Answer:
[tex]a=0.96\ m/s^2[/tex]
Explanation:
Given that,
Mass of cabinet, m = 45 kg
Two horizontal force of 25 newtons and 18 newtons respectively in the same direction.
When the forces are acting in same direction, the net force is equal to the sum of forces i.e.
F = 25 N + 18 N = 43 N
Let a is the aceleration of the cabinet
So,
F = ma
[tex]a=\dfrac{F}{m}\\\\a=\dfrac{43}{45}\\\\a=0.96\ m/s^2[/tex]
So, the acceleration of the cabinet is [tex]0.96\ m/s^2[/tex].
Question 1 of 10
What is a neutron
A. A particle inside the nucleus that has no charge
B. A positive particle inside the nucleus
O C. A negative particle outside the nucleus
O D. A negative particle inside the nucleus
Answer:
The neutron is a subatomic particle, symbol
n
or
n0
, with no electric charge and a mass slightly greater than that of a proton. Protons and neutrons constitute the nuclei of atoms. Since protons and neutrons behave similarly within the nucleus, and each has a mass of approximately one atomic mass unit, they are both referred to as nucleons.[6] Their properties and interactions are described by nuclear physics.
Explanation:
pls mark me as BRAINLIEST
which of the following best describes the function of a generator?
[tex]\mathfrak{\huge{\pink{\underline{\underline{AnSwEr:-}}}}}[/tex]
Actually Welcome to the Concept of the Mechanics.
So basically here, we know the fact that,
A Generator is opposite of the motor that Converts the Mechanical Energy to Electrical Energy.
C. Concerts Mechanical Energy to Electrical Energy.
Iron man wears an awesome ironsuit.He is flying over high current carrying wire. Will he be affected?
Answer:
According to super hero logic , nothing will happen to him.
But according to science , yes he will get current shock but good news is that he wouldn't get elected until he is in contact with the wires.
He may / may not be affected but his suit will be damaged for sure as it is made of metal.
HOPE THIS HLEP AND PLSSSSS MARK AS BRAINLIEST AND THNXX :)
An alternating source drives a series RLC circuit with an emf amplitude of 6.34 V, at a phase angle of +32.6°. When the potential difference across the capacitor reaches its maximum positive value of +5.08 V, what is the potential difference across the inductor (sign included)?
Answer:
V=-8.35v
Explanation:
Pls see attached file
hey help me plzzzzz i will mark brainliest
Answer:
The answer to your question is given below.
Explanation:
Mechanical advantage (MA) = Load (L)/Effort (E)
MA = L/E
Velocity ratio (VR) = Distance moved by load (l) / Distance moved by effort (e)
VR = l/e
Efficiency = work done by machine (Wd) /work put into the machine (Wp) x 100
Efficiency = Wd/Wp x100
Recall:
Work = Force x distance
Therefore,
Work done by machine (wd) = load (L) x distance (l)
Wd = L x l
Work put into the machine (Wp) = effort (E) x distance (e)
Wp = E x e
Note: the load and effort are measured in Newton (N), while the distance is measured in metre (m)
Efficiency = Wd/Wp x100
Efficiency = (L x l) / (E x e) x 100
Rearrange
Efficiency = L/E ÷ l/e x 100
But:
MA = L/E
VR = l/e
Therefore,
Efficiency = L/E ÷ l/e x 100
Efficiency = MA ÷ VR x 100
Efficiency = MA / VR x 100
What must be the tension of a string 202 cm long and 3 x 10-2 Kg / m per unit length (µ) for it to emit a sound of 78 Hz?
Answer:
2980 N
Explanation:
The fundamental frequency of a string is:
f₁ = √(T/µ) / (2L)
78 Hz = √(T / (0.03 kg/m)) / (2 (2.02 m))
315.12 m/s = √(T / (0.03 kg/m))
99,300 m²/s² = T / (0.03 kg/m)
T = 2980 N
A Rac variant, in which the residue at position 61 was replaced with an alanine (Rac61A), was synthesized. Wild-type Rac and Rac61A were incubated separately with VopC. To obtain data to support that VopC modifies Rac at residue 61, the samples should be analyzed for the presence of which compound?
Answer:
Ammonia, [tex](NH_{3})[/tex]
Explanation:
If a Rac variant, in which the residue at position 61 was replaced with an alanine (Rac-61A), was synthesized. Also, Wild-type Rac and Rac-61A were incubated separately with VopC.
In order to obtain data to support that VopC modifies Rac at residue 61, the samples should be analyzed for the presence of Ammonia ([tex]NH_{3}[\tex]) compound.
In chemistry, the Vanadyl Phthalocyanine (VopC) comprises of a catalytic domain that activates host cells GTPase Rac irreversibly, through the deamide side chain of the residue existing at position 61.
Also, we know that a deamidation reaction gives off Ammonia, [tex](NH_{3})[/tex] and as such when ammonia is present in the sample containing Wild-type Rac but not in Rac-61A, this simply proves or provide the data to support that VopC modifies Rac at residue 61.
Additionally, deamidation can be defined as the chemical conversion (hydrolysis) of an amide functional group such as glutamine, asparagine, in a polypeptide to another functional group such as glutamic acid or isoaspartic acid respectively by treating it with a strong acid (deamidate, transamidase).
In a Little League baseball game, the 145 g ball reaches the batter with a speed of 15.0 m/s. The batter hits the ball, and it leaves his bat with a speed of 20.0 m/s in exactly the opposite direction. What is the magnitude of the impulse delivered by the bat to the ball
Answer: 5.075Ns
Explanation:
Given the following :
Mass of ball = 145g
Initial Speed of ball = 15m/s
Final speed of ball when hit by the batter = - 20m/s ( Opposite direction)
The impulse of a body is represented using the relation:
Force(f) * time(t) = mass (m) * (final Velocity(V) - initial velocity(u))
Therefore, using:
m(v - u) = impulse
Mass of ball = 145 / 1000 = 0.145kg
Impulse = 0.145(- 20 - 15)
Impulse = 0.145(-35)
Impulse = 5.075Ns
PLEASE HELP. i will mark brainliest. and do more!
I would suggest an easier route. So for example, I would research light energy, where it comes from, and what causes it. I won't be able to do the work for you though.
A 1550 kg car is initially at rest on level ground when the engine does 450 000 J of work on it. If all of the work is converted into kinetic energy, what is the final speed of the car?
Answer:
24.1 m/s
Explanation:
Work = change in KE = [tex]KE_{f} - KE_{i}[/tex]
KE initial is 0 because the initial velocity is 0. Therefore, work is equal to final kinetic energy. With that:
[tex]W = KE_{f} = \frac{1}{2}mv_{f}^{2}[/tex]
With some algebraic manipulations,
[tex]v_{f} = \sqrt{\frac{2W}{m} } = \sqrt{\frac{2*450000 J}{1550 kg} } = 24.1 m/s[/tex]
A submarine is 58.8 m from a whale. The sub sends out a sonar ping to locate the whale. The speed of sound underwater is 1520 m/s. How much time does it take for the sound wave to travel to the whale and back? (Unit = s)
Answer:
0.08
Explanation:
this problem assume that both of whale and submarine are in rest position or in constant linier motion in same direction and same speed.
The sound will travel from Submarine to the whale and back again to submarine. so the time will be like this
t = 2d/v
t = 2*58.8/1520
t = 117.6/1520
t = 0.077368 s
t ≈ 0.08 s (less then 1 s)
Answer:
0.0774s
Explanation:
The formula to find the time of an echo is time = (2)(distance) / (velocity)
So plug in the numbers to the formula
(2)(58.8) / (1520)
= 0.0774s
A vertical bar consists of three prismatic segments A1, A2, and A3 with cross-sectional areas of 6000 mm2 , 5000 mm2 , and 4000 mm2 , respectively. The bar is made of steel with E 5 200 GPa. Calculate the displacements at points B, D
Answer and Explanation:
For computing the displacement at point B and D we need to determine the following calculations
[tex]P_Net = P_C + P_E + P_B[/tex]
= 250 + 350 - 50
= 550 N
Now the deflection for bar AB is
[tex]\delta_{AB} = \frac{PL_{AB}}{AE} \\\\ = \frac{550 \times 500}{6,000 \times 200 \times 10^{3}}[/tex]
[tex]= 2.292 \times 10^{-4} mm[/tex]Now for bar BC it is
[tex]\delta_{BC} = \frac{PL_{BC}}{AE} \\\\ = \frac{(550 + 50) \times 250}{5,000 \times 200 \times 10^{3}} \\\\ = 1.5 \times 10^{-04} mm[/tex]
And for bar CD it is
[tex]\delta_{CD} = \frac{PL_{CD}}{AE} \\\\ = \frac{(550 -250 + 50) \times 250}{5,000 \times 200 \times 10^{3}} \\\\ = 0.875 \times 10^{-4} mm[/tex]
Now the displacement is as follows
For B
2.292 × 10^{-4} mm
For D, it is
[tex]= 2.292 \times 10^{-4} + 1.5 \times 10^{-4} + 0.875 \times 10^{-4} mm \\\\ = 4.667 \times 10^{-4} mm[/tex]
We simply applied the above formulas for determining the displacements at points B, D and the same is to be considered
How much heat is required to convert 500g of liquid water at 28°C into steam at 150 °C? Take the specific heat capacity of water to be 4183 J/Kg °C and the latent heat of vaporization to be 2.26 × 10^6 J/Kg.
