A 1.10-kg object slides to the right on a surface having a coefficient of kinetic friction 0.250 (Figure a). The object has a speed of vi = 2.60 m/s when it makes contact with a light spring (Figure b) that has a force constant of 50.0 N/m. The object comes to rest after the spring has been compressed a distance d (Figure c). The object is then forced toward the left by the spring (Figure d) and continues to move in that direction beyond the spring's unstretched position. Finally, the object comes to rest a distance D to the left of the unstretched spring (Figure e).
The right end of a horizontal spring labeled k is attached to a wall. Five images show five configurations as a block labeled m approaches, compresses, and then moves away from the spring.
In figure a, the block is to the left of the spring, and an arrow above the block points to the right.
In figure b, the block is just touching the uncompressed spring, and an arrow labeled vector vi above the block points to the right.
In figure c, the block has compressed the spring by a distance d, and a label indicates vector vf = 0.
In figure d, the block is just touching the uncompressed spring, and an arrow labeled vector v above the block points to the left.
In figure e, the block is a distance D away from the spring, and a label indicates vector v = 0.
(a)
Find the distance of compression d (in m).
m
(b)
Find the speed v (in m/s) at the unstretched position when the object is moving to the left (Figure d).
m/s
(c)
Find the distance D (in m) where the object comes to rest.
m
(d)
What If? If the object becomes attached securely to the end of the spring when it makes contact, what is the new value of the distance D (in m) at which the object will come to rest after moving to the left?
m
Answers
Answer:
(a) Approximately [tex]0.335\; \rm m[/tex].
(b) Approximately [tex]1.86\; \rm m\cdot s^{-1}[/tex].
(c) Approximately [tex]0.707\; \rm m[/tex].
(d) Approximately [tex]0.228\; \rm m[/tex].
Explanation:
[tex]v_i[/tex] denotes the velocity of the object in the first diagram right before it came into contact with the spring. Let [tex]m[/tex] denote the mass of the block. Let [tex]\mu[/tex] denote the constant of kinetic friction between the object and the surface. Let [tex]g[/tex] denote the constant of gravitational acceleration.Let [tex]k[/tex] denote the spring constant of this spring.(a)Consider the conversion of energy in this object-spring system.
First diagram: Right before the object came into contact with the spring, the object carries kinetic energy [tex]\displaystyle \frac{1}{2}\, m \cdot {v_{i}}^2[/tex].
Second diagram: As the object moves towards the position in the third diagram, the spring gains elastic potential energy. At the same time, the object loses energy due to friction.
Third diagram: After the velocity of the object becomes zero, it has moved a distance of [tex]D[/tex] and compressed the spring by the same distance.
Energy lost to friction: [tex]\underbrace{(\mu \cdot m \cdot g)}_{\text{friction}} \cdot D[/tex]. Elastic potential energy that the spring has gained: [tex]\displaystyle \frac{1}{2}\,k\, D^2[/tex].The sum of these two energies should match the initial kinetic energy of the object (before it comes into contact with the spring.) That is:
[tex]\displaystyle \frac{1}{2}\, m \cdot {v_{i}}^{2} = (\mu\cdot m \cdot g) \cdot D + \frac{1}{2}\, k \cdot D^2[/tex].
Assume that [tex]g = 9.81\; \rm m \cdot s^{-2}[/tex]. In the equation above, all symbols other than [tex]D[/tex] have known values:
[tex]m =1.10\; \rm kg[/tex].[tex]v_i = 2.60\; \rm m \cdot s^{-1}[/tex].[tex]\mu = 0.250[/tex].[tex]g = 9.81\; \rm m \cdot s^{-2}[/tex].[tex]k = 50.0\; \rm N \cdot m^{-1}[/tex].Substitute in the known values to obtain an equation for [tex]D[/tex] (where the unit of [tex]D\![/tex] is [tex]m[/tex].)
[tex]3.178 = 2.69775\, D + 25\, D^2[/tex].
[tex]2.69775\, D + 25\, D^2 + 3.178 = 0[/tex].
