Answer:
Sound and light are similar in that both are forms of energy that travel in waves. They both have properties of wavelength, freqency and amplitude. Here are some differences: Sound can only travel through a medium (substance) while light can travel through empty space.
Rank the following objects by their accelerations down an incline (assume each object rolls without slipping) from least to greatest:
a. Hollow Cylinder
b. Solid Cylinder
c. Hollow Sphere
d. Solid Sphere
Answer:
acceleration are
hollow cylinder < hollow sphere < solid cylinder < solid sphere
Explanation:
To answer this question, let's analyze the problem. Let's use conservation of energy
Starting point. Highest point
Em₀ = U = m g h
Final point. To get off the ramp
Em_f = K = ½ mv² + ½ I w²
notice that we include the kinetic energy of translation and rotation
energy is conserved
Em₀ = Em_f
mgh = ½ m v² +1/2 I w²
angular and linear velocity are related
v = w r
w = v / r
we substitute
mg h = ½ v² (m + I / r²)
v² = 2 gh [tex]\frac{m}{m+ \frac{I}{r^2} }[/tex]
v² = 2gh [tex]\frac{1}{1 + \frac{I}{m r^2} }[/tex]
this is the velocity at the bottom of the plane ,, indicate that it stops from rest, so we can use the kinematics relationship to find the acceleration in the axis ax (parallel to the plane)
v² = v₀² + 2 a L
where L is the length of the plane
v² = 2 a L
a = v² / 2L
we substitute
a = [tex]g \ \frac{h}{L} \ \frac{1}{1+ \frac{I}{m r^2 } }[/tex]
let's use trigonometry
sin θ = h / L
we substitute
a = g sin θ \ \frac{h}{L} \ \frac{1}{1+ \frac{I}{m r^2 } }
the moment of inertia of each object is tabulated, let's find the acceleration of each object
a) Hollow cylinder
I = m r²
we look for the acerleracion
a₁ = g sin θ [tex]\frac{1}{1 + \frac{mr^2 }{m r^2 } }[/tex]1/1 + mr² / mr² =
a₁ = g sin θ ½
b) solid cylinder
I = ½ m r²
a₂ = g sin θ [tex]\frac{1}{1 + \frac{1}{2} \frac{mr^2}{mr^2} }[/tex] = g sin θ [tex]\frac{1}{1+ \frac{1}{2} }[/tex]
a₂ = g sin θ ⅔
c) hollow sphere
I = 2/3 m r²
a₃ = g sin θ [tex]\frac{1}{1 + \frac{2}{3} }[/tex]
a₃ = g sin θ [tex]\frac{3}{5}[/tex]
d) solid sphere
I = 2/5 m r²
a₄ = g sin θ [tex]\frac{1 }{1 + \frac{2}{5} }[/tex]
a₄ = g sin θ [tex]\frac{5}{7}[/tex]
We already have all the accelerations, to facilitate the comparison let's place the fractions with the same denominator (the greatest common denominator is 210)
a) a₁ = g sin θ ½ = g sin θ [tex]\frac{105}{210}[/tex]
b) a₂ = g sinθ ⅔ = g sin θ [tex]\frac{140}{210}[/tex]
c) a₃ = g sin θ [tex]\frac{3}{5}[/tex]= g sin θ [tex]\frac{126}{210}[/tex]
d) a₄ = g sin θ [tex]\frac{5}{7}[/tex] = g sin θ [tex]\frac{150}{210}[/tex]
the order of acceleration from lower to higher is
a₁ <a₃ <a₂ <a₄
acceleration are
hollow cylinder < hollow sphere < solid cylinder < solid sphere
Determine the potential difference between the ends of the wire of resistance 5 Ω if 720 C passes through it per minute.
Answer:
The potential difference between the ends of a wire is 60 volts.
Explanation:
It is given that,
Resistance, R = 5 ohms
Charge, q = 720 C
Time, t = 1 min = 60 s
We know that the charge flowing per unit charge is called current in the circuit. It is given by :
I = 12 A
Let V is the potential difference between the ends of a wire. It can be calculated using Ohm's law as :
V = IR
V = 60 Volts
So, the potential difference between the ends of a wire is 60 volts. Hence, this is the required solution.
d. What is the net force on the bowling ball rolling lane
Answer:
Friction.
Explanation:
Light from the Sun is converted into
.... by plants.
Answer:
Explanation:
The answer is chemical energy
A 10Ω and a 15Ω resistor are connected in series across a 110V potential difference. (Can you find them) please help
A) what is the total resistance of the circuit?
B) what is the current through each resistor?
C) what is the voltage drop across each resistor
Answer:
(A) The total resistance of the circuit is 25 Ω
(B) The current through each resistor is 4.4 A
(C) For 10Ω: Potential drop = 44 V
For 15Ω: Potential drop = 66 V
Explanation:
Given;
potential difference, V = 110V
resistors in series, = 10Ω and a 15Ω
(A) The total resistance of the circuit is calculated as follows;
Rt = 10Ω + 15Ω = 25Ω
(B) The current through each resistor;
Same current will flow through the two resistors since they are in series.
I = V/Rt
I = 110 / 25
I = 4.4 A
(C) The voltage drop across each resistor;
For 10Ω: Potential drop = IR₁ = 4.4 x 10 = 44 V
For 15Ω: Potential drop = IR₂ = 4.4 x 15 = 66 V
The concept of photons applies to which regions of the electromagnetic spectrum?
A. visible light only
B. infrared light, visible light, and UV light only
C. X-rays and gamma rays only
D. all regions of the spectrum
Answer:
D. all regions of the spectrum
Explanation:
I did some research ; )
Lee and Leigh are twins. At their first birthday party, Lee is placed on a spaceship that travels away from the earth and back at a steady 0.714 c . The spaceship eventually returns, landing in the swimming pool at Leigh's eleventh birthday party. When Lee emerges from the ship, how old is he?
a. He is still only 1 year old
b. He is 8 years old
c. He is also 11 years old
d. He is 18 years old
Answer:
b. He is 8 years old
Explanation:
We will use Einstein's formula for time dilation, to calculate the age of Lee. Because Lee was traveling comparable to the speed of light, his age must be lesser than Leigh.
[tex]T = \frac{T_o}{\sqrt{1-\frac{v^2}{c^2} } }[/tex]
where,
T₀ = Time on Earth = ?
T = Relative Time = 10 years
v = relativistic speed of Lee = 0.714 c
c = speed of light = 3 x 10⁸ m/s
Therefore,
[tex]10\ years = \frac{T_o}{\sqrt{1-\frac{(0.714\ c)^2}{c^2} } } \\\\[/tex]
T₀ = 7 years
Hence, the age of Lee will be:
[tex]Lee's\ Age = 1\ year + 7\ years = 8\ years[/tex]
b. He is 8 years old
Particles q1 = +8.0 UC, 92 = +3.5 uc, and
q3 = -2.5 uC are in a line. Particles qi and q2 are
separated by 0.10 m and particles q2 and q3 are
separated by 0.15 m. What is the net force on
particle q2?
Answer:
-22.3
Explanation:
Acellus
The net force on particle q2 is 21.7 N.
The given parameters:
Charge on particle 1, q1 = 8 μCCharge on particle 2, q2 = 3.5 μCCharge on particle 3, q3 = -2.5 μCDistance between particle 1 and particle 2 = 0.1 mDistance between particle 2 and particle 3, = 0.15 mThe net force on particle q2 is calculated as follows;
[tex]Q_2_{net} = q_{12} \ + \ q_{23}\\\\Q_2_{net} = \frac{kq_1q_2}{r_{12} ^2} \ + \ \frac{kq_2q_3}{r_{23}^2} \\\\Q_2_{net} = \frac{9\times 10^9 \times 8 \times 10^{-6} \times 3.5 \times 10^{-6} }{0.1^2} \ + \ \frac{9 \times 10^9 b\times 3.5 \times 10^{-6} \times (-2.5) \times 10^{-6} }{0.15^2} \\\\Q_2_{net} = 25.2 \ N \ - \ 3.5 \ N\\\\Q_2_{net} = 21.7 \ N[/tex]
Thus, the net force on particle q2 is 21.7 N.
