Answer:
G. No applied force is needed.
Explanation:
Force is any motion which changes or intends to change the position of an object. When a force is applied to an object it causes acceleration to increase and velocity changes. The sled is moving at a constant velocity, there is no acceleration in the force therefore force is not applied nor needed.
(a) According to Hooke's Law, the force required to hold any spring stretched x meters beyond its natural length is f(x)=kx. Suppose a spring has a natural length of 20 cm. If a 25-N force is required to keep it stretched to a length of 30 cm, how much work is required to stretch it from 20 cm to 25 cm?
(b) Find the area of the region enclosed by one loop of the curve r=2sin(5θ).
Answer:
a) The work required to stretch the spring from 20 centimeters to 25 centimeters is 0.313 joules, b) The area of the region enclosed by one loop of the curve [tex]r(\theta) = 2\cdot \sin 5\theta[/tex] is [tex]4\pi[/tex].
Explanation:
a) The work, measured in joules, is a physical variable represented by the following integral:
[tex]W = \int\limits^{x_{f}}_{x_{o}} {F(x)} \, dx[/tex]
Where
[tex]x_{o}[/tex], [tex]x_{f}[/tex] - Initial and final position, respectively, measured in meters.
[tex]F(x)[/tex] - Force as a function of position, measured in newtons.
Given that [tex]F = k\cdot x[/tex] and the fact that [tex]F = 25\,N[/tex] when [tex]x = 0.3\,m - 0.2\,m[/tex], the spring constant ([tex]k[/tex]), measured in newtons per meter, is:
[tex]k = \frac{F}{x}[/tex]
[tex]k = \frac{25\,N}{0.3\,m-0.2\,m}[/tex]
[tex]k = 250\,\frac{N}{m}[/tex]
Now, the work function is obtained:
[tex]W = \left(250\,\frac{N}{m} \right)\int\limits^{0.05\,m}_{0\,m} {x} \, dx[/tex]
[tex]W = \frac{1}{2}\cdot \left(250\,\frac{N}{m} \right)\cdot [(0.05\,m)^{2}-(0.00\,m)^{2}][/tex]
[tex]W = 0.313\,J[/tex]
The work required to stretch the spring from 20 centimeters to 25 centimeters is 0.313 joules.
b) Let be [tex]r(\theta) = 2\cdot \sin 5\theta[/tex]. The area of the region enclosed by one loop of the curve is given by the following integral:
[tex]A = \int\limits^{2\pi}_0 {[r(\theta)]^{2}} \, d\theta[/tex]
[tex]A = 4\int\limits^{2\pi}_{0} {\sin^{2}5\theta} \, d\theta[/tex]
By using trigonometrical identities, the integral is further simplified:
[tex]A = 4\int\limits^{2\pi}_{0} {\frac{1-\cos 10\theta}{2} } \, d\theta[/tex]
[tex]A = 2 \int\limits^{2\pi}_{0} {(1-\cos 10\theta)} \, d\theta[/tex]
[tex]A = 2\int\limits^{2\pi}_{0}\, d\theta - 2\int\limits^{2\pi}_{0} {\cos10\theta} \, d\theta[/tex]
[tex]A = 2\cdot (2\pi - 0) - \frac{1}{5}\cdot (\sin 20\pi-\sin 0)[/tex]
[tex]A = 4\pi[/tex]
The area of the region enclosed by one loop of the curve [tex]r(\theta) = 2\cdot \sin 5\theta[/tex] is [tex]4\pi[/tex].
A beam of light is propagating in the x direction. The electric-field vector Group of answer choices can oscillate in any arbitrary direction in space. must oscillate in the z direction. must oscillate in the yz plane. must oscillate in the x direction. must have a steady component in the x direction.
Answer:
Option C is correct.
The electric-field vector must oscillate in the yz plane.
Explanation:
Light, in waveform, is an electromagnetic wave.
And electromagnetic waves are known to have their electric and magnetic field perpendicular to each other and also simultaneously perpendicular to the direction of propagation of the wave.
If the velocity of direction of propagation of the wave is in one direction, the electric-field vector must be in a direction we are sure is perpendicular to this direction of wave propagation and the wave's magnetic field.
Of the options provided, only option B (z-direction) and option C (yz-plane) show a direction that is indeed perpendicular to the direction of propagation of the wave (x-axis).
And truly, the electric-field vector for this wave can be in any of the two directions without breaking the laws of physics, but the electric-field vector oscillating in the yz-plane is a more general answer as it covers all the possible directions that the electric-field can oscillate in, including the one specified by option B (z-direction).
Hence, the correct answer is option C.
Hope this Helps!!!
1. A ski-plane with a total mass of 1200 kg lands towards the west on a frozen lake at 30.0
m/s. The coefficient of kinetic friction between the skis and the ice is 0.200. How far does
the plane slide before coming to a stop?
Answer:
d = 229.5 m
Explanation:
It is given that,
Total mass of a ski-plane is 1200 kg
It lands towards the west on a frozen lake at 30.0 m/s.
The coefficient of kinetic friction between the skis and the ice is 0.200.