Simplify and solve for [tex]D[/tex]. Note that [tex]D > 0[/tex] because the energy lost to friction should be greater than zero.
[tex]D \approx 0.335\; \rm m[/tex].
(b)The energy of the object-spring system in the third diagram is the same as the elastic potential energy of the spring:
[tex]\displaystyle \frac{1}{2}\,k\, D^2 \approx 2.81\; \rm J[/tex].
As the object moves to the left, part of that energy will be lost to friction:
[tex](\mu \cdot m \cdot g) \, D \approx 0.905\; \rm J[/tex].
The rest will become the kinetic energy of that block by the time the block reaches the position in the fourth diagram:
[tex]2.81\; \rm J - 0.905\; \rm J \approx 1.91\; \rm J[/tex].
Calculate the velocity corresponding to that kinetic energy:
[tex]\displaystyle v =\sqrt{\frac{2\, (\text{Kinetic Energy})}{m}} \approx 1.86\; \rm m \cdot s^{-1}[/tex].
(c)As the object moves from the position in the fourth diagram to the position in the fifth, all its kinetic energy ([tex]1.91\; \rm J[/tex]) would be lost to friction.
How far would the object need to move on the surface to lose that much energy to friction? Again, the size of the friction force is [tex]\mu \cdot m \cdot g[/tex].
[tex]\displaystyle (\text{Distance Travelled}) = \frac{\text{(Work Done by friction)}}{\text{(Size of the Friction Force)}} \approx0.707\; \rm m[/tex].
(d)Similar to (a), solving (d) involves another quadratic equation about [tex]D[/tex].
Left-hand side of the equation: kinetic energy of the object (as in the fourth diagram,) [tex]1.91\; \rm J[/tex].
Right-hand side of the equation: energy lost to friction, plus the gain in the elastic potential energy of the spring.
[tex]\displaystyle {1.91\; \rm J} \approx (\mu\cdot m \cdot g) \cdot D + \frac{1}{2}\, k \cdot D^2[/tex].
[tex]25\, D^2 + 2.69775\, D - 1.90811\approx 0[/tex].
Again, [tex]D > 0[/tex] because the energy lost to friction is greater than zero.
[tex]D \approx 0.228\; \rm m[/tex].
The energy transferred between the object and the spring as a closed system, therefore, conserved are;
(a) The distance of compression, d ≈ 0.3354 meters
(b) The speed in the un-stretched position wen the object is sliding to the left, v ≈ 1.8623 m/s
(c) The distance where the object comes to rest, D ≈ 0.7071 m
(d) The distance the object will come to rest attached to the spring, D ≈ 0.2278 m
The reason the above values are correct are as follows;
The known parameters are;
Mass of the object, m₁ = 1.10 kg
Coefficient of friction, μ = 0.250
The initial speed of the object, [tex]v_i[/tex] = 2.60 m/s
Force constant of the spring, K = 50.0 N/m
Distance the spring is compressed by the object = d
(a) Conservation of energy principle
[tex]Kinetic \ energy = \dfrac{1}{2} \cdot m\cdot v^2[/tex]
Work done = Force × Distance
Friction force, [tex]F_f[/tex] = W × μ
Weight, W = m·g
Weight = Mass × Acceleration
Energy transferred by object = Work done by spring + Work done by friction
[tex]Energy \ transferred \ by \ object = Kinetic \ energy = \dfrac{1}{2} \times 1.10\times 2.60^2 = 3.718[/tex]
Energy transferred by object = 3.718 J
[tex]Work \ done \ by \ spring = \dfrac{1}{2} \cdot k\cdot x^2[/tex]
[tex]Work \ by \ spring \ to \ bring \ object \ to \ rest, \ W_{spring} = \dfrac{1}{2} \times 50\times d^2[/tex]
