Answer:
eˣ-8.
Explanation:
(eˣ-8x+7)'=eˣ-8.
A 3.42 kg steel ball strikes a massive wall at 14.3 m/s at an angle of α = 61.5° with respect to the plane of the wall. It bounces off with the same speed and angle as seen in the figure.
If the ball is in contact with the wall for 0.207 s, what is the average force exerted on the ball by the wall?
The magnitude of the average force exerted on the ball by the wall is 225.469 N.
What is impulse?The change in momentum is equal to the product of impact force applied while colliding and time for that impact.
Impulse F. t = m (Vf -Vi)
where, Vf is the final velocity and Vi is the initial velocity.
Given is a 3.42 kg steel ball strikes a massive wall at 14.3 m/s at an angle of α = 61.5° with respect to the plane of the wall. It bounces off with the same speed and angle. The ball is in contact with the wall for 0.207 s
Substitute the values into the expression, we get
Impulse = 2 x mvcosθ
Impulse= 2 x 3.42 x 14.3 x cos 61.5°
Impulse = 46.672 kg.m/s
The impact force can be written as
F.t = I
Put the given values, we have
F = 46.672 / 0.207
F = 225.469 N
Thus, the magnitude of force exerted by the wall is 225.469 N
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A force of looN acts upwards on a mass of 50 kg, another of
80N acts on the Same Mass downwards. Calculate the acceleration?
of the body and in which direction will be th acceleration?
Answer:
a force of acts on a body of mass 50 kg for 10 sec when the first of acting on the body the body covers 80 and in the next 10 second
Answer:
a = 0.4 ms⁻²
Explanation:
• First calculate the resultant force on the body:
If we consider upwards to be the positive direction and downwards to be the negative direction, then the resultant force will be:
F = +100 + (-80)
= +20 N
The resultant force is positive, so it will act upwards.
• Next calculate the acceleration:
F = ma
a = F/m
a = 20/ 50
a = 0.4 ms⁻²
As the force is acting upwards, the acceleration will also be upwards.
According to the __________perspective, people who suffer from PTSD are classically conditioned.
According to the pavlovian perspective, people who suffer from PTSD are classically conditioned.
What is PSTD?
This is referred to as post traumatic stress disorder which is a mental disorder from a terrifying incident.
The pavlovian perspective however view this disorder as being a type of classical conditioned one.
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Determine the speed at which the medicine leaves the needle
The speed at which the medicine leaves the needle is 2.462 m/s
What is Bernoulli's theorem?When an incompressible, ideal fluid is flowing through a tube or pipe, the total energy remains constant.
p₁ /ρg + v₁²/2g +z₁ = p₂ /ρg + v₂²/2g +z₂
Where, p/ρg = pressure energy
v²/2g = kinetic energy
z = potential energy
Given is during an injection, pressure in the barrel of syringe is 1.03 atm while pressure in the needle section is 1.00 atm. Assuming the syringe lays horizontally and mass density of the liquid medicine ρ =1000 kg/m³.
The fluid is initially at rest.
Using the Bernoulli's equation, we have
v₁² - v₂²= 2 x (p₂ -p₁) / ρ
Substituting the values, we get
0 - v₂² = 2 x (1.00 -1.03) x 1.01 x 10⁵ /1000
v₂² = 6.06
v₂ = 2.462 m/s
Thus, the speed at which the medicine leaves the needle is 2.462 m/s
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A local municipality placed an advert in the local newspaper for a competition of local high schools to participate to build a security system in the community hall to avoid vandalism and break ins
We will use the design process to meet the requirements of the competition by the given explained in the main answer.
What is the competition?The initiative was taken by two or more parties operating independently to win the business of a third party by competing to hire contractors on the best terms possible.
The complete question is
"A local businessperson placed an advert in the local newspaper for a competition for girls only from different schools to participate in the competition. The competition is about drawing ideas for a shopping complex for the local community. Girls from the different schools are excited about the competition. you are a group of boys in the local schools who saw the advert and feel it is unfair for you to be excluded from the competition.
consider yourself as one of the learners in the local schools.
Explain why the project should be given to boys also who have better skills for this project than the girls do.
1.2 think about the design process skills, investigates, design, make, evaluate and communicate."
Explain how you will use the design process to meet the requirements of the competition.
[ 1.1 ]Boys should also be given the project because some of them have more artistic talent than girls do when it comes to sketching designs for local shopping centers.
