a) The mass flow rate through the nozzle is 0.27 kg/s.
b) The exit area of the nozzle is 23.6 cm².
a) The mass flow rate through the nozzle can be calculated with the following equation:
[tex] \dot{m_{i}} = \rho_{i} v_{i}A_{i} [/tex]
Where:
[tex]v_{i}[/tex]: is the initial velocity = 20 m/s
[tex]A_{i}[/tex]: is the inlet area of the nozzle = 60 cm²
[tex]\rho_{i}[/tex]: is the density of entrance = 2.21 kg/m³
[tex] \dot{m} = \rho_{i} v_{i}A_{i} = 2.21 \frac{kg}{m^{3}}*20 \frac{m}{s}*60 cm^{2}*\frac{1 m^{2}}{(100 cm)^{2}} = 0.27 kg/s [/tex]
Hence, the mass flow rate through the nozzle is 0.27 kg/s.
b) The exit area of the nozzle can be found with the Continuity equation:
[tex] \rho_{i} v_{i}A_{i} = \rho_{f} v_{f}A_{f} [/tex]
[tex] 0.27 kg/s = 0.762 kg/m^{3}*150 m/s*A_{f} [/tex]
[tex] A_{f} = \frac{0.27 kg/s}{0.762 kg/m^{3}*150 m/s} = 0.00236 m^{2}*\frac{(100 cm)^{2}}{1 m^{2}} = 23.6 cm^{2} [/tex]
Therefore, the exit area of the nozzle is 23.6 cm².
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a) Mass flow rate through the nozzle: 0.265 kilograms per second, b) Exit area of the nozzle: 23.202 square centimeters.
We determine the Mass Flow Rate through the nozzle and the Exit Area of the nozzle by means of the Principle of Mass Conservation. A nozzle is a device that works at Steady State, so that Mass Balance can be reduced into this form:
[tex]\dot m_{in} = \dot m_{out}[/tex] (1)
Where:
[tex]\dot m_{in}[/tex] - Inlet mass flow, in kilograms per second.
[tex]\dot m_{out}[/tex] - Outlet mass flow, in kilograms per second.
Given that air flows at constant rate, we expand (1) by dimensional analysis:
[tex]\rho_{in} \cdot A_{in}\cdot v_{in} = \rho_{out}\cdot A_{out}\cdot v_{out}[/tex] (2)
Where:
[tex]\rho_{in}, \rho_{out}[/tex] - Air density at inlet and outlet, in kilograms per cubic meter.
[tex]A_{in}, A_{out}[/tex] - Inlet and outlet area, in square meters.
[tex]v_{in}, v_{out}[/tex] - Inlet and outlet velocity, in meters per second.
a) If we know that [tex]\rho_{in} = 2.21\,\frac{kg}{m^{3}}[/tex], [tex]A_{in} = 60\times 10^{-4}\,m^{2}[/tex] and [tex]v_{in} = 20\,\frac{m}{s}[/tex], then the mass flow rate through the nozzle is:
[tex]\dot m = \rho_{in}\cdot A_{in}\cdot v_{in}[/tex]
[tex]\dot m = \left(2.21\,\frac{kg}{m^{3}} \right)\cdot (60\times 10^{-4}\,m^{2})\cdot \left(20\,\frac{m}{s} \right)[/tex]
[tex]\dot m = 0.265\,\frac{kg}{s}[/tex]
The mass flow rate through the nozzle is 0.265 kilograms per second.
b) If we know that [tex]\rho_{in} = 2.21\,\frac{kg}{m^{3}}[/tex], [tex]A_{in} = 60\times 10^{-4}\,m^{2}[/tex], [tex]v_{in} = 20\,\frac{m}{s}[/tex], [tex]\rho_{out} = 0.762\,\frac{kg}{m^{3}}[/tex] and [tex]v_{out} = 150\,\frac{m}{s}[/tex], then the exit area of the nozzle is:
[tex]\rho_{in} \cdot A_{in}\cdot v_{in} = \rho_{out}\cdot A_{out}\cdot v_{out}[/tex]
[tex]A_{out} = \frac{\rho_{in}\cdot A_{in}\cdot v_{in}}{\rho_{out}\cdot v_{out}}[/tex]
[tex]A_{out} = \frac{\left(2.21\,\frac{kg}{m^{3}} \right)\cdot (60\times 10^{-4}\,m^{2})\cdot \left(20\,\frac{m}{s} \right)}{\left(0.762\,\frac{kg}{m^{3}} \right)\cdot \left(150\,\frac{m}{s} \right)}[/tex]
[tex]A_{out} = 2.320\times 10^{-3}\,m^{2}[/tex]
[tex]A_{out} = 23.202\,cm^{2}[/tex]
The exit area of the nozzle is 23.202 square centimeters.
Vặt nhỏ được ném lên từ điểm A trên mặt đất với vận tốc đầu 20m/s theo phương thẳng đứng. Xác định độ cao của điểm O mà vật đạt được. Bỏ qua ma sát
Explanation:
mặt đất với vận tốc ban đầu 20m/s. Bỏ qua mọi ma sát, lấy g = 10 m/s2. Độ cao cực đại mà vật đạt được là.
A Ball A and a Ball B collide elastically. The initial momentum of Ball A is -2.00kgm/s and the initial momentum of Ball B is -5.00kgm/s. Ball A has a mass of 4.00kg and is traveling at 2.50 m/s after the collision. What is the velocity of ball B if it has a mass of 6.50kg?
The velocity of B after the collision is obtained as -2.6 m/s.
What is the principle of conservation of momentum?Now we now that the principle of conservation of momentum states that the momentum before collision is equal to the momentum after collision.
Thus;
(-2.00kgm/s) + ( -5.00kgm/s) = ( 4.00kg * 2.50 m/s) + ( 6.50kg * v)
-7 = 10 + 6.5v
-7 - 10 = 6.5v
v = -7 - 10 /6.5
v = -2.6 m/s
Hence, the velocity of B after the collision is obtained as -2.6 m/s.
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A bus moving on a straight road increases its speed uniformly from rest to 20m's over a time period of 1 min. The distance travelled during the time is (a) 150 m (b) 300 m (c) 600 m (d) 900 m
Explanation:
Given that,
Initial velocity (u) = 0 m/sFinal velocity (v) = 20 m/sTime taken (t) = 1 minute = 60 secondsIn order to find the distance travelled, firstly we need calculate the acceleration.
→ v = u + at
→ 20 = 0 + 60a
→ 20 = 60a
→ 20 ÷ 60 = a
→ ⅓ m/s² = a
Now, by using the 2nd equation of motion :
→ s = ut + ½at²
→ s = 0(60) + ½ × ⅓ × (60)²
→ s = ⅙ × 3600
→ s = 1 × 600
→ s = 600 m
Hence, the distance travelled is 600 m.
The graph below shows a cycle of a heat engine. Add the following labels to the graph. Some labels are used more than once.
Labels: Isobaric process; W= 0J; Work done on the system; Work done by the system.
I will give brainliest!
P.S. AL2006 if you see this please help!
I'm not very good at this material. I'll try it, but if I were you, I wouldn't bet money on these answers.
"Isobaric" means constant pressure. So those are the horizontal lines, where every point on the line is at the same pressure. Those are the processes 1>2 and 3>4 .
I'm going around and around in my mind with the other labels, and I can't decide. So I'm afraid I can't answer any more of them ... they might be wrong.
Answer:
1 -> 2 & 3 -> 4: Isobaric process
4 -> 1: Work done BY the system
2 -> 3: Work done ON the system
W(total): W = 0J
Explanation:
The two horizontal lines (1 -> 2 & 3 -> 4) are "Isobaric" since isobaric processes take place at constant pressure. I believe 4 -> 1 is "Work done BY the system" since pressure increases when there is an increase of thermal energy, in other words, the system is absorbing heat. This is why the volume increases from 1 -> 2 after the system has absorbed heat in 4 -> 1. Following the directions of the arrows, 2 -> 3 would be "Work done ON the system" since pressure is DECREASING, meaning temperature is also exiting the system. That's why the next step (3 -> 4) shows a decrease in volume. This model depicts a process that has a W(total) of 0 J because this is a cycle.
I hope this helps :))
En la siguiente expresión matemáticas w=mg el peso w con relación a se relaciona con la masa m en una proporción
a) Directamente proporcional b) Inversamente proporcional c) Es constante
d) Ninguna de las anteriore
Answer:
a) Directamente proporcional
Explanation:
El peso se puede definir como la fuerza que actúa sobre un cuerpo o un objeto como resultado de la gravedad.
Matemáticamente, el peso de un objeto viene dado por la fórmula;
[tex] Peso = mg [/tex]
Donde;
m es la masa del objeto.
g es la aceleración debida a la gravedad.
De la expresión matemática, podemos deducir que el valor del peso de un objeto es directamente proporcional a la masa del objeto.
Por lo tanto, un aumento en la masa de un objeto provocaría un aumento en el peso del objeto y viceversa.
A silicon solar cell behaves like a battery with a 0.46 V terminal voltage. Suppose that 1.0 W of light of wavelength 620 nm falls on a solar cell and that 50%% of the photons give their energy to charge carriers, creating a current. What is the solar cell's efficiency that is, what percentage of the energy incident on the cell is converted to electric energy?
We have that the percentage of the energy incident on the cell that is converted to electric energy is
[tex]n=11\%[/tex]
From the question we are told that:
Voltage [tex]V=0.46V[/tex]
Power of light [tex]P=1.0W[/tex]
Wavelength [tex]w=620nm[/tex]
50 \% of the photons give their energy to charge carriers,
Generally, the equation for number of Protons is mathematically given by
[tex]N_p=\frac{P}{E}[/tex]
[tex]N_p=\frac{P \lambda}{hc}[/tex]
[tex]N_p=\frac{1}{(6.62*10^(-34)}*\frac{3*10^8}{(570*10^{-9}))}[/tex]
[tex]N_p=2.87*10^{18}[/tex]
Generally, the equation for Number of electron is mathematically given by
[tex]N_e=50 \% *n_3[/tex]
[tex]N_e=0.5*2.87*10^{18}[/tex]
[tex]N_e=1.43*10^{18}[/tex]
Therefore
Total current
[tex]I= e*N_e[/tex]
Where
e=electron Charge
Therefore
[tex]I=1.43*10^{18}*1.6*10^-{19}[/tex]
[tex]I=0.230A[/tex]
Generally, the equation for Power is mathematically given by
[tex]P=VI[/tex]
[tex]P=0.46*0.230[/tex]
[tex]P=0.1058W[/tex]
Therefore
Efficiency
[tex]n=\frac{0.1058}{1}[/tex]
[tex]n=0.1058[/tex]
[tex]n=11\%[/tex]
In conclusion
The percentage of the energy incident on the cell that is converted to electric energy is
[tex]n=11\%[/tex]
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Light of the same wavelength passes through two diffraction gratings. One grating has 4000 lines/cm, and the other one has 6000 lines/cm. Which grating will spread the light through a larger angle in the first-order pattern
Answer:
6000 lines/cm
Explanation:
From the question we are told that:
Grating 1=4000 lines/cm
Grating 2=6000 lines/cm
Generally The Spread of fringes is Larger when the Grating are closer to each other
Therefore
Grating 2 will spread the the light through a larger angle in the first-order pattern because its the closest with 6000 lines/cm
why meter cube is called derived unit
Answer:
Because it is the result of two more fundamental units, a derived unit is termed that. For volume, the cubic meter (m³) is the fundamental unit of area. Any number that cannot be measured directly with any equipment is referred to as a derived unit. For example, we can't quantify a substance's density using a rule, scale, or bucket.
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Consider a piston filled with 3 mols of an ideal gas, kept at a constant temperature 290 K. We slowly compress the gas starting at 2 m3 and ending at 1 m3. How much work do we need to do on the gas to perform this operation
Answer: [tex]-5013.65\ J[/tex]
Explanation:
Given
No of moles [tex]n=3[/tex]
Temperature [tex]T=290\ K[/tex]
Initial volume [tex]V_1=2\ m^3[/tex]
Final volume [tex]V_2=1\ m^3[/tex]
Work done in constant temperature process is
[tex]W=nRT\ln \left(\dfrac{V_2}{V_1}\right)[/tex]
Insert the values
[tex]\Rightarrow W=3\times 8.314\times 290\ln \left (\dfrac{1}{2}\right)\\\\\Rightarrow W=-870\times 8.314\times \ln (2)\\\Rightarrow W=-5013.65\ J[/tex]
A bullet 2cm log is fired at 420m/s and passes straight a 10cm thick board exiting at 280m/s
Complete question:
A bullet 2 cm long is fired at 420m/s and passes straight through a 10.0 cm thick board, exiting at 280m/s? What is the average acceleration of the bullet through the board?
Answer:
The average acceleration of the bullet through the board is -4.083 x 10⁵ m/s²
Explanation:
Given;
initial velocity of the bullet, u = 420 m/s
final velocity of the bullet, v = 280 m/s
length of the bullet, d₁ = 2 cm
thickness of the board, d₂ = 10 cm
total distance penetrated by the bullet through the board;
d = d₁ + d₂ = 2 cm + 10 cm = 12 cm = 0.12 m
The average acceleration of the bullet through the board is calculated as;
[tex]v^2 = u^2 + 2ad\\\\2ad = v^2 - u^2\\\\a = \frac{v^2 - u^2}{2d} \\\\a = \frac{(280^2) - (420^2)}{2(0.12)} = -4.083 \times 10^{5} \ m/s^2[/tex]
Therefore, the average acceleration of the bullet through the board is -4.083 x 10⁵ m/s²
A hydraulic vane pump has a real flow of 20 liters per minute, a pressure of 230 bar, a pump speed of 1400 rpm. Know that the input power is 10kW and the mechanical efficiency is 88%.
a) Calculate the volumetric efficiency of the pump
b) Calculate the specific volume of the pump (cm/rev). Question 3 (2,5d): Design a pneumatic transmission system to control 02 single-acting cylinders, using 02 reversing valves 3/2 acting by push button, 02 throttle valves - one-way. Describe the working principle of the system.
Explanation:
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Which of the following is true about resistivity of any given metal? depends on its temperature. varies nearly linearly with temperature. has units of ohm-meter. A. II and III only B. I and II only C. I and III only D. I, II and III E. III only
___________________
[tex]\huge{\underline{\sf{\blue{Answer}}}}[/tex]
___________________
[tex]\sf{C. \:I\: and \:III}[/tex]
___________________
The correct statements about resistivity of any given metal are The resistivity of metal is more than that of insulators and Metals can carry electricity more easily than insulators. Option a and c are correct answer.
Resistivity is a property that quantifies how strongly a material opposes the flow of electric current. Metals have lower resistivity compared to insulators. This means that metals allow electric current to flow more easily than insulators.
Due to their lower resistivity, metals have higher electrical conductivity and can carry electric current more easily compared to insulators. Insulators, on the other hand, have high resistivity and hinder the flow of electric current. Resistivity is a material-specific property and varies for different substances. Metals, such as copper or aluminum, have low resistivity and are often used as conductors for transmitting electricity. Insulators, such as rubber or plastic, have high resistivity and are used to prevent the flow of electricity.
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The complete question is, "Which of the following statements are true about resistivity of any given metal?
A. The resistivity of metal is more than that of insulators.
B. The insulators and metals have same resistivity.
C. Metals can carry electricity more easily than insulators.
D. The resistivity of insulators is more than that of metals.
Cuando el pistón tiene un volumen de 2x10^-4 m^3, el gas en el pistón está a una presión de 150 kPa. El área del pistón es 0.00133 m^2. Calcular la fuerza que el gas ejerce sobre el embolo del pistón.
Answer:
F = 1.128 10⁸ Pa
Explanation:
Pressure is defined by
P = F / A
If the gas is ideal for equal force eds on all the walls, so on the piston area we have
F = P A
We reduce the pressure to the SI system
P = 150 kpa (1000 Pa / 1kPa = 150 103 Pa
we calculate
F = 150 10³ / 0.00133
F = 1.128 10⁸ Pa
The time delay between transmission and the arrival of the reflected wave of a signal using ultrasound traveling through a piece of fat tissue was 0.13 ms. At what depth did this reflection occur? (The average propagation speed for sound in body tissue is 1540 m/s)
Answer:
10.01 cm
Explanation:
Given that,
The time delay between transmission and the arrival of the reflected wave of a signal using ultrasound traveling through a piece of fat tissue was 0.13 ms.
The average propagation speed for sound in body tissue is 1540 m/s.
We need to find the depth when the reflection occur. We know that, the distance is double when transmitting and arriving. So,
[tex]v=\dfrac{2d}{t}\\\\d=\dfrac{vt}{2}\\\\d=\dfrac{1540\times 0.13\times 10^{-3}}{2}\\\\d= $$0.1001\ m[/tex]
or
d = 10.01 cm
So, the reflection will occur at 10.01 cm.
A ball is thrown upward from the edge of a cliff with an initial velocity of 6 m/s. (a) How fast is it moving 0.5 s later? In what direction? (b) How fast is it moving 2 s later? In what direction?
Answer:
Explanation:
Kinematic equation
v = u + at
If UP is assumed to be the positive direction and we let gravity be 10 m/s² which will be in the downward direction so will be negative.
a) v = 6 + (-10)(0.5) = 1 m/s the result is positive, so upward
b) v = 6 + (-10)(2) = -14 m/s the result is negative, so downward
Light takes 1.2 sec to get from the moon to the Earth. Assume you are looking at the moon with noticeable earth shine. If the Sun burned out, you would eventually see the crescent of the moon disappear. The earth shine part of the moon would disappear Answer 2.4 s after the crescent disappeared.
Answer:
1.2 seconds
Explanation:
Answer to the following question is 1.2 seconds
Because light from the moon takes 1.2 seconds to reach Earth, the light released from the crescent immediately before it vanishes will also take 1.2 seconds to reach Earth. As a result, the earth-shine portion of the moon will vanish 1.2 seconds after the crescent has vanished.
In first case a mass M is split into two parts with one part being 1/6.334 th of the original mass. In second case M is split into two equal parts. In both the cases the two parts are separated by same distance. What ratio of the magnitude of the gravitational force in first case to the magnitude of the gravitational force in the second case
Answer:
[tex]F_r=0.132:0.25[/tex]
Explanation:
From the question we are told that:
[tex]M_1=M*\frac{1}{6.334}[/tex]
Therefore
[tex]M_2=M-M*\frac_{1}{6.334}[/tex]
[tex]M_2=M*\frac{5.334}{6.334}[/tex]
Generally the equation for Gravitational force of attraction is mathematically given by
For Unequal split
[tex]F=\frac{GM_1M_2}{d^2}[/tex]
[tex]F=\frac{G(M*\frac_{1}{6.334})(M*\frac{5.334}{6.334})}{d^2}[/tex]
[tex]F=\frac{GM^2}{d^2}*(0.132)[/tex]
For equal split
[tex]F=\frac{GM_1M_2}{d^2}[/tex]
[tex]F=\frac{G(\frac{M}{2})((\frac{M}{2}}{d^2}[/tex]
[tex]F=0.25 \frac{GM^2}{d^2}[/tex]
Therefore the ratio of the gravitational force is
[tex]F_r=0.132:0.25[/tex]
A converging lens is used to focus light from a small bulb onto a book. The lens has a focal length of 10.0 cm and is located 40.0 cm from the book. Determine the distance from the lens to the light bulb.
Answer:
[tex]u=13.3cm[/tex]
Explanation:
From the question we are told that:
Focal Length [tex]F=10.0cm[/tex]
Distance [tex]d=40cm[/tex]
Generally the equation for Focal length is mathematically given by
[tex]\frac{1}{f}=\frac{1}{u}+\frac{1}{v}[/tex]
[tex]\frac{1}{10}=\frac{1}{u}+\frac{1}{40}[/tex]
[tex]\frac{1}{u}=\frac{3}{40}[/tex]
[tex]u=13.3cm[/tex]
Focal length is the distance from the center of the lens to principle foci. The distance of the from the lens to the light bulb is 13.3 cm.
The distance can be determined by the formula,
[tex]\bold {\dfrac 1{f} = \dfrac 1{u} + \dfrac 1{v} }[/tex]
Where,
f - focal length = 10 cm
u - distance of object = ?
v = distance of image = 40 cm
Put the values in the equation,
[tex]\bold {\dfrac 1{10} = \dfrac 1{u} + \dfrac 1{40} }\\\\\bold {\dfrac 1{u} = \dfrac 3{40}}\\\\\bold {\dfrac 1{u} = 13.3 cm}[/tex]
Therefore, the distance of the from the lens to the light bulb is 13.3 cm.
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A person runs up the stairs elevating his 102 kg body a vertical distance of 2.29 meters in a time of 1.32 seconds at a constant speed.
Determine the work done by the person climbing the stair case.
Answer:
Work done = 2289.084 Joules
Explanation:
Given the following data;
Mass = 102 Kg
Height = 2.29
Time = 1.32 seconds
We know that acceleration due to gravity, g = 9.8 m/s²
a. To find the work done by the person;
Here, work would be done in the form of gravitational potential energy.
Gravitational potential energy (GPE) is an energy possessed by an object or body due to its position above the earth.
Mathematically, gravitational potential energy is given by the formula;
G.P.E = mgh
Where;
G.P.E represents potential energy measured in Joules.
m represents the mass of an object.
g represents acceleration due to gravity measured in meters per seconds square.
h represents the height measured in meters.
Substituting into the formula, we have;
Work done = 102 * 2.29 * 9.8
Work done = 2289.084 Joules
If ∆H = + VE , THEN WHAT REACTION IT IS
1) exothermic
2) endothermic
Answer:
endothermic
Explanation:
An endothermic is any process with an increase in the enthalpy H (or internal energy U) of the system. In such a process, a closed system usually absorbs thermal energy from its surroundings, which is heat transfer into the system.
Which of the following is a form of mechanical energy?
A. Chemical energy
B. Gravitational potential energy
C. Thermal energy
D. Nuclear energy
Answer:
B
Explanation:
no reason for this answer
A 1-cm long wire carrying 15 A is inside a solenoid 4 cm in radius with 800 turns/m carrying a current of 40 mA. The wire segment is oriented perpendicularly to the axis of the solenoid. What is the magnitude of the magnetic force on this wire segment in ???? N?
Answer:
the magnitude of the magnetic force on the wire segment is 6.03 x 10⁻⁶ N
Explanation:
Given;
length of the conductor, L = 1 cm = 0.01 m
current carried by the solenoid, I₁ = 15 A
radius of the solenoid, r = 4 cm
number of turns per length of the solenoid, n = 800 turns/m
current carried by the solenoid, I₂ = 40 mA = 0.04 A
The magnetic field of the solenoid is calculated as;
B = μnI₂
where;
μ is the permeability of free space = 4π x 10⁻⁷ Tm/A
B = ( 4π x 10⁻⁷) x (800) x (0.04)
B = 4.022 x 10⁻⁵ T
The magnitude of the magnetic force on the wire segment is calculated as;
F = BI₁L sinθ
where
θ is the angle made by the wire segment against the solenoid = 90⁰
F = (4.022 x 10⁻⁵) x (15) x (0.01) x sin(90)
F = 6.03 x 10⁻⁶ N
Therefore, the magnitude of the magnetic force on the wire segment is 6.03 x 10⁻⁶ N
b. Block on an incline
A block of mass mı = 3.9 kg on a smooth inclined plane of angle 38° is connected by a cord over a small frictionless
pulley to a second block of mass m2 = 2.6 kg hanging vertically. Take the positive direction up the incline and use 9.81
m/s2 for g.
What is the tension in the cord to the nearest whole number?
Block on the incline:
• net force parallel to the incline
∑ F (para.) = m₁ g sin(38°) - T = m₁ a
where T is the magnitude of tension in the cord.
Notice that we take down-the-incline to be the positive direction, so that if the 3.9-kg block pulls the 2.6-kg block upwards, then the acceleration of the system is positive.
Suspended block:
• net vertical force
∑ F (vert.) = T - m₂ g = m₂ a
Solve both equations for the acceleration a, set the results equal to one another, and solve for T :
a = g sin(38°) - T/m₁
a = T/m₂ - g
==> g sin(38°) - T/m₁ = T/m₂ - g
==> T (1/m₂ + 1/m₁) = g (sin(38°) + 1)
==> T = g (sin(38°) + 1) / (1/m₂ + 1/m₁)
==> T = (9.81 m/s²) (sin(38°) + 1) / (1/(2.6 kg) + 1/(3.9 kg)) ≈ 25 N
Question 5 of 10
What must be the same for two resistors that are connected in parallel?
Answer:
in parallel combination : potential difference between two terminal of resistors are always constant. ... hence, potential difference ( voltage ) must be same across each resistor .
Explanation:
An observer on Earth sees Planet X to be stationary and also sees a rocket traveling toward Planet X at 0.5c. The rocket emits a pulse of light that travels outward in all directions. According to an observer on Planet X, how fast is the light pulse traveling toward them?
a) 2c/3
b) c/2
c) 2c/3
d) 5c/6
e) c
(E) c
Explanation:
The speed of light is always equal to c regardless of the relative motion of the light source.