[tex]W_{spring}[/tex] = 25·d²
Work done by friction, [tex]W_{friction}[/tex] = 1.10×9.81×0.250×d = 2.69775·d
Therefore;
3.718 = 25·d² + 2.69775·d
25·d² + 2.69775·d - 3.718 = 0
Solving gives
The distance of the compression d ≈ 0.3354 m
(b) The energy given by the spring = 25·d²
The work done by friction, [tex]W_{friction}[/tex] = 2.69775·d
Kinetic energy given to object = 0.55·v²
0.55·v² = 25·d² - 2.69775·d
0.55·v² = 25×0.3354² - 2.69775×0.3354
∴ v = √(3.4682) = 1.8623
The velocity of the object at the un stretched position, v ≈ 1.8623 m/s
(c) The kinetic energy, K.E. of the object on the way left is given as follows;
K.E. = 0.5 × 1.10 kg × 3.4682 m²/s² = 1.90751 J
The work done by friction before object comes to rest = 2.69775·D
[tex]D = \dfrac{1.90751 \, J}{2.69775 \, N} \approx 0.7071 \, m[/tex]
The distance where the object comes to rest, D ≈ 0.7071 m
(d) The work done on spring, [tex]W_{spring}[/tex] = 25·D'²
Work done on friction, [tex]W_{friction}[/tex] = 2.69775·D'
Kinetic energy of object, K.E. ≈ 1.90751 J
K.E. = [tex]W_{spring}[/tex] + [tex]W_{friction}[/tex]
1.90751 ≈ 25·D'² + 2.6775·D'
25·D'² + 2.6775·D' - 1.90751 = 0
Solving with a graphing calculator gives;
D' ≈ 0.2278 m
The new value of the distance D = 0.2278 m
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Related Questions
A 1500 kg car traveling at 35 m/s hits its brakes and comes to rest in 5 seconds. Calculate the force applied by the brakes. (hint: solve for a first) Show all steps!!
Answers
Answer:
We are given:
m = 1500 kg
initial velocity (u) = 35 m/s
final velocity (v) = 0 m/s
time (t) = 5 seconds
acceleration = a m/s/s
Force (f) = F newtons
Solving for acceleration:
Using the first equation of motion:
v = u + at
replacing the variables
0 = 35 + a(5)
a = -35/5
a = -7 m/s/s
Force applied by the brakes:
From newton's second law of motion:
F = ma
replacing the variables
F = (1500)(-7)
F = -10500 N
Therefore, the force applied by the brakes is -10500 N, we have a negative sign since the force is being applied opposite to the direction of motion
2. Which of the following is a true statement.
Sound travels faster than light.
Sound travels slower than light
Answers
Answer:
Sound travels slower than light
Explanation:
Sound travels at around 1/3 km per second, while light travels around 300,000 km per second.
Also, this is the reason you see the lightning before you hear the thunder.
I hope this helped.
Answer:
Sound travels slower than light
Explanation:
Bc light is faster than sound
One statement of the first law of thermodynamics is that:___________.
a. the amount of work done on a system is dependent of pathway.
b. the total work done on a system must equal the heat absorbed by the system.
c. the heat flow in or out of a system is independent of pathway.
d. the total energy flow in or out of a system is equal to the sum of the heat transferred to or from the system and the work done by or on the system.
e. in any chemical process the heat flow must equal the change in enthalpy.
Answers
Answer:
a. the amount of work done on a system is dependent of pathway
Explanation:
The first law of thermodynamics states that the change in internal energy of a system equals the net heat transfer into the system minus the net work done by the system.
