what is the smallest amount of time in which the person can accelerate the car from rest to 23 m/s and still keep the coffee cup on the roof. The coefficient of the static friction is 0.21. The maximum acceleration of the car that is allowed so that the cup does not fall is 2.1 m/s^2
Answers
Given:
The coefficient of the static friction, μ=0.21
The maximum acceleration of the car so that the cup does not fall, a=2.1 m/s²
The initial velocity of the car, u=0 m/s
The final velocity of the car, v=23 m/s
To find:
The smallest amount of the time in which the car can accelerate so that the coffee cup will still be on the roof.
Explanation:
From the equation of motion,
[tex]v=u+at[/tex]Where t is the smallest amount of time in which the person can accelerate and still keep the cup on the car.
On rearranging the above equation,
[tex]t=\frac{v-u}{a}[/tex]On substituting the known values,
[tex]\begin{gathered} t=\frac{23-0}{2.1} \\ =10.95\text{ s} \end{gathered}[/tex]Final answer:
Thus the smallest amount of the time in which the person can accelerate the car at the given rate and still keep the cup on the roof of the car is 10.95 s
Related Questions
if the dolphin is moving horizontally when it goes through the hoop how high above the water is the center of the hoop
Answers
We are given that a dolphin moves describing a projectile motion. This can be represented in the following graph of position vs time:
Since the dolphin moves horizontally as he goes through the hoop this means that the hoop is at the maximum height of the motion. The maximum height of a projectile motion is given by:
[tex]h_{\max }=\frac{v^2\sin ^2\theta}{2g}[/tex]Where:
[tex]\begin{gathered} h_{\max }=\text{ max}imum\text{ height} \\ v=velocity_{} \\ \theta=\text{ initial angle} \\ g=\text{ acceleration of gravity} \end{gathered}[/tex]Now, we plug in the values:
[tex]h_{\text{max}}=\frac{(10\frac{m}{s})^2(\sin (41))^2}{2(9.8\frac{m}{s^2})}[/tex]Solving the operations:
[tex]h_{\max }=2.2m[/tex]Therefore, the hoop is at 2.2 meters above the water.
5. Draw a transverse wave with two wavelengths and label amplitude, crest, trough, and
equilibrium position.
Answers
A wave is considered to be transverse if its oscillations run counterclockwise to the wave's direction of advance. A longitudinal wave, on the other hand, moves in the direction of its oscillations. Transverse waves include water waves.
A waveform signal's wavelength, which is the distance between two identical locations (adjacent crests) in the succeeding cycles, determines whether it is sent through space or via a wire. This length is typically defined in wireless systems in metres (m), centimetres (cm), or millimetres (mm).
The largest displacement or distance made by a point on a wave or vibrating body relative to its equilibrium position is its amplitude. It is equal to one-half of the vibration path.
The crest and trough of a wave, respectively, are its highest and lowest surface portions. The wave height is the vertical distance between the peak and trough. The wavelength is the horizontal separation between two consecutive crests or troughs.
The horizontal line at the wave's centre stands in for balance. A period is the length of time it takes to complete a cycle, which includes travelling from one peak to another, from one trough to another, or from one equilibrium point to another (both equilibrium points same direction).
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In terms of area, about how much more pizza is given if the diameter is 12 inches compared to one with a diameter of 8 inches?
Answers
B. 2.3 times more
Explanation:The pizza is circular in shape
The diameter of the large-sized pizza, d₁ = 12 inches
Tha area of the large sized pizza is calculated as:
[tex]\begin{gathered} A_1=\frac{\pi{d^2_1}}{4} \\ A_1=\frac{\pi{12^2}}{4} \\ A_1=\frac{\pi{144^{}}}{4} \\ A_1=36\pi\text{ in}^{2} \end{gathered}[/tex]The diameter of the small-sized pizza, d₂ = 8 inches
The area of the small-sized pizza is calculated as:
[tex]\begin{gathered} A_2=\frac{\pi{d^2_2}}{4} \\ A_2=\frac{\pi{8^2}}{4} \\ A_2=\frac{64\pi{}}{4} \\ A_2=16\pi\text{ in}^{2} \end{gathered}[/tex]Ratio of A₁ to A₂
[tex]\begin{gathered} \frac{A_1}{A_2}=\frac{36\pi{}}{16\pi} \\ \frac{A_1}{A_2}=2.25 \\ \frac{A_1}{A_2}=2.3(to\text{ the nearest 1 dp)} \end{gathered}[/tex]The 12 inches pizza is 2.3 times more than the 8 inches pizza
Find the volume of a cuboid with the following measurments
Length:4.5 cm Breadth: 2 cm Height: 50 cm
ans is :45,000
But I am not getting the correct ans
please send working
Answers
Answer:
The answer should be 450 cm³
Cannot be 4500 unless one of the measurements is in meters. See detailed explanation below
Explanation:
This is pretty much straightforward
V = l x w x h
where
l = length, w = width, h = height
Given the values for l, w, h
V = 4.5 x 2 x 50
= 450 cm³
The only way you can get 45,000 is if one of the measurements is in meters and others in cm. Please check your question carefully to see what units were used and what units the answer should be in
A particular cookie provides 54.0 kcal of energy. An athlete does an exercise that involves repeatedly lifting (without acceleration) a 103-kilogram weight 2.45-decimeters above the ground with an energy efficiency of 25%. How many repetitions of this exercise can the athlete do with the energy supplied from one of these cookies?
