Answer:
groups
Explanation:
Answer: Groups
Explanation: They are in the same group! Like the Alkaline Metals are all the group. They all lose an electron. :)
Leticia leaves the grocery store And walks 150 M’s to parking lot then she turns 90° to the right and walks an additional 70 M’s to her car what is the magnitude of displacement of her car from the grocery store at exit
Answer:
Explanation:
Its 165.5m
An unstrained horizontal spring has a length of 0.29 m and a spring constant of 180 N/m. Two small charged objects are attached to this spring, one at each end. The charges on the objects have equal magnitudes. Because of these charges, the spring stretches by 0.021 m relative to its unstrained length. Determine (a) the possible algebraic signs and (b) the magnitude of the charges.
Answer:
A) they both have the same algebraic sign
B)6.377×10^-6 C
Explanation:
From columb's law, the force acting on both charges can be expressed as
F=( kq1*q2)/r^2
Where F= electrostatic force
r= distance between the charges
q1 and q2= charges
The force acting on a spring can be expressed as
F= kx..................eqn(2)
Where
K= spring constant = 180 N/m.
x= stretch of the string= 0.021m
Substitute the values into eqn (2)
F= (180×0.021)
F= 3.78N
If we compare with spring force,
Hence, F( electrostatic) = 3.78N
From
F=( kq1*q2)/r^2 ..............eqn(1)
Where
r= (0.29 m + 0.021m)= 0.311m
K= the electrostatic constant= 8.99×10^9 kg⋅m3⋅s−2⋅C−2.
If we substitute the values we have
Since the charges are the same, then
kq1 and q2 equals "q"
3.78= (8.99×10^9 ×q^2)/(0.311)^2
Making q^2 subject of the formula
3.78× (0.311)^2 = 8.99×10^9 × q^2
q^2= [(0.311)^2 × 3.78]/8.99×10^9
q^2= 40.668×10^-12
q=√40.668×10^-12
q= 6.377×10^-6 C
(a) the possible algebraic signs
They have the same algebraic sign
(b) the magnitude of the charges.
6.377×10^-6 C
A salmon jumps up a waterfall 2.4 m high. With what minimum speed did the salmon leave the water below to reach the top?
Answer:
6.86 m/s
Explanation:
The minimal velocity needed is when we have only vertical motion, then i will think in the problem only in one axis.
I suppose that the only force, in this case, is the gravitational force acting on the fish.
Then the gravitational equation of the fish will be:
a(t) = -9.8m/s^2
For the velocity equation we need to integrate over time to get:
v(t) = (-9.8m/s^2)*t + v0
Where v0 is the initial velocity of the fish and is what we want to find.
For the position equation we need to integrate over time again to get:
p(t) = (1/2)*(-9.8m/s^2)*t^2 + v0*t + p0
p0 is the initial position of the fish, and because he starts one the water, the initial position is p0 = 0 m
Then the equation is:
p(t) = (1/2)*(-9.8 m /s^2)*t^2 + v0*t
p(t) = (-4.9 m/s^2)*t^2 + v0*t
We know that the maximum height is 2.4m
The value of time at which the fish gets his maximum height is when the velocity of the fish is equal to zero, then we first need to solve:
v(t) = (-9.8m/s^2)*t + v0 = 0
t = v0/9.8m/s^2
Now we replace this in the position equation to get the maxmimum height, which is equal to 2.4m
2.4m = p( v0/9.8m/s^2) = (1/2)*(-9.8 m /s^2)*(v0/9.8m/s^2)^2 + v0*(v0/9.8m/s^2)
2.4m = (1/2)(-v0)^2(-9.8 m /s^2) + v0^2/(9.8m/s^2))
2.4m = (1 - 1/2)*v0^2/(9.8m/s^2)
2.4m = 0.5*v0^2/(9.8m/s^2)
2.4m/0.5 = v0^2/(9.8m/s^2)
4.8m*(9.8m/s^2) = v0^2
√(4.8m*(9.8m/s^2)) = v0 = 6.86 m/s
What do you think about the attached scenario?
If a person weighs 140 lb'on Earth, their mass in kilograms is
Answer:
70 kg
Explanation:
divide it by 2
Hope this helped!
Answer:
63.502932 Kilograms
Explanation:
A 3 kg ball moving to the right at 4 m/sec collides with a 4 kg ball moving to the right at 2 m/sec. Find the final velocities of the balls in m/sec if the coefficient of restitution is 0.6.
A. 2.2, 3.4
B. 1, 2
C. 4, 5
D. 6, 8
Answer:
Option A
Explanation:
To solve this problem we need to apply the momentum conservation, and analyze the data.
For this problem, I will call the initial velocities as V₁ and V₂, while the final velocities will be V₃ and V₄.
According to the momentum principle, this states the following:
m₁V₁ + m₂V₂ = m₁V₃ + m₂V₄ (1)
From this equation we can write an expression in function of V₃ and V₄. We also know that coefficient of restitution is 0.6. Knowing this, we can write the expression that will help us to solve for the final velocities:
e = V₄ - V₃ / 2 (2)
With both expressions we can solve for the final velocities. Let's use (1) first and see what we can simplify first by replacing the given data:
(3*4) + (4*2) = 3V₃ + 4V₄
12 + 8 = 3V₃ + 4V₄
20 = 3V₃ + 4V₄ (3)
This is all we can do for now. Let's use (2) now:
0.6 = V₄ - V₃ / 2
1.2 = V₄ - V₃
V₄ = 1.2 + V₃ (4)
Now, we can replace (4) into (3), and then, solve for V₃:
20 = 3V₃ + 4(1.2 + V₃)
20 = 3V₃ + 4.8 + 4V₃
15.2 = 7V₃
V₃ = 15.2 / 7
V₃ = 2.17 m/sWe have the value of one final velocity, let's see the other one.
V₄ = 1.2 + V₃
V₄ = 1.2 + 2.17
V₄ = 3.37 m/sThe closest values to these results are in option A, so this will be the correct option.
Hope this helps
he potential energy between two atoms in a particular molecule has the form U(s) = 2.6/x^8 - 4.3/x^4 where the units of x are length and the numbers 2.G and 4.3 have appropriate units so that U(x) has units of energy. What b the equilibrium separation of the atoms (that is the distance at which the force between the atoms is zero)?
Answer:
x = 1.04866
Explanation:
Force can be defined from power energy by the expressions
F = [tex]- \frac{ dU}{ dx}[/tex]
in this case we are the expression of the potential energy
U = [tex]\frac{2.6}{x^{8} } - \frac{4.3}{ x^{4} }[/tex]
let's find the derivative
dU / dx = 2.6 ( [tex]\frac{-8}{x^{9} }[/tex]) - 4.3 ([tex]\frac{-4}{ x^{5} }[/tex])
dU / dx = [tex]- \frac{20.8}{ x^{9} } + \frac{17.2 }{ x^{5} }[/tex]
we substitute
F = + \frac{20.8}{ x^{9} } - \frac{17.2 }{ x^{5} }
at the equilibrium point the force is zero, so
[tex]\frac{20.8}{ x^{9} } = \frac{17.2}{ x^{5} }[/tex]
20.8 / 17.2 = x⁴
x⁴ = 1.2093
x = [tex]\sqrt[4]{ 1.2093}[/tex]
x = 1.04866
Calculate the work done to raise a charge of 25 coulombs through an emf of 8 volts.
1) 3
2) 200
Corrected, it's 2) 200
What energy store is in the torch
BEFORE it gets switched on?
Answer:
Chemical energy
Explanation:
The energy in the torch is stored as chemical energy before the torch gets switch on.
The chemical energy energy in the battery of cell will power the cell and allows it to produce light.
Chemical energy is a form of potential energy. The electrolytes within the battery are capable of producing electric current. So the chemical energy is transformed into electrical energy which is used to produce the light of the torch.A turbofan operates at 25,000 ft and moves at 815 ft/s. It ingests 1.2 times the amount of air into the fan than into the core, which all exits through the fan exhaust. The fuel-flow-to-core airflow ratio is 0.0255. The exit densities of the fan and core are 0.00154 and 0.000578 slugs/ft3, respe~tively. The exit pressures from the fan and core are 10.07 and 10.26 psia, respectively. The developed thrust is 10,580 !bf, and the exhaust velocities from the fan and core are 1147 and 1852 ft/s, respectively. (a) Find the ingested air mass flow rate for the core and TSFC. (b) What are the exit areas of the fan and core nozzles
Answer:
a)
Mass flow rate of core = [tex]m_{e}[/tex] = 60.94 Kg/s
Mass flow rate of fan = [tex]m_{s}[/tex] = 73.12 kg/s
TSFC = 3.301 x [tex]10^{-5}[/tex]
b)
Exit Area of Fan = [tex]A_{e}[/tex] = 0.3624 [tex]m^{2}[/tex]
Exit Area of Core = [tex]A_{s}[/tex] = 0.2635 [tex]m^{2}[/tex]
Explanation:
Data Given:
Height = 25000 ft
Vehicle velocity = [tex]u_{a}[/tex] = 815 ft/s = 248.41 m/s
[tex]m_{s} = 1.2m_{e}[/tex]
[tex]m_{f}[/tex] = 0.0255[tex]m_{e}[/tex]
Where,
[tex]m_{s}[/tex] = Mass flow rate of fan
[tex]m_{e}[/tex] = Mass flow rate of core
F = Thrust
Density of core = [tex]D_{e}[/tex] = 0.000578 slugs/[tex]ft^{3}[/tex] = 0.2979 kg/[tex]m^{3}[/tex]
Density of fan = [tex]D_{s}[/tex] = 0.00154 slugs/[tex]ft^{2}[/tex] = 0.7937 kg/[tex]m^{3}[/tex]
Ambient Pressure of Fan = [tex]P_{s}[/tex] = 10.07 Psi = 69430.21 Pa
Ambient Pressure of core = [tex]P_{e}[/tex] = 10.26 Psi = 70740.2 Pa
Thrust = F = 10580 lbf = 47062.2 N
Velocity of fan = [tex]u_{s}[/tex] = 1147 ft/s = 349.6 m/s
Velocity of core = [tex]u_{e}[/tex] = 1852 ft/s = 564.5 m/s
At the height of 25000 ft, P = 37600 [tex]P_{a}[/tex]
Now,
we have:
[tex]m_{e}[/tex] = [tex]u_{e}[/tex] x [tex]D_{e}[/tex] x [tex]A_{e}[/tex]
Plugging in the values, we get:
[tex]m_{e}[/tex] = 168.16 [tex]A_{e}[/tex] Equation 1
And,
[tex]m_{s}[/tex] = [tex]D_{s}[/tex] x [tex]A_{s}[/tex] x [tex]u_{s}[/tex]
[tex]m_{s}[/tex] = 277.5 [tex]A_{s}[/tex] Equation 2
As, we know,
[tex]m_{s} = 1.2m_{e}[/tex]
[tex]m_{s}[/tex] = 277.5 [tex]A_{s}[/tex]
And now for Thrust, we have:
F = [tex]A_{e}[/tex] x ([tex]P_{e}[/tex] - [tex]P_{a}[/tex] ) + [tex]A_{s}[/tex] x ([tex]P_{s}[/tex] - [tex]P_{a}[/tex] ) + [tex]m_{e}[/tex]x ([tex]u_{e}[/tex] - [tex]u_{a}[/tex] ) + [tex]m_{s}[/tex] x ([tex]u_{s}[/tex] - [tex]u_{a}[/tex] ) Equation 3
Now, substitute equation 1 and 2 in equation 3, we get:
Exit Area of Fan = [tex]A_{e}[/tex] = 0.3624 [tex]m^{2}[/tex]
Exit Area of Core = [tex]A_{s}[/tex] = 0.2635 [tex]m^{2}[/tex]
Mass flow rate of core = [tex]m_{e}[/tex] = 60.94 Kg/s
Mass flow rate of fan = [tex]m_{s}[/tex] = 73.12 kg/s
TSFC = Thrust Specific Fuel Consumption = fuel mass flow rate / Thrust
TSFC = [tex]m_{f}[/tex]/F
And,
[tex]m_{f}[/tex] = 0.0255[tex]m_{e}[/tex]
[tex]m_{e}[/tex] = 60.94
[tex]m_{f}[/tex] = 0.0255 x 60.94
[tex]m_{f}[/tex] = 1.55397
TSFC = [tex]m_{f}[/tex]/F
TSFC = 1.55397/47062.2
TSFC = 3.301 x [tex]10^{-5}[/tex]
Low TSFC = High efficiency
High TSFC = Low efficiency
a)
Mass flow rate of core = [tex]m_{e}[/tex] = 60.94 Kg/s
Mass flow rate of fan = [tex]m_{s}[/tex] = 73.12 kg/s
TSFC = 3.301 x [tex]10^{-5}[/tex]
b)
Exit Area of Fan = [tex]A_{e}[/tex] = 0.3624 [tex]m^{2}[/tex]
Exit Area of Core = [tex]A_{s}[/tex] = 0.2635 [tex]m^{2}[/tex]
a wooden block is cut into two pieces, one with three times the mass of the other. a depression is made in both faces of the cut so that a fire cracker can be placed in it and the block is reassembled. the reassembled block is set on rough surface and the fuse is lit. when the fire cracker explodes, the two blocks separate. what is the ratio of distances traveled by blocks?
Answer:
1/9
Explanation:
Let A denote the bigger piece and let B denote the smaller piece.
We are told that one with three times the mass of the other.
Therefore, we have;
M_a = 3M_b
Firecracker is placed in the block and it explodes and thus, momentum is conserved.
Thus;
V_ai = V_bi = 0
Where V_ai is initial velocity of piece A and V_bi is initial velocity of piece B.
Since initial momentum equals final momentum, we have;
P_i = P_f
Thus;
0 = (M_a × V_af) + (M_b × V_bf)
Since M_a = 3M_b, we have;
(3M_b × V_af) + (M_b × Vbf) = 0
Making V_af the subject, we have;
V_af = -⅓V_bf
The kinetic energy gained by each block during the explosion will later be lost due to the negative work done by friction. Thus;
W_f = -½M_b•(v_bf)²
Now, let's express the work is in terms of the force and the distance.
Thus;
W_f = F_f × Δx × cos 180°
Frictional force is also expressed as μmg
Thus;
W_f = -μM_b × g × Δx
Earlier, we saw that;
W_f = -½M_b•(v_bf)²
Thus;
-½M_b•(v_bf)²= -μM_b × g × Δx
Δx = (v_bf)²/2μg
Let the distance travelled by block A be Δx_a and that travelled by B be Δx_b
Thus;
Δx_a/Δx_b = ((v_ba)²/2μg)/((v_bf)²/2μg)
Δx_a/Δx_b = ((v_af)²/((v_bf)²)
Δx_a/Δx_b = (-⅓V_bf)²/(V_bf)²
Δx_a/Δx_b = 1/9
QUESTION 4.
If
you have 2 randomly selected vectors like R and R;
Show that R. RX 5) = 0
(102)
Answer:
Follows are the solution to this question:
Explanation:
Please find the correct question in the attachment file.
Let:
[tex]\overrightarrow{R}= R_i\hat{i}+R_j\hat{j}+R_k\hat{k}\\\\\overrightarrow{S}= S_i\hat{i}+S_j\hat{j}+S_k\hat{k}\\\\[/tex]
Calculating the value of [tex]\overrightarrow{R} \times \overrightarrow{S}:[/tex]
[tex]\to \left | \begin{array}{ccc}\hat{i}&\hat{j}&\hat{K}\\R_i&R_j&R_k\\S_i&S_j&S_k\end{array}\right | = \hat{i}[R_j S_k-S_jR_k]-\hat{j}[R_i S_k-S_iR_k]+\hat{k}[R_i S_j-S_iR_j][/tex]
Calculating the value of [tex]\overrightarrow{R} \cdot (\overrightarrow{R} \times \overrightarrow{S}):[/tex]
[tex]\to (R_i\hat{i}+R_j\hat{j}+R_k\hat{k}) \cdot ( \hat{i}[R_j S_k-S_jR_k]-\hat{j}[R_i S_k-S_iR_k]+\hat{k}[R_i S_j-S_iR_j])[/tex]
by solving this value it is equal to 0.
If an ice cube with the mass of 5.0 grams melts in a closed system such as a closed glass jar what is the mass of the liquid water after the ice cube completely melts
a plane passes over Point A with a velocity of 8,000 m/s north. Forty seconds later it passes over Point B with a velocity of 10,000 m/s north. Which is the plane's acceleration from A to B ?
Acceleration = (change in velocity) / (time for the change)
Change in velocity = (ending velocity) - (starting velocity)
Change in the plane's velocity = (10,000 m/s north) - (8,000 m/s north)
Change in the plane's velocity = 2,000 m/s north
Time for the change = 40 seconds
Acceleration = (2,000 m/s north) / (40 seconds)
Acceleration = 50 m/s² north
Which has the most mass?
O The Moon
O A Pencil
O Your teacher.
O Earth
Answer:
Earth
lol... ....
A primary coil has 360 turns, while secondary has 120 turns in a transformer. The output voltage is 220V. What is the input voltage, and what type of transformer is it
Answer:
550V
Step - Down transformer
Explanation:
Given parameters:
Number of turns in primary coil = 300 turns
Secondary turns = 120 turns
Output voltage = 220V
Unknown:
Input voltage = ?
Type of transformer = ?
Solution:
To solve this problem, we use the expression below:
[tex]\frac{V_{out} }{V_{in} }[/tex] = [tex]\frac{Ns}{Np}[/tex]
So insert the parameters and find Vin;
[tex]\frac{220}{Vin}[/tex] = [tex]\frac{120}{300}[/tex]
120Vin = 220 x 300
Vin = [tex]\frac{220 x 300}{120}[/tex] = 550V
Since the input voltage is greater than the output voltage, this is step - down transformer.
Match each vocabulary word with the correct definition. 1. measure of how quickly velocity is changing 2. speed in a given direction 3. force that resists moving one object against another 4. measure of the pull of gravity on an object 5. tendency of an object to resist a change in motion 6. size friction acceleration velocity magnitude inertia weight friction
1. measure of how quickly velocity is changing . . . acceleration
2. speed in a given direction . . . velocity
3. force that resists moving one object against another . . . friction
4. measure of the pull of gravity on an object . . . weight
5. tendency of an object to resist a change in motion . . . inertia
6. size . . . magnitude
1. measure of how quickly velocity is changing is acceleration.
2. speed in a given direction is velocity.
3. force that resists moving one object against another is friction.
4. measure of the pull of gravity on an object is weight.
5. tendency of an object to resist a change in motion is inertia.
What is acceleration?Acceleration has the term used in mechanics to describe the pace at which the velocity of an object varies over time. Acceleration is a vector quantity (in that they have magnitude and direction). The direction of an object's acceleration is determined by the direction of the net force acting on it.
It is a vector quantity with an SI unit is m/s² and the dimension formula is LT⁻². A massive body will accelerate or alter its velocity at a constant rate when a constant force is applied to it, according to Newton's second law. In the simplest case, when a force is applied to an object at rest, it accelerates in the force's direction.
Therefore, 1. measure of how quickly velocity is changing is acceleration.
2. speed in a given direction is velocity.
3. force that resists moving one object against another is friction.
4. measure of the pull of gravity on an object is weight.
5. tendency of an object to resist a change in motion is inertia.
Learn more about friction on:
https://brainly.com/question/13000653
#SPJ5
* Psychology
Match the types of psychoactive drugs to their functions,
depressants
stimulants
amphetamines
hallucinogens
to excite neural activity and temporarily
elevate awareness
to increase dopamine activity and produce
schizophrenic-like paranoid symptoms
>
to inhibit the function of the central nervous
system and neural activity
to distort perceptions and effects on thinking
Answer:
See explanation below
Explanation:
Psychoactive drugs are drugs that affect the central nervous system. They alter cognitive function by changing mood and consciousness.
Examples;
Depressants: Inhibit the function of the central nervous system and neural activity.
Stimulants: Excite neural activity and temporarily elevate awareness.
Amphetamines: Increase dopamine activity and produce schizophrenic-like symptoms.
Hallucinogens: Distort perceptions and effects on thinking.
A drug is any substance that alters how the body functions.
What is a drug?A drug is any substance that alters how the body functions. There are different types of drugs that affect different parts of the body.
We shall now explain the following classifications of drugs;
depressants - to inhibit the function of the central nervousstimulants - elevate awarenesshallucinogens - to distort perceptions and effects on thinkingamphetamines - schizophrenic-like paranoid symptomsLearn more about drugs: https://brainly.com/question/6022349
12
Select the correct answer.
What creates an electric force field that moves electrons through a circuit?
ОА.
energy source
B.
load
O c.
metal wires
OD.
resistance
Answer:
A
Explanation:
greyhound pursues a hare and takes 5 leaps for every 6 leaps of the hare, but
3 leaps of the hound are equal to 5 leaps of the hare. Compare the speed of the
hound and the hare,
need full solution:-
[tex]{\large{\bold{\rm{\underline{Given \; that}}}}}[/tex]
★ A grey hound pursues a hare and takes 5 leaps for every 6 leaps of the hare, but 3 leaps of the hound are equal to 5 leaps of the hare.
[tex]{\large{\bold{\rm{\underline{To\; find}}}}}[/tex]
★ The speed of the hound and the hare
[tex]{\large{\bold{\rm{\underline{Solution}}}}}[/tex]
★ The speed of the hound and the hare = 25:18
[tex]{\large{\bold{\rm{\underline{Full \; Solution}}}}}[/tex]
[tex]\dashrightarrow[/tex] As it's given that a grey hound pursues a hare and takes 5 leaps for every 6 leaps of the hare, but 3 leaps of the hound are equal to 5 leaps of the hare.
So firstly let us assume a metres as the distance covered by the hare in one leap.
Ok now let's talk about 5 leaps,.! As it's cleared that the hare cover the distance of 5a metres.
But 3 leaps of the hound are equal to 5 leaps of the hare.
Henceforth, (5/3)a meters is the distance that is covered by the hound.
Now according to the question,
Hound pursues a hare and takes 5 leaps for every 6 leaps of the hare..! (Same interval)
Now the distance travelled by the hound in it's 5 leaps..!
(5/3)a × 525/3a metresNow the distance travelled by the hare in it's 6 leaps..!
6a metresNow let us compare the speed of the hound and the hare. Let us calculate them in the form of ratio..!
25/3a = 6a25/3 = 625:18An unmarked police car traveling a constant 38.6 m/s is passed by a speeder traveling 53.4 m/s. Precisely 2.2 seconds after the speeder passes, the police officer steps on the accelerator; if the police car's acceleration is a constant 1.6 m/s2, how much time passes before the police car overtakes the speeder after the speeder passes (assumed moving at constant speed).
Answer:
The unmarked police car needs approximately 71.082 seconds to overtake the speeder.
Explanation:
Let suppose that speeder travels at constant velocity, whereas the unmarked police car accelerates at constant rate. In this case, we need to determine the instant when the police car overtakes the speeder. First, we construct a system of equations:
Unmarked police car
[tex]s = s_{o}+v_{o,P}\cdot (t-t') + \frac{1}{2}\cdot a\cdot (t-t')^{2}[/tex] (1)
Speeder
[tex]s = s_{o} + v_{o,S}\cdot t[/tex] (2)
Where:
[tex]s_{o}[/tex] - Initial position, measured in meters.
[tex]s[/tex] - Final position, measured in meters.
[tex]v_{o,P}[/tex], [tex]v_{o,S}[/tex] - Initial velocities of the unmarked police car and the speeder, measured in meters per second.
[tex]a[/tex] - Acceleration of the unmarked police car, measured in meters per square second.
[tex]t[/tex] - Time, measured in seconds.
[tex]t'[/tex] - Initial instant for the unmarked police car, measured in seconds.
By equalizing (1) and (2), we expand and simplify the resulting expression:
[tex]v_{o,P}\cdot (t-t')+\frac{1}{2}\cdot a\cdot (t-t')^{2} = v_{o,S}\cdot t[/tex]
[tex]v_{o,P}\cdot t -v_{o,P}\cdot t' +\frac{1}{2}\cdot a\cdot t^{2}-a\cdot t'\cdot t+\frac{1}{2}\cdot a\cdot t'^{2} = v_{o,S}\cdot t[/tex]
[tex]\frac{1}{2}\cdot a\cdot t^{2}+[(v_{o,P}-v_{o,S})-a\cdot t']\cdot t -\left(v_{o,P}\cdot t'-\frac{1}{2}\cdot a\cdot t'^{2}\right) = 0[/tex]
If we know that [tex]a = 1.6\,\frac{m}{s^{2}}[/tex], [tex]v_{o,P} = 0\,\frac{m}{s}[/tex], [tex]v_{o,S} = 53.4\,\frac{m}{s}[/tex] and [tex]t' = 2.2\,s[/tex], then we solve the resulting second order polynomial:
[tex]0.8\cdot t^{2}-56.92\cdot t +3.872 = 0[/tex] (3)
[tex]t_{1} \approx 71.082\,s[/tex], [tex]t_{2}\approx 0.068[/tex]
Please notice that second root is due to error margin for approximations in coefficients. The required solution is the first root.
The unmarked police car needs approximately 71.082 seconds to overtake the speeder.
describe why people are better off not consuming an additional good or service if the marginal cost is greater than the marginal benefit.
If a person visits an exercise facility, buys a new piece of fitness/sporting equipment,
or just starts planning to be active, which of the five stages of change for physical
activity are they at?
Planning
Maintenance
Precontemplation
Contemplation
Answer:planning
Explanation:
The person is in the stage of planning due to its action of planning to be active.
What is planning stage?The person is in the planning stage among the five stages of change for physical activity because the person just started planning to be active not yet started the activity. If a person is in the state of looking thoughtfully at something for a very long time then it is said to be Contemplation.
While on the other hand, if a person is in a stage in which there is no intention to change behavior in the foreseeable future then it is called precontemplation so we can conclude that the person is in the stage of planning due to its action of planning to be active.
Learn more about physical activity here: https://brainly.com/question/1561572
a 250.0 g snowball of radius 4.00 cm starts from rest at the top of the peak of a roof and rolls down a section angled at 30.0 degrees
Answer:
The response to this question is as follows:
Explanation:
The whole question and answer can be identified in the file attached, please find it.
The force diagram of all the forces acting on the snowball include the normal force acting upwards, the weight of the snowball acting downwards and the frictional force acting horizontal.
The given parameters;
mass of the snow ball, m = 250 gradius of the snow ball, r = 4 cmangle of inclination of the plane, θ = 30 ⁰The force diagram of all the forces acting on the snowball is calculated as follows;
↑ N
⊕ → F
↓ W
Where;
N is the normal force on the snowballF is the frictional force on the snowballW is the weight of the ballThus, the force diagram of all the forces acting on the snowball include the normal force acting upwards, the weight of the snowball acting downwards and the frictional force acting horizontal.
Learn more here:https://brainly.com/question/3624253
A piece of aluminum with a mass of 3.05 g initially at a temperature of 10.8 °C is heated to a temperature of 20.
Assume that the specific heat of aluminum is 0.901 J/(g°C).
How much heat was needed for this temperature change to take place?
Answer:
25.3J
Explanation:
Given parameters:
Mass of aluminum = 3.05g
Initial temperature = 10.8 °C
Final temperature = 20 °C
Specific heat = 0.9J/g °C
Unknown:
Amount of heat needed for the temperature to change = ?
Solution:
To solve this problem, we use the expression:
H = m C Ф
H is the amount of heat
m is the mass
C is the specific heat capacity
Ф is the change in temperature
H = 3.05 x 0.901 x (20 - 10.8) = 25.3J
if you watch football let me know who you think is going to win super bowl 55 and what do you think the score going to be Kansas city chiefs or tampa bay buccaneers
Answer:
I think the bucs are gonna win because Tom Brady is on their team and it's rigged
but maybe I'm just thinking negatively lol
Sam heaves a 16lb shot straight upward, giving it a constant upward acceleration from rest of 35 m/s^2 for 64.0 cm. He releases it 2.20m above the ground. You may ignore air resistance.
(a) What is the speed of the shot when Sam releases it?
(b) How high above the ground does it go?
(c ) How much time does he have to get out of its way before it returns to the height of the top of his head, 1.83 m above the ground?
Answer:
6.69 m/s
4.483 m
1.42s
Explanation:
Given that:
Initial Velocity, u = 0
Final velocity, v =?
Acceleration, a = 35m/s²
1.) using the relation :
v² = u² + 2as
v² = 0 + 2(35) * 64*10^-2m
v² = 70 * 0.64
v = sqrt(44.8)
v = 6.693
v = 6.69 m/s
B.) height from the ground, h0 = 2.2
How high ball went , h:
Using :
v² = u² + 2as
Upward motion, g = - ve
0 = 6.69² + 2(-9.8)*(h - 2.2)
0= 6.69² - 19.6(h - 2.2)
44.7561 + 43.12 - 19.6h = 0
19.6h = 44.7561 - 43.12
h = 87.8761 / 19.6
h = 4.483 m
C.)
vt - 0.5gt² = h - h0
6.69t - 0.5(9.8)t²
6.69t - 4.9t² = 1.83 - 2.2
-4.9t² + 6.69t + 0.37 = 0
Using the quadratic equation solver :
Taking the positive root:
1.4185 = 1.42s
1. A particle is projected vertically upwards with a velocity of 30 ms from a point 0. Find (a) the maximum height reached(b) the time taken for it to return to 0 (c) the taken for it to be 35m below 0
Assuming the particle is in free fall once it is shot up, its vertical velocity v at time t is
v = 30 m/s - g t
where g = 9.8 m/s² is the magnitude of the acceleration due to gravity, and its height y is given by
y = (30 m/s) t - 1/2 g t ²
(a) At its maximum height, the particle has 0 velocity, which occurs for
0 = 30 m/s - g t
t = (30 m/s) / g ≈ 3.06 s
at which point the particle's maximum height would be
y = (30 m/s) (3.06 s) - 1/2 g (3.06 s)² ≈ 45.9184 m ≈ 46 m
(b) It takes twice the time found in part (a) to return to 0 height, t ≈ 6.1 s.
(c) The particle falls 35 m below its starting point when
-35 m = (30 m/s) t - 1/2 g t ²
Solve for t to get a time of about t ≈ 7.1 s
Two parallel conducting plates are separated by 10.0 cm, and one of them is taken to be at zero volts. (a) What is the magnitude of the electric field strength between them, if the potential 7.35 cm from the zero volt plate (and 2.65 cm from the other) is 533 V
Answer:
[tex]E=6.8Kv/m[/tex]
Explanation:
From the question we are told that
Distance b/w plate [tex]d=10cm=>0.1m[/tex]
P_1 Potential at 7.35 [tex]V=533v[/tex]
Generally the equation for electric field at a distance is mathematically given as
[tex]E=\frac{v}{d}[/tex]
[tex]E=\frac{533}{7.85*10^-^2}[/tex]
[tex]E=6789.808917[/tex]
[tex]E=6.8*10^3[/tex]
[tex]E=6.8Kv/m[/tex]
Consider two points in an electric field. The potential at point 1, V1, is 24V. The potential at point 2, V2, is 154V. A proton is moved from point 1 to point 2.
(a) Write an equation for the change of electric potential energy AU of the proton, in terms of the symbols given and the charge of the proton e.
(b) Find the numerical value of the change of the electric potential energy in electron volts (eV).
(c) Express v2, the speed of the electron at point 2, in terms of AU, and the mass of the electron me.
(d) Find the numerical value of v2 in m/s
Answer:
[tex]\triangle U=-e (V_2-V_1)[/tex]
[tex]\triangle U=130eV[/tex]
[tex]V_2=\sqrt{ \frac{2}{me}(\frac{1}{2}meV_1^2+e(V_2-V_1)}[/tex]
Explanation:
From the question we are told that
The potential at point 1, [tex]V_1 = 24V[/tex]
The potential at point 2, [tex]V_2 = 154V[/tex]
a)Generally work done by proton is given as
[tex]w=-\triangle U[/tex]
[tex]e\triangle V=-\triangle U[/tex]
[tex]\triangle U=-e (V_2-V_1)[/tex]
Generally the Equation for the change of electric potential energy AU of the proton, in terms of the symbols given and the charge of the proton e is mathematically given as
[tex]\triangle U=-e (V_2-V_1)[/tex]
b)Generally the electric potential energy in electron volts (eV). is mathematically given as
[tex]\triangle U=-e (154-24)V[/tex]
[tex]|\triangle U| =|-e (130)V|[/tex]
[tex]\triangle U=130eV[/tex]
c) Generally according to the law of conservation of energy
[tex](K.E+P.E)_1=(K.E+P.E)_2[/tex]
[tex]\frac{1}{2}meV_1^2+eV_1 =\frac{1}{2}mev_2^2+eV_2[/tex]
[tex]V_2^2=\frac{2}{me}(\frac{1}{2}meV_1^2+e(V_2-V_1)[/tex]
[tex]V_2=\sqrt{ \frac{2}{me}(\frac{1}{2}meV_1^2+e(V_2-V_1)}[/tex]