Answer:
3985 s or 66.42 mins
Explanation:
Given:-
- The length of the rod, L = 83.0 cm
- The cross sectional diameter of rod , d = 2.4 cm
- The temperature of reservoir, Tr = 540°C
- The amount of ice in chamber, m = 1.43 kg
- The temperature of ice, Ti = 0°C
- Thermal conductivity of silver, k = 406 W / m.K
- The latent heat of fusion of water, Lf = 3.33 * 10^5 J / kg
Find:-
How much time is required for the ice to melt completely
Solution:-
- We will first determine the amount of heat ( Q ) required to melt 1.43 kg of ice.
- The heat required would be used as latent heat for which we require the latent heat of fusion of ice ( Lf ). We will employ the first law of thermodynamics assuming no heat is lost from the chamber ( perfectly insulated ):
[tex]Q = m*L_f\\\\Q = ( 1.43 ) * ( 3.33 * 10 ^5 )\\\\Q = 476190 J[/tex]
- The heat is supplied from the hot reservoir at the temperature of 540°C by conduction through the silver rod.
- We will assume that the heat transfer through the silver rod is one dimensional i.e along the length ( L ) of the rod.
- We will employ the ( heat equation ) to determine the rate of heat transfer through the rod as follows:
[tex]\frac{dQ}{dt} = \frac{k.A.dT}{dx}[/tex]
Where,
A: the cross sectional area of the rod
dT: The temperature difference at the two ends of the rod
dx: The differential element along the length of rod ( 1 - D )
t: Time ( s )
- The integrated form of the heat equation is expressed as:
[tex]Q = \frac{k*A*( T_r - T_i)}{L}*t[/tex]
- Plug in the respective parameters in the equation above and solve for time ( t ):
[tex]476190 = \frac{406*\pi*0.024^2 * ( 540 - 0 ) }{0.83*4}*t \\\\t = \frac{476190}{119.49619} \\\\t = 3985 s = 66.42 mins[/tex]
Answer: It would take 66.42 minutes to completely melt the ice
Find another example of separation that is used to extract a material made useful by humans. Describe the process of separation and what we use the separated component for. (4-6 sentences)
If anyone would answer this I’ll answer ur questions for return!
Please and thank you!
Answer:
Salt
Explanation:
Salt plays a crucial role in maintaining human health. It is the main source of sodium and chloride ions in the human diet. Sodium is essential for the nerve and muscle function and is involved in the regulation of fluids in the body. Sodium also plays a role in the body's control of blood pressure and volume. Salt is harvested by seawater or brine is fed into large ponds of water and is drawn out through natural evaporation which allows the salt to be subsequently harvested.
Have a good day and stay safe!
Which of these boxes will not accelerate!
30 Newtons
40 Newtons
50 kg
15 Newton
B.
10 kg
30 Newtons
C.
30 Newtons
80 kg
20 Newtons
20 Newtons
20 Newtons
D.
75 kg
Answer:
(possibly) Box D
Explanation:
The one that has balanced forces will not accelerate. The forces are unbalanced in figures A, B, C. We cannot tell about figure D, because the downward force is not shown. If that force is 20 N, box D will not accelerate.
1. In 214 BC, Archimedes invented a large spherical-type mirror used to focus the sun's intense rays onto far away enemy boats, which would eventually light them on fire. If the boats were travelling in a nearby channel approximately 1,000 m from the river bank, what would the radius of curvature of his mirror need to be? Show your work.
Answer:
2000 m
Explanation:
since the boat is 1000 m from the river bank, the beam must be focused at this point. This indicates that the focal length is 1000 m
for a spherical mirror, the focal length is given by
f = R/2
where R is the radius of curvature
1000 = R/2
R = 2000 m
R = 2000 m
this means that the radius of curvature must be 2000 m
Suppose that 7.4 moles of a monatomic ideal gas (atomic mass = 1.39 × 10-26 kg) are heated from 300 K to 500 K at a constant volume of 0.74 m3. It may help you to recall that CV = 12.47 J/K/mole and CP = 20.79 J/K/mole for a monatomic ideal gas, and that the number of gas molecules is equal to Avagadros number (6.022 × 1023) times the number of moles of the gas.
1) How much energy is transferred by heating during this process?2) How much work is done by the gas during this process?3) What is the pressure of the gas once the final temperature has been reached?4) What is the average speed of a gas molecule after the final temperature has been reached?5) The same gas is now returned to its original temperature using a process that maintains a constant pressure. How much energy is transferred by heating during the constant-pressure process?6) How much work was done on or by the gas during the constant-pressure process?
Answer:
Explanation:
1 ) Since it is a isochoric process , heat energy passed into gas
= n Cv dT , n is no of moles of gas , Cv is specific heat at constant volume and dT is rise in temperature .
= 7.4 x 12.47 x ( 500 - 300 )
= 18455.6 J.
2 ) Since there is no change in volume , work done by the gas is constant.
3 ) from , gas law equation
PV = nRT
P = nRT / V
= 7.4 x 8.3 x 500 / .74
= .415 x 10⁵ Pa.
4 ) Average kinetic energy of gas molecules after attainment of final temperature
= 3/2 x R/ N x T
= 1.5 x 1.38 x 10⁻²³ x 500
= 1.035 x 10⁻²⁰ J
1/2 m v² = 1.035 x 10⁻²⁰
v² = 2 x 1.035 x 10⁻²⁰ / 1.39 x 10⁻²⁶
= 1.49 x 10⁶
v = 1.22 x 10³ m /s
5 ) In this process , pressure remains constant
gas is cooled from 500 to 300 K
heat will be withdrawn .
heat withdrawn
= n Cp dT
= 7.4 x 20.79 x 200
= 30769.2 J .
6 )
gas will have reduced volume due to cooling
reduced volume = .74 x 300 / 500
= .444 m³
change in volume
= .74 - .444
= .296 m³
work done on the gas
= P x dV
pressure x change in volume
= .415 x 10⁵ x .296
= 12284 J.
Tech A says that as engines gain miles, the spark plug gap increases, which raises the ignition system’s available voltage. Tech B says that misfire occurs when required voltage is higher than available voltage. Who is correct? Group of answer choices
Answer: Tech A is correct
Explanation:
Every vehicle has ignition system and without this system,it will not work. The battery of everything vehicle contain energy that start the vehicle and ignore it to start working. Electrical current move from the vehicle's battery and get to the induction coil, the induction coil increases the voltage in it so that the plug will be ignited. The spark plugs produce fire. The spark plug is connected to the ignition system. Once voltage is produced from the induction coil, electrical impulses move from induction coil to insulated plug wires. The spark plug need a very high voltage from the small voltage battery. Once the high voltage exceed the dielectric strength of the gases, spark jump the gap between the plug's fire end.
A small rock with mass 0.12 kg is fastened to a massless string with length 0.80 m to form a pendulum. The pendulum is swinging so as to make a maximum angle of 45 ∘ with the vertical. Air resistance is negligible. Part A What is the speed of the rock when the string passes through the vertical position
Answer:
v = 3.33 m/s
Explanation:
In the position of 45 degrees, all the energy of the rock is gravitational, then we have:
E = m*g*L*cos(angle)
and in the vertical position of the string, all the energy is kinetic, so we have:
E = m*v^2/2
If there is no dissipation, both energies are equal, so we have:
m*g*L*cos(45) = m*v^2/2
9.81 * 0.8 * 0.7071 * 2 = v^2
v^2 = 11.0986
v = 3.33 m/s
At an instant when a soccer ball is in contact with the foot of the player kicking it, the horizontal or x component of the ball's acceleration is 950 m/s2 and the vertical or y component of its acceleration is 750 m/s2. The ball's mass is 0.35 kg. What is the magnitude of the net force acting on the soccer ball at this instant?
Answer:
F = 423.63 N
Explanation:
Since, the x-component and y-components of the acceleration of ball are given. Therefore, we need to find the resultant or net acceleration of the soccer ball first. For that purpose we use to the formula for the resultant of rectangular components of a vector:
a = √(ax² + ay²)
where,
a = net acceleration = ?
ax = x - component of acceleration = 950 m/s²
ay = y - component of acceleration = 750 m/s²
Therefore,
a = √[(950 m/s²)² + (750 m/s²)²]
a = 1210.4 m/s²
Now, from Newton's Second Law, we know that:
F = ma
where,
m = mass of ball = 0.35 kg
F = Net force acting on ball = ?
F = (0.35 kg)(1210.4 m/s²)
F = 423.63 N
A student has derived the following nondimensionally homogeneous equation: a=x/t2-vt+F/m where v is a velocity's magnitude , a is an acceleration's magnitude, t is a time, m is a mass, F is a force's magnitude , and x is a distance (or length). Which terms are dimensionally homogeneous? .
a) x/t
b) vt
c) a
d) F/m
Answer:
Letter C) and D) is the correct answer.
Explanation:
We know that the a is an acceleration's magnitude, so the units of a are m/s².
Now, let's analyze each terms. If we want that each term will be dimensionally homogeneous, all of them must have the same units of a.
[tex][\frac{x}{t}]=[\frac{m}{s}][/tex]
[tex][vt]=[m][/tex]
[tex][\frac{F}{m}]=[\frac{N}{kg}]=[kg\frac{m}{s^{2}kg}]=[\frac{m}{s^{2}}][/tex]
Therefore, the term F/m is the correct answer.
I hope it helps you!
We can see that a and F/M are dimensionally homogeneous.
In solving dimensions, we try to express a quantity in terms of the fundamental quantities;
MassLengthTimeFor the term a, its dimension is LT^-2
For the term F/m, its dimension is LT^-2
Hence, it follows that a and F/M are dimensionally homogeneous.
Learn more about dimensions: https://brainly.com/question/944206
an aluminium bar 600mm long, with diameter 40mm, has a hole drilled in the center of the bar. the hole is 30mm in diameter and is 30mm and is 100mm long. if modulus of elasticity for the aluminium is 85GN/m2, calculate the total contraction on the bar due to a compressive load of 180KN
Answer:
ΔL = 1.011 mm
Explanation:
Let's begin by listing out the given information:
Length (L) = 600 mm = 0.6 m,
Diameter (D) = 40 mm = 0.04 m ⇒ Radius (r) = 20 mm = 0.2 m,
Area (cross sectional) = πr² = 3.14 x .02² = 0.001256 m²,
Modulus of Elasticity (E) = 85 GN/m²,
Compressive load (F) = 180 KN
Using the formula, Stress = Load ÷ Area
Mathematically,
σ = F ÷ A = 180 x 10³ ÷ 0.001256
σ = 143312.1 KN/m²
Modulus of elasticity = stress ÷ strain
E = σ ÷ ε
ε = ΔL/L
85 x 10⁹ = 143312.1 x 10³ ÷ (ΔL/L)
ΔL = 143312.1 x 10³ ÷ 85 X 10⁹ = 1686.02 * 10⁻⁶
ΔL = L x 1686.02 * 10⁻⁶
ΔL = 0.6 * 1686.02 * 10⁻⁶ = 1011.61 x 10⁻⁶
ΔL = 1.011 x 10⁻³ m
ΔL = 1.011 mm
∴The bar contracts by 1.011 mm
A boy and a girl are on a spinning merry-go-round. The boy is at a radial distance of 1.2 m from the central axis; the girl is at a radial distance of 1.8 m from the central axis. Which is true?A- Boy and girl have zero tangential and angular accelerations.B- The girl has a larger angular acceleration than the boy.C- The boy has a larger tangential acceleration than the girl.D- The boy has a larger angular acceleration than the girl.E- The girl has a larger tangential acceleration than the boy.
Answer:
E) True. The girl has a larger tangential acceleration than the boy.
Explanation:
In this exercise they do not ask us to say which statement is correct, for this we propose the solution to the problem.
Angular and linear quantities are related
v = w r
a = α r
the boy's radius is r₁ = 1.2m the girl's radius is r₂ = 1.8m
as the merry-go-round rotates at a constant angular velocity this is the same for both, but the tangential velocity is different
v₁ = w 1,2 (boy)
v₂ = w 1.8 (girl)
whereby
v₂> v₁
reviewing the claims we have
a₁ = α 1,2
a₂ = α 1.8
a₂> a₁
A) False. Tangential velocity is different from zero
B) False angular acceleration is the same for both
C) False. It is the opposite, according to the previous analysis
D) False. Angular acceleration is equal
E) True. You agree with the analysis above,
4. A neutrally charged conductor has a negatively charged rod brought close to it, and thus has an induced positive charge on the surface closest to the rod. What can we say about the overall charge on the conductor
Answer:
Overall charge still remains zero on conductor until touched by charged rod.
Explanation:
Here, we want to know what has happened to the overall charge on the conductor.
Since the conductor is neutral, the overall charge on the conductor must remain zero because positive charge is induced on close end to rod then equal and negaitve charge is induced on far end to rod.
Thus, overall charge still remains zero on conductor until touched by charged rod.
A proton moving along the x axis has an initial velocity of 4.0 × 106 m/s and a constant acceleration of 6.0 × 1012 m/s2. What is the velocity of the proton after it has traveled a distance of 80 cm? Group of answer choices
Answer:
5.06*10^6 m/s
Explanation:
Given that
Initial velocity, u = 4*10^6 m/s
Acceleration, a = 6*10^12 m/s²
Distance traveled, s = 80 cm
Final velocity, v = ?
We can find the final velocity by using one of the equations of motion.
v² = u² + 2as
On substituting the values, we have
v² = (4*10^6)² + 2 * 6*10^12 * 0.8
v² = 2.56*10^13
v = √2.56*10^13
v = 5.06*10^6 m/s
Therefore, the final velocity of the proton is adjudged to be 5.06*10^6 m/s
The final velocity of the proton over the given distance is [tex]5.06 \times 10^6 \ m/s[/tex].
The given parameters;
initial velocity of the proton, u = 4 x 10⁶ m/sacceleration of the proton, a = 6 x 10¹² m/s²distance traveled by the proton, s = 80 cm = 0.8 mThe final velocity of the proton over the given distance is calculated as follows;
[tex]v^2 = u^2 + 2as\\\\v^2 = (4\times 10^6)^2 \ + \ 2(6.0 \times 10^{12})(0.8)\\\\v^2 = 2.56 \times 10^{13} \\\\v = \sqrt{2.56 \times 10^{13} } \\\\v = 5.06 \times 10^6 \ m/s[/tex]
Thus, the final velocity of the proton over the given distance is [tex]5.06 \times 10^6 \ m/s[/tex]
Learn more here:https://brainly.com/question/13613973
Representar con una escala de 1cm = 10N dos fuerzas que tengan igual dirección, distinto sentido y sus intensidades son de 40n y 60n, respectivamente.
Alguien que me lo hagaaaaaaa
Answer:
To solve this problem we just need to graph two forces with same direction, pointing to different sides with intensities of 40 N and 60 N.
The image attached shows these forces.
Notice that the vectors are parallel, that's because they have the same direction, but they point to different sides, and their magnitudes have a difference of 20 N.
Assuming 100% efficient energy conversion, how much water stored behind a 50
centimeter high hydroelectric dam would be required to charge the battery?
Answer:
Explanation:
The power rating of the battery isn't provided. But let us assume that it is one of the common batteries with ratings of 12 V and 50 A.h
Potential energy possessed by water at that height = mgh
m = mass of the water = ρV
ρ = density of water = 1000 kg/m³
V = volume of water = ?
g = acceleration due to gravity = 9.8 m/s²
h = height of water = 50 cm = 0.5 m
Potential energy = ρVgh = 1000 × V × 9.8 × 0.5 = (4900V) J
Energy of the battery = qV
q = 50 A.h = 50 × 3600 = 180,000 C
V = 12 V
qV = 180,000 × 12 = 2,160,000 J
Energy = 2,160,000 J
At a 100% conversion rate, the energy of the water totally powers the battery
(4900V) = (2,160,000)
4900V = 2,160,000
V = (2,160,000/4900)
V = 440.82 m³
Hence, with our assumed power ratings for the battery (12 V and 50 A.h), 440.82 m³ of water at the given height of 50 cm would power the battery.
Incase the power ratings of the battery in the complete question is different, this solution provides you with how to obtain the correct answer, given any battery power rating.
Hope this Helps!!!
A potential difference of 71 mV is developed across the ends of a 12.0-cm-long wire as it moves through a 0.27 T uniform magnetic field at a speed of 6.0 m/s. The magnetic field is perpendicular to the axis of the wire.
Required:
What is the angle between the magnetic field and the wire's velocity?
Answer:
Explanation: please see attached file I attached the answer to your question.
The angle between the magnetic field and the wire's velocity is 33.2 degrees.
Calculation of the angle:Since the potential difference = 71mv = 71 *10 ^-3 V
The length is 12 cm = 0.12m
The magnetic field i.e. B = 0.27T
The speed or v = 4 m/s
here we assume [tex]\theta[/tex] be the angle
So,
e = Bvl sin[tex]\theta[/tex]
So,
[tex]Sin\theta[/tex] = e/bvl
= 71*10^-3 / 0.27 *4*0.12
= 0.5478
= 33.2 degrees
Therefore, the angle should be 33.2 degrees
Learn more about an angle here: https://brainly.com/question/14661707
A student throws a set of keys vertically upward to his fraternity brother, who is in a window 3.60 m above. The brother's outstretched hand catches the keys 1.60 s later. (Take upward as the positive direction. Indicate the direction with the sign of your answer.)With what initial velocity were the keys thrown?
Answer:
[tex]v_{i}=10.10 m/s[/tex]
Explanation:
The equation of the position is:
[tex]y=y_{i}+v_{i}t-0.5gt^{2}[/tex]
Where:
v(i) is the initial velocity
The initial position y(i) will be zero and the final position y = 3.60 m.
So, we just need to solve this equation for v(i).
[tex]v_{i}=\frac{y+0.5gt^{2}}{t}[/tex]
[tex]v_{i}=\frac{3.6+0.5*9.81*1.6^{2}}{1.6}[/tex]
[tex]v_{i}=10.10 m/s[/tex]
Therefore, the initial velocity is 10.10 m/s upwards.
I hope it helps you!
the heat capacity of 0.125Kg of water is measured to be 523j/k at a room temperature.Hence, calculate the heat capacity of water
(a) per unit mass
(b) per unit volume
Answer:
A. 4148 J/K/Kg
B. 4148 J/K/L
Explanation:
A. Heat capacity per unit mass is known as the specific heat capacity, c.
C = Heat capacity/mass(kg)
C = (523 J/K) / 0.125 Kg = 4148 J/K/Kg
B. Volume of water = mass/density
Density of water = 1 Kg/L
Volume of water = 0.125 Kg/ 1Kg/L
Volume of water = 0.125 L
Heat capacity per unit volume = (523 J/K) / 0.125 L
Heat capacity per unit volume = 4148 J/K/L
A ball rolls over the edge of a platform with only a horizontal velocity. The height of the platform is 1.60m and the horizontal range of the ball from the base of the platform is 20.0m. What is the velocity of the ball just before it touches the ground
Answer:
v = 46.99 m/s
Explanation:
The velocity of the ball just before it touches the ground, is given by the following formula:
[tex]v=\sqrt{v_x^2+v_y^2}[/tex] (1)
vx: horizontal component of the velocity
vy: vertical component of the velocity
The vertical component vy is calculated by using the following formula:
[tex]v_y^2=v_{oy}^2+2gh[/tex] (2)
vy: final velocity
voy: initial vertilal velocity = 0m/s (because it is a semi parabolic motion)
g: gravitational acceleration = 9.8 m/s^2
h: height = 1.60m
You replace the values of the parameters in the equation (2):
[tex]v_y=2(9.8m/s^2)(1.60m)=31.36\frac{m}{s}[/tex]
vx is calculated by using the information about the horizontal range of the ball:
[tex]R=v_o\sqrt{\frac{2h}{g}}[/tex] (3)
R: horizontal range of the ball = 20.0 m
You solve the previous equation for vo, the initial horizontal velocity:
[tex]v_o=R\sqrt{\frac{g}{2h}}=(20.0m)\sqrt{\frac{9.8m/s^2}{2(1.60m)}}\\\\v_o=35\frac{m}{s}[/tex]
The horizontal component of the velocity is constant in the complete trajectory, hence, you have that
vx = vo = 35 m/s
Finally, you replace the values of vx and vy in the equation (1):
[tex]v=\sqrt{(35m/s)^2+(31.36m/s)^2}=46.99\frac{m}{s}[/tex]
The velocity of the ball just before it touches the ground is 46.99 m/s
Olaf is standing on a sheet of ice that covers the football stadium parking lot in Buffalo, New York; there is negligible friction between his feet and the ice. A friend throws Olaf a ball of mass 0.400 kg that is traveling horizontally at 11.3 m/s. Olaf's mass is 75.0 kg. (a) If Olaf catches the ball, with what speed v_f do Olaf and the ball move afterward
Answer:
v = 0.059 m/s
Explanation:
To find the final speed of Olaf and the ball you use the conservation momentum law. The momentum of Olaf and the ball before catches the ball is the same of the momentum of Olaf and the ball after. Then, you have:
[tex]mv_{1i}+Mv_{2i}=(m+M)v[/tex] (1)
m: mass of the ball = 0.400kg
M: mass of Olaf = 75.0 kg
v1i: initial velocity of the ball = 11.3m/s
v2i: initial velocity of Olaf = 0m/s
v: final velocity of Olaf and the ball
You solve the equation (1) for v and replace the values of all variables:
[tex]v=\frac{mv_{1i}}{m+M}=\frac{(0.400kg)(11.3m/s)}{0.400kg+75.0kg}=0.059\frac{m}{s}[/tex]
Hence, after Olaf catches the ball, the velocity of Olaf and the ball is 0.059m/s
A subatomic particle X spontaneously decays into two particles, A and B, each of rest energy 1.40 × 10^2 MeV. The particles fly off in opposite directions, each with speed 0.827c relative to an inertial reference frame S. What is the total energy of particle A?
Answer:
E = 389 MeV
Explanation:
The total energy of particle A, will be equal to the sum of rest mass energy and relative energy of particle A. Therefore,
Total Energy of A = E = Rest Mass Energy + Relative Energy
Using Einstein's Equation: E = mc²
E = m₀c² + mc²
From Einstein's Special Theory of Relativity, we know that:
m = m₀/[√(1-v²/c²)]
Therefore,
E = m₀c² + m₀c²/[√(1-v²/c²)]
E = m₀c²[1 + 1/√(1-v²/c²)]
where,
m₀c² = rest mass energy = 140 MeV
v = relative speed = 0.827 c
Therefore,
E = (140 MeV)[1 + 1/√(1 - (0.827c)²/c²)]
E = (140 MeV)(2.78)
E = 389 MeV
Some runners train with parachutes that trail behind them to provide a large drag force. These parachutes are designed to have a large drag coefficient. One model expands to a square 1.8 m on a side, with a drag coefficient of 1.4. A runner completes a 200 m run at 5.0 m/s with this chute trailing behind. Part A How much thermal energy is added to the air by the drag force
Answer:
13.9 kJ
Explanation:
Given that
Length of the side, l = 1.8 m
Drag coefficient, C(d) = 1.4
Distance of run, d = 200 m
Velocity of run, v = 5 m/s
Density, ρ = 1.23
Using the Aerodynamics Drag Force formula. We have
F(d) = 1/2.ρ.A.C(d).v²
The Area, A needed is 1.8 * 1.8 = 3.24 m². So that,
F(d) = 1/2 * 1.23 * 3.24 * 1.4 * 5²
F(d) = 139.482/2
F(d) = 69.74
recall that, energy =
W = F * d
W = 69.74 * 200
W= 13948
W = 13.9kJ
Therefore, the thermal energy added to the air by the drag force is 13.9kJ
A 1200 kg car reaches the top of a 100 m high hill at A with a speed vA. What is the value of vA that will allow the car to coast in neutral so as to just reach the top of the 150 m high hill at B with vB
Complete question is:
A 1200 kg car reaches the top of a 100 m high hill at A with a speed vA. What is the value of vA that will allow the car to coast in neutral so as to just reach the top of the 150 m high hill at B with vB = 0 m/s. Neglect friction.
Answer:
(V_A) = 31.32 m/s
Explanation:
We are given;
car's mass, m = 1200 kg
h_A = 100 m
h_B = 150 m
v_B = 0 m/s
From law of conservation of energy,
the distance from point A to B is;
h = 150m - 100 m = 50 m
From Newton's equations of motion;
v² = u² + 2gh
Thus;
(V_B)² = (V_A)² + (-2gh)
(negative next to g because it's going against gravity)
Thus;
(V_B)² = (V_A)² - (2gh)
Plugging in the relevant values;
0² = (V_A)² - 2(9.81 × 50)
(V_A) = √981
(V_A) = 31.32 m/s
A person stands on a scale in an elevator. As the elevator starts, the scale has a constant reading of 592 N. As the elevator later stops, the scale reading is 400 N. Assume the magnitude of the acceleration is the same during starting and stopping. (a) Determine the weight of the person. N (b) Determine the person's mass. kg (c) Determine the magnitude of acceleration of the elevator. m/s2
Answer:
a) 496Nb) 50.56kgc) 1.90m/s²Explanation:
According to newton's secomd law, ∑F = ma
∑F is the summation of the force acting on the body
m is the mass of the body
a is the acceleration
Given the normal force when the elevator starts N1 = 592N
Normal force after the elevator stopped N2 = 400N
When the elevator starts, its moves upward, the sum of force ∑F = Normal (N)force on the elevator - weight of the person( Fg)
When moving up;
N1 - Fg = ma
N1 = ma + Fg ...(1)
Stopping motion of the elevator occurs after the elevator has accelerates down. The sum of forces in this case will give;
N2 - Fg = -ma
N2 = -ma+Fg ...(2)
Adding equation 1 and 2 we will have;
N1+N2 = 2Fg
592N + 400N = 2Fg
992N 2Fg
Fg = 992/2
Fg = 496N
The weight of the person is 496N
\b) To get the person mass, we will use the relationship Fg = mg
g = 9.81m/s
496 = 9.81m
mass m = 496/9.81
mass = 50.56kg
c) To get the magnitude of acceleration of the elevator, we will subtract equation 1 from 2 to have;
N1-N2 = 2ma
592-400 = 2(50.56)a
192 = 101.12a
a = 192/101.12
a = 1.90m/s²
Our Sun shines bright with a luminosity of 3.828 x 1025 Watt. Her energies
responsible for many processes and the habitable temperatures on the earth that
make our life possible.
a) Calculate the amount of energy arriving on the Earth in a single day
b) To how many litres of heating oil (energy density 37.3 x 10^6 J/litre is the equivalent?
C) The Earth reflects 30% of this energy : Determine the temperature on Earth's sufact
d) what other factors should be considered to get an even more precisa temperature postiache
Note: The Earth's radius is 6370km; the Sun's sadius is 696 ×10^3km, I AU is 1.495 × 10^8km)
Answer:
a) E = 1.58 10²¹ J , b) Oil = 4,236 107 liter , e) T = 54.3 C
Explanation:
a) To calculate the energy that reaches Earth, let us combine that the power emitted by the Sun is distributed uniformly on a spherical surface
I = P / A
A = 4π r²
in this case the radius of the sphere is the distance from the Sun to Earth r = 1.5 10¹¹ m
I = P / A
I = P / 4π r²
let's calculate
I = 3,828 10²⁵/4 pi (1.5 10¹¹)²
I = 1.3539 10²W / m² = 135.4 W / m2
the energy that reaches the disk of the Earth is
E = I A
the area of a disc
A = π r²
E = I π r²
where r is the radius of the Earth 6.37 10⁶ m
E = 135.4 π(6.37 10⁶)
E = 1,726 10¹⁶ W
This is the energy per unit of time that reaches Earth
t = 1 dai (24h / 1day) (3600s / 1h) = 86400 s
E = 1,826 10¹⁶ 86400
E = 1.58 10²¹ J
b) for this part we can use a direct proportions rule
Oil = 1.58 10²¹ (1 / 37.3 10⁶)
Oil = 4,236 10⁷ liter
c) to silence the surface temperature of the Earth we use the Stefan-Bolztman Law
P = σ A e T⁴
T = [tex]\sqrt[4]{P/Ae}[/tex]
nos indicate the refect, therefore the amount of absorbencies
P_absorbed = 0.7 P
let's calculate
T = REA (0.7 1.58 1021 / [pi (6.37 106) 2 1)
T = RER (8,676 106)
T = 54.3 C
b) Among the other factors that must be taken into account is the greenhouse effect, due to the absorption of gases from the atmosphere
help yall 13 points!!
Answer:
Explanation:
12.)
A. Opposite poles attract
B. Same poles repel
13.)
IDK
If an instalment plan quotes a monthly interest rate of 4%, the effective annual/yearly interest rate would be _____________. 4% Between 4% and 48% 48% More than 48%
Answer:
More than 48%
Explanation:
If the interest is computed monthly on the outstanding balance, it has an effective annual rate of ...
(1 +4%)^12 -1 = 60.1% . . . . more than 48%
The effective annual or yearly interest rate would be=30.56% which is Between 4% and 48%
Calculation of Annual Interest rateThe formula used to calculate annual Interest rate =
[tex](1+ \frac{i}{n} ) {}^{n} - 1[/tex]
where i= nominal interest rate = 4%
n= number of periods= 12 months
Annual Interest rate=
[tex](1 + \frac{4\%}{12} ) {}^{12} - 1[/tex]
= (1+0.333)^12 -1
= (1.333)^12-1
= 31.56 - 1
= 30.56%
Therefore, the effective annual or yearly interest rate would be= 30.56%
Learn more about interest rate here:
https://brainly.com/question/25793394
The mass of a particular eagle is twice that of a hunted pigeon. Suppose the pigeon is flying north at ,2=17.1 m/s when the eagle swoops down, grabs the pigeon, and flies off. At the instant right before the attack, the eagle is flying toward the pigeon at an angle =52.7 ° below the horizontal and a speed of ,1=41.5 m/s.
Answer:
31.4 m/s
44.4°
Explanation:
Momentum is conserved in the horizontal direction:
pₓᵢ = pₓ
m vᵢ₂ + 2m vᵢ₁ cos θ = (m + 2m) vₓ
vᵢ₂ + 2 vᵢ₁ cos θ = 3 vₓ
17.1 m/s + 2 (41.5 m/s) (cos -52.7°) = 3 vₓ
vₓ = 22.5 m/s
Momentum is conserved in the vertical direction:
pᵧᵢ = pᵧ
2m vᵢ₁ sin θ = (m + 2m) vᵧ
2 vᵢ₁ sin θ = 3 vᵧ
2 (41.5 m/s) (sin -52.7°) = 3 vᵧ
vᵧ = -22.0 m/s
The speed is:
v = √(vₓ² + vᵧ²)
v = √((22.5 m/s)² + (-22.0 m/s)²)
v = 31.4 m/s
The direction is:
θ = atan(vᵧ / vₓ)
θ = atan(-22.0 m/s / 22.5 m/s)
θ = -44.4°
The speed of the eagle at that instant is 31.4 m/s while it moves off in the direction of 44.4°.
Since momentum is conserved horizontally;
17.1 m/s + 2 (41.5 m/s) (cos -52.7°) = 3 vx
vx = 17.1 m/s + 2 (41.5 m/s) (cos -52.7°)/3
vx = 22.5 m/s
Also, momentum is conserved vertically hence;
2 (41.5 m/s) (sin -52.7°) = 3 vy
vy = 2 (41.5 m/s) (sin -52.7°) /3
vy = -22.0 m/s
The effective speed therefore, is;
v = √((22.5 m/s)² + (-22.0 m/s)²)
v = 31.4 m/s
The direction of this effective speed is;
θ = tan-1(22.0 m/s / 22.5 m/s)
θ = 44.4°
Learn more: https://brainly.com/question/13322477
The electron gun in an old CRT television accelerates electrons between two charged parallel plates (the cathode is negative; the anode is positive) 1.2 cm apart. The potential difference between them is 25 kV. Electrons enter through a small hole in the cathode, are accelerated, and then exit through a small hole in the anode. Assume the plates act as a capacitor.
a. What is the electric field strength and direction between the plates?
b. With what speed does an electron exit the electron gun if its entry speed is close to zero? [Note: ignore relativity]
c. If the capacitance of the plates is 1 nF, how much charge is stored on each plate? How many extra electrons does the cathode have?
d. If you wanted to push an electron from the anode to the cathode, how much work would you have to do?
Answer:
A. 2.083 MV/m from anode to cathode.
B. 93648278.15 m/s
C. 2.5x10^-5 C and there are about 1.56x10^14 electrons
D. 4x10^-15 Joules
Explanation:
Voltage V across plate is 25 kV = 25x10^3 V
Distance apart x = 1.2 cm = 1.2x10^-2 m
A. Electric field strength is the potential difference per unit distance
E = V/x = 25x10^3/1.2x10^-2 = 2083333.3 V/m
= 2.083 MV/m
B. Energy of electron is electron charge times the voltage across
i.e eV
Charge on electron = 1.6x10^-19 C
Energy of electron = 1.6x10^-19 x 25x10^3 = 4x10^-15 Joules
Mass of electron m is 9.12x10^-31 kg
Kinetic energy of electron = 0.5mv^2
Where v is the speed
4x10^-15 = 0.5 x 9.12x10^-31 x v^2
v^2 = 8.77x10^15
v = 93648278.15 m/s
C. From Q = CV
Q = charge
C = capacitance = 1 nF 1x10^-9 F
V = voltage = 25x10^3 V
Q = 1x10^-9 x 25x10^3 = 2.5x10^-5 C
Total number of electrons = Q/e
= 2.5x10^-5/1.6x10^-19 = 1.56x10^14 electrons
D. To push electron from cathode to anode, I'll have to do a work of about
4x10^-15 Joules
0.92 kg of R-134a fills a 0.14-m^3 weighted piston–cylinder device at a temperature of –26.4°C. The container is now heated until the temperature is 100°C. Determine the final volume of R-134a.
Answer:
The final volume of R-134a is 0.212m³Explanation:
Using one of the general gas equation to find the final volume of the R-134a.
According to pressure law; The volume of a given mas of gas is directly proportional to its temperature provided that the pressure remains constant.
VαT
V = kT
k = V/T
V1/T1 = V2/T2 = k
Given V1 = 0.14-m³ at T1 = –26.4°C = –26.4° + 273 = 246.6K
V2 = ? at T = 100°C = 100+273 = 373K
On substituting this values for T2;
0.14/246.6 = V2/373
373*0.14 = 246.6V2
V2 = 373*0.14 /246.6
V2 = 0.212m³
The final volume of R-134a is 0.212m³
An automobile being tested on a straight road is 400 feet from its starting point when the stopwatch reads 8.0 seconds and is 550 feet from the starting point when the stopwatch reads 10.0 seconds.
A. What was the average velocity of the automobile during the interval from t = 10.0 seconds to t = 8.0 seconds
B. What was the average velocity of the automobile during the interval from t - Ostot - 10.0 s? (Assume that the stopwatch read t = 0 and started at the same time as the auto.)
C. If the automobile averages 100 ft/s from t - 10.0 stot - 20.0 s, what distance does it travel during this interval?
D. The automobile has a special speedometer calibrated in feet/s instead of in miles/hour. Att 85 the speedometer reads 65 ft/s; and at t = 10 s it reads 80 ft/s. What is the average acceleration during this interval?
Answer:
a) v = 75 ft / s , b) v = 55 ft / s , c) Δx = 1000 ft
Explanation:
We can solve this exercise with the expressions of kinematics
a) average speed is defined as the distance traveled in a given time interval
v = (x₂-x₁) / (t₂-t₁)
v = (550 - 400) / (10 -8)
v = 75 ft / s
b) we repeat the calculations for this interval
v = (550 - 0) / (10 -0)
v = 55 ft / s
c) we clear the distance from the average velocity equation
Δx = v (t₂ -t₁)
Δx = 100 (20-10)
Δx = 1000 ft