Answers
A maximum of about 229 repetitions of something like the exercise can be performed by that of the athlete utilizing the energy provided by each of the biscuits.
The proportion of input to produced energy can be used to define energy consumption.
A cookie, therefore, therefore has 54.0 kcal of calories. The 54.0 kcal throughout this croissant is used as power input by the athlete.
Efficiency = output energy / input energy
It can be written as:
Output energy = efficiency × input energy
Puting the values of efficiency and input energy.
Output energy = 0.25 × 54 kcal = 13.5 kcal.
The weightlifting exercise can be done n times for the output energy. This outgoing energy comes from mgh in the shape of potential energy. So,
Energy per repetition = [tex]mgh[/tex]
Put the values of m, g and h in above equation.
Energy per repetition = 103 kg × 9.8 m/ × (2.45 × 0.1m) = 247.303 J
Energy per repetition = 0.059 kcal.
So,
amount of repetitions = sum of output energy / energy per repetition
amount of repetitions = 13.5 kcal / 0.059 kcal = 229 repetitions.
Therefore, amount of repetition can be be 229.
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A Tour de France cyclist at a rate of 6.5 m/s and is 333 m
ahead of a crew van with powerdrink refills. The van is
traveling at 15 m/s and is accelerating at a constant rate of
0.4 m/s2
How much time will it take for the crew van to catch up
with the cyclist?
Answers
Given data:
* The speed of the cyclist is 6.5 m/s.
* The initial distance of the cyclist from the van is 333 m.
* The initial velocity of the van is 15 m/s.
* The acceleration of the van is,
[tex]a=0.4ms^{-2}[/tex]Solution:
Let x be the distance from the van initial position at which the cyclist and van meet.
Let the cyclist meet the van at time t.
By the kinematics equation, the position of the cyclist at time t is,
[tex]x-333=u_ct+\frac{1}{2}a_ct^2[/tex]where u_c is the speed of the cyclist, a_c is the acceleration of the cyclist, t is the time taken and 333-x is the distance traveled by the cyclist at time t,
The acceleration of the cyclist is zero.
Substituting the known values,
[tex]\begin{gathered} x-333=6.5t+0 \\ x-333=6.5t \end{gathered}[/tex]By the kinematics equation, the position of the van after time t is,
[tex]x=u_vt+\frac{1}{2}a_vt^2[/tex]where u_v is the velocity of the van, a_v is the acceleration of the van, and t is the time taken,
Substituting the known values,
[tex]\begin{gathered} x=15t+\frac{1}{2}\times0.4\times t^2 \\ x=15t+0.2t^2 \end{gathered}[/tex]Substituting this value of x in the kinematics equation of the cyclist,
[tex]\begin{gathered} (15t+0.2t^2)-333=6.5t \\ 15t+0.2t^2-6.5t-333=0 \\ 0.2t^2+8.5t-333=0 \end{gathered}[/tex]By solving the quadratic equation,
[tex]\begin{gathered} t=\frac{-8.5\pm\sqrt[]{8.5^2-(4\times0.2\times(-333))}}{2\times0.2} \\ t=\frac{-8.5\pm\sqrt[]{^{}8.5^2+(4\times0.2\times333)}}{2\times0.2} \\ t=\frac{-8.5\pm18.4}{0.4} \\ t=24.8\text{ s or-67.25 s} \end{gathered}[/tex]As the value of time cannot be negative.
