Answer:
1.84m
Explanation:
Given parameters:
Initial velocity = 6m/s
Unknown:
height of jump = ?
Solution:
To solve this problem, we have to apply the right motion equation:
V² = U² - 2gH
V is the final velocity
U is the initial velocity
g is the acceleration due to gravity = 9.8m/s²
H is the height
Final velocity is 0
Solve;
0² = 6² - 2x9.8xH
-36 = -19.6H
H = 1.84m
The amount of energy necessary to remove an electron from an atom is a quantity called the ionization energy, Ei. This energy can be measured by a technique called photoelectron spectroscopy, in which light of wavelength λ is directed at an atom, causing an electron to be ejected. The kinetic energy of the ejected electron (Ek) is measured by determining its velocity, υ (Ek= mυ2/2), and Ei is then calculated using the conservation of energy principle. That is, the energy of the incident light equals Ei plus Ek. What is the ionization energy of selenium atoms in kJ/mol if light with λ = 48.2 nm produces electrons with a velocity of 2.371x106 m/s? The mass, m, of an electron is 9.109x10-31 kg. (Round to the ones place.)
Answer:
The value is [tex]E_i = 1.5596 *10^{-18} \ J[/tex]
Explanation:
From the question we are told that
The wavelength is [tex]\lambda = 48.2 nm = 48.2 *10^{- 9 }\ m[/tex]
The velocity is [tex]v = 2.371*10^6 \ m/s[/tex]
The mass of electron is [tex]m_e = 9.109*10^{-31} \ kg[/tex]
Generally the energy of the incident light is mathematically represented as
[tex]E = \frac{h * c}{\lambda}[/tex]
Here c is the speed of light with value [tex]c = 3.0 *10^{8} \ m/s[/tex]
h is the Planck constant with value [tex]h = 6.62607015 * 10^{-34 } J\cdot s[/tex]
So
[tex]E = \frac{6.62607015 * 10^{-34 }* 3.0 *10^{8}}{48.2 *10^{- 9 }}[/tex]
=> [tex]E = 4.12 *10^{-18} \ J [/tex]
Generally the kinetic energy is mathematically represented as
[tex]E_k = \frac{1}{2} * m_e * v^2[/tex]
=> [tex]E_k = \frac{1}{2} * 9.109*10^{-31} * (2.371*10^6 )^2[/tex]
=> [tex]E_k = 2.56 *0^{-18} \ J [/tex]
Generally the ionization energy is mathematically represented as
[tex]E_i = 4.12 *10^{-18} - 2.56 *0^{-18}[/tex]
=> [tex]E_i = 1.5596 *10^{-18} \ J[/tex]
A book sitting on a table has potential energy because of the gravitational pull of Earth. Which of the following would increase the potential energy of the book? A) Sliding it across the table C) Dropping it on the floor B) Opening it to the middle D) Lifting it to a high shelf
Answer:
D) Lifting it to a high shelf
Explanation:
Potential energy is where energy could be released at any moment. For example, if a cat gets on a tree it's creating potential energy, once that cat jumps from the tree the energy releases. D is the answers because if you set the book in a high shelf, you are only creating more potential energy.
hope I explain my self well enough
cheers man and good luck
why bananas are curved
they are bananas and bananas are curved because they grow curved
As the chief design engineer for a major toy company, you are in charge of designing a loop-the-loop toy for youngsters. The idea is that a ball of mass m and radius r will roll down an inclined track and around the loop without slipping. The ball starts from rest at a height h above the tabletop that supports the whole track. The loop radius is R. Determine the minimum height h, in terms of R and r, for which the ball will remain in contact with the track during the whole of its loop-the-loop journey
Answer: h = 2.7 R - 1.7 r
Explanation:
normally, force is 0 at the top ;
-N - mg = - m v^2 / ( R - r )
-0 - mg = (- mv^2) / ( R-r )
mg = (mv^2) / (R - r)
g = v^2 / ( R - r ) ; ----------------equation 1
conservation of energy ;
ΔK + ΔP = 0 ;
1/2 I ω^2 + 1/2 m v^2 + mg ( h2 - h1 ) = 0 ;
0.5 * ( 2/5 ) m r^2 * ( v / r )^2 + 0.5 m v^2 + mg ( ( 2R - r ) -h ) = 0 ;
0.5 * ( 2/5 ) m r^2 * ( v / r )^2 + 0.5 m v^2 = mg ( - 2R + r + h ) ;
0.5 * ( 2/5 )r^2 * ( v / r )^2 + 0.5 v^2 = g ( - 2R + r + h ) ;
0.5 * ( 2/5 ) v^2 + 0.5 v^2 = g ( - 2R + r + h ) ;
[ 0.5 * ( 2/5 ) + 0.5 ] v^2 = g ( - 2R + r + h ) ;-------------equation 2
from equation 1 , v^2 = g ( R - r ), input in equation 2
[ 0.5 * ( 2/5 ) + 0.5 ] [ g ( R - r ) ] = g ( - 2R + r + h )
[ 0.5 * ( 2/5 ) + 0.5 ] [ ( R - r ) ] = ( - 2R + r + h )
0.7 ( R - r ) = h - 2R + r
0.7R - 0.7r = h - 2R + r
solve for h
h = 0.7R + 2R - 0.7r - r
h = 2.7 R - 1.7 r
A car traveling from rest to a velocity of 7.0 m/s accelerates uniformly at the rate of 0.80 m/s2. What is its tr. 31
time?
Answer:
54mph
Explanation:
please help me with this physics problem
Answer:
It says the answer. (5 marks.)
Explanation:
The FitnessGram Pacer Test is a multistage aerobic capacity test that progressively gets more difficult as it continues. The 20 meter pacer test will begin in 30 seconds. Line up at the start. The running speed starts slowly, but gets faster each minute after you hear this signal. A single lap should be completed each time you hear this sound. Remember to run in a straight line, and run as long as possible.
A package is dropped from an air balloon two times. In the first trial the distance between the balloon and the surface is Hand in the second trial 4H. Compare the FINAL VELOCITY'S for the packages right as they hit the
ground (or right before)?
Answer:
The final speed of the second package is twice as much as the final speed of the first package.
Explanation:
Free Fall Motion
If an object is dropped in the air, it starts a vertical movement with an acceleration equal to g=9.8 m/s^2. The speed of the object after a time t is:
[tex]v=gt[/tex]
And the distance traveled downwards is:
[tex]\displaystyle y=\frac{gt^2}{2}[/tex]
If we know the height at which the object was dropped, we can calculate the time it takes to reach the ground by solving the last equation for t:
[tex]\displaystyle t=\sqrt{\frac{2y}{g}}[/tex]
Replacing into the first equation:
[tex]\displaystyle v=g\sqrt{\frac{2y}{g}}[/tex]
Rationalizing:
[tex]\displaystyle v=\sqrt{2gy}[/tex]
Let's call v1 the final speed of the package dropped from a height H. Thus:
[tex]\displaystyle v_1=\sqrt{2gH}[/tex]
Let v2 be the final speed of the package dropped from a height 4H. Thus:
[tex]\displaystyle v_2=\sqrt{2g(4H)}[/tex]
Taking out the square root of 4:
[tex]\displaystyle v_2=2\sqrt{2gH}[/tex]
Dividing v2/v1 we can compare the final speeds:
[tex]\displaystyle v_2/v_1=\frac{2\sqrt{2gH}}{\sqrt{2gH}}[/tex]
Simplifying:
[tex]\displaystyle v_2/v_1=2[/tex]
The final speed of the second package is twice as much as the final speed of the first package.
Playing with a stress ball you squeeze it as
hard as you can. How does this affect the
density of the ball?
Answer:
it dies
Explanation:
MWUAHHAHAHHA
what are the answers to the question
Answer: the link isnt loading
Protons and ____ have electric charge?
100 POINTS. PLEASE PROVIDE EXPLANATION
Answer:
60 kg
80 kg
Explanation:
Work is equal to the change in energy.
W = ΔE = E − E₀
Let's start with block B. The work done by the tension force is equal to the change in energy. Initially, the block has potential energy. Finally, the block has kinetic energy.
W = ΔE
FΔy = ½ mv² − mgh
T (-2.0 m) = ½ m (6.00 m/s)² − m (10 m/s²) (2.0 m)
T (-2.0 m) = m (-2 m²/s²)
T = m (1 m/s²)
Now let's look at block A. The work done by tension and against friction is equal to the change in energy. Initially, the block has no energy. Finally, it has both kinetic and potential energy.
W = ΔE
Fd = ½ mv² + mgh − 0
(T − Nμ) (2.0 m) = ½ (4.00 kg) (6.00 m/s)² + (4.00 kg) (10 m/s²) (⅗ × 2.0 m)
(T − Nμ) (2.0 m) = 120 J
T − Nμ = 60 N
Draw a free body diagram of block A and sum the forces in the perpendicular direction to find the normal force N.
N = mg cos θ
N = (4.00 kg) (10 m/s²) (⅘)
N = 32 N
Substitute:
T − 32μ = 60 N
If μ = 0, then T = 60 N and m = 60 kg.
If μ = ⅝, then T = 80 N and m = 80 kg.
Answer:did you find you answer also i answered so you can give him brainiest
Explanation:
An African Swallow can travel at an average velocity of 11 m/s. How far can an African Swallow carry a coconut in 120 seconds?
Answer:
The distance will be x = 1320 [m]
Explanation:
To solve this problem we must use the expression of physics that relates space to time, which is defined as speed.
v = Speed = 11 [m/s]
t = time = 120 [s]
x = distance [m]
v = x/t
x = v*t
x = 11*120
x = 1320 [m]
A 2.5 μF capacitor consists of 2 large, thin metal plates separated by a plexiglas sheet. The plexiglas sheet has a dielectric constant of 3.4. The capacitor is connected to a 60 V battery and fully charged. Then, whilst still connected to the battery, the plexiglas sheet is removed (assume that the spacing between the metal plates does not change). Calculate the amount of charge that flows to (or from) the capacitor plates when the sheet is removed. Additional charge flowing onto the plates is represented with a + sign and charge leaving the plates is represented with a - sign.
Answer:
Q / Q₀ = 3.4 , ΔQ = 31.14 10⁻⁶ C
Explanation:
For this exercise we will calculate the charge of the capacitor with and without dielectric
the capacitance is given by
C = Q /ΔV
we also use the ratio of capacitances
C = k C₀
Q /ΔV = kQ₀ /ΔV
Q = k Q₀
Q / Q₀ = k
Q / Q₀ = 3.4
If we want a numerical value
C = k Q / DV
Q = C ΔV / k
Q = 2.5 10⁻⁶ 60 / 3.4
Q = 44.12 10⁻⁶ C
whereby the charge without dielectric is
Q₀ = Q / 3.4
Q₀ = 44.12 / 3.4 10⁻⁶
Q₀ = 12.98 10⁻⁶ C
therefore the charge when removing the dielectric is reduced, the amount of charge that flows is
ΔQ = Q - Q₀
ΔQ = (44.12 - 12.98) 10⁻⁶
ΔQ = 31.14 10⁻⁶ C
The amount of charge should be Q / Q₀ = 3.4 , ΔQ = 31.14 10⁻⁶ C
Capacitor:For this given situation, we will determine the charge of the capacitor with and without dielectric
We know that
the capacitance is given by
C = Q /ΔV
Now here we can applied the ratio of capacitances
So,
C = k C₀
Q /ΔV = kQ₀ /ΔV
Q = k Q₀
Q / Q₀ = k
Q / Q₀ = 3.4
In case of numerical value
C = k Q / DV
Q = C ΔV / k
Q = 2.5 10⁻⁶ 60 / 3.4
Q = 44.12 10⁻⁶ C
whereby the charge without dielectric is
Q₀ = Q / 3.4
Q₀ = 44.12 / 3.4 10⁻⁶
Q₀ = 12.98 10⁻⁶ C
Thus the charge at the time when removing the dielectric is decreased , the amount of charge that flows is
ΔQ = Q - Q₀
ΔQ = (44.12 - 12.98) 10⁻⁶
ΔQ = 31.14 10⁻⁶ C
Learn more about battery here: https://brainly.com/question/24841364
A man throws a ball straight up to his friend on a balcony who catches it at its highest point. The ball was thrown with an initial velocity of 17.9 m/s2. What height does the ball reach?
Answer:
The maximum height reached by the ball is 16.35 m.
Explanation:
Given;
initial velocity of the ball, u = 17.9 m/s
the final velocity of the ball at the maximum height, v = 0
The maximum height reached by the ball is given by;
v² = u² + 2gh
During upward motion, gravity is negative
v² = u² + 2(-g)h
v² = u² - 2gh
0 = u² - 2gh
2gh = u²
h = u² / 2g
h = (17.9)² / (2 x 9.8)
h = 16.35 m
Ttherefore, the maximum height reached by the ball is 16.35 m.
A truck is parked at the bottom of a hill it moves from the bottom of the hill and parks at the top compared to when it was parked at the bottom of the hill at its new position in the truck now has
A. more kinetic energy
B. more potential energy
C. Less kinetic energy
D. less potential energy
If a car moves at a velocity of 1.45 x102 m/s for 1.8 x103
seconds. How far does the car go?
Answer:
The answer is
[tex]2.61 \times {10}^{5} \: \: m[/tex]Explanation:
The distance covered by an object given it's velocity and time taken can be found by using the formula
distance = velocity × timeFrom the question
velocity = 1.45 × 10² m/s
time = 1.8 × 10³ s
We have
distance = 1.45 × 10² × 1.8 × 10³
We have the final answer as
[tex]2.61 \times {10}^{5} \: \: m[/tex]
Hope this helps you
A pile of bricks of mass M is being raised to the tenth floor of a building of height H = 4y above the ground by a crane that is on top of the building. During the first part of the lift, the crane lifts the bricks a vertical distance h1=3y in a time t1=4T. During the second part of the lift, the crane lifts the bricks a vertical distance h2=y in t2=T. Which of the following correctly relates the power P1 generated by the crane during the first part of the lift to the power P2 generated by the crane during the second part of the lift?
A. P2=4P1
B. P2=43P1
C. P2=P1
D. P2=34P1
E. P2=13P1
Complete Question
A pile of bricks of mass M is being raised to the tenth floor of a building of height H = 4y above the ground by a crane that is on top of the building. During the first part of the lift, the crane lifts the bricks a vertical distance h1=3y in a time t1=4T. During the second part of the lift, the crane lifts the bricks a vertical distance h2=y in t2=T. Which of the following correctly relates the power P1 generated by the crane during the first part of the lift to the power P2 generated by the crane during the second part of the lift?
[tex]A.\ \ P_2=4P_1[/tex]
[tex]B.\ \ P_2=\frac{4}{3} P1[/tex]
[tex]C.\ \ P_2=P_1[/tex]
[tex]D. \ \ P_2=\frac{3}{4} P_1[/tex]
[tex]E. \ \ \ P_2=\frac{1}{3} P_1[/tex]
Answer:
The correct option is B
Explanation:
From the question we are told that
The mass of the brick is M
The height height of the 10th floor is H = 4y
The height attained during the first part of the lift is [tex]h_1 = 3y[/tex]
The time taken is [tex]t_1 = 4T[/tex]
The height attained during the second part of the lift is [tex]h_2 = y[/tex]
The time taken is [tex]t_2 = T[/tex]
Generally the velocity of the crane during the first lift is mathematically represented as
[tex]v_1 = \frac{h_1}{t_1}[/tex]
=> [tex]v_1 = \frac{3y}{4T}[/tex]
Generally the velocity of the crane during the first lift is mathematically represented as
[tex]v_1 = \frac{h_2}{t_2}[/tex]
=> [tex]v_1 = \frac{y}{T}[/tex]
Generally the power generated during the first lift is
[tex]P_1 = F_1 * v_1[/tex]
Here [tex]F_1[/tex] force applied during the first lift which is mathematically represented as
[tex]F_1 = M * g[/tex] here g is the acceleration due to gravity
So
[tex]P_1 = Mg * \frac{3y}{4T}[/tex]
Generally the power generated during the second lift is
[tex]P_2 = F_2 * v_2[/tex]
Here [tex]F_2[/tex] force applied during the second lift which is mathematically represented as
[tex]F_2 = M * g[/tex] here g is the acceleration due to gravity
So
[tex]P_2 = Mg * \frac{y}{T}[/tex]
So the ratio of the first power to the second power is
[tex]\frac{P_1}{P_2} = \frac{Mg * \frac{3y}{4T}[}{Mg * \frac{y}{T}}[/tex]
=> [tex]\frac{P_1}{P_2} = \frac{3}{4}[/tex]
=> [tex]P_2 = \frac{4}{3} P_1[/tex]
A truck slows from a velocity of 34 m/s to a stop in 50 m. What was the
truck's acceleration? *
The acceleration/deceleration of the truck is 11.56 m/s^2
Velocity and acceleration Velocity is the rate of change of displacement of a body.Velocity = change in displacement/ time
Acceleration is the rate of change of velocity with time.Acceleration = change in velocity/ time.
The acceleration, velocity and distance travelled of a body is related by the equation below:
v^2 = u^2 + 2aswhere v is the initial velocity
u is the final velocity
a is the acceleration, and
s is the distance travelled
From the data provided:
v = 0
u = 34 m/s
s = 50 m
a = ?
making a subject of the formula
a = -u^2/2s
a = - 34^2 / 2 × 50
a = -11.56 m/s^2
Therefore, the acceleration/deceleration of the truck is 11.56 m/s^2
Learn more about velocity and acceleration at: https://brainly.com/question/742413
Displacement consists of
Answer:
The displacement of a body has two components: rigid-body displacement and deformation. A rigid-body displacement consists of a simultaneous translation and rotation of the body without changing its shape or size.
Hope this helps :)
The rate at which heat enters an air conditioned building is often roughly proportional to the difference in temperature between inside and outside.
(a) Under this assumption, show that the cost of operating an air con- ditioner is proportional to the square of the temperature difference.
(b) Give a numerical example for a typical house and discuss implications for your electric bill.
(c) Suppose instead that heat enters the building at a rate proportional to the square-root of the temperature difference between inside and outside. How would the operating cost now depend on the temperature difference?
Answer:
Considering first question
Generally the coefficient of performance of the air condition is mathematically represented as
[tex]COP = \frac{T_i}{T_o - T_i}[/tex]
Here [tex]T_i[/tex] is the inside temperature
while [tex]T_o[/tex] is the outside temperature
What this coefficient of performance represent is the amount of heat the air condition can remove with 1 unit of electricity
So it implies that the air condition removes [tex] \frac{T_i}{T_o - T_i}[/tex] heat with 1 unit of electricity
Now from the question we are told that the rate at which heat enters an air conditioned building is often roughly proportional to the difference in temperature between inside and outside. This can be mathematically represented as
[tex]Q \ \alpha \ (T_o - T_i)[/tex]
=> [tex]Q= k (T_o - T_i)[/tex]
Here k is the constant of proportionality
So
since 1 unit of electricity removes [tex] \frac{T_i}{T_o - T_i}[/tex] amount of heat
E unit of electricity will remove [tex]Q= k (T_o - T_i)[/tex]
So
[tex]E = \frac{k(T_o - T_i)}{\frac{T_i}{ T_h - T_i} }[/tex]
=> [tex]E = \frac{k}{T_i} (T_o - T_i)^2[/tex]
given that [tex]\frac{k}{T_i}[/tex] is constant
=> [tex]E \ \alpha \ (T_o - T_i)^2[/tex]
From this above equation we see that the electricity required(cost of powering and operating the air conditioner) is approximately proportional to the square of the temperature difference.
Considering the second question
Assuming that [tex]T_i = 30 ^oC[/tex]
and [tex]T_o = 40 ^oC[/tex]
Hence
[tex]E = K (T_o - T_i)^2[/tex]
Here K stand for a constant
So
[tex]E = K (40 - 30)^2[/tex]
=> [tex]E = 100K [/tex]
Now if the [tex]T_i = 20 ^oC[/tex]
Then
[tex]E = K (40 - 20)^2[/tex]
=> [tex]E = 400 \ K[/tex]
So from this see that the electricity require (cost of powering and operating the air conditioner)when the inside temperature is low is much higher than the electricity required when the inside temperature is higher
Considering the third question
Now in the case where the heat that enters the building is at a rate proportional to the square-root of the temperature difference between inside and outside
We have that
[tex]Q = k (T_o - T_i )^{\frac{1}{2} }[/tex]
So
[tex]E = \frac{k (T_o - T_i )^{\frac{1}{2} }}{\frac{T_i}{T_o - T_i} }[/tex]
=> [tex]E = \frac{k}{T_i} * (T_o - T_i) ^{\frac{3}{2} }[/tex]
Assuming [tex]\frac{k}{T_i}[/tex] is a constant
Then
[tex]E \ \alpha \ (T_o - T_i)^{\frac{3}{2} }[/tex]
From this above equation we see that the electricity required(cost of powering and operating the air conditioner) is approximately proportional to the square root of the cube of the temperature difference.
