Answer:
[tex]v=15.9499m/s[/tex]
Explanation:
From the question we are told that:
Mass of wood [tex]m=1kg[/tex]
force constants [tex]k= 200N-m[/tex]
Coefficient of kinetic friction [tex]\mu= 0.2[/tex]
Bullet mass [tex]m_b= 20 \approx 0.02kg[/tex]
Spring compresion [tex]y=15cm \approx 0.15 m[/tex]
Generally the equation for kinetic energy of bullet [tex]K>E_b[/tex] is mathematically given by
Complete Question
[tex]K.E_b=spring potential energy+work done against friction[/tex]
[tex]K.E_b=\frac{1}{2} mbv^2[/tex]
[tex]\frac{1}{2} m_b v^2=\frac{1}{2} ky^2+\mu my[/tex]
[tex]\frac{1}{2} (0.02)v^2=\frac{1}{2} (0.2)(0.15)^2+0.2(1)(0.15)[/tex]
[tex]v=15.9499m/s[/tex]
[tex]v\approx16m/s[/tex]
One way to probe the nucleus is to bombard a sample with high-energy electrons. To learn about the nuclear structures in a sample, the de Broglie wavelengths of these electrons would need to be a little smaller than a nuclear radius. Estimate the energy of such electrons. Give your answer in electron-volts. (Assume that the wavelength used is about 9.0 fm.) eV
Answer:
E = 1.38 x 10⁸ eV = 138 MeV
Explanation:
The energy associated with the given wavelength can be found from the following formula:
[tex]E = \frac{hc}{\lambda}[/tex]
where,
E = Energy of electron = ?
h = Plank's Constant = 6.625 x 10⁻³⁴ J.s
c = Speed of Light = 3 x 10⁸ m/s
λ = wavelength = 9 fm = 9 x 10⁻¹⁵ m
Therefore,
[tex]E = \frac{(6.625\ x\ 10^{-34}\ J.s)(3\ x\ 10^8\ m/s)}{9\ x\ 10^{-15}\ m}\\\\E = (2.21\ x\ 10^{-11}\ J)(\frac{1\ eV}{1.6\ x\ 10^{-19}\ J})[/tex]
E = 1.38 x 10⁸ eV = 138 MeV
The activity of a radioisotope is found to decrease 40% of its original value in 2.59 x 10 s.
Calculate the decay constant of the radioisotope.
Answer: [tex]0.0353\ s^{-1}[/tex]
Explanation:
Given
Radioactive material is found to decrease 40% of its original value in [tex]2.59\times 10\ s[/tex]
Sample at any time is given by
[tex]N=N_oe^{-\lambda t}[/tex]
where, [tex]\lambda=\text{decay constant}[/tex]
Put values
[tex]\Rightarrow 0.4N_o=N_oe^{-\lambda\cdot 2.59\times 10}\\\Rightarrow 0.4=e^{-\lambda\cdot 2.59\times 10[/tex]
Taking natural logarithm both side
[tex]\Rightarrow \lambda=\dfrac{\ln 2.5}{25.9}\\\\\Rightarrow \lambda =0.0353\ s^{-1}[/tex]
Object X of mass 4 kg travels with a speed of 3 ms toward object Y of mass 2 kg that is initially at rest. Object X then collides with and sticks to object Y . After the collision, object X and object Y remain stuck together. How much mechanical energy is converted into nonmechanical energy during the collision?
The nonmechanical energy or the loss of energy after the collision will be equal to [tex]E=6\ J[/tex]
What is the conservation of momentum?The conservation of the momentum is defined as when two bodies collide with each other then the total energy of the masses will remain constant.
It is given in the question that
Mass of the body X is [tex]M_X=4\ kg[/tex]
The velocity of the body X is =[tex]V_x=3 \ \frac{m}{s}[/tex]
Mass of the body Y is [tex]M_Y= 2\ kg[/tex]
The velocity of body Y is [tex]V_y= 0[/tex]
Now to find out the energy converted after the collision we will first find the final velocity of both the bodies.
