Answer:
velocity is the speed of something in a particular direction while time is what is measured in hours , minutes and seconds
A 85.0kg man and a 65, 0kg woman are riding a Ferris wheel with a radius of 20.0m. What is the Ferris wheels tangential velocity if the net centripetal force on the woman is 115N
Answer:
The Ferris wheel's tangential (linear) velocity if the net centripetal force on the woman is 115 N is 3.92 m/s.
Explanation:
Let's use Newton's 2nd Law to help solve this problem.
F = maThe force acting on the Ferris wheel is the centripetal force, given in the problem: [tex]F_c=115 \ \text{N}[/tex].
The mass "m" is the sum of the man and woman's masses: [tex]85+65= 150 \ \text{kg}[/tex].
The acceleration is the centripetal acceleration of the Ferris wheel: [tex]a_c=\displaystyle \frac{v^2}{r}[/tex].
Let's write an equation and solve for "v", the tangential (linear) acceleration.
[tex]\displaystyle 115=m(\frac{v^2}{r} )[/tex][tex]\displaystyle 115 = (85+65)(\frac{v^2}{20})[/tex][tex]\displaystyle 115=150(\frac{v^2}{20} )[/tex][tex].766667=\displaystyle(\frac{v^2}{20} )[/tex][tex]15.\overline{3}=v^2[/tex][tex]v=3.9158[/tex]The Ferris wheel's tangential velocity is 3.92 m/s.
Describe what will happen to
the molecules below if the
temperature is increased by
20°C.
any form of stored energy is described as
Given the information in the velocity vs. time graph, what is the displacement of the object after 1 second?
In the velocity vs. time graph, the displacement of the object after 1 second is 3 m.
What is displacement?The displacement of an object is the change in position of the object.
From the graph, the displacement of the object after 1 second is calculated as follows;
x = vt
where;
v is velocity = 3 m/st is time = 1 secondx = 3 x 1 = 3 m
Thus, in the velocity vs. time graph, the displacement of the object after 1 second is 3 m.
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How long will it take you to ride your bike 120 miles if you are riding at 5mph?
Determine the net force required to accelerate a 2,160 kg truck from 0 to 27 m/s in 10 s
∑F=m.a = m.(Δv/Δt)
∑F=2.16 x (27-0/10)
∑F=5.832 N
Which best ilustrates the way in which radiation transfers thermal energy?
Answer:
B) Warm radiates to Cool
Explanation:
Radiation transfers thermal energy without need for contact, where areas with higher thermal energy (hot) transfer energy to areas with lower thermal energy (cold)
5. For the following pairs of masses, which one has the largest gravitational attraction between masses? Which one has the smallest gravitational attraction between masses? Explain.
Answer:
MOST : E LEAST : D
Explanation:
Grav attraction = G m1 m1 / r^2
G and 'r' are the same for all of the pictures, so the one picture with the biggest mass has the MOST attraction and the one with the least mass has the least attraction.
The pair with the largest gravitational attraction is the Earth and the Moon.
The pair with the smallest gravitational attraction is the Earth and a grain of sand.
The gravitational force between two masses is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. In other words, the more massive the objects, the stronger the gravitational force between them. The closer the objects are, the stronger the gravitational force between them.
The Earth and the Moon are the most massive objects in the pair. The Earth has a mass of about 5.972 × 10²⁴ kg, and the Moon has a mass of about 7.34 × 10²² kg. The distance between the Earth and the Moon is about 384,400 km.
The grain of sand is much less massive than the Earth or the Moon. The mass of a grain of sand is about 10⁻⁶ kg. The distance between the Earth and a grain of sand is much greater than the distance between the Earth and the Moon.
Therefore, the gravitational attraction between the Earth and the Moon is much stronger than the gravitational attraction between the Earth and a grain of sand.
Here is a table of the gravitational attractions for each pair of masses:
Pair of masses Gravitational attraction
Earth and Moon 1.98 × 10²⁰ N
Earth and a grain of sand 1.0 × 10⁻¹⁷ N
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If the trend changed toward traditional (pre-World War II) families, how would that affect women’s rights?
If the trend changed toward traditional (pre-World War II) families, the women’s rights are employment in manufacturing sector.
What is World war?
The war between two countries to take over each other's kingdom using weapons to kill each other.
Before the world war, the army needs armor, weapons, guns and tanks. Their manufacturing is only possible with many workers to work for long hours. If the men are not enough, then women are given opportunities to work with them.
Thus, the women’s rights are employment in manufacturing sector when trend changed traditional families.
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How many times is the motion of the object stopped?
Answer:
4
Explanation:
We can say the motion of the object is stopped when its velocity is 0. That means we just have to check how many times the line of the graph intersects the x-axis. Hence, we can say that the object of the motion is stopped 4 times.
A (1.28×10-5 C) charge is moving at (5.63×107 m/s) perpendicular to a magnetic field
of (8.91×10-4 T). What is the magnitude of the magnetic force on the charge?
A. 6.42×10 -1 N
B. 5.02×10 4 N
C. 7.21×10 2 N
D. 1.14×10 -8 N
Hello!
We can use the following equation for magnetic force on a moving particle:
[tex]F_B = qv \times B[/tex]
[tex]F_B[/tex] = Magnetic Force (? N)
q = Charge of particle (1.28 × 10⁻⁵C)
v =velocity of particle (5.63× 10⁷m/s)
B = Strength of magnetic field (8.91 × 10⁻⁴T)
This is a cross-product, so the magnetic force depends on the SINE of the angle between the particle's velocity vector and the magnetic field vector.
Since the charge is moving PERPENDICULAR to the field, the angle between the velocity and magnetic field is 90°. The sine of 90° = 1, so we can simplify the equation to:
[tex]F_B = qvB(1) = qvB\\[/tex]
Plug in the values and solve.
[tex]F_B = (1.28*10^{-5})(5.63*10^7)(8.91*10^{-4}) = \boxed{\text{A. } 0.6421 N}[/tex]
can someone help with the vocabulary of the relative age of rocks
Waldo gets stopped by the police.
Answer:
At least they found him.
Explanation:
in the R-L-C series circuit shown, suppose R = 200 Ω, L = 70 mH, C = 2.0 μF, Vmax = 80V and the angular frequency of the voltage source ω = 2,000 rad/s. (a) Find the reactances and , and the impedance . (3 marks) (b) Find the maximum current amplitude, Imax and the maximum voltage amplitude across each circuit element. (4 marks). (c) Explain why the maximum voltage across each element add up to more than 80 V. (3 marks)
Hello!
A)
We can solve for the reactances of each element using the following equations.
Capacitive reactance:
[tex]X_C = \frac{1}{\omega C}[/tex]
[tex]X_C[/tex] = Capacitive reactance (Ω)
ω = Angular frequency (2000 rad/sec)
C = Capacitance (2 μF)
Plug in the given values and solve.
[tex]X_C = \frac{1}{(2000)(0.000002)} = \boxed{250 \Omega}[/tex]
Inductive reactance:
[tex]X_L = \omega L[/tex]
[tex]X_L[/tex] = Inductive reactance (Ω)
L = Inductance (70 mH)
Solve:
[tex]X_L = 2000 * 0.07 = \boxed{140 \Omega}[/tex]
We can solve for impedance using the following equation:
[tex]Z = \sqrt{R^2 + (X_L - X_C)^2}[/tex]
Z = Impedance (Ω)
R = Resistance (200Ω)
Find the impedance using the values above and the given resistance:
[tex]Z = \sqrt{200^2 + (140 - 250)^2} = \boxed{228.254 \Omega}[/tex]
B)
The maximum current amplitude is found using Ohm's law:
[tex]I_{Max} = \frac{V_{Max}}{Z}[/tex]
[tex]V_{Max}[/tex] = 80V
Z = 228.254 Ω
Solve.
[tex]I_{Max} = \frac{80}{228.254} = \boxed{0.3505 A}[/tex]
To find the max voltages across each element, we can also use Ohm's Law.
For the resistor:
[tex]V_{R, max} = i_{Max} R\\\\V_{R, max} = (0.3505)(200) = \boxed{70.097 V}[/tex]
For the capacitor:
[tex]V_{C, max} = i_{Max}X_C\\\\V_{C, max} = (0.3505)(250) = \boxed{87.622 V}[/tex]
For the inductor:
[tex]V_{L, max} = i_{Max}X_L\\\\V_{L, max} = (0.3505)(140) = \boxed{49.068 V}[/tex]
C)
The maximum voltage across each element is greater than 80 V because of the transfer of energy between the capacitor and inductor that occur at different times (energy is stored and released) and being in addition to the voltage of the power source.
A sprinter starting to run from rest, reaches a speed of 8m/s in 2 second, what is the average power needed to accelerate like this ?
Answer:
okay
Explanation:
power is rate of doing work
power is workdone over time taken
but you don't have information concerning the force
Power = work/time
Work = force x distance
Force = mass x acceleration
So Power = (mass x acceleration x distance) / (time)
acceleration = 8m/s / 2sec = 4 m/s²
distance = (avg speed) x (time) = 8 m
time = 2 sec
we don't know the runner's mass
Power = (mass) x (4 m/s²) x (8 m) / (2 sec)
Power = (16 x mass in kg) watts
=============================================
Slightly easier way to do it:
Power = (runner's kinetic energy at the end) / (time)
Power = (1/2 m v²) / (time)
Power = (1/2 mass) (64 m²/s²) / (2 sec)
Power = (16 x runner's mass in kg) watts
9
Robin chases a squirrel moving 6 meters from the porch to a tree in 1.2 seconds. Wha
was his rate of movement?
Answer:
5 m/s
Explanation:
Rate of movement is distance / time
6 m / 1.2 s = 5 m/s
Throughout the reflection, make sure you have a
down menus to complete the statements.
copy of the Student Guide and your data tables. Use the drop-
In Part I of this experiment, the
was intentionally manipulated. This was the independent
variable. The dependent variable measured was the
In Part II of this experiment, the
was intentionally manipulated. This was the
independent variable. The dependent variable measured was the
Answer:
The dependent variable is the variable that is studied while the independent variable is the variable that is being manipulated
Answer: In this experiment, the
✔ fan speed
was intentionally manipulated. This was the independent variable.
The dependent variable measured was the
✔ acceleration
.
Explanation:
Legolas shoots 333 arrows at once from his bow. He has 177177177 arrows.
How many times can Legolas shoot his bow before he needs more arrows?
Answer:
532063
Explanation:
If he's shooting 333 arrows per shot and has 177177177 arrows, you need to divide to find the number of arrows he can shoot total, ignoring the decimal since you can't shoot a partial arrow.
177177177/333=532063
At left A red ball in a box with arrows pointing away from the ball in all directions. In the middle, a blue ball in a box with arrows pointing toward the ball from all directions. At right, a grey ball in a box. Which object represents a negatively charged particle? Which object represents a positively charged molecule? Which object represents an uncharged molecule? Which object will not move when in an electric field?
Answer: The answers are, 1) B. 2) A. 3) C. 4) C.
here is the proof maybe u will understand it more than the numbers and letters lm.ao
Explanation:
The red ball with arrows pointing away is a negatively charged particle, the blue ball is a positively charged molecule, and the grey ball is an uncharged molecule. Electric fields don't move uncharged molecules.
Analyse each scenario to identify representations:
A negatively charged particle is the red ball in a box with all arrows pointing away from it. The red ball generates an electric field, and negatively charged particles radiate electric field lines. The arrows indicate how a positive test charge would move in the electric field.
A positively charged molecule is the blue ball in a box with all arrows pointing to it. An electric field converges on the blue ball, signifying its destination. The electric field lines point towards positively charged particles, indicating the direction a positive test charge would move.
A boxed grey ball depicts an uncharged molecule. The ball has no arrows. Uncharged molecules do not emit or experience electric fields. The grey ball is electrically neutral because it lacks electric pitch lines.
An electric field will not move the grey ball in a box (uncharged molecule). Electric fields do not interact with uncharged molecules. Thus, an electric field will keep the grey ball still.
Science and engineering require understanding these representations and their electric field behaviour. Scientists and engineers can create and manipulate electrical systems, electronics, and other technologies that use electric charge interactions by recognising charged particles, uncharged molecules, and electric field effects.
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A ball rolls 60 centimeters along a sidewalk in 5 seconds what is the speed of the ball.
Answer:
12 cm/s
Explanation:
Quite simply, you are looking for cm/s
so 60 cm / 5s = 12 cm/s
Answer:
12 cm/second
Explanation:
To calculate the speed of an object, simply obtain the distance traveled by the ball and divide it by the total time it took to travel 60 centimeters.
⇒ Distance/Time = Speed
In this case, the ball traveled 60 centimeters in 5 seconds.
Therefore, the speed of the ball is;
60 centimeters (Distance)/5 seconds (Time) = 12 cm/second
someone help me answer this.
Bug Walking on Pivoted Ring A ring of radius R and mass my lies on its side on a frictionless table. It is pivoted to the table at its rim. A bug of mass my walks on the ring with constant speed v relative to the ring, starting at the pivot, when the ring is initially at rest. Take k to point out of the page. top view frictionless table ring bug Pluot Point (a) What is the angular velocity of the ring when the bug is halfway around? Express you answer in terms of some or all of the following: mi, m2, u, R and k. (b) What is the angular velocity of the ring when the bug is back at the pivot? Express you answer in terms of some or all of the following: mı, m2, v, R and k.
The angular velocity of the ring when the bug is halfway around and the angular velocity of the ring when the bug is back at the pivot is [m₂v / {(2m₁ +m₂)R}].
What is angular velocity?The velocity of a particle when moving in the circular path.
Let speed of the bug with respect to ground is u.
Speed of bag with respect to ring will be
v = u - (- Rω) =
Then, u = v- Rω...............(1)
Angular momentum of ring and bug will remain conserved.
Initial momentum: L ring + Lbug =0
Final momentum: -2m₁ R²ω + m₂uR =0...............(2)
Using equation (1) and (2), the angular velocity expression will be
ω =[m₂v / {(2m₁ +m₂)R}] in positive z direction
Thus, the angular velocity of the ring when the bug is halfway around and the angular velocity of the ring when the bug is back at the pivot is [m₂v / {(2m₁ +m₂)R}].
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Find the velocity using the d/t graph below
Answer:
-10/3
Explanation:
the slope of the graph is negative, the velocity is slowing. so we know the positive answers are not correct. next we just need to figure out how much of a slope the line has. it's down 10 and over 3... soo
-10/3 looks good .. the format for the fraction is rise / run . in this case rise is negative, but other wise it's the same.
How do the stomach and the small intestines work together to digest the food? (2 points)
The stomach breaks down the food and passes it to the small intestine to be digested further.
The stomach passes the digested food from the small intestine to the large intestine.
The small intestine turns the solid food into a liquid mixture and sends it to the stomach.
The stomach digests the food by using digestive juices made by the small intestine.
Answer:
A) The stomach breaks down the food and passes it to the small intestine to be digested further.
Explanation:
I am pretty sure it is the first option because there is no way the stuff you eat can be turned into a liquid, it wouldn't make sense for the stomach to pass the digested food from small intestine to large intestine, so it would have to be the first one. A out of all of the options makes more sense and the stomach has to pass the digested food somewhere.
I hope I helped you
3. Which vehicle has more momentum? *
1 point
A.
B
VA= 25 m/s
Ug = 20 m/s
ma= 15000 Kg me = 15000 Kg
Ο Α
О.
B
Answer:
the one with v = 25 m/s
Explanation:
Momentum = m * v
if they both have the same mass (15000 kg) , then the one with the higher v has more momentum...I think A= 25 m/s
(See picture) may I have help!!?
A ball of mass 0.4kg is dropped from a height of 30m. Determine the potential energy on impact to the ground
Find the wavelength of light given off by a hydrogen atom when its electron drops from the n = 4 to n = 1 energy level
Answer: 1.55 x 10 -19J on top
Explanation:
Find energy
[tex]\\ \rm\Rrightarrow E=-13.6\left(\dfrac{1}{6^2}-\dfrac{1}{4^2}\right)[/tex]
[tex]\\ \rm\Rrightarrow E=-13.6\left(\dfrac{1}{36}-\dfrac{1}{16}\right)[/tex]
[tex]\\ \rm\Rrightarrow E=-13.6(0.0278-0.0625)[/tex]
[tex]\\ \rm\Rrightarrow E=-13.6(-0.347)[/tex]
[tex]\\ \rm\Rrightarrow E=0.47eV[/tex]
Now
[tex]\\ \rm\Rrightarrow E=h\nu[/tex]
[tex]\\ \rm\Rrightarrow \nu=\dfrac{E}{h}[/tex]
[tex]\\ \rm\Rrightarrow \nu=\dfrac{0.47\times 10^{-19}}{6.626\times 10^{-34}}[/tex]
[tex]\\ \rm\Rrightarrow \nu=0.071\times 10^{15}[/tex]
[tex]\\ \rm\Rrightarrow \nu=7.1\times 10^{13}Hz[/tex]
If you drive 2 hours at 50mph and then drive 4 hours at 65mph, how far did you drive in all?
Answer:
i think its 400 miles driven
Explanation:
attempted to do it...
QUESTION 4 A ball is dropped from a height. If it takes 0.2 s to cross the last 6 m before hitting the ground, find the approximate height from which it is dropped.
Answer:
S = V t + 1/2 g t^2 where the ball has a speed of V and falls for .2 sec
6 = V t + 4.9 * .04 = V * .2 + .2
V = (6 - .2) / .2 = 29 m/s speed entering area
T = V / g = 29 / 9.8
T = 2.96 sec time to reach speed V
H = 1/2 g t^2 time to fall a distance H
H = 4.9 * 2.96^2 = 42.9 m height from which ball was dropped
Check using first equation:
S = 29 * .2 + 4.9 + .2^2 = 6 m
Two spherical objects have masses of 7.5 x 10 kg and 9.2 x 10'kg. Their centers are
separated by a distance of 2.5 x 103 m. What is the magnitude of the gravitational force
between them?
Answer: 7.364 x 10^-4 N
Explanation:
[tex]$$Given:$$\begin{aligned}m_{1} &=7.5 \times 10^{5} \mathrm{~kg} \\m_{2} &=9.2 \times 10^{7} \mathrm{~kg} \\r &=2.5 \times 10^{3} \mathrm{~m}\end{aligned}$$Here, $m_{1}, m_{2}$, and $r$ are the first mass, the second mass, and the distance[/tex]
[tex]$$The gravitational force exerted between the masses is determined as follows.$$F=G \frac{m_{1} m_{2}}{r^{2}}$$Here, $G$ is the gravitational constant.Substitute the known values.$$\begin{aligned}F &=\left(6.67 \times 0^{-11} \mathrm{~N} \mathrm{~m}^{2} / \mathrm{kg}^{2}\right) \frac{\left(7.5 \times 10^{5} \mathrm{~kg}\right)\left(2.5 \times 10^{7} \mathrm{~kg}\right)}{\left(2.5 \times 10^{3} \mathrm{~m}\right)^{2}} \\&\bold{=7.364 \times 10^{-4} \bold{\mathrm{~N}}}\end{aligned}$$[/tex]