We need to find the distance covered by the plane before coming to rest. In this case,
[tex]\mu mg=ma\\\\a=\mu g\\\\a=0.2\times 9.8\\\\a=1.96\ m/s^2[/tex]
It is decelerating, a = -1.96 m/s²
Now using the third equation of motion to find the distance covered by the plane such that :
[tex]v^2-u^2=2ad\\\\d=\dfrac{-u^2}{2a}\\\\d=\dfrac{-(30)^2}{2\times -1.96}\\\\d=229.59\ m[/tex]
So, the plane slide a distance of 229.5 m.
Find the terminal velocity (in m/s) of a spherical bacterium (diameter 1.81 µm) falling in water. You will first need to note that the drag force is equal to the weight at terminal velocity. Take the density of the bacterium to be 1.10 ✕ 103 kg/m3. (Assume the viscosity of water is 1.002 ✕ 10−3 kg/(m · s).)
Answer:
The terminal velocity of a spherical bacterium falling in the water is 1.96x10⁻⁶ m/s.
Explanation:
The terminal velocity of the bacterium can be calculated using the following equation:
[tex] F = 6\pi*\eta*rv [/tex] (1)
Where:
F: is drag force equal to the weight
η: is the viscosity = 1.002x10⁻³ kg/(m*s)
r: is the radium of the bacterium = d/2 = 1.81 μm/2 = 0.905 μm
v: is the terminal velocity
Since that F = mg and by solving equation (1) for v we have:
[tex] v = \frac{mg}{6\pi*\eta*r} [/tex]
We can find the mass as follows:
[tex] \rho = \frac{m}{V} \rightarrow m = \rho*V [/tex]
Where:
ρ: is the density of the bacterium = 1.10x10³ kg/m³
V: is the volume of the spherical bacterium
[tex] m = \rho*V = \rho*\frac{4}{3}\pi*r^{3} = 1.10 \cdot 10^{3} kg/m^{3}*\frac{4}{3}\pi*(0.905 \cdot 10^{-6} m)^{3} = 3.42 \cdot 10^{-15} kg [/tex]
Now, the terminal velocity of the bacterium is:
[tex] v = \frac{mg}{6\pi*\eta*r} = \frac{3.42 \cdot 10^{-15} kg*9.81 m/s^{2}}{6\pi*1.002 \cdot 10^{-3} kg/(m*s)*0.905 \cdot 10^{-6} m} = 1.96 \cdot 10^{-6} m/s [/tex]
Therefore, the terminal velocity of a spherical bacterium falling in the water is 1.96x10⁻⁶ m/s.
I hope it helps you!
A 150 V battery is connected across two parallel metal plates of area 28.5 cm2 and separation 0.00820 m. A beam of alpha particles (charge +2e, mass 6.64Ã10â27 kg) is accelerated from rest through a potential difference of 1.75 kV and enters the region between the plates perpendicular to the electric field.What magnitude and direction of magnetic field are needed so that the alpha particles emerge undeflected from between the plates?
Answer:
B = 4.45mT in the +^k direction
Explanation:
In order to calculate the required magnitude of the magnetic force, to achieve that the beam of particles emerge undeflected of the parallel plates, the electric force between the plates and the magnetic field in that region must be equal.
[tex]F_E=F_B\\\\qE=qvB[/tex] (1)
q: charge of the particles beam = +2e = 2*1.6*10^-19C
v: speed of the particles = ?
B: magnitude of the magnetic field = ?
E: electric field between the plates = V/d
V: potential difference between the parallel plates = 150V
d: distance of separation of the plates = 0.00820m
If you assume that the below plate is negative, the electric force on the particles has a direction upward (+^j). Then, the direction of the magnetic force must be downwards (-^j).
To obtain a downward magnetic force, it is necessary that the magnetic field point out of the page. In fact, if the direction of motion of the particles is toward east (+^i) and the magnetic field points out of the page (+^k), you have:
^i X ^k = -^j
Furthermore, it is necessary to calculate the sped of the particles. It is made by using the information about the charge, the potential difference that accelerates the particles and the kinetic energy.