Now from the conservation of the momentum
[tex]M_X V_X +M_YV_Y=(M_X+M_Y)V_F[/tex]
[tex](4\times 3)+(2\times0)=(4+2)\times V_F[/tex]
[tex]V_F= \dfrac{12}{6}[/tex]
[tex]V_F = 2 \frac{m}{s}[/tex]
Now to find the Change in the energy of the body
[tex]\Delta E=E_i-E_f[/tex]
[tex]\Delta E=\dfrac{1}{2} M_XV_X^2+\dfrac{1}{2} M_YV_Y^2-\dfrac{1}{2} (M_X+M_Y)V_F^2[/tex]
[tex]\Delta E=\dfrac{1}{2} (4)(3^2)+\dfrac{1}{2} (2)(0^2)-\dfrac{1}{2} ( 4+2) 2^2[/tex]
[tex]\Delta E= 6\ J[/tex]
Thus the nonmechanical energy or the loss of energy after the collision will be equal to [tex]E=6\ J[/tex]
To know more about the collision follow
https://brainly.com/question/7694106
Types of telescope
for Space
observation
Answer:The three main types are reflecting telescopes, refracting telescopes, and catadioptric telescopes. Radio telescopes collect and focus radio waves from distant objects. Space telescopes orbit Earth, collecting wavelengths of light that are normally blocked by the atmosphere.
Answer:
oii me manda mensagem fofo vc tem namorada fofo
No file link please
please help with both questions I’m giving all my points :) it’s 23 and 24
A uniform ladder of length 10.8 m is leaning against a vertical frictionless wall. The weight of the ladder is 323 N, and it makes an angle of 1.16 radians with the floor. A man weighing 734 N climbs slowly up the ladder. When he has climbed to a point that is 7.46 m from the base of the ladder (measured along the ladder), the ladder starts to slip. Find the coefficient of static friction between the floor and the ladder.
Answer:
0.3625
Explanation:
From the given information:
Consider the equilibrium conditions;
On the ladder, net torque= 0
Thus,
[tex]\tau_{net} = 0[/tex]; and
[tex]-fL \s in \theta +m_L g \dfrac{L}{2} cos \theta + mg (7.46\ m) cos \theta = 0[/tex]
However, by rearrangement;
[tex]fL \s in \theta =m_L g \dfrac{L}{2} cos \theta + mg (7.46\ m) cos \theta \\ \\ \mu(m_L + m) gL \ sin \theta = (323 \ N) ( 10.8 \ meters) \ cos 56^0 + (734 \ N) (7.46 \ m) \ cos \ 66.46^0[/tex]
[tex]\mu= \dfrac{ (323 \ N) ( 10.8 \ m) \ cos 56^0 + (734 \ N) (7.46 \ m) \ cos \ 66.46^0}{\Big [(323 \ N)+(734 \ N) \Big] (10.8 \ m)}[/tex]
[tex]\mathbf{\mu= 0.3625 }[/tex]
In an NMR experiment, the RF source oscillates at 34 MHz and magnetic resonance of the hydrogen atoms in the sample being in- vestigated occurs when the external field Bext has magnitude 0.78 T. Assume that Bint and Bext are in the same direction and take the pro- ton magnetic moment component u, to be 1.41 X 10-26 J/T. What is the magnitude of Bint?
Answer:
[tex]B_{int}=-0.015T[/tex]
Explanation:
From the question we are told that:
RF source oscillation speed [tex]\sigma= 34 MHz[/tex]
The external field [tex]Bext =0.78 T[/tex].
Pro- ton magnetic moment component [tex]\mu=1.41 X 10-26 J/T[/tex]
Generally the equation for magnitude of [tex]B_{int}[/tex] is mathematically given by
[tex]B_{int}=B_{ext}-\frac{h\triangle \sigma}{2 \mu}[/tex]
[tex]B_{int}=0.78-\frac{6.6*10^{-34}*34*10^6}{2*1.41*10^{26}}[/tex]
[tex]B_{int}=0.78-0.7957[/tex]
[tex]B_{int}=-0.015T[/tex]
PLEASE HELP! I'LL GIVE BRAINLEST
Answer:
a - is the amount of matter in this object
Which of the following types of electromagnetic radiation has waves with the highest frequency ? A. Infrared radiation B. Visible light c. Microwaves D. X-rays
A sound wave has a wavelength of 2M and a frequency of 100 Hz. The speed of the wave is
Answer:
200 meters per second
Explanation:
wave speed = wavelength x frequency
2 meters x 100 hertz = 200 meters per second
In this equation, what shows that transmutation has taken place?
242 Cm – 328 Pu + He
A. There is conservation of both nucleons and atoms.
B. The nucleus of an atom changes
C. It involves more than one element
D. The number of atoms is conserved,but the number of nucleons is not
Another word for kinetic energy
could be
energy.
A. Safe
B. Moving
C. Stored
D. Potential
Answer:
moving
Explanation:
hope it helped!!!
Answer:
B
Explanation:
What would the current be for a circuit that has a voltage of 0.8 V and a resistance of 0.01 Q?
0 1 = 0.01 A
0 1 = 0.8 A
0 1 = 80 A
O I = 0.08 A
Answer:
80 A
Explanation:
Hi there!
Ohm's law states that [tex]V=IR[/tex] where V is the voltage, I is the current and R is the resistance.
Plug the given information into Ohm's law (V=0.8, R=0.01) and solve for I
[tex]V=IR\\0.8=I(0.01)[/tex]
Divide both sides by 0.01 to isolate I
[tex]0.8=I(0.01)\\\frac{0.8}{0.01}= \frac{I(0.01)}{0.01} \\80=I[/tex]
Therefore, the current for this circuit would be 80 A.
I hope this helps!
An experiment is performed on an unknown material and produces the given heat curve. The temperature of the material is shown as a function of heat added. Other experiments determine that the material has a temperature of fusion of fusion=235 °C and a temperature of vaporization of vapor=471 °C.
If the sample of material has a mass of =9.80 g, calculate the specific heat when this material is a solid, s, and when it is liquid, l.
The specific heat of the solid phase is 0.333 joules per gram-degree Celsius.
The specific heat of the liquid phase is 0.593 joules per gram-degree Celsius.
In this case, we need to determine that specific heat for solid and liquid states of matter. By Heat Physics, we understand that specific heat is contained in the slopes of the two sensible phases in the following form:
[tex]\frac{\Delta T}{\Delta Q} = \frac{1}{m\cdot c}[/tex] (1)
Where:
[tex]\Delta T[/tex] - Temperature change, in degrees Celsius.[tex]\Delta Q[/tex] - Heat received, in joules.[tex]m[/tex] - Mass of the sample, in grams.[tex]c[/tex] - Specific heat of the sample, in joules per kilogram-degrees Celsius.Solid phase
If we know that [tex]m = 9.80\,g[/tex], [tex]T_{1} = 40\,^{\circ}C[/tex], [tex]T_{2} = 235\,^{\circ}C[/tex], [tex]Q_{1} = 183\,J[/tex] and [tex]Q_{2} = 819\,J[/tex], then the specific heat of the solid phase is:
[tex]c = \frac{\Delta Q}{m\cdot \Delta T}[/tex]
[tex]c = \frac{819\,J-183\,J}{(9.80\,g)\cdot (235\,^{\circ}C - 40\,^{\circ}C)}[/tex]
[tex]c = 0.333\,\frac{J}{g\cdot ^{\circ}C}[/tex]
The specific heat of the solid phase is 0.333 joules per gram-degree Celsius.
Liquid phase
If we know that [tex]m = 9.80\,g[/tex], [tex]T_{3} = 230\,^{\circ}C[/tex], [tex]T_{4} = 471\,^{\circ}C[/tex], [tex]Q_{3} = 1470\,J[/tex] and [tex]Q_{4} = 2870\,J[/tex], then the specific heat of the liquid phase is:
[tex]c = \frac{\Delta Q}{m\cdot \Delta T}[/tex]
[tex]c = \frac{2870\,J - 1470\,J}{(9.80\,g)\cdot (471\,^{\circ}C - 230\,^{\circ}C)}[/tex]
[tex]c = 0.593\,\frac{J}{g\cdot ^{\circ}C}[/tex]
The specific heat of the liquid phase is 0.593 joules per gram-degree Celsius.
We kindly invite you to see this question related to specific heat: https://brainly.com/question/11194034
Infrared radiation from young stars can pass through the heavy dust clouds surrounding them, allowing astronomers here on Earth to study the earliest stages of star formation, before a star begins to emit visible light. Suppose an infrared telescope is tuned to detect infrared radiation with a frequency of 1.61THz. Calculate the wavelength of the infrared radiation. Round your answer to 3 significant digits.
Answer:
λ = 1.86 x 10⁻⁴ m = 186 μm
Explanation:
The relationship between the wavelength and the frequency of a wave is given by the following equation:
[tex]c = f\lambda\\\\\lambda = \frac{c}{f}[/tex]
where,
λ = wavelength of infrared radiation = ?
c = speed of infrared radiation = speed of light = 3 x 10⁸ m/s
f = frequency of infrared radiation = 1.61 THz = 1.61 x 10¹² Hz
Therefore,
[tex]\lambda = \frac{3\ x\ 10^8\ m/s}{1.61\ x\ 10^{12}\ Hz}[/tex]
λ = 1.86 x 10⁻⁴ m = 186 μm
PLS HELP.
A rope breaks when the speed of a 0.309 kg mass moving in a circle of radius of 0.429 m reaches 12.9 m/s
how much tension is in the strings when it breaks?
unti=n
120 works for acellus
Answer: 119.9
Explanation:
F = (mv^2)/r
Here we know m (0.309kg), v (12.9m/s) and r (0.429m)
So F = (0.309*12.9^2)/0.429 = 119.861748
Conservation equations are powerful equations, because of how versatile and flexible they are to different physical situations. Conservation of mass is a simple one that can even be applied to fluids with care. The conservation of momentum also applies to any closed system, such as collisions. We've examined the nuances of energy conservation in our labs with total mechanical energy. Does kinetic energy remain conserved in all collisions too?
a) true
b) false
Answer:
b) false
Explanation:
Since in the given situation it is mentioned that the mass conversation would be simple and same applied to the fluids with care also the conservation of momentum would be applied to any type of closed system like collisions
But as we know that
In the inelastic collisions, the total kinetic energy would not be remain conserved
So the given statement is false
A compact fluorescent bulb is 17.0% efficient. How much energy input would be required for the bulb to produce 252 J of light energy?
What time of energy is the waste output and how much would be created?
Help
Answer:
1482.35J
Explanation:
Efficency=(output energy/input energy) x100
17.0 =(252/input energy) x100
input energy =100x252/17=1482.35J
wasted energy=1482.35-252=1230.35J
Two vectors have magnitudes 3 and 4 . how are the directions of the two vectors related if: a/the sum has magnitude 7.0
Which of the following best describes Earth's crust, according to the theory of plate tectonics?
Answer:
the Earth's crust is broken into about 12 plates that float on hotter, softer rocks in the underlying mantle
Explanation:
Jack has a weight of 300 N and sits 2.0 m from the pivot of see - saw. Jill has a weight of 450 N and sits 1.5 m from pivot. Who will move down?
Answer:
Jill will move down first
Explanation:
A football is thrown down off of a building with a force of 70
Newtons. Gravity is pulling down the football with a force of
19 Newtons. What is the net force on the football?
Answer:
89 N downward.
Explanation:
Both forces are working in the same direction -- down.
Therefore you can add them together.
F = F1 + F2
F is the total force
F1 is the gravitational force
F2 is the force the person provides
F1 = 19
F2 = 70
F = ?
F = 19 + 70
F = 89 Newtons downward.
Answer:
wE dONt taLk aBt brunooooOoooOoOo ;')
Explanation:
What is the frequency of a wave traveling at 300,000,000 m/s with a wavelength of .0025 m/cycle? Please help me !!