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What units must be used for mass in
the calculation of kinetic energy?
Answer:
The units of kinetic energy are mass times the square of speed, or kg⋅m2/s2 kg ⋅ m 2 /s 2. But the units of force are mass times acceleration, kg⋅m/s2 kg ⋅ m/s 2, so the units of kinetic energy are also the units of force times distance, which are the units of work
Explanation:
EDGE 2021
Consider two identical objects of mass m = 0.250 kg and charge q = 4.00 μC. The first charge is held in place at the origin of a coordinate system, unable to move at all times. The second object is initially placed 3.00 cm along the positive x-axis and is free to move. The moment the second object is released at x = 3.00 cm, what is the acceleration of this second object? This experiment is done far away from other massive objects, in outer space.
Answer:
a = 640 m/s²
Explanation:
From work-kinetic energy principles,
The net force acting on the second object is the gravitational force and the electric force due to the first object.
So, the gravitational force on the mass is F₁ = Gm₁m₂/r² since m₁ = m₂ = m, U = -Gm²/r²
Also, the electric force on the charge is F₂ = kq₁q₂/r² since q₁ = q₂ = q, U = kq²/r²
The net Force F = ma
So, -F₁ + F₂ = F (F₁ is negative since it is an attractive force in the negative x -direction and F₂ is positive since it is a repulsive force in the positive x- direction)
-Gm²/r² + kq²/r² = ma
ma = -Gm²/r² + kq²/r²
a = (-Gm²/r² + kq²/r²)/m
a = (-G + kq²/m²)m/r²
Since m = 0.250 kg, q = 4.00 μC = 4.00 × 10⁻⁶ C, r = 3.00 cm = 3.00 × 10⁻² m, G = 6.67 × 10⁻¹¹ Nm²/kg², k = 9 × 10⁹ Nm²/C² and a = acceleration of second mass.
Substituting the variables into the equation, we have
a = (m/r²)(-G + k(q/m)²)]
a = (0.250 kg/{3.00 × 10⁻² m}²)(-6.67 × 10⁻¹¹ Nm²/kg² + 9 × 10⁹ Nm²/C²(4.00 × 10⁻⁶ C/0.250 kg)²)
a = (0.250 kg/9.00 × 10⁻⁴ m)(-6.67 × 10⁻¹¹ Nm²/kg² + 9 × 10⁹ Nm²/C²(16 × 10⁻⁶ C/kg)²)]
a = (0.250 kg/9.00 × 10⁻⁴ m)(-6.67 × 10⁻¹¹ Nm²/kg² + 9 × 10⁹ Nm²/C²(256 × 10⁻¹² C²/kg²)]
a = (0.250 kg/9.00 × 10⁻⁴ m)(-6.67 × 10⁻¹¹ Nm²/kg² + 2304 × 10⁻³ Nm²/kg² ]
a = (0.250 kg/9.00 × 10⁻⁴ m)(2.304 Nm²/kg²)
a = 0.576 Nm²/kg /9.00 × 10⁻⁴ m²
a = 0.064 × 10⁴N/kg
a = 64 × 10 N/kg)
a = 640 m/s²
You desire to observe details of the Statue of Freedom, the sculpture by Thomas Crawford that is the crowning feature of the dome of the United States Capitol in Washington, D.C. For this purpose, you construct a refracting telescope, using as its objective a lens with focal length 86.3 cm. In order to acheive an angular magnification of magnitude 5.01, what focal length fe should the eyepiece have?