Boys can also present a clear picture of how local communities can expand their shopping options.
For instance, they can encourage local merchants to offer seasonal items because doing so will benefit their businesses.
They can also suggest that retailers encourage customers to bring a friend to participate in competitions, which will be highly regarded.
[1.2] In order to fulfill the needs of the competition, I would make sure to explore how a shopping center for the neighborhood functions, for instance, by making sure I am aware of the strategies one may use to enhance his or her firm in the neighborhood.
In order to win the competition, I would make sure that I was confident in what I was saying and that I was audible enough to be heard during the competition.
Finally, I would make sure I met the competition's conditions by staying on-topic and solely addressing the issues raised in the advertisement
Hence will use the design process to meet the requirements of the competition.
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What is the energy of a photon of blue light with a frequency of 6.53 × 10¹4 Hz?
Answer:
V = 6.53 × 10^14
h = 6.626 × 10^-34
E = hV
E = (6.626 × 10^-34) (6.53 × 10^14)
E = 4.33 × 10^-19
6. A metal block increases in temperature from 15 °C to 60 °C when supplied with 13 500 J of heat energy.
a) Calculate the heat capacity of the metal.
b) Calculate the specific heat capacity of the metal, if this sample has a mass of 0.75 kg.
Answer:
Since 1 cal = 4.19 J the heat Q received by the metal is
Q = 13,500 J * 1 C / 4.19 J = 3,222 calories
S = ΔQ / (ΔT * ΔM) = 3222 cal / (45 deg C / 750 gm)
S = .095 cal / gm deg C
Note that specific heat capacity for Cu is .093 cal / gm deg C
Lester plays middle linebacker for dices football team. During one play, he made the following movements. First, he back-pedaled in the southern direction for 2.6 meters
Answer:
Explanation:
missing info .....
A bike accelerates uniformly from rest to a speed of 10 m/s over a distance of 50 m. (a) Determine the acceleration of the bike. (b) how long will take to do that?
Answer:
a) 0.2m/s
b) 5m/s
Explanation:
a) acceleration=∆v/∆t
v=10
t=50
10/50=1/5
=0.2m/s
b) time= d/s
d=50m
s=10
50/10
=5m/s
Answer:
See below
Explanation:
Average speed = (0+ 10)/2 = 5 m/s
then to cover 50 m will take 50 m / 5 m/s = 10 seconds
change in velocity/ change in time = acceleration = 10/10 = 1 m/s^2
suppose a uniform electric field 4 N/C is in the positive x-direction .
The magnitude of the electric field at x =4m on the x axis at this time 1 N/C.
Electric field at position 4 mElectric field at a given distance is calculated as follows;
E = kq/r²
E₂ = (9 x 10⁹ x q)/(2²)
E₂ = 2.25 x 10⁹q
E₂ + E₀ = 0
2.25 x 10⁹q + 4 = 0
2.25 x 10⁹q = - 4
q = -4 / (2.25 x 10⁹)
q = -1.78 x 10⁻⁹
E₄ = (9 x 10⁹ x (-1.78 x 10⁻⁹) ) / (4²)
E₄ = - 1 N/C
|E₄| = 1 N/C
Thus, the magnitude of the electric field at x =4m on the x axis at this time 1 N/C.
The complete question is below:
Suppose a uniform electric field of 4N/C is in the positive x direction. When a charge is placed and at a fixed to the origin, the resulting electric field on the x axis at x =2m becomes zero. What is the magnitude of the electric field at x =4m on the x axis at this time?
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What must the charge (sign and magnitude) of a particle of mass 1.45 g be for it to remain stationary when placed in a downward-directed electric field of magnitude 600 N/C ? Use 9.80 m/s2 for the magnitude of the free-fall acceleration.
[tex]q = -21 * 10^{-6} C[/tex]
What is Free-fall acceleration?The acceleration is constant and equal to the gravitational acceleration g which is 9.8 m/s at sea level on the Earth.
As we know that charge and its sign that remains in equilibrium is under gravity must be such that it will balance the gravitational force by electric force,
mg =qE
[tex]1.45 * 10^{-3}(9.80)=q(600)\\\\q = 21 *10^{6} C\\\\[/tex]
and its sign must be negative so that it will have upward electric force
so it is
[tex]q = -21 * 10^{-6} C[/tex]
The charge of a particle of mass is [tex]-21 * 10^{-6} C[/tex]
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Which of the following statements is false?