A force of 1000N is used to kick a football of mass 0.8kg find the velocity with which the ball moves if it takes 0.8 sec to be kicked.
The velocity of the ball is 100m/s
The first step is to write out the parameters;
The force used to kick the ball is 1000N
The mass of the ball is 0.8 kg
Time is 0.8 seconds
Therefore the velocity can be calculated as follows
F= Mv-mu/t
1000= 0.8(v) - 0.8(0)/0.8
1000= 0.8v- 0.8/0.8
Cross multiply both sides
1000(0.8) = 0.8v
800= 0.8v
divide both sides by the coefficient of v which is 8
800/0.8= 0.8v/0.8
v= 1000m/s
Hence the velocity is 1000m/s
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Two forces A and B act at a point. If their resultant is [given by] (B - A) in the direction of B, then
A. A and B are equal
B. A is greater than B
C. the angle between A and B is 0°
D. the angle between A and B is 90°
E. the angle between A and B is 180°
A refrigerator has a coefficient of performance equal to 4.00. The refrigerator takes in 110 J of energy from a cold reservoir in each cycle. (a) Find the work required in each cycle. J (b) Find the energy expelled to the hot reservoir. J
Answer:
The correct answer is:
(a) 27.5 Joules
(b) 141.5 Joules
Explanation:
Given:
Energy,
[tex]Q_c = 110 \ J[/tex]
Coefficient of performance refrigerator,
[tex]Cop(refrig)=4[/tex]
(a)
As we know,
⇒ [tex]Cop(refrig) = \frac{Q_c}{Work}[/tex]
or,
⇒ [tex]Work=\frac{Q_c}{Cop(refrig)}[/tex]
[tex]=\frac{110}{4}[/tex]
[tex]=27.5 \ Joules[/tex]
(b)
⇒ [tex]Heat \ expelled = Heat \ removed +Work \ done[/tex]
or,
⇒ [tex]Q_h = Q_c+Work[/tex]
[tex]=114+27.5[/tex]
[tex]=141.5 \ Joules[/tex]
Maximum range of a projectile is 1.6 m. Then the velocity of projection will be..... (g=10m/s)
Answer:
4 m/s
Explanation:
From the question given above, the following data were obtained:
Maximum range (Rₘₐₓ) = 1.6 m
Acceleration due to gravity (g) = 10 m/s²
Initial velocity (u) =?
The initial velocity of the projectile can be obtained as follow:
Rₘₐₓ = u² / g
1.6 = u² / 10
Cross multiply
u² = 1.6 × 10
u² = 16
Take the square root of both side
u = √16
u = 4 m/s
Therefore, the velocity of the projectile is 4 m/s
b. Projectile on cliff (range)
An object of mass 5 kg is projected at an angle of 25° to the horizontal with a speed of 22 ms-1 from the top of the cliff.
The height of the cliff is 21 m. Take g, the acceleration due to gravity, to be 9.81 ms2
How far horizontally (to 1 decimal place) from the base of the cliff does the object land?
Answer:
x = 41.28 m
Explanation:
This is a projectile launching exercise, let's find the time it takes to get to the base of the cliff.
Let's start by using trigonometry to find the initial velocity
cos 25 = v₀ₓ / v₀
sin 25 = Iv_{oy} / v₀
v₀ₓ = v₀ cos 25
v_{oy} = v₀ sin 25
v₀ₓ = 22 cos 25 = 19.94 m / s
v_{oy} = 22 sin 25 = 0.0192 m / s
let's use movement on the vertical axis
y = y₀ + v_{oy} t - ½ g t²
when reaching the base of the cliff y = 0 and the initial height is y₀ = 21 m
0 = 21 + 0.0192 t - ½ 9.81 t²
4.905 t² - 0.0192 t - 21 = 0
t² - 0.003914 t - 4.2813 =0
we solve the quadratic equation
t = [tex]\frac{ 0.003914\ \pm \sqrt{0.003914^2 + 4 \ 4.2813 } }{2}[/tex]
t = [tex]\frac{0.003914 \ \pm 4.13828}{2}[/tex]
t₁ = 2.07 s
t₂ = -2.067 s
since time must be a positive scalar quantity, the correct result is
t = 2.07 s
now we can look up the distance traveled
x = v₀ₓ t
x = 19.94 2.07
x = 41.28 m
