Answer:
hand saws
power saws
Circular Saw
Explanation:
that is all that i know
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
A circular coil of wire of 200 turns and diameter 2.0 cm carries a current of 4.0 A. It is placed in a magnetic field of 0.70 T with the plane of the coil making an angle of 30° with the magnetic field. What is the magnetic torque on the coil?
Answer:
0.087976 Nm
Explanation:
The magnetic torque (τ) on a current-carrying loop in a magnetic field is given by;
τ = NIAB sinθ --------- (i)
Where;
N = number of turns of the loop
I = current in the loop
A = area of each of the turns
B = magnetic field
θ = angle the loop makes with the magnetic field
From the question;
N = 200
I = 4.0A
B = 0.70T
θ = 30°
A = π d² / 4 [d = diameter of the coil = 2.0cm = 0.02m]
A = π x 0.02² / 4 = 0.0003142m² [taking π = 3.142]
Substitute these values into equation (i) as follows;
τ = 200 x 4.0 x 0.0003142 x 0.70 sin30°
τ = 200 x 4.0 x 0.0003142 x 0.70 x 0.5
τ = 200 x 4.0 x 0.0003142 x 0.70
τ = 0.087976 Nm
Therefore, the torque on the coil is 0.087976 Nm
A car moving at a speed of 25 m/s enters a curve that traces a circular quarter turn of radius 129 m. The driver gently applies the brakes, slowing the car with a constant tangential acceleration of magnitude 1.2 m/s2.a) Just before emerging from the turn, what is the magnitudeof the car's acceleration?
b) At that same moment, what is the angle q between the velocity vector and theacceleration vector?
I am having trouble because this problem seems to have bothradial and tangential accleration. I tried finding the velocityusing V^2/R, but then that didnt take into account thedeceleration. Any help would be great.
Answer:
8.7 m/s^2
82.15°
Explanation:
Given:-
- The initial speed of the car, vi = 25 m/s
- The radius of track, r = 129 m
- Car makes a circular " quarter turn "
- The constant tangential acceleration, at = 1.2 m/s^2
Solution:-
- We will solve the problem using rotational kinematics. Determine the initial angular velocity of car ( wi ) as follows:
[tex]w_i = \frac{v_i}{r} \\\\w_i = \frac{25}{129}\\\\w_i = 0.19379 \frac{rad}{s}[/tex]
- Now use the constant tangential acceleration ( at ) and determine the constant angular acceleration ( α ) for the rotational motion as follows:
at = r*α
α = ( 1.2 / 129 )
α = 0.00930 rad/s^2
- We know that the angular displacement from the initial entry to the exit of the turn is quarter of a turn. The angular displacement would be ( θ = π/2 ).
- Now we will use the third rotational kinematic equation of motion to determine the angular velocity at the exit of the turn (wf) as follows:
[tex]w_f^2 = w_i^2 + 2\alpha*theta\\\\w_f = \sqrt{0.19379^2 + 0.00930\pi } \\\\w_f = 0.25840 \frac{rad}{s}[/tex]
- We will use the evaluated final velocity ( wf ) and determine the corresponding velocity ( vf ) as follows:
[tex]v_f = r*w_f\\\\v_f = 129*0.2584\\\\v_f = 33.33380 \frac{x}{y}[/tex]
- Now use the formulation to determine the centripetal acceleration ( ac ) at this point as follows:
[tex]a_c = \frac{v_f^2}{r} \\\\a_c = \frac{33.3338^2}{129} \\\\a_c = 8.6135 \frac{m}{s^2}[/tex]
- To determine the magnitude of acceleration we will use find the resultant of the constant tangential acceleration ( at ) and the calculated centripetal acceleration at the exit of turn ( ac ) as follows:
[tex]|a| = \sqrt{a^2_t + a_c^2} \\\\|a| = \sqrt{1.2^2 + 8.6135^2} \\\\|a| = 8.7 \frac{m}{s^2}[/tex]
- To determine the angle between the velocity vector and the acceleration vector. We need to recall that the velocity vector only has one component and always tangential to the curved path. Hence, the velocity vector is parallel to the tangential acceleration vector ( at ). We can use the tangential acceleration ( at ) component of acceleration ( a ) and the centripetal acceleration ( ac ) component of the acceleration and apply trigonometric ratio as follows:
[tex]q = arctan \frac{a_c}{a_t} = arctan \frac{8.7}{1.2} \\\\q = 82.15 ^.[/tex]
Answer: The angle ( q ) between acceleration vector ( a ) and the velocity vector ( v ) at the exit of the turn is 82.15° .
An interference pattern is produced by light with a wavelength 550 nm from a distant source incident on two identical parallel slits separated by a distance (between centers) of 0.500 mm .
a. If the slits are very narrow, what would be the angular position of the second- order, two-slit interference maxima?
b. Let the slits have a width 0.300 mm. In terms of the intensity lo at the center of the central maximum, what is the intensity at the angular position in part "a"?
Answer:
a
[tex]\theta = 0.0022 rad[/tex]
b
[tex]I = 0.000304 I_o[/tex]
Explanation:
From the question we are told that
The wavelength of the light is [tex]\lambda = 550 \ nm = 550 *10^{-9} \ m[/tex]
The distance of the slit separation is [tex]d = 0.500 \ mm = 5.0 *10^{-4} \ m[/tex]
Generally the condition for two slit interference is
[tex]dsin \theta = m \lambda[/tex]
Where m is the order which is given from the question as m = 2
=> [tex]\theta = sin ^{-1} [\frac{m \lambda}{d} ][/tex]
substituting values
[tex]\theta = 0.0022 rad[/tex]
Now on the second question
The distance of separation of the slit is
[tex]d = 0.300 \ mm = 3.0 *10^{-4} \ m[/tex]
The intensity at the the angular position in part "a" is mathematically evaluated as
[tex]I = I_o [\frac{sin \beta}{\beta} ]^2[/tex]
Where [tex]\beta[/tex] is mathematically evaluated as
[tex]\beta = \frac{\pi * d * sin(\theta )}{\lambda }[/tex]
substituting values
[tex]\beta = \frac{3.142 * 3*10^{-4} * sin(0.0022 )}{550 *10^{-9} }[/tex]
[tex]\beta = 0.06581[/tex]
So the intensity is
[tex]I = I_o [\frac{sin (0.06581)}{0.06581} ]^2[/tex]
[tex]I = 0.000304 I_o[/tex]
The average density of the body of a fish is 1080kg/m^3 . To keep from sinking, the fish increases its volume by inflating an internal air bladder, known as a swim bladder, with air.
By what percent must the fish increase its volume to be neutrally buoyant in fresh water? Use 1.28kg/m^3 for the density of air at 20 degrees Celsius. (change in V/V)
Answer:
Increase of volume (F) = 8.01%
Explanation:
Given:
Density of fish = 1,080 kg/m³
Density of water = 1,000 kg/m³
density of air = 1.28 kg/m³
Find:
Increase of volume (F)
Computation:
1,080 kg/m³ + [F × 1.28 kg/m³ ] = (1+F) × 1,000 kg/m³
1,080 + 1.28 F =1,000 F + 1,000
80 = 998.72 F
F = 0.0801 (Approx)
F = 8.01% (Approx)
How do I find an apparent weight in N for a metal connected to a string submerged in water if a scale shows the mass 29.52 g when it is submerged ? Also how do I measure its density
The Tension of the string is going to be less when submerged in water by a value called the buoyancy force, so below in the attached file is explanation on how to calculate the apparent weight and density of the submerged object
key points that can be found in the realist philosophical position
Answer:
Key points that can be found in the realist philosophical position are as follows:
The view that we observe or identify is real, truly out there.The objects which are identified are independent of someone's perceptions, linguistic practices, conceptual scheme, and beliefs.Quantum mechanics is an example of philosophical realism that claims world is mind-independent.An airplane flies in a horizontal circle of radius 500 m at a speed of 150 m/s. If the radius were changed to 1000 m, but the speed remained the same, by what factor would its centripetal acceleration change?
Answer:
The centripetal acceleration changed by a factor of 0.5
Explanation:
Given;
first radius of the horizontal circle, r₁ = 500 m
speed of the airplane, v = 150 m/s
second radius of the airplane, r₂ = 1000 m
Centripetal acceleration is given as;
[tex]a = \frac{v^2}{r}[/tex]
At constant speed, we will have;
[tex]v^2 =ar\\\\v = \sqrt{ar}\\\\at \ constant\ v;\\\sqrt{a_1r_1} = \sqrt{a_2r_2}\\\\a_1r_1 = a_2r_2\\\\a_2 = \frac{a_1r_1}{r_2} \\\\a_2 = \frac{a_1*500}{1000}\\\\a_2 = \frac{a_1}{2} \\\\a_2 = \frac{1}{2} a_1[/tex]
a₂ = 0.5a₁
Therefore, the centripetal acceleration changed by a factor of 0.5
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
A slender rod of length L has a varying mass-per-unit-length from the left end (x=0) according to dm/dx=Cx where C has units kg/m2. Find the total mass in terms of C and L, and then calculate the moment of inertia of the rod for an axis at the left end note: you need the total mass in order to get the answer in terms of ML^2
Answer:
ML²/6
Explanation:
Pls see attached file
The total mass is M = CL²/2, and the moment of inertia is I = ML²/2,
Moment of inertia:The length of the rod is L. It has a non-uniform distribution of mass given by:
dm/dx = Cx
where C has units kg/m²
dm = Cxdx
the total mass M of the rod can be calculated by integrating the above relation over the length:
[tex]M =\int\limits^L_0 {} \, dm\\\\M=\int\limits^L_0 {Cx} \, dx\\\\M=C[x^2/2]^L_0\\\\M=C[L^2/2]\\\\[/tex]
Thus,
C = 2M/L²
Now, the moment of inertia of the small element dx of the rod is given by:
dI = dm.x²
dI = Cx.x²dx
[tex]dI = \frac{2M}{L^2}x^3dx\\\\I= \frac{2M}{L^2}\int\limits^L_0 {x^3} \, dx \\\\I= \frac{2M}{L^2}[\frac{L^4}{4}][/tex]
I = ML²/2
Learn more about moment of inertia:
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Use Kepler's third law to determine how many days it takes a spacecraft to travel in an elliptical orbit from a point 6 590 km from the Earth's center to the Moon, 385 000 km from the Earth's center.
Answer:
1.363×10^15 seconds
Explanation:
The spaceship travels an elliptical orbit from a point of 6590km from the earth center to the moon and 38500km from the earth center.
To calculate the time taken from Kepler's third Law :
T^2 = ( 4π^2/GMe ) r^3
Where Me is the mass of the earth
r is the average distance travel
G is the universal gravitational constant. = 6.67×10-11 m3 kg-1 s-2
π = 3.14
Me = mass of earth = 5.972×10^24kg
r =( r minimum + r maximum)/2 ......1
rmin = 6590km
rmax = 385000km
From equation 1
r = (6590+385000)/2
r = 391590/2
r = 195795km
From T^2 = ( 4π^2/GMe ) r^3
T^2 = (4 × 3.14^2/ 6.67×10-11 × 5.972×10^24) × 195795^3
= ( 4×9.8596/ 3.983×10^14 ) × 7.5059×10^15
= 39.4384/ 3.983×10^14 ) × 7.5059×10^15
= (9.901×10^14) × 7.5059×10^15
T^2 = 7.4321× 10^30
T =√7.4321× 10^30
T = 2.726×10^15 seconds
The time for one way trip from Earth to the moon is :
∆T = T/2
= 2.726×10^15 /2
= 1.363×10^15 secs
6a. A special lamp can produce UV radiation. Which two statements
describe the electromagnetic waves emitted by a UV lamp? *
They have a higher frequency than X-rays.
They have the same wave speed as visible light
They have a longer wavelength than microwaves.
They have a lower frequency than gamma rays.
They have a greater wave speed than radio waves.
Answer:
The correct options are:
B) They have the same wave speed as visible light
D) They have a lower frequency than gamma rays.
Explanation:
B) Ultraviolet rays, commonly known as UV rays, are a type of electromagnetic ways. As electromagnetic waves, in the layman's term, are all kinds of life that can be identified, all electromagnetic waves (UV rays, visible light, infrared, radio etc) all travel with the same velocity, that is the speed of light, given as v = 3 × 10⁸ m/s
D) The frequency of all electromagnetic rays can be found by electromagnetic spectrum (picture attached below).
We can clearly see in the picture that the frequencies of UV rays lie at about 10¹⁵ - 10¹⁶ Hz which is lower than the frequency of Gamma ray, which lie at about 10²⁰ Hz.
⦁ A 68 kg crate is dragged across a floor by pulling on a rope attached to the crate and inclined 15° above the horizontal. (a) If the coefficient of static friction is 0.5, what minimum force magnitude is required from the rope to start the crate moving? (b) If µk= 0.35, what is the magnitude of the initial acceleration of the crate?
Answer:
303.29N and 1.44m/s^2
Explanation:
Make sure to label each vector with none, mg, fk, a, FN or T
Given
Mass m = 68.0 kg
Angle θ = 15.0°
g = 9.8m/s^2
Coefficient of static friction μs = 0.50
Coefficient of kinetic friction μk =0.35
Solution
Vertically
N = mg - Fsinθ
Horizontally
Fs = F cos θ
μsN = Fcos θ
μs( mg- Fsinθ) = Fcos θ
μsmg - μsFsinθ = Fcos θ
μsmg = Fcos θ + μsFsinθ
F = μsmg/ cos θ + μs sinθ
F = 0.5×68×9.8/cos 15×0.5×sin15
F = 332.2/0.9659+0.5×0.2588
F =332.2/1.0953
F = 303.29N
Fnet = F - Fk
ma = F - μkN
a = F - μk( mg - Fsinθ)
a = 303.29 - 0.35(68.0 * 9.8- 303.29*sin15)/68.0
303.29-0.35( 666.4 - 303.29*0.2588)/68.0
303.29-0.35(666.4-78.491)/68.0
303.29-0.35(587.90)/68.0
(303.29-205.45)/68.0
97.83/68.0
a = 1.438m/s^2
a = 1.44m/s^2
A water-balloon launcher with mass 5 kg fires a 1 kg balloon with a velocity of
8 m/s to the east. What is the recoil velocity of the launcher?
Answer:
1.6 m/s west
Explanation:
The recoil velocity of the launcher is 1.6 m/s west.
What is conservation of momentum principle?When two bodies of different masses move together each other and have head on collision, they travel to same or different direction after collision.
A water-balloon launcher with mass 5 kg fires a 1 kg balloon with a velocity of 8 m/s to the east.
Final momentum will be zero, so
m₁u₁ +m₂u₂ =0
Substitute the values for m₁ = 5kg, m₂ =1kg and u₂ =8 m/s, then the recoil velocity will be
5 x v +1x8 = 0
v = - 1.6 m/s
Thus, the recoil velocity of the launcher is 1.6 m/s (West)
Learn more about conservation of momentum principle
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5) What is the weight of a body in earth. if its weight is 5Newton
in moon?
Answer:
8.167
Explanation:
A 1.20 kg water balloon will break if it experiences more than 530 N of force. Your 'friend' whips the water balloon toward you at 13.0 m/s. The maximum force you apply in catching the water balloon is twice the average force. How long must the interaction time of your catch be to make sure the water balloon doesn't soak you
Answer:
t = 0.029s
Explanation:
In order to calculate the interaction time at the moment of catching the ball, you take into account that the force exerted on an object is also given by the change, on time, of its linear momentum:
[tex]F=\frac{\Delta p}{\Delta t}=m\frac{\Delta v}{\Delta t}[/tex] (1)
m: mass of the water balloon = 1.20kg
Δv: change in the speed of the balloon = v2 - v1
v2: final speed = 0m/s (the balloon stops in my hands)
v1: initial speed = 13.0m/s
Δt: interaction time = ?
The water balloon brakes if the force is more than 530N. You solve the equation (1) for Δt and replace the values of the other parameters:
[tex]|F|=|530N|= |m\frac{v_2-v_1}{\Delta t}|\\\\|530N|=| (1.20kg)\frac{0m/s-13.0m/s}{\Delta t}|\\\\\Delta t=0.029s[/tex]
The interaction time to avoid that the water balloon breaks is 0.029s
A total electric charge of 2.00 nC is distributed uniformly over the surface of a metal sphere with a radius of 26.0 cm . The potential is zero at a point at infinity.
a) Find the value of the potential at 45.0 cm from the center of the sphere.
b) Find the value of the potential at 26.0 cm from the center of the sphere.
c) Find the value of the potential at 16.0 cm from the center of the sphere.
Answer:
a) 40 V
b) 69.23 V
c) 69.23 V
Explanation:
See attachment for solution
(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].
The electric field must be zero inside a conductor in electrostatic equilibrium, but not inside an insulator. It turns out that we can still apply Gauss's law to a Gaussian surface that is entirely within an insulator by replacing the right-hand side of Gauss's law, Qin/eo, with Qin/e, where ε is the permittivity of the material. (Technically, Eo is called the vacuum permittivity.) Suppose that a 70 nC point charge is surrounded by a thin, 32-cm-diameter spherical rubber shell and that the electric field strength inside the rubber shell is 2500 N/C.
What is the permittivity of rubber?
Answer:
The permittivity of rubber is [tex]\epsilon = 8.703 *10^{-11}[/tex]
Explanation:
From the question we are told that
The magnitude of the point charge is [tex]q_1 = 70 \ nC = 70 *10^{-9} \ C[/tex]
The diameter of the rubber shell is [tex]d = 32 \ cm = 0.32 \ m[/tex]
The Electric field inside the rubber shell is [tex]E = 2500 \ N/ C[/tex]
The radius of the rubber is mathematically evaluated as
[tex]r = \frac{d}{2} = \frac{0.32}{2} = 0.16 \ m[/tex]
Generally the electric field for a point is in an insulator(rubber) is mathematically represented as
[tex]E = \frac{Q}{ \epsilon } * \frac{1}{4 * \pi r^2}[/tex]
Where [tex]\epsilon[/tex] is the permittivity of rubber
=> [tex]E * \epsilon * 4 * \pi * r^2 = Q[/tex]
=> [tex]\epsilon = \frac{Q}{E * 4 * \pi * r^2}[/tex]
substituting values
[tex]\epsilon = \frac{70 *10^{-9}}{2500 * 4 * 3.142 * (0.16)^2}[/tex]
[tex]\epsilon = 8.703 *10^{-11}[/tex]
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!
Charge of uniform surface density (0.20 nC/m2) is distributed over the entire xy plane. Determine the magnitude of the electric field at any point having z
The question is not complete, the value of z is not given.
Assuming the value of z = 4.0m
Answer:
the magnitude of the electric field at any point having z(4.0 m) =
E = 5.65 N/C
Explanation:
given
σ(surface density) = 0.20 nC/m² = 0.20 × 10⁻⁹C/m²
z = 4.0 m
Recall
E =F/q (coulumb's law)
E = kQ/r²
σ = Q/A
A = 4πr²
∴ The electric field at point z =
E = σ/zε₀
E = 0.20 × 10⁻⁹C/m²/(4 × 8.85 × 10⁻¹²C²/N.m²)
E = 5.65 N/C
At what minimum speed must a roller coaster be traveling when upside down at the top of a 7.4 m radius loop-the-loop circle so the passengers will not fall out?
Answer:
v = 8.5 m/s
Explanation:
In order for the passengers not to fall out of the loop circle, the centripetal force must be equal to the weight of the passenger. Therefore,
Weight = Centripetal Force
but,
Weight = mg
Centripetal Force = mv²/r
Therefore,
mg = mv²/r
g = v²/r
v² = gr
v = √gr
where,
v = minimum speed required = ?
g = 9.8 m/s²
r = radius = 7.4 m
Therefore,
v = √(9.8 m/s²)(7.4 m)
v = 8.5 m/s
Minimum speed for a roller coaster while travelling upside down so that the person will not fall out = 8.5 m/s
For a roller coaster be traveling when upside down the Force balance equation can be written for a person of mass m.
In the given condition the weight of the person must be balanced by the centrifugal force.
and for the person not to fall out centrifugal force must be greater than or equal to the weight of the person
According to the Newton's Second Law of motion we can write force balance
[tex]\rm mv^2/r -mg =0 \\\\mg = mv^2 /r (Same\; mass) \\\\\\g = v^2/r\\\\v = \sqrt {gr}......(1)[/tex]
Given Radius of loop = r = 7.4 m
Putting the value of r = 7.4 m in equation (1) we get
[tex]\sqrt{9.8\times 7.4 } = \sqrt{72.594} = 8.5\; m/s[/tex]
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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
In a double-slit interference experiment you are asked to use laser light of different wavelengths and determine the separation between adjacent maxima. You observe that this separation is greatest when you illuminate the double slit with In a double-slit interference experiment you are asked to use laser light of different wavelengths and determine the separation between adjacent maxima. You observe that this separation is greatest when you illuminate the double slit with:_________.
1. yellow light.
2. red light.
3. blue light.
4. green light.
5. The separation is the same for all wavelengths.
Answer:
Red light
Explanation:
This because All interference or diffraction patterns depend upon the wavelength of the light (or whatever wave) involved. Red light has the longest wavelength (about 700 nm)
A proton with an initial speed of 400000 m/s is brought to rest by an electric field.
Part A- Did the proton move into a region of higher potential or lower potential?
Part B - What was the potential difference that stopped the proton?
?U = ________V
Part C - What was the initial kinetic energy of the proton, in electron volts?
Ki =_________eV
Answer:
moves into a region of higher potential
Potential difference = 835 V
Ki = 835 eV
Explanation:
given data
initial speed = 400000 m/s
solution
when proton moves against a electric field so that it will move into higher potential region
and
we know Work done by electricfield W is express as
W = KE of proton K
so
q × V = 0.5 × m × v² ......................1
put here va lue
1.6 × [tex]10^{-19}[/tex] × V = 0.5 × 1.67 × [tex]10^{-27}[/tex] × 400000²
Potential difference V = 1.336 × 10-16 / 1.6 × 10-19
Potential difference = 835 V
and
KE of proton in eV is express as
Ki = V numerical
Ki = 835 eV
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.
A spherical shell rolls without sliding along the floor. The ratio of its rotational kinetic energy (about an axis through its center of mass) to its translational kinetic energy is:
Answer:
The ratio is [tex]\frac{RE}{TE} = \frac{2}{3}[/tex]
Explanation:
Generally the Moment of inertia of a spherical object (shell) is mathematically represented as
[tex]I = \frac{2}{3} * m r^2[/tex]
Where m is the mass of the spherical object
and r is the radius
Now the the rotational kinetic energy can be mathematically represented as
[tex]RE = \frac{1}{2}* I * w^2[/tex]
Where [tex]w[/tex] is the angular velocity which is mathematically represented as
[tex]w = \frac{v}{r}[/tex]
=> [tex]w^2 = [\frac{v}{r}] ^2[/tex]
So
[tex]RE = \frac{1}{2}* [\frac{2}{3} *mr^2] * [\frac{v}{r} ]^2[/tex]
[tex]RE = \frac{1}{3} * mv^2[/tex]
Generally the transnational kinetic energy of this motion is mathematically represented as
[tex]TE = \frac{1}{2} mv^2[/tex]
So
[tex]\frac{RE}{TE} = \frac{\frac{1}{3} * mv^2}{\frac{1}{2} * m*v^2}[/tex]
[tex]\frac{RE}{TE} = \frac{2}{3}[/tex]
Consider the Earth and the Moon as a two-particle system.
Find an expression for the gravitational field g of this two-particle system as a function of the distance r from the center of the Earth. (Do not worry about points inside either the Earth or the Moon. Assume the Moon lies on the +r-axis. Give the scalar component of the gravitational field. Do not substitute numerical values; use variables only. Use the following as necessary: G, Mm, Me, r, and d for the distance from the center of Earth to the center of the Moon.)"
sorry but I don't understand
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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at the temperature at which we live, earth's core is solid or liquid?
Explanation:
The Earth has a solid inner core