A Heat is a form of energy while temperature is a measure of degree of hotness
B Heart and temperature are linearly related
C Heat can be measured by physical changes due to change in temperature
D Temperature can be measured by physical changes due to change in heat
d. is wrong...the temperature cannot be measured due to physical changes
Which graph shows constant acceleration with correctly placed independent and dependent variables?
Answer:
a = (V2 - V1) / t equation for constant acceleration
if t the independent variable is placed on the abcissa or x-axis and V the dependent variable is placed on the ordinate or y-axis then the acceleration would be the slope at the point considered.
A force of 20N is applied to the end of a wire of length 5m to produce an extension of 0.20mm
calculate: the stress on the wire.
The stress on the wire is determined as 6.37 x 10⁶ N/m².
Stress on the wireThe stress on the wire is calculated as follows;
σ = F/A
where;
F is the force applied on the wireA is area of the wireLet the diameter of the wire = 2 mm
Area = πr²
Area = π(1 x 10⁻³)²
Area = 3.142 x 10⁻⁶ m²
Stress = 20/(3.142 x 10⁻⁶)
Stress = 6.37 x 10⁶ N/m²
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(i). A ball of mass 1.500 kg is attached to the end of a cord 1.50 m long. The ball moves in a horizontal circle. If the cord can withstand a maximum tension of 64.0 N, a. What is the maximum speed at which the ball can move before the cord breaks? Assume the string remains horizontal during the motion. (5 marks) b. Suppose the ball moves in a circle of larger radius at the same speed v. Is the cord more likely or less likely to break? ( i ) . A ball of mass 1.500 kg is attached to the end of a cord 1.50 m long . The ball moves in a horizontal circle . If the cord can withstand a maximum tension of 64.0 N , a . What is the maximum speed at which the ball can move before the cord breaks ? Assume the string remains horizontal during the motion . ( 5 marks ) b . Suppose the ball moves in a circle of larger radius at the same speed v . Is the cord more likely or less likely to break ?
(a) Let [tex]v[/tex] be the maximum linear speed with which the ball can move in a circle without breaking the cord. Its centripetal/radial acceleration has magnitude
[tex]a_{\rm rad} = \dfrac{v^2}R[/tex]
where [tex]R[/tex] is the radius of the circle.
The tension in the cord is what makes the ball move in its plane. By Newton's second law, the maximum net force on it is
[tex]F = (1.500\,\mathrm{kg}) a_{\rm rad}[/tex]
so that
[tex](1.500\,\mathrm{kg}) \dfrac{v^2}{1.50\,\rm m} = 64.0\,\mathrm N[/tex]
Solve for [tex]v[/tex] :
[tex]v^2 = \dfrac{(64.0\,\mathrm N)(1.50\,\mathrm m)}{1.500\,\rm kg} \\\\ \implies \boxed{v = 8.00 \dfrac{\rm m}{\rm s}}[/tex]
(b) The net force equation in part (a) leads us to the relation
[tex]F = \dfrac{mv^2}R \implies v = \sqrt{\dfrac{FR}m}[/tex]
so that [tex]v[/tex] is directly proportional to the square root of [tex]R[/tex]. As the radius [tex]R[/tex] increases, the maximum linear speed [tex]v[/tex] will also increase, so the cord is less likely to break if we keep up the same speed.
11. Trait theory claims that
O A. people from the same locations
share the same personality type.
B. you always behave the way your
personality type says you will.
O C. your personality is made up of a
number of traits.
O D. you have one characteristic that
defines your entire personality.
The action force is the ballon pushing the air out. What is the magnitude of the reaction force of the air pushing the ballon
The magnitude of the reaction force of the air pushing the balloon would be equal and opposite.
How is the reaction force equal and opposite?A push or a pull that an object experiences as a result of interacting with another item is known as a reaction force.
Newton's third law is officially expressed as follows: There is an equal and opposite reaction to every action. The implication of the statement is that there are always two forces acting on the two interacting objects. The force acting on the first object is equal in size to the force acting on the second. The force acting on the first object is acting in the opposite direction to the force acting on the second object. Force pairs—equal and opposing action-reaction force pairs—always exist in pairs.
Knowing that everything has an equal and opposite reaction according to Newton's second law. The preasured air that a balloon had to push out into the free air acts as the reaction force.
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During World War II, it was found that r, the radius of the shockwave produced during an atomic bomb explosion, depends on the energy released, W, the elapsed time, t, and the air density, .
The atomic bomb explosion, depends on the energy released, W, the elapsed time, t, and the air density which altogether lead to the vaporisation of the contents within the mushroom cloud created.
What is an atomic bomb?An atomic bomb is type of nuclear weapon that releases a vast amount of energy upon explosion in the form of fission reaction.
During an explosion of atomic bomb, a cloud of mushroom fire is formed which vaporises anything found within it. The extent of destruction depends on:
energy released, W, the elapsed time, t, and the air density.Therefore, the atomic bomb explosion, depends on the energy released, W, the elapsed time, t, and the air density which altogether lead to the vaporisation of the contents within the mushroom cloud created.
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Calculate the magnitude of the linear momentum for the following cases.
(a) a proton with mass 1.67 x 10-27 kg, moving with a speed of 5.05 x 106 m/s
(d) the Earth (mass - 5.98 x 1024 kg) moving with an orbital speed equal to 2.98 x 10 m/s.
Answer:
8.4335 x 10^-²¹, 1.78204 x 10²⁶
Explanation:
p = mv
p = (5.05 x 10⁶)(1.67 x 10^-²⁷)
p = 8.4335 x 10^-²¹
p = mv
p = (5.98 x 10²⁴)(2.98 x 10)
p = 1.78204 x 10²⁶
What is the magnitude of the force exerted on an electron in an electric field with a strength of 6.5 *10^ 8 N/C? Remember, the charge on an electron is 1.6 * 10 ^ - 19 * C
All work must be shown to earn credit
The field exerts 6.5 x 10⁸ Newton of force on EACH COULOMB of charge in the field.
We're putting 1.6 x 10⁻¹⁹ Coulomb of charge into the field.
The force on it will be
(1.6 x 10⁻¹⁹ Coulomb) x (6.5 x 10⁻⁸ Newton/Coulomb).
That's 1.04 x 10⁻¹⁰ Newton.
what's the difference between coplanar forces and resultant forces?
Answer:
coplanar When all forces are acting in the same
resultunt the single force and associated torque obtained by combining a system of forces and torques acting on a rigid body via vector addition
Exam Which is a true statement about the energy level of gas? A. Gases have very little energy. B. Gases have more energy than liquids and solids. C. Gases have no energy. D. Gases have less energy than solids. Copyright © 2003-2022 International Academy of Science. All Rights Reserved. Exam Which is a true statement about the energy level of gas ? A. Gases have very little energy . B. Gases have more energy than liquids and solids . C. Gases have no energy . D. Gases have less energy than solids . Copyright © 2003-2022 International Academy of Science . All Rights Reserved .
Statement B is true about the energy level of gas, Gases have more energy than liquids and solids.
What is gas?A gas is a sample of matter that adopts the shape of the container in which it is housed and develops a uniform density inside the container.
Liquid molecules are more energetic than solid molecules. Further heating will cause the molecules to move so quickly that they won't stick together at all. The gas molecules have the highest energy content.
Gases have more energy than liquids and solids.
Hence statement B is true about the energy level of gas.
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A spaceprobe in outer space is flying with a constant speed of 1.795 km/s. The probe has a payload of 1635.0 kg and it carries 4092.0 kg of rocket fuel. The rocket engines of the probe are capable of expelling propellant at a speed of 4.161 km/s. Then the rocket engines are fired up. How fast will the spaceprobe travel when all the rocket fuel is used up?
The speed by which the spaceprobe travels when all the rocket fuel is used up will be 29.262 m/sec.
What is the law of conservation of linear momentum?
According to the law of conservation of linear momentum before the collision is equal to the momentum after the collision. These laws state how momentum gets conserved.
Unit conversion;
1 km/sec = 1000 m/sec
Given data;
Spaceprobe speed = 1.795 km/s = 1795 m /sec
Probe mass = 635.0 kg
Fuel mass = 4092.0 kg
Expelled propellent velocity = 4.161 km/s = 41461 m/sec
From the momentum conservation principle;
[tex]\rm P_i = P_f \\\\ (m_p+m_f)v_i = m_pV - m_fv_p \\\\ V = \frac{(635+4092)1795+4092 \times 41461}{635} \\\\ V = 280540.7 \ m/sec \\\\ V = 28.05 m/sec[/tex]
Hence, the speed by which the spaceprobe travels when all the rocket fuel is used up will be 29.262 m/sec.
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What is the wavelength of a compression wave with a speed of 500 m/s and a frequency of 250 Hz?
Use this formula: λ = s/f
Select one:
a. 2 m
b. 1.25 m
c. 250 m
d. 750 m
Answer:
a (2m)
Explanation:
according to wavelenght = speed / frequency,
the wavelength is 2 m
The following equation shows the position of a particle in time t, x=at2i + btj where t is in second and x is in meter. A=2m/s2, b=1m/s.
Find
A, the average velocity of the particle in the time interval t₁=2sec and t₂=3sec
B, the velocity and acceleration at any time t.
C, the average acceleration in the time interval given in part (a)
(a) The average velocity of the particle in the time interval t₁=2sec and t₂=3sec is 10 m/s.
(b) The velocity and acceleration at any time t is v = (4ti + j) m/s and a = a = 4i m/s²
(c) The average acceleration in the time interval given in part (a) is 3.98 m/s².
Position of the particlex = at²i + btj
x = 2t²i + tj
Average velocity, at t₁=2sec and t₂=3secΔv = Δx/Δt
x(2) = 2(2)²i + 2j
x(2) = 8i + 2j
|x(2)| = √(8² + 2²) = 8.246
x(3) = 2(3)²i + 3j
x(3) = 18i + 3j
|x(3)| = √(18² + 3²) = 18.248
Δv = (18.248 - 8.246)/(3 - 2)
Δv = 10 m/s
Velocity and acceleration at any time, tv = dx/dt
v = (4ti + j) m/s
a = dv/dt
a = 4i m/s²
Average accelerationv(2) = 4(2)i + j
v(2) = 8i + j
|v(2)| = 8.06 m/s
v(3) = 4(3)i + j
v(3) = 12i + j
|v(3)| = 12.04 m/s
a = (12.04 - 8.06)/(3 - 2)
a = 3.98 m/s²
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water
0.6m
water wave
tank
How long does it take for the wave to return to the
position XY, but moving to the right?
[3]
b A man is cutting down a tree with an axe. He
hears the echo of the impact of the axe hitting
the tree after 1.6 s.
i What sort of obstacle could have caused the
echo?
ii The speed of sound is 330 m/s. How far is
the tree from the obstacle?
c Distinguish between the nature of the sound
wave in b and the water wave in a.
[2]
ii the amplitude of
b The cone of a louds
diagram shows how
out in front of the c
loudspeaker
P is a compression,
i
Describe how
changes from
ii Describe the
the sound w
iii Copy the di
and mark a
wavelength
5 a The first diagrams
Answer:
Discrimination is the worst thing in the world you can't even do a thing so you will do such a physical thing or do a mathematics problem ok done it's ok
A spring is compressed by 0.0880 m and is used to launch an object horizontally with a speed of 2.76 m/s. If the object were attached
to the spring, at what angular frequency (in rad/s) would it oscillate?
Answer:
Approximately [tex]3.14\; {\rm rad \cdot s^{-1}}[/tex].
Explanation:
Fact: the angular velocity [tex]\omega[/tex] of a simple harmonic oscillator is the ratio between the maximum velocity [tex]v_{\text{max}}[/tex] and the maximum displacement [tex]x_\text{max}[/tex] of this oscillator. In other words:
[tex]\begin{aligned} \omega &= \frac{v_{\text{max}}}{x_{\text{max}}}\end{aligned}[/tex].
Derivation of the previous equation:
Let [tex]A[/tex] denote the amplitude of this oscillation, and let [tex]\omega[/tex] denote the angular velocity.
The displacement of the oscillator at time [tex]t[/tex] would be:
[tex]x(t) = A\, \sin(\omega\, t)[/tex].
The maximum displacement of this oscillator would be [tex]x_\text{max} = A[/tex].
The velocity of this oscillator at time [tex]t[/tex] is the derivative of displacement with respect to time:
[tex]\begin{aligned} v(t) &= \frac{d}{d t}\, [x(t)] \\ &= \frac{d}{d t} [A\, \sin(\omega\, t)] \\ &= A\, \omega\, \cos(\omega\, t)\end{aligned}[/tex].
The maximum velocity of this oscillator would be [tex]v_\text{max} = A\, \omega[/tex].
Notice that dividing [tex]v_\text{max} = A\, \omega[/tex] by [tex]x_\text{max} = A[/tex] would give:
[tex]\displaystyle \frac{v_\text{max}}{x_\text{max}} = \frac{A\, \omega}{A} = \omega[/tex].