Answer:
its 00.0035474
Explanation:
..... .. . . .. . . . . .. .
helpppppppppppppppppppppppppppp............
an object is moving with initial velocity of 5 m/s. After 10 seconds final velocity is 10 m/s. Calculate its acceleration.
Answer:
0.5 m/s 2 is the acceleration
Explanation:
hope it helped!!!
What makes astronomers think that impact rates for the Moon must have been higher earlier than 3.8 billion years ago?
Answer: See explanation
Explanation:
The reason why astronomers think that the rates of impact for the Moon must have been higher earlier than 3.8 billion years ago is because on the older highlands, there are ten times more craters than on the younger maria.
It is believed that the impact rate was higher earlier and thus can be seen when the numbers of the craters that can be seen on the lunar highlands is being compared to that on the maria. It should be noted that there are about 10 times more craters that can be found on the highlands than those on the maria.
If there was a constant rate of impact throughout the history of the Moon, then the highlands be about 10 times older and therefore will have been formed about 38 billion years ago.
Two window washers, Bob and Joe, are on a 3.00 m long, 395 N scaffold supported by two cables attached to its ends. Bob weighs 805 N and stands 1.00 m from the left end. Two meters from the left end is the 500 N washing equipment. Joe is 0.500 m from the right end and weighs 820 N. Given that the scaffold is in rotational and translational equilibrium, what are the forces on each cable
Answer:
- the forces on the left hand side is 1.038 kN
- the forces on the right hand side is 1.483 kN
Explanation:
Given the data in the question, as illustrated in the image below;
Length of the scaffold = 3 m
weight of the scaffold = 395 N
Weight of Bob = 805 N and stands 1 m from the left end
weight of washing equipment = 500N and on sits 2 m from the left end
Weight of Joe = 820 N and stand 0.500 m from the right end
so the force on the left cable will be;
[tex]T_{left[/tex] = [tex]\frac{1}{3m}[/tex][ (805 N)( (3-1) m) + ( 395 N )( [tex]\frac{3}{2} m[/tex]) + ( 500 N )(1m ) + ( 820 N)( 0.500m ) ]
[tex]T_{left[/tex] = [tex]\frac{1}{3m}[/tex][ 1610 + 592.5 + 500 + 410 ]
[tex]T_{left[/tex] = [tex]\frac{1}{3m}[/tex][ 3112.5 ]
[tex]T_{left[/tex] = 1037.5 N
[tex]T_{left[/tex] = 1.038 kN
Therefore, the forces on the left hand side is 1.038 kN
On the right hand side;
[tex]T_{Right[/tex] = ( 805 N + 395 N + 500 N + 820 N ) - 1037.5 N
[tex]T_{Right[/tex] = 2520 N - 1037.5 N
[tex]T_{Right[/tex] = 1482.5 N
[tex]T_{Right[/tex] = 1.483 kN
Therefore, the forces on the right hand side is 1.483 kN
A 10 kg migratory swan cruises at 20m/s. A calculation that takes into ac-count the necessary forces shows that this motion requires 200W of mechanical power. If we assume an efficiency similar to humans (say, 25%), a reasonableassumption, then the metabolic power of the swan is significantly higher thanthis. The swan does not stop to eat during a long day of flying; it get theenergy it needs from fat stores. Assuming an efficiency similar to humans, after12 hours of flight.
Required:
a. How far has the swan traveled?
b. How much metabolic energy has it used?
c. What fraction of its body mass does it lose?
