Answer:
simple machines such as ramps lessen the moment required to do work. if a triangle has a base of 5 and the height is 7, a ramp would make the hypotenuse of this triangle lessoning the total distance. using a²+b²=c² 25+49=c² 74≈8.6 and it is obvious that 8.6 is less than 12 in every unit. other simple machines such as pulleys make it lighter making it simply easier for an object to be lifted.
Explanation:
Answer:
Simple machines are useful because they reduce effort or extend the ability of people to perform tasks beyond their normal capabilities. Simple machines that are widely used are the wheel and axle, pulley, inclined plane, screw, wedge and lever.
We recommend that our students get at least _____ hours of behind-the-wheel instruction.
A. 6
B. 10
C. 25
D. 50
A Labrador retriever runs 50 m in 7.2 s to retrieve a toy bird. The dog then runs half way
back in 3.85 s. Determine the average speed and velocity of the dog
Answer:
The average velocity and average speed of the dog are 2.262 meters per second and 6.787 meters per second, respectively.
Explanation:
From Physics we must remember the definitions of average speed and average velocity, both measured in meters per second. Velocity is a vectorial quantity, that is, it has both magnitude and direction, whereas speed is an scalar quantity, which is a quantity that is represented solely by its magnitude. We assume that dog moves at constant speed.
For the case of the dog, we get that average speed and average velocity of the animal are, respectively:
Average velocity:
[tex]\vec v_{avg} = \frac{1}{\Delta t}\cdot (\vec r_{B}-\vec r_{A})[/tex] (Eq. 1)
Where:
[tex]\Delta t[/tex] - Travelling time of the dog, measured in seconds.
[tex]\vec r_{A}[/tex] - Initial vector position of the dog, measured in meters.
[tex]\vec r_{B}[/tex] - Final vector position of the dog, measured in meters.
Average speed:
[tex]v_{avg} = \frac{1}{\Delta t} \cdot (s_{A}+s_{B})[/tex] (Eq. 2)
Where [tex]s_{A}[/tex] and [tex]s_{B}[/tex] are the travelled distances of each stage, measured in meters.
If we know that [tex]\Delta t = 11.05\,s[/tex], [tex]\vec r_{A} = 0\,\hat{i}\,\,\,[m][/tex] and [tex]\vec r_{B} = 25\,\hat{i}\,\,\,[m][/tex], [tex]s_{A} = 50\,m[/tex] and [tex]s_{B} = 25\,m[/tex], average velocity and average speed are, respectively:
[tex]\vec v_{avg} = \frac{1}{11.05\,s}\cdot (25\,\hat{i})\,\,\,[m][/tex]
[tex]\vec v_{avg} = 2.262\,\hat{i}\,\,\,\left[\frac{m}{s} \right][/tex]
[tex]v_{avg} = \frac{75\,m}{11.05\,s}[/tex]
[tex]v_{avg} = 6.787\,\frac{m}{s}[/tex]
The average velocity and average speed of the dog are 2.262 meters per second and 6.787 meters per second, respectively.
A 6.13-g bullet is moving horizontally with a velocity of 361 m/s, where the sign indicates that it is moving to the right (see part a of the drawing). The bullet is approaching two blocks resting on a horizontal frictionless surface. Air resistance is negligible. The bullet passes completely through the first block (an inelastic collision) and embeds itself in the second one, as indicated in part b. Note that both blocks are moving after the collision with the bullet. The mass of the first block is 1233 g, and its velocity is 0.741 m/s after the bullet passes through it. The mass of the second block is 1646 g. (a) What is the velocity of the second block after the bullet imbeds itself
Answer:
v₃ = 0.786 m/s
Explanation:
Here, we will use the law of conservation of momentum, which states the following:
Total Momentum of System Before Collision =
Total Momentum of System After Collision
m₁u₁ + m₂u₂ + m₃u₃ = m₁v₁ + m₂v₂ + m₃v₃
where,
m₁ = mass of bullet = 6.13 g = 0.00613 kg
m₂ = mass of 1st block = 1233 g = 1.233 kg
m₃ = mass of 2nd block = 1646 g = 1.646 kg
u₁ = speed of first bullet before collision = 361 m/s
u₂ = speed of first block before collision = 0 m/s
u₃ = speed of 2nd block before collision = 0 m/s
v₁ = speed of bullet after collision
v₂ = speed of 1st block after collision = 0.741 m/s
v₃ = speed of 2nd block after collision = ?