Answer:
Q = 1.404 × 10^(5) KJ
Explanation:
We are given:
Mass;m = 500 g = 0.5kg
Temperature 1;T1 = 28 °C
Temperature:T2 = 150 °C
Specific heat capacity;c_p = 4183 J/Kg °C
Latent heat of vaporization;L = 2.26 × 10^(6) J/Kg.
The heat energy needed is given by;
Q = sensible heat energy + Latent heat
Formula for sensible heat is;
Sensible heat energy = mc(t2 - t1)
Formula for Latent heat is ;
Latent heat = mL
Thus:
Q = mc(t2 - t1) + mL
Q = m[c(t2 - t1) + L]
Q = 0.5((4183(159 - 28) + (2.26 × 10^(6)))
Q = 1.404 × 10^(8) J = 1.404 × 10^(5) KJ
48 POINTS
If a transverse wave vibrates up and down three times each second, and the distance
between wave crests is 2 meters, what is the speed of the wave?
1.5 m/s
0.66 m/s
5 m/s
O 6 m/s
Answer:
6 m/s
Explanation:
s=2*3
=6
So, the speed of trqnsverse wave is 6 m/s.
Answer:
6 m/s
Explanation:
2 x 3=6
define the unit of current
Answer:
The unit of current is defined as the flow of 1 coulomb of charge in one second
pleasee help me helppppp
light travels from a region of air into a region of air making an angel of incidence of 60 degree which of the following best describes the path of the light as it moves into the air
a)the light will bend toward normal
b)the light will bend away from the normal
c) the light will continue without bending
d)the light will move in some manner not determined by the information here
Answer:
B
Explanation:
If the light is pointing down ,and then hits a surface of a 60 degree angle, then the direction will be away from the light since the light is touching a surface that is flat of a 60 degree angle then it will reflect off of it Perpendicular.
An open container contained 150g of lead carbonate. After heating for 5 hours, the contents of the container were measured to have a mass of 98g.
How much mass was “lost” from the container? Where did it go?
Answer:
52 g was lost as carbon dioxide to the atmosphere
Explanation:
When lead carbonate is heated, it decomposes into two components:
1. Lead oxide
2. carbon dioxide
While the lead oxide remains a yellow solid in the heating container, the carbon dioxide escapes into the atmosphere as a gas. The equation of the reaction is as below:
[tex]PbCO_3_{(s)} --> PbO_{(s)} + CO_2_{(g)}[/tex]
Hence, if a 150 g lead oxide is heated in a container and the final mass is 98 g, it means 52 g (150 - 98) of the total mass has been lost as carbon dioxide to the atmosphere.
Find the mass. 10 points. Will give brainliest.
Answer:
3.94 kgExplanation:
Given,
Force ( f ) = 30 N
Acceleration(a) = 7.6 m/s
Now, Let's find the mass of the ball
Using the Newton's second law of motion:
We get:
[tex]force \: = mass \: \times acceleration[/tex]
plug the value
[tex]30 \: = m \: \times 7.6[/tex]
Use the commutative property to reorder the terms
[tex] 30 = 7.6 \: m[/tex]
Swap the sides of the equation
[tex]7.6m = 30[/tex]
Divide both sides of the equation by 7.6
[tex] \frac{7.6 \: m}{7.6} = \frac{30}{7.6} [/tex]
Calculate
[tex]m = 3.94 \: kg[/tex]
Hope this helps..
