Answer:
the answer is d
Explanation:
the answer is d
what is a hyponyms and a what is its meaning
A hyponym is a word or phrase whose semantic field is more specific than its hypernym. ... For example, verbs such as stare, gaze, view and peer can also be considered hyponyms of the verb look, which is their hypernym. Hypernyms and hyponyms are asymmetric.
How do these terms relate to what is being attempted when making a nuclear
bomb (quick, explosive effect) as opposed to a nuclear power plant (controlled
steady production of energy)?.
Answer:
More than four-fifths of the single points of light we observe in the night sky are actually two or more stars orbiting together. The most common of the multiple star systems are binary stars, systems of only two stars together.o
A parallel-plate capacitor is constructed of two disks spaced 2.00 mm apart. It is charged to a potential difference of 500. V. A proton is shot through a small hole in the negative plate with a speed of 2.0 × 105 m/s. What is the farthest distance from the negative plate that the proton reaches?
Answer:
[tex]0.000835\ \text{m}[/tex]
Explanation:
d = Distance between plates = 2 mm
V = Potential difference = 500 V
v = Velocity of proton = [tex]2\times 10^5\ \text{m/s}[/tex]
a = Acceleration
m = Mass of proton = [tex]1.67\times 10^{-27}\ \text{kg}[/tex]
Electric field is given by
[tex]E=\dfrac{V}{d}\\\Rightarrow E=\dfrac{500}{2\times 10^{-3}}\\\Rightarrow E=250000\ \text{V/m}[/tex]
Force balance is given by
[tex]ma=qE\\\Rightarrow a=\dfrac{qE}{m}\\\Rightarrow a=\dfrac{1.6\times 10^{-19}\times 250000}{1.67\times 10^{-27}}\\\Rightarrow a=2.395\times 10^{13}\ \text{m/s}^2[/tex]
We have the relation
[tex]v^2=u^2+2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{(2\times 10^5)^2-0}{2\times 2.395\times 10^{13}}\\\Rightarrow s=0.000835\ \text{m}[/tex]
The farthest distance from the negative plate that the proton reaches is [tex]0.000835\ \text{m}[/tex].
The farthest distance from the negative plate that the proton reaches will be s=0.000835 m
What is electric field?The electric field is defined as the force across the charged particles which attract or repel the other charged particles.
Now it is given in the question that
d = Distance between plates = 2 mm
V = Potential difference = 500 V
v = Velocity of proton = [tex]2\times 10^5\ \dfrac{m}{s}[/tex]
a = Acceleration
m = Mass of proton = [tex]1.67\times10^{-27} kg[/tex]
The Electric field will be calculated as
[tex]E=\dfrac{V}{d}[/tex]
[tex]E=\dfrac{500}{2\times10^{-3}} =250000\ \frac{V}{m}[/tex]
Force balance is given by
[tex]ma=qe[/tex]
[tex]a=\dfrac{qE}{m}[/tex]
[tex]a=\dfrac{1.6\times10^{-19}\times250000}{1.67\times10^{-27}}[/tex]
[tex]a=2.395\times 10^{13}\frac{m}{s^2}[/tex]
Now from equation of motion
[tex]v^2=u^2+2as[/tex]
[tex]s=\dfrac{v^2-u^2}{2a}[/tex]
[tex]s=\dfrac{(2\times10^5)^2-0}{2\times 2.395\times 10^{13}}[/tex]
[tex]s=0.000875 m[/tex]
Thus the farthest distance from the negative plate that the proton reaches will be s=0.000835 m
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Two vehicles are approaching an intersection. One is a 2600 kgkg pickup traveling at 17.0 m/sm/s from east to west (the −x−x- direction), and the other is a 1300 kgkg sedan going from south to north (the +y−+y− direction at 24.0 m/sm/s ). Part A Find the xx -component of the net momentum of this system. pxpx = nothing kg⋅m/skg⋅m/s SubmitRequest Answer Part B Find the yy-component of the net momentum of this system. pypy = 0 kg⋅m/skg⋅m/s SubmitPrevious AnswersRequest Answer Incorrect; Try Again; 4 attempts remaining Part C What is the magnitude of the net momentum?