Answer:
Part A:
Distance=864000 m=864 km
Part B:
Energy Used=ΔE=8638000 Joules
Part C:
[tex]\frac{\triangle m}{m}=0.004998=0.49985\%[/tex]
Explanation:
Given Data:
v=20m/s
Time =t=12 hours
In Secs:
Time=12*60*60=43200 secs
Solution:
Part A:
Distance = Speed**Time
Distance=v*t
Distance= 20*43200
Distance=864000 m=864 km
Part B:
Energy Used=ΔE= Energy Required-Kinetic Energy of swans
Energy Required to move= Power Required*time
Energy Required to move=200*43200=8640000 Joules
Kinetic Energy=[tex]\frac{1}{2}mv^2[/tex]
[tex]K.E\ of\ Swans=\frac{1}{2} *10*(20)^2=2000\ Joules[/tex]
Energy Used=ΔE=8640000 -2000
Energy Used=ΔE=8638000 Joules
Part C:
Fraction of Mass used=Δm/m
For This first calculate fraction of energy used:
Fraction of energy=ΔE/Energy required to move
ΔE is calculated in part B
Fraction of energy=8638000/8640000
Fraction of energy=0.99977
Kinetic Energy=[tex]\frac{1}{2}mv^2[/tex]
Now, the relation between energies ratio and masses is:
[tex]\frac{\triangle E}{E}=\frac{\triangle m}{2m}v^2[/tex]
[tex]\frac{\triangle m}{m}=\frac{2}{v^2} *\frac{\triangle E}{E}\\\frac{\triangle m}{m}=\frac{2}{20^2} *0.99977[/tex]
[tex]\frac{\triangle m}{m}=0.004998=0.49985\%[/tex]
One method for determining the amount of corn in early Native American diets is the stable isotope ratio analysis (SIRA) technique. As corn photosynthesizes, it concentrates the isotope carbon-13, whereas most other plants concentrate carbon-12. Overreliance on corn consumption can then be correlated with certain diseases, because corn lacks the essential amino acid lysine. Archaeologists use a mass spectrometer to separate the 12 C and 13 C isotopes in samples of human remains. Suppose you use a velocity selector to obtain singly ionized (missing one electron) atoms of speed 8.50 km/s, and you want to bend them within a uniform magnetic field in a semicircle of diameter 25.0 cm for the 12 C. The measured masses of these isotopes are 1.99×10−26kg(12C) and 2.16×10−26kg(13C).
(a) What strength of magnetic field is required?
(b) What is the diameter of the 13 C semicircle?
(c) What is the separation of the 12 C and 13 C ions at the detector at the end of the semicircle? Is this distance large enough to be easily observed?
Answer:
[tex]0.0084575\ \text{T}[/tex]
[tex]0.272\ \text{m}[/tex]
2.2 cm easily observable
Explanation:
[tex]m_1[/tex] = Mass of 12 C = [tex]1.99\times 10^{-26}\ \text{kg}[/tex]
[tex]m_2[/tex] = Mass of 13 C = [tex]2.16\times 10^{-26}\ \text{kg}[/tex]
[tex]r_1[/tex] = Radius of 12 C = [tex]\dfrac{25}{2}=12.5\ \text{cm}[/tex]
B = Magnetic field
v = Velocity of atom = 8.5 km/s
[tex]r_2[/tex] = Radius of 13 C
The force balance of the system is
[tex]qvB=\dfrac{m_1v^2}{r}\\\Rightarrow B=\dfrac{m_1v}{rq}\\\Rightarrow B=\dfrac{1.99\times 10^{-26}\times 8500}{12.5\times 10^{-2}\times 1.6\times 10^{-19}}\\\Rightarrow B=0.0084575\ \text{T}[/tex]
The required magnetic field is [tex]0.0084575\ \text{T}[/tex]
Radius is given by
[tex]r=\dfrac{mv}{qB}[/tex]
[tex]r\propto m[/tex]
So
[tex]\dfrac{r_2}{r_1}=\dfrac{m_2}{m_1}\\\Rightarrow r_2=\dfrac{m_2}{m_1}r_1\\\Rightarrow r_2=\dfrac{2.16\times 10^{-26}}{1.99\times 10^{-26}}\times 12.5\times 10^{-2}\\\Rightarrow r_2=0.136\ \text{m}[/tex]
The required diameter is [tex]2\times 0.136=0.272\ \text{m}[/tex]
Separation is given by
[tex]2(r_2-r_1)=2(0.136-0.125)=0.022\ \text{m}[/tex]
The distance of separation is 2.2 cm which is easily observable.
What is the answer to this?
Answer:
B
Explanation:
Answer:
B
Explanation: