d. Fill in the blanks in the following paragraph to correctly identify the conductive properties of materials. (1 point)
A metal such as copper is a(n) _______________ because it provides a pathway for electric charges to move easily. A material such as rubber is a(n) _______________ because it _______________ the flow of electric charges. A material that partially conducts electric current is a(n) _______________. These materials include _______________ elements.
Answer:
conductor, insulator, resists, semiconductor, 3 and 5 group
Explanation:
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.
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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:
9.
A 2.85-cm diameter coin is placed a distance of 31.4 cm from a diverging lens that has a focal length
of-11.6 cm. Determine the image distance and the diameter of the image.
Answer:
image distance = -8.47 cm
diameter of the image = 0.769 cm
Which two criteria are least important for engineers to consider when developing a process to produce an important chemical?
A. The process should be located in a major city.
B. The process should use local, organic reactants.
C. The process should produce a significant amount of the product.
D. The process should be cost-effective.
Answer:
two least important criteria are A and B.
Explanation:
In an industrial production process there are criteria on the location of the production plants. Among the most important criteria we have that a good quantity is produced from the producer and that the process is cost-beneficial - cats to increase the profit of the company.
There are other desirable but minor criteria
A) The size of the city is of minor importance
b) Finding raw material locally is interesting, but its importance is less, the raw material can be sent from other locations.
Consequently the two least important criteria are A and B.
Answer:
A. The process should be located in a major city.
B. The process should use local, organic reactants.
Explanation:
ap.ex
The head scale of a screw gauge contains 100 divisions and it's pitch is 1mm. what is the least count of the screw gauge
Answer:
0.01 mm
Explanation:
The least count of the screw gauge = pitch/number of divisions on circular scale
Pitch = 1 mm and number of divisions = 100
Substituting the values of the variables into the equation, we have
least count of the screw gauge = pitch/number of divisions on circular scale = 1 mm/100 = 0.01 mm
When measuring volume, your final
result must be in
A. linear units.
B. single units.
C. squared units.
D. cubed units.
6. What is the wavelength of a wave that is traveling at 30 m/s and has a frequency of 3.2 Hz?
Answer:
6. 9.4 m
7. 1050 m/s.
Explanation:
6. Determination of the wavelength
Velocity (v) = 30 m/s
Frequency (f) = 3.2 Hz
Wavelength (λ) =?
v = λf
30 = λ × 3.2
Divide both side by 3.2
λ = 30 / 3.2
λ = 9.4 m
Thus, the wavelength of the wave is 9.4 m.
7. Determination of the speed of the wave.
Wavelength (λ) = 350 m
Frequency (f) = 3 Hz
Velocity (v) =.?
v = λf
v = 350 × 3
v = 1050 m/s
Thus, the speed of the wave is 1050 m/s
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]
Focal distance of a concave lens is always:
O A. smaller than the image distance.
O B. positive.
O C. negative.
O D. larger than the object distance.
is warming up important before doing an intense exercise or work-out? Wh?
Answer:
Yes, warming up is necessary before intense exercise, work-out or sports because they help regulate temperatures of your body. Warming up also makes your muscles ready for intense physical activities and prevents muscle cramps.
Explanation:
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 would be the mass of an atom with 2 protons and 3 neutrons?
Answer:
5
Explanation:
From the question given above, the following data were obtained:
Proton number = 2
Neutron number = 3
Mass number =.?
The mass number of a given atom is simply the sum of the protons and neutrons in the nucleus of the atom. Mathematically, it is expressed as:
Mass number = proton + neutron
With the above formula, we can obtain the mass number of the atom as follow:
Proton number = 2
Neutron number = 3
Mass number =.?
Mass number = proton + neutron
Mass number = 2 + 3
Mass number = 5
Thus, the mass number the atom is 5.
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].
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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.
the gravitational attraction between two masses of 3kg that are separated by a distance of 1cm is
Answer:
6.003×10¯⁶ N
Explanation:
We'll begin by converting 1 cm to m. This can be obtained as follow:
100 cm = 1 m
Therefore,
1 cm = 1 cm × 1 m / 100 cm
1 cm = 0.01 m
Finally, we shall determine the gravitational attraction. This can be obtained as follow:
Mass 1 (M₁) = 3 Kg
Mass 2 (M₂) = 3 Kg
Distance apart (r) = 0.01 m
Gravitational constant (G) = 6.67×10¯¹¹ Nm²/Kg²
Force of attraction (F) =?
