Answer:
3.333333333333333333333333333333333333333
Explanation:
3.3333333333333333333333333333333333
According to information found in an old hydraulics book, the energy loss per unit weight of fluid flowing through a nozzle connected to a hose can be estimated by the formula where h is the energy loss per unit weight, D the hose diameter, d the nozzle tip diameter, V the fluid velocity in the hose, and g the acceleration of gravity. Do you think this equation is valid in any system of units
This question is incomplete, the complete question is;
According to information found in an old hydraulics book, the energy loss per unit weight of fluid flowing through a nozzle connected to a hose can be estimated by the formula; h= (0.04 to 0.09)(D/d)⁴V²/2g
where h is the energy loss per unit weight, D the hose diameter, d the nozzle tip diameter, V the fluid velocity in the hose, and g the acceleration of gravity.
Do you think this equation is valid in any system of units
Answer:
YES, the equation is a general equation that is valid in any system of units
Explanation:
Given the data in the question;
h = (0.04 to 0.09)(D/d)⁴ × [tex]\frac{V^{2} }{2g}[/tex]
so
[ N.m/N ] = (0.04 to 0.09) ( m/m)² × (m²/s²)1/2 × (s²/m)
[ N.L/N ] = (0.04 to 0.09) ( L⁴/L⁴) × (L²/T²)1/2 × (T²/L)
∴ [ L ] = (0.04 to 0.09) [L]
So as each term in the equation must have the same dimensions, the constant term (0.04 to 0.09) must be without dimension.
Therefore, YES, the equation is a general equation that is valid in any system of units
Two objects travel the same distance. The one that is moving faster will:
Take more time to go the distance
Take less time to go the same distance
Take the same time as the slower object
None of the above
Answer: take less time to go the same distance
Explanation:
Because if it is going faster let’s say mph 60 mph is 60 miles per hour if you are going 40 miles per hour it will take you longer to get to your destination.
A roller coaster rider traveling in a straight line changes from a speed of 4 m/s to 16 m/s in 3 seconds. What is the acceleration of the rider?
A. 1.33 m/s2
B. 3 m/s2
C. 5.33 m/s2
D. 4 m/s2
Answer:27
Explanation:
If a roller coaster rider traveling in a straight line changes from a speed of 4 m/s to 16 m/s in 3 seconds, then the acceleration of the rider would be 4 m / s² , therefore the correct answer is option D.
What are the three equations of motion?There are three equations of motion given by Newton,
v = u + at
S = ut + 1/2 × a × t²
v² - u² = 2 × a × s
As given in the problem, A roller coaster rider traveling in a straight line changes from a speed of 4 m/s to 16 m/s in 3 seconds.
By using the first equation of the motion,
v = u + at
16 = 4 + 3a
a = 16 -4 / 3
= 12 / 3
= 4 m / s²
Thus, the acceleration of the rider would be 4 m / s².
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Write the properties of Non Metals and the families containig non Metals.
Non-Malleable and Ductile: Non-metals are very brittle, and cannot be rolled into wires or pounded into sheets. Conduction: They are poor conductors of heat and electricity. Luster: These have no metallic luster and do not reflect light.
Group 15, the nitrogen family, contains two nonmetals: nitrogen and phosphorus. These non-metals usually gain or share three electrons when reacting with atoms of other elements. Group 16, the oxygen family, contains three nonmetals: oxygen, sulfur, and selenium.
Elements: Nitrogen; Oxygen; Phosphorus; Selenium...
What is the period of an objects motion?
The friction force that acts on objects that are at rest is___________
Answer:
static friction
Explanation:
static friction is the friction force that acts on objects at rest
What must the charge (sign and magnitude) of a particle of mass 1.40 gg be for it to remain stationary when placed in a downward-directed electric field of magnitude 640 N/CN/C
Answer:
the charge of the particle is -2.144 x 10⁻⁵ C.
Explanation:
The force acting on the particle is calculated as;
F = EQ = mg
[tex]Q = \frac{mg}{E}[/tex]
where;
Q is magnitude of the charge of the particle
[tex]Q = \frac{(1.4\times 10^{-3})(9.8)}{640} \\\\Q = 2.144 \ \times \ 10^{-5} \ C[/tex]
since the magnetic field is acting downward, the force must be acting upward in opposite direction.
Thus, the charge of the particle will be -2.144 x 10⁻⁵ C.
A car enters a 105-m radius flat curve on a rainy day when the coefficient of static friction between its tires and the road is 0.4. What is
the maximum speed which the car can travel around the curve without sliding
Static friction (magnitude Fs) keeps the car on the road, and is the only force acting on it parallel to the road. By Newton's second law,
Fs = m a = W a / g
(a = centripetal acceleration, m = mass, g = acceleration due to gravity)
We have
a = v ² / R
(v = tangential speed, R = radius of the curve)
so that
Fs = W v ² / (g R)
Solving for v gives
v = √(Fs g R / W)
Perpendicular to the road, the car is in equilibrium, so Newton's second law gives
N - W = 0
(N = normal force, W = weight)
so that
N = W
We're given a coefficient of static friction µ = 0.4, so
Fs = µ N = 0.4 W
Substitute this into the equation for v. The factors of W cancel, so we get
v = √((0.4 W) g R / W) = √(0.4 g R) = √(0.4 (9.80 m/s²) (105 m)) ≈ 20.3 m/s
The motion of a piston of a car engine approximates simple harmonic motion. Given that the stroke (twice the amplitude) is 0.100 m, the engine runs at 2,800 r/min, and the piston starts at the middle of its stroke, find the equation for the displacement d as a function of t. Sketch two cycles.
Answer:
y = - 0.050 sin (131.59t )
Explanation:
In this exercise we are told to approximate the movement of a piston to the simple harmonic movement
y = A cos (wt + Ф)
in this case they indicate that the stroke (C) of the piston is twice the amplitude
C = 2A
A = C / 2
angular velocity is related to frequency
w = 2π f
let's substitute
y = [tex]\frac{C}{2}[/tex] cos (2π f t +Ф)
To find the phase (fi) we will use the initial conditions, the piston starts at the midpoint of the stroke, if we create a reference system where the origin is at this point
y = 0 for t = 0
we substitute in the equation
0 = \frac{C}{2} cos (0 + Ф)
The we sew zero values for the angles of Ф = π/2 rad
we substitute in the initial equation
y = \frac{C}{2} cos (2π f t + π/2)
let's use the double angle relationship
cos ( a +90) = cos a cos 90 - sin a sin 90
cos (a+90) = - sin a
y = -\frac{C}{2} sin (2πf t )
let's reduce the frequency to SI units
f = 200 rpm (2π rad / 1rev) (1 min / 60s) = 20.94 rad / s
we substitute the given values
y = - [tex]\frac{0.100}{2}[/tex] sin (2π 20.94 t )
y = - 0.050 sin (131.59t )
A roller coaster car is released from rest as shown in the image below. If
friction is neglected, the car will oscillate back and forth across the "dip" in
the roller coaster. What is the approximate velocity of the roller coaster car
each time it reaches the bottom of the roller coaster in the image? (Recall
that g = 9.8 m/s2.)
TAS
81 m
O A. 40 m/s
B. 25 m/s
C. 30 m/s
D. 15 m/s
Answer:
40m/s
Explanation:
a=g
u=0
s=81
v²=u²+2as
v²=2(9.81)(81)
v=√1587.6=39.8446985181≈40m/s
The velocity of the roller coaster car each time it reaches the bottom is 40 ms⁻¹. The correct option is (A).
The rate at which the position of an object changes with respect to time is described by the physical quantity known as velocity. It has both magnitude and direction because it is a vector quantity.
Given:
Initial velocity, u = 0 m/s
Acceleration, a = -9.8 ms⁻²
Distance, d = 81 m
From the third equation of motion:
v² = u² - 2as
v² = 0 - 2×(-9.8)×81
v = 40 ms⁻¹
Hence, the velocity of the roller coaster car is 40 ms⁻¹. The correct option is (A).
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An archer shoots an arrow toward a 300-g target that is sliding in her direction at a speed of 2.30 m/s on a smooth, slippery surface. The 22.5-g arrow is shot with a speed of 38.0 m/s and passes through the target, which is stopped by the impact. What is the speed of the arrow after passing through the target
Answer:
4.79m/s
Explanation:
According to law of conservation of momentum;
The sum of momentum of the bodies before collision is equal to the momentum after collision.
m1u1 + m2u2 = (m1+m2)v
Given;
m1 = 0.3kg
u1 = 2.30m/s
m2 = 0.0225kg
u2 = 38m/s
Required
speed of the arrow after passing through the target v
Substituting the given data into the formula
0.3(2.3) + 0.0225(38) = (0.3 + 0.0225)v
0.69 + 0.855 = 0.3225v
1.545 = 0.3225v
v = 1.545/0.3225
v = 4.79m/s
Hence the speed of the arrow after passing through the target is 4.79m/s
According to the work-energy theorem, if work is done on an object, its potential and/or kinetic energy changes. Consider a car that accelerates from rest on a flat road. What force did the work that increased the car’s kinetic energy?