Answer:
a) the x-component of the net momentum is 44200 kgm/s [tex](-x)[/tex]
b) the y-component of the net momentum is 31200 kgm/s [tex](y)[/tex]
c) the magnitude of the net momentum is 54102.5 kgm/s
Explanation:
Given the data in the question;
a) x-component of the net momentum of this system.
the second vehicle ( sedan ) doesn't have momentum along x-axis, the momentum along x-axis is strictly contributed by the pick up
so;
Px = 2600 kg × 17.0 m/s [tex](-x)[/tex]
Px = 44200 kgm/s [tex](-x)[/tex]
Therefore, the x-component of the net momentum is 44200 kgm/s [tex](-x)[/tex]
b) y-component of the net momentum of this system
Also, momentum along y-axis is entirely provided by the sedan
Py = 1300 kg × 24.0 m/s [tex](y)[/tex]
Py = 31200 kgm/s [tex](y)[/tex]
Therefore, the y-component of the net momentum is 31200 kgm/s [tex](y)[/tex]
c) magnitude of the net momentum?
magnitude of the net momentum P = √( Px² + Py² )
so we substitute
P = √( (44200)² + (31200)² )
P = √( 2927080000 )
P = 54102.5 kgm/s
Therefore, the magnitude of the net momentum is 54102.5 kgm/s
1. Lucky Larry was in a car crash. He hit a brick wall going 40 mph. But his airbag
inflated and saved him because
A. it reduced the time of the collision.
B. it reduced his initial velocity.
C. it reduced his change in momentum.
D.it reduced his average
force.
Answer:
The answer is D (It reduced his average
force.)
2. What is the total effect of sound produced in an enclosed space called?
Acoustics
Overtones
sound quality
Color
Answer:
Acoustics
Explanation:
Which graph shows acceleration? A graph of position (meters) versus time (seconds) has a straight line running from 0 seconds 0 meters upward. A graph of position (meters) versus time (seconds) has a concave line running from 0 seconds 0 meters upward. A graph of position (meters) versus time (seconds) has a straight line running from 0 seconds positive number of meters downward to some later time 0 meters.
It's the one that has a concave line running upward.
Answer:
its the middle one
Explanation:
i did it on edge 2021
A train is travelong at 22 m/s when the condutor gets a radio call about a car stalled on the tracks. He hits the emergancy brake and the train comes to a halt in 135 sevonds several meters before the car. What was the train's accleration during the emergency?
Answer:
final velocity = 0
because the train stoped
so,
acceleration = (v - u) ÷ tacceleration = (0 - 22) ÷ 135acceleration = -22 ÷ 135acceleration = -0.162 m/s²The heat of fusion of copper is 205000 J/kg. What is the mass of a copper rod that takes 4950000000 J of energy to melt the rod. Be sure to put the correct unit at the end of your answer and also for your final answer, place your final numbers two places after the decimal and round.
Answer:
m = 24146.34 kg
Explanation:
The heat of fusion is defined as the amount of heat required to melt a unit mass of a substance completely. Hence, it can be calculated by the following formula:
[tex]H = \frac{Q}{m}\\\\m = \frac{Q}{H}[/tex]
where,
m = mass of copper = ?
Q = Heat absorbed by rod = 4950000000 J
H = Heat of fusion of copper = 205000 J/kg
Therefore,
[tex]m = \frac{4950000000\ J}{205000\ J/kg}[/tex]
m = 24146.34 kg
How fast (in rpm) must a centrifuge rotate if a particle7.0cm from the axis of rotation is to experience an acceleration of 100000 m/s2?
Answer:
w = 3.6087 10⁶ rpm
Explanation:
The centripetal acceleration is
a = v²/r
where r is the distance from the center of rotation and v is the magnitude of the velocity
let's reduce to the SI system
r = 7.0 cm (1m / 100cm) = 0.070 m
the angular and linear variables are related
v = w r
we substitute
a = w² r
w = [tex]\sqrt{\frac{a}{r} }[/tex]
w = [tex]\sqrt{ \frac{100000^2}{0.07} }[/tex]
w = 3.779 10⁵ rad / s
let's reduce to rpm
w = 3.779 10⁵ rad / s (1rev / 2pi rad) (60 s / 1min)
w = 3.6087 10⁶ rpm
suppose you and your two lab partners each read the volume of a sample of ñiquid in a graduate cylinder the three measurements you come up with are 211 mL,212mL and 212ML. the teacher tells you that the actual volume of the sample is 211 mL which of the following best describes your results?