ΔU = Q - W
Where;
Q, the net heat transfer into the system depends on the pathway
W, the net work done by the system also depends on the pathway
But, ΔU, the change in internal energy is independent of pathway
Therefore, the correct option is "A"
a. the amount of work done on a system is dependent of pathway
1) Construct a graph showing the relationship between and independent variable and a dependent variable. 2) Demonstrate potential and kinetic energy. 3) Design a pendulum. 4) Investigate which variables influence the period of a pendulum. 5) Solve a problem using the scientific method. 6) Use graphs to characterize gow drop height responds to bounce height. (Next answer questions 1 - 6). 1) Write the goal of the lab or the question you tried to answer? 2) which observations, experiences, or lesson materials helped you form a hypothesis? 3) State your hypothesis? 4) Use the data from your experiment to complete these data tables? [Data Table 1: Period vs. Mass] Mass(g) / Time for 10 periods (s). [ Data Table 2: Period vs. String Length] String Length (cm) / Time for 10 periods (s). ( Question 5) Graph your data: A) Determine the values in seconds (s) for the y-axis based on your time data. B) Plot mass vs. time for 10 periods on the first graph. C) Plot string length vs. time for 10 periods on the second graph. ( question 6) Compare your data with your hypothesis. Explain how the data supports or does not support your hypothesis. If your data does not support your hypothesis, use the data you collected to answer the original question. { Truly Need Help and will mark brainlest} Please Please be very honest with your response.
Answers
The number of
• in the atom of an element determines its chemical properties.
Answers
Answer:
Yes, the number of electrons determines the chemical properties of the atom.
Explanation:
Suppose a rocket-propelled motorcycle is fi red from rest horizontally across a canyon 1.00 km wide. (a) What minimum constant acceleration in the x-direction must be provided by the engines so the cycle crosses safely if the opposite side is 0.750 km lower than the starting point? (b) At what speed does the motorcycle land if it maintains this constant horizontal component of acceleration? Neglect air drag, but remember that gravity is still acting in the negative y-direction.
Answers
x
=
a
t
2
2
x=
2
at
2
y
=
−
g
t
2
2
y=−
2
gt
2
Initial conditions
g
=
9.8
m
s
2
g=9.8
s
2
m
the cycle crosses safely if the...
1000
=
a
t
2
2
1000=
2
at
2
−
750
=
−
g
t
2
2
−750=−
2
gt
2
From this we have acceleration
a
=
1000
⋅
9.8
750
=
13.1
m
s
2
a=
750
1000⋅9.8
=13.1
s
2
m
WHat is the correct definnition for recovery heart rate ?
Answers
Answer:
when your heart rate is not high than normal or lower than normally
Explanation:
Recovery heart rate is when your heart recovers from a certain heart rate eg high heart rate or even low heart rate this means you heart is now functioning normally .
Answer:
Number of beats per minute the heart drops after exercise.
A speed skater moving across frictionless ice at 8.8 m/s hits a 6.0 m -wide patch of rough ice. She slows steadily, then continues on at 5.8 m/s . What is her acceleration on the rough ice?
Answers
Recall that
v² - u² = 2 a ∆x
where u and v are initial and final velocities, respectively; a is acceleration; and ∆x is the distance traveled (because the skater moves in only one direction).
So we have
(5.8 m/s)² - (8.8 m/s)² = 2 a (6.0 m)
a = ((5.8 m/s)² - (8.8 m/s)²) / (12 m)
a = -3.65 m/s²
List some of the most common injuries that can occur from sports. Please hurry!
Answers
Will ice melt faster in salt, sugar or water?
Answers
Answer:
Water because the salt will just make it colder and the sugar won't do anything
The lungs are large organs which contain smaller, expandable sacs.
These sacs greatly increase the surface area of the lungs.
Why is a large surface area important to the function of the lungs? A. Large amounts of liquid wastes and fatty tissue must be stored in the lungs. B. Oxygen in air taken in by the lungs must move quickly through the lung tissue and into the blood. C. The lungs must be very sturdy and rigid so they cannot move. D. The lungs must be able to filter oxygen from water while the organism is swimming.
Answers
Answer:
B. It is B because without this much space for oxygen and blood to flow and go through, we would not survive.
Explanation:
A rectangular loop with an area of 2 m2 is placed perpendicular to a uniform magnetic field of 1 Tesla. The field’s magnitude is increased to 6 Tesla in 4 seconds. The magnitude of the induced emf is equal to:
Answers
Answer:
Induced emf = 0
Explanation:
An emf can be induced due to the change in magnetic field. It can be given by :
[tex]\epsilon=\dfrac{d\phi}{dt}\\\\\because \phi=BA\cos\theta\\\\\epsilon=\dfrac{d(BA\cos\theta)}{dt}\\\\\epsilon=A\cos\theta\dfrac{dB}{dt}[/tex]
As the loop is placed perpendicular to a uniform magnetic field of 1 Tesla. It means that [tex]\theta=90^{\circ}[/tex] and cos(90) = 0. Hence, the induced emf is equal to 0.