Thus, the time at which the cyclist and van meet is 24.8 seconds.
There is no _________ movement in a longitudinal wave.A. HorizontalB. Back and forthC. VerticalD. Parallel
Answers
Explanation
A longitudinalwave is in which the particles of the medium vibrate in the direction of the line of advance of the wave.Longitudinal waves cause the medium to move parallel to the direction of the wave.
A longitudinal wave can be set up for example in a streched spring by compressing the coils in a small region, and releasing the compressed region,
the back and forth motions of the coils of the spring is in the same direction that the wave travels
so, in a longitudinal wave there is not Vertical movement, so the answer is
C. Vertical
A) the frictional force F newtonsB)The resultant normal reaction of the surface on the metal block
Answers
Given:
The mass of the block is.
[tex]m=10\text{ kg}[/tex]The tension on the rope is,
[tex]T=100\text{ N}[/tex]The angle with the horizontal is,
[tex]\theta=60^{\circ}[/tex]The block is moving with constant speed.
as the block is moving with constant speed, the net force on the block will be zero.
Part (A)
we can write in the horizontal direction the component of the tension will be equal to the frictional force and we write,
[tex]\begin{gathered} T\cos 60^{\circ}=F \\ F=100\cos 60^{\circ} \\ F=50\text{ N} \end{gathered}[/tex]Hence the frictional force is 50 N.
\\
Part(B)
The resultant normal reaction will be,
[tex]\begin{gathered} N=T\sin 60^{\circ}-mg \\ =100sin60^{\circ}-10\times9.8 \\ =-11.4\text{ N} \end{gathered}[/tex]hence the resultant normal reaction is -11.4 N.
The period of a simple harmonic oscillator is the time it takes for one complete cycle of oscillation to be completed. Is this true or false?
Answers
Yes, the given statement is true.
The period of a simple harmonic oscillator is the time it takes for one complete cycle of oscillation to be completed.
Which shape fits a position vs. time graph of an object that is slowing down? Which shape fits a position vs. time graph of an object that is speeding up?
Answers
1.
In a position x time graph, if the velocity is constant, so the position increases at the same rate over the time, so we have a linear relation between the position and the time.
Therefore the shape that represents this relation is C.
2.
In a velocity x time graph, if the velocity is constant, its value is always the same over the time, it doesn't change. That is represented graphically by a horizontal line, therefore the shape that represents this relation is B.
.
According to the law of conservation of charge which statement can be true?A. A silk cloth gained charge. B. A metal rod lost charge.C. A peice of glass transferred electrons to felt. D. A balloon remains neutrally charged when rubbed.
Answers
Option D
Explanation:The law of conservation of charge states that the amount of charge in a system is constant
This means that as time changes, the amount of charge in a system does not change
By careful consideration of the options stated:
Each of options A to C either shows that charge is lost or gained
Only option D typifies the law of conservation of conservation of charges because charges are not lost or gained by the ballon so described.
A roller coaster has a vertical loop with radius 22.8 m. With what minimum speed should the roller-coaster car be moving at the top of the loop so that the passengers do not lose contact with the seats?
Answers
Given,
The radius of the loop of the roller coaster, r=22.8 m
The forces that are acting on the roller coaster when it is at the top of the loop are the centripetal force directed upwards and the weight of the roller coaster including the passengers directed downwards.
For the passengers to stay in the seat, the centripetal force must be, at the least, equal to the weight of the passengers and the rollercoaster.
That is,
[tex]\frac{Mv^2}{r}=Mg[/tex]Where M is the combined mass of the rollercoaster and the passengers, v is the minimum speed of the roller coaster when it is at the top of the loop, and g is the acceleration due to gravity.
On simplifying the above equation,
[tex]v=\sqrt[]{gr}[/tex]On substituting the known values,
[tex]\begin{gathered} v=\sqrt[]{9.8\times22.8} \\ =14.95\text{ m/s} \end{gathered}[/tex]Thus the minimum speed that the roller coaster must have when it is at the top of the loop so that the passengers stay in contact with the seats is 14.95 m/s.
You observe waves on the beach and measure that a wave hits the beach every 5 seconds. What is the period and the frequency of the waves ?
Answers
Answer:
Period = 5 seconds
Frequency = 0.2 Hz
Explanation:
The period is the time per cycle. So, if a wave hits the beach every 5 seconds, the period will be 5 seconds.
