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.
What is the net force acting on the airplane?
740 N right -->
700 N right -->
100 N left <--
760 N right -->
[tex]\huge{ \mathrm{ \underline{ Answer }\: \: ✓ }}[/tex]
Total force acting on right side = 800 N
Total force acting on left side :
60 N + 40 N100 NNow, equivalent force acting on the plane is :
greater force - minor force 800 N - 100 N 700 NewtonsAnd the direction of equivalent force will be the direction of greater force, that is right direction.
Hence, Correct option is :
700 N right -->_____________________________
[tex]\mathrm{ \:TeeNForeveR\:}[/tex]
. A tennis ball rolls off the lab bench with an initial velocity of 3.0 m/s. The top of the lab bench is 1.5 m above the floor. How long will the tennis ball be in the air before it hits the ground
Answer:
[tex] { \huge{s}} = ut + \frac{1}{2} g {t}^{2} \\ 1.5 = 3t + \frac{1}{2} \times 10\times {t}^{2} \\ 1.5 = 3t + 5 {t}^{2} \\ 5 {t}^{2} + 3t - 1.5 = 0 \\ t = 0.3 \: seconds[/tex]
1. If you use an applied force of 45N to slide a 12Kg wooden crate across a floor at a constant velocity, what is the coefficient of kinetic friction between the crate and the floor?
Answer:
Coefficient of kinetic friction = 0.38 (Approx.)
Explanation:
Given:
Applied force = 45 N
Mass of wooden crate = 12 kg
Find:
Coefficient of kinetic friction
Computation:
Coefficient of kinetic friction = Applied force / (Mass)(Acceleration due to gravity)
Coefficient of kinetic friction = 45 / (12)(9.8)
Coefficient of kinetic friction = 45 / 117.6
Coefficient of kinetic friction = 0.3826
Coefficient of kinetic friction = 0.38 (Approx.)
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
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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:
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.
Hello, can you help me? I have this question and I don't know how to answer it. Is a tire a conductor or an insulator? Thank you!
Answer: They're Conductors.
Explanation:
plsss plsss plsss helppppp ASAP thank u ❤️
Answer:
7. (D) uniformly accelerated vertical motion
8. (A) zero
9. (A) zero
10. (C) parabolic
Answer:
7.Uniformly accelerated vertical motion
8.0m/s²
9.9.8m/s
10.parabolic
11.vertical component.
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.
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:
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 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.
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.
5. A 905 kg test car travels around a 3.04 km circular track. If the magnitude of the centripetal force is 2100 N. What is the car's speed?
Answer:
Explanation:
The equation for centripetal force is
[tex]F=\frac{mv^2}{r}[/tex]. We have all the values we need except for the radius. We have the circumference of the circle, though, so we will solve for the radius using that and the fact that C = 2πr:
3.04 = 2(3.1415)r and
r = .484 m, to the correct number of sig fig's.
Now that we have everything we need and isolating the v NOT squared:
[tex]v=\sqrt{\frac{rF}{m} }[/tex] and filling in:
[tex]v=\sqrt{\frac{(.484)(2100)}{905} }[/tex] . This answer will need 2 sig fig's since 2100 has 2 sig fig's in it. That means that the velocity of the test car is
1.1 m/sec
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]
'
'
ANSWER FAST PLEASE HELP
Answer:
B. 175 N
Explanation:
Net force can be defined as the vector sum of all the forces acting on a body or an object i.e the sum of all forces acting simultaneously on a body or an object.
Mathematically, net force is given by the formula;
[tex] Fnet = Fapp + Fg[/tex]
Where;
Fnet is the net force
Fapp is the applied force
Fg is the force due to gravitation
In this scenario, we observed that both forces are acting in the same direction.
Therefore:
Net force = 100 N + 75 N
Net force = 175 Newton
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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A powerful motorcycle can produce an acceleration of 3.00 m/s2 while traveling at 106.0 km/h. At that speed, the forces resisting motion, including friction and air resistance, total 432.0 N. (Air resistance is analogous to air friction. It always opposes the motion of an object.) What is the magnitude of the force that motorcycle exerts backward on the ground to produce its acceleration if the mass of the motorcycle with rider is 241 kg
Answer:
"1155 N" is the appropriate solution.