8. Using Newton's second law, solve the following problems using
GFFS.
a. A force of 35N is applied to a 25kg object, what is its
acceleration??
b. What force is required to accelerate a 14kg object at 4.5
m/s2?
c. Calculate the mass of an object that accelerated at a rate of
2.7 m/s2 by a force of 18N.
Answers
Answer:
a) 1.4 m/s2
b) 63N
c) 6.66kg
Question 4 of 10
A 6.2 kg bowling ball that weighs 60.76 N is accelerating at 1.8 m/s2. After 2
seconds, it reaches a speed of 3.6 m/s. What is its momentum at this point?
A. 11.2 kg-m/s
B. 1.7 kg.m/s
C. 22.3 kg-m/s
D. 16.9 kg.m/s
Answers
Answer:
22.3 kg•m/s
Explanation:
Apex;)
Question 3 of 10
A 0.057 kg tennis ball and a tennis racket collide. The racket has an initial
momentum of -1.75 kg-m/s and a final momentum of -1.25 kg-m/s. The ball
has an initial momentum of 0.00684 kg-m/s. If you assume the collision is
elastic, what is the final velocity of the ball?
A. -0.50 m/s
B. -52.48 m/s
C. -2.99 m/s
d
D. -8.65 m/s
Answers
If we assume the collision to be elastic the final velocity of the ball will be -8.65 m/s, therefore the correct option is D.
What is elastic collision?It is a type of Collison for which the momentum, as well as the kinetic energy after and before the collision, is constant. there is no loss of energy in a perfectly elastic collision.
By using the law of conservation of momentum
momentum before collision = momentum after the collision
the initial momentum of rocket +initial momentum, of ball = final momentum of rocket + final momentum of the ball
By substituting the respective values
-1.75 + 0.00684 = -1.25 + m*v
-1.75 + 0.00684 = -1.25 + 0.057*v
v = -8.65 m/s
Thus, the final velocity of the ball is -8.65 m/s
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three charges are located at 100m intervals along a horizontal line: a charge of -3.0C on the left, 2.0C in the middle, and 1.0C on the right. What is the resultant force on the 1.0C charge due to the other two
Answers
We know, force is given by :
[tex]F=\dfrac{kq_1q_2}{r^2}[/tex]
Here, k in a constant.
[tex]Net \ Force = F_1+F_2\\\\Net \ Force = k(\dfrac{Qq_1}{r_1^2}+\dfrac{Qq_2}{r_2^2})\\\\Net \ Force = k( \dfrac{-3}{50}+\dfrac{2}{50})\\\\Net \ Force = 9\times 10^9\times \dfrac{-1}{50}\ N\\\\Net\ Force = -1.8\times 10^8\ N[/tex]
Hence, this is the required solution.
Calculate the maximum energy that a 3-MeV alpha particle can transfer to an electron in a single collision.
Answers
Answer:
The maximum energy that can be transferred to an electron in a single collision of alpha particle is 1.63 keV
Explanation:
let mass of electron, Me = 1
let mass of proton, Mp = 1836Me = 1836
let mass of alpha particle, Ma = 4Mp = 7344
The maximum energy that can be transferred to an electron in a single collision is given by;
[tex]Q_{max} = \frac{4mME}{(M+m)^2}\\\\Q_\alpha_{max} = \frac{4(1)(7344)(3\ MeV)}{(1+7344)^2} = 0.00163 \ MeV = 1.63 \ KeV[/tex]
Therefore, the maximum energy that can be transferred to an electron in a single collision of alpha particle is 1.63 keV
A cheetah can run at a maximum speed
91 km/h and a gazelle can run at a maximum speed of 72.7 km/h.