Answer:
the focal length of the eyepiece is 17.23 cm
Explanation:
The computation of the focal length of the eyepiece is shown below:
= Focal length of objective lens ÷ angular magnification magnitude
= 86.3 ÷ -5.01
= 17.23 cm
Hence, the focal length of the eyepiece is 17.23 cm
We simply divided the angular magnification magnitude from the focal length of objective lens so that the focal length of the eyepiece could come
Although the use of absorbances at 450 nm provided you with maximum sensitivity, the absorbances at, say, 400 nm or 500 nm are not zero and could have been used throughout this experiment. Would the same value of K be obtained at one of these wavelengths
Answer:
Yes, the value will be the same.
Explanation:
Yes, or at least to some degree, that value of K will remain the same. You're looking for a difference in absorbance, and the difference should be visible at all wavelengths, not only at the limit. That being said, resolution varies, and if we don't read the value to the maximum, we can get a less accurate reading.
As you look out of your dorm window, a flower pot suddenly falls past. The pot is visible for a time t, and the vertical length of your window is Lw. Take down to be the positive direction, so that downward velocities are positive and the acceleration due to gravity is the positive quantity g. Assume that the flower pot was dropped by someone on the floor above you (rather than thrown downward). If the bottom of your window is a height hb above the ground, what is the velocity vground of the pot as it hits the ground? You may introduce the new variable vb, the speed at the bottom of the window, defined by
vb = Lwt + gt2.
Answer:
[tex]\mathbf{v_{ground} = \sqrt{{v^2+2ghb}}}[/tex]
Explanation:
From the information given:
The avg. velocity post the window is;
[tex]v_{avg} = \dfrac{L_w}{t}[/tex]
[tex]v_b[/tex] = velocity located at the top of the window
[tex]v_b[/tex] = velocity situated at the bottom of the window
Using the equation of kinematics:
[tex]v_b = v_t + gt[/tex]
Hence,
[tex]v_t = v_b - gt[/tex]
To determine the average velocity as follows:
[tex]v_{avg} = \dfrac{1}{2} (v_t + v_b)\dfrac{L_w}{t}= \dfrac{1}{2}(v_b - gt +v_b) \\ \\\dfrac{L_w}{t} = v_b - \dfrac{1}{2}gt \\ \\ v_b = \dfrac{L_w}{t }+ \dfrac{1}{2} gt\\ \\ = \dfrac{1}{t} \Bigg(L_w + \dfrac{1}{2}gt^2 \Bigg) \\ \\[/tex]
where;
[tex]v_b[/tex] = velocity gained when fallen through the height h.
Similarly, using the equation of kinematics, we have;
[tex]v_b^2 = 2gh \\ \\h = \dfrac{v_b^2}{2g}[/tex]
[tex]\implies \dfrac{(L_w + \dfrac{1}{2} gt^2_^2}{2gt^2}[/tex]
Thus, the velocity at the ground is;
[tex]v^2_{grround} = v_b^2 + 2ghb[/tex]
[tex]\mathbf{v_{ground} = \sqrt{{v^2+2ghb}}}[/tex]
What has to happen for a star to join the main sequence ?
1) Nuclear Fusion
2) Shell Heating
3) Use up most of its available fuel
4) Hydrostatic Equilibrium
Answer:
Nuclear fusion
Explanation:
This is because main sequence of star is powered by fusion of hydrogen to helium atoms together and this process releases energy. The energy released when the gas collapse into a protostar make the center of the protostar to be extremely hot. When the core becomes very hot, nuclear fusion can start.
Two balloons become equally charged once they are rubbed against each other. If the force between the balloons is 6.2 * 10^23 N, what would happen to the force if the charge were to triple on one of the balloons?
A) the force would triple
B) the force would become one-nineth
if the dissipated power between a and b equal 210 watt then VB equal
Answer:
correct answer is A
Explanation:
The diagram shows a series circuit with three resistors and two power sources.
In a series circuit the current through the entire circuit is constant and the resistance is the sum of the resistances in the circuit.
When the power sources are placed in opposite position the voltage between them is subtracted.
V_b - 30 = I (10 + 4 + 6)
V_b = I (20) - 30
V_b = 30 - 2 20
V_b = 10 V
the correct answer is A
Ac soucre change it polarity how many time
After your school's team wins the regional championship, students go to the dorm roof and start setting off fireworks rockets. The rockets explode high in the air and the sound travels out uniformly in all directions. If the sound intensity is 1.67 10-6 W/m2 at a distance of 233 m from the explosion, at what distance from the explosion is the sound intensity half this value
Answer:
the distance is 315.3696 m
Explanation:
The computation of the distance is given below:
Given that
Sound intensity = 1.67 × 10^-6 W/m^2
And, the distance = 233 m
Now as we know that
Power = Intensity × surface area
1.67 × 10^-6 × 4π(233)^2 = 1.67 × 10^-6 ÷ 2× 4π × d^2
d^2 = 2 × (223)^2
= √2 × 223
= 315.3696 m
Hence, the distance is 315.3696 m
For an adiabatic process, the change in T is determined by the change in V. In this problem you will compute the contributions to S from the V and T terms separately, then add them up to find the total entropy change for an adiabatic process. Argon gas, initially at pressure 100 kPa and temperature 300 K, is allowed to expand adiabatically from 0.01 m3 to 0.026 m3 while doing work on a piston.
This question is incomplete, the complete question is;
The entropy of an ?-ideal gas changes in the following way as a function of temperature and volume:
ΔS = nRln(V[tex]_f[/tex]/V[tex]_i[/tex]) + ∝nRln(T[tex]_f[/tex]/T[tex]_i[/tex])
For an adiabatic process, the change in T is determined by the change in V. In this problem you will compute the contributions to S from the V and T terms separately, then add them up to find the total entropy change for an adiabatic process.
Argon gas, initially at pressure 100 kPa and temperature 300 K, is allowed to expand adiabatically from 0.01 m³ to 0.026 m³ while doing work on a piston.
1) What is the change in entropy due to the volume change alone, ignoring any effects of changing internal energy? ΔS = ? J/K
2) For this adiabatic expansion, what is the final temperature? T[tex]_f[/tex] = ? K
Answer:
1) the change in entropy due to the volume change alone, ignoring any effects of changing internal energy is 3.185 J/K.
2) the final temperature is 158.66 K
Explanation:
Given the data in the question;
ΔS = nRln(V[tex]_f[/tex]/V[tex]_i[/tex]) + ∝nRln(T[tex]_f[/tex]/T[tex]_i[/tex])
P[tex]_i[/tex] = 100 kPa = 100000 Pa
V[tex]_i[/tex] = 0.01 m³
V[tex]_f[/tex] = 0.026 m³
T[tex]_i[/tex] = 300 K
1) the change in entropy due to the volume change alone
from the question; ΔS = nRln(V[tex]_f[/tex]/V[tex]_i[/tex]) + ∝nRln(T[tex]_f[/tex]/T[tex]_i[/tex])
so change in entropy due to the volume change alone is;
ΔS = nRln(V[tex]_f[/tex]/V[tex]_i[/tex])
we know that, from ideal gas law; PV = nRT
so, nR = P[tex]_i[/tex]V[tex]_i[/tex]/T[tex]_i[/tex] ---- let this be equation 1
∴ ΔS = P[tex]_i[/tex]V[tex]_i[/tex]/T[tex]_i[/tex] × ln(V[tex]_f[/tex]/V[tex]_i[/tex])
we substitute
ΔS = [( 100000 Pa × 0.01 m³) / 300 K ] × ln(0.026m³ / 0.01m³ )
ΔS = 3.185 J/K
Therefore, the change in entropy due to the volume change alone, ignoring any effects of changing internal energy is 3.185 J/K.