Therefore,
(0.00613 kg)(361 m/s) + (1.233 kg)(0 m/s) + (1.646 kg)(0 m/s) = (0.00613 kg)(v₁) + (1.233 kg)(0.741 m/s) + (1.646 kg)(v₃)
2.2129 kg m/s + 0 kg m/s + 0 kg m/s - 0.9136 kg m/s = (0.00613 kg)(v₁) + (1.233 kg)(0.741 m/s) + (1.646 kg)(v₃)
1.2992 kg m/s = (0.00613 kg)(v₁) + (1.646 kg)(v₃)
since, the bullet is embedded in 2nd block after collision. Thus, there velocities will become same. (v₁ = v₃)
Therefore,
1.2992 kg m/s = (0.00613 kg)(v₃) + (1.646 kg)(v₃)
v₃ = (1.2992 kg m/s)/(1.6521 kg)
v₃ = 0.786 m/s
why the bodies of water important for recreation
Explanation:
Recreational water activities can have substantial benefits to health and well-being. Swimming pools, beaches, lakes, rivers and spas provide environments for rest and relaxation, physical activity, exercise, pleasure and fun. Yet they also present risks to health.
pls make it the brainliest of it has helped you !!!!
Please help me with this
Answer:
7 Newton's East
Explanation:
when the force is going in the same direction in this case east, you add the forces.
A spring has natural length 16 cm. A force of 3 N is required to holdthe spring compressed compressed to 11 cm. Find the amount ofwork instretching the spring from 17 cm to 19 cm.
Answer:
W = 0.012 J
Explanation:
For this exercise let's use Hooke's law to find the spring constant
F = K Δx
K = F / Δx
K = 3 / (0.16 - 0.11)
K = 60 N / m
Work is defined by
W = F. x = F x cos θ
in this case the force and the displacement go in the same direction therefore the angle is zero and the cosine is equal to 1
W = ∫ F dx
W = k ∫ x dx
we integrate
W = k x² / 2
W = ½ k x²
let's calculate
W = ½ 60 (0.19 -0.17)²
W = 0.012 J
PLEASE HELP WILL MARK BRAINLIEST
Answer:
also choose D and E i belive those are also correct
7. A 1,500-N force is applied to a 1,000-kg car. What is the car's acceleration?
Answer:
1.5m/s^2
Explanation:
Answer:
1.5 m/s2. accerelation =force ÷mass
Gibbons, small Asian apes, move by brachiation, swinging below a handhold to move forward to the next handhold. A 9.0 kg gibbon has an arm length (hand to shoulder) of 0.60 m. We can model its motion as that of a point mass swinging at the end of a 0.60-m-long, massless rod. At the lowest point of its swing, the gibbon is moving at 3.5 m/s. What upward force must a branch provide to support the swinging gibbon
Answer:
The correct solution will be "271.95 N".
Explanation:
The given values are:
velocity
v = 3.5 m/s
mass
m = 9.0 kg
r = 0.6 m
According to the question:
⇒ [tex]F_{branch}=F_{gravity}+F_{centrifugal}[/tex]
⇒ [tex]=mg+\frac{mv^2}{r}[/tex]
On substituting the values, we get
⇒ [tex]=9\times 9.8+\frac{9\times (3.5)^2}{0.6}[/tex]
⇒ [tex]=88.2+\frac{110.35}{0.6}[/tex]
⇒ [tex]=271.95 \ N[/tex]
To obtain your Class E learner's license, you'll need to _____.
A. pass a vision and hearing test
B. pass a literacy test
C. submit proof of employment
D. submit proof of insurance
Answer:
This answer was wrong
Explanation:
I took the test and I missed this question. So it is not answer B: pass a literacy test.
It will need an A. pass a vision and hearing test
Class E license:It is the standard driver's license for people that drive personal vehicles. It permits for drive a noncommercial vehicle that weighs less than 26,001 pounds.So the vision and hearing test should be required.Learn more about the insurance here: https://brainly.com/question/989103?referrer=searchResults
Consider a river flowing toward a lake at an average speed of 3 m/s at a rate of 550 m3/s at a location 58 m above the lake surface. Determine the total mechanical energy of the river water per unit mass (in kJ/kg) and the power generation potential of the entire river at that location (in MW). The density of water is 1000 kg/m3, and the acceleration due to gravity is 9.81 m/s2. The total mechanical energy of the river per unit mass is kJ/kg. The power generation potential of the entire river at that location is MW..
Answer:
1. 0.574 kJ/kg
2. 315.7 MW
Explanation:
1. The mechanical energy per unit mass of the river is given by:
[tex] E_{m} = E_{k} + E_{p} [/tex]
[tex] E_{m} = \frac{1}{2}v^{2} + gh [/tex]
Where:
Ek is the kinetic energy
Ep is the potential energy
v is the speed of the river = 3 m/s
g is the gravity = 9.81 m/s²
h is the height = 58 m
[tex] E_{m} = \frac{1}{2}(3 m/s)^{2} + 9.81 m/s^{2}*58 m = 0.574 kJ/Kg [/tex]
Hence, the total mechanical energy of the river is 0.574 kJ/kg.
2. The power generation potential on the river is:
[tex] P = m(t)E_{m} = \rho*V(t)*E_{m} = 1000 kg/m^{3}*550 m^{3}/s*0.574 kJ/kg = 315.7 MW [/tex]
Therefore, the power generation potential of the entire river is 315.7 MW.
I hope it helps you!
If you are driving to see your cousins at a speed of 84.6 km/h and it took you 6.5 h to get there, how far did you travel?
Answer: 549.9 km
Explanation: 84.6km every hour so 84.6*6.5= 549.9
What are two different ways that resistors can be connected together in a circuit? Explain the difference between the two in terms of calculating equivalent resistance.
Explanation:
Resistors in circuits can either be connected in series or in parallel.
For a series connected resistors, the same current flows through the resistors while different voltages is passed across them. The formula for calculating effective resistances in a circuit is expressed as
[tex]R_T = R_1+R_2+R_3...+R_n[/tex] where R1, R2, R3 etc. are the resistances connected in series.
The effective resistances for series connected resistors is the sum of the individual resistors
For a parallel connected resistors, different current flows through the resistors while same voltage is passed across them. The formula for calculating effective resistances in a circuit is expressed as:
[tex]\frac{1}{R_T} = \frac{1}{R_1} +\frac{1}{R_2} +\frac{1}{R_3} +...\frac{1}{R_n}[/tex]
The effective resistances for parallel connected resistors is the sum of the reciprocal of the individual resistors
how can philosophy help you become a productive citizen
Answer:
Philosophy is a study that involves the nature of knowledge and truth. It serves as a guide that helps an individual seek which things are valuable and essential in life. ... It gives you a sense of direction, knowing the weight of things, therefore making you more productive.
a tiger leaps with an initial velocity of 55 km/hr at an angle of 13° with respect to the horizontal. what are the components of the tigers velocity?
Answer:
vₓ = 53.6 km/h
vy = 12.4 km/h
Explanation:
if we define two axis perpendicular each other with origin in the point represented by the tiger leaping (assuming we can treat it as a point mass) coincidently with the horizontal (x-axis) and vertical (y-axis) directions, we can obtain the components of the velocity in both independent directions.We can do it simply getting the projections of the velocity vector on both axes, using simple trigonometry, as follows:[tex]v_{x} = v_{o} * cos \theta = 55 km/h * cos 13 = 53.6 km/h[/tex]
[tex]v_{y} = v_{o} * sin\theta = 55 km/h * sin 13 = 12.4 km/h[/tex]
What is the probability that a junior non-Physics major and then a freshman non-Physics major are chosen at random?
Answer:
Probability = 0.0244
Explanation:
Probability that Junior Non Physics Major & then a Freshman Non Physics Major are chosen:
Prob (Jr No-Ph Mjr) = Jr No-Ph Mjr / Total
= 18 / 82 = 0.2195
Prob (Fr No-Ph Mjr) = Fr No-Ph Mjr / Total (remaining)
= 9 / 81 = 0.1111
Prob [ Jr No-Ph Mjr & Fr No-Ph Mjr ] = 0.2195 x 0.1111 = 0.02439
≈ 0.0244
You are driving your car at 45 m/s, when a raccoon runs into the street in front of you. You slam on the brakes and come to a stop in 5 seconds. What is the acceleration of your car?
Answer:
-9m/s²
Explanation:
Given parameters:
Initial velocity = 45m/s
Final velocity = 0
duration = 5s
Unknown:
acceleration = ?
Solution:
Acceleration is the rate of change of velocity with time;
Acceleration = [tex]\frac{v- u}{t}[/tex]
v is the final velocity
u is the initial velocity
t is the time taken
Input the parameters and solve;
Acceleration = [tex]\frac{0 - 45}{5}[/tex] = -9m/s²
The car accelerates at a rate of -9m/s² which is a deceleration
g A particle moves according to a law of motion s = f(t), t ≥ 0, where t is measured in seconds and s in feet. f(t) = 0.01t4 − 0.03t3 (a) Find the velocity at time t (in ft/s). v(t) = 0.04t^3−0.09t^2 Correct: Your answer is correct. (b) What is the velocity after 1 second(s)? v(1) = -0.05 Correct: Your answer is correct. ft/s (c) When is the particle at rest? t = 0 Correct: Your answer is correct
Answer:
Explanation:
If a particle move with time and expressed according to the formula:
f(t) = 0.01t⁴ − 0.03t³
a) Velocity is the change in motion of the particle with respect to time and it is expressed as;
[tex]v(t) =\frac{d(f(t))}{dt}[/tex]
[tex]v(t) = 4(0.01)t^{4-1} - 3(0.03)t^{3-1}\\v(t) = 0.04t^3 - 0.09t^2[/tex]
Hence the velocity of the particle at time t is [tex]v(t) = 0.04t^3 - 0.09t^2[/tex]
b) To calculate the velocity after 1 second, we will substitute t = 1 into the function v(t) in (a) as shown:
[tex]v(t) = 0.04t^3 - 0.09t^2\\v(1) = 0.04(1)^3 - 0.09(1)^2\\v(t) = 0.04 - 0.09\\v(t) = -0.05[/tex]
Hence the velocity after 1second is -0.05
c) The particle is at rest when when the time is zero.
Initially, the body is not moving and the time during this time is 0. Hence the particle is at rest when t = 0second
21. Prediction: If you were to measure the current at points A, B and C, how do you think the values would compare? Why? 22. Prediction: If you were to measure the potential differences across these bulbs (what the voltmeter measures) how do you think the values will compare to each other and to the potential difference across the battery pack or the power supply? Why?
Answer:
hello your question is incomplete attached below is the complete question
21) The current at points B and C would be the same ( identical bulbs) while the current at Point A will be greater than the currents at point B and C. i.e. twice the current at either point B or point C
22) The potential difference across the bulbs will be the same and this is because the bulbs are connected in parallel to the the power source ( battery)
hence the voltage in the battery will be equal to the voltage across each bulb
Explanation:
The current at points B and C would be the same ( identical bulbs) while the current at Point A will be greater than the currents at point B and C. i.e. twice the current at either point B or point C
The potential difference across the bulbs will be the same and this is because the bulbs are connected in parallel to the the power source ( battery)
hence the voltage in the battery will be equal to the voltage across each bulb
It takes 525 J of work to compress a spring 25 cm. What is the force constant of the spring (in kN/m)?
Answer:
1.680kN/m
Explanation:
Work done by the spring is expressed as shown:
[tex]W = \frac{1}{2}ke^2[/tex] where:
k is the spring constant
e is the extension
Given
W = 525Joules
extension = 25cm = 0.25m
Substitute into the formula:
[tex]525 = \frac{1}{2}k(0.25)^{2} \\525 = \frac{0.0625k}{2}\\ 525 = 0.03125k\\k = \frac{525}{0.3125}\\k = 1680N/m\\k = 1.680kN/m[/tex]
Hence the force constant of the spring is 1.680kN/m
Suppose a star the size of our Sun, but with mass 9.0 times as great, were rotating at a speed of 1.0 revolution every 7.0 days. If it were to undergo gravitational collapse to a neutron star of radius 13 km , losing three-quarters of its mass in the process, what would its rotation speed be
Answer:
Its rotation will be 3.89x10⁴ rad/s.
Explanation:
We can find the rotation speed by conservation of the angular momentum:
[tex] L_{i} = L_{f} [/tex]
[tex] I_{i}\omega_{i} = I_{f}\omega_{f} [/tex] (1)
The initial angular speed is:
[tex] \omega_{i} = \frac{1 rev}{7 d} = 0.14 \frac{rev}{d} [/tex]
The moment of inertia (I) of a sphere is:
[tex] I = \frac{2}{5}mr^{2} [/tex] (2)
Where m is 9 times the sun's mass and r is the sun's radius
By entering equation (2) into (1) we have:
[tex] \frac{2}{5}m_{i}r_{i}^{2}\omega_{i} = \frac{2}{5}m_{f}r_{f}^{2}\omega_{f} [/tex]
[tex]9m_{sun}(696342 km)^{2}0.14\frac{rev}{d} = \frac{3}{4}9m_{sun}(13 km)^{2}\omega_{f}[/tex]
[tex]\omega_{f} = \frac{4}{3}*0.14 \frac{rev}{d}(\frac{696342 km}{13 km})^{2} = 5.36 \cdot 10^{8} \frac{rev}{d}*\frac{1 d}{24 h}*\frac{1 h}{3600 s}*\frac{2\pi rad}{1 rev} = 3.89 \cdot 10^{4} rad/s[/tex]
Hence, its rotation will be 3.89x10⁴ rad/s.
I hope it helps you!
A 2-m3 rigid insulated tank initially containing saturated water vapor at 1 MPa is connected through a valve to a supply line that carries steam at 400°C. Now the valve is opened, and steam is allowed to flow slowly into the tank until the pressure in the tank rises to 2 MPa. At this instant the tank temperature is measured to be 300°C. Determine the mass of the steam that has e
Answer:
5.6449
9 mpa
Explanation:
we are to determine mass of steam that has entered and also the pressure of steam.
After solving
Mass of steam = m2 - m1
= 15.925-10.2901
= 5.6449kg
Then the enthalpy of steam was calculated to be 3109.26
Using steam table, tl = 400⁰c
Hl = 3109.26
Supply line pressure = 9mpa
Please refer to attachment for all calculations
A pole-vaulter just clears the bar at 5.53 m and falls back to the ground. The change in the vaulter's potential energy during the fall is -3200 J. What is his weight?
Answer:
578.66 N
Explanation:
The first step is to calculate the mass
mgh= 3200J
3200/9.8×5.53
3200/54.194
m = 59.047 kg
Therefore the weight can be calculated as follows
Weight = m × g
= 59.047 × 9.8
= 578.66 N
Complete each statement by dragging the forms of energy into their appropriate boxes.
wind turbine
roller coaster going downhill
toaster
car
A
converts electrical energy into thermal energy.
A
converts rotational energy into electrical energy.
A
converts gravitational energy into mechanical energy.
A
converts rotational energy into mechanical energy.
Statements 1,2,3 and 4 match statements B, C, A, and D respectively.A wind turbine converts rotational energy into electrical energy.
What is the law of conservation of energy?According to the Law of conservation of energy. Energy can not be created nor be destroyed, it can transfer from one to another form.
1.A wind turbine converts rotational energy into electrical energy.
2.A roller coaster going downhill converts gravitational energy into mechanical energy
3. Toaster converts electrical energy into thermal energy
4.A car converts rotational energy into mechanical energy.
Hence,statements 1,2,3 and 4 match statements B, C, A, and D respectively.
To learn more about the law of conservation of energy refer:
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3- For given three vectors a, b and c, c = a x b, then the vector c is:
Answer:
VB
Explanation:
You are studying circular motion by placing pennies on a turntable and then turning it on so that it will spin. You keep increasing the speed until one of the pennies slips off. You repeat this procedure and observe that the penny close to the outer edge always slip off first. What is the best inference?
1.The penny near the edge has a greater tangential velocity than the one in the center, so it experiences more air resistance. It’s the effect of the air “blowing” it off.
2.The centripetal force required to keep the pennies in place increases with the distance from the center. Eventually, as the turntable spins faster, the friction force between the turntable and the penny near the edge is not enough to supply the required centripetal force.
3.The centrifugal force acting on the pennies is stronger on the one near the edge than the one near the center.
Answer:
2.The centripetal force required to keep the pennies in place increases with the distance from the center. Eventually, as the turntable spins faster, the friction force between the turntable and the penny near the edge is not enough to supply the required centripetal force.
Explanation:
centripetal force = m ω² R
here m is mass , ω is angular velocity and R is distance of penny from centre
So this force depends upon R
penny on the outer edge will require greater centripetal force to move in circular path .
The centripetal force will be provided by frictional force of table which is same for both the coin . Hence the penny on the outer edge will slip off first the moment , frictional force reach its maximum value for it . But it will be sufficient to keep in balance the penny nearer to the centre .
Question 1-1: In each case, lifting or pushing, why must you exert a force to keep the object moving at a constant velocity?
Answer:
We must apply a force to keep the object moving at a constant velocity due to gravitational force or weight (in case of lifting), and due to frictional force (in case of pushing).