It is given that [tex]v_\text{max} = 2.76\; {\rm m\cdot s^{-1}}[/tex] while [tex]x_\text{max} = 0.0880\; {\rm m}[/tex]. Therefore:
[tex]\begin{aligned} \omega &= \frac{v_{\text{max}}}{x_{\text{max}}} \\ &= \frac{2.76\; {\rm m\cdot s^{-1}}}{0.0880\; {\rm m}} \\ &\approx 3.14\; {\rm s^{-1}}\end{aligned}[/tex].
(Radians per second.)
Erbium-165 has a half-life of 10.4 hours. If you start with 1,000 grams of
erbium-165, how much time will it take to have 125 grams of erbium-165 left
in the sample?
A. 41.6 hours
OB. 20.8 hours
C. 31.2 hours
D. 10.4 hours
The amount of time it will take to have 125 grams of erbium-165 left
in the sample is; C: 31.2 hours
How to calculate decay time?The formula for amount of substance after decay is;
N(t) = N₀(¹/₂)^(t/t_¹/₂)
We are given;
Half life; t_¹/₂ = 10.4 hours
Initial amount; N₀ = 1000 g
Amount left; N(t) = 125 g
Thus;
125 = 1000 * (¹/₂)^(t/10.4)
125/1000 = (¹/₂)^(t/10.4)
In 0.125 = (t/10.4) In 0.5
-2.08 = -0.693(t/10.4)
t/10.4 = -2.08/-0.693
t = 10.4 * 3
t = 31.2 hours
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a car is traveling along a straight road at a velocity of 30m/s when its engine cuts off. For the next ten seconds, the car slows down, and its average acceleration is a1. For the next five seconds, the car slows down further at a velocity of 24m/s, and its average acceleration is a2. The ratio of the average acceleration values is a1/a2=1.5. Find the velocity of the car at the end of the initial ten-second interval.
Answer:
See below....the question is unclear....pick the answer you think they want
Explanation:
Car goes from 30 m/s to 24 m/s in 10 seconds
acceleration = change in velocity/change in time
= -6 m/s / 10 s = - 3/5 m/s^2
The wording of the question is not very good.....perhaps they meant to say that after the 5 seconds the velocity is now 24 m/s ?
a1 *10 + a2 *5 = -6 m/s
( 6 m/s is the change in velocity from 30 to 24 m/s)
a1/a2 = 1.5 so a2 = a1/1.5 (substitute this in)
10 a1 + 5 ( a1/1.5) =-6
15 a1 + 5 a1 = -9
a1 = - .45 m/s^2
The velocity of the car at the end of the initial ten-second interval, given that the car further slows down for the next five seconds, is 25.5 m/s
How to determine the velocity of the car at the end of the 10 s interval?From the first statement, we have:
Initial velocity (u₁) = 30 m/sTime (t₁) = 10 secondsVelocity at the end of 10 s (v₁) = ?v₁ = u₁ + a₁t₁
v₁ = 30 + (a₁ × 10)
v₁ = 30 + 10a₁ ......(1)
From the second statement, we have:
Time (t₂) = 5 sFinal velocity (v₂) = 24 m/sRation of acceleration (a₁/a₂) = 1.5Initial velocity (u₂) = Velocity at the end of 10 sv₂ = u₂ + a₂t₂
24 = u₂ + (a₂ × 5)
24 = u₂ + 5a₂ .......(2)
But,
u₂ = v₁ = 30 + 10a₁
Thus, we have:
24 = 30 + 10a₁ + 5a₂
But,
a₁/a₂ = 1.5
a₁ = 1.5a₂ ..... (3)
Thus, we have:
24 = 30 + 10a₁ + 5a₂
24 = 30 + 10(1.5a₂) + 5a₂
24 = 30 + 15a₂ + 5a₂
24 = 30 + 20a₂
Collect like terms
24 - 30 = 20a₂
-6 = 20a₂
Divide both sides by 20
a₂ = -6 / 20
= -0.3 m/s²
Substituting the value of a₂ into equation 3, we have
a₁ = 1.5a₂
= 1.5 × -0.3
= -0.45 m/s²
Substitute the value of a₁ into equation 1 to obtain the velocity, v₁ at the end of the initial to 10 s
v₁ = 30 + 10a₁
= 30 + (10 × -0.45)
= 30 - 4.5
= 25.5 m/s
Thus, velocity of the car at the end of the initial ten-second interval is 25.5 m/s
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the question in photo
Answer:
a
Explanation:
