Answers
Answer:
i believe its D because he is not sliding and he is applying force on the wall
Explanation:
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Related Questions
A cloud mass moving across the ocean at an altitude of 2000 m encounters a coastal mountain range. As it rises to a height of 3500 m to pass over the mountains, it undergoes an adiabatic expansion. The pressure at 2000 m is 0.802 atm and at 3500 is 0.602 atm. If the initial temperature of the cloud mass is 288 K, what is the cloud temperature as it passes over the mountains? Assume that Cp,m for air is 28.86 J K-1 mol-1 and that the air obeys the ideal gas law. If you are on the mountain, should you expect rain or snow?
Answers
Answer:
snow
Explanation:
Since the process undergoes adiabatic expansion, hence q = 0 and ΔU = w.
We can sole this problem using the following derivation:
[tex]ln(\frac{T_2}{T_1} )=-(\gamma -1)ln(\frac{V_f}{V_i} )=-(\gamma -1)ln(\frac{T_2}{T_1}\frac{P_i}{P_f} )\\=-(\gamma -1)ln(\frac{T_2}{T_1})-(\gamma -1)ln(\frac{P_i}{P_f})\\=-(\frac{\gamma -1}{\gamma})ln(\frac{P_i}{P_f})\\=-(\frac{\frac{C_{p,m}}{C_{p,m}-R} -1}{\frac{C_{p,m}}{C_{p,m}-R}})ln(\frac{P_i}{P_f})\\\\ln(\frac{T_2}{T_1} )==-(\frac{\frac{C_{p,m}}{C_{p,m}-R} -1}{\frac{C_{p,m}}{C_{p,m}-R}})ln(\frac{P_i}{P_f})\\\\Substituting\ values:\\\\[/tex]
[tex]ln(\frac{T_2}{T_1} )=-(\frac{\frac{28.86}{28.86-8.314} -1}{\frac{28.86}{28.86-8.314}})ln(\frac{0.802\ atm}{0.602\ atm})=-0.0826\\\\ln(\frac{T_2}{T_1} )=-0.0826\\\\Taking\ exponential\ of\ both \ sides:\\\\\frac{T_2}{T_1} =e^{-0.0826}\\\\T_2=0.9207T_1\\\\T_2=0.9207*288\\\\T_2=265\ K\\[/tex]
Since T2 = 265 K, we should expect a snow
When a ball is thrown straight up with no air Resistance, the acceleration is in what direction ?
Answers
Answer:
The ball has a velocity upwards but the acceleration is downwards. Gravity is giving the ball a downwards acceleration from the moment the ball leaves the hands.
A .What is the difference between fossils fuels and nuclear fuel ?
B .Explain why it is important for us to find alternatives to fossil fuels to meet pur energy?
C .What are renewable energy sources?give three examples.
Answers
Answer:
A. Nuclear and fossil fuel-burning power plants differ mainly in where their energy comes from; a nuclear reactor produces heat from radioactive metals, and a fossil-fuel plant burns coal, oil or natural gas.
B. Environmental and economic benefits of using renewable energy include: Generating energy that produces no greenhouse gas emissions from fossil fuels and reduces some types of air pollution. Diversifying energy supply and reducing dependence on imported fuels
C. These are energy sources that are constantly being replenished, such as sunlight, wind, and water.
The most popular renewable energy sources currently are:
Solar energy.
Wind energy.
Hydro energy.
Tidal energy.
Geothermal energy.
Biomass energy.
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A snapshot of three racing cars is shown in the diagram. All three cars start the race at the same time, at the same place, and move along a straight track. As they approach the finish line, which car has the lowest average speed?
Answers
Answer:
The car furthest from the finish line: Car III (Choice C).
Explanation:
It's asking for lowest average speed throughout the entire race. Therefore, whoever is last technically has the lowest average speed.
Car III is far behind Car I and Car II so Choice A and B aren't correct. Choice D is incorrect since the three cars aren't in the same position. Choice E is incorrect because there is enough information to see that Choice C has the lowest average speed.
2. A fish swimming at a constant speed of 18 m/s spots a sta-
tionary barracuda. Just as the fish passes the barracuda, the
predator begins swimming with a constant acceleration of
magnitude 2.2 m/s². The fish and barracuda are moving in the
same direction.
(a) How far does the barracuda swim before catching the fish?
(b) At what time will this occur? (Hint: Graphing may help you
visualize this problem.)
пип
Answers
Answer:
a) 294.55 meters
b) 16.4 seconds
Explanation:
The suggested graph shows ...
(a) The barracuda swims 294.55 meters before the it catches the fish.
__
(b) This occurs after about 16.4 seconds.
Metal reactivity
a. increases
b. decreases
from left to right in the periodic table.
C. stays the same
d. can increase or decrease depending on the
element
Answers
Answer:
i think its A
Explanation:
015
10.0 points
A car traveling in a straight line has a velocity
of 5.25 m/s at some instant. After 4.27 s, its
velocity is 9.38 m/s.
What is its average acceleration in this time
interval?
Answer in units of m/s.
Answers
Divide the change in speed by the given time:
a = (9.38 m/s - 5.25 m/s) / (4.27 s) ≈ 0.967 m/s²
The distance, x, covered by a particle in time, t, is given as x=a +bc+ct^2 +dt^3
.find the dimension of the constants a, b, c and d
Answers
Answer:
[tex]a[/tex] has units of distance
[tex]b[/tex] has units of distance over time
[tex]c[/tex] has units of distance over [tex]time^2[/tex]
[tex]d[/tex] has units of distance over [tex]time^3[/tex]
Explanation:
Since the expression for the distance is:
[tex]x = a+b\,t+c\,t^2+d\,t^3[/tex]
then:
[tex]a[/tex] has units of distance
[tex]b[/tex] has units of distance over time
[tex]c[/tex] has units of distance over [tex]time^2[/tex]
[tex]d[/tex] has units of distance over [tex]time^3[/tex]
because we are supposed to be able to add all of the terms and get a distance. So the products on each term that contains factors of time (t) should be cancelling those time units with units in the denominator of the multiplicative constant s that accompany them.
The coordinate of a particle in meters is given by x(t)=1 6t- 3.0t , where the time tis in seconds. The
particle is momentarily at rest at t is:
Select one:
a. 9.3s
b. 1.3s
C. 0.75s
d.5.3s
e. 7.3s
Answers
Complete question:
The coordinate of a particle in meters is given by x(t)=1 6t- 3.0t³ , where the time tis in seconds. The
particle is momentarily at rest at t is:
Select one:
a. 9.3s
b. 1.3s
C. 0.75s
d.5.3s
e. 7.3s
Answer:
b. 1.3 s
Explanation:
Given;
position of the particle, x(t)=1 6t- 3.0t³
when the particle is at rest, the velocity is zero.
velocity = dx/dt
dx /dt = 16 - 9t²
16 - 9t² = 0
9t² = 16
t² = 16 /9
t = √(16 / 9)
t = 4/3
t = 1.3 s
Therefore, the particle is momentarily at rest at t = 1.3 s
An rescue airplane flies horizontally over level terrain with a speed of 251 m/s and at an altitude of 2.73 km. The airplane releases one package. (a) How far does the package travel horizontally between its release and its impact on the ground? Ignore air resistance. m (b) If the pilot maintains the airplane's original course, altitude, and speed, where will the plane be when the package hits the ground? Ignore air resistance. behind the bomb directly above the bomb ahead of the bomb (c) The package hits the target seen in the airplane's telescopic "bombsight" at the moment of the package's release. At what angle from the vertical was the bombsight set? ° from the vertical
Answers
Answer:
a) x = 5923.6 m , b) directly above the bomb ahead of the bomb,
c) θ = 65.3º
Explanation:
This is a projectile launching exercise
a) As the package is released from the plane at its same speed, vox 0 251 m / s, they indicate that the plane flies horizontally, therefore the initial vertical speed is zero, let's calculate the time of arrival to the ground
y = y₀ + [tex]v_{oy}[/tex] t - ½ g t²
to the ground y = 0
0 = y₀ + 0 - ½ g t²
t = √ (2y₀ / g)
t = √ (2 2730 / 9.8)
t = 23.60 s
the horizontal distance traveled is
x = v₀ₓ t
x = 251 23.6
x = 5923.6 m
b) since there is no acceleration on the x-axis and it is indicated that the air resistance is zero in part it must be just below the plane
directly above the bomb ahead of the bomb
c) for this part we can use trigonometry
since the angle with respect to the vertical is requested, the opposite leg is the horizontal distance
tan θ = x / y
θ = tan⁻¹ (x / y)
θ = tan⁻¹ (5923.6 / 2730)
θ = 65.3º
A car in a roller coaster moves along a track that consists of a sequence of ups and downs. Let the
x axis be parallel to the ground and +y axis point upward. In the time interval from t=0 to t=4 s the trajectory of the caralong a certain section of the track is given by
→r=A(1 m/s)t^i+A((1 m/s3)t3−6(1 m/s2)t2)^j, where A is positive dimensionless constant.a. At t=2.0s is the roller coaster car ascending or descending?
b. Derive a general expression for the speed v of the car. Make sure that your expression would give the correct value for the speed (in m/s), but you don't need to put in anything explicitly about units (unlike what you see in the original expressionfor →r. Express your answer in meters per second in terms of A and t.
Answers
Answer:
Explanation:
r=A(1 m/s)t^i+A((1 m/s3)t3−6(1 m/s2)t2)^j,
dr / dt = A i + (3 A t² - 6 x 2 t ) j
At t = 2
dr / dt = A i + (12A - 24 A ) j = A i - 12 A j .
Since slope is negative so velocity is downwards , hence it is descending
b ) velocity = dr / dr = . A i + (3 A t² - 6 x 2 t ) j
Vx = A
Vy = 3A t² - 12 t
The items in a mixture can be returned to their original form.
True
False
Answers
Answer:
I believe it is true
If not, pls let me know. :)
3. How does the resistance of the light bulbs differ when the bulbs are cold and when the bulbs are hot? Why do you think this happens?
Answers
Answer:
the hot bulb will have high resistance to the flow of current. While the cold bulb will have a low resistance to the flow of current.
Explanation:
A conductor that does not obey Ohm's law is described as non - ohmic. An example is a filament lamp. It glows as the current passes through it.
How does the resistance of the light bulbs differ when the bulbs are cold and when the bulbs are hot ?
The resistance of the light bulbs increase gradually as its temperature is increased.
So, the hot bulb will have high resistance to the flow of current. While the cold bulb will have a low resistance to the flow of current.
Because the resistance of an impure metal wire is greater than the resistance of a pure metal wire of the same dimension.
What is the acceleration at the apex of a vertical up and down problem?
Answers
Answer:
The correct option is +9.81 m/s²
Explanation:
The acceleration of a vertical (up and down) plane depends on if the object is going up or down. Acceleration due to gravity is 9.81 m/s/s or 9.81 m/s². When an object falls/comes down (vertically) without interference, the acceleration of such an object is same as acceleration due to gravity (+9.81 m/s²). However, when an object is thrown/goes up, the acceleration of such objects goes against the gravity (of earth) hence the acceleration is -9.81 m/s².
At the top/apex of a vertical up and down problem, the object will be pulled back down (because of gravity) and hence, it's acceleration becomes +9.81 m/s² (changing from negative while coming up to positive).
What must be the units for the gravitational constant G in order for gravitational force to have units of newtons?
m3/(kg⋅s)
m3/(kg2⋅s2)
m3/(kg2⋅s)
m3/(kg⋅s2)
Answers
Answer:
m³/(kg⋅s²)
Explanation:
Hello.
In this case, since the involved formula is:
[tex]F=G*\frac{m_1m_2}{r^2}[/tex]
By writing a dimensional analysis with the proper algebra handling, we obtain:
[tex]N[=]G*\frac{kg*kg}{m^2}\\ \\kg*\frac{m}{s^2}[=]G *\frac{kg*kg}{m^2}\\\\G[=]\frac{kg*m*m^2}{kg^2*s^2}\\ \\G[=]\frac{m^3}{kg*s^2}[/tex]
Thus, answer is:
m³/(kg⋅s²)
Note that the [=] is used to indicate the units of G.
Best regards
The electric field everywhere on the surface of a thin, spherical shell of radius 0.710 m is of magnitude 936 N/C and points radially toward the center of the sphere. (a) What is the net charge within the sphere's surface? nC (b) What is the distribution of the charge inside the spherical shell? The positive charge has an asymmetric charge distribution. The negative charge has an asymmetric charge distribution. The positive charge has a spherically symmetric charge distribution. The negative charge has a spherically symmetric charge distribution.
Answers
Explanation:
Given that,
Radius of a spherical shell, r = 0.71 m
Electric field that points radially toward the center of the sphere is 936 N/C
(a) Let q is the net charge within the sphere's surface. Using Gauss's law, we can find it :
[tex]\dfrac{1}{q}\times \epsilon_0=EA\\\\q=\dfrac{\epsilon_0}{EA}\\\\q=\dfrac{8.85\times 10^{-12}}{936\times \pi \times (0.71)^2}\\\\q=5.97\times 10^{-15}\ C[/tex]
(b) Gauss's law is used to find the amount of charge enclosed within a surface. It doesn't say anything about the distribution of the charge inside the spherical shell.
A 25.0 kg bag of peat moss sits in the back of a flatbed truck, driving up a hill. The bag experiences a 225N normal force. The maximum acceleration the truck can have so the bag does not slip is 2.40 m/s2 . Calculate the (a) angle of the hill relative to horizontal and (b) coefficient of static friction between the bag and the truck. (c) The truck is now travelling on level ground at constant speed. The sand bag is tossed forward sliding along the truck bed with a horizontal speed of 2.55 m/s. If the coefficient of kinetic friction is 0.350, how far does the bag slide before coming to rest
Answers
Answer:
a
[tex]\theta = 23.32^o [/tex]
b
[tex] \mu_s = 0.27 [/tex]
c
[tex] s = 0.948 \ m [/tex]
Explanation:
From the question we are told that
The mass of the bag is [tex]m_b = 25.0 \ kg[/tex]
The normal force experienced is [tex]F_n = 225 \ N[/tex]
The maximum acceleration of the bag is [tex]a = 2.40 \ m/s^2[/tex]
Generally this normal force experience by the bag is mathematically represented as
[tex]F_n = mg cos \theta[/tex]
=> [tex]225 = (25 * 9.8) cos \theta[/tex]
=> [tex] 0.9183 = cos \theta[/tex]
=> [tex]\theta = cos^{-1}[0.9183][/tex]
=> [tex]\theta = 23.32^o [/tex]
Generally for the bag not to slip , it means that the frictional force is equal to the sliding force
[tex]F_f = F_s[/tex]
Hence [tex]F_f [/tex] is mathematically represented as
[tex]F_f = \mu_s * F_n [/tex]
While [tex]F_s [/tex] is mathematically represented as
[tex]F_s = m * a [/tex]
So
[tex] \mu_s * F_n = m * a [/tex]
=> [tex] \mu_s * 225 = 25 * 2.40 [/tex]
=> [tex] \mu_s = 0.27 [/tex]
Generally from the workdone equation we have that
[tex]KE_f - KE_i = W_f[/tex]
Here [tex]W_f[/tex] is the work done by friction which is mathematically represented as
[tex]W_f = m * g * \mu_k * s[/tex]
Here s is the distance covered by the bag
[tex]KE_f[/tex] is zero given that velocity at rest is zero
and
[tex]KE_i = \frac{1}{2} * m* v_i^2[/tex]
so
[tex] \frac{1}{2} * m* v_i^2 = m * g * \mu_k * s [/tex]
=> [tex] \frac{1}{2} * v_i^2 = g * \mu_k * s [/tex]
substituting 2.55 m/s for v_i and 0.350 for \mu_k we have that
[tex] \frac{1}{2} * 2.55^2 = 9.8 * 0.350 * s [/tex]
=> [tex] s = 0.948 \ m [/tex]
a piano dropped from a plane in the air. As it falls, upward force of air resistance gets greater as the piano picks up speed eventually, the resistance forces equal to the downward force of gravity. the piano now
Answers
Answer:
When the piano is dropped, the first force acting on it will be the gravitational force, that accelerates the piano at 9.8m/s^2 downwards.
As the piano accelerates, the velocity increases, now appears other force, the air resistance, that opposes to the motion of the piano.
As the velocity of the piano increases, also does the force that the air applies on the piano.
There is a point where the velocity of the piano is such that the air resistance is equal in magnitude, but in the opposite direction, to the force of gravity.
Then the net force on the piano is zero, which means that there is no acceleration, so the piano will keep falling down at constant velocity after this point.
A student pushes a 0.5 kg lab cart with a force of 3 Newtons. Determine the acceleration of the cart in
m/s2.
Answers
Answer:
The acceleration is 6 [tex]\frac{m}{s^2}[/tex]
Explanation:
Use Newton's second law to solve the problem:
[tex]Force = mass\,\,*\,\,acceleration\\3\,N = 0.5\,\,kg\,*\,a\\a = \frac{3}{0.5} \frac{m}{s^2} \\a=6\,\,\frac{m}{s^2}[/tex]
the acceleration of the cart given its mass and force exerted is 6m/s².
Given the data in the question;
Mass of lab cart; [tex]m = 0.5kg[/tex]
Force applied; [tex]F = 3N[/tex]
Acceleration of cart; [tex]a = \ ?[/tex]
To determine the acceleration of the cart, we use the expression from Newton's second law of Motion:
[tex]F = m* a[/tex]
Where F is force exerted, m is mass and a is acceleration
We substitute our given values into the equation
[tex]3N = 0.5kg \ * \ a\\\\3kgm/s^2 = 0.5kg\ *\ a\\\\a = \frac{3kgm/s^2}{0.5kg} \\\\a = 6m/s^2[/tex]
Therefore, the acceleration of the cart given its mass and force exerted is 6m/s².
Learn more: https://brainly.com/question/15839425
give five characteristics of a metal
Answers
Answer:
are good conductors of heat,are you used to make doors utensils e.t.c.
PLEASE HELP!!! THANKS I GIVE BRAINLIEST !!A student examines the effect of the number of D batteries in a closed circuit on the brightness of a light bulb. In the experiment, four circuits were built with 1, 2, 3, and 4 batteries respectively. For each trial, the brightness of the light bulb was measured using a light meter. Which variable should the student keep constant
Answers
Answer:
The batteries
Explanation:
What the density of pure water?
a)1.0kg/l
b) 0.1 kg/l
c)100kg/l
d)10kg/l
Answers
Answer:
a) 1.0(kg/l) = 1 [kg/Lt]
Explanation:
The density of pure water is taken as 1000 [kg/m3], now we must convert the units to (kg/liters)
[tex]1000[\frac{kg}{m^{3}}]*\frac{1m^{3} }{1000Lt} = 1000[\frac{kg}{lt} ][/tex]
A car is traveling north with a velocity of 18.1 m/s. Find the velocity of the car after 7.50 seconds if the acceleration is 2.4 m/s^2. *
Answers
Vx = V0x + Ax*t
Vx = 18.1 + 2.4t
Let’s take time as 7.50 seconds:
Vx = 18.1 + 2.4*7.50
Vx = 18.1 + 18 = 36.1 m/s
Then, the final velocity of the car is 36.1 m/s.
Atoms are spherical in shape. Therefore, the Pt atoms in the cube cannot fill all the available space. If only 74.0 percent of the space inside the cube is taken up by Pt atoms, calculate the radius in picometers of a Pt atom. The mass of a single Pt atom is 3.240 × 10−22 g. [The volume of a sphere of radius r is (4/3) πr3. The volume of a cube is l3, where l is the length of a side. Avogadro's number is 6.022 × 1023.]
Answers
Answer:
A)6.6×10^22atoms of Pt in the cube
B)1.4×10^-8m
Explanation:
(a) Calculate the number of Pt atoms in the cube.
an edge length of platinum (Pt) = 1.0 cm.
Then Volume= 1.0 cm×1.0 cm×1.0 cm=1cm^3
Then we have volume of the cube as 1cm^3
Given:
The density Pt = 21.45 g/cm3
the mass of a single Pt atom =3.240 x 10^-22 g
Then with 1atom of the platinum element, we can calculate the number of Pt atoms in the cube as
Density of pt/mass of a single Pt atom
=(21.45 /=3.240 x 10^-22)
=6.6×10^22atoms of Pt in the cube
B)Volume of cube V=4/3πr^3
V= 4/3 ×π×r^3
V= 4.19067r^3
r^3= V/4.19067
But volume is not total volume but just 74% of it, then With 74% of the space inside the cube is taken up by Pt atoms, then we need to find 74% of volume of the cube which is 1cm^3
74/100 ×1= 0.74cm^3
Then our new volume V is 0.74cm^3
r^3=0.74/4.19067×6.620 x 10^22
r^3=2.6674×10^-24
r= 3√2.6674×10^-24
r=1.4×10^-8m
A physics student rolls a ball down a hill with an initial velocity of 2.5m/s. If it accelerates down the hill at a rate of 6.3m/s^2, How long will it take to reach the bottom of the hill, 150 meters away?
Please help!!!
Answers
I'll assume you mean the hill is 150 m along its slope, since that's the direction the ball rolls.
The distance x it rolls after time t is
x = (2.5 m/s) t + 1/2 (6.3 m/s²) t²
Set x = 150 m and solve for t :
150 m = (2.5 m/s) t + 1/2 (6.3 m/s²) t²
(3.15 m/s²) t² + (2.5 m/s) t - 150 m = 0
t = 1/2 [(-2.5 m/s) + √((2.5 m/s)² - 4 * (3.15 m/s²) * (-150 m))]
(Use the quadratic formula, and take the positive root)
t ≈ 6.52 s
Calculate the distance moved by a runner who runs with a speed of 5 km/h for a period of 1.5 hours.
7.5 km
10 km
2.5 km
5 km
Answers
Answer:
7.5 km
Explanation:
h5 per hour means that he traveled 5 km in 1 our. And then half of the hour, which means half an hour 5 km which is 2.5.
5 + 2.5 = 7.5
or just 1.5 x 5 = 7.5
Sandra shot a rocket so that it moved with an initial velocity of 9.81 m/s straight upward. The rocket leaves and returns to ground level. What is the total time the rocket was in the air before it strikes the ground? (Ignore air resistance.
Answers
Answer:
2.00 seconds.
Explanation:
The following data were obtained from the question:
Initial velocity (u) = 9.81 m/s
Total time in air (T) =..?
Next, we shall determine the time taken for the rocket to reach it's maximum height. This can be obtained as follow:
Note: At maximum height, the final velocity is zero.
Initial velocity (u) = 9.81 m/s
Final velocity (v) = 0 m/s
Acceleration due to gravity (g) = 9.8 m/s²
Time to reach the maximum height (t) =.?
v = u – gt (since the rocket is going against gravity)
0 = 9.81 – 9.8t
Rearrange
9.8t = 9.81
Divide both side by 9.8
t = 9.81/9.8
t = 1.00 s
Therefore the time taken to reach the maximum height is 1.00 second.
Finally, we shall determine the total time spent by the rocket in the air as follow:
Time to reach the maximum height (t) = 1.00 s
Total time in air (T) =..?
T = 2t
T = 2 × 1.00
T = 2.00 s
Therefore, the total time spent by the rocket in the air is 2.00 seconds.
Imagine you are standing in the hallway. If the gravitational field strength were to suddenly double, what would change?
Answers
Answer:
If the gravitational field strength were to suddenly double while I was standing in the hallway, I'd suddenly feel like heavier.
Explanation:
The weight of a matter is given as the production between it's mass and the acceleration of gravity.
Mathematically, this is expressed as:
m x g that is m.g or mg.
Where m = mass and
g = gravity.
If the value of gravity was to double then weight (w) will become
w = m×2g
Assume for a moment that before the increase m was 5 and g was 5, it means that w would be 25.
But if g doubled, then we would have
w = 5 x 10
which is equals 50.
Thus, with the increase in gravitational field strenght, I'd find myself exerting more effort to stand upright, jump or run.
Cheers!
The kinetic energy of an object depends upon its speed and its... what?
Answers
Answer:
Mass
Explanation:
kinetic energy (KE) is equal to half of an object's mass (1/2*m) multiplied by the velocity (speed) squared.
An object experiences a net acceleration to the left. Which of the following statements about this object are true? There may be more than one true statement below. Group of answer choices If there is one force to the left on the object, there must at least be one force on it to the right as well. There must be two or more forces on the object. The object cannot be slowing down. The net force on the object is to the right. If there is one force down on the object, there must at least be one force up on it as well. If the mass of the object was doubled, it would experience an acceleration of half the magnitude. Assume the forces on the object are unchanged. The object might be in equilibrium. If the mass of the object were halved, it would experience an acceleration to the right. Assume the forces on the object are unchanged. The net force on the object is to the left.
Answers
Answer:
When an object experiences acceleration to the left, the net force acting on this object will also be to the left.If the mass of the object was doubled, it would experience an acceleration of half the magnitudeExplanation:
When an object experiences acceleration to the left, the net force acting on this object will also be to the left.
From Newton's second law of motion, the acceleration of the object is given as;
a = ∑F / m
a = -F / m
The negative value of "a" indicates acceleration to the left
where;
∑F is the net force on the object
m is the mass of the object
At a constant force, F = ma ⇒ m₁a₁ = m₂a₂
If the mass of the object was doubled, m₂ = 2m₁
a₂ = (m₁a₁) / (m₂)
a₂ = (m₁a₁) / (2m₁)
a₂ = ¹/₂(a₁)
Therefore, the following can be deduced from the acceleration of this object;
When an object experiences acceleration to the left, the net force acting on this object will also be to the left.If the mass of the object was doubled, it would experience an acceleration of half the magnitude