F = GM₁M₂ / r²
F = 6.67×10¯¹¹ × 3 × / 0.01²
F = 6.003×10¯¹⁰ / 1×10¯⁴
F = 6.003×10¯⁶ N
Thus the gravitational attraction is 6.003×10¯⁶ N
Future space rockets might propel themselves by firing laser beams, rather than exhaust gases, out the back. The acceleration would be small, but it could continue for months or years in the vacuum of space. Consider a 1200 kg uncrewed space probe powered by a 15 MW laser. After one year, how far will it have traveled and how fast will it be going
Answer:
Acceleration = 0.0282 m/s^2
Distance = 13.98 * 10^12 m
Explanation:
we will apply the energy theorem
work done = ΔK.E ( change in Kinetic energy ) ---- ( 1 )
where :
work done = p * t
= 15 * 10^6 watts * ( 1 year ) = 473040000 * 10^6 J
( note : convert 1 year to seconds )
and ΔK.E = 1/2 mVf^2 given ; m = 1200 kg and initial V = 0
back to equation 1
473040000 * 10^6 = 1/2 mv^2
Vf^2 = 2(473040000 * 10^6 ) / 1200
∴ Vf = 887918.92 m/s
i) Determine how fast the rocket is ( acceleration of the rocket )
a = Vf / t
= 887918.92 / ( 1 year )
= 0.0282 m/s^2
ii) determine distance travelled by rocket
Vf^2 - Vi^2 = 2as
Vi = 0
hence ; Vf^2 = 2as
s ( distance ) = Vf^2 / ( 2a )
= ( 887918.92 )^2 / ( 2 * 0.0282 )
= 13.98 * 10^12 m
If a wave is traveling at 60 cm/second and has a wavelength of 15 cm, what is the frequency?
A. 900 Hz
B. 60 Hz
C. 8 Hz
D. 4 Hz
Answer: B
G v = 60 cm/s λ = 15 cm E f = V / λ S f = 4 Hz U f = ? S f = (60 cm/s) / 15 cm 5.
Explanation:
A 2-kg ball is thrown at a speed of ____, exhibits 25j of kinetic energy.
Answer:
A 2-kg ball is thrown at a speed of 5 m/s, exhibits 25 J of kinetic energy.
Explanation:
Given that,
The mass of a ball, m = 2 kg
Kinetic energy of the ball, K = 25 J
We need to find the speed of the ball. The formula for the kinetic energy is given by :
[tex]K=\dfrac{1}{2}mv^2\\\\v=\sqrt{\dfrac{2K}{m}} \\\\v=\sqrt{\dfrac{2\times 25}{2}} \\\\v=5\ m/s[/tex]
So,
A 2-kg ball is thrown at a speed of 5 m/s, exhibits 25 J of kinetic energy.
do you think you will benefit from the discoveries of Archimedes’ and Pascal’s Principles?
Answer:
I hate to say it is a bit of the holiday but
Water flows through a garden hose which is attached to a nozzle. The water flows through hose with a speed of 1.81 m/s and through the nozzle with a speed of 18.3 m/s. Calculate the maximum height (in m) to which water could be squirted if it emerges from the nozzle and emerges with the nozzle removed. (a) Emerges from the nozzle. m (b) Emerges with the nozzle removed, assuming the same flow rate. m
Answer:
a) 17.086m
b) 0.1671 m
Explanation:
Given data: speed of water through the hose = 1.81 m/s
through the nozzle = 18.3 m/s
We know that maximum height of an object with upward velocity v is given by,
a) H = v^2/2g
where H is the maximum height water emerges
= 18.3^2/(2×9.8) = 17.086 m answer
b) Again,
H = v^2/2g
= 1.81^2/(2×9.8) = 0.1671 m
choose the correct answer 14: which of the following force follows the inverse square law of distance A: gravitaional force B: electromagnetic force C: both (a) and (b) D: none of these
Answer:
C. both A and B
Explanation:
Sana nakatulong
The 6kg box is pushed to the left at a constant speed. The coefficient of friction is 0.78. Solve for the amount of force with which the hand pushes the box.
Answer:
45.864N
Explanation:
Using the formula
F = nR
n is the coefficient of friction
R is the normal reaction
R = mg
F = nmg
F = 0.78 * 6 * 9.8
F = 45.864N
Hence the amount of force with which the hand pushes the box is 45.864N
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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a stone on the ground has zero energy why
Explanation:
it has no energy when considered with respect to earth ,as it has neither height (i e potential energy) nor velocity (i.e kinetic energy).
Three advantages of working as a nuclear physicist
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
An unknoAn unknown additional charge q3 is now placed at point B, located at coordinates (0 m, 15.0 m ). Find the magnitude and sign of q3 needed to make the total electric field at point A equal to zero.wn additional charge q3 is now placed at point B, located at coordinates (0 m, 15.0 m ). Find the magnitude and sign of q3 needed to make the total electric field at point A equal to zero.\
Answer:
hello your question has some missing information attached below is the missing information
answer : q3 = 0.300 nC
Explanation:
Given that Eay is in the Y axis ( upward ) , E3 will be downward and this will make the sign on the the charge ( q3 ) to be positive
E3 = Eay. ( for an electric field to be neutral/zero the electric field in the opposite direction will have same magnitude )
To calculate the value of q3 we will apply the relation below
K*q3/( 15 - 12 )^2 = 0.300
= (9 * 10^9) * (q3 / 9 ) = 0.300
∴ q3 = 0.300 / 10^9
= 0.300 nC
Note : The value of Eay is been calculated outside this solution as it is not part of the question asked
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.)