Answer:
The rotational kinetic energy is [tex]K = 2116.3 \ J[/tex]
Explanation:
From the question we are told that
The moment of inertia is [tex]I = 0.35 \ kg \cdot m^2[/tex]
The number of revolution is N = 70 revolution
The time taken is t = 4.0 s
Generally the angular velocity is mathematically represented as
[tex]w = \frac{2 \pi N }{t }[/tex]
substituting values
[tex]w = \frac{2* 3.142 * 70 }{4 }[/tex]
[tex]w = 109.97 \ rad/s[/tex]
The rotational kinetic energy K i mathematically represented as
[tex]K = \frac{1}{ 2} * I * w^2[/tex]
substituting values
[tex]K = \frac{1}{ 2} * 0.35 * (109.97)^2[/tex]
[tex]K = 2116.3 \ J[/tex]
light bulb is connected to a 110-V source. What is the resistance of this bulb if it is a 100-W bulb
Answer:
121ohmsExplanation:
Formula used for calculating power P = current * voltage
P = IV
From ohms law, V = IR where R is the resistance. Substituting V = IR into the formula for calculating power, we will have;
P = IV
P =(V/R)V
P = V²/R
Given parameters
Power rating of the bulb P = 100 Watts
Source voltage V = 110V
Required
Resistance of the bulb R
Substituting the given parameters into the formula for calculating power to get Resistance R;
P = V²/R
100 = 110²/R
R = 110²/100
R = 110 * 110/100
R = 12100/100
R = 121 ohms
Hence, the resistance of this bulb is 121 ohms
Water pressurized to 3.5 x 105 Pa is flowing at 5.0 m/s in a horizontal pipe which contracts to 1/3 its former area. What are the pressure and velocity of the water after the contraction
Answer:
the pressure after contraction is 2×10^5 Pa
the speed after contraction is 15m/s
Explanation:
We were given Pressure P to be 3.5 x 10^5 that is Flowing with speed of 5.0 m/s,
For us to calculate pressure we need to calculate the area first as;
Let initial Area = A₁
And Final area A₂
We were told that in a horizontal pipe it contracts to 1/3 its former area. Which means
A₂= A₁/3.................
V₁ is the speed
the pressure and speed of the water after the contraction can be calculated using equation of continuity below
A₂V₂ = A₁V₁
But
If we substitute given value in the expresion we have
V₂ = (3A *5)/A
V₂ = 15m/s
Therefore, the speed after contraction is 15m/s
Now we can calculate the pressure using
Bernoulli's equation
p₁ + ½ρv₁² + ρgh₁ = p₂ + ½ρv₂² + ρgh₂
But we know that the pipe is horizontal, then "h" terms cancel out then
p₁ + ½ρv₁² = p₂ + ½ρv₂²
Making P₂ subject of formula we have
p₂ = 0.5ρ( V ₁² - v₂² ) + P₁
P₂=. 0.5 × 1000 (5² -15² ) + 3*10^5
=2×10^5 Pa
Therefore, the pressure after contraction is 2×10^5 Pa
(a) the final speed of the water after contraction is 15 m/s.
(b) The final pressure of the water after contraction is 2.5 x 10⁵ Pa.
The given parameters;
initial pressure, P₁ = 3.5 x 10⁵ Painitial speed, v₁ = 5 m/sdensity of water, ρ = 1000 kg/m³Let the initial area of the pipe = A₁
Apply the continuity equation to determine the final speed of the water after contraction as follows;
[tex]A_1 V_1 = A_2 V_2\\\\V_2 = \frac{A_1V_1}{A_2} \\\\V_2 = \frac{A_1 \times 5}{\frac{1}{3} A_1 } \\\\V_2 = 15 \ m/s[/tex]
The final pressure of the water after contraction is determined by applying Bernoulli's equation for horizontal pipe;
[tex]P_1 + \frac{1}{2} \rho V_1^2= P_2 + \frac{1}{2} \rho V_2^2\\\\P_2 = \frac{1}{2} \rho (V_1^2 - V_2^2) + P_1\\\\P_2 = \frac{1}{2} \times 1000(5^2 - 15^2) + 3.5 \times 10^5\\\\P_2 = 2.5 \times 10^5 \ Pa[/tex]
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A baseball (m=145g) traveling 35 m/s moves a fielder's glove backward 23 cm when the ball is caught. What was the average force exerted by the ball on the glove?
Answer:
386.13 N
Explanation:
The kinetic energy of the baseball is converted into workdone in moving the glove backward( work energy theorem).
Therefore, KE of the ball
[tex]\frac{1}{2} mv^2 =\frac{1}{2}(0.145)35^2\\ = 88.81 \text{J}[/tex]
Now, workdone in moving the glove
W= Fd
where F = Force applied, d = displacement of the glove= 0.23 cm.
88.81 = F×0.23
F= 88.81/0.23 = 386.13 N
Consider a series RLC circuit where R=25.0 Ω, C=35.5 μF, and L=0.0940 H, that is driven at a frequency of 70.0 Hz. Determine the phase angle ϕ of the circuit in degrees.
Answer:
137.69°Explanation:
The phase angle of an RLC circuit ϕ is expressed as shoen below;
ϕ = [tex]tan^{-1} \dfrac{X_l-X_c}{R}[/tex]
Xc is the capacitive reactance = 1/2πfC
Xl is the inductive reactance = 2πfL
R is the resistance = 25.0Ω
Given C = 35.5 μF, L = 0.0940 H, and frequency f = 70.0Hz
Xl = 2π * 70*0.0940
Xl = 41.32Ω
For the capacitive reactance;
Xc = 1/2π * 70*35.5*10⁻⁶
Xc = 1/0.0156058
Xc = 64.08Ω
Phase angle ϕ = [tex]tan^{-1} \frac{41.32-64.08}{25} \\\\[/tex]
ϕ = [tex]tan^{-1} \frac{-22.76}{25} \\\\\\\\[/tex]
[tex]\phi = tan^{-1} -0.9104\\\\\phi = -42.31^0[/tex]
Since tan is negative in the 2nd quadrant;
[tex]\phi = 180-42.31^0\\\\\phi = 137.69^0[/tex]
Hence the phase angle ϕ of the circuit in degrees is 137.69°
The phase angle ϕ of the series RLC circuit that is driven at a frequency of 70.0 Hz is ϕ = 137.69°
Phase angle:Given that:
capacitance C = 35.5 μF,
Inductance L = 0.0940 H,
The resistance R = 25.0Ω
and frequency f = 70.0Hz
The capacitive reactance is given by:
Xc = 1/2πfC
Xc = 1/2π × 70 × 35.5× 10⁻⁶
Xc = 1/0.0156058
Xc = 64.08Ω
The inductive reactance is given by:
Xl = 2πfL
Xl = 2π × 70 × 0.0940
Xl = 41.32Ω
The phase angle of an RLC circuit ϕ is given by:
[tex]\phi=tan^{-1}\frac{X_l-X_c}{R}\\\\\phi=tan^{-1}\frac{41.32-64.08}{25}[/tex]
Ф = -42.31°
Since tan is negative in the 2nd quadrant, thus:
ϕ = 180° - 42.31°
ϕ = 137.69°
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An electric train operates on 800 V. What is its power consumption when the current flowing through the train's motor is 2,130 A?
Answer:
1704 kWExplanation:
To solve for the power consumed by the trains motor we have to employ the formula for power which is
Power= current * voltage
Given that
voltage V= 800 V
current I= 2130 A
Substituting in the formula for power we have
Power= 2130*800= 1704000 watt
Power = 1704 kW
This is the amount of energy consumed, transferred or converted per unit of time
Hence the power consumed by the trains motor is 1704 kW
An electric device delivers a current of 5.0 A to a circuit. How many electrons flow through this circuit in 5 s?
Answer:
1.6×10²⁰
Explanation:
An ampere is a Coulomb per second.
1 A = 1 C / s
The amount of charge after 5 seconds is:
5.0 A × 5 s = 25 C
The number of electrons is:
25 C × (1 electron / 1.6×10⁻¹⁹ C) = 1.6×10²⁰ electrons
15. Food chain always start with
a. Photosynthesis
Decay
b. Respiration
d. N2 Fixation
C.Photosynthesis
Convert 76.2 kilometers to meters?
Answer
76200meters
Explanation:
we know that 1km=1000meters
to convert km into meters we we divide km by meters
=76.2/1000
=76200meters
A 58 g firecracker is at rest at the origin when it explodes into three pieces. The first, with mass 12 g , moves along the x axis at 37 m/s in the positive direction. The second, with mass 22 g , moves along the y axis at 34 m/s in the positive direction. Find the velocity of third piece.