Best regards!!
Answer:
[tex]\displaystyle \boxed{\mathrm{3.95 \: kg }}[/tex]
Explanation:
[tex]\mathrm{force \: (N) = mass \: (kg) \times acceleration \: (m/s^2)}[/tex]
[tex]\mathrm{force = 30N}[/tex]
[tex]\mathrm{acceleration = 7.6 \: m/s^2 }[/tex]
[tex]\mathrm{Find \: the \: mass.}[/tex]
[tex]\mathrm{30 = m \times 7.6}[/tex]
[tex]\displaystyle \mathrm{m =\frac{30}{7.6} }[/tex]
[tex]\displaystyle \mathrm{m = 3.947... }[/tex]
An object weighs 0.250 kgf in air, 0.150 kgf in water and 0.125 kgf in an oil.
Find out the density of the object and the oil.
Answer: Upthrust = Weight - Apparent weight
= 0.250 kgf - 0. 150 kgf
= 0.100 kgf
Density = mass / volume
volume = mass / density
= 0.100 kg / (1000 kg / m³)
= 0.0001 m³
density of object = mass / volume
= 0.250 kg / 0.0001 m³
= 2500 kg / m³
upthrust of oil = Weight - Apparent weight
= 0.250 kgf - 0.125 kgf
= 0.125 kgf
density = mass / volume
= 0.125 kg / 0.0001 m³
= 1250 kg/m³
density of the object = 2500 kg / m³
density of oil = 1250 kg / m³
Four identical point charges (+6.0 nC) are placed at the corners of a rectangle which measures 6.0 m×8.0 m. If the electric potential is taken to be zero at infinity, what is the potential at the geometric center of this rectangle
The electric potential at the geometric center of this rectangle is determined as 43.2 V.
Potential at the center of the rectanglePotential at the center of the rectangle is calculated as follows;
Let the distance from each corner to the center = xLet the length = aLet the breadth = bDistance from each corner to the center is calculated as follows;
[tex]x = \sqrt{(a/2)^2 + (b/2)^2}[/tex]
Potential due to four point charges is calculated
[tex]V = \frac{kq}{x} \\\\V =4 (\frac{kq}{x} )\\\\V = 4(\frac{kq}{\sqrt{(a/2)^2 + (b/2)^2} } )\\\\V = \frac{4 \times 9\times 10^{9}\times 6\times 10^{-9}}{\sqrt{(6/2)^2 + (8/2)^2} } \\\\V = \frac{4 \times 9\times 10^{9}\times 6\times 10^{-9}}{5} \\\\V = 43.2 \ Volts[/tex]
Learn more about electric potential here: https://brainly.com/question/14306881
#SPJ2
Earthquake damage causes two rabbits to be separated from the rest of the rabbits in their large habitat. They have no way to get back to their original habitat. The two rabbits mate with each other. Over time, all the offspring in the new habitat are descendants of the original two rabbits. What are the outcomes of this situation? A.The rabbits in the new habitat will have lower genetic variation than the rabbits in the original habitat. B. The rabbits in the new habitat will have a higher risk of random genetic mutations than the rabbits in the original habitat. C. The rabbits in the original habitat have a greater likelihood of choosing an unrelated mate than the rabbits in the new habitat. D.The rabbits in the original habitat will be less likely to reproduce than the rabbits in the new habitat.
Answer:
A.The rabbits in the new habitat will have lower genetic variation than the rabbits in the original habitat.
Explanation:
If two animals of opposite sex are isolated from a larger group of animal, and made to reproduce. They will produce offspring with similar genetic makeup. If this offspring still remain isolated, and continue to interbreed within themselves for a number of consecutive generations, their offspring will all be very closely related genetically. Situations like this just as with the two rabbits in the question leads to a lower genetic variation within the offspring of the two animals.
Animals need to reproduce within a larger group in order to increase genetic variation. Increasing genetic variation reduces the risk of been sucked into a gene pool. A lower genetic variation reduces the fitness of the animals involved. It is only an advantage in cases in which the the original pair are resistant to a deadly disease. In this case all the offspring also develop this immunity. Mostly the effects of a lower genetic variation leaves negative impacts, and animals try to avoid this by preferring to interbreed with unrelated partner
Answer:
I got this and got the answer correct.