If both animals are running at full speed,
with the gazelle 87.5 m ahead, how long before
the cheetah catches its prey?
Answer in units of s.
Answers
Answer:
Approximately 17.21 seconds
Explanation:
With subtraction, we have the gazelle 18.3 km/h slower than the cheetah, which is about 5.08333 m/s. As the gazelle is 87.5 meters ahead of the cheetah, 87.5 divided by 5.083333333 is about 17.21 seconds.
An underwater sound source emits waves of frequency 30 kHz in all directions. How does the intensity of the waves (in Watts/m2) vary with distance r from the source?
a) 1/r^3
b) 1/r^2
c) 1/r
d) None of above
Answers
When an underwater sound source emits waves of frequency 30 kHz in all directions, the intensity of the waves (in Watts/m2) vary with distance r from the source by the relation 1/r²
As the intensity mechanical sound wave is inversely proportional to the square of the distance from the source, therefore the correct option is B.
What is the Wavelength?Wavelength can be understood in terms of the distance between any two similar successive points across any wave for example wavelength can be calculated by measuring the distance between any two successive crests.
It is the total length of the wave for which it completes one cycle.
The intensity of a mechanical wave is inversely proportional to the square of the distance from the source.
An underwater sound source emits waves of frequency of 30 kHz in all directions, the intensity of the waves (in Watts/m2) varies with distance r from the source by the relation 1/r², therefore the correct option is B.
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Which of the following does not discribe a mineral
Answers
Answer:
give us some further context to answer your question as well
Explanation:
If the breaking strength of the string is 120 N, what is the minimum angle the string can make with the horizontal
Answers
Complete Question
A 940-g rock is whirled in a horizontal circle at the end of a 1.30-m-long string, If the breaking strength of the string is 120 N, what's the minimum angle the string can make with the horizontal?
Answer:
The value is [tex]\theta = 4.41^o[/tex]
Explanation:
From the question we are told that
The mass of the rock is [tex]m_r = 940 \ g = 0.94 \ kg[/tex]
The length of the string is [tex]l = 1.30 \ m[/tex]
The breaking strength(i.e the maximum tension) on the string is [tex]T = 120 \ N[/tex]
Gnerally the vertical component of the tension experienced by the string is mathematically represented as
[tex]T_v = T sin(\theta)[/tex]
Generally this vertical component of tension is equivalent to the weight of the rock
So
[tex]Tsin (\theta) = mg[/tex]
=> [tex]\theta = sin^{-1} [\frac{mg}{ T} ][/tex]
=> [tex]\theta = sin^{-1} [\frac{0.940 *9.8 }{ 120} ][/tex]
=> [tex]\theta = 4.41^o[/tex]
Consider a wire of length L =0.30 m that runs north-south on a horizontal surface. There is a current I = 0.50 A flowing north in tie wire. The Earth's magnetic field at this location has a magnitude of 0.50 gauss (or, in SI units. 0.5 x 10^—4 and points north and 38 degrees down from the horizontal, toward the ground. What is the size of the magnetic force on the wire due to the Earth's magnetic field?
Answers
Answer:
The size of the magnetic force on the wire due to the Earth's magnetic field is 4.62 × 10⁻⁶ N.
Explanation:
To determine the size of the magnetic force on the wire due to the Earth's magnetic field,
The magnetic force is given by the formula
F = ILB sinθ
Where F is the magnetic force on the wire
I is the electric current in Amperes (A)
L is is the length of wire in meters (m)
B is the magnetic field in Tesla (T)
and θ is the angle between current and magnetic field
From the question,
L = 0.30 m
I = 0.50 A
B = 0.50 gauss = 0.5 × 10⁻⁴ T (NOTE: 1 Gauss = 10⁻⁴ Tesla)
θ = 38°
Now, putting the values into the equation
F = ILB sinθ
F = 0.50 × 0.30 × 0.5 × 10⁻⁴ sin38°
F = 7.5 × 10⁻⁶ (0.61566)
F = 4.62 × 10⁻⁶ N
Hence, the size of the magnetic force on the wire due to the Earth's magnetic field is 4.62 × 10⁻⁶ N.
A baseball's resultant velocity is 25.79 m/s, at 68.64° from the level ground.
Determine the vertical component of the baseball's velocity.
Answers
Answer:
Vy = 24.01 [m/s]
Explanation:
To solve this problem we must understand that velocity is a vector and therefore has magnitude and direction, in this way we can decompose the velocity in the components x & y, by means of the trigonometric functions of the sine and cosine of the angle.
For the vertical component, the sine of the angle is used.
Vy = 25.79*sin(68.64)
Vy = 24.01 [m/s]
Which example illustrates Newton's first law?
Answers
Answer: The motion of a ball falling down through the atmosphere, or a model rocket being launched up into the atmosphere are both examples of Newton's first law. The motion of a kite when the wind changes can also be described by the first law.
Explanation:
show answer No Attempt 50% Part (b) Calculate the non-relativistic speed of these electrons v in m/s.
Answers
Complete Question
The complete question is shown on the first uploaded image
Answer:
The value is [tex]v = 9.18 *10^{6} \ m/s[/tex]
Explanation:
From question we are told that
The potential difference is [tex]\Delta V = 0.24 kV = 0.24 *10^{3} \ V[/tex]
Generally the the non-relativistic speed of these electrons is mathematically represented as
[tex]v = \sqrt{\frac{2 * e * \Delta V}{m} }[/tex]
Here m is the mass of an electron with value [tex]m = 9.11 *10^{-31 } \ kg[/tex]
e is the charge on an electron with value [tex]e = 1.60 *10^{-19} \ C[/tex]
So
[tex]v = \sqrt{\frac{2 * 1.60 *10^{-19} * 0.24 *10^{3} }{9.11*10^{-31}} }[/tex]
=> [tex]v = 9.18 *10^{6} \ m/s[/tex]
Sand is made of tiny pieces of rock that have been worn
down by wind and water. Explain why the formation of
sand is a physical change.
Answers
Answer:
Sand is a form of physical change because it shows how one object is being worn out (broken down) into another object by nature.
Explanation:
Nora notices water droplets on the grass in the morning. It did not rain during the night. Which statement is true about this change of state?
Mass was added to the water particles, resulting in deposition.
Energy was added to the water particles, resulting in evaporation.
Mass was removed from the water particles, resulting in sublimation.
Energy was removed from the water particles, resulting in condensation.
Answers
Answer:
Mass was added to the water particles, resulting in deposition.
Explanation:
Answer: The energy was removed from the water particles resulting in condensation
Explanation Think about when you leave a cup of ice of and after a while you see water droplets thats what happened to the grass in the morning and thats called condensation.
Jack is buying a subscription to a magazine club. The function c=45+3.75m
c
=
45
+
3
.
75
m
describes the cost of the yearly subscription, c
c
, in terms of the number of magazines, m
m
, he orders.
What is the appropriate domain for the function?
the set of integers
the set of whole numbers
the set of all real numbers
the set of rational numbers
Answers
Answer:
the set of all real numbers
Explanation:
Given the cost of the magazine expressed as:
c=45+3.75m where:
c describes the cost of the yearly subscription.
m is the amount of order.
You must understand that domain are the values of the input variable m that will make the function exist.
The value of m that will produce a positive output since we can only have a positive value of the cost. According to the option, we can say that the domain is on the set of all real numbers (some are negative but all positive real numbers)
2.9030 cm has 4 significant digits.
TRUE OR FALSE
Answers
Answer: false
Explanation:
Are renewable resources always renewable, or can they become non-renewable
Answers
Answer:
Can become non-renewable but it depends on the object
Explanation:
Answer:
Renewable resources are so abundant or are replaced so rapidly that, for all practical purposes, they can't run out. Fossil fuels are the most commonly used non-renewable resources
Explanation:
Hope this helps (: