Answer:
Explanation:
moment of inertia of each blade which is similar to rod rotating about its one end
= 1/3 ml²
moment of inertia of 3 blades = ml²
= 5500 x 46²
I = 11638 x 10³ kg m²
angular velocity = 2πn where n is rotation per second
n = 11 / 60
angular velocity = 2π x 11/60
= 1.1513 rad /s
angular momentum
= moment of inertia x angular velocity
= 11638 x 10³ x 1.1513
= 13399 x 10³ kg m² per second.
When we apply the energy conversation principle to a cylinder rolling down an incline without sliding, we exclude the work done by friction because: A. there is no friction present B. the angular velocity of the center of mass about the point of contact is zero C. the coefficient of kinetic friction is zero D. the linear velocity of the point of contact (relative to the inclined surface) is zero E. the coefficient of static and kinetic friction are equal
Answer:
D. the linear velocity of the point of contact (relative to the inclined surface) is zero
Explanation:
The force of friction emerges only when there is relative velocity between two objects . In case of perfect rolling , there is no sliding so relative velocity between the surface and the point of contact is zero . In other words the velocity of point of contact becomes zero , even though , the whole body is in linear motion . It happens due point of contact having two velocities which are equal and opposite . One of the velocity is in forward direction and the other velocity which is due to rotation is in backward direction . So net velocity of point of contact becomes zero . Due to absence of sliding , displacement due to friction becomes zero . Hence work done by friction becomes zero.
A cobalt-60 source with activity 2.60×10-4 Ci is embedded in a tumor that has
mas 0.20 kg. The source emits gamma photons with average energy 1.25 MeV.
Half the photons are absorbed in the tumor, and half escape.
i. What energy is delivered to the tumor per second? [4 marks]
ii. What absorbed dose, in rad, is delivered per second? [2 marks]
iii. What equivalent dose, in rem, is delivered per second if the RBE for
these gamma rays is 0.70? [2 marks]
Page 6 of 7
iv. What exposure time is required for an equivalent dose of 200 rem? [2
marks]
B. A laser with power output of 2.0 mW at a wavelength of 400 nm is projected
onto a Calcium metal. The binding energy is 2.31 eV.
i. How many electrons per second are ejected? [6 marks]
ii. What power is carried away by the electrons? [4 marks]
C. A hypodermic needle of diameter 1.19 mm and length 50 mm is used to
withdraw blood from a patient? How long would it take for 500 ml of blood to be
taken? Assume a blood viscosity of 0.0027 Pa.s and a pressure in the vein of
1,900 Pa. [10 marks]
D. A person with lymphoma receives a dose of 35 gray in the form of gamma
radiation during a course of radiotherapy. Most of this dose is absorbed in 18
grams of cancerous lymphatic tissue.
i. How much energy is absorbed by the cancerous tissue? [2 marks]
ii. If this treatment consists of five 15-minute sessions per week over the
course of 5 weeks and just one percent of the gamma photons in the
gamma ray beam are absorbed, what is the power of the gamma ray
beam? [4 marks]
iii. If the gamma ray beam consists of just 0.5 percent of the photons
emitted by the gamma source, each of which has an energy of 0.03
MeV, what is the activity, in Curies, of the gamma ray source? [4 marks]
E. A water heater that is connected across the terminals of a 15.0 V power supply
is able to heat 250 ml of water from room temperature of 25°C to boiling point
in 45.0 secs. What is the resistance of the heater? The density of water is 1,000
kg/m2 and the specific heat capacity of water is 4,200 J/kg/°C. [10 marks]
Answer:
A i. E = 9.62 × 10⁻⁷ J/s
ii. The absorbed dose is 4.81 × 10⁻⁶ Gy
iii. The equivalent dose is 3.37 × 10⁻⁴ rem/s
iv. t = 593471.81 seconds
B. i. 4.025 × 10¹⁵/s
ii. 0.512 mW
C. 7218092.2 seconds
D. i. 6.3 × 10⁻¹ J
ii. 1.4 × 10⁻² W
iii. 1.57 × 10³ Curie
E. 0.129 Ω
Explanation:
The given parameters are;
Mass of tumor = 0.20 kg
Activity of Cobalt-60 = 2.60 × 10⁻⁴ Ci
Photon energy = 1.25 MeV
(i) The energy, E, delivered to the tumor is given by the relation;
[tex]E = \frac{1}{2}\left (Number \, of \, decay / seconds \right )\times \left (Energy \, of \, photon \right )[/tex]
[tex]E = \frac{1}{2}\left (2.6\times 10^{-4}Ci )\times \left (\frac{3.70\times 10^{10}decays/s}{1 Ci} \right )\times 1.25\times 10^{6}eV\times \frac{1.6\times 10^{-19}J}{1eV}[/tex]
E = 9.62 × 10⁻⁷ J/s
(ii) The equation for absorbed dose is given as follows;
Absorbed dose, D, in Grays Gy = (Energy Absorbed Joules J)/Mass kg
Therefore, absorbed dose = (9.62 × 10⁻⁷ J/s)/( kg) = 4.81 × 10⁻⁶ Gy
1 Gray = 100 rad
4.81 × 10⁻⁷ Gy = 100 × 4.81 × 10⁻⁶ = 4.81 × 10⁻⁴ rad/s
(iii) Equivalent dose, H, is given by the relation;
H = D × Radiation factor, [tex]w_R[/tex]
∴ H = 0.7 × 4.81 × 10⁻⁴ rad/s = 3.37 × 10⁻⁴ Sv = 3.37 × 10⁻⁴ rem/s
(iv) The exposure time required for an equivalent dose of 200 rem is given as follows;
[tex]\dot{H} = \dfrac{H}{t}[/tex]
Therefore;
[tex]t= \dfrac{200}{{3.37 \times 10^{-4}} } = 593471.81 \, s[/tex]
∴ t = 6.9 days
B. The number of electrons ejected is given by the relation;
[tex]N = \frac{P}{E} = \frac{P \times \lambda}{hc}[/tex]
[tex]N = \dfrac{2.0 \times 10^{-3} \times 400 \times 10^{-9}}{6.626 \times 10^{-34} \times 3 \times 10^8} = 4.025 \times 10^{15}/s[/tex]
(ii) The power carried by the electron
The energy carried away by the electrons is given by the relation;
[tex]KE_e = hv - \Phi[/tex]
[tex]KE_e = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{400 \times 10^{-9}} - 2.31 \times \frac{1.6 \times 10 ^{-19} }{1}[/tex]
[tex]KE_e = 4.9695 \times 10^{-19} - 3.696 \times 10 ^{-19} = 1.2735 \times 10^{-19} J[/tex]
Power, P[tex]_e[/tex], carried away by the electron = 4.025 × 10¹⁵ × 1.2735 × 10⁻¹⁹ = 0.512 mW
C. The given parameters are;
d = 1.19 mm, ∴ r = 1.19/2 = 0.595 × 10⁻³ m
l = 50 mm = 5 × 10⁻³ m
V = 500 ml = 5 × 10⁻⁴ m³
η = 0.0027 Pa
p = 1,900 Pa.
[tex]\dfrac{V}{t} = \dfrac{\pi }{8} \times \dfrac{P/l}{\eta } \times r^4[/tex]
[tex]t = \dfrac{8\times \eta\times V\times l }{\pi \times P \times r^4}[/tex]
[tex]t = \dfrac{8\times 0.0027 \times 5 \times 10^{-4} \times 5 \times 10^{-2} }{\pi \times 1900 \times (0.595 \times 10^{-4} )^4}[/tex]
t = 7218092.2 seconds
D) i. Energy absorbed is given by the relation;
E = m×D
Where:
D = 35 Gray = 35 J/kg
m = 18 g = 18 × 10⁻³ kg
∴ E = 35 × 18 × 10⁻³ = 6.3 × 10⁻¹ J
ii. Total time for treatment = 15 × 5 = 75 minutes
Energy absorbed = 6.3 × 10⁻¹ × 100 = 63 J
Power = Energy(in Joules)/Time (in seconds)
∴ Power = 63/(75×60) = 1.4 × 10⁻² W
iii. Whereby the power is provided by 0.5% of the photons emitted by the source, we have;
[tex]P_{source}= \frac{P_{beam}}{0.005} =\frac{0.0014}{0.005} =0.28 \, W[/tex]
1 MeV = 1.60218 × 10⁻¹³ J
0.03 MeV = 0.03 × 1.60218 × 10⁻¹³ J = 4.80654 × 10⁻¹⁵ J/photon
Therefore, the number of disintegration per second = 0.28 J/s ÷ 4.80654 × 10⁻¹⁵ J/photon = 5.83 × 10¹³ disintegrations per second
1 Curie = 3.7 × 10¹⁰ disintegrations per second
Hence, 5.83 × 10¹³ disintegrations per second = (5.83 × 10¹³)/(3.7 × 10¹⁰) Curie
= 1.57 × 10³ Curie
E. The parameters given are;
Density of water = 1000 kg/m³
Volume of water = 250 ml = 0.00025 m³
Initial temperature, T₁, = 25°C
Final temperature, T₂, = 100°C
Change in temperature, ΔT = 100 - 25 = 75°
Specific heat capacity of the water = 4200 J/kg/°C
Mass of water = Density × Volume = 1000 × 0.00025 = 0.25 kg
∴ Heat supplied = 4200 × 0.25 × 75 = 78,750 J
Time to heat the water = 45.0 sec
Therefore, power = Energy/time = 78750/45 = 1750 W
The formula for electrical power = I²R =VI = V²/R
Therefore, where V = 15.0 V, we have;
15²/R = 1750
R = 15²/1750 = 0.129 Ω.
The resistance of the heater = 0.129 Ω.
a 1200 kg trailer is hitched to a 1400 kg car. the car and trailer are traveling at 72 km.h when the driver applies the brakes on both the car and the trailer. knowing that the braking forces exerted on the car and the trailer are 5000 N and 4000 N respectively, determine (a) the distance traveled by the car and trailer before they come to a stop and (b) the horizontal component of the force exerted by the trailer hitch
Answer:
a) 8.67m
b) 1000N
Explanation:
(a) To find the distance you use the second Newton Law for both car and trailer, in order to calculate the dis-acceleration of the system:
[tex]F=ma\\\\a_=\frac{F}{m}=\frac{5000N+4000N}{1400kg+1200kg}=3.46\frac{m}{s^2}[/tex]
once you have this value, you use the the following kinematic equation to calculate the distance traveled by both car and trailer:
[tex]v^2=v_o^2-2ax\\\\x=\frac{-v^2+v_o^2}{2a}[/tex]
v: final velocity=0
vo: initial velocity = 72km/h = 60 m/s
by replacing the values of these parameters you obtain for x:
[tex]x=\frac{-0m/s+60m/s}{2(3.46m/s^2)}\\\\x=8.67m[/tex]
(b) The horizontal component of the force exerted by the trailer hitch is given by:
[tex]F_T=5000N-4000N=1000N[/tex]
A spherical, non-conducting shell of inner radius = 10 cm and outer radius = 15 cm carries a total charge Q = 13 μC distributed uniformly throughout the volume of the shell. What is the magnitude of the electric field at a distance r = 11.2 cm from the center of the shell? (ε0 = 8.85 × 10-12 C2/N ∙ m2) (Give your answer to the nearest 0.01 MN/C)
Answer:
E = 1580594.95 N/C
Explanation:
To find the electric field inside the the non-conducting shell for r=11.2cm you use the Gauss' law:
[tex]\int EdS=\frac{Q_{in}}{\epsilon_o}[/tex] (1)
dS: differential of the Gaussian surface
Qin: charge inside the Gaussian surface
εo: dielectric permittivity of vacuum = 8.85 × 10-12 C2/N ∙ m2
The electric field is parallel to the dS vector. In this case you have the surface of a sphere, thus you have:
[tex]\int EdS=ES=E(4\pi r^2)[/tex] (2)
Qin is calculate by using the charge density:
[tex]Q_{in}=V_{in}\rho=\frac{4}{3}(r^3-a^3)\rho[/tex] (3)
Vin is the volume of the spherical shell enclosed by the surface. a is the inner radius.
