Q: How much heat must be absorbed by 125 g of ethanol to change its temperature from 21.5 oC to 34.8 oC? The specific heat of ethanol is 2.44 J/(gC).
Answer:
4056.5 J
Explanation:
The formula for the specific heat capacity of ethanol is given as
Q = cm(t₂-t₁)..................... Equation 1
Where q = quantity of heat, c = specific heat capacity of ethanol, m = mass of ethanol, t₁ = initial temperature of ethanol, t₂ = final temperature of ethanol.
Given: m = 125 g, t₁ = 25.5 °C, t₂ = 34.8 °C
Constant; c = 2.44 J/g.°C
Substitute into equation 1
Q = 125(2.44)(34.8-21.5)
Q = 125(2.44)(13.3)
Q = 4056.5 J.
Hence the amount of heat absorbed = 4056.5 J
The diffuser in a jet engine is designed to decrease the kinetic energy of the air entering the engine compressor without any work or heat interactions. Calculate the velocity at the exit of a diffuser when air at 100 kPa and 30°C enters it with a velocity of 350 m/s and the exit state is 200 kPa and 90°C.
Explanation:
Expression for energy balance is as follows.
[tex]\Delta E_{system} = E_{in} - E_{out}[/tex]
or, [tex]E_{in} = E_{out}[/tex]
Therefore,
[tex]m(h_{1} \frac{v^{2}_{1}}{2}) = m (h_{2} \frac{V^{2}_{2}}{2})[/tex]
[tex]h_{1} + \frac{V^{2}_{1}}{2} = h_{2} + \frac{V^{2}_{2}}{2}[/tex]
Hence, expression for exit velocity will be as follows.
[tex]V_{2} = [V^{2}_{1} + 2(h_{1} - h_{2})^{0.5}[/tex]
= [tex]V^{2}_{1} + 2C_{p}(T_{1} - T_{2})]^{0.5}[/tex]
As [tex]C_{p}[/tex] for the given conditions is 1.007 kJ/kg K. Now, putting the given values into the above formula as follows.
[tex]V_{2} = V^{2}_{1} + 2C_{p}(T_{1} - T_{2})]^{0.5}[/tex]
= [tex][(350 m/s)^{2} + 2(1.007 kJ/kg K) (30 - 90) K \frac{1000 m^{2}/s^{2}}{1 kJ/kg}]^{0.5}[/tex]
= 40.7 m/s
Thus, we can conclude that velocity at the exit of a diffuser under given conditions is 40.7 m/s.
A charge of 12.6 µC is at the geometric center of a cube. What is the electric flux through one of the faces? The permittivity of a vacuum is 8.85419 × 10−12 C 2 /N · m2 . Answer in units of N · m2 /C.
The electric flux through one of the faces of the cube with the charge at its geometric center can be found using Gauss's Law. The electric flux enclosed by the surface divided by the permittivity of the medium is 1.424 N·m2/C.
Explanation:We can use Gauss's Law to determine the electric flux through one of the faces of the cube with the charge at its geometric center. Gauss's Law states that the electric flux through a closed surface is equal to the charge enclosed by that surface divided by the permittivity of the medium.
The charge enclosed by one face of the cube is 12.6 µC. The permittivity of a vacuum is 8.85419 × 10-12 C2/N·m2.
Therefore, the electric flux through one face of the cube is (12.6 µC) / (8.85419 × 10-12 C2/N·m2) = 1.424 N·m2/C.
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Four point charges each having charge Q are located at the corners of a square having sides of length a. (a) Find an expression for the total electric potential at the center of the square due to the four charges. (Use any variable or symbol stated above along with the following as necessary: ke.)
Answer:
[tex]\displaystyle V_t=36\sqrt{2}\times 10^9 \frac{Q}{a}[/tex]
Explanation:
Electric Potential of Point Charges
The electric potential from a point charge Q at a distance r from the charge is
[tex]\displaystyle V=k\frac{Q}{r}[/tex]
Where k is the Coulomb's constant. The total electric potential for a system of point charges is equal to the scalar sum of their individual potentials. The potential is not a vector, so there is no direction or vectors to deal with.
We are required to compute the total electric potential in the center of the square. We need to know the distance from each corner to the center. The diagonal of the square is
[tex]d=\sqrt2 a[/tex]
where a is the length of the side.
The distance from any corner to the center is half that diagonal, thus
[tex]\displaystyle r=\frac{d}{2}=\frac{a}{\sqrt{2}}[/tex]
The total potential in the center is
[tex]V_t=V_1+V_2+V_3+V_4[/tex]
Please note all the potentials must be calculated including the sign of the charges. Since all the charges are equal to Q, and the distances are the same, the total potential is 4 times the individual potential of each charge.
[tex]V_t=4\times V[/tex]
[tex]\displaystyle V=9\times 10^9 \frac{Q}{\frac{a}{\sqrt{2}}}[/tex]
Operating
[tex]\displaystyle V=9\sqrt{2}\times 10^9 \frac{Q}{a}[/tex]
Thus:
[tex]\displaystyle V_t=4\times 9\sqrt{2}\times 10^9 \frac{Q}{a}[/tex]
[tex]\boxed{\displaystyle V_t=36\sqrt{2}\times 10^9 \frac{Q}{a}}[/tex]
A dog running in an open field has components of velocity vx = 2.6 m/s and vy = -1.8 m/s at t1 = 10.0 s. For the time interval from t1 = 10.0 s to t2 = 20.0 s, the average acceleration of the dog has magnitude 0.45 m/s2 and direction 31.0o measured from the + x-axis toward the + y-axis. At t2 = 20.0 s, (a) what are the x- and y-components of the dog’s velocity? (b) What are the magnitude and direction of the dog’s velocity? (c) Sketch the velocity vectors at t1 and t2. How do these two vectors differ?
