Answer:
Average speed = 0, Average velocity = 0.02 m/s
Explanation:
Given that,
A car travels from A to B and from B back to A.
The distance between A and B is 12 km
It took 20 minutes to travel full.
We need to find the average speed and the average velocity of the car.
As the car reaches the point from where it was started, the displacement of the car is equal to 0. As a result, the average velocity is equal to 0.
For average speed,
[tex]s=\dfrac{12+12}{60\times 20}\\\\s=0.02\ m/s[/tex]
Hence, this is the required solution.
A 0.2 kg hockey park is sliding along the eyes with an initial velocity of -10 m/s when a player strikes it with his stick, causing it to reverse its direction and giving it a velocity of +25 m/s the impulse the stick applies to the park is most nearly
Answer:
The impulse applied by the stick to the hockey park is approximately 7 kilogram-meters per second.
Explanation:
The Impulse Theorem states that the impulse experimented by the hockey park is equal to the vectorial change in its linear momentum, that is:
[tex]I = m\cdot (\vec{v}_{2} - \vec{v_{1}})[/tex] (1)
Where:
[tex]I[/tex] - Impulse, in kilogram-meters per second.
[tex]m[/tex] - Mass, in kilograms.
[tex]\vec{v_{1}}[/tex] - Initial velocity of the hockey park, in meters per second.
[tex]\vec{v_{2}}[/tex] - Final velocity of the hockey park, in meters per second.
If we know that [tex]m = 0.2\,kg[/tex], [tex]\vec{v}_{1} = -10\,\hat{i}\,\left[\frac{m}{s}\right][/tex] and [tex]\vec {v_{2}} = 25\,\hat{i}\,\left[\frac{m}{s} \right][/tex], then the impulse applied by the stick to the park is approximately:
[tex]I = (0.2\,kg)\cdot \left(35\,\hat{i}\right)\,\left[\frac{m}{s} \right][/tex]
[tex]I = 7\,\hat{i}\,\left[\frac{kg\cdot m}{s} \right][/tex]
The impulse applied by the stick to the hockey park is approximately 7 kilogram-meters per second.
Which of the following statements are true about covalent bonding between two atoms? Select all that apply.
A. Electrons are shared.
B. The electronegativities of the two atoms are close to each other.
C. The two atoms can be of the same element.
D. Electrons transfer from one atom to the other.
Answer:
A and C
Explanation:
Covalent bonding involves sharing by the atoms involved
Statements that can be considered as true statement about covalent bonding between two atoms are:
A. Electrons are shared.
B. The electronegativities of the two atoms are close to each other.
C. The two atoms can be of the same element.
Covalent bond can be regarded as chemical bond in which electrons pairs are been shared between atoms, these atoms can be of the same element.These electron pairs are called pairs or bonding pairs.Therefore, option A,B,C are correct.
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A 6.93*10-4 C charge has a
potential energy U = -3.09 J at a
point in space. What is the electric
potential V at that point?
Include the sign, + or -
(Unit = V)
Answer:
P = V * Q potential energy = potential * charge
V = =3.09 J / 6.93 * 10E-4 C = 4460 Joules / Coulomb
The electric potential, V at the point given the data from the question is –4458.87 V
What is electric potential?The electric potential or electromotive force (EMF) is defined as the energy supplied by a battery per unit charge. Mathematically, it can be expressed as:
Electromotive force (EMF) = Work (W) / charge (Q)
V = EMF = W / Q
How to determine the Electric potentialwork (W) = –3.09 JCharge on electron = 6.93×10⁻⁴ CElectric potential (V) =?V = W / Q
V = –3.09 / 6.93×10⁻⁴
V = –4458.87 V
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Parallel Circuits:
A) are rarely used in the wiring in homes
B) always need more voltage than series circuits
C) will have positive charges flowing in one branch, negative charges in the other
D) provide more than one path fo current flow
Answer:
D.)
Explanation:
the current separates on each branch according to the resistance it experience.
Answer:D
Explanation:
Define emf of a battery?
Answer:
Electromotive force or EMF is equal to the terminal potential difference when no current flows. EMF (ϵ) is the amount of energy (E) provided by the battery to each coulomb of charge (Q) passing through.
A circuit has a current of 3 amps and is using a 9 volt battery. The circuit has a resistance of ____
ohms.
Answer:
so 9/3=3 current is 3 amperes
Explanation:
The fomula to calculate resistance is:
voltage/cutrent
9 V/3 A= 3 ohms
Determine the applied force required to accelerate a 2.25 kg object rightward with a
constant acceleration of 1.50 m/s2 if the force of friction opposing the motion is 18.2 N.
(Neglect air resistance.)
Answer:
Explanation:
Im going to be using the rules for significant digits properly so I hope you're quite familiar with them. The equation we need for this is
F - f = ma where F is the applied force (our unknown), f is the frictional force, m is the mass, and a is the acceleration. Filling in:
F - 18.2 = 2.25(1.50) and
F = 2.25(1.50) + 18.2 Do the multiplication first and round to get
F = 3.38 + 18.2 The addition rules tell us that we will be rounding to the tenths place after we add to get
F = 21.6 N
A solid cylinder has a mass of 5 kg and radius of 2 m and is fixed so that it is able to rotate freely around its center without friction. A 0.02 kg bullet is moving from right to left with an angular momentum of 9 kgm2s just before it strikes the cylinder near its bottom and gets stuck at the outer radius. What is the angular velocity (magnitude and direction) of the cylinder bullet system after the impact
Answer:
0.893 rad/s in the clockwise direction
Explanation:
From the law of conservation of angular momentum,
angular momentum before impact = angular momentum after impact
L₁ = L₂
L₁ = angular momentum of bullet = + 9 kgm²/s (it is positive since the bullet tends to rotate in a clockwise direction from left to right)
L₂ = angular momentum of cylinder and angular momentum of bullet after collision.
L₂ = (I₁ + I₂)ω where I₁ = rotational inertia of cylinder = 1/2MR² where M = mass of cylinder = 5 kg and R = radius of cylinder = 2 m, I₂ = rotational inertia of bullet about axis of cylinder after collision = mR² where m = mass of bullet = 0.02 kg and R = radius of cylinder = 2m and ω = angular velocity of system after collision
So,
L₁ = L₂
L₁ = (I₁ + I₂)ω
ω = L₁/(I₁ + I₂)
ω = L₁/(1/2MR² + mR²)
ω = L₁/(1/2M + m)R²
substituting the values of the variables into the equation, we have
ω = L₁/(1/2M + m)R²
ω = + 9 kgm²/s/(1/2 × 5 kg + 0.02 kg)(2 m)²
ω = + 9 kgm²/s/(2.5 kg + 0.02 kg)(4 m²)
ω = + 9 kgm²/s/(2.52 kg)(4 m²)
ω = +9 kgm²/s/10.08 kgm²
ω = + 0.893 rad/s
The angular velocity of the cylinder bullet system is 0.893 rad/s in the clockwise direction-since it is positive.
Two astronauts push off each other in space. The 90 kg astronaut ends up with a speed of 5 ms. What is the mass of the other astronaut it he ended up with
a speed of 7 m/s?
Answer:
jsjshs
Explanation:
nsnajsajsnzbzbzbaba
A voltage source provides ____ required for electric current. A. the electrons B. the potential difference C. the resistance D. the pathway
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Solutions, Solubility, Acids/Bases 20-2121 of 35
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Question 32 of 35
The models above show particle arrangement. Read each statement below.
image
1. The substances must undergo a chemical change to transform into a different model
2. The substance in diagram A has the highest energy
3. The substances in diagrams B and C would flow.
4. The substance in diagram C has the lowest temperature.
Which statement above is true?
A.
1
B.
3
C.
4
D.
2
Answer:
hard2
Explanation:
Jenna flips a 10p coin four times. Here are her results:
Trial
Result
1. heads
2. tails
3. heads
4. heads
How can Jenna best improve the reliability of her results?
A
flip the 10p coin one more time
B
flip the 10p coin six more times
C
use a 50p coin instead of a 10p coin
D
use a different 10p coin for each of the four trials
Answer:
Option B
Explanation:
The reliability of an experiment can be increased by conducting experiment sufficient number of times keeping other things constant. Hence, the 10 p coin needs to be flipped again to measure reliability for more than one times.
Hence, option B is correct
3 - An object is being pushed with a net force of 15 N. If the net force is cut in third to 5 N, how will the acceleration be changed?
Answer:
Explanation:
F = ma is a linear equation. This means that the Force change as the accleration changes. And vice versa. If the Force is cut in thirds, then the acceleration is also cut in thirds. Let's do some math on this just to prove it, shall we?
We know that at first, the F = 15. Let's give this object a mass of 5kg. That means that
15 = 5a so
a = 3
Then the F is cut into thirds, so
5 = 5a so
a = 1
The second acceleration is one-third of the first one, where the Force is 3 times greater.
La distancia por carretera de Chitré a Parita es de 12 km; exprese en pies ésta distanciaLa distancia por carretera de Chitré a Parita es de 12 km; exprese en pies ésta distancia
Answer:
La distancia por carretera de Chitré a Parita es de 12 km o 39370.08 pies.
Explanation:
La regla de tres es una forma de resolver problemas de proporcionalidad entre tres valores conocidos y un valor desconocido, estableciendo una relación de proporcionalidad entre todos ellos.
Si la relación entre las magnitudes es directa, es decir, cuando una magnitud aumenta, también lo hace la otra (o cuando una magnitud disminuye, también lo hace la otra), se debe aplicar la regla directa de tres. Para resolver una regla directa de tres, se debe seguir la siguiente fórmula, siendo a, b y c los valores conocidos y x el valor a determinar:
a ⇒ b
c ⇒ x
Entonces [tex]x=\frac{c*b}{a}[/tex]
La regla directa de tres es la regla que se aplica en este caso donde hay un cambio de unidades. Para realizar esta conversión de unidades, primero debes saber que 1 km = 3280,84 pies. Entonces, si 1 km son 3280,84 pies, ¿cuántos pies son 12 km?
1 km ⇒ 3280.84 pies
12 km ⇒ x
[tex]x=\frac{12 km*3280.84 pies}{1 km}[/tex]
x= 39370.08 pies
La distancia por carretera de Chitré a Parita es de 12 km o 39370.08 pies.
Two charges, each q, are separated by a distance r, and exert mutual attractive forces of F on each other. If both charges become 2q and the distance becomes 3r, what are the new mutual forces
Answer:
F = ⅔ F₀
Explanation:
For this exercise we use Coulomb's law
F = k q₁q₂ / r²
let's use the subscript "o" for the initial conditions
F₀ = k q² / r²
now the charge changes q₁ = q₂ = 2q and the new distance is r = 3 r
we substitute
F = k 4q² / 9 r²
F = k q² r² 4/9
F = ⅔ F₀
The two scientists who gave us a better understanding of the universe are?
Answer:
I'm pretty sure it is Edwin Powell Hubble and Albert Einstein
Explanation:
A vector points -43.0 units
along the x-axis, and 11.1 units
along the y-axis.
Find the direction of the vector.
Answer:
Explanation:
To find the direction of this vector we need o find the angle that has a tangent of the y-component over the x-component:
[tex]tan^{-1}(\frac{11.1}{-43.0})=-14.5[/tex] but since we are in Q2 we have to add 180 degrees to that angle giving us 165.5 degrees
Transformar las siguientes unidades al Sistema Internacional: 30 km/h ; 37 Dm ; 750 g ; 4x10-6 km2 ; 7500 cm ; 600000 cm2 ; 520700000 mm3 ; 3,4 años.
Answer:
a) 3.0 10⁴ m / s, b) 3.7 10¹ m, c) 0.750 kg, d) 4 10¹² m², e) 75 m, f) 60 m²
g) 5.207 10³ m², e) 4.847 10⁷ s
Explanation:
The international system (SI) of measurements has as fundamental units the meter for length, the second for time and kilogram for mass.
Let's reduce the different magnitudes to the SI system
a) 30 km / h (1000m / 1 km) (1 h / 3600 s) = 3.0 10⁴ m / s
b) 37 Dm (10 m / 1 Dm) = 3.7 10¹ m
c) 750 g (1 kg / 10,000 g) = 0.750 kg
d) 4 10⁶ km² (1000 m / 1km) ² = 4 10¹² m²
e) 7500 cm (1 m / 100 cm) = 75 m
f) 600000 cm² (1m / 10² cm) ² = 60 m²
g) 520700000 mm³ (1 m / 10³ mm) ³ = 5.20700000 109/10 ^ 6
= 5.207 10³ m²
e) 3.4 years (l65 days / 1 yr) (24 h / 1 day) (3600 s / 1h) = 4.847 10⁷ s
The equation provided (from the textbook) first defines the elastic potential energy of a spring as ΔUsp = −(WB + WW), where WB is work the spring does on an attached block and WW is work the spring does on the wall to which it is attached. But WW is ignored in the next step. Why?
Answer:
The given potential energy of the spring is expressed as follows;
ΔUsp = -(WB + WW)
Where;
WB = Th work done by the spring on the block to which it is attached
WW = The work done by the spring on the wall
We recall that work done, W = Force applied × Distance moved in the direction of the force
The work done by the spring on the block, WB = The spring force × The distance the block moves
The work done by the spring on the wall, WW = The spring force × The distance the wall moves
However, given that the wall does not move, we have;
The distance the wall moves = 0
∴ The work done by the spring on the wall, WW = The spring force × 0 = 0 J
Therefore, WW = 0 J, and the spring does not do work on the wall, and WW can be ignored in the next subsequent) steps
Explanation:
Which of the following represent units of capacitance? You may choose more than one correct answer:
Ampere
coulomb/volt
coulomb/second
Farad
volt/coulomb
Answer:
Capacitance is a derived physical quantity measured in farad
Answer:
"Farad" is another term for the coulomb/volt measurement of capacitance, so both of those options are the correct answer. Amperes and coulomb/second are measurements of electric flow, in other words, how strong a current is.
In a warehouse, the workers sometimes slide boxes along the floor to move them. Two boxes were sliding toward each other and crashed. The crash caused both boxes to change speed. Based on the information in the diagram, which statement is correct? In your answer, explain what the forces were like and why the boxes changed speed.
Box 1 has more mass than Box 2.
Box 1 and Box 2 are the same mass.
Box 1 has less mass than Box 2
Answer:
The second one.
Explanation:
It caused both to change speed because they have both the same mass.
Steph uses 18 Newtons of force to push a piano up a ramp. She uses a total of 900 Joules getting this piano into the moving truck. Calculate the
distance of the ramp.
[tex]\huge{ \mathcal{ \underline{ Answer }: }}[/tex]
[tex] \boxed{ \mathrm{Work done = Force × displacement}}[/tex]
[tex]900 = 18 \times d[/tex][tex]d = \dfrac{900}{18} [/tex][tex]d = 50[/tex]Therefore the distance covered :
[tex] \large \boxed{50 \: m}[/tex]
_____________________________
[tex]\mathrm{ ☠ \: TeeNForeveR \:☠ }[/tex]
a rocket with an initial velocity of 20 m/s fires another engine that gives it an acceleration of 4 m/s2 over 10 seconds. How far did the rocket travel during this time?
Answer: 400 m
Explanation:
Vf= 20 + (4*10)
Vf= 60 [m/s]
x= (60^2 - 20^2) / (2*4)
x= 400 m
The series circuit depicts three resistors connected to a voltage
source. The voltage source (AVtot) is a 110-V source and the resistor
values are 7.2 (R1), 6.2 A2 (R2) and 8.6 22 (R3).
b. Determine the current in the circuit.
A
c. Determine the voltage drops across each individual resistor.
Answer:
B. Current in the circuit is 5.
Ci. Voltage across 7.2 Ω (R₁) is 36 V
Cii. Voltage across 6.2 Ω (R₂) is 31 V
Ciii. Voltage across 8.6 Ω (R₃) is 43 V
Explanation:
We'll begin by calculating the number equivalent resistance in the circuit. This can be obtained as follow:
Resistor 1 (R₁) = 7.2 Ω
Resistor 2 (R₂) = 6.2 Ω
Resistor 3 (R₃) = 8.6 Ω
Equivalent Resistance (R) =?
Since the resistors are in series connection, the equivalent resistance can be obtained as follow:
R = R₁ + R₂ + R₃
R = 7.2 + 6.2 + 8.6
R = 22 Ω
B. Determination of the current.
Voltage (V) = 110 V
Resistance (R) = 22 Ω
Current (I) =?
V = IR
110 = I × 22
Divide both side by 22
I = 110 / 22
I = 5 A
Therefore, the current in the circuit is 5.
Ci. Determination of the voltage across 7.2 Ω (R₁)
Resistor 1 (R₁) = 7.2 Ω
Current (I) = 5 A
Voltage 1 (V₁) =?
V₁ = IR₁
V₁ = 5 × 7.2
V₁ = 36 V
Therefore, the voltage across 7.2 Ω (R₁) is 36 V
Bii. Determination of the voltage across 6.2 Ω (R₂)
Resistor 2 (R₂) = 6.2 Ω
Current (I) = 5 A
Voltage 2 (V₂) =?
V₂ = IR₂
V₂ = 5 × 6.2
V₂ = 31 V
Therefore, the voltage across 6.2 Ω (R₂) is 31 V
Ciii. Determination of the voltage across 8.6 Ω (R₃)
Resistor 3 (R₃) = 8.6 Ω
Current (I) = 5 A
Voltage 3 (V₃) =?
V₃ = IR₃
V₃ = 5 × 8.6
V₃ = 31 V
Therefore, the voltage across 8.6 Ω (R₃) is 43 V
A tank has the shape of an inverted circular cone with height 16m and base radius 3m. The tank is filled with water to a height of 9m. Find the work required to empty the tank by pumping all of the water over the top of the tank. Use the fact that acceleration due to gravity is 9.8 m/sec2 and the density of water is 1000kg/m3. Round your answer to the nearest kilojoule.
Answer:
[tex]W=17085KJ[/tex]
Explanation:
From the question we are told that:
Height [tex]H=16m[/tex]
Radius [tex]R=3[/tex]
Height of water [tex]H_w=9m[/tex]
Gravity [tex]g=9.8m/s[/tex]
Density of water [tex]\rho=1000kg/m^3[/tex]
Generally the equation for Volume of water is mathematically given by
[tex]dv=\pi*r^2dy[/tex]
[tex]dv=\frac{\piR^2}{H^2}(H-y)^2dy[/tex]
Where
y is a random height taken to define dv
Generally the equation for Work done to pump water is mathematically given by
[tex]dw=(pdv)g (H-y)[/tex]
Substituting dv
[tex]dw=(p(=\frac{\piR^2}{H^2}(H-y)^2dy))g (H-y)[/tex]
[tex]dw=\frac{\rho*g*R^2}{H^2}(H-y)^3dy[/tex]
Therefore
[tex]W=\int dw[/tex]
[tex]W=\int(\frac{\rho*g*R^2}{H^2}(H-y)^3)dy[/tex]
[tex]W=\rho*g*R^2}{H^2}\int((H-y)^3)dy)[/tex]
[tex]W=\frac{1000*9.8*3.142*3^2}{9^2}[((9-y)^3)}^9_0[/tex]
[tex]W=3420.84*0.25[2401-65536][/tex]
[tex]W=17084965.5J[/tex]
[tex]W=17085KJ[/tex]
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are Car Travels certain distance with a speed of 50 kilometre per hour and returns with a speed of 40 kilometre per hour what is the average speed for the whole journey
Best answer branlist and NO LINKS
Answer:
The average speed for the whole journey is 44.[tex]\overline 4[/tex] kilometers per hour
Explanation:
The average speed with which the car travels the distance, v₁ = 50 kilometer per hour
The average speed with which the car returns, v₂ = 40 kilometer per hour
Average speed, v = (Total distance, d)/(Total time, t)
Let 'd' represent the distance travelled, we have;
The time it takes the car to travel to the distance = d/50 hours
The time it takes the car to return = d/40 hours
The total time = (d/50 hours + d/40 hours) = d·(40 + 50)/(40 × 50) hours= 9·d/200 hours
The total distance = d kilometers+ d kilometers = 2·d kilometers
The average speed for the whole journey, v = 2·d kilometers/(9·d/200 hours) = 400/9 kilometers per hours = 44.[tex]\overline 4[/tex] kilometer/hour.
Fish are hung on a spring scale to determine their mass (most fishermen feel no obligation to truthfully report the mass). What is the force constant of the spring in such a scale if it the spring stretches 8.50 cm for a 8.1 kg load
Answer:
k = 933.88 N/m
Explanation:
Given that,
The mass of a load, m = 8.1 kg
The displacement in the spring, x = 8.5 cm
We need to find the force constant of the spring. We know that Hooke's law is given by :
F = -kx
Also,
F = mg (Force due to its weight)
So,
[tex]k=\dfrac{mg}{x}\\\\k=\dfrac{8.1\times 9.8}{0.085}\\\\k=933.88\ N/m[/tex]
So, the force constant of the spring is equal to 933.88 N/m.
A car start moving from the rest.If the acceleration of the car is 2m/2 for 10 seconds what will be it final velocity
Answer:
20 m/s
Explanation:
Applying,
a = (v-u)/t.................... Equation 1
Where a = acceleration of the car, v = final velocity of the car, u = initial velocity of the car, t = time.
make v the subject of the equation
v = u+at.............. Equation 2
From the question,
Given: u = 0 m/s(start from rest), a = 2 m/s², t = 10 seconds
Substitute these values into equation 2
v = 0+(2×10)
v = 20 m/s
A ball is dropped from rest out of a high window in a tall building for 5 seconds. Assuming the we ignore air resistance and assume upwards to be positive. A) What will be the final velocity of the ball B) What is the height of the building if it hits the ground after those 5 seconds. *
Answer:
I am not sure if this is the answer
(B) what is the height of the building if it hits the ground after those 5 seconds.
Suppose the height of object is +3cm and height of image is -12 cm. What is its magnification?
Answer:
magnification is 4
Explanation:
m= image height / object height
m= 12/3
m= 4