The charge density is given by:
[tex]\rho=\frac{Q}{V}=\frac{13*10^{-6}C}{\frac{4}{3}\pi((0.15m)^3-(0.10m)^3)}\\\\\rho=1.30*10^{-3}\frac{C}{m^3}[/tex]
Next, you use the results of (3), (2) and (1):
[tex]E(4\pi r^2)=\frac{4}{3\epsilon_o}(r^3-a^3)\rho\\\\E=\frac{\rho}{3\epsilo_o}(r-\frac{a^3}{r^2})[/tex]
Finally, you replace the values of all parameters, and for r = 11.2cm = 0.112m you obtain:
[tex]E=\frac{1.30*10^{-3}C/m^3}{3(8.85*10^{-12}C^2/Nm^2)}((0.112m)-\frac{(0.10)^3}{(0.112m)^2})\\\\E=1,580,594.95\frac{N}{C}[/tex]
hence, the electric field is 1580594.95 N/C
a) When we were examining the Electromagnetic Tab, we saw that a flow of electrons or a current as we say it, creates a magnetic field. What about the converse, can a magnetic field be involved in the creation of a flow of electrons/current? Therefore is it reasonable to suggest that we can create a magnetic field by having a flow of current and this can be used to make more current? Explain how this can occur
Answer:
Magnetic field can be used to produce current, infact a changing magnetic field can produce current.
A changing magnetic field in a loop causes the flux linked with the loop to change in turn generating a emf in the loop and therefore a current.
For a loop of area A and resistance R.
I =dPhi/dt/R
В. А
I = AcosФ/R .dB /dt
But it isn't reasonable to say that we can create a magnetic field by having a flow of current and this can be used to make more current because the current generated due to change in magnetic field created by increase/decrease in flow of current will be in a direction such that it will counter act the change in magnetic field caused by increase/decrease in current flow.(lenz's law).
We were unable to transcribe this image
Ф= В. А
I = Acos dB Rd
What is believed to be at the center of the Milky Way galaxy or virtually all large galaxies?
Answer:
A black hole
Explanation:
I'm not sure, i just know that's the right answer
An astronaut visiting Jupiter's satellite Europa leaves a canister of 1.20 mol of nitrogen gas (28.0 g/mol) at 50.0 ∘C on the satellite's surface. Europa has no significant atmosphere, and the acceleration due to gravity at its surface is 1.30 m/s2. The canister springs a leak, allowing molecules to escape from a small hole. Neglect the interaction with surrounding atmosphere. (a) What is the maximum height (in km) above Europa's surface that is reached by a nitrogen molecule whose speed equals the rms speed? Assume that the molecule is shot straight up out of the hole in the canister, and ignore the variation in g with altitude. (b) The escape speed from Europa is 2025 m/s. Can any of the nitrogen molecules escape from Europa and into space?
Answer:
the answer is a
Explanation:
What is the resistance of a resistor attached to a 1.5 V battery if the current in the circuit is 0.15 A?
Answer:10 ohms
Explanation:
Current=0.15A
Voltage=1.5v
Resistance=voltage ➗ current
Resistance=1.5 ➗ 0.15
Resistance=10 ohms
A 1000-kg car is driving toward the north along a straight horizontal road at a speed of 20.0 m/s. The driver applies the brakes and the car comes to a rest uniformly in a distance of 160 m. What are the magnitude and direction of the net force applied to the car to bring it to rest?
Answer:
Force applied to stop the car = 1,250 N
Explanation:
Given:
Mass of car (M) = 1,000 kg
Initial velocity (U) = 20 m/s
Final velocity (V) = 0 m/s
Distance (S) = 160 m
Find:
Force applied to stop the car.
Computation:
[tex]v^2 = u^2 + 2as\\\\0^2=20^2+2(a)(160)\\\\0=400+320(a)\\\\Acceleration = a = -1.25m/s^2\\\\Force = ma \\\\Force= 1,000(1.25)\\\\Force = 1,250 N[/tex]
Force applied to stop the car = 1,250 N
Which has the deer’s image? Why?
Answer: what ( younger brothers of hers )
Explanation:
K bye
A turntable has a moment of inertia of 3.00 x 10-2 kgm2 and spins freely on a frictionless bearing at 25.0 rev/min. A 0.300 kg ball of putty is dropped vertically on the turntable and sticks at a point 0.10m from the center. The total moment of inertia of the system increases, and the turntable slows down. But by what factor does the angular momentum of the system change after the putty is dropped onto the turntable
Answer:
There will be no change in the angular momentum of the system.
Explanation:
Total angular momentum of the system will remain unchanged . We can apply law of conservation of momentum because no external torque is acting on the system . There is increase in the momentum of inertia due to dropping of ball of putty . In order to conserve angular momentum , the system decreases its angular velocity . Hence the final angular momentum remains unchanged .
Geophysicists have estimated that the temperature at the center of the Earth's
core is at least 5000°C. Express this temperature in Fahrenheit degrees.
CA 8703 °F
B 9032°F
C 10,087°F
D 5609°F
Answer: B. 9032°F
Explanation:
First we take down the data made available to us.
Estimated temperature of the Earth's core = 5000° Celsius
Now to express this temperature in Fahrenheit, we make use of the formula;
[°F] = [°C] × 9/5 + 32
Where
[°F] is temperature in Fahrenheit and
[°C] is temperature in Celsius
So we just input our figure into the formula
[°F] = [5000°] × 9/5 + 32
°F = 5000 × 1.8 + 32
°F = 9000 + 32
°F = 9032°
So 5000° Celsius is expressed as 9032° Fahrenheit
A convex mirror of focal length 33 cm forms an image of a soda bottle at a distance of 19 cm behind the mirror.If the height of the image is 7.0 cm,where is the object located,and how tall is it? What is the magnification of the image? Is the image virtual or real? Is the image inverted or upright? Draw a ray diagram to confirm your results.
Answer:
Image distance = 44.8cm, Image height = 16.5cm, Magnification = 0.42
The image is a virtual and upright image.
Explanation:
The nature of image formed by an object placed in front of a convex mirror is always diminished, virtual and erect.
The focal length f and the image distance are always NEGATIVE beacause the image is formed behind the mirror.
Given f = -33.0cm, v = -19.0cm
using thr mirror formula to get the object distance u, we have;
[tex]\frac{1}{f}=\frac{1}{u} + \frac{1}{v}\\ \frac{1}{u}=\frac{1}{f} - \frac{1}{v}\\\frac{1}{u}=\frac{1}{-33} - \frac{1}{-19}\\\frac{1}{u}=\frac{-19+33}{627} \\\frac{1}{u}=\frac{14}{627} \\u=\frac{627}{14} \\u = 44.8cm[/tex]
To calculate the image height, we will use the magnification formula
M = [tex]\frac{image\ height}{object\ height}=\frac{image\ distance}{object\ distance} \\[/tex]
M = [tex]\frac{Hi}{HI}=\frac{v}{u}[/tex]
Given Hi = 7.0cm
v = 19.0cm
u = 44.8cm
HI = 7*44.8/19
HI = 16.5cm
The object height is 16.5cm
Magnification = v/u = 19.0/44.8 = 0.42
SInce the image is formed behind the mirror, the image is a VIRTUAL and UPRIGHT image
Four unequal resistors are connected in a series circuit. Which one of the following statements is correct about this circuit? The total resistance is equal to average of the four resistors. The total resistance is less than the smallest resistor. The total resistance is less than the largest resistor. The total resistance is equal to any one of the resistors. The total resistance is more than the largest resistor.
Answer:
The total resistance is more than the largest resistor.
Explanation:
In series connection of resistors, the total resistance of the connection is the sum of all the resistance of the resistors connected.
For example; let assume three resistors with resistance 3,4 and 5 ohms, the total resistance is;
Rt = 3+4+5 = 12 ohms
So, the total resistance is greater than the largest resistor since 12 ohms is greater than 5 ohms(largest resistor).
A car speeds up from 18.54 m/s to
29.52 m/s in 13.84 s.
The acceleration of the car is:
Answer:
.7934[tex]m/s^{2}[/tex]
Explanation:
Acceleration = change in velocity / change in time
A = 10.98[tex]m/s[/tex] / 13.84[tex]s[/tex]
A = .7934[tex]m/s^{2}[/tex]
Answer:0.8 m/s^2
Explanation:
initial velocity(u)=18.54m/s
Final velocity(v)=29.52m/s
Time(t)=13.84 sec
Acceleration =(v-u)/t
acceleration =(29.52-18.54)/13.84
Acceleration =10.98/13.34
Acceleration=0.8 m/s^2
A 25kg box in released on a 27° incline and accelerates down the incline at 0.3 m/s2. Find the friction force impending its motion? What is the coefficient of kinetic friction?
A block is given an initial speed of 3m/s up a 25° incline. Coefficient of friction
Answer:
a) μ = 0.475 , b) μ = 0.433
Explanation:
a) For this exercise of Newton's second law, we create a reference system with the x-axis parallel to the plane and the y-axis perpendicular to it
X axis
Wₓ - fr = m a
the friction force has the expression
fr = μ N
y Axis
N - [tex]W_{y}[/tex] = 0
let's use trigonometry for the components the weight
sin 27 = Wₓ / W
Wₓ = W sin 27
cos 27 = W_{y} / W
W_{y} = W cos 27
N = W cos 27
W sin 27 - μ W cos 27 = m a
mg sin 27 - μ mg cos 27 = m a
μ = (g sin 27 - a) / (g cos 27)
very = tan 27 - a / g sec 27
μ = 0.510 - 0.0344
μ = 0.475
b) now the block starts with an initial speed of 3m / s. In Newton's second law velocity does not appear, so this term does not affect the result, the change in slope does affect the result
μ = tan 25 - 0.3 / 9.8 sec 25
μ = 0.466 -0.03378
μ = 0.433
How does a concave mirror form an image
Answer:
a magnified and erect virtual is found to be formed.
During an earthquake, _______ travels through the Earth's interior as _______ waves.
Answer:
During an earthquake, seismic waves travels through the Earth's interior as body or p waves.
Explanation:
If neither of the bold words look familiar from your lesson feel free to ignore this answer
What is an independent variable?
A. A variable that is intentionally changed during an experiment
B. A variable that depends on the experimental variable
C. A variable that is not used in an experiment
D. A variable that is unknown during the experiment
Answer:
The answer is A
Explanation:
Independent variables don't have to depend on other factors of the experiment because they're independent
Answer:
A.
Explanation:
Independent variables don't have to depend on other factors of the experiment because they're independent.
Which two factors affect the force between two masses, according to the universal law of gravitation? the masses of the objects the radii of the two objects the speed of the two objects the distance between the objects the weight of the two objects
Answer:1. The masses Of The objects 2.the distance between the objects
Explanation:
newton's law of universal gravitation states that the force of attraction between two masses is directly proportional to the product Of The masses, and inversely proportional to thermal square Of The distance between them.therefore we can conclude that the two factors affecting the force between masses are:(1)the masses of the objects (2)the distance between the objects
The two factors that affect the force between two masses, according to the universal law of gravitation are
the masses of the objectsthe distance between the objectsWhat is newton's Universal law of gravitation?Newton's Universal law of gravitation states that the force of attraction that exists between particle or objects is directly proportion to the product of their masses and inversely proportional to the distances between them .
Therefore, The two factors that affect the force between two masses, according to the universal law of gravitation are
the masses of the objects.the distance between the objects.Learn more about Newton's universal law of gravitation below.
https://brainly.com/question/10454047
#SPJ2
Two atomic particles approach each other in a head-on collision. Each particle has a mass of 2.97 × 10-25 kg. The speed of each particle is 2.19 × 108 m/s when measured by an observer standing in the laboratory. (a) What is the speed of one particle as seen by the other particle? (b) Determine the relativistic momentum of one particle, as it would be observed by the other.
Answer:
a) [tex]\vec{v}_{12}=2.86*10^{8} m/s[/tex]
b) [tex]p=2.81*10^{-16} kg*m/s[/tex]
Explanation:
a) When we have two particles traveling in parallel directions, the formula for relative velocity is:
[tex]\vec{v}_{12}=\frac{\vec{v}_{1}-\vec{v}_{2}}{1-\frac{\vec{v}_{1}\vec{v}_{2}}{c^{2}}}[/tex]
Here we have that v(1) = -v(2), the speed of the of the second particle is the negative of the first one.
If we use these equivalence we have:
[tex]\vec{v}_{12}=\frac{2\vec{v}_{1}}{1+\frac{\vec{v}_{1}^{2}}{c^{2}}}[/tex]
[tex]\vec{v}_{12}=\frac{2*2.19*10^{8}}{1+\frac{2.19*10^{16}}{3*10^{16}}}[/tex]
[tex]\vec{v}_{12}=2.86*10^{8} m/s[/tex]
And, [tex]\vec{v}_{21}=-2.86*10^{8} m/s[/tex]
b) The relativistic momentum equation to one particle observed by the other particle, is:
[tex]p=\gamma mv[/tex]
Where gamma is:
[tex]\gamma = \frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}[/tex]
v is the speed of the first particle relative to the second particle ([tex]2.86*10^{8}[/tex])m is the mass of the particle [tex]2.97*10^{-25} kg[/tex]Then gamma will be:
[tex]\gamma=\frac{1}{\sqrt{1-\frac{(2.86*10^{8})^{2}}{(3*10^{8})^{2}}}}[/tex]
[tex]\gamma=3.31[/tex]
Finally, the value of the momentum will be:
[tex]p=3.31*2.97*10^{-25}*2.86*10^{8}[/tex]
[tex]p=2.81*10^{-16} kg*m/s[/tex]
I hope it helps you!
Exercise should challenge your body and be at a greater intensity than your usual bif daily activity. Discuss
Answer:
A fan pushes hot air out of a vent and into a room. The hot air displaces cold air in the room, causing the cold air to move closer to the floor.
The hot air displacing the cold air is an example of transfer by
Explanation:
You could use an analytical or triple beam balance to determine a ___ called ____
A)
physical property; mass.
B)
chemical property, mass.
C)
physical property; weight.
D)
physical property; density.
Answer:
a and b are the correct answers
Explanation:
Answer:
A) physical property; mass.
Explanation:
took the test
25. You lift a book from the floor to a bookshelf 1.0 m above the ground. How much power is used if the
upward force is 15.0 N and you do the work in 2.0 ?
Explanation:
P=E/T
E=15N
T=2s
P=15/2
P=7.5
A small object begins a free-fall from a height of =81.5 m at 0=0 s . After τ=2.20 s , a second small object is launched vertically up from the ground with an initial velocity of 0=40.0 m/s . At what height from the ground will the two objects first meet?
Answer:
33.2 m
Explanation:
For the first object:
y₀ = 81.5 m
v₀ = 0 m/s
a = -9.8 m/s²
t₀ = 0 s
y = y₀ + v₀ t + ½ at²
y = 81.5 − 4.9t²
For the second object:
y₀ = 0 m
v₀ = 40.0 m/s
a = -9.8 m/s²
t₀ = 2.20 s
y = y₀ + v₀ t + ½ at²
y = 40(t−2.2) − 4.9(t−2.2)²
When they meet:
81.5 − 4.9t² = 40(t−2.2) − 4.9(t−2.2)²
81.5 − 4.9t² = 40t − 88 − 4.9 (t² − 4.4t + 4.84)
81.5 − 4.9t² = 40t − 88 − 4.9t² + 21.56t − 23.716
81.5 = 61.56t − 111.716
193.216 = 61.56t
t = 3.139
The position at that time is:
y = 81.5 − 4.9(3.139)²
y = 33.2
If the acceleration of the projective is: a = c s m/s 2 Where c is a constant that depends on the initial gas pressure behind the projectile. The initial position of the projectile is s=1.5m and the projectile is initially at rest. The projectile accelerates until it reaches the end of the barrel at s=3m. What is the value of the constant c such that the projectile leaves the barrel with a velocity of 200 m/s?
Answer:
c = 4,444.44
Explanation:
You have the following expression for the acceleration of the projectile:
[tex]a=cs[/tex] (1)
s: distance to the ground of the projectile
To find the value of the constant c you use the following formula:
[tex]v^2=v_o^2+2a \Delta s[/tex] (2)
vo: initial velocity = 0 m/s
v: final speed = 200 m/s
Δs: distance traveled by the projectile = 3m - 1.5m = 1.5m
You replace the expression (1) into the expression (2):
[tex]v^2=2(cs)\Delta s[/tex]
You do the constant c in the last equation, then you replace the values of v, s and Δs:
[tex]c=\frac{v^2}{2s\Delta s}=\frac{(200m/s)^2}{2(3m/s^2)(1.5m)}=4444.44[/tex]
The strength of the force of friction depends on which two factors?
Answer:
coefficient of friction (μ) and normal force (N)
Answer: How hard the surfaces push together and the types of surfaces involved
Explanation:
A mutation causes a dog to be born with a tail that is shorter than normal.
Which best describes this mutation?
Answer:
A mutation causes a dog to be born with a tail that is shorter than normal. Which best describes this mutation? It is harmful because it obviously affects the dog’s survival. It is harmful because it affects the dog’s physical appearance. It is neutral because it does not obviously affect the dog’s survival. It is beneficial because it affects the dog’s physical appearance.
Explanation:
Answer:
C
Explanation:
:)))
Newton's law of cooling states that the temperature of an object changes at a rate proportional to the difference between its temperature and that of its surroundings. Suppose that the temperature of a cup of coffee obeys Newton's law of cooling. If the coffee has a temperature of 210 degrees Fahrenheit when freshly poured, and 2.5 minutes later has cooled to 191 degrees in a room at 64 degrees, determine when the coffee reaches a temperature of 156 degrees.
Answer:
Explanation:
The problem is based on Newton's law of cooling .
According to Newton's law
dQ / dt = k ( T - T₀ ) ,
dT / dt = k' ( T - T₀ ) ; dT / dt is rate of fall of temperature.
T is average temperature of hot body , T₀ is temperature of surrounding .
In the first case rate of fall of temperature = (210 - 191) / 2.5
= 7.6 degree / s
average temperature T = (210 + 191) /2
= 200.5
Putting in the equation
7.6 = k' ( 200.5 - 64 )
k' = 7.6 / 136.5
= .055677
In the second case :---
In the second case, rate of fall of temperature = (191 - 156) / t
= 35 / t , t is time required.
average temperature T = (156 + 191) /2
= 173.5
Putting in the equation
35 / t = .05567 ( 173.5 - 64 )
t = 5.74 minute .
A bird flies 1.3 m with an average speed of 2.4m/s How many seconds did the bird fly?
Answer:
0.54 sec.
Explanation:
The Speed: 2.4 m/seconds
The Distance: 1.3 m
So time = Distance/Speed
Time = 1.3 / 2.4
THE ANSWER WILL BE 0.54 seconds.