1. the force of the car engine
2. air resistance
3. the friction between the road and the tires
4. gravity
Answer:
1. The force of the car engine.
Explanation:
We shall see the effect and role played by each force, one by one, as follows:
1. The force of car engine:
The engine produces a force through combustion that is converted to mechanical work through the shaft. This work is then transmitted to the wheels of the car that cause the motion in the car and increase its kinetic energy.
2. Air Resistance:
It is the opposing force of air that tries to reduce the motion of the car and as a result, reduce its kinetic energy.
3. Frictional Force between road and tire:
It is also the opposing force of air that tries to reduce the motion of the car and as a result, reduce its kinetic energy.
4. Gravity:
Gravity pulls everything towards the center of Earth so it does not have much significant role in horizontal motion like this.
Hence, the force of the car engine did the work that increased the car's kinetic energy.
46) Recoil is noticeable if we throw a heavy ball while standing on roller skates. If instead we go through the motions of throwing the ball but hold onto it, our net recoil will be
Answer:
Zero
Explanation:
Appearing to throw the ball but still holding on to it means the recoil velocity will be zero because the recoil velocity is defined as the backward velocity as a result of throwing an object or shooting a bullet. In this case the object was not thrown and thus there is no recoil velocity.
The radius of the Sun is 6.96 x 108 m and the distance between the Sun and the Earth is roughtly 1.50 x 1011 m. You may assume that the Sun is a perfect sphere and that the irradiance arriving on the Earth is the value for AMO, 1,350 W/m2. Calculate the temperature at the surface of the Sun.
Answer:
5766.7 K
Explanation:
We are given that
Radius of Sun , R=[tex]6.96\times 10^{8} m[/tex]
Distance between the Sun and the Earth, D=[tex]1.50\times 10^{11}m[/tex]
Irradiance arriving on the Earth is the value for AMO=[tex]1350W/m^2[/tex]
We have to find the temperature at the surface of the Sun.
We know that
Temperature ,T=[tex](\frac{K_{sc}D^2}{\sigma R^2})^{\frac{1}{4}}[/tex]
Where [tex]K_{sc}=1350 W/m^2[/tex]
[tex]\sigma=5.67\times 10^{-8}watt/m^2k^4[/tex]
Using the formula
[tex]T=(\frac{1350\times (1.5\times 10^{11})^2}{5.67\times 10^{-8}\times (6.96\times 10^{8})^2})^{\frac{1}{4}}[/tex]
T=5766.7 K
Hence, the temperature at the surface of the sun=5766.7 K
The Burj Khalifa is the tallest building in the world at 828 m. How much work would a man with a weight of 700 N do if he climbed to the top of the building
Answer:
579600J
Explanation:
Given parameters:
Height of the building = 828m
Weight of the man = 700N
Unknown:
Work done by the man = ?
Solution:
The work done by the man is the same as the potential energy expended.
Work done:
Work done = Weight x height = 700 x 828
Work done = 579600J
Fluids
A = 2804 cm3 B = 2862 cm2 C = 2916 cm3
Three separate fluids, A, B, and C have been selected at random and each initially fills a 3000 cm3 volume at atmospheric pressure. A gage pressure of 6 x 107 N/m2 is then applied to each fluid. The final volume is given below. Determine which fluids were selected from the given list.
Acetone E = 0.92 GPa Glycerin E = 4.35 GP
Water E = 2.15 GPa Mercury E = 28.5 GPa
Benzene E = 1.05 GPa Sulfuric Acid E = 3.0 GPa
Ethyl Alcohol E = 1.06 GPa Gasoline E = 1.3 GPa
Petrol E = 1.45 GPa Seawater E = 2.34 GPa
Answer:
Explanation:
Fluid A :
Δ V = Change in volume = (3000 - 2804) x 10⁻⁶ m³ = 196 x 10⁻⁶ m³
volume strain = Δ V / V = 196 x 10⁻⁶ / 3000 x 10⁻⁶
= .06533
Δ P = increase in pressure = 6 x 10⁷ Pa
E = Δ P / volume strain = 6 x 10⁷ / .06533 = 91.84 x 10⁷ Pa = .92 GPa .
It is Acetone .
Fluid B :
Δ V = Change in volume = (3000 - 2862) x 10⁻⁶ m³ = 138 x 10⁻⁶ m³
volume strain = Δ V / V = 138 x 10⁻⁶ / 3000 x 10⁻⁶
= .046
Δ P = increase in pressure = 6 x 10⁷ Pa
E = Δ P / volume strain = 6 x 10⁷ / .046 = 130.43 x 10⁷ Pa = 1.3 GPa .
It is Gasoline .
Fluid C :
Δ V = Change in volume = (3000 - 2916) x 10⁻⁶ m³ = 84 x 10⁻⁶ m³
volume strain = Δ V / V = 84 x 10⁻⁶ / 3000 x 10⁻⁶
= .028
Δ P = increase in pressure = 6 x 10⁷ Pa
E = Δ P / volume strain = 6 x 10⁷ / .028 = 214.28 x 10⁷ Pa = 2.14 GPa .
It is Water .
A vertical piston-cylinder device contains a gas at a pressure of 100 kPa. The piston has a mass of 10 kg and a diameter for 14 cm. Pressure of the gas is to be increased by placing some weights on the piston. Determine the local atmospheric pressure and the mass of the weights that will doublethe pressure of the gas inside the cylinder.
Answer:
the local atmospheric pressure is 93.63 kPa
the mass of the weights is 156.9 kg
Explanation:
Given that;
Initial pressure of gas = 100 kPa
mass of piston = 10 kg and diameter = 14 cm = 0.14 m
g = 9.81 m/s²
Now,
P_gas = P_atm + P_piston
100 = P_atm + P_piston --------- let this equation 1
P_piston = M_piston × g / A = (10 × 9.81) / π/4×d²
P_piston = 98.1 / (π/4×( 0.14 )²)
P_piston = 98.1 / 0.01539 = 6374,269 Pa = 6.37 kPa
now, from equation 1
100 = P_atm + P_piston
we substitute
100 = P_atm + 6.37
P_atm = 100 - 6.37
P_atm = 93.63 kPa
Therefore, the local atmospheric pressure is 93.63 kPa
Now for pressure of the gas in the cylinder ⇒ 2×initial pressure
Pgas_2 = 2 × 100 = 200 kPa
Pgas_2 = P_atm + P_piston + P_weight
Pgas_2 = P_gas + P_weight
we substitute
200 kPa = 100 kPa + P_weight
P_weight = 200 kPa - 100 kPa
P_weight = 100 kPa = 100,000 Pa
Also;
P_weight = M×g / A
100,000 Pa = ( M × 9.81 ) / (π/4 × (0.14)²)
100,000 × 0.01539 = M × 9.81
1539 = M × 9.81
M = 1539 / 9.81
M = 156.9 kg
Therefore, the mass of the weights is 156.9 kg
6. What is the lowest temperature on the Kelvin scale? What happens to matter when it
reaches this temperature?
7. What is different about the degrees on the Fahrenheit and Kelvin scales and the Celsius
and Kelvin scales?
A long, straight wire carries a current of 5.20 A. An electron is traveling in the vicinity of the wire. At the instant when the electron is 4.40 cm from the wire and traveling at a speed of 6.20 * 104 m>s directly toward the wire, what are the magnitude and direction (relative to the direction of the current) of the force that the magnetic field of the current exerts on the electron
Answer:
Explanation:
Magnetic field due to current at a distance of 4.4 cm
B = 10⁻⁷ x 2 x 5.2 / 4.4 x 10⁻² [ B = 10⁻⁷ x 2i / r = ]
= 2.36 x 10⁻⁵ T.
Force on moving electron = Bqv , B is magnetic field , q is charge and v is velocity of charge .
Force = 2.36 x 10⁻⁵ x 1.6 x 10⁻¹⁹ x 6.2 x 10⁴
= 23.41 x 10⁻²⁰ N .
This force will be perpendicular to the direction of current .
The nose of an ultralight plane is pointed south, and its airspeed indicator shows 44 m/s. The plane is in a 18 m/s wind blowing toward the southwest relative to earth.
a. letting x be east and y be north, find the components of \vec v_{\rm P/E} (the velocity of the plane relative to the earth.
b.Find the magnitude of \vec v_{\rm P/E}.
c.Find the direction of \vec v_{\rm P/E}.
Answer:
a) vx = -12.7 m/s vy = -56.7m/s
b) v= 58.1 m/s
c) θ = 77.4º S of W
Explanation:
a)
In order to get the components of the velocity of the plane relative to the earth, we need just to get the components of both velocities first:Since the nose of the plane is pointing south, if we take y to be north, and positive, this means that the velocity of the plane can be written as follows:[tex]v_{ps} = -44m/s (1)[/tex]
Since the wind is pointing SW, it's pointing exactly 45º regarding both directions, so we can find its components as follows (they are equal each other in magnitude)[tex]v_{we} = - 18m/s * cos (45) = -12.7 m (2)[/tex]
[tex]v_{ws} = - 18m/s * cos (45) = -12.7 m (3)[/tex]
The component of v along the x-axis is simply (2), as the plane has no component of velocity along this axis:[tex]v_{e} = v_{x} = -12.7 m/s (4)[/tex]
The component of v along the -y axis is just the sum of (1) and (3)[tex]v_{y} = -44 m/s + (-12.7m/s) = -56.7 m/s (5)[/tex]b)
We can find the magnitude of the velocity vector, just applying the Pythagorean Theorem to (4) and (5):[tex]v = \sqrt{(-12.7m/s)^{2} + (-56.7m/s)^{2}} = 58.1 m/s (6)[/tex]
c)
Taking the triangle defined by vx, vy and v, we can find the angle that v does with the negative x-axis, just using the definition of tangent, as follows:[tex]tg_{\theta} =\frac{v_{y} }{v_{x} } = \frac{(-56.7m/s)}{(-12.7m/s} = 4.46 (7)[/tex]
Taking tg⁻¹ from (7), we get:tg⁻¹ θ = tg⁻¹ (4.46) = 77.4º S of W. (8)
Which characteristic involves cleavage and fracture?
the way a mineral breaks apart
the color of a mineral’s powder
the light that is reflected from a mineral’s surface
the number and angle of crystal faces of a mineral
Answer:
the way a mineral breaks apart
Explanation:
The way a mineral breaks apart involves fracture and cleavage.
These characteristics are very important in mineral identification especially during physical observations. Minerals have different cleavage properties and fracture planes.
Cleavage of a mineral is the ability of a mineral to split along preferred weakness planes. These planes are usually ingrained within the mineral during its formation. Fracture is the plane along through which minerals are able to break.Answer:
A:the way a mineral breaks apart
Explanation:
A daredevil is attempting to jump his motorcycle over a line of buses parked end to end by driving up a ramp of 64.8 degrees at a speed of 25.4 m/s. What would be the largest number of buses he can clear if the top of the takeoff ramp is at the same height as the bus tops and the buses are 10.0 m long
Answer: he can only make it over 5 buses
Explanation:
Given the data in the question;
we know that range is expressed as;
R = (V₀²sin2∅₀)/g
V₀ is the initial velocity( 25.4 m/s), ∅₀ is the angle of projection( 64.8°), g is acceleration due to gravity( 9.8 m/s²),
so we substitute
R = ((25.4)²sin2(64.8))/9.8
R = 50.7 m
now, them number of buses will be;
n = R / bus length
given that bus length is 10.0 m
we substitute
n = 50.7 m / 10.0
n = 5.07 ≈ 5
Therefore, he can only make it over 5 buses
The distance between the ruled lines on a diffraction grating is 1900 nm. The grating is illuminated at normal incidence with a parallel beam of white light in the 400 nm to 700 nm wavelength band. What is the angular width of the gap between the first order spectrum and the second order spectrum
Answer:
3.28 degree
Explanation:
We are given that
Distance between the ruled lines on a diffraction grating, d=1900nm=[tex]1900\times 10^{-9}m[/tex]
Where [tex]1nm=10^{-9} m[/tex]
[tex]\lambda_2=400nm=400\times10^{-9}m[/tex]
[tex]\lambda_1=700nm=700\times 10^{-9}m[/tex]
We have to find the angular width of the gap between the first order spectrum and the second order spectrum.
We know that
[tex]\theta=sin^{-1}(\frac{m\lambda}{d})[/tex]
Using the formula
m=1
[tex]\theta_1=sin^{-1}(\frac{1\times700\times 10^{-9}}{1900\times 10^{-9}})[/tex]
[tex]\theta=21.62^{\circ}[/tex]
Now, m=2
[tex]\theta_2=sin^{-1}(\frac{2\times400\times 10^{-9}}{1900\times 10^{-9}})[/tex]
[tex]\theta_2=24.90^{\circ}[/tex]
[tex]\Delta \theta=\theta_2-\theta_1[/tex]
[tex]\Delta \theta=24.90-21.62[/tex]
[tex]\Delta \theta=3.28^{\circ}[/tex]
Hence, the angular width of the gap between the first order spectrum and the second order spectrum=3.28 degree
The electric field between two parallel plates is uniform, with magnitude 628 N/C. A proton is held stationary at the positive plate, and an electron is held stationary at the negative plate. The plate separation is 4.22 cm. At the same moment, both particles are released.
A. Calculate the distance (in cm) from the positive plate at which the two pass each other.
B. Repeat part (a) for a sodlum lon (Nat) and a chlorlde lon (CI).
Answer:
Answer is explained in the explanation section below.
Explanation:
Solution:
Data Given:
Electric Field between two parallel plates = 628 N/C
Separation = 4.22 cm
a) In this part, we are asked to calculate the distance from positive plate at which the electron and proton pass each other.
Solution:
First of all:
Force on proton due to the Electric field between the plates is:
[tex]F_{p}[/tex] = [tex]q_{p}[/tex]E
and, we know that, F = ma
So,
[tex]m_{p}[/tex]a = [tex]q_{p}[/tex]E
a = [tex]\frac{q_{p}.E }{m_{p} }[/tex] Equation 1
So,
The distance covered by the electron is:
S = ut + 1/2[tex]at^{2}[/tex]
Here, u = 0.
S = 1/2[tex]at^{2}[/tex]
Put equation 1 into the above equation:
S = 1/2 x ([tex]\frac{q_{p}.E }{m_{p} }[/tex] )[tex]t^{2}[/tex] Equation 2
So,
Similarly, the distance covered by electron will be:
(D-S) = 1/2 x ([tex]\frac{q_{e}.E }{m_{e} }[/tex] )[tex]t^{2}[/tex] Equation 3
We know that the charge of electron is equal to the charge of proton so,
[tex]q_{p}[/tex] = [tex]q_{e}[/tex] = q
By dividing the equation 2 by equation 3, we get:
[tex]\frac{S}{D-S}[/tex] = [tex]\frac{m_{e} }{m_{p} }[/tex]
Solve the above equation for S,
S[tex]m_{p}[/tex] = [tex]m_{e}[/tex]D - [tex]m_{e}[/tex]S
So,
S = [tex]\frac{m_{e}.D }{(m_{e} + m_{p}) }[/tex]
Plugging in the values,
As we know the mass of electron is 9.1 x [tex]10^{-31}[/tex] and the mass of proton is 1.67 x [tex]10^{-27}[/tex]
S = [tex]\frac{9.1 . 10^{-31} . 4.22 }{(9.1 . 10^{-31} + 1.67 . 10^{-27} }[/tex]
S = 0.002298 cm (Distance from the positive plate at which the two pass each other)
b) In this part, we to calculate distance for Sodium ion and chloride ion as above.
So,
we already have the equation, we need to put the values in it.
So,
S = [tex]\frac{m_{Cl}.D }{(m_{Cl} + m_{Na}) }[/tex]
As we know the mass of chlorine is 35.5 and of sodium is 23
S = [tex]\frac{35.5 . 4.22}{(35.5 + 23)}[/tex]
S = 2.56 cm
Why are soft materials used in theaters and auditoriums?
Answer:
The roof and walls of the auditorium or cinema hall are generally covered with sound absorbent materials like draperies or compressed fibreboard to reduce reverberation. These materials reduce the formation of echoes by absorbing sound waves.
Explanation:
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A group of 25 particles have the following speeds: two have speed 11 m/s, seven have 16 m/s , four have 19 m/s, three have 26 m/s, six have 31 m/s, one has 37 m/s, and two have 45 m/s.
Requiredd:
a. Determine the average speed.
b. Determine the rms speed.
c. Determine the most probable speed.
Answer:
a) Average speed is 24.04 m/s
b) the rms speed is 25.84 m/s
c) the most probable speed is 16 m/s
Explanation:
Given the data in the question;
a) Determine the average speed.
To determine the average speed, we simply divide total some of speed by number of particles;
Average speed = [(2×11 m/s)+(7×16 m/s)+(4×19 m/s)+(3×26 m/s)+(6×31 m/s)+(1×37 m/s)+(2×45 m/s)] / 25
= 601 / 25
= 24.04 m/s
Therefore, Average speed is 24.04 m/s
b) Determine the rms speed
we know that (rms speed)² = sum of square speed / total number of particles
so
(rms speed)² = [(2×11²)+(7×16²)+(4×19²)+(3×26²)+(6×31²)+(1×37²)+(2×45²)] / 25
(rms speed)² = 16691 / 25
(rms speed)² = 667.64
(rms speed) = √ 667.64
(rms speed) = 25.84 m/s
Therefore, the rms speed is 25.84 m/s
c) Determine the most probable speed.
Most particles (7) have velocity 16 m/s
i.e 7 is the maximum number of particle for a particular speed ,
Therefore, the most probable speed is 16 m/s
In an engine, an almost ideal gas is compressed adiabatically to half its volume. In doing so, 2820 J of work is done on the gas.
Required:
a. How much heat flows into or out of the gas?
b. What is the change in internal energy of the gas?
c. Does its temperature rise or fall?
Answer:
[tex]Q=0[/tex][tex]U=2820[/tex]Energy increasesExplanation:
From the question we are told that
Work done [tex]W=2820[/tex]
a)Generally the heat flow for an adiabatic process is 0 (zero)
[tex]Q= U + W =>0[/tex]
[tex]Q=0[/tex]
b)Generally Change in internal energy of gas is mathematically given by
Since [tex]W=-2820J[/tex]
Therefore
[tex]U=2820[/tex]
Giving
[tex]Q= 2820 -2820[/tex]
[tex]Q=O[/tex]
c)With increases in internal energy brings increase in temperature
Therefore
Energy increases
How to find average speed in physics
Answer: you divide total distance by time. To get the time, divide total distance by speed. To get distance, multiply speed times the amount of time.
Explanation:
I hope this helps
A piece of aluminum with a mass of 3.05 g initially at a temperature of 10.8 °C is heated to a temperature of 20.
Assume that the specific heat of aluminum is 0.901 J/(g°C).
How much heat was needed for this temperature change to take place?
Answer:
25.3J
Explanation:
Given parameters:
Mass of aluminum = 3.05g
Initial temperature = 10.8 °C
Final temperature = 20 °C
Specific heat = 0.9J/g °C
Unknown:
Amount of heat needed for the temperature to change = ?
Solution:
To solve this problem, we use the expression:
H = m C Ф
H is the amount of heat
m is the mass
C is the specific heat capacity
Ф is the change in temperature
H = 3.05 x 0.901 x (20 - 10.8) = 25.3J
Two metal bricks are held off the edge of a balcony from the same height above the ground. The bricks are the same size but one is made of Titanium (density of 4.5 g/cm%) and one is made of Lead (density of 11.3 g/cm3) so the Lead is about twice as heavy as the Titanium. The time it takes the bricks to reach the ground will be:________.
a. less but not necessarily half as long for the heavier brick
b. about half as long for the lighter brick
c. less but not necessarily half as long for the lighter brick
d. about half as long for the heavier brick
e. about the same time for both bricks
Answer:
e.
Explanation:
Assuming that the air resistance is neglectable, both bricks are only accelerated by gravity, which produces a constant acceleration on both bricks, which is the same, according Newton's 2nd Law, as we can see below:[tex]F_{g} = m*g = m*a (1)[/tex]⇒a = g = 9.8m/s² (pointing downward)Since acceleration is constant, if both fall from the same height, we can apply the following kinematic equation:[tex]\Delta y = v_{o} * t - \frac{1}{2} *g*t^{2} (2)[/tex]
Since both bricks are held off the edge, the initial speed is zero, so (2) reduces to the following equation:[tex]h =\frac{1}{2} *g*t^{2} (3)[/tex]
Since h (the height of the balcony) is the same, we conclude that both bricks hit ground at exactly the same time.If the air resistance is not negligible, due both bricks have zero initial speed, and have the same shape, they will be affected by the drag force in similar way, so they will reach the ground at approximately the same time.