Explanation:
LIFTING:
When an object is lifted, we first need to overcome the force exerted on it by the field of gravity. Due to this force, which is also called the weight of object, we must apply a force on the object to keep it moving at constant speed, otherwise the gravity force will cause the object to slow down and eventually fall back on ground.
PUSHING:
When pushing an object the person must apply the force to first overcome the frictional force. The frictional force acts in opposite direction of motion. Thus, to move the object at constant speed we must apply force to it.
Hence, we must apply a force to keep the object moving at a constant velocity due to gravitational force or weight (in case of lifting), and due to frictional force (in case of pushing).
Why does the brightness of a bulb not change noticeably when you use longer copper wires to connect it to the battery?
a. All the current is used up in the bulb, so the connecting wires don't matter.
b. Very little energy is dissipated in the thick connecting wires.
c. Electric field in the connecting wires is zero, so emf = E_bulb * L_bulb.
d. Current in the connecting wires is smaller than current in the bulb.
e. The electric field in connecting wires is very small, so emf almost = E_ bulb * L_bulb.
Answer:
Options B & E are correct
Explanation:
Looking at all the options, B & E are the correct ones.
Option B is correct because the thicker the wire per unit length, the lesser resistance it will posses and the lesser the energy that will be dissipated by the wire and in return more energy will be dissipated by the bulb.
Option E is also correct because the resistance of the copper wires is low enough to ensure that there's not much drop in voltage across the copper wires. Thus, there will not be any noticeable differences in the voltage across the bulb.
Option A is not correct because the current is not used up and thus the charge is conserved, and it will circulate just through the circuit.
Option C is not correct because although the Electric field along the wire is not zero, it is very small.
Option D is not correct because the wires and the light bulb are connected in series and as such, the current in both the wires and the light bulb will be identical.
The brightness of a bulb that not change noticeably when you use longer copper wires to connect it to the battery is :
b. Very little energy is dissipated in the thick connecting wires.
e. The electric field in connecting wires is very small, so emf almost = E_ bulb * L_bulb.
"Energy"The brightness of a bulb that not change noticeably when you use longer copper wires to connect it to the battery is very little energy is dissipated in the thick connecting wires and the electric field in connecting wires is very small, so emf almost = E_ bulb * L_bulb.
The thicker the wire per unit length, the lesser resistance it'll posses and the lesser the vitality that will be scattered by the wire and in return more vitality will be disseminated by the bulb.
The resistance of the copper wires is low sufficient to guarantee that there's not much drop in voltage over the copper wires. Hence, there will not be any noticeable contrasts within the voltage over the bulb.
Thus, the correct answer is B and E.
Learn more about "Circuit":
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You are working out on a rowing machine. Each time you pull the rowing bar (which simulates the oars) toward you, it moves a distance of 1.1 m in a time of 1.8 s. The readout on the display indicates that the average power you are producing is 90 W. What is the magnitude of the force that you exert on the handle?
Answer:
147.27N
Explanation:
Power = workdone/time
Power = Force*distance/time
Given
Power = 90Watts
Distance = 1.1m
Time = 1.8secs
Force = ?
Substitute the given parameters into the formula:
[tex]90 = \frac{1.1d}{1.8}\\cross \ multiply\\ 90 \times 1.8 = 1.1F\\162 = 1.1F\\1.1F = 162\\F = \frac{162}{1.1} \\F = 147.27N[/tex]
Hence the magnitude of the force that you exert on the handle is 147.27N
Question C) needs to be answered, please help (physics)
(a) Differentiate the position vector to get the velocity vector:
r(t) = (3.00 m/s) t i - (4.00 m/s²) t² j + (2.00 m) k
v(t) = dr/dt = (3.00 m/s) i - (8.00 m/s²) t j
(b) The velocity at t = 2.00 s is
v (2.00 s) = (3.00 m/s) i - (16.0 m/s) j
(c) Compute the electron's position at t = 2.00 s:
r (2.00 s) = (6.00 m) i - (16.0 m) j + (2.00 m) k
The electron's distance from the origin at t = 2.00 is the magnitude of this vector:
||r (2.00 s)|| = √((6.00 m)² + (-16.0 m)² + (2.00 m)²) = 2 √74 m ≈ 17.2 m
(d) In the x-y plane, the velocity vector at t = 2.00 s makes an angle θ with the positive x-axis such that
tan(θ) = (-16.0 m/s) / (3.00 m/s) ==> θ ≈ -79.4º
or an angle of about 360º + θ ≈ 281º in the counter-clockwise direction.