A.neither accurate nor precise
B. Precise but not accurate
C.accurate but not precise
D.both accurate and precise
I need help please
A 14.6 g mass is attached to a horizontal spring with a spring constant of 15.7 N/m and released from rest with an amplitude of 29.8 cm. What is the speed of the mass when it is halfway to the equilibrium position if the surface is frictionless
Answer:
v = 8.46 m/s
Explanation:
Given that,
Mass of the object, m = 14.6 g
Spring constant of the spring, k = 15.7 N/m
It is released from rest with an amplitude of 29.8 cm.
We need to find the speed of the mass when it is halfway to the equilibrium position if the surface is frictionless.
We can use the conservation of energy in this case. Let v be the speed of the mass. So,
[tex]EPE_i=\dfrac{1}{2}mv^2+EPE_f\\\\\dfrac{1}{2}kx_i^2=\dfrac{1}{2}mv^2+\dfrac{1}{2}kx_f^2[/tex]
Substitue all the values in the above expression.
[tex]\dfrac{1}{2}\times 15.7\times (0.298)^2=\dfrac{1}{2}\times 0.0146\times v^2+\dfrac{1}{2}\times 15.7\times (\dfrac{0.298}{2})^2\\\\v=8.46\ m/s[/tex]
So, the speed of the mass is equal to 8.46 m/s.
You have been called as an expert witness for a trial in which a driver has been charged with speeding but is claiming innocence. He claims to have slammed on his brakes to avoid rear-ending another car, but tapped the back of the other car just as he came to rest. You have been hired by the prosecution to prove that the driver was indeed speeding. You have received data as follows from the police: skid marks left by the driver are 60.0 m long and the roadway is level. Tires matching those on the car of the driver have been dragged over the same roadway to determine that the coefficient of kinetic friction between the tires and the roadway is 0.620 at all points along the skid mark. The speed limit on the road is 35 mi/h. Construct an argument to be used in court to show that the driver was indeed speeding. (Enter the speed of the driver, in mi/h, just before braking.)
Answer:
v₀ = 60.38 mi / h
Explanation:
For this exercise let's start by using Newton's second law
Y axis
N-W = 0
N = W
X axis
fr = m a
the expression for the friction force is
fr = μ N
we substitute
μ mg = m a
μ g = a
calculate us
a = 0.620 9.8
a = 6.076 m / s²
now we can use the kinematics relations
v² = v₀² - 2 a x
suppose v = 0
v₀ = [tex]\sqrt{2ax}[/tex]
let's calculate
v₀ = [tex]\sqrt{2 \ 6.075 \ 60.0}[/tex]
v₀ = 27.00 m / s
let's slow down to the english system
v₀ = 27.0 m / s (3.28 ft / 1m) (1 mile / 5280 ft) (3600s / 1h)
v₀ = 60.38 mi / h
With this stopping distance, the starting speed should have been 60.38, which is much higher than the maximum speed allowed.
The amplitude of a wave function representing a moving particle can change from positive to negative values in the domain (0, a) over which the wave function is defined. It must therefore pass through zero at some value x0, where 0 < x0 < a. Therefore the probability of the particle being at x0 is zero and the particle can't get from a position x < x0 to a position x > xo.
Required:
Is this reasoning correct?
Answer:
Explanation:
In a standing wave function[tex]\psi (x,t) = A sin(kx)[/tex] characterized for x between (0.a). on the off chance that the amplitude of the wave interchange from positive to negative at the interval. there probably been a node at [tex]x_0[/tex], among 0 and a to such an extent that [tex]0<x_0 <a[/tex]. The reasoning is right that the likelihood of discovering the particle at the node [tex]x_0[/tex] is 0 in light of the fact that by definition, the nodes of the wave are the place where the wave function falls and is equivalent to 0. Since the likelihood of discovering a particle at a position [tex]x_0[/tex] at time [tex]t_0[/tex], is provided by [tex]P=|\psi(x_0,t_0)|^2 dx[/tex], this implies that at the nodes of a standing wave,
[tex]P = | \psi (x_0,t_0)|^2 \ dx \\ \\ P = |0|^2 dx \\ \\ P = 0[/tex]
So the reasoning that the likelihood of the particle being at [tex]x_0[/tex] is 0 is right.