Explanation:
Given:
Acceleration,
[tex]a=3 \ m/s^2[/tex]
Forces resisting motion,
[tex]F_f=432 \ N[/tex]
Mass,
[tex]m = 241 \ kg[/tex]
By using Newton's second law, we get
⇒ [tex]F-F_f=ma[/tex]
Or,
⇒ [tex]F=ma+F_f[/tex]
By putting the values, we get
⇒ [tex]=(3\times 241)+432[/tex]
⇒ [tex]=723+432[/tex]
⇒ [tex]=1155 \ N[/tex]
particles that are found in the sun's plasma
Answer:
This plasma mostly consists of electrons
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.
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
Numa máquina térmica uma parte da energia térmica fornecida ao sistema(Q1) é transformada em trabalho mecânico (τ) e o restante (Q2) é dissipado, perdido para o ambiente.
sendo:
τ: trabalho realizado (J) [Joule]
Q1: energia fornecida (J)
Q2: energia dissipada (J)
temos: τ = Q1 - Q2
O rendimento (η) é a razão do trabalho realizado pela energia fornecida:
η= τ/Q1
Exercícior resolvido:
Uma máquina térmica cíclica recebe 5000 J de calor de uma fonte quente e realiza trabalho de 3500 J. Calcule o rendimento dessa máquina térmica.
solução:
τ=3500 J
Q1=5000J
η= τ/Q1
η= 3500/5000
η= 0,7 ou seja 70%
Energia dissipada será:
τ = Q1 - Q2
Q2 = Q1- τ
Q2=5000-3500
Q2= 1500 J
Exercicio: Qual seria o rendimento se a máquina do exercicio anterior realizasse 4000J de trabalho com a mesma quantidade de calor fornecida ? Quanta energia seria dissipada agora?
obs: Entregar foto da resolução ou o cálculo passo a passo na mensagem
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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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
Which factor contributes the most to the process of water erosion?
A) evaporation
B) gravity
C) pressure
D) temperature
HELP
Answer:
Evaporation
Explanation:
Erosion is a natural process, but human activity can make it happen more quickly. Human activity altering the vegetation of an area is perhaps the biggest human factor contributing to erosion. Trees and plants hold soil in place.
Factor that contributes the most to the process of water erosion is Evaporation. The correct option is A.
What is evaporation?Evaporation is the phenomenon of converting the liquid into gas phase by the addition of heat energy.
Erosion is also a natural phenomenon. Trees and plants hold soil in place and prevents the soil erosion. Just like that, the water erosion happens when more than natural evaporation happens due to increased heat by various factors.
Thus, Evaporation contributes the most to the process of water erosion. The correct option is A.
Background Information: Energy can not be created or destroyed. Stored energy is called
potential energy, and the energy of motion is called kinetic energy. Due to gravity, potential
energy changes as the height of an object changes, this is called gravitational potential energy.
Objective: to determine the relationship between height and gravitational potential energy.
Problem: How does the drop height (gravitational potential energy) of a ball affect the bounce
height (kinetic energy) of the ball?
Hypothesis: If the gravitational potential energy (drop height) of a ball is increased, then the
kinetic energy (bounce height) will (increase/decrease/remain the same) because
Variables: Independent variable (known information) is
Dependent variable (unknown information) is
Constants (variables kept the same for accuracy) are
Materials: List all the materials used in this experiment.
Procedure: Follow the steps below to conduct your experiment. Be sure to record all data and
any observations during the experiment. Follow all safety rules.
1. Tape the meter stick to the side of the lab table with the 0-cm end at the bottom and the 100-cm end at the
top. Be sure that the meter stick is resting flat on the floor and is standing straight up.
2. Choose a ball type and record the ball type in the data table.
3. Use the triple beam balance to determine the mass of the ball and record the ball’s mass in the data table.
4. Calculate the gravitational potential energy (GPE) for the ball at each drop height. Record GPE in data table.
a. GPE = ball mass x drop height
5. For Trial 1, hold the ball at a height of 40 cm, drop the ball carefully and observe the bounce height. Record
the bounce height in the data table.
6. Drop the ball 4 more times from 40 cm, recording the bounce height each time, for a total of 5 drops.
7. For Trial 2, repeat steps 5 and 6 but drop the ball from a height of 50 cm. Record the 5 bounce heights in
the data table.
8. For Trial 3, drop the ball five times from 60 cm and record the 5 bounce heights in the data table.
9. For Trial 4, drop the ball five times from 70 cm and record the 5 bounce heights in the data table.
10. For Trial 5, drop the ball five times from 80 cm and record the 5 bounce heights in the data table.
11. For Trial 6, drop the ball five times from 90 cm and record the 5 bounce heights in the data table.
12. For Trial 7, drop the ball five times from 100 cm and record the 5 bounce heights in the data table.
13. Repeat steps 2 through 12 for a different type of ball.
14. Calculate the average bounce height of the 5 drops for each drop height. Record the average bounce height
in the data table. Calculate the average bounce height for all Trials.
a. To calculate average: Add the 5 bounce heights for a trial then divide the total by 5 drops. Example
for Trial 1: drop1 + drop2 + drop3 + drop4 +drop5 = total; total divided by 5 = average bounce height.
i. Sample: Trial 1: 5+6+5+5+7 = 28 28/5 = 5.6 average bounce height.
15. Plot the average bounce heights on a line graph. Place the independent variable of drop height on the x-axis
and place the dependent variable of bounce height on the y-axis. Label the line with the ball type.
16. CHALLENGE: Repeat all necessary steps for a third ball type and include this data on the graph.
17. Answer discussion questions and write your lab summary.
Data:
Ball Type: Ball Mass (g) = .
Gravitational
Potential Energy
(GPE)
Bounce Height (cm)
Drop
Height (cm)
GPE = mass x height Drop 1 Drop 2 Drop 3 Drop 4 Drop 5 Average
Bounce
Height
40
50
60
70
80
90
100
Ball Type: Ball Mass (g) = .
Gravitational
Potential Energy
(GPE)
Bounce Height (cm)
Drop
Height (cm)
GPE = mass x height Drop 1 Drop 2 Drop 3 Drop 4 Drop 5 Average
Bounce
Height
40
50
60
70
80
90
100
Discussion Questions: Answer using complete sentences.
1. Describe the relationship between drop height and the bounce height.
Was the relationship the same for both ball types that you tested?
2. Compare your gravitational potential energy to your bounce height for each trial. Describe
the relationship between GPE and bounce height.
3. Look at the results of both ball types you tested.
a. Which ball type had the most gravitational potential energy?
b. Which ball type has the most mass?
c. Describe the relationship between mass and GPE.
4. What are the variables that affect gravitational potential energy of an object?
Conclusion: Write a conclusion, using complete sentences, that states the following: if your
hypothesis was supported or negated; and what the real answer to the problem is.
Summary: Write a three paragraph summary using our standard format (1. what you were doing
and why you were doing it, 2. what you learned while doing this experiment, and 3. how what you
learned relates to your life).
Answer:
this question is very lengthy and even nit at all understandable so pls can u explain it in breif
Answer:
Explanation:
I did it and this is what i got
A voltage source provides ____ required for electric current. A. the electrons B. the potential difference C. the resistance D. the pathway
Lisa made the electromagnet shown. A nail with wire coiled around it has its head labeled S to the right and its point labeled N to the left. The end of the wire leading to the S is attached to the positive terminal of a battery. The end of the wire leading to the N is attached to the negative terminal of the battery. What can Lisa do to increase the strength of the electromagnet? She can use a nail with weaker magnetic properties. She can change the direction of the nail. She can increase the number of wire loops. She can reduce the current in the wire.
Answer:
C. She can increase the number of wire loops.
Explanation:
The more wire loops the more energy.
For a coil of wire, the magnetic field strength is increased by increasing the number of coils around the nail.
What is electromagnet?An electromagnet is a soft metal core shaped into a magnet by the passing the electric current through a coil surrounding it.
The end of the wire leading to the S is attached to the positive terminal of a battery. The end of the wire leading to the N is attached to the negative terminal of the battery. The current begins to flow. Current cant be changed to increase magnetic field strength, but the no of coils will definitely increase it.
Thus, To increase the strength of the electromagnet, Lisa can increase the number of wire loops.
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Quanto tempo deve ficar ligado um ferro eletrico de 1000 w para que tenha o mesmo consumo de energia que um chuveiro de 4400 w que fica ligado 10 minutos
Answer:
Thus, the time for the first lamp is 44 minutes.
Explanation:
Power of first lamp, P' = 1000 W
Power of second lamp, P'' = 4400 W
time for second lamp, t'' = 10 minutes
Let the time for first lamp is t'.
As the energy is same, so,
P' x t' = P'' x t''
1000 x t' = 4400 x 10
t' = 44 minutes
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.