Answer:
I HOPE THIS IS CORRECT
Explanation:
Power of water =2 kw=2000w
Mass of water =200kg
difference in temperature ΔT=70−10=60oC
Concept
energy required to heat the water = energy given by water in time t=pt
energy required to increase tempeature of water by 60oC,Q=msΔT
S= specific heat =4200J/kgoC
pt=msΔT
2000×t=200×4200×60
t=25200
or t=25.2×103sec.
The quantity of heat generated is 720,000 Joules or 720 kiloJoules.
Given the following data:
Time = 1 hourPower = 200 WattsTo find the quantity of heat generated when the electrical coil is used:
Note: the quantity of heat generated is energy.
First of all, we would convert the time in hour to seconds.
Conversion:
1 hour = 3600 seconds
Mathematically, the energy of a device is given by the formula;
[tex]Energy = power[/tex] × [tex]time[/tex]
Substituting the values into the formula, we have;
[tex]Energy = 200[/tex] × [tex]3600[/tex]
Energy = 720,000 Joules or 720 kiloJoules.
Therefore, the quantity of heat generated is 720,000 Joules or 720 kiloJoules.
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the 200 g baseball has a horizontal velocity of 30 m/s when it is struck by the bat, B, weighing 900 g, moving at 47 m/s. during the impact with the bat, how many impules of importance are used to find the final velocity of the bat
Solution :
Given :
Mass of the baseball, m = 200 g
Velocity of the baseball, u = -30 m/s
Mass of the baseball after struck by the bat, M = 900 g
Velocity of the baseball after struck by the bat, v = 47 m/s
According to the conservation of momentum,
[tex]Mv+mu=Mv_1+mv_2[/tex]
(900 x 47) + (200 x -30) = (900 x [tex]v_1[/tex]) + (200 x [tex]v_2[/tex])
36300 = (900 x [tex]v_1[/tex]) + (200 x [tex]v_2[/tex])
[tex]9v_1 + 2v_2 = 363[/tex] ..............(i)
[tex]9v_1 = 363 - 2v_2[/tex]
[tex]v_1=\frac{363 - 2v_2}{9}[/tex]
The mathematical expression for the conservation of kinetic energy is
[tex]\frac{1}{2}Mv^2+\frac{1}{2}mu^2 = \frac{1}{2}Mv_1^2+\frac{1}{2}mv_2^2[/tex]
[tex]\frac{1}{2}(900)(47)^2+\frac{1}{2}(200)(-30)^2 = \frac{1}{2}(900)v_1^2+\frac{1}{2}(200)v_2^2[/tex] ................(ii)
[tex]$(9)(14)^2+(2)(-30)^2 = (9)v_1^2+2v_2^2$[/tex]
[tex]21681 = 9v_1^2+2v_2^2[/tex]
Substituting (i) in (ii)
[tex]21681= 9\left( \frac{363-2v_2}{9}\right)^2+2v_2^2[/tex]
[tex](363-2v_2)^2+18v_2^2=195129[/tex]
[tex](363)^2+18v_2^2-2(363)(2v_2)+(363)^2-195129=0[/tex]
[tex]22v_2^2-145v_2-63360=0[/tex]
Solving the equation, we get
[tex]v_2=96 \ m/s, -30 \ m/s[/tex]
The negative velocity is neglected.
Therefore, substituting 96 m/s for [tex]v_2[/tex] in (i), we get
[tex]v_1=\frac{363-(2 \times 96)}{9}[/tex]
= 19
Thus, only impulse of importance is used to find final velocity.
Hi there! I'm not quite sure on how to solve this....