However, to examine whether the particle can travel from a position [tex]x <x_0[/tex] to a position of [tex]x_0>x[/tex]. All together words, can the molecule be found on one or the other side of the node?
The appropriate response is yes.
Recall that in quantum mechanics. wave functions at most present with the likelihood of discovering a particle at a specific time inside a time frame. The wave function doesn't present with an old classical actual trajectory that a particle should follow to go in space: all things being equal, it simply yields chances of whether a particle can be found in a specific spot at a specific time. So the reasoning that a particle can't get from a position [tex]x <x_0[/tex] to a position of [tex]x>x_0[/tex], is incorrect.
Which of these is the best example of Jackson doing science by inquiry?
Answer:
D.
Explanation:
I'm assuming this is the question:
Which of these is the best example of Jackson doing science by inquiry?
A. Jackson watches a video on objects sinking and floating. He
writes down all the main points of the video.
B. Jackson researches the history of soda can use, including how the
size of the cans has changed over the last three decades.
C. Jackson finds calculations on sinking and floating. He copies one
of the calculations.
D. Jackson asks himself whether unopened soda cans sink or float.
He designs an experiment to find out.
a u tube contains a liquid of an unknown density an oil of density is poured into the right arm of the tube until the oil column is 16.0 cm high the oil air interfdace is 4.5 cm above the liquid level in the left arm of the u tube algrebraic expression to deter,ine the density of the unknown fluid
Answer:
The answer is "[tex]1155\ \frac{kg}{m^3}[/tex]"
Explanation:
Please find the complete question in the attached file.
[tex]p = p_0 + ?gh[/tex]
pi = pressure only at two liquids' devices
PA = pressure atmosphere.
1 = oil density
2 = uncertain fluid density
[tex]h_1 = 11 \ cm\\\\h_2= 3 \ cm[/tex]
The pressures would be proportional to the quantity [tex]11-3 = 8[/tex] cm from below the surface at the interface between both the oil and the liquid.
[tex]\to p_A + ?2g(h_1 - h_2) = p_A + ? 1gh_1\\\\\to ?2 = \frac{?1h_1}{(h_1 - h_2)} \\\\[/tex]
[tex]= \frac{840 \frac{kg}{m^3}}{\frac{11}{8}} \\\\= 1155\ \frac{kg}{m^3}[/tex]
Which conversion can take place in a transformer
A. An electric current into a magnetic field
B. Electric energy into mechanical energy
C. Mechanical energy into electric energy
D. A lower voltage into a higher voltage
Answer:
D. A lower Voltage into a higher
n moles of CO2 gas (T1, V1) obey a van der Waals equation of state are contained in an insulated piston-cylinder arrangement. A reversible expansion of the gas is carried out until the volume become 2V1.
[CV.m = 28.80 J mol−1K−1, a = 3.610 atm. dm6. mol−2, b = 4.29 x 10−2 dm3. mol−1]
(a) Find the final temperature of gas as a function of n, . , T1, V1 and the van
der Waals parameter a and b.
(b) If two moles of CO2 gas involved and T1 = 350K, V1 = 40L, compute final
temperature of the gas.
(c) Find work done and internal energy of the gas.
Answer:
one or two three or four
Explanation:
animals of CO2 gas find the final temperature of gas
The final temperature of the 2 moles of gas has been 0.0014 K.
The gas has been assumed to be the ideal gas. The moles of the has been n. According to the ideal gas equation:
(a) PV= nRT.
For the reversible expansion of 1 mole of gas:
P = [tex]\rm \dfrac{RT}{V - b}\;\times\;\dfrac{a}{V^2}[/tex]
For n moles of gas, the reversible expansion will be:
P = [tex]\rm \dfrac{nRT}{V - b}\;\times\;\dfrac{a}{V^2}[/tex]
The final temperature of the reversible expansion has been:
[tex]\rm T_f\;=\;T_i\;+\;\dfrac{an^2\;[\frac{1}{Vf\;-\;\frac{1}{Vi} } ]}{Cv}[/tex]
(B) The ideal gas equation can be given as:
[tex]\rm \dfrac{P1V1}{nT1}\;=\; \dfrac{P2V2}{nT2}[/tex]
Substituting the values:
[tex]\rm \dfrac{40}{2\;\times\;350}\;=\;\dfrac{80}{2\;\times\;T2}[/tex]
T2 = 0.0014 K.
The final temperature of the 2 moles of gas has been 0.0014 K.
(c) The work done can be given by:
W = nRT In[tex]\rm \dfrac{V2}{V1}[/tex]
The internal energy of the system has been given by:
[tex]\Delta[/tex]U = [tex]\rm C_v[/tex][[tex]\rm T_f\;-\;T_i[/tex]] - a[tex]\rm n^2[/tex] [tex]\rm [\dfrac{1}{Vf}\;-\;\dfrac{1}{Vi} ][/tex]
For more information about the reversible expansion of gas, refer to the link:
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Write any two differences between echo and reverberation.
Why the speed of sound in hot air is more than that in the cold air
Answer:
1) Because of the size and time the sound wave moves, an echo is typically transparent and easy to discern. Since reverberations typically don't have enough distance or time to fly, they will pile up on top of each other, making it impossible to understand.
People will detect an echo if the distance between the source of the sound and the reflected body is greater than 50 feet. When a sound wave is bounced off a nearby surface, it may create a reverberation.
2) The speed of sound in hot air is more than that in cold air because air molecules are traveling faster in hot air.
Explanation:
Suppose you design an apparatus in which a uniformly charged disk of radius R is to produce an electric field. The field magnitude is most important along the central perpendicular axis of the disk, at a point P at distance 2.00R from the disk. Cost analysis suggests that you switch to a ring of the same outer radius R but with inner radius R/2.00 . Assume that the ring will have the same surface charge density as the original disk. If you switch to the ring, by what percentage will you decrease the electric field magnitude at P?
Answer:
The electric field will be decreased by 29%
Explanation:
The distance between point P from the distance z = 2.0 R
Inner radius = R/2
Outer raidus = R
Thus;
The electrical field due to disk is:
[tex]\hat {K_a} = \dfrac{\sigma}{2 \varepsilon _o} \Big( 1 - \dfrac{z}{\sqrt{z^2+R_i^2}} \Big)[/tex])
[tex]\implies \dfrac{\sigma}{2 \vaepsilon _o} \Big ( 1 - \dfrac{2.0 \ R}{\sqrt{ (2.0\ R)^2+(R)^2}} \Big)[/tex]
Similarly;
[tex]\hat {K_b} = \hat {k_a} - \dfrac{\sigma}{2 \varepsilon_o} \Big( 1 - \dfrac{2.0 \ R}{\sqrt{(2.0 \ r)^2 + (\dfrac{R}{2}^2)}}\Big)[/tex]
However; the relative difference is: [tex]\dfrac{\hat {k_a} - \hat {k_b}}{\hat {k_a} }= \dfrac{E_a -E_a + \dfrac{\sigma}{2 \varepsilon_o \Big[1 - \dfrac{2.0 \ R}{\sqrt{(2.0 \ R)^2 + (\dfrac{R}{2})^2}} \Big] } } { \dfrac{\sigma}{2 \varepsilon_o \Big [ 1 - \dfrac{2.0 \ R}{\sqrt{ (2.0 \ R)^2 + (R)^2}} \Big] }}[/tex]
[tex]\dfrac{\hat {k_a} - \hat {k_b}}{\hat {k_a} }= \dfrac{1 - \dfrac{2.0}{\sqrt{(2.0)^2 + \dfrac{1}{4}}} }{1 - \dfrac{2.0 }{\sqrt{(2.0)^2 + 1}}}[/tex]
[tex]= 0.2828 \\ \\ \mathbf{\simeq 29\%}[/tex]
If an alpha particle (two protons and twoneutrons) is given an initial (nonrelativistic) velocityvat a very far distance and is aimed directly at a gold nucleus(Z=79), what is the closest distance d the alpha particle will come to the nucleus?
In this problem you can estimate that the mass of the proton m_p is equal to the mass of the neutron and only consider theeffects of a single gold nucleus. Assume that the alpha particle comes close enough so that the nucleus is not substantiallyscreened by inner-shell electrons.
Express your answer in terms ofm_p, the permittivity of free space epsilon_0, the magnitude of the electron charge e, and v.
Answer:
r = [tex]\frac{4\pi \epsilon_o}{79} \frac{m_proton \ v^2}{e^2}[/tex]
Explanation:
To solve this problem we can use conservation of energy,
starting point. With the alpha particle too far from the gold nucleus
Em₀ = K = ½ m v²
final point. The point of closest approach, whereby the speed of the alpha particle is zero.
Em_f = U = k q₁q₂ / r
where q₁ is the charge of the alpha particle and q₂ the charge of the Gold nucleus.
Energy is conserved
Em₀ = Em_f
½ m v² = k q₁q₂ / r
r = [tex]\frac{1}{2} \ \frac{m v^2}{k \ q_1q_2}[/tex]
the mass of the particular alpha is
m_particle = 2 m_proton + 2M_neutron
m_particle = 4 m_proton
the charge of the alpha and the gold particle are
q₁ = 2e
q₂ = 79 e
we substitute
r = [tex]\frac{1}{2} \frac{4 m_proton \ v^2 }{k \ 2 \ 79 \ e^2}[/tex]
r = [tex]\frac{1}{79} \ \frac{m_proton \ v^2}{ k \ e^2}[/tex]
k = [tex]\frac{1}{4\pi \epsilon_o }[/tex]
we substitute
r = [tex]\frac{4\pi \epsilon_o}{79} \frac{m_proton \ v^2}{e^2}[/tex]
The closest distance the alpha particle will come to the nucleus will be [tex]\frac{4 \pi \epsilon_0 m_pv^2}{79e^2}[/tex].
What is the law of conservation of energy?According to the law of conservation of energy, energy can not be created nor be destroyed can be transferred from one form to another form.
The given data in the problem is;
q₁ is the charge of the alpha particle =2e
q₂ the charge of the Gold nucleus. = 79 e
We can employ energy conservation to fix this problem.to begin with, the gold nucleus is too far away from the alpha particle. The kinetic energy of the particle;
[tex]\rm KE= \frac{1}{2} mv^2[/tex]
The potential energy of the particle;
[tex]\rm U= \frac{Kq_1q_2}{r}[/tex]
The energy is conserved;
[tex]\rm KE = PE\\\\ \frac{1}{2}mv^2=\frac{ kq_1q_2}{r} \\\\ r= \frac{1}{2}\frac{mv^2 }{q_1q_2}[/tex]
[tex]\rm r= \frac{1}{2}\frac{4m_pv^2}{k \times 2 \times 79 \times e^2} \\\\ \rm r= \frac{1}{79}\frac{m_pv^2}{k \times \times 79 \times e^2} \\\\ k= \frac{1}{4 \pi \epsilon_0} \\\\ \rm r= \frac{4 \pi \epsilon_0 m_pv^2}{79e^2}[/tex]
Hence the closest distance the alpha particle will come to the nucleus will be [tex]\frac{4 \pi \epsilon_0 m_pv^2}{79e^2}[/tex].
To learn more about the law of conservation of energy refer to link;
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A knife thrower throws a knife toward a 300 g target that is sliding in her direction at a speed of 2.30 m/s on a horizontal frictionless surface. She throws a 22.5 g knife at the target with a speed of 40.0 m/s. The target is stopped by the impact and the knife passes through the target. Determine the speed of the knife (in m/s) after passing through the target.
Answer:
The speed of the knife after passing through the target is 9.33 m/s.
Explanation:
We can find the speed of the knife after the impact by conservation of linear momentum:
[tex] p_{i} = p_{f} [/tex]
[tex] m_{k}v_{i_{k}} + m_{t}v_{i_{t}} = m_{k}v_{f_{k}} + m_{t}v_{f_{t}} [/tex]
Where:
[tex] m_{k}[/tex]: is the mass of the knife = 22.5 g = 0.0225 kg
[tex] m_{t}[/tex]: is the mass of the target = 300 g = 0.300 kg
[tex] v_{i_{k}}[/tex]: is the initial speed of the knife = 40.0 m/s
[tex] v_{i_{t}} [/tex]: is the initial speed of the target = 2.30 m/s
[tex]v_{f_{k}}[/tex]: is the final speed of the knife =?
[tex] v_{f_{t}} [/tex]: is the final speed of the target = 0 (it is stopped)
Taking as a positive direction the direction of the knife movement, we have:
[tex] m_{k}v_{i_{k}} - m_{t}v_{i_{t}} = m_{k}v_{f_{k}} [/tex]
[tex] v_{f_{k}} = \frac{m_{k}v_{i_{k}} - m_{t}v_{i_{t}}}{m_{k}} = \frac{0.0225 kg*40.0 m/s - 0.300 kg*2.30 m/s}{0.0225 kg} = 9.33 m/s [/tex]
Therefore, the speed of the knife after passing through the target is 9.33 m/s.
I hope it helps you!
what are the materials that you have illustrated?
Answer:
An illustration is a decoration, interpretation or visual explanation of a text, concept ... such as posters, flyers, magazines, books, teaching materials, animations, video games and films.
Explanation:
Three advantages of working as a nuclear physicist
calculate the tension in a horizontal string that whirls a 2kg toy in a circle of radius 2.5 meters when it moves 3m/s on icy surface
Answer:
The tension in the horizontal string is 7.2 N
Explanation:
With the assumption that the ice surface has no friction, the tension in the horizontal string is equal to the centripetal force, 'F', acting on the string to keep it in circular motion
[tex]The \ centripetal \ force, \ F = m \times \dfrac{v^2}{r}[/tex]
Where;
m = The mass of the object in the circular
v = The tangential velocity of the object
r= The radius of the circle through whose path the object is moved
The given parameters are;
The mass of the toy that is whirled in a circle by the string, m = 2 kg
The radius of the circle the toy's path turns, r = 2.5 meters
The velocity of the toy at the instant the tension is sought, v = 3 m/s
Therefore, we get;
[tex]The \ centripetal \ force, \ F = 2 \, kg \times \dfrac{(3 \, m/s)^2}{2.5 \, m} = 7.2 \, N[/tex]
The centripetal force, F = 7.2 N = The tension in the horizontal string
∴ The tension in the horizontal string = 7.2 N.
unit 3 test waves sps4 physics need all answers
Answer:
What is the question on the testtt
What do all clouds contain the some amount of
Answer:
A cloud is made up of liquid water droplets. A cloud forms when air is heated by the sun. As it rises, it slowly cools it reaches the saturation point and water condenses, forming a cloud.
Explanation:
8. Cart 1 with mass of 2kg moving at +6m/s collides with cart 2 with a mass of 1kg,
which is at rest. If cart 1 moves at +2m/s after the collision, what is the velocity of
cart 2 after the collision? (include correct units, or you get no credit)
Answer:its 2kg
Explanation:
A 50 kg crateis being pushed on a horizontal floor at constant velocity. Given that the coefficient of kenitic friction between crate and floor is hk= 0.1, what is push force f ( friction)?
Someone please help me
What effect would lowering the temperature have on this reaction?
3H2 + N2 + 2NH3 + energy
A. H2 and N2 would react to produce more NH3.
B. The reaction would proceed more slowly in both directions.
C. Decreasing the temperature would not have an effect on this
reaction.
D. NH3 would react to produce more H2 and N2.
On lowering the temperature of the given reaction, H₂ gas and N₂ gas would react to produce more NH₃. Therefore, option (A) is correct.
What is the effect of temperature on equilibrium?When the chemical reaction is exothermic then increasing the temperature will cause the backward reaction to occur, decreasing the amounts of the products while increasing the amounts of reactants. Lowering the temperature will produce more reactants and cause the reaction to occur in the forward direction.
The formation of the ammonia is an exothermic reaction that can be represented as:
N₂ (g) + 3 H₂ (g) ⇄ 2NH₃ (g)
If the temperature will be decreased, the chemical reaction will proceed forward to produce more heat. The effect of temperature on equilibrium will change the value of the equilibrium constant of the chemical reaction. The production of more ammonia on lowering the temperature.
Learn more about equilibrium, here:
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