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.
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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:
We observe that a small sample of material placed in a non-uniform magnetic field accelerates toward a region of stronger field. What can we say about the material?
Answer:
C) It is either ferromagnetic or paramagnetic
Explanation:
The complete question is given below
We observe that a small sample of material placed in a non-uniform magnetic field accelerates toward a region of stronger field. What can we say about the material?
A) It must be ferromagnetic.
B) It must be paramagnetic.
C) It is either ferromagnetic or paramagnetic.
D) It must be diamagnetic.
A ferromagnetic material will respond towards a magnetic field. They are those materials that are attracted to a magnet. Ferromagnetism is associated with our everyday magnets and is the strongest form of magnetism in nature. Iron and its alloys is very good example of a material that readily demonstrate ferromagnetism.
Paramagnetic materials are weakly attracted to an externally applied magnetic field. They usually accelerate towards an electric field, and form internal induced magnetic field in the direction of the external magnetic field.
The difference is that ferromagnetic materials can retain their magnetization when the externally applied magnetic field is removed, unlike paramagnetic materials that do not retain their magnetization.
In contrast, a diamagnetic material is repelled away from an externally applied magnetic field.
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]
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.
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
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.
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⦁ 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
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]
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
Two identical pendulums have the same period when measured in the factory. While one pendulum swings on earth, the other is taken on a spaceship traveling at 95%% the speed of light. Assume that both pendulums operate under the influence of the same net force and swing through the same angle.
When observed from earth, how many oscillations does the pendulum on the spaceship undergo compared to the pendulum on earth in a given time interval?
a. more oscillations
b. fewer oscillations
c. the same number of oscillations
Answer:
Explanation:
As a result of impact of time widening, a clock moving as for an observer seems to run all the more gradually than a clock that is very still in the observer's casing.
At the point when observed from earth, the pendulum on the spaceship takes more time to finish one oscillation.
Hence, the clock related with that pendulum will run more slow (gives fewer oscillations as observed from the earth) than the clock related with the pendulum on earth.
Ans => B fewer oscillations
A loop of wire with cross-sectional area 1 m2 is inserted into a uniform magnetic field with initial strength 1 T. The field is parallel to the axis of the loop. The field begins to grow with time at a rate of 2 Teslas per hour. What is the magnitude of the induced EMF in the loop of wire
Answer:
The magnitude of the EMF is 0.00055 volts
Explanation:
The induced EMF is proportional to the change in magnetic flux based on Faraday's law:
[tex]emf\,=-\,N\, \frac{d\Phi}{dt}[/tex]
Since in our case there is only one loop of wire, then N=1 and we get:
[tex]emf\,=-\,N\, \frac{d\Phi}{dt}[/tex]
We need to express the magnetic flux given the geometry of the problem;
[tex]\Phi=B\,\,A[/tex]where A is the area of the coil that remains unchanged with time, and B is the magnetic field that does change with time. Therefore the equation for the EMF becomes:
[tex]emf\,=-\,N\, \frac{d\Phi}{dt} = \frac{d\Phi}{dt} =-\frac{d\,(B\,A)}{dt} =-\,A\,\frac{d\,(B)}{dt}=- 1\,m^2(2\,\,T/h})= -2\,\,m^2\,T/(3600\,\,s)= -0.00055\,Volts[/tex]
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
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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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) 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].
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.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
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
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
Value of g in CGS system
Answer:
in CGS system G is denoted as gram
An ideal gas in a cubical box having sides of length L exerts a pressure p on the walls of the box. If all of this gas is put into a box having sides of length 0.5L without changing its temperature, the pressure it exerts on the walls of the larger box will be...
p.
2p.
4p.
8p.
12p.
Answer:
2P
Explanation:
See attached file
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)
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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.
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
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Blue light (λ = 475 nm) is sent through a single slit with a width of 2.1 µm. What is the maximum possible number of bright fringes, including the central maximum, produced on the screen? (Hint: What is the largest angle that can be used?)
Answer:
m = 4
Explanation:
The expression that explains the constructive interference of a diffraction pattern is
a sin θ = m λ
where a is the width of the slit and λ the wavelength
sin θ = m λ / a
The maximum value is for when the sine is 1, let's substitute
1 = m λ/a
m = a /λ
let's reduce the magnitudes to the SI system
a = 2.1 um = 2.1 10⁻⁶
lam = 475 nm = 475 10⁻⁹ m
let's calculate
m = 2.1 10⁻⁶ / 475 10⁻⁹
m = 4.42
with m must be an integer the highest value is
m = 4
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
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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A 2.0-kg object moving 5.0 m/s collides with and sticks to an 8.0-kg object initially at rest. Determine the kinetic energy lost by the system as a result of this collision.
Answer:
20J
Explanation:
In a collision, whether elastic or inelastic, momentum is always conserved. Therefore, using the principle of conservation of momentum we can first get the final velocity of the two bodies after collision. This is given by;
m₁u₁ + m₂u₂ = (m₁ + m₂)v ---------------(i)
Where;
m₁ and m₂ are the masses of first and second objects respectively
u₁ and u₂ are the initial velocities of the first and second objects respectively
v is the final velocity of the two objects after collision;
From the question;
m₁ = 2.0kg
m₂ = 8.0kg
u₁ = 5.0m/s
u₂ = 0 (since the object is initially at rest)
Substitute these values into equation (i) as follows;
(2.0 x 5.0) + (8.0 x 0) = (2.0 + 8.0)v
(10.0) + (0) = (10.0)v
10.0 = 10.0v
v = 1m/s
The two bodies stick together and move off with a velocity of 1m/s after collision.
The kinetic energy(KE₁) of the objects before collision is given by
KE₁ = [tex]\frac{1}{2}[/tex]m₁u₁² + [tex]\frac{1}{2}[/tex]m₂u₂² ---------------(ii)
Substitute the appropriate values into equation (ii)
KE₁ = ([tex]\frac{1}{2}[/tex] x 2.0 x 5.0²) + ([tex]\frac{1}{2}[/tex] x 8.0 x 0²)
KE₁ = 25.0J
Also, the kinetic energy(KE₂) of the objects after collision is given by
KE₂ = [tex]\frac{1}{2}[/tex](m₁ + m₂)v² ---------------(iii)
Substitute the appropriate values into equation (iii)
KE₂ = [tex]\frac{1}{2}[/tex] ( 2.0 + 8.0) x 1²
KE₂ = 5J
The kinetic energy lost (K) by the system is therefore the difference between the kinetic energy before collision and kinetic energy after collision
K = KE₂ - KE₁
K = 5 - 25
K = -20J
The negative sign shows that energy was lost. The kinetic energy lost by the system is 20J
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