[tex]\frac{dx}{dt} = 2.5 \: \frac{cm}{sec} [/tex]
Explanation:
The volume of a cube is V = x^3. Taking the time derivative of this expression, we get
[tex] \frac{dV}{dt} = 3 {x}^{2} \frac{dx}{dt} [/tex]
or
[tex]\frac{dx}{dt} = \frac{1}{3 {x}^{2}} \frac{dV}{dt} [/tex]
We know that dV/dt = 30 cm^3/sec so the value of dx/dt when x = 2 cm is
[tex]\frac{dx}{dt} = \frac{1}{3 {(2 \: cm)}^{2}}(30 \: \frac{ {cm}^{3} }{sec} ) = 2.5 \: \frac{cm}{sec} [/tex]
Answer:
Explanation:
[tex]V=x^3\\\\\frac{dV}{dt}=3x^2\frac{dx}{dt}\\\\30\frac{cm^3}{s}=3x^2\frac{dx}{dt}\\\\\frac{dx}{dt}=\frac{30\frac{cm^3}{s}}{3x^2}~at~x=2cm,~\frac{dx}{dt}=\frac{30\frac{cm^3}{s}}{3*(2cm)^2}=\frac{5}{2}\frac{cm}{s}[/tex]
Page
E QON
1 What is force ? Write its unit and mention
any
three effects of the force.
Force is a push or a pull that changes or trends to change the state of rest or uniform motion of an object or changes the direction or shape of an object. It causes objects to accelerate. SI unit is Newton.
1) Can change the state of an object : For example, pushing a heavy stone in order to move it.
2) May change the speed of an object if it is already moving. For example, catching a ball hit by a batsman.
3) May change the direction of motion of an object.
An object that sinks in water has a mass in air of 0.0675 kg. Its apparent mass when submerged in water is 0.0424 kg. What is the specific gravity SG of the object? What material is the object probably made?
Answer:
1. SG
true
=2.689
2. The object is probably some sort of minerals and rocks such as Feldspar, Corals, Beryl, etc.
Explanation:
Given:
mass in the air= 0.0675 kg
mass in water= 0.0424 kg
The specific gravity of the object will be 2.6892. It is the ratio of the density of the given fluid and the standard fluid.
What is density?Density is specified as the mass divided by the volume. It is represented by the unit of measurement as kg/m³.
The mass of the object in air;
m=Vρ₀
m=0.0675 kg
Buoyant force on the object;
B= Vρₐg
For equilibrium;
N+B=m₀g
n=m₀g-Vρₓg
N/g=m₀-Vρₓ
N/g=0.0424 kg
[tex]\rm \frac{V\rho_0}{V\rho_x} =\frac{0.0675 }{m_0-0.0424 \ kg} \\\\ \frac{\rho_0}{\rho_x} =\frac{0.0675}{0.0675-0.0424} \\\\ \frac{\rho_0}{\rho_x} =2.6892[/tex]
Hence, the specific gravity of the object will be 2.6892.
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A pulley has a mechanical advantage of 1. What does this tell you about the size and direction of the input and output forces?
Answer:
The number of input force is the same as output. ... If it equals once, then both numbers are equal making it the same.Explanation:
Does this helps
Answer:
The number of input force is the same as output. Formula for MA (Mechanical Advantage) is Input Force/Output Force. If it equals once, then both numbers are equal making it the same. In order to raise MA, you must lower efficiency, something you learn around grade 8. Good luck!
P.S. Direction is the same for both, meaning if you pull something, the object you pull will come towards you.
Tres ladrillos idénticos están atados entre sí por medio de cuerdas y penden de una balanza que marca en total 24 N. ¿Cuál es la tensión de la cuerda que soporta al ladrillo inferior? ¿Cuál es la tensión en la cuerda que se encuentra entre el ladrillo de en medio y el superior?
ayuda!!!!!
answer 8N 8N
la tensión del ladrillo inferior es 8N ya que los ladrillos son idénticos, decido 24 por 3 para darme 8
la tensión entre el ladrillo superior y el medio también es 8N
Determine the magnitude of the minimum acceleration at which the thief can descend using the rope. Express your answer to two significant figures and include the appropriate units.
Answer: hello your question is incomplete below is the missing part
A 69-kg petty thief wants to escape from a third-story jail window. Unfortunately, a makeshift rope made of sheets tied together can support a mass of only 58 kg.
answer:
To 2 significant Figures = 1.6 m/s^2
Explanation:
Calculate the magnitude of minimum acceleration at which the thief can descend
we apply the relation below
Mg - T = Ma --- ( 1 )
M = 69kg
g = 9.81
T = 58 * 9.81
a = ? ( magnitude of minimum acceleration)
From equation 1
a = [ ( 69 * 9.81 ) - ( 58 * 9.81 ) ] / 69
= 1.5639 m/s^2
To 2 significant Figures = 1.6 m/s^2
A shooting star is actually the track of a meteor, typically a small chunk of debris from a comet that has entered the earth's atmosphere. As the drag force slows the meteor down, its kinetic energy is converted to thermal energy, leaving a glowing trail across the sky. A typical meteor has a surprisingly small mass, but what it lacks in size it makes up for in speed. Assume that a meteor has a mass of 1.5
Answer:
A. Power generated by meteor = 892857.14 Watts
Yes. It is obvious that the large amount of power generated accounts for the glowing trail of the meteor.
B. Workdone = 981000 J
Power required = 19620 Watts
Note: The question is incomplete. A similar complete question is given below:
A shooting star is actually the track of a meteor, typically a small chunk of debris from a comet that has entered the earth's atmosphere. As the drag force slows the meteor down, its kinetic energy is converted to thermal energy, leaving a glowing trail across the sky. A typical meteor has a surprisingly small mass, but what it lacks in size it makes up for in speed. Assume that a meteor has a mass of 1.5 g and is moving at an impressive 50 km/s, both typical values. What power is generated if the meteor slows down over a typical 2.1 s? Can you see how this tiny object can make a glowing trail that can be seen hundreds of kilometers away? 61. a. How much work does an elevator motor do to lift a 1000 kg elevator a height of 100 m at a constant speed? b. How much power must the motor supply to do this in 50 s at constant speed?
Explanation:
A. Power = workdone / time taken
Workdone = Kinetic energy of the meteor
Kinetic energy = mass × velocity² / 2
Mass of meteor = 1.5 g = 0.0015 kg;
Velocity of meteor = 50 km/s = 50000 m/s
Kinetic energy = 0.0015 × (50000)² / 2 = 1875000 J
Power generated = 1875000/2.1 = 892857.14 Watts
Yes. It is obvious that the large amount of power generated accounts for the glowing trail of the meteor.
B. Work done by elevator against gravity = mass × acceleration due to gravity × height
Work done = 1000 kg × 9.81 m/s² × 100 m
Workdone = 981000 J
Power required = workdone / time
Power = 981000 J / 50 s
Power required = 19620 Watts
Therefore, the motor must supply a power of 19620 Watts in order to lift a 1000 kg to a height of 100 m at a constant speed in 50 seconds.
If the frequency of a sound wave is 250 hertz and its wavelength is 1.36 meters, what is the wave's velocity
OA. 250 meters/second
ОВ.
340 meters/second
O c.
200 meters/second
OD.
120 meters/second
Answer:
B
Explanation:
V=frequency*wavelength V=? W=1.36mF=250hertzAnswer:
B
Explanation:
Bc i say
A 4-kW resistance heater in a water heater runs for 3 hours to raise the water temperature to the desired level. Determine the amount of electric energy used in both kWh and kJ.
Answer:
12kWhr
Explanation:
Energy = Power * Time
Power = 4kW
Time = 3hrs
Substitute into the formula
Energy used up = 4kW * 3hrs
Energy used up = 12kWhr
A 70-kg boy is surfing and catches a wave which gives him an initial speed of 1.6 m/s. He then drops through a height of 1.60 m, and ends with a speed of 8.5 m/s. How much nonconservative work (in kJ) was done on the boy
Answer:
3.6 KJ
Explanation: Given that a 70-kg boy is surfing and catches a wave which gives him an initial speed of 1.6 m/s. He then drops through a height of 1.60 m, and ends with a speed of 8.5 m/s. How much nonconservative work (in kJ) was done on the boy
The workdone = the energy.
There are two different energies in the scenario - the potential energy (P.E ) and the kinetic energy ( K.E )
P.E = mgh
P.E = 70 × 9.8 × 1.6
P.E = 1097.6 J
P.E = 1.098 KJ
K.E = 1/2mv^2
K.E = 1/2 × 70 × 8.5^2
K.E = 2528.75 J
K.E = 2.529 KJ
The non conservative workdone = K.E + P.E
Work done = 1.098 + 2.529
Work done = 3.63 KJ
Therefore, the non conservative workdone is 3.6 KJ approximately
13. What type of lens bends light outwards and away from a point?
concave
Answer:
No,it isn't concave. The correct answer is convex lens.
Explanation:
A lens is a piece of transparent material bound by two surfaces of which at least one is curved. A lens bound by two spherical surfaces bulging outwards is called a bi-convex lens or simply a convex lens. A single piece of glass that curves outward and converges the light incident on it is also called a convex lens.
Convex lens is the answer.
See the attached diagram.
The attraction of a person's body toward the Earth is called weight. The reaction to this force is
Answer:
Person's body pulling on the earth
Explanation:
The weight of an object is the attraction of a person's body toward the Earth. The weight of an object is given by :
Weight (W) = mass (m) × acceleration due to gravity (g)
We know that, for an action there is an equal and opposite reaction.
So, there must be a reaction force for the weight of an object. The reaction force must be the person's body pulling on the earth.
A 3 5m container is filled with 900 kg of granite (density of 2400 3 kg m/ ). The rest of the volume is air, with density equal to 3 1.15 / kg m . Find the mass of air and the overall (average) specific volume
Complete question:
A 5-m³ container is filled with 900 kg of granite (density of 2400 kg/m3). The rest of the volume is air, with density equal to 1.15 kg/m³. Find the mass of air and the overall (average) specific volume.
Answer:
The mass of the air is 5.32 kg
The specific volume is 5.52 x 10⁻³ m³/kg
Explanation:
Given;
total volume of the container, [tex]V_t[/tex] = 5 m³
mass of granite, [tex]m_g[/tex] = 900 kg
density of granite, [tex]\rho _g[/tex] = 2,400 kg/m³
density of air, [tex]\rho_a[/tex] = 1.15 kg/m³
The volume of the granite is calculated as;
[tex]V_g = \frac{m_g}{ \rho_g}\\\\V_g = \frac{900 \ kg}{2,400 \ kg/m^3} \\\\V_g = 0.375 \ m^3[/tex]
The volume of air is calculated as;
[tex]V_a = V_t - V_g\\\\V_a = 5 \ m^3 \ - \ 0.375 \ m\\\\V_a = 4.625 \ m^3[/tex]
The mass of the air is calculated as;
[tex]m_a = \rho_a \times V_a\\\\m_a = 1.15 \ kg/m^3 \ \times \ 4.625 \ m^3\\\\m_a = 5.32 \ kg[/tex]
The specific volume is calculated as;
[tex]V_{specific} = \frac{V_t}{m_g \ + \ m_a} = \frac{5 \ m^3}{900 \ kg \ + \ 5.32\ kg} = 5.52 \times 10^{-3} \ m^3/kg[/tex]
A 2000-kg truck traveling at a speed of 6.0 m/s slows down to 4.0 m/s along a straight road. What
is the magnitude of the impulse?
The magnitude of the impulse of the truck is equal to 4000 Kg.m/s.
What is impulse?Impulse can be described as the integral of a force over the time interval for which it acts. Impulse is also a vector quantity since force is a vector quantity. Impulse can be applied to an object that generates an equivalent vector change in its linear momentum.
The S.I. unit of impulse is N⋅s and the dimensionally equivalent unit of momentum is kg⋅m/s. A resultant force gives acceleration and changes the velocity of an object for as long as it acts.
Given the mass of the truck, m= 2000 Kg
The initial speed of the truck, u = 6 m/s
The final speed of the truck, v = 4 m/s
The change in the linear momentum is equal to the impulse.
I = ΔP = mv - mu
I = 2000 ×4 - 2000 × 6
I = 8000 - 12000
I = - 4000 Kg.m/s²
Therefore, the magnitude of the impulse is 4000 Kg.m/s².
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A pilot flies an airplane at a constant speed of 590 km/h in the vertical circle of radius 970 m. Calculate the force exerted by the seat on the 82-kg pilot at point A and at point B.
Answer:
[tex]F = \frac{m {v}^{2} }{r} \\ = \frac{82 \times { (\frac{590 \times 1000}{3600} )}^{2} }{970} \\ = 2270.6 \: newtons[/tex]
A 5.0-kg mass is placed at (3.0, 4.0) m, and a 6.0-kg mass is placed at (3.0, -4.0) m. What is the moment of inertia of this system of masses about the y-axis?
Answer:
the moment of inertia of this system of masses about the y-axis is 99 kgm²
Explanation:
Given the data in the question;
mass m₁ = 5.0 kg at point ( 3.0, 4.0 )
mass m₂ = 6.0 kg at point ( 3.0, -4.0 )
Now, Moment of inertia [tex]I[/tex] of this system of masses about the y-axis will be;
Moment of inertia [tex]I[/tex]ₓ = mixi²
Moment of inertia [tex]I[/tex] = m₁x₁² + m₂x₂²
we substitute
Moment of inertia [tex]I[/tex] = [ 5.0 × ( 3 )² ] + [ 6.0 × ( 3 )² ]
Moment of inertia [tex]I[/tex] = [ 5.0 × 9 ] + [ 6.0 × 9 ]
Moment of inertia [tex]I[/tex] = 45 + 54
Moment of inertia [tex]I[/tex] = 99 kgm²
Therefore, the moment of inertia of this system of masses about the y-axis is 99 kgm²
An irregular shape object has a mass of 19 oz. A graduated cylinder with and initial volume of 33.9 mL. After the object was dropped in the graduated cylinder, it had a volume of 92.8 mL. What is the density of object( g/mL)
Explanation:
m = 19 oz × (28.3 g/1 oz) = 537.7 g
V = 92.8 mL
[tex]\rho = \dfrac{m}{V}= \dfrac{537.7\:g}{92.8\:mL} = 5.79\:\frac{g}{mL}[/tex]
why are you teachers regarded as professionals
Answer:
coz teaching is their profession.
Who stated that man is an animal
Electromagnetic waves from the sun carry what to the earth
Answer:
Solar radiation
Explanation:
Visible light, ultraviolet light, infrared, radio waves, X-rays, and gamma rays.
==> Energy
==> Radio noise, heat, visible light, ultraviolet radiation, X-rays, gamma rays
==> They carry all these kinds of energy wherever they go. Not only to the Earth.
Find the final velocity if the initial velocity of 8 m/s with an acceleration of 7 m/s2 over a 3 second interval?
I don't know about it your answer will give another people
Answer: Let the final velocity be v.
Given,
Initial velocity(u)=8m/s
Acceleration(a)=7m/s2
Time(t)=3 sec
Then,
v=u+at
=8+7*3 m/s
=29m/s
Therefore, the final velocity is 29m/s.
What is the law of conservation of energy
A pilot drops a package from a plane flying horizontally at a constant speed. Neglecting air resistance, when the package hits the ground the horizontal location of the plane will Group of answer choices be behind the package. be over the package. be in front of the package depend of the speed of the plane when the package was released.
Answer:
The location of helicopter is behind the packet.
Explanation:
As the packet also have same horizontal velocity as same as the helicopter, and also it has some vertical velocity as it hits the ground.
The horizontal velocity remains same as there is no force in the horizontal direction. The vertical velocity goes on increasing as acceleration due to gravity acts.
So, the helicopter is behind the packet.
help me with this question
Explanation:
Let's set the x-axis to be parallel to the and positive up the plane. Likewise, the y-axis will be positive upwards and perpendicular to the plane. As the problem stated, we are going to assume that m1 will move downwards so its acceleration is negative while m2 moves up so its acceleration is positive. There are two weight components pointing down the plane, [tex]m_1g \sin \theta[/tex] and [tex]m_2g \sin \theta[/tex] and two others pointing up the plane, the two tensions T along the strings. There is a normal force N pointing up from the plane and two pointing down, [tex]m_1g \sin \theta[/tex] and [tex]m_2g \sin \theta[/tex]. Now let's apply Newton's 2nd law to this problem:
x-axis:
[tex]m1:\:\:\:\displaystyle \sum_i F_i = T - m_1g \sin \theta = - m_1a\:\:\:\:(1)[/tex]
[tex]m2:\:\:\:\displaystyle \sum_i F_i = T - m_2g \sin \theta = m_2a\:\:\:\:(2)[/tex]
y-axis:
[tex]\:\:\:\displaystyle \sum_i F_i = N - m_1g \cos \theta - m_2g \cos \theta = 0[/tex]
Use Eqn 1 to solve for T,
[tex]T = m_1(g \sin \theta - a)[/tex]
Substitute this expression for T into Eqn 2,
[tex]m_1g \sin \theta - m_1a - m_2g \sin \theta = m_2a[/tex]
Collecting all similar terms, we get
[tex](m_1 + m_2)a = (m_1 - m_2)g \sin \theta[/tex]
or
[tex]a = \left(\dfrac{m_1 - m_2}{m_1 + m_2} \right)g \sin \theta[/tex]
Determine the tension in the string that connects M2 and M3.
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each of the following conversions contains an error. In each case, explain what the error is and find the correct answer to make a true statement .
a 1000 kg mg (1kg/1000g) = 1g.
b. 50m (1cm/100m)=0.5 cm
c. "Nano" is 10^-9 , so there are 10^-9 nm in a meter.
d. micro is 10^-6, so 1kg is 10^6 ug
Answer:
a. [tex]1000\ kg(\frac{1000\ g}{1\ kg}) = 1\ x\ 10^{6}\ g[/tex]
b. [tex]50\ m (\frac{100\ cm}{1\ m} ) = 5000\ cm[/tex]
c. "Nano" is 10⁻⁹ , so there are 10⁻⁹ meter in a nm (OR) "Nano" is 10⁻⁹ , so there are 10⁹ nm in a meter.
d. micro is 10⁻⁶, so 1kg is 10⁹ ug
Explanation:
a.
The conversion factor is written inverted. The correct statement will be:
[tex]1000\ kg(\frac{1000\ g}{1\ kg}) = 1\ x\ 10^{6}\ g[/tex]
b.
The values in the conversion factor used are wrong. The correct statement will be:
[tex]50\ m (\frac{100\ cm}{1\ m} ) = 5000\ cm[/tex]
c.
Change of units is the mistake here. The correct statement will be:
"Nano" is 10⁻⁹ , so there are 10⁻⁹ meter in a nm (OR) "Nano" is 10⁻⁹ , so there are 10⁹ nm in a meter.
d.
the conversion will be as follows:
[tex]1\ kg(\frac{1000\ g}{1\ kg})(\frac{1\ \mu g}{10^{-6}\ g}) = 10^9 \mu g[/tex]
therefore, the correct statement will be:
micro is 10⁻⁶, so 1kg is 10⁹ ug
Why does the moon appear dark from space?
But why does it appear bright when observed from earth, especially when it is full moon?
Answer:
The moon is actually quite dim.
Explanation:
compared to other astronomical bodies. The moon only seems bright in the night sky because it is so close to the earth and because the trees, houses, and fields around you are so dark at night. In fact, the moon is one of the least reflective objects in the solar system.
Answer:
It reflects the light send from the sun.
Explanation:
If the moon is between you and the sun, you will see the back of it which doesnt reflect light.
what is the velocity of a 1.3 kg puppy with a forward momentum of 6 kg m/s
Answer:
by using p = mv equation we can find v,
6 = 1.3 v
4.615 = v
A car accelerates uniformly from rest to a speed of 55.0 mi/h in 13.0 s. (a) Find the distance the car travels during this time. m (b) Find the constant acceleration of the car. m/s2
Answer:
(a) 159.84 m
(b) 1.89 m/s²
Explanation:
Applying,
(a)
s = (v+u)t/2.................. Equation 1
Where s = distance traveled by the car, u = initial velocity, v = final velocity, t = time.
From the question,
Given: u = 0 m/s ( from rest), v = 55 mi/h = (55/2.237) m/s = 24.59 m/s, t = 13 s
Substitute these values into equation 1
s = (24.59+0)13/2
s = 159.84 m
(b)
Also applying,
a = (v+u)/t................. Equation 2
Where a = acceleration of the car.
substituting into equation 2,
a = (24.59+0)/13
a = 1.89 m/s²