Explanation:
Hope it helps :)
explain why the term Nano is used to describe nanotechnology
Answer:
Nanotechnology means working with materials at the scale of one billionth of a metre.
Explanation:
The prefix “nano” refers to one billionth: it’s part of the scientific scale of measurement. Its science, engineering, and technology are conducted at the nanoscale, which is about 1 to 100 nanometers. Nanoscience and nanotechnology are the study and application of extremely small things. Thus, this describes why the term nano is used to describe nanotechnology.
I hope this helps :)
Answer:
Nanotechnology is technology that is about a nanometer in size, or between 1 nm and 100 nm.
Explanation:
its the sample response on edge 2021
You can chew through very tough objects with your incisors because they exert a large force on the small area of a pointed tooth. What pressure in Pa can you create by exerting a force of 360 N with your tooth on an area of 1.08 mm2
Answer:
3.33×10⁸ Pa
Explanation:
Pressure: This can be defined as the force acting normally (perpendicular) per unit area. The S.I unit of force is N/m² or Pa.
P = F/A.................. Equation 1
Where F = force created with the tooth, A = Area.
Given: F = 360 N, A = 1.08 mm² = 1.08/1000000 = 1.08×10⁻⁶m.
Substitute this values into equation 1
P = 360/1.08×10⁻⁶
P = 3.33×10⁸ Pa.
Hence the pressure created = 3.33×10⁸ Pa
Two long, straight wires are parallel and 10 cm apart. One carries a current of 2.0 A, the other a current of 5.0 A. If the two currents flow in opposite directions, what is the magnitude and direction of the force per unit length of one wire on the other
Answer:
The magnitude and direction of the force per unit length of one wire on the other is 2 x 10⁻⁵ N/m, attractive force.
Explanation:
Given;
distance between the two parallel wires, r = 10 cm = 0.1 m
current in the first wire, I₁ = 2A
current in the second wire, I₂ = 5 A
The force per unit length on each wire can be calculated as;
[tex]\frac{F}{L} = \frac{\mu_oI_1I_2}{2\pi r}[/tex]
where;
μ₀ is permeability of free space = 4π x 10⁻⁷ m/A
[tex]\frac{F}{L} = \frac{\mu_oI_1I_2}{2\pi r} \\\\\frac{F}{L} = \frac{4\pi*10^{-7}*2*5}{2\pi *0.1} \\\\\frac{F}{L} = 2 *10^{-5} \ N[/tex]/ m
The direction of the force between the two wires is attractive since the current in the two wires are in opposite direction.
Therefore, the magnitude and direction of the force per unit length of one wire on the other is 2 x 10⁻⁵ N/m, attractive force.
Two large rectangular aluminum plates of area 180 cm2 face each other with a separation of 3 mm between them. The plates are charged with equal amount of opposite charges, ±17 µC. The charges on the plates face each other. Find the flux (in N · m2/C) through a circle of radius 3.5 cm between the plates when the normal to the circle makes an angle of 4° with a line perpendicular to the plates. Note that this angle can also be given as 180° + 4°.
Answer:
Electric flux;
Φ = 30.095 × 10⁴ N.m²/C
Explanation:
We are given;
Charge on plate; q = 17 µC = 17 × 10^(-6) C
Area of the plates; A_p = 180 cm² = 180 × 10^(-4) m²
Angle between the normal of the area and electric field; θ = 4°
Radius;r = 3 cm = 3 × 10^(-2) m = 0.03 m
Permittivity of free space;ε_o = 8.85 × 10^(-12) C²/N.m²
The charge density on the plate is given by the formula;
σ = q/A_p
Thus;
σ = (17 × 10^(-6))/(180 × 10^(-4))
σ = 0.944 × 10^(-3) C/m²
Also, the electric field is given by the formula;
E = σ/ε_o
E = (0.944 × 10^(-3))/(8.85 × 10^(-12))
E = 1.067 × 10^(8) N/C
Now, the formula for electric flux for uniform electric field is given as;
Φ = EAcos θ
Where A = πr² = π × 0.03² = 9π × 10^(-4) m²
Thus;
Φ = 1.067 × 10^(8) × 9π × 10^(-4) × cos 4
Φ = 30.095 × 10⁴ N.m²/C
Two trains run in the opposite direction with speeds of v1 = 15 m / s and v2 = 20 m / s. A passenger on the first train (the one on v1) notes that train 2 takes 6 s to pass on its side. What is the length of the second train? (The passenger is supposed to be immobile looking through the window)
Answer:
210 m
Explanation:
The speed of train 2 relative to train 1 is 15 m/s + 20 m/s = 35 m/s.
It takes 6 seconds for the train to pass, so the length of the train is:
(35 m/s) (6 s) = 210 m
Complete all four parts. 15 points. Will give brainliest! Show work!
Answer:
A. 5.08 secs.
B. 10.16 secs.
C. 126.50 m.
D. 373.36 m
Explanation:
Data obtained from the question include the following:
Initial velocity (u) = 65 m/s
Angle of projection θ = 50°
A. Determination of the time taken to reach the peak.
Initial velocity (u) = 65 m/s
Angle of projection θ = 50°
Acceleration due to gravity (g) = 9.8 m/s²
Time (t) =.?
t = u•Sine θ/g
t = (65 × Sine 50) /9.8
t = 5.08 secs.
B. Determination of the total time spent by the ball in air.
Time (t) taken to reach the peak = 5.08 secs.
Total time (T) spent by the ball in air =?
T = 2t
T = 2 × 5.08
T = 10.16 secs
Therefore, the total time spent by the ball in air is 10.16 secs.
C. Determination of the maximum height.
Initial velocity (u) = 65 m/s
Angle of projection θ = 50°
Acceleration due to gravity (g) = 9.8 m/s²
Maximum height (H) =..?
H = u²•Sine² θ / 2g
H = 65² × (Sine 50)² / 2 × 9.8
H = 4225 × (Sine 50)² /19.6
H = 126.50 m
Therefore, the maximum height reached by the ball is 126.50 m.
D. Determination of the horizontal distance travelled by the ball.
Initial velocity (u) = 65 m/s
Angle of projection θ = 50°
Acceleration due to gravity (g) = 9.8 m/s²
Horizontal distance (R) =..?
R = u²•Sine 2θ / g
R = 65² × Sine (2×30) / 9.8
R = (4225 × Sine 60) / 9.8
R = 373.36 m
Therefore, the horizontal distance travelled by the ball is 373.36 m
The length, breadth and height of a box is 2m, 3m and 1m respectively. What is its volume?
Answer:
6m^3
Explanation:
volume of box= length×breadth×height
= 2×3×1
=6m^3
Answer:
i n k ok
Explanation:
A 0.675 kg mass is attached to a
spring of spring constant 72.4 N/m,
pulled, and released. What is the
period of the resulting oscillation?
(Unit = s)
Answer:
T= 0.6 secExplanation:
This problem bothers on the simple harmonic motion of a loaded spring
Given data
mass attached, m= 0-.675 kg
spring constant, k= 72.4 N/m
the period of oscillation can be solved for using the formula bellow
[tex]T= 2\pi \sqrt{\frac{m}{k} }[/tex]
Substituting the given data into the expression above we have
[tex]T= 2*3.142\sqrt{\frac{0.675}{72.4} }\\T= 6.284*\sqrt{0.0093 }\\T= 0.6[/tex]
T= 0.6 sec
Answer:
0.607
Explanation:
Trust me
Diagram 7 shows the velocity-time graph of an
object for 12 s.
When is the object at a distance of one-quarter of
the total distance travelled in the 12 s?
A 3 s
B 4 s
C 5 s
D 6 s
*The answer is D but can anyone tell me the solving steps? Or perhaps the answer is wrong?*
Answer:
t = 6
Explanation:
Displacement is equal to the area under a velocity vs time graph.
In this case, the area is a triangle. At time t, the base of the triangle is t. The height of the triangle can be found with similar triangles:
h / t = 8 / 12
h = ⅔ t
So the distance traveled at time t is:
d = ½ (t) (⅔ t)
d = ⅓ t²
The distance traveled at time 12 is:
D = ½ (12) (8)
D = 48
We want to find when d = D/4.
d = D/4
⅓ t² = 48/4
⅓ t² = 12
t² = 36
t = 6
Alternatively, since the acceleration is constant here, we could use a constant acceleration equation.
Δx = v₀ t + ½ at²
Given v₀ = 0 m/s and a = ⅔ m/s²:
Δx = (0) t + ½ (⅔) t²
Δx = ⅓ t²
When t = 12, Δx = 48.
⅓ t² = 48/4
t = 6