[tex]K=qV=\frac{1}{2}mv^2\\\\v=\sqrt{\frac{2qV}{m}}[/tex] (2)
You replace the expression (2) into the equation (1) and solve for B:
[tex]B=\frac{E}{v}=E\sqrt{\frac{m}{2qV}}[/tex]
[tex]B=\frac{V}{d}\sqrt{\frac{m}{2qV}}\\\\B=\frac{150V}{0.0820m}\sqrt{\frac{6.64*10^{-27}kg}{2(2(1,6*10^{-19}C))(1.75*10^3V)}}\\\\B=4.45*10^{-3}T=4.45mT[/tex]
The required magnitude of the magnetic field is 4.45mT and has a direction out of the page +^k
Following are the solution to the given question:
In order to emerge using reflecting means, use the following formula:
[tex]\to F_E = F_B ..............(1)\\\\ \to F_E = \text{electric force}\\\\ \to F_B = \text{magnetic force}\\\\[/tex]
Calculating the Lorent's force:
[tex]\to F=qE+qv \times B \ \ also,\ \ K_{E} =U_{E} \\\\[/tex]
[tex]\to K_{E}[/tex][tex]= \text{kinetic energy} = -\frac{1}{2} \ mv^2 \\\\[/tex]
[tex]\to U_{E} = \text{potential energy} = q_V[/tex]
Calculating the value of v: \\\\
[tex]\to v= \sqrt{\frac{2qV}{m}} \\\\ \to q = 2e^{+} = 2 (1.6 \times 10^{-19}\ C) = 3.2 \times 10^{-19} C \\\\\to V = 1.75 \times 10^{3} \V \\\\\to m = 6.64 \times 10^{-27} \ kg\\\\ \to v = 410,700.25 \ \frac{m}{s}\\\\[/tex]
It's the particle's velocity, but the velocity also is defined as:
[tex]\to v=\frac{E}{B}[/tex]
solving for B:
[tex]\to B= \frac{E}{v}\\\\[/tex]
[tex]= \frac{\frac{V}{d}}{v}\\\\ =\frac{V}{vd} \\\\= \frac{150\ V}{(410,700.25 \ \frac{m}{s}) (8.2 \times 10^{-3} m)} \\\\= 0.045\ T\\\\[/tex]
When indicated in the diagram, the direction is parallel to "v" and E.
Learn more:
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An electron has an initial velocity of (17.1 + 12.7) km/s, and a constant acceleration of (1.60 × 1012 m/s2) in the positive x direction in a region in which uniform electric and magnetic fields are present. If = (529 µT) find the electric field .
Answer:
Explanation:
Since B is perpendicular, it does no work on the electron but instead deflects it in a circular path.
q = 1.6 x 10-19 C
v = (17.1j + 12.7k) km/s = square root(17.1² + 12.7²) = 2.13 x 10⁴ m/s
the force acting on electron is
F= qvBsinΦ
F= (1.6 x 10⁻¹⁹C)(2.13.x 10⁴ m/s)(526 x 10⁻⁶ T)(sin90º)
F = 1.793x 10⁻¹⁸ N
The net force acting on electron is
F = e ( E+ ( vXB)
= ( - 1.6 × 10⁻¹⁹) ( E + ( 17.1 × 10³j + 12.7 × 10³ k)X( 529 × 10⁻⁶ ) (i)
= ( -1.6 × 10⁻¹⁹ ) ( E- 6.7k + 9.0j)
a= F/m
1.60 × 10¹² i = ( -1.6 × 10⁻¹⁹ ) ( E- 6.9 k + 7.56 j)/9.11 × 10⁻³¹
9.11 i = - ( E- 6.7 k + 9.0 j)
E = -9.11i + 6.7k - 9.0j
what is mean by the terminal velocity
Terminal Velocity is the constant speed that a falling thing reaches when the resistence of a medium prevents the thing to reach any further speed.
Best of Luck!
Oceanographers often express the density of sea water in units of kilograms per cubic meter. If the density of sea water is 1.025 g/cm3 at 15ºC, what is its density in kilograms per cubic meter?
Answer: The density in kilograms per cubic meter is 1025
Explanation:
Density is defined as mass contained per unit volume.
Given : Density of sea water = [tex]1.025g/cm^3[/tex]
Conversion : [tex]1.025g/cm^3=?kg/m^3[/tex]
As 1 g = 0.001 kg
Thus 1.025 g =[tex]\frac{0.001}{1}\times 1.025=0.001025kg[/tex]
Also [tex]1cm^3=10^{-6}m^3[/tex]
Thus [tex]1.025g/cm^3=\frac{0.001025}{10^{-6}kg/m^3}=1025kg/m^3[/tex]
Thus density in kilograms per cubic meter is 1025
Which of the following gives the magnitude of the average velocity (over the entire run) of an athlete running on a circular track with a circumference of 0.5 km, if that athlete runs a total length of 1.0 km in a time interval of 4 minutes?
a. O m/s
b. 2 m/s
c. 4.2 m/s
d. 16.8 m/s
Answer:
c. 4.2 m/s
Explanation:
The definition of the average velocity, measured in meters per second, is given by the following expression:
[tex]\bar v = \frac{x_{f}-x_{o}}{t_{f}-t_{o}}[/tex]
Where:
[tex]x_{o}[/tex], [tex]x_{f}[/tex] - Initial and final positions, measured in meters.
[tex]t_{o}[/tex], [tex]t_{f}[/tex] - Initial and final instants, measured in seconds.
Positions and instants must be written in meters and seconds, respectively:
[tex]x_{o} = 0\,m[/tex], [tex]x_{f} = 1000\,m[/tex].
[tex]t_{o} = 0\,s[/tex], [tex]t_{f} = 240\,s[/tex].
Finally, the average velocity of the athlete that runs a total length of 1.0 kilometer in a time interval of 4 minutes is:
[tex]\bar v = \frac{1000\,m-0\,m}{240\,s-0\,s}[/tex]
[tex]\bar v = 4.167\,\frac{m}{s}[/tex]
Hence, the best option is C.
Legacy issues $570,000 of 8.5%, four-year bonds dated January 1, 2019, that pay interest semiannually on June 30 and December 31. They are issued at $508,050 when the market rate is 12%.
1. Determine the total bond interest expense to be recognized.
Total bond interest expense over life of bonds:
Amount repaid:
8 payments of $24,225 $193,800
Par value at maturity 570,000
Total repaid 763,800
Less amount borrowed 645 669
Total bond interest expense $118.131
2. Prepare a straight-line amortization table for the bonds' first two years.
Semiannual Period End Unamortized Discount Carrying Value
01/01/2019
06/30/2019
12/31/2019
06/30/2020
12/31/2020
3. Record the interest payment and amortization on June 30. Note:
Date General Journal Debit Credit
June 30
4. Record the interest payment and amortization on December 31.
Date General Journal Debit Credit
December 31
Answer:
1) Determine the total bond interest expense to be recognized.
Total bond interest expense over life of bonds:
Amount repaid:
8 payments of $24,225: $193,800
Par value at maturity: $570,000
Total repaid: $763800 (193,800 + 570,000)
Less amount borrowed: $508050
Total bond interest expense: $255750 (763800 - 508,050)
2)Prepare a straight-line amortization table for the bonds' first two years.
Semiannual Interest Period End; Unamortized Discount; Carrying Value
01/01/2019 61,950 508,050
06/30/2019 54,206 515,794
12/31/2019 46,462 523,538
06/30/2020 38,718 531,282
12/31/2020 30,974 539,026
3) Record the interest payment and amortization on June 30:
June 30 Bond interest expense, dr 31969
Discount on bonds payable, Cr (61950/8) 7743.75
Cash, Cr ( 570000*8.5%/2) 24225
4) Record the interest payment and amortization on December 31:
Dec 31 Bond interest expense, Dr 31969
Discount on bonds payable, Cr 7744
Cash, Cr 24225
Q 6.30: What is the underlying physical reason for the difference between the static and kinetic coefficients of friction of ordinary surfaces
Answer:
the coefficient of static friction is larger than kinetic coefficients of friction
Explanation:
The coefficient of static friction is usually larger than the kinetic coefficients of friction because an object at rest has increasingly settled agreements with the surface it's resting on at the molecular level, so it takes more force to break these agreement.
Until this force is been overcome, kinetic coefficient of friction is not going to surface.
Note: coefficient of static friction is the friction between two bodies when the bodies aren't moving. Meanwhile, kinetic coefficient is the ratio of frictional force of a moving body to the normal reaction.
[tex]F_{s}[/tex] ≤μ[tex]_{s}[/tex]N(coefficient of static friction)
where [tex]F_{s}[/tex] is the static friction, μ[tex]_{s}[/tex] is the coefficient of static friction and N is the normal reaction
μ = [tex]\frac{F}{N}[/tex](kinetic coefficient of friction)
attached is diagram illustrating the explanation
An infinitely long line of charge with uniform density, rho???????? lies in y-z plane parallel to the zaxis at y=1m. (a) Find the potential VAB at point A (4m, 2m, 4m) in Cartesian coordinates with respect to point B (0,0,0). (b) Find E filed at point B.
Answer with Explanation:
We are given that
Density=[tex]\rho l[/tex]
A(4m,2m,4m) and B(0,0,0)
y=1 m
a. Linear charge density=[tex]\frac{\rho l}{l}=\rho C/m[/tex]
Let a point P (0,1,4) on the line of charge and point Q (0,1,0)
Therefore,
Distance AP=[tex]\sqrt{(4-0)^2+(2-1)^2+(4-4)^2}=\sqrt{17}[/tex]
Distance,BQ=[tex]\sqrt{(0-0)^2+(1-0)^2+(0-0)^2}=1[/tex]
Electric field for infinitely long line
[tex]E=\frac{\rho}{2\pi \epsilon_0 r}\cdot \hat{r}[/tex]
Therefore, potential
[tex]V_{BA}=-\int_{a}^{b}E\cdot dl[/tex]
[tex]V_{BA}=-\int_{\sqrt{17}}^{1}\frac{\rho}{2\pi \epsilon_0 r}\hat{r}\cdot \hat{r} dr[/tex]
[tex]V_{BA}=-\int_{\sqrt{17}}^{1}\frac{\rho}{2\pi \epsilon_0 r}dr[/tex]
[tex]V_{BA}=-\frac{\rho}{2\pi \epsilon_0}[\ln r]^{1}_{\sqrt{17}}[/tex]
[tex]V_{BA}=-\frac{\rho}{2\pi \epsilon_0}(ln 1-ln(\sqrt{17})=\frac{\rho}{2\pi \epsilon_0}(ln(\sqrt{17})[/tex]
[tex]V_{BA}=V_B-V_A[/tex]
[tex]V_{AB}=V_A-V_B=-V_{BA}=-\frac{\rho}{2\pi \epsilon_0}(ln(\sqrt{17})[/tex]
b.Electric field at point B
[tex]E=\frac{\rho}{2\pi \epsilon_0 r}\cdot \hat{r}[/tex]
Unit vector r=[tex]-\hat{j}[/tex]
Therefore,
[tex]E=\frac{\rho}{2\pi \epsilon_0 r}\cdot \hat{-j}[/tex]
Calculate the angular momentum of a solid uniform sphere with a radius of 0.150 m and a mass of 13.0 kg if it is rotating at 5.70 rad/s about an axis through its center.
Answer:
The angular momentum of the solid sphere is 0.667 kgm²/s
Explanation:
Given;
radius of the solid sphere, r = 0.15 m
mass of the sphere, m = 13 kg
angular speed of the sphere, ω = 5.70 rad/s
The angular momentum of the solid sphere is given;
L = Iω
Where;
I is the moment of inertia of the solid sphere
ω is the angular speed of the solid sphere
The moment of inertia of solid sphere is given by;
I = ²/₅mr²
I = ²/₅ x (13 x 0.15²)
I = 0.117 kg.m²
The angular momentum of the solid sphere is calculated as;
L = Iω
L = 0.117 x 5.7
L = 0.667 kgm²/s
Therefore, the angular momentum of the solid sphere is 0.667 kgm²/s
What is the work done in stretching a spring by a distance of 0.5 m if the restoring force is 24N?
Answer:
3Nm
Explanation:
work = 0.5 x 12 x 0.5 = 3
The work done in stretching the spring by a distance of 0.5 m, with a restoring force of 24 N, is 6 joules.
To calculate the work done in stretching a spring, we can use the formula for work done by a spring:
Work = (1/2) * k *[tex]x^2[/tex]
where:
k = spring constant
x = distance the spring is stretched
Given that the restoring force (F) acting on the spring is 24 N, and the distance the spring is stretched (x) is 0.5 m, we can find the spring constant (k) using Hooke's law:
F = k * x
k = F / x
k = 24 N / 0.5 m
k = 48 N/m
Now, we can calculate the work:
Work = (1/2) * 48 N/m * [tex](0.5 m)^2[/tex]
Work = (1/2) * 48 N/m * [tex]0.25 m^2[/tex]
Work = 6 joules
Therefore, the work done in stretching the spring by a distance of 0.5 m, with a restoring force of 24 N, is 6 joules.
To know more about work done, here
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A soap bubble of radius R is situated in a uniform electric field of magnitude E. The electric flux through the surface of the soap bubble is
Answer:
The electric flux around the soap bubble will be
Ф = q/ε
Explanation:
The radius of the soap bubble = R
The electric field around the soap bubble = E
The electric flux = ?
The soap can be approximated to be a sphere, so we find the surface area of the sphere
For the soap bubble, the surface area will be
A = [tex]4\pi R^{2}[/tex]
Recall that electric flux is given as
Ф = EA
substituting value of A from above, we'll have
Ф = E[tex]4\pi R^{2}[/tex]..... equ 1
Also recall that the electric field E is given as
E = q/(4πε[tex]R^{2}[/tex])
substitute the value of E into equ 1, to get
Ф = q/(4πε[tex]R^{2}[/tex]) * [tex]4\pi R^{2}[/tex]
The electric flux around the soap bubble reduces to
Ф = q/ε
A coil has resistance of 20 W and inductance of 0.35 H. Compute its reactance and its impedance to an alternating current of 25 cycles/s.
Answer:
Reactance of the coil is 55 WImpedance of the coil is 59 WExplanation:
Given;
Resistance of the coil, R = 20 W
Inductance of the coil, L = 0.35 H
Frequency of the alternating current, F = 25 cycle/s
Reactance of the coil is calculated as;
[tex]X_L=[/tex] 2πFL
Substitute in the given values and calculate the reactance [tex](X_L)[/tex]
[tex]X_L =[/tex] 2π(25)(0.35)
[tex]X_L[/tex] = 55 W
Impedance of the coil is calculated as;
[tex]Z = \sqrt{R^2 + X_L^2} \\\\Z = \sqrt{20^2 + 55^2} \\\\Z = 59 \ W[/tex]
Therefore, the reactance of the coil is 55 W and Impedance of the coil is 59 W
When a hydrometer (see Fig. 2) having a stem diameter of 0.30 in. is placed in water, the stem protrudes 3.15 in. above the water surface. If the water is replaced with a liquid having a specific gravity of 1.10, how much of the stem would protrude above the liquid surface
Answer:
5.79 in
Explanation:
We are given that
Diameter,d=0.30 in
Radius,r=[tex]\frac{d}{2}=\frac{0.30}{2}=0.15 in[/tex]
Weight of hydrometer,W=0.042 lb
Specific gravity(SG)=1.10
Height of stem from the water surface=3.15 in
Density of water=[tex]62.4lb/ft^3[/tex]
In water
Volume of water displaced [tex]V=\frac{mass}{density}=\frac{0.042}{62.4}=6.73\times 10^{-4} ft^3[/tex]
Volume of another liquid displaced=[tex]V'=\frac{V}{SG}=\frac{6.73\times 10^{-4}}{1.19}=5.66\times 10^{-4}ft^3[/tex]
Change in volume=V-V'
[tex]V-V'=\pi r^2 l[/tex]
Substitute the values
[tex]6.73\times 10^{-4}-5.66\times 10^{-4}=3.14\times (\frac{0.15}{12})^2l[/tex]
By using
1 ft=12 in
[tex]\pi=3.14[/tex]
[tex]l=\frac{6.73\times 10^{-4}-5.66\times 10^{-4}}{3.14\times (\frac{0.15}{12})^2}[/tex]
l=2.64 in
Total height=h+l=3.15+2.64= 5.79 in
Hence, the height of the stem protrude above the liquid surface=5.79 in
A typical arteriole has a diameter of 0.080 mm and carries blood at the rate of 9.6 x10-5 cm3/s. What is the speed of the blood in (cm/s) the arteriole
Answer:
v= 4.823 × 10⁻⁹ cm/s
Explanation:
given
flow rate = 9.6 x10-5 cm³/s, d = 0.080mm
r = d/2= 0.080/2= 0.0040 cm
speed = rate of blood flow × area
v = (9.6 x 10⁻⁵ cm³/s) × (πr²)
v = (9.6 x 10⁻⁵ cm³/s) × π(0.0040 × cm)²
v= 1.536 × 10⁻⁹π cm/s
v= 4.823 × 10⁻⁹ cm/s
2. A 2.0-kg block slides down an incline surface from point A to point B. Points A and B are 2.0 m apart. If the coefficient of kinetic friction is 0.26 and the block is starting at rest from point A. What is the work done by friction force
Answer:a
Explanation:
In this problem we will consider the collision of two cars initially moving at right angles. We assume that after the collision the cars stick together and travel off as a single unit. The collision is therefore completely inelastic. Two cars of masses m1 and m2 collide at an intersection. Before the collision, car 1 was traveling eastward at a speed of v1, and car 2 was traveling northward at a speed of v2. After the collision, the two cars stick together and travel off in the direction.
Required:
a. Write the momentum conservation equation for the east-west components.
b. Write the momentum conservation equation for the north-south components.
c. Find the tangent of the angle.
Answer:
a) vfₓ = m₁ / (m₁ + m₂) v₁, b) tan θ = m₂ / m₁ v₂ / v₁, c)
Explanation:
Momentum is a vector quantity, so the consideration must be fulfilled in all axes
a) conservation of the moment east-west direction
the system is formed by the two cases, so that the forces during the sackcloth have been internal and therefore the mummer remains
before the crash
p₀ = m₁ v₁
after the crash
[tex]p_{f}[/tex]= (m1 + m2) vfₓ
p₀ = pf
m₁ v₁ = (m₁ + m₂) vfₓ
vfₓ = m₁ / (m₁ + m₂) v₁
b) conservation of the North-South axis moment
before the shock
p₀ = m₂ v₂
after the crash
p_{f} = ( m₁ +m₂) [tex]vf_{y}[/tex]
p₀ = p_{f}
me 2 v₂ = (m₁ + m₂) vfy
[tex]vf_{y}[/tex] = m₂ / (m₁ + m₂) v₂
c) the angle with which the car moves is
tan θ = Vfy / Vfₓ
tan θ = [m₂ / (m₁ + m₂) v] / [m₁ / (m₁ + m₂) v₁]
tan θ = m₂ / m₁ v₂ / v₁
The momentum conservation equation for the north-south components is [tex]m_1u_1 = v(m_1 + m_2)[/tex]
The momentum conservation equation for the north-south components is [tex]m_2u_2 = v(m_1 + m_2)[/tex]
The tangent of the angle is 1.
The given parameters;
angle between the initial velocity of the cars, θ = 90Apply the principle of conservation of linear momentum of inelastic collision as shown below;
[tex]m_1u_1 + m_2u_2 = v(m_1 + m_2)[/tex]
The momentum conservation equation for the east-west components is written as follows;
[tex]m_1(u_1cos \ 0) + m_2(u_2 cos 90)= v(m_1 + m_2)\\\\m_1u_1 = v(m_1 + m_2)[/tex]
The momentum conservation equation for the north-south components is written as follows;
[tex]m_1(u_1sin 0) + m_2(u_2sin90) = v(m_1 + m_2)\\\\m_2u_2 = v(m_1 + m_2)[/tex]
The tangent of the angle is calculated as follows;
[tex]tan \ \theta = \frac{p_y}{p_x} = \frac{v(m_1 + m_2)}{v(m_1 + m_2)} \\\\tan \ \theta = 1\\\\\theta = tan^{-1} (1) \\\\\theta = 45\ ^0[/tex]
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Given small samples of three liquids, you are asked to determine their refractive indexes. However, you do not have enough of each liquid to measure the angle of refraction for light retracting from air into the liquid. Instead, for each liquid, you take a rectangular block of glass (n= 1.52) and Place a drop of the liquid on the top surface f the block. you shine a laser beam with wavelength 638 nm in vacuum at one Side of the block and measure the largest angle of incidence for which there is total internal reflection at the interface between the glass and the liquid. Your results are given in the table.
Liquid A B C
θ 52.0 44.3 36.3
Required:
a. What is the refractive index of liquid A at this wavelength?
b. What is the refractive index of liquid B at this wavelength?
c. What is the refractive index of liquid C at this wavelength?
Answer:
A — 1.198B — 1.062C — 0.900Explanation:
The index of refraction of the liquid can be computed from ...
[tex]n_i\sin{(\theta_t)}=n_t[/tex]
where ni is the index of refraction of the glass block (1.52) and θt is the angle at which there is total internal refraction. nt is the index of refraction of the liquid.
For the given incidence angles, the computed indices of refraction are ...
A: n = 1.52sin(52.0°) = 1.198
B: n = 1.52sin(44.3°) = 1.062
C: n = 1.52sin(36.3°) = 0.900
Does there appear to be a simple mathematical relationship between the acceleration of an object (with fixed mass and negligible friction) and the force applied to the object (measured by the force probe mounted on the object)? Describe the mathematical relationship in words.
Answer:
the net force applied to an object is directly proportional to the acceleration undergone by that object
Explanation:
This verbal statement can be expressed in equation form as follows:
a = Fnet / m
Which statement describes one feature of a mineral's definite chemical composition?
It always occurs in pure form.
It always contains certain elements.
It cannot form from living or once-living materials.
It cannot contain atoms from more than one element.
N
Answer:
It always contains certain elements
Explanation:
Minerals can be defined as natural inorganic substances which possess an orderly internal structural arrangement as well as a particular, well known chemical composition, crystal structures and physical properties. Minerals include; quartz, dolomite, basalt, etc. Minerals may occur in isolation or in rock formations.
Minerals contain specific, well known chemical elements in certain ratios that can only vary within narrow limits. This is what we mean by a mineral's definite chemical composition. The structure of these minerals are all well known as well as their atom to atom connectivity.
The statement describes one feature of a mineral's definite chemical composition - It always contains certain elements.
A mineral is a naturally occurring chemical compound, usually of a crystalline form.
A mineral has one specific chemical composition.chemical composition that varies within a specific limited range and the atoms that make up the mineral must occur in specific ratiosthe proportions of the different elements and groups of elements in the mineral.Thus, The statement describes one feature of a mineral's definite chemical composition - It always contains certain elements.
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Two lenses of focal length 4.5cm and 1.5cm are placed at a certain distance apart, calculate the distance between the lenses if they form an achromatic combination
3.0cm
Explanation:
For lenses in an achromatic combination, the following condition holds, assuming the two lenses are of the same materials;
d = [tex]\frac{f_1 + f_2}{2}[/tex] ---------(i)
Where;
d= distance between lenses
f₁ = focal length of the first lens
f₂ = focal length of the second lens
From the question;
f₁ = 4.5cm
f₂ = 1.5cm
Substitute these values into equation (i) as follows;
d = [tex]\frac{4.5+1.5}{2}[/tex]
d = [tex]\frac{6.0}{2}[/tex]
d = 3.0cm
Therefore, the distance between the two lenses is 3.0cm
your washer has a power of 350 watts and your dryer has a power of 1800 watts how much energy do you use to clean a load of clothes in 1 hour of washing and 1 hour of drying?
A. 1.29 x 10^3 J
B. 2.58 x 10^3 J
C. 1.55 x 10^7 J
D. 7.74 x 10^6 J
Answer:
7.74 x 10⁶ Joules
Explanation:
recall that "Watts" is the SI unit used for "energy per unit time"
Hence "Watts" may also be expressed as Joules / Second (or J/s)
We are given that the washer is rated at 350W (i.e. 350 Joules / s) and the dryer is rated at 1800W (i.e. 1800 Joules / s).
We are also given that the appliances are each run for 1 hour
1 hour = 60 min = (60 x 60) seconds = 3600 seconds
Hence the total energy used,
= Energy used by Washer in 1 hour + Energy used by dryer in 1 hour
= (350 J/s x 3600 s) + (1800 J/s x 3600 s)
= 3600 ( 350 + 1800)
= 3600 (2150)
= 7,740,000 Joules
= 7.74 x 10⁶ Joules
A basketball rolls across a classroom floor without slipping, with its center of mass moving at a certain speed. A block of ice of the same mass is set sliding across the floor with the same speed along a parallel line. (i) Which object has more kinetic energy
Answer:
The two objects encounter a ramp sloping upward.
Explanation:
The basketball will travel farther up theramp
A spring with a 3.15kg weight hanging from it measures 13.40cm, and without the weight 12.00cm. If you hang a weight on it so as to store 10.0J potential energy in it, how long will the spring be?
Answer:
21.52 cm
Explanation:
Given that
mass of the spring, m = 3.15 kg
Length of the spring l2, = 13.4 cm = 0.134 m
Length of the spring l1 = 12 cm = 0.12 m
change in extension, x = 0.134 - 0.12 = 0.014 m
Acceleration due to gravity, g = 9.8 m/s²
Potential Energy, U = 10 J
See attachment for calculation
Answer:
Final Length = 12.45 cm
Explanation:
First we need to find the spring constant. From Hooke's Law:
F = kΔx
where,
F = Force Applied on Spring = Weight = mg = (3.15 kg)(9.8 m/s²) = 30.87 N
k = spring constant = ?
Δx = change in length of spring = 13.4 cm - 12 cm = 1.4 cm = 0.014 m
Therefore,
30.87 N = k(0.014 m)
k = (30.87 N)/(0.014 m)
k = 2205 N/m
Now, for the Potential Energy of 10 J:
P.E = (1/2)KΔx²
where,
P.E = Potential Energy of Spring = 10 J
Δx = ?
Therefore,
10 J = (2205 N/m)Δx²
Δx = √[10 J/(2205 N/m)
Δx = Final Length - Initial length = 0.0045 m = 0.45 cm
Final Length = 0.45 cm + 12 cm
Final Length = 12.45 cm
A 40.0 kg ballet dancer stands on her toes during a performance with 25.0 cm2 in contact with the floor. What is the pressure exerted by the floor over the area of contact if the dancer is stationary
Explanation:
40×10ms^-2
400N.
25/100m^2
0.25m^2
1.P=F/A
=400N/0.25m^2
=100Nm^-2
=100Pa
PLS HELP ILL MARK U BRAINLIEST I DONT HAVE MUCH TIME!!
A football player of mass 103 kg running with a velocity of 2.0 m/s [E] collides head-
on with a 110 kg player on the opposing team travelling with a velocity of 3.2 m/s
[W]. Immediately after the collision the two players move in the same direction.
Calculate the final velocity of the two players.
Answer:
The final velocity of the two players is 0.69 m/s in the direction of the opposing player.
Explanation:
Since the players are moving in opposite directions, from the principle of conservation of linear momentum;
[tex]m_{1} u_{1}[/tex] - [tex]m_{2}u_{2}[/tex] = [tex](m_{1} + m_{2} )[/tex] v
Where: [tex]m_{1}[/tex] is the mass of the first player, [tex]u_{1}[/tex] is the initial velocity of the first player, [tex]m_{2}[/tex] is the mass of the second player, [tex]u_{2}[/tex] is the initial velocity of the second player and v is the final common velocity of the two players after collision.
[tex]m_{1}[/tex] = 103 kg, [tex]u_{1}[/tex] = 2.0 m/s, [tex]m_{2}[/tex] = 110 kg, [tex]u_{2}[/tex] = 3.2 m/s. Thus;
103 × 2.0 - 110 × 3.2 = (103 + 110)v
206 - 352 = 213 v
-146 = 213 v
v = [tex]\frac{-146}{213}[/tex]
v = -0.69 m/s
The final velocity of the two players is 0.69 m/s in the direction of the opposing player.
How can global warming lead to changes to the Earth’s surface? a. Global warming can lead to an increased number of earthquakes, which change the Earth’s surface. b. Global warming can lead to glaciers melting, causing flooding to areas and the decrease of glacial land masses. c. Global warming leads to a decrease in water levels of coastal wetlands. d. Global warming cannot lead to changes to the Earth’s surface.
Answer:
Option: b. Global warming can lead to glaciers melting, causing flooding to areas and the decrease of glacial land masses.
Explanation:
Global warming is the reason for the changes in environment and climate on earth. Melting of glaciers leads to an increase in water level and a decrease in landmass. One of the most climactic consequences is the decrease in Arctic sea ice. Melting polar ice along with ice sheets and glaciers across Greenland, North America, Europe, Asia, and South America suspected to increase sea levels slowly. There is an increase in the glacial retreat due to global warming, which leaves rock piles that covered with ice.
Answer:
B: Global warming can lead to glaciers melting, causing flooding to areas and the decrease of glacial land masses.
Explanation:
Global warming is primarily caused by the increase in greenhouse gases, such as carbon dioxide, in the Earth's atmosphere. This leads to a rise in global temperatures, which has various impacts on the Earth's surface. One significant effect is the melting of glaciers and ice caps in polar regions and mountainous areas.
As temperatures increase, glaciers and ice sheets start to melt at a faster rate. This melting results in the release of massive amounts of water into rivers, lakes, and oceans. Consequently, there can be an increase in the frequency and intensity of flooding events in regions downstream from these melting glaciers.
Moreover, the melting of glaciers and ice caps contributes to a rise in sea levels. As the melted ice enters the oceans, it adds to the overall volume of water, leading to a gradual increase in sea levels worldwide. This rise in sea levels poses a threat to coastal areas, as they become more vulnerable to coastal erosion, storm surges, and saltwater intrusion into freshwater sources.
Additionally, the loss of glacial land masses due to melting can have long-term effects on ecosystems. Glaciers act as freshwater reservoirs, releasing water gradually throughout the year. With their decline, the availability of freshwater for agriculture, drinking water, and other human needs can be significantly affected.
Therefore, global warming can indeed lead to changes in the Earth's surface, particularly through the melting of glaciers and subsequent impacts on sea levels, flooding, and glacial land masses.
E23 verified.