Answer:
Explanation:
We shall apply conservation of momentum law in vector form to solve the problem .
Initial momentum = 0
momentum of 12 g piece
= .012 x 37 i since it moves along x axis .
= .444 i
momentum of 22 g
= .022 x 34 j
= .748 j
Let momentum of third piece = p
total momentum
= p + .444 i + .748 j
so
applying conservation law of momentum
p + .444 i + .748 j = 0
p = - .444 i - .748 j
magnitude of p
= √ ( .444² + .748² )
= .87 kg m /s
mass of third piece = 58 - ( 12 + 22 )
= 24 g = .024 kg
if v be its velocity
.024 v = .87
v = 36.25 m / s .
A professor, with dumbbells in his hands and holding his arms out, is spinning on a turntable with an angular velocity. What happens after he pulls his arms inwards
Answer:
His angular velocity will increase.
Explanation:
According to the conservation of rotational momentum, the initial angular momentum of a system must be equal to the final angular momentum of the system.
The angular momentum of a system = [tex]I[/tex]'ω'
where
[tex]I[/tex]' is the initial rotational inertia
ω' is the initial angular velocity
the rotational inertia = [tex]mr'^{2}[/tex]
where m is the mass of the system
and r' is the initial radius of rotation
Note that the professor does not change his position about the axis of rotation, so we are working relative to the dumbbells.
we can see that with the mass of the dumbbells remaining constant, if we reduce the radius of rotation of the dumbbells to r, the rotational inertia will reduce to [tex]I[/tex].
From
[tex]I[/tex]'ω' = [tex]I[/tex]ω
since [tex]I[/tex] is now reduced, ω will be greater than ω'
therefore, the angular velocity increases.
A windmill on a farm rotates at a constant speed and completes one-half of a rotation in 0.5 seconds. What is its rotation speed
Answer:
v = 6.28 m/s
Explanation:
It is given that,
A windmill on a farm rotates at a constant speed and completes one-half of a rotation in 0.5 seconds,
Number of revolution is half. It means angular velocity is 3.14 radians.
Let v is the angular speed. So,
[tex]v=\dfrac{\omega}{t}\\\\v=\dfrac{3.14}{0.5}\\\\v=6.28\ m/s[/tex]
So, the rotation speed is 6.28 m/s.
The angular velocity is the rotation speed, which is the angle of rotation
of the windmill per second, which is 2·π radians.
Response:
The rotation speed is 2·π rad/sHow can the rotational speed of the windmill be calculated?The given parameter are;
The angle of rotation the windmill rotates in 0.5 seconds = One-half a
rotation.
Required:
The rotational speed (angular velocity)
Solution:
The angle of one rotation = 2·π radians
Angle of one-half ration = [tex]\frac{1}{2}[/tex] × 2·π radians = π radians
[tex]Rotational \ speed = \mathbf{\dfrac{Angle \ of \ rotation}{Time}}[/tex]
Which gives;
[tex]Rotational \ speed, \omega = \dfrac{\pi}{0.5 \ s} = \mathbf{2 \cdot \pi \ rad/s}[/tex]
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A red card is illuminated by red light. Part A What color will the card appear? What color will the card appear? a. Red b. Black c. White d. Green
The color that is reflected when a red card is illuminated by red light is white.
The color an object is perceived to have, depends on the frequency of light it reflects.
If white light incidents on a red filter, red is transmitted while blue and green are absorbed.
Consequently, when a red card is illuminated by red light, the red card will reflect back almost all the incident light on it, causing it to appear brighter which creates an illusion of white color to the eyes.
Thus, we can conclude the color that is reflected when a red card is illuminated by red light is white.
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front wheel drive car starts from rest and accelerates to the right. Knowing that the tires do not slip on the road, what is the direction of the friction force the road applies to the rear tire
Answer:
The frictional force the road applies to the rear tire is static friction and it acts opposite to the direction in which the car is traveling.
Explanation:
This question suggests that the car is accelerating forward. Thus, the easiest way for us to know what friction is doing is for us to know what happens when we turn friction off.
Now, if there is no friction and the car is stopped, if we push down on the accelerator, it will make the front wheels to spin in a clockwise manner. This spin occurs on the frictionless surface with the rear wheels doing nothing while the car doesn't move.