11. What is the momentum of a truck with a mass of 3000 kg traveling north at a
speed of 25 m/s?
Answer:
We are given:
mass of the truck (m) = 3000 kg
velocity of the truck (v) = 25 m/s
Calculating the Momentum:
We know that the momentum of an object is equal to its mass times it's velocity
P = mv (where P is the momentum of the truck)
Momentum of the truck:
replacing with the values of the truck
P = mv
P = 3000 * 25
P = 75000 kg m/s
Therefore, the momentum of the truck is 75000 kg m/s
Without a battery , why would a circuit not work
Answer:
The cell will have no electromotive force in it.
Explanation:
The battery provides the chemical energy that produces the force needed to move electron around the circuit and to generate electrical energy.
electrochemical cells are devices in which chemical reactions produce electric current. The electromotive force inherent in a cell provides the driving force for the current to flow. Without this force, there would not be enough energy to move the electrons.Answer and i will give you branilest
Answer:
nice cool math and jim jake jr as a baby and just jim jake baby and nice pfp d00d XD
Explanation:
how can you observe the law of conservation of energy in action at the skatepark?
Explanation:
When the skater is dropped onto the ramp from above, the potential energy decreases and the kinetic energy increases.
Every time the skater bounces from the impact, thermal energy is gained, and both potential and kinetic energy are lost.
What is the elapsed time between the 0-m mark and the 40-m mark
Answer:
a) 4.0 s
b) 16 m/s
c) The distance covered between 4 - 5 s is four times the distance covered between 4 - 5 s
d) Equal distance are covered between 0 - 4 s and 4 -5 s
Note: The question is incomplete. The complete question is as follows;
Refer to the chart below that has data about a moving object to answer the following questions.
Time Elapsed 0.0s 1.0s 2.0s 3.0s 4.0s 5.0 s
Distance Traveled 0.0m 10.0m 20.0m 30.0m 40.0m 80.0 m
a. What is the elapsed time between the 0-m mark and the 40-m mark?
b. How large is the average velocity of the object for the interval from 0-5 s ?
c. How does the interval of 3-4 s compare with the interval from 4-5 s in terms of distance?
d. How does the interval of 0-4 s compare with the interval from 4-5 s in terms of distance?
Explanation:
a. From the data provide, time elapsed between the 0 m - 40 m mark is 4.0 s
b. Average velocity = total distance/ total time
average velocity = 80 m/ 5.0 s = 16.0 m/s
c. Distance covered between 3 - 4 s = 40 m - 30 m = 10 m
Distance covered between 4 - 5 s = 80 m - 40 m = 40 m
The distance covered between 4 - 5 s is four times the distance covered between 4 - 5 s
d. Distance covered between 0 - 4 s = 40 m - 0 m = 40 m
Distance covered between 4 - 5 s = 80 m - 40 m = 40 m
Equal distance are covered between 0 - 4 s and 4 -5 s
I need help with these 2 questions pls help
the measure of how many a wave passes on a certain point in a certain time is
frequency
Which is faster +25 m/s or -250 m/s
Reset Answers
Practice Energy transformations (Ouestion 5 of 10)
Q5
A penny is dropped from the top of the Statue of Liberty. The Statue of Liberty is 93 m tall If a penny has about 3 Joules of gravitational potential energy at the top, how much
potential energy will have transformed into kinetic energy at the half way point (46.5m) ?
Explanation:
At a height of 93 m, the gravitational potential energy is given by :
P = mgh
Where, m is the mass of a penny
We can find its mass.
[tex]m=\dfrac{P}{gh}\\\\m=\dfrac{3}{9.8\times 93}\\\\m=0.00329\ kg[/tex]
We need to find how much potential energy will have transformed into kinetic energy at the half way point (46.5m). It can be calculated as :
[tex]E=mgh'\\\\E=0.00329\times 9.8\times 46.5\\\\E=1.499\ J[/tex]
Hence, a penny will have transferred 1.499 J of potential energy into kinetic energy.
Pretty Easy question please answer only 20 minutes left:
A summary of the results of a scientific investigation is called a:
observation
research
hypothesis
conclusion
Answer:
D. Conclusion.
Explanation:
2. What is the normal force acting a 800-kg car if there are two 55-kg people sitting inside the car?