Answer:
a) vₓ = 6,457 m / s , v_{y} = 0.518 m / s , b) v = 6.478 m / s, θ = 4.9°
Explanation:
a) This is a kinematic problem, let's use trigonometry to find the components of acceleration
sin 31 = [tex]a_{y}[/tex] / a
cos 31 = aₓ = a
a_{y} = a sin31
aₓ = a cos 31
Now let's use the kinematic equation for each axis
X axis
vₓ = v₀ₓ + aₓ (t-t₀)
vₓ = v₀ₓ + a cos 31 (t-t₀)
vₓ = 2.6 + 0.45 cos 31 (20-10)
vₓ = 6,457 m / s
Y Axis
v_{y} = v_{oy} + a_{y} t
v_{y} = v_{oy} + a_{y} sin31 (t-to)
v_{y} = -1.8 + 0.45 sin31 (20-10)
v_{y} = 0.518 m / s
b) let's use Pythagoras' theorem to find the magnitude of velocity
v = √ (vₓ² + v_{y}²)
v = √ (6,457² + 0.518²)
v = √ (41.96)
v = 6.478 m / s
We use trigonometry for direction
tan θ = v_{y} / vₓ
θ = tan⁻¹ v_{y} / vₓ
θ = tan⁻¹ 0.518 / 6.457
θ = 4.9°
c) let's look for the vector at the initial time
v₁ = √ (2.6² + 1.8²)
v₁ = 3.16 m / s
θ₁ = tan⁻¹ (-1.8 / 2.6)
θ₁ = -34.7
We see that the two vectors differ in module and direction, and that the acceleration vector is responsible for this change.
a = (v₂ -v₁) / (t₂-t₁)
A proton with a speed of 3.000×105 m/s has a circular orbit just outside a uniformly charged sphere of radius 7.00 cm. What is the charge on the sphere?
To solve this problem we will apply the principles of energy conservation. The kinetic energy in the object must be maintained and transformed into the potential electrostatic energy. Therefore mathematically
[tex]KE = PE[/tex]
[tex]\frac{1}{2} mv^2 = \frac{kq_1q_2}{r}[/tex]
Here,
m = mass (At this case of the proton)
v = Velocity
k = Coulomb's constant
[tex]q_{1,2}[/tex] = Charge of each object
r= Distance between them
Rearranging to find the second charge we have that
[tex]q_2 = \frac{\frac{1}{2} mv^2 r}{kq_1}[/tex]
Replacing,
[tex]q_2 = \frac{\frac{1}{2}(1.67*10^{-27})(3*10^5)^2(7*10^{-2})}{(9*10^9)(1.6*10^{-19})}[/tex]
[tex]q_2 = 3.6531nC[/tex]
Therefore the charge on the sphere is 3.6531nC
A muon has a rest mass energy of 105.7 MeV, and it decays into an electron and a massless particle. If all the lost mass is converted into the electron’s kinetic energy, what is the electron’s velocity?
Answer:
The electron’s velocity is 0.9999 c m/s.
Explanation:
Given that,
Rest mass energy of muon = 105.7 MeV
We know the rest mass of electron = 0.511 Mev
We need to calculate the value of γ
Using formula of energy
[tex]K_{rel}=(\gamma-1)mc^2[/tex]
[tex]\dfrac{K_{rel}}{mc^2}=\gamma-1[/tex]
Put the value into the formula
[tex]\gamma=\dfrac{105.7}{0.511}+1[/tex]
[tex]\gamma=208[/tex]
We need to calculate the electron’s velocity
Using formula of velocity
[tex]\gamma=\dfrac{1}{\sqrt{1-(\dfrac{v}{c})^2}}[/tex]
[tex]\gamma^2=\dfrac{1}{1-\dfrac{v^2}{c^2}}[/tex]
[tex]\gamma^2-\gamma^2\times\dfrac{v^2}{c^2}=1[/tex]
[tex]v^2=\dfrac{1-\gamma^2}{-\gamma^2}\times c^2[/tex]
Put the value into the formula
[tex]v^2=\dfrac{1-(208)^2}{-208^2}\times c^2[/tex]
[tex]v=c\sqrt{\dfrac{1-(208)^2}{-208^2}}[/tex]
[tex]v=0.9999 c\ m/s[/tex]
Hence, The electron’s velocity is 0.9999 c m/s.
The efficiency of a squeaky pulley system is 73 percent. The pulleys are usedto raise a mass to a certain height. What force is exerted on the machine if arope is pulled 18.0 m in order to raise a 58 kg mass a height of 3.0 m
The efficiency of the machine is defined as
[tex]\eta = \frac{W_{out}}{W_{in}}[/tex]
Here
Work out is the work output and Work in is the work input
To find the Work in we have then
[tex]W_{in} = \frac{W_{out}}{\eta}[/tex]
[tex]W_{in} = \frac{mgh}{\eta}[/tex]
Replacing with our values
[tex]W_{in} = \frac{(58)(9.8)(3)}{73\%}[/tex]
[tex]W_{in} = 2335.89J[/tex]
The work done by the applied force is
W = Fd
Here,
F = Force
d = Distnace
Rearranging to find F,
[tex]F = \frac{W}{d}[/tex]
[tex]F = \frac{2335.89J }{18}[/tex]
F = 129.77N
Therefore the force exerted on the machine after rounding off to two significant figures is 130N
The force exerted on the pulley system when a rope is pulled 18.0 m in order to raise a 58 kg mass a height of 3.0 m with an efficiency of 73 percent is about 129.42 Newtons.
Explanation:To solve this problem, we need to understand the concept of machine efficiency and work. The efficiency of a machine is the ratio of output work to input work.
In this case, the efficiency of the pulley system is given as 73%. Meaning output work is 73% of the input work. The force exerted on the machine is equivalent to the input work divided by the distance pulled.
The output work (W_out) can be determined using the formula W_out = mass * gravity * height = 58 kg * 9.8 m/s^2 * 3.0 m = 1701.6 Joules.
Then, we can find the input work (W_in) using the efficiency formula: W_in = W_out / efficiency = 1701.6 J / 0.73 = 2329.59 Joules.
Finally, we can find the force exerted on the pulley system (F_in) by dividing the input work by the distance pulled: F_in = W_in / distance = 2329.59 J / 18.0 m = 129.42 Newtons.
So, the force exerted on the machine would be approximately 129.42 Newtons.
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The wavelength of green light is about the size of an atom. (T/F)
Explanation:
The wavelength of green light is about 500 nanometers, or two thousandths of a millimeter. The typical wavelength of a microwave oven is about 12 centimeters, which is larger than a baseball.
The statement is falls, because the the wavelength of green light is about 500 nm or 500 × 10⁻⁹ m. But the size of an atom is about 1.2 × 10⁻¹⁰ m. Hence atomic size not equals the wavelength of green light.
What is wavelength?The wavelength of an electromagnetic wave is the distance between two consecutive crests or troughs. The upward peak in the wave is called crests and the downward peaks are called troughs.
The wavelength of high energy waves will be shorter. Visible region is in between IR and UV rays in the electromagnetic spectrum. Green light is in the exact middle region of the VIBGYOR. Thus, it is having a wavelength of 500 -520 nm or 500 × 10⁻⁹ m.
The size of an atom is estimated in the range of 1.2 × 10⁻¹⁰ m and it varies from element to element. However the atomic size is not comparable with the wavelength of green light.
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t requires1200 kg of coal to produce the energy needed to make 1.0 kg of aluminum metal. If a single soda can requires approximately 15 g of Al, what mass of coal would be needed to produce a 6-pack of cans?
Answer:
The mass of coal is 108 kg.
Explanation:
Given that,
Energy of coal = 1200 kg
Mass of aluminum = 1.0 kg
Energy required for single soda can = 15 g of Al
Energy required for 6 pack of cans = [tex]6\times15=90\ g\ of\ Al[/tex]
We need to calculate the mass of coal
Using formula of mass
[tex]\text{mass of coal}=\dfrac{\text{Energy of coal}\times\text{Energy required for 6 pack of cans}}{\text{Mass of aluminum}}[/tex]
Put the value into the formula
[tex]m=\dfrac{1200\times90\times10^{-3}}{1.0}[/tex]
Put the value into the formula
[tex]m=108\ Kg[/tex]
Hence, The mass of coal is 108 kg.
If the plane is flying in a horizontal path at an altitude of 98.0 m above the ground and with a speed of 73.0 m/s, at what horizontal distance from the target should the pilot release the canister? Ignore air resistance.
Explanation:
The given data is as follows.
height (h) = 98.0 m, speed (v) = 73.0 m/s,
Formula of height in vertical direction is as follows.
h = [tex]\frac{gt^{2}}{2}[/tex],
or, t = [tex]\sqrt{\frac{2h}{g}}[/tex]
Now, formula for the required distance (d) is as follows.
d = vt
= [tex]v \sqrt{\frac{2h}{g}}[/tex]
= [tex]73.0 m/s \sqrt{\frac{2 \times 98.0 m}{9.8 m/s^{2}}}[/tex]
= 326.5 m
Thus, we can conclude that 326.5 m is the horizontal distance from the target from where should the pilot release the canister.
According to the Stefan-Boltzmann law, how much energy is radiated into space per unit time by each square meter of the Sun’s surface? If the Sun’s radius is 696,000 km, what is the total power output of the Sun?
Answer:
[tex]3.8469943828\times 10^{26}\ W[/tex]
Explanation:
[tex]\sigma[/tex] = Stefan-Boltzmann constant = [tex]5.67\times 10^{-8}\ W/m^2K^4[/tex]
T = Surface temperature of the Sun = 5778 K
r = Radius of the Sun = 696000 km
From Stefan-Boltzmann law
[tex]F=\sigma T^4\\\Rightarrow F=5.67\times 10^{-8}\times 5778^4\\\Rightarrow F=63196526.546\ W/m^2K[/tex]
Power is given by
[tex]P=F4\pi r^2\\\Rightarrow P=63196526.546\times 4\pi\times (696000000)^2\\\Rightarrow P=3.8469943828\times 10^{26}\ W[/tex]
The power output of the Sun is [tex]3.8469943828\times 10^{26}\ W[/tex]
A star much cooler than the sun would appear (a) red; (b) blue; (c) smaller; (d) larger.
A star that is cooler than the sun would appear red (a), because the color of a star is associated with its temperature, according to Wien's Law.
Explanation:The color of a star is closely related to its temperature. In the case of a star that's cooler than the sun, it would appear (a) red. This is due to a principle in astrophysics known as Wien's Law, which states that the peak wavelength (color) of the light emitted by an object (such as a star) shifts to longer, redder wavelengths as the object gets cooler. So, cool stars like red giants appear red, while hotter stars appear blue or white.
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A drunken sailor stumbles 600 meters north, 550 meters northeast, then 500 meters northwest. What is the total displacement and the angle of the displacement?
We will make a graph to better understand the displacement of the individual. From the trigonometric properties we will find the required distances.
[tex]d_1 = \frac{450}{\sqrt{2}} = 318.198[/tex]
[tex]d_2 = \frac{400}{\sqrt{2}} = 282.843[/tex]
[tex]D = 500+d_1+d_2 = 1101.041[/tex]
Displacement ,
[tex]x = \sqrt{1101.041^2+(318.198-282.843)^2}}[/tex]
[tex]x = 1101.608m[/tex]
The angle would be
[tex]\theta = cos^{-1} (\frac{1101.041}{1101.608})[/tex]
[tex]\theta = E 1.838\° N[/tex]
Therefore the displacement was of 1101.608m to a angle of 1.838° from East to North.
The sailor's total displacement is approximately 1118.16 meters, at an angle of approximately 88.13 degrees north of east.
Explanation:This is a problem of physics specifically in the field of mechanics, about displacement and vectors. To find the total displacement, the directions in which the sailor traveled need to be considered. A north direction has an angle of 90 degrees, northeast is 45 degrees, and northwest is 135 degrees.
Using the Pythagorean theorem and some trigonometry, we can calculate the displacement in the x and y directions. The total displacement in the x direction (East-West) is 550*cos(45) - 500*cos(45) = 35.35 meters (approximately). The total displacement in the y direction (North-South) is 600 + 550*sin(45) + 500*sin(45) = 1116.57 meters (approximately).
The total displacement would then be the square root of (35.35 ^2 + 1116.57 ^2) = 1118.16 meters (approximately).
To find the angle of the displacement, we can use the arctangent function (atan function in most calculators). The angle is atan (1116.57 / 35.35) = 88.13 degrees (approximately).
Therefore, the sailor's total displacement is approximately 1118.16 meters, at an angle of approximately 88.13 degrees north of east.
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A student decides to measure the muzzle velocity of a pellet shot from his gun. He points the gun horizontally. He places a target on a vertical wall a distance x away from the gun. The pellet hits the target a vertical distance y below the gun.
(a) Show that the position of the pellet when traveling through the air is given by y = Ax^2, where A is a constant.
(b) Express the constant A in terms of the initial velocity v and the free-fall acceleration g.
(c) If x = 3 m and y = 0.21 m, what is the initial speed of the pellet?
Answer:
a) y = A x² , b) A = - ½ g / v₀², c) v₀ = 15.46 m / s
Explanation:
For this problem of two-dimensional kinematics, we will use that the time to reach the wall is the same
X axis
x = v₀ₓ t
t = x / v₀ₓ
Y Axis
y = [tex]v_{oy}[/tex] t - ½ g t²
As it shoots horizontally the vertical speed is zero
y = - ½ g t²
We replace
y = - ½ g (x / v₀ₓ)²
The initial speed is all horizontal
v₀ₓ = v₀
y = - ½ g / v₀² x²
y = A x²
b) the expression for the constant is
A = - ½ g / v₀²
c) we look for the initial speed
v₀² = - ½ g x² / y
As the object falls below the exit point its height is negative
v₀ = √ (- ½ 9.8 3²/ (-0.21))
v₀ = 15.46 m / s
An electron and a proton are separated by a distance of 1 m. What happens to the size of the force on the first electron if a second electron is placed next to the proton?
Answer:
The force on the electron will become zero.
Explanation:
As electron has negative charge and proton has positive charge. The magnitude of charge on both particles is equal. Therefore, there will be a force of attraction between electron and proton. When another electron is brought near the proton the net charge in that area will become equal to zero. Therefore, first electron will not experience any force.
Final answer:
The original force between the first electron and proton remains unchanged when a second electron is placed next to the proton, but the first electron will experience an additional repulsive force from the second electron. This alters the net force acting on the first electron but does not directly change the force between it and the proton.
Explanation:
The question inquires about the effect on the force on an electron when a second electron is placed next to a proton, with all particles separated by a distance of 1 meter.
According to Coulomb's Law, which governs the electrostatic force between two charged particles, the force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.
When a second electron is placed next to the proton, this setup introduces additional forces: a repulsive force between the two electrons and an attractive force between the new electron and the proton.
However, the original question seems to focus on how the force on the first electron changes with this new arrangement. The presence of a second electron does not alter the magnitude of the force between the first electron and the proton directly since the forces here are evaluated pairwise, and the distance between them remains unchanged.
What changes, though, is the overall electrical environment, introducing a new set of forces that need to be considered for the complete system. The addition of the second electron introduces a repulsive force on the first electron, which is separate from but concurrent with the attractive force from the proton.
It is important to consider the net force acting on any charge in such scenarios, which would involve adding vectorially the attractive force from the proton and the repulsive force from the second electron.
Thus, while the force due to the proton-electron pair remains constant, the first electron experiences an additional repulsive force due to the second electron, affecting the net force on it but not the force between it and the proton directly.
A fence post is 52.0 m from where you are standing, in a direction 37.0° north of east. A second fence post is due south from you. How far are you from the second post if the distance between the two posts is 68.0° m?
The distance from the starting point is approximately 26.3 m and the compass direction is 25.3° west of north.
Explanation:To find the distance between the starting point and the final position, we can use the Pythagorean theorem. We can calculate the distances in the north and west directions using the given lengths and angles. Using the given values, the distance from the starting point is approximately 26.3 m. The compass direction of the line connecting the starting point to the final position is 25.3° west of north.
Final answer:
To find the distance between the starting point and final position, use the Pythagorean theorem. The total distance is approximately 31.3 m. To determine the compass direction, use trigonometry. The compass direction is approximately 36.9° north of west.
Explanation:
To find the distance between your starting point and final position, you can use the Pythagorean theorem. First, calculate the horizontal distance by adding the horizontal components of the two legs of the walk: A = 18.0 m west and B = 0 m south. The horizontal distance is 18.0 m. Then, calculate the vertical distance by summing up the vertical components of the two legs of the walk: A = 0 m north and B = 25.0 m north. The vertical distance is 25.0 m. Apply the Pythagorean theorem to find the total distance by taking the square root of the sum of the squares of the horizontal and vertical distances. The total distance is approximately 31.3 m.
To determine the compass direction of a line connecting your starting point to your final position, you can use trigonometry. First, calculate the angle θ between the horizontal distance (18.0 m) and the hypotenuse (31.3 m). You can use the inverse tangent function: θ = arctan(vertical distance/horizontal distance). Calculate θ to be approximately 53.1°. Since you walked west and then north, the compass direction is 90° less than θ. Therefore, the compass direction is approximately 36.9° north of west.
What is the normal condition for atoms? What is an excited atom? What are orbitals?
An atom in normal conditions refers when electrons are in the fundamental state. When you leave the atom, an electron absorbs energy from an external source and moves to a higher energy state.
Energy states or energy levels are called orbitals. The difference between the energy states in an atom is responsible for the emission of photons when an electron transition occurs between these two energy states.
Because the energy levels are discrete, the emitted photons also possess different energies.
The ozone molecule O3 has a permanent dipole moment of 1.8×10−30 Cm. Although the molecule is very slightly bent-which is why it has a dipole moment-it can be modeled as a uniform rod of length 2.5×10−10 m with the dipole moment perpendicular to the axis of the rod. Suppose an ozone molecule is in a 8000 N/C uniform electric field. In equilibrium, the dipole moment is aligned with the electric field. But if the molecule is rotated by a small angle and released, it will oscillate back and forth in simple harmonic motion.
What is the frequency f of oscillation?
Answer:
934701926.438 Hz
Explanation:
Mass of molecule
[tex]m=3\times 16\times 1.67\times 10^{-27}\ kg[/tex]
Moment of inertia is given by
[tex]I=\dfrac{1}{12}ml^2\\\Rightarrow I=\dfrac{1}{12}\times 3\times 16\times 1.67\times 10^{-27}\times (2.5\times 10^{-10})^2\\\Rightarrow I=4.175\times 10^{-46}\ kgm^2[/tex]
E = Electric field = 8000 N/C
p = Dipole moment = [tex]1.8\times 10^{-30}\ Cm[/tex]
l = Length of rod = [tex]2.5\times 10^{-10}\ m[/tex]
Frequency of oscillations is given by
[tex]f=\dfrac{1}{2\pi}\sqrt{\dfrac{pE}{I}}\\\Rightarrow f=\dfrac{1}{2\pi}\sqrt{\dfrac{1.8\times 10^{-30}\times 8000}{4.175\times 10^{-46}}}\\\Rightarrow f=934701926.438\ Hz[/tex]
The frequency of oscillations is 934701926.438 Hz
A 9-hp (shaft) pump is used to raise water to an elevation of 15 m. If the mechanical efficiency of the pump is 82 percent, determine the maximum volume flow rate of water. Take the density of water to be rho = 1000 kg/m3.
Answer:
The maximum volume flow rate is 0.03745m^3/s
Explanation:
Power input (Pi) = 9-hp = 9×746W = 6,714W
Elevation (h) = 15m
Efficiency (E) = 82% = 0.82
Density (D) of water = 1000kg/m^3
E = Po/Pi
Po (power output) = E×Pi = 0.82×6,714W = 5505.48W
Po = mgh/t
m/t (mass flow rate) = Po/gh = 5505.48/9.8×15 = 37.45kg/s
Volume flow rate = mass flow rate ÷ density = 37.45kg/s ÷ 1000kg/m^3 = 0.03745m^3/s
Why did you measure 20 cycles of the pendulum motion to determine the period, rather than just one cycle?
Final answer:
Measuring 20 cycles of the pendulum motion to determine the period instead of just one cycle allows for a more accurate and representative measurement. By averaging out any errors or variations in the measurement process, the value for the period of the pendulum is more precise. Measuring multiple cycles also helps identify any potential changes in the period over time.
Explanation:
The reason we measure 20 cycles of the pendulum motion to determine the period, rather than just one cycle, is to obtain a more accurate and representative measurement of the period. By measuring multiple cycles, we can average out any errors or variations in the measurement process, resulting in a more precise value for the period of the pendulum.
For example, if we were to measure only one cycle of the pendulum, any small errors in the timing or counting of the cycles could significantly affect our measurement. However, by measuring 20 cycles and then dividing the total time by 20, we can minimize these errors and obtain a more reliable measurement of the period.
Furthermore, measuring multiple cycles allows us to observe any potential changes in the period over time. In certain situations, the period of a pendulum may vary slightly due to factors such as air resistance or changes in the length of the string. By measuring multiple cycles, we can identify and account for any such variations, providing a more comprehensive understanding of the pendulum's behavior.
A bar extends perpendicularly from a vertical wall. The length of the bar is 2 m, and its mass is 10 kg. The free end of the rod is attached to a point on the wall by a light cable, which makes an angle of 30° with the bar. Find the tension in the cable.
Answer:
T = 98.1 N
Explanation:
Given:
- mass of bar m = 10 kg
- Length of the bar L = 2 m
- Angle between Cable and wall Q = 30 degres
Find:
- Find the tension in the cable.
Solution:
- Take moments about intersection of bar and wall, to be zero (static equilibrium)
(M)_a = 0
T*sin( 30 )*L - m*g*L/2 = 0
T*sin(30) - m*g / 2 = 0
T = m*g / 2*sin(30)
T = 10*9.81 / 2*sin(30)
T = 98.1 N
An iron block with mass mB slides down a frictionless hill of height H. At the base of the hill, it collides with and sticks to a magnet with mass mM.
Now assume that the two masses continue to move at the speed v from Part A until they encounter a rough surface. The coefficient of friction between the masses and the surface is μ. If the blocks come to rest after a distance s, which of the following equations would you use to find s?View Available Hint(s)Now assume that the two masses continue to move at the speed from Part A until they encounter a rough surface. The coefficient of friction between the masses and the surface is . If the blocks come to rest after a distance , which of the following equations would you use to find ?((mB)2mB+mM)gH=μmBgs((mB)2mB+mM)gH=μmMgs((mB)2mB+mM)gH=μ(mB+mM)gs((mB)2mB+mM)gH=−μ(mB+mM)gs((mB)2mB+mM)gH=μ(mB+mM)g
The correct equation to use to find the distance 's' when the iron block and magnet come to a rest is ((mB)2mB+mM)gH=μ(mB+mM)gs. This equation represents the conversion of potential energy into work done against friction.
Explanation:((mB)^2 + mM)gH = μ(mB + mM)gs.
This question represents a problem of mechanics in Physics, specifically involving both potential energy and work-energy theorem. The relevant equation to use in finding the distance 's' in this scenario is: ((mB)2mB+mM)gH=μ(mB+mM)gs. This equation derives from setting the initial potential energy of the system equal to the final kinetic energy when it rests, including the energy dissipated due to friction.
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The correct equation to find 's' when an iron block slides down a hill and collides with a magnet before encountering a rough surface, is ((mB)^2 + mB + mM)gH = μ(mB + mM)gs. This equation is derived from equating the kinetic energy at the base of the hill with the work done by friction.
Explanation:The correct equation to use to define s in this scenario would be: ((mB)2mB+mM)gH=μ(mB+mM)gs. This equation stems from equating the energy at the top of the hill (kinetic + potential energy) with the work done against friction (which eventually stops the two masses). Let's understand it in a detailed step-by-step manner.
Step 1: At the base of the hill, the kinetic energy of the block is equal to the potential energy at the top of the hill, which can be represented as (mB + mM)gH = 0.5(mB + mM)v^2. From this, we can turn the speeds into velocities, which in turn gives us v = sqrt(2gH).
Step 2: When the two masses encounter the rough surface, they will lose their kinetic energy due to friction. The total kinetic energy lost, which is equal to work done by friction, can be represented as 0.5(mB + mM)v^2 = μ(mB + mM)gs. Substituting the value of v from step 1 into this equation gives us the final formula: ((mB)^2 + mB + mM)gH = μ(mB + mM)gs.
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A model of a helicopter rotor has four blades, each of length 3.0 m from the central shaft to the blade tip. The model is rotated in a wind tunnel at a rotational speed of 560 rev/min.part A :What is the linear speed of the bladetip?part B :What is the radial acceleration of the bladetip expressed as a multiple of the acceleration of gravity,g?
Answer with Explanation:
We are given that
r=3 m
Angular frequency=[tex]\omega[/tex]=560rev/min
A.1 revolution=[tex]2\pi[/tex] radian
560 revolutions=[tex]560\times 2\pi[/tex] rad
Angular frequency=[tex]2\times 3.14\times \frac{560}{60}[/tex]rad/s
1 min=60 s
[tex]\pi=3.14[/tex]
Angular frequency=[tex]\omega=58.6rad/s[/tex]
Linear speed=[tex]\omega r[/tex]
Using the formula
Linear speed=[tex]58.6\times 3=175.8m/s[/tex]
Hence, the linear speed of the blade tip=175.8m/s
B.Radial acceleration=[tex]a_{rad}=\frac{v^2}{r}[/tex]
By using the formula
Radial acceleration=[tex]a_{rad}=\frac{(175.8)^2}{3}= 10.301\times 10^3m/s^2[/tex]
Radial acceleration=[tex]\frac{10.301}{9.8}g\times 10^3=1.05\times 10^3 g[/tex]
Where [tex]g=9.8m/s^2[/tex]
Hence, the radial acceleration[tex]=1.05\times 10^3 g[/tex]
A bullet is fired from a rifle that is held 1.60 m above the ground in a horizontal position. The initial speed of the bullet is 1058 m/s. Calculate the time it takes for the bullet to strike the ground.
Answer:
time takes for the bullet to strike the ground is 0.5711 second
Explanation:
given data
height h = 1.60 m
initial speed u = 1058 m/s
solution
we get here time that is express here as
time = [tex]\sqrt{\frac{2h}{g}}[/tex] ......................1
put here value and we will get here time that is
time = [tex]\sqrt{\frac{2*1.60}{9.81}}[/tex]
time = 0.5711 second
so time takes for the bullet to strike the ground is 0.5711 second
Final answer:
The time it takes for a bullet to strike the ground when fired horizontally from 1.60 m is calculated using the equation t = √(2h/g). Substitute 1.60 m for h and 9.81 m/s² for g to obtain t ≈ 0.571 seconds.
Explanation:
Calculating the Time for a Bullet to Strike the Ground
To calculate the time it takes for a bullet to strike the ground when fired from a horizontal position 1.60 meters above the ground with an initial speed of 1058 m/s, we can use the equations of motion for uniformly accelerated motion (in this case, the acceleration due to gravity). Since the bullet is fired horizontally, there is no initial vertical velocity component, so the vertical motion can be treated as a free-fall problem.
Using the formula for the time of free fall (t) from a height (h):
t = √(2h/g), where g is the acceleration due to gravity (approximately 9.81 m/s²).
Plugging in the given height of 1.60 m and solving for t gives us:
t = √(2 * 1.60 m / 9.81 m/s²) = √(0.3261 s²) ≈ 0.571 seconds.
Therefore, the bullet will take approximately 0.571 seconds to hit the ground.
You and your roommate are moving to a city360 mi away. Your roommate drives a rental truck at a constant 50 mi/h , and you drive your car at 60 mi/h . The two of you begin the trip at the same instant. An hour after leaving, you decide to take a short break at a rest stop
If you are planning to arrive at your destination a half hour before your roommate gets there, how long can you stay at the rest stop before resuming your drive?
Answer:
42 minutes
Explanation:
First, let us find out the time required by the roommate, who is driving the truck, to reach the city.
We know that,
[tex]time=\frac{distance}{speed} =\frac{360}{50} hrs=7.2hrs[/tex]
Now, you are planning to rest a bit after traveling for an hour. So,
distance covered in that 1 hour = [tex]60mi/h\times1h=60 miles[/tex]
time required by you to cover the total distance = [tex]\frac{360mi}{60mi/h} =6hours[/tex]
If you wish reach at the destination half an hour (0.5 h) before your roommate, you can expend a total of [tex]7.2-0.5=6.7hours[/tex] throughout your journey.
Hence, you can rest for [tex]6.7-6=0.7hours=(0.7\times60)minutes=42minutes[/tex]
After driving 60 miles in the first hour and intending to arrive half an hour before your roommate who travels at 50 mph, you can afford a break time of 30 minutes at the rest stop.
Explanation:The subject of your question is
relative speed
and
time management
, which falls under Mathematics. Your overall travelling speed is 60 mph. After driving for an hour, you decide to take a break at a rest stop. In that one hour, you've driven 60 miles, so there are 300 miles left to the destination. Your roommate, who doesn't stop for a break, continues to drive at 50 mph. You want to arrive half an hour before she does, which means you essentially have her travel time less 30 minutes to reach the destination. Because her speed in relation to the remaining distance is constant, her remaining travel time is 300 miles divided by 50 mph which equals to 6 hours. So, you actually have 6 - 0.5 = 5.5 hours to travel the remaining 300 miles including your rest stop. Considering your speed of 60 mph, it will take you 300/60 = 5 hours to reach the destination. Therefore, the length of the break you can take while still beating your roommate to the destination is 5.5 hours minus 5 hours, which is
30 minutes
.
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A student throws a set of keys vertically upward to her sorority sister, who is in a window 4.00 m above. The keys are caught 1.50 s later by the sister's outstretched hand. (a) With what initial velocity were the keys thrown?(b) What was the velocity of the keys just before they were caught?
Answer:
a)The keys were thrown with an initial velocity of 10.0 m/s.
b)The velocity of the keys just before they were caught is -4.72 m/s (the keys were caught on their way down).
Explanation:
Hi there!
The equations for the height and velocity of the keys are the following:
h = h0 + v0 · t + 1/2 · g · t²
v = v0 + g · t
Where:
h =height of the keys after a time t.
h0 = initial height.
v0 = initial velocity.
t = time.
g = acceleration due to gravity (-9.81 m/s² considering the upward direction as positive)
v = velocity of the keys at time t.
a) We know that at t = 1.50 s, h = 4.00 m and let´s consider that the origin of the frame of reference is located at the point where the keys are thrown so that h0 = 0. Then, using the equation of height, we can obtain the initial velocity.
h = h0 + v0 · t + 1/2 · g · t²
4.00 m = v0 · 1.50 s - 1/2 · 9.81 m/s² · (1.50 s)²
4.00 m + 1/2 · 9.81 m/s² · (1.50 s)² = v0 · 1.50 s
v0 = ( 4.00 m + 1/2 · 9.81 m/s² · (1.50 s)²) / 1.50 s
v0 = 10.0 m/s
The keys were thrown with an initial velocity of 10.0 m/s.
b) Now, using the equation of velocity we can calculate the velocity at t = 1.50 s:
v = v0 + g · t
v = 10.0 m/s - 9.81 m/s² · 1.50 s
v = -4.72 m/s
The velocity of the keys just before they were caught is -4.72 m/s (the keys were caught on their way down).
A firefighting crew uses a water cannon that shoots water at 25.0 m/s at a fixed angle of 53.0° above the horizontal. The fire-fighters want to direct the water at a blaze that is 10.0 m above ground level. How far from the building should they position their cannon? There are two possibilities; can you get them both? (Hint: Start with a sketch showing the trajectory of the water.)
To find out how far the firefighting crew should position their cannon from the building, we use the equations of projectile motion separately for the vertical and the horizontal components. Since there are two viable times that satisfy the conditions of motion, there are two potential positions of the cannon. This question combines the principles of physics, which are specifically related to projectile motion.
Explanation:This question is about the physics of projectile motion, specifically under gravity. To find out how far from the building they should position their cannon, we have to deal with the vertical and horizontal components of the projectile motion separately.
First, we determine the time it takes for the water to reach the desired height. We get this by using the equation of motion: H = V*sin(theta)*t - 0.5*g*t^2, where H is the height above the ground level (10m), V is speed (25.0 m/s), sin(theta) is the vertical component of initial velocity, g is acceleration due to gravity (approximated as 10 m/s^2 for ease of calculation), and t is time. We can rearrange this equation to find t.
Next, we use the time obtained and the horizontal component of initial velocity (V*cos(theta)) to find the horizontal distance made by using the equation: s = V*cos(theta)*t.
The hint here is that there are two potential time values, a smaller one and a larger one, leading to two potential distances from the building that the cannon could be positioned. Both times satisfy the conditions of the motion.
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The firefighting crew should place their cannon either 8.0 meters or 56.1 meters from the building to reach a blaze 10.0 meters above ground.
To determine how far the firefighting crew should position their cannon, we need to analyze the projectile motion of the water. We'll break this down into x-direction (horizontal) and y-direction (vertical) components.
Step-by-Step Solution
1. Break down initial velocity into components
Initial speed, v₀ = 25.0 m/s
Angle, θ = 53.0°
v₀x = v₀ x cos(θ) = 25.0 x cos(53.0°) ≈ 15.0 m/s
v₀y = v₀ x sin(θ) = 25.0 x sin(53.0°) ≈ 20.0 m/s
2. Write the equations of motion
Vertical (y-direction): y = v₀y x t - 0.5 x g x t²
Given y = 10.0 m, v₀y = 20.0 m/s, g = 9.8 m/s²
Using the quadratic formula, we find the time (t) it takes for the water to reach 10.0 m:
10.0 = 20.0 x t - 0.5 x 9.8 x t²
4.9t² - 20.0t + 10.0 = 0
Solving for t, we get t ≈ 0.54 s or t ≈ 3.74 s
Both times are valid because water can take two paths: one on the way up (0.54 s) and one on the way down (3.74 s).
3. Calculate horizontal distance
Distance (x) = v₀x x t
For t ≈ 0.54 s: x = 15.0 x 0.54 ≈ 8.0 m
For t ≈ 3.74 s: x = 15.0 x 3.74 ≈ 56.1 m
Therefore, the cannon can be placed either 8.0 meters or 56.1 meters from the building.
A block with a mass of 8.7 kg is dropped from rest from a height of 8.7 m, and remains at rest after hitting the ground. 1)If we consider the system to consist of the block, the ground, and the surrounding air, what is the change in the internal energy of the system
To solve this problem we will apply the concepts related to gravitational potential energy.
This can be defined as the product between mass, gravity and body height.
Mathematically it can be expressed as
[tex]\Delta P = mgh[/tex]
[tex]\Delta P = (8.7)(9.8)(3)[/tex]
[tex]\Delta P = 255.78J[/tex]
Therefore the change in the internal energy of the system is 255.78
The thermal conductivity of a sheet of rigid insulation is reported to be 0.029 W/(m·K). The temperature difference across a 20 mm thick sheet of insulation is 10˚C. (a) What is the heat flux through a 2m x 2m sheet of this insulation? (b) What is the total rate of heat transfer through the sheet?
Answer:
a)[tex]q=14.5\ W/m^2[/tex]
b)Q= 58 W
Explanation:
Given that
Thermal conductivity ,K = 0.029 W/m.k
The temperature difference ,ΔT= 10°C
The thickness ,L = 20 mm
We know that
[tex]Q=\dfrac{KA}{L}\times \Delta T[/tex]
Now by putting the values
[tex]Q=\dfrac{0.029\times 4}{0.02}\times 10\ W[/tex]
Q= 58 W
The heat flux through the sheet is given as
[tex]q=\dfrac{Q}{A}\ W/m^2[/tex]
[tex]q=\dfrac{58}{2\times 2}\ W/m^2[/tex]
[tex]q=14.5\ W/m^2[/tex]
a)[tex]q=14.5\ W/m^2[/tex]
b)Q= 58 W
What is the freezing point of an aqueous solution that boils at 101 degrees C? Kfp = 1.86 K/m and Kbp = 0.512 K/m. Enter your answer to 2 decimal places.
Answer:
-3.63 degree Celsius
Explanation:
We are given that
Boiling point of solution=[tex]T_b=101^{\circ}[/tex] C
Boiling point water=100 degree Celsius
[tex]K_f=1.86K/m[/tex]
[tex]K_b=0.512 K/m[/tex]
[tex]\Delta T_b=T-T_0[/tex]
Where [tex]T[/tex]=Boiling point of solution
[tex]T_0=[/tex]Boiling point of pure solvent
[tex]\Delta T_b=101-100=1^{\circ}[/tex]C
[tex]\Delta T_b=k_bm[/tex]
Using the formula
[tex]1=0.512\times m[/tex]
Molality,[tex]m=\frac{1}{0.512}[/tex] m
[tex]\Delta T_f=k_fm[/tex]
Using the formula
[tex]\Delta T_f=\frac{1}{0.512}\times 1.86[/tex]
[tex]\Delta T_f=3.63 C[/tex]
We know that
[tex]\Delta T_f=T_0-T_1[/tex]
Where [tex]T_0[/tex] =Freezing point of solvent
[tex]T_1=[/tex] Freezing point of solution
Using the formula
[tex]3.63=0-T_1[/tex]
Freezing point of water=0 degree Celsius
[tex]T_1=0-3.63=-3.63 C[/tex]
Hence, the freezing point of solution=-3.63 degree Celsius
Final answer:
The freezing point of an aqueous solution can be calculated using the freezing point depression equation ΔTf = Kf * m. In this case, the solution boils at 101°C, indicating the boiling point elevation constant (Kb) is provided. Assuming complete dissociation, we can calculate the freezing point depression to be -0.512°C.
Explanation:
The freezing point depression can be calculated by using the equation:
ΔTf = Kf * m
Where ΔTf is the change in freezing point, Kf is the cryoscopic constant, and m is the molality of the solution.
In this case, since we have the boiling point elevation constant (Kb), we need to use the equation:
ΔTf = Kb * m
Given that Kb is 0.512 K/m and assuming complete dissociation, a 1.0 m aqueous solution of a solute will contain 1.0 mol of particles per kilogram of water. Therefore, the freezing point depression (ΔTf) will be:
ΔTf = 0.512 K/m * 1.0 m = 0.512 K
Since the freezing point of pure water is 0°C, the freezing point of the aqueous solution will be:
0°C - 0.512 K = -0.512°C