2) Final temperature
we know that, in an adiabatic expansion;
[tex]PV^Y[/tex] = K
where Y = 5/3
so
[tex]P_i[/tex][tex]V_i^{(5/3)[/tex] = [tex]P_f[/tex][tex]V_f^{(5/3)[/tex]
[tex]P_f[/tex] = [tex]P_i[/tex][tex]( \frac{V_i}{V_f})^{(5/3)[/tex]
we substitute
[tex]P_f[/tex] = ( 100000 Pa) [tex]( \frac{0.01 m^3}{0.026 m^3})^{(5/3)[/tex]
[tex]P_f[/tex] = 20341.255 Pa
Also from ideal gas law;
PV = nRT
T = PV / nR
so
T[tex]_f[/tex] = P[tex]_f[/tex]V[tex]_f[/tex] / nR
but from equation 1; nR = PV/T
so
T[tex]_f[/tex] = (P[tex]_f[/tex]V[tex]_f[/tex]) / (P[tex]_i[/tex]V[tex]_i[/tex]/T[tex]_i[/tex] )
T[tex]_f[/tex] = ( P[tex]_f[/tex]V[tex]_f[/tex]T[tex]_i[/tex] / P[tex]_i[/tex]V[tex]_i[/tex] )
we substitute
T[tex]_f[/tex] = ( 20341.255 Pa × 0.026 m³ × 300 K) / 100000 Pa × 0.01 m³ )
T[tex]_f[/tex] = 158.66 K
Therefore, the final temperature is 158.66 K
A pendulum is constructed from a heavy metal rod and a metal disk, both of uniform mass density. The center of the disk is bolted to one end of the rod, and the pendulum hangs from the other end of the rod. The rod has a mass of =1.0 kg and a length of =49.8 cm. The disk has a mass of =4.0 kg and a radius of =24.9 cm. The acceleration due to gravity is =9.8 m/s2.
The pendulum is held with the rod horizontal and then released. What is the magnitude of its angular acceleration at the moment of release?
The magnitude of the angular acceleration of the pendulum at the moment of release is; α = 18.45 rad/s²
We are given;
Mass of rod; m = 1 kg
Length of rod; L = 49.8 cm = 0.498 m
Mass of Disk; M = 4 kg
Radius of disk; r = 24.9 cm = 0.249 m
Let us first calculate the torque acting from the formula;
τ = mg(L/2) + MgL
Thus;
τ = (1 × 9.8 × (0.498/2)) + (4 × 9.8 × 0.498)
τ = 21.96 N.m
Using parallel axis theorem, we can find the moment of inertia about the given axis as;
I = (mL²/3) + ½MR² + ML²
Plugging in the relevant values gives;
I = (1 * 0.498²/3) + ½(4 * 0.249²) + (4 * 0.498²)
I = 1.19 kg.m²
The angular acceleration is given by the formula;
α = I/τ
α = 21.96/1.19
α = 18.45 rad/s²
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Two in-phase loudspeakers that emit sound with the same frequency are placed along a wall and are separated by a distance of 8.00 m. A person is standing 12.0 m away from the wall, equidistant from the loudspeakers. When the person moves 3.00 m parallel to the wall, she experiences destructive interference for the second time. What is the frequency of the sound
Answer: [tex]278\ Hz[/tex]
Explanation:
Given
Distance between two speakers is 8 m
Man is standing 12 away from the wall
When the person moves 3 parallel to the wall
the parallel distances from the speaker become 4+3, 4-3
Now, the difference of distances from the speaker is
[tex]\Delta d=\sqrt{12^2+(4+3)^2}-\sqrt{12^2-(4-3)^2}\\\Delta d=1.85\ m[/tex]
Condition for destructive interference is
[tex]\Delta d=(2n-1)\dfrac{\lambda }{2}=(2n-1)\dfrac{\nu }{2f}\\\\\Rightarrow f=(2n-1)\dfrac{v}{2\Delta d}[/tex]
for second destructive interference; n=2
[tex]\Rightarrow f=(2\times 2-1)\dfrac{343}{2\times 1.85}=278.10\approx 278\ Hz[/tex]
Franny drew a diagram to compare images produced by concave and convex lenses.
2 overlapping circles, the left circle labeled Concave lenses and the right circle labeled Convex lenses. An X in the overlap.
Which belongs in the area marked X?
Answer:
Virtual
Explanation:
Answer:
B. Virtual
Good Luck!
form
bonds with each other.
There are many kinds of mixtures. Some mixtures are
chunky like a mixture of peanuts and raisins. These
mixtures are called
I
mixtures.
Answer:
Homogeneous mixtures
Explanation:
I think so because homogeneous means mixed mixtures
A 3.00 x 10^2-W electric immersion heater is
used to heat a cup of water. The cup is made
of glass and its mass is 3.00 10^2 g. It con-
tains 250 g of water at 15° C. How much time
is needed for the heater to bring the water to
the boiling point? Assume the temperature of
the cup to be the same as the temperature of
the water at all times and no heat is lost to
the air.
Answer
t = 367.77 s = 6.13 min
Explanation:
According to the law of conservation of energy:
[tex]Heat\ Supplied\ By \ Heater = Heat\ Absorbed\ by\ Glass + Heat\ Absorbed\ by\ Water\\Pt = m_gC_g\Delta T_g + m_wC_w\Delta T_w\\[/tex]
where,
P = Electric Power of Heater = 300 W
t = time required = ?
m_g = mass of glass = 300 g = 0.3 kg
m_w = mass of water = 250 g = 0.25 kg
C_g = speicific heat of glass = 840 J/kg.°C
C_w = specific heatof water = 4184 J/kg.°C
ΔT_g = ΔT_w = Change in Temperature of Glass and water = 100°C - 15°C
ΔT_g = ΔT_w = 85°C
Therefore,
[tex](300\ W)(t) = (0.3\ kg)(840\ J/kg.^oC)(85^oC)+(0.25\ kg)(4184\ J/kg.^oC)(85^oC)\\[/tex]
t = 367.77 s = 6.13 min
During a soccer game, a player grabs and holds an opponent's shirt outside of the penalty box. After the foul is called, what kick is awarded to put the ball back into play?
a
Penalty Kick
b
Indirect Free Kick
c
Kickoff
d
Direct Free Kick
which letter represents the way the wave is moving?
Wouldn't it be B because it's a majority pointing to it?
Sorry if i'm wrong.
What are the benefits when you engage in physical fitness?
Answer:
manage your weight better, have stronger bones, have lower blood pressure, less risk of a heart attack, etc.
Answer:
You become healthier, your body starts regulating better, you get stronger bones and muscles, and you lower the risk of diabetes,heart problems and other diseases.
The unit called the____
is defined based on the amount of
work a horse can do in 1 minute.
The unit called horsepower is defined based on the amount of work a horse can do in 1 minute.
What is horsepower?James Watt used the horsepower unit for the first time in 1782. According to the account, James Watt used a pony to charge coal from the mines. According to the narrative, he needed a unit to measure the force of one of these animals.
He discovers that they can move 22.000 lbs per minute and decides to (arbitrarily) boost this metric by 50%. Having been the unit 33.000 lb/ft per minute, horsepower is an engine's output horsepower rating, whereas brake horsepower is an engine's input brake horsepower.
Brake horsepower measures an engine's power without accounting for power losses, whereas HP accounts for power losses. Horsepower of the brakes
Therefore, the horsepower unit is defined by the amount of work a horse can do in one minute.
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What is the average speed of the bicyclist's ride?
A.45m/s
B.7.5m/s
C45mi/hr
D.7.5mi/hr
If the length of the standing wave below is 2 meters, what is the wavelength of the standing
wave? *
Answer:
fffffgggggggggggggghhh