Additionally, the frequency is the inverse of the period, so the frequency of the waves can be calculated as:
[tex]\text{frequency = }\frac{1}{Period}=\frac{1}{5}=0.2\text{ Hz}[/tex]So, the answers are:
Period = 5 seconds
Frequency = 0.2 Hz
10. ABC.Per=1200 NNet Force:Pit=600 NEngen-SONFrid=20 NPry=800 NPrax=800 NF50NWhich situation above would best describe free fall velocity?Which situation above would best describe a crane lifting an object?If situation C had a Fapp of 40N to the right, the net force on the object would beIf situation Chad Fapp of 20N to the right, the forces would be (balanced, unbalanced) and thehorizontal velocity would be (constant, + accelerating, - accelerating). Circle the correct terms
Answers
When the body is under free fall its apperaent weight will be zero.
Therefore
why free-fall acceleration can be regarded as a constant for objects falling within a few hundred miles of Earth’s surface.
Answers
Megan kicks a soccer ball with a mass of 2 kg. The ball leaves the ground moving 50 meters per second. What is the kinetic energy of the ball?
Answers
ANSWER
[tex]2500J[/tex]EXPLANATION
The kinetic energy of a body is the energy it possesses due to its motion and it can be found by applying the formula:
[tex]E=\frac{1}{2}mv^2[/tex]where m = mass, v = velocity
From the question:
[tex]\begin{gathered} m=2\text{ kg} \\ v=50\text{ m/s} \end{gathered}[/tex]Therefore, the kinetic energy of the ball is:
[tex]\begin{gathered} E=\frac{1}{2}\cdot2\cdot50^2 \\ E=2500J \end{gathered}[/tex]Will mark as brainlist
Which of the following best represents R= A - B ?
Please help, it’s due soon!
Answers
Option C is the best representation for the resultant vector representing the resultant R=A+B because the result is represented by the closing side of the triangle, so it is the correct answer.
What exactly is a vector quantity?A measure with both magnitude and direction It is typically represented by an arrow with the same direction as the quantity and a length proportional to the magnitude of the quantity. A vector, while including magnitude and demand, does not have a position. Vector quantities are physical quantities that are distinguished by the presence of both magnitude and direction. Displacement, force, torque, momentum, acceleration, velocity, and so on. Vector quantities are physical quantities that have distinct magnitude and direction definitions. For example, suppose a boy is riding a bike at 30 km/hr in the north-east direction. In nature, vectors have velocity, acceleration, force, electromagnetic fields, and importance. (Weight is the force produced by gravity's acceleration acting on a mass.) A Scalar is a quantity or phenomenon that has only magnitude and no specific direction.To learn more about vector quantity, refer to:
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What is the efficiency of a block and tackle if you pull 20 m of rope with a force of 600 N to raise 200kg piano 5 m?
Answers
W out = mass * distance * gravity
E in = Force * distance
Efficiency = W out / E in = (200 kg * 5m*9.8N) / (600N * 20m) = 0.81
0.81 x 100 = 81 %
I need help with some questions to study them for midterms!
Answers
The correct answer is option C.
The acceleration of an object is defined as the time rate of change of velocity. Velocity is a vector and has magnitude and direction. The change velocity may mean either a change in magnitude or change in direction or both.
Thus an object can have acceleration even when the speed is constant. For example, an object in a circular motion.
Thus the correct answer is option C.
A car starts from rest and travels for 9.0 s with a uniform acceleration of +2.4 m/s²?. The driver then applies the brakes, causing a uniform acceleration of -2.5 m/s². If the brakes areapplied for 2.0 s, determine each of the following(a) How fast is the car going at the end of the braking period?m/s(b) How far has the car gone?m
Answers
a) At the end of the braking period, the car is moving at a speed of 16.6m/s
b) The car has traveled a total distance of 135.4m
Explanations:The car starts from rest
The initial velocity, u = 0 m/s
Uniform acceleration, a = 2.4 m/s²
time, t = 9.0 seconds
Find the final velocity when the car accelerates at 2.4m/s² using the equation
v = u + at
v = 0 + 2.4(9)
v = 21.6 m/s
The distance covered when the car accelerates at 2.4m/s²
s = ut + 0.5at²
s = 0(9) + 0.5(2.4)(9²)
s = 97.2 m
The distance covered when the car accelerates at 2.4m/s² is 97.2 m
The driver then applied a brake for 2.0 s and accelerates at -2.5m/s²
The initial velocity now becomes 21.6 m/s
That is, u = 21.6 m/s
t = 2 seconds
a = -2.5m/s²
The final speed v is calculated as:
v = u + at
v = 21.6 + (-2.5)(2)
v = 21.6 - 5
v = 16.6m/s
At the end of the braking period, the car is moving at a speed of 16.6m/s
The distance covered during the braking period is calculated as:
[tex]\begin{gathered} s\text{ = (}\frac{u+v}{2})t \\ s\text{ = }\frac{21.6+16.6}{2}\times2 \\ s\text{ = }\frac{38.2}{2}\times2 \\ s\text{ = 38.2 m} \end{gathered}[/tex]The car traveled a distance of 38.2 m during the braking period
Total distance covered = 97.2m + 38.2m
Total distance covered = 135.4m
The car has gone a distance of 135.4m
How much force must be applied to push a 253.2 kg crate across the floor at a constant velocity if the coefficient of kinetic friction is 0.55?
Answers
Given,
The mass of the crate, m= 253.2 kg
The coefficient of the kinetic friction between the floor and the crate, μ=0.55
Given that the crate is pushed with a constant velocity. That is the net force on the crate is zero.
The only two forces acting on the crate are the force with which it is being pushed and the friction that is opposing the applied force.
The net force on the crate is given by,
[tex]\begin{gathered} F_n=0=F-f \\ =F-mg\mu \end{gathered}[/tex]Where f is the frictional force between the floor and the crate, F is the applied force, and g is the acceleration due to gravity.
On substituting the known values,
[tex]\begin{gathered} F-253.2\times9.8\times0.55=0 \\ \Rightarrow F=253.2\times9.8\times0.55 \\ =1364.75\text{ N} \end{gathered}[/tex]Thus the force with which the crate must be pushed is 1364.75 N
Please help with Question(ii). I don't understand the shown step of calculating the momentum of ball B. Especially after the third line 12+Pb=15.
Answers
Given:
m1 = mass 1 = 1kg
v1= initial velocity 1 = 12 m/s
m2= mass 2 = 3 kg
P after = momentum after collision = 15 kgm/s
(i)
Momentum of Ball A before collision
Momentum = mass x velocity
Pa = m1 v1
Replacing with the values given:
Pa = (1 kg) (12 m/s) = 12 kg m/s
(ii)
Momentum before = momentum after
Pa + Pb = P after
12 + Pb = 15
Since The ball B is travelling North, the distances travelled form a right triangle:
Apply pythagorean theorem:
c^2 = a^2 + b^2
Where c is the hypotenuse= P after = 15
a & b are the other 2 legs of the triangle = Pa and Pb
Replacing:
15^2 = 12^2 + Pb^2
Solve for Pb
15^2 - 12^2 = pb^2
√15^2 -12^2 = Pb
pb= 9 kgms^2
Need help with this question Short straight forward answers please :)
Answers
We will have the following:
a. The gravitational potential energy will be:
[tex]P_C=(15kg)(9.8m/s^2)(6m)\Rightarrow P_C=882J[/tex]So, the gravitational potential energy of C is 882 J.
b. The velocity of C right before it hits the ground will be:
[tex]\begin{gathered} 882J=\frac{1}{2}(15kg)v^2\Rightarrow\frac{1764J}{15kg}=v^2 \\ \\ \Rightarrow v=\frac{14\sqrt{15}}{5}m/s\Rightarrow v\approx10.84m/s \end{gathered}[/tex]So, the velocity will be approximately 10.84 m/s.
c.
1. We will have that Eg at the initial position will be: B < C
2. Vfinal upon impact with ground: B = C
3. Ek right before hitting he ground: B < C
f.
1. Eg: A > B
2. V final: A > B
3. Ek: A > B
4. V at 2 meters above the ground: A > B
5. Total energy at 2 m above the ground: A > B.
Shown in the figure, are electric field lines around two point charges. The two charges might be which of these? A)two protons B)a proton and an electron C)two electrons D)two croutons E)two neutrons
Answers
Given:
A figure of the electric field lines.
To find:
The nature of the charges.
Explanation;
The electric field lines always start at a positive charge. And the electric field lines always end at a negative charge. If there is a single charge, the field lines may start at infinity or end at infinity.
Electric field lines also do not form closed loops. They do not cross each other.
In the figure, the field lines start at the blue charge and end at the charge represented by the color yellow.
Therefore the blue charge is positive and the yellow charge is negative.
Final answer:
Therefore the correct answer is option B.
An airplane traveling at 1008 meters above the ocean at 135 km/h is going to drop a box of supplies to shipwrecked victims below. How many seconds before the plane is directly overhead should the box be dropped?
Answers
The horizontal velocity of the airplane is,
[tex]v=135\text{ km/h}[/tex]The height of the airplane is,
[tex]h=1008\text{ m}[/tex]The vertical initial velocity of the box is zero as the airplane is moving in the horizontal direction.
let the time to reach the victim is t.
we can write,
[tex]\begin{gathered} h=\frac{1}{2}gt^2 \\ t=\sqrt[]{\frac{2h}{g}} \end{gathered}[/tex]Substituting the values we get,
[tex]\begin{gathered} t=\sqrt[]{\frac{2\times1008}{9.8}} \\ =14.3\text{ s} \end{gathered}[/tex]Hence the required time is 14.3 s
An object has an excess charge of −1.6 × 10−17 C. How many excess electrons does it have?
Answers
Given:
The charge on the object is
[tex]q=-1.6\times10^{-17}\text{ C}[/tex]Required: Number of electrons
Explanation:
The number of electrons can be calculated using the quantization of charge
[tex]q\text{ = ne}[/tex]Here, n is the number of electrons
e is the charge on the electron whose value is
[tex]e\text{ = -1.6}\times10^{-19}\text{ C}[/tex]
On substituting the values, the number of electrons will be
[tex]\begin{gathered} n=\frac{q}{e} \\ =\frac{-1.6\times10^{-17}}{-1.6\times10^{-19}} \\ =100 \end{gathered}[/tex]Final Answer: The object has an excess of 100 electrons.
The radioactive isotope 14C has a half-life of approximately 5715 years. Now there are 50g of 14C.(1) How much of it remains after 1600 years? (Round your answer to three decimal places.)
Answers
We know that the amount of matter is given by:
[tex]N=N_0e^{-\lambda t}[/tex]where λ is the decay constant. The decay constant is related to the half-life of the element by the equation:
[tex]\lambda=\frac{\ln2}{t_{\frac{1}{2}}}[/tex]Then we can express our first equation as:
[tex]N=N_0e^{-\frac{\ln2}{t_{\frac{1}{2}}}t}[/tex]Plugging the initial amount, 50 g, the half-life of 5715 years and the time we want to know we have that:
[tex]\begin{gathered} N=50e^{-\frac{\ln2}{5715}(1600)} \\ N=41.181 \end{gathered}[/tex]Therefore, after 1600 years there are 41.181 g
Given v = 520 sin (30t - 5π/4), what is the phase angle?Question 4 options:-225 degrees90 degrees-135 degrees-90 degrees
Answers
Given data:
The voltage can be expressed as,
v = 520 sin (30t - 5π/4) ...... (1)
Now, the general equation of the sine wave can be given as,
[tex]v=V_m\text{ sin}(\omega t+\phi)\ldots\ldots\text{ }(2)[/tex]Here,
[tex]\phi[/tex]is the phase angle,
[tex]V_m[/tex]is the maximum voltage, and
[tex]\omega[/tex]is the angular frequency.
Compare equations (1) and (2), we get:
[tex]\begin{gathered} \phi=-\frac{5\pi\text{ rad}}{4}(\frac{180\degree}{\pi}) \\ =-225\degree \end{gathered}[/tex]Thus, the phase angle is
[tex]-225\degree[/tex]and the first option (-225 degrees) is correct.
A 0.327-kg model rocket accelerates at 35.7 m/s/s on takeoff. Determine the upward thrust experienced by the rocket
Answers
The upward thrust of the rocket is determined as 11.67 N.
What is the upward thrust of the rocket?The upward thrust of the rocket is calculated by applying Newton's second law of motion as shown below;
F = ma
where;
m is the mass of the rocketa is the upward acceleration of the rocketSubstitute the given parameters and solve for the upward thrust of the rocket.
F = (0.327 kg) x (35.7 m/s²)
F = 11.67 N
Thus, the upward thrust of the rocket is determined as 11.67 N.
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Energy transformations always produce a wasteful amount of energy called ?
Answers
Energy transformation always produce a wasteful amount of energy called heat energy.
Hence, the answer is heat energy.
