g(earth), the period would change by sqrt[1/(1/6)] = sqrt(6) ~ 2.5 times longer on the moon. Since the period & frequency are inverses, the frequency would be 1/2.5 or 0.4 times shorter on the moon.
CAN SOMEONE HELP ME PLEASE ITS INTEGRATED SCIENCE AND I AM STUCK
Answer:
[tex]\huge \boxed{\mathrm{Option \ D}}[/tex]
Explanation:
Two forces are acting on the object.
Subtracting 2 N from both forces.
2 N → Object ← 5 N
- 2 N - 2N
0 N → Object ← 3 N
The force 3 N is pushing the object to the left side.
The mass of the object is 10 kg.
Applying formula for acceleration (Newton’s Second Law of Motion).
a = F/m
a = 3/10
a = 0.3
A solenoid is designed to produce a magnetic field of 3.50×10^−2 T at its center. It has a radius of 1.80 cm and a length of 46.0 cm , and the wire can carry a maximum current of 13.0 A.
Required:
a. What minimum number of turns per unit length must the solenoid have?
b. What total length of wire is required?
Answer:
a. 2143 turns/m
b. 111.5 m
Explanation:
a. The minimum number of turns per unit length (N/L) can be found using the following equation:
[tex] B = \frac{\mu_{0}NI}{L} [/tex]
[tex] \frac{N}{L} = \frac{B}{\mu_{0}I} = \frac{3.50 \cdot 10^{-2} T}{4\pi \cdot 10^{-7} Tm/A*13.0 A} = 2143 turns/m [/tex]
Hence, the minimum number of turns per unit length is 2143 turns/m.
b. The total length of wire is the following:
[tex] N = 2143 turns/m*L = 2143 turns/m*46.0 \cdot 10^{-2} m = 986 turns [/tex]
Since each turn has length 2πr of wire, the total length is:
[tex] L_{T} = N*2\pi r = 986 turn*2*\pi*1.80 \cdot 10^{-2} m = 111.5 m [/tex]
Therefore, the total length of wire required is 111.5 m.
I hope it helps you!
hat a 15 kg body is pulled along a horizontal fictional table by a force of 4N what is the acceleration of the body
Answer:
Acceleration of the body is:
[tex]a=0.27\,\,m/s^2[/tex]
Explanation:
Use Newton's second Law to solve for the acceleration:
[tex]F=m\,\,a\\a=\frac{F}{m} \\a=\frac{4\,N}{15\,\,kg} \\a=0.27\,\,m/s^2[/tex]