Now, if we apply friction to just the front wheels, the car will accelerate forward while the back wheels would be dragging along the road and not be spinning. Thus, friction opposes the motion and as such, it must act im a direction opposite to where the car is going. This must be static friction.
The frictional force the road applies to the rear tire is static friction and it acts opposite to the direction in which the car is traveling.
You have a horizontal grindstone (a disk) that is 95 kg, has a 0.38 m radius, is turning at 87 rpm (in the positive direction), and you press a steel axe against the edge with a force of 16 N in the radial direction.
(a) Assuming the kinetic coefficient of friction between steel and stone is 0.20, calculate the angular acceleration of the grindstone.
(b) How many turns will the stone make before coming to rest?
Answer:
Explanation:
The moment of inertia of the disk I = 1/2 m R² where R is radius of the disc and m is its mass .
putting the values
I = .5 x 95 x .38²
= 6.86 kg m²
n = 87 rpm = 87 / 60 rps
n = 1.45 rps
angular velocity ω = 2π n , n is frequency of rotation .
= 2 x 3.14 x 1.45
= 9.106 radian /s
frictional force = 16 x .2
= 3.2 N
torque created by frictional force = 3.2 x .38
= 1.216 N.m
angular acceleration = torque / moment of inertia
= - 3.2 / 6.86
α = - 0.4665 rad /s²
b ) ω² = ω₀² + 2 α θ , where α is angular acceleration
0 = 9.106² - 2 x .4665 θ
θ = 88.87 radian
no of turns = 88.87 / 2π
= 14.15 turns
Which of the following explains why a “control” is important in a case-control study of a disease? The researchers need to control the bias that those who contracted the disease may create when they talk to others. The researchers need to compare those who contracted the disease to those who did not. The researchers need to compare those who contracted the disease to those who contracted previous diseases. The researchers need to control the disease so that it is not spread further.
The researchers need to compare those who contracted the disease to those who did not.
A fish is 80 cm below the surface of a pond. What is the apparent depth (in cm) when viewed from a position almost directly above the fish
Answer:
Apparent depth (Da) = 60.15 cm (Approx)
Explanation:
Given:
Distance from fish (D) = 80 cm
Find:
Apparent depth (Da)
Computation:
We know that,
Refractive index of water (n2) = 1.33
So,
Apparent depth (Da) = D(n1/n2)
Apparent depth (Da) = 80 (1/1.33)
Apparent depth (Da) = 60.15 cm (Approx)
The apparent depth of the fish is 60 cm.
To calculate the apparent depth of the fish, we use the formula below.
Formula:
R.F(water) = Real depth(D)/Apparent depth(D')R.F = D/D'.................... Equation 1Where:
R.F = Refractive index of waterMake D' The subject of the equation.
D' = D/R.F................... Equation 2From the question,
Given:
D = 80 cmR.F = 1.333Substitute these values into equation 2
D' = 80/1.33D' = 60.01D' = 60 cmHence, the apparent depth of the fish is 60 cm
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A student holds a bike wheel and starts it spinning with an initial angular speed of 7.0 rotations per second. The wheel is subject to some friction, so it gradually slows down.
In the 10.0 s period following the inital spin, the bike wheel undergoes 60.0 complete rotations. Assuming the frictional torque remains constant, how much more time Δ????s will it take the bike wheel to come to a complete stop?
The bike wheel has a mass of 0.625 kg0.625 kg and a radius of 0.315 m0.315 m. If all the mass of the wheel is assumed to be located on the rim, find the magnitude of the frictional torque ????fτf that was acting on the spinning wheel.
Answer:
a) Δt = 24.96 s , b) τ = 0.078 N m
Explanation:
This is a rotational kinematics exercise
θ = w₀ t - ½ α t²
Let's reduce the magnitudes the SI system
θ = 60 rev (2π rad / 1 rev) = 376.99 rad
w₀ = 7.0 rot / s (2π rad / 1 rpt) = 43.98 rad / s
α = (w₀ t - θ) 2 / t²
let's calculate the annular acceleration
α = (43.98 10 - 376.99) 2/10²
α = 1,258 rad / s²
Let's find the time it takes to reach zero angular velocity (w = 0)
w = w₀ - alf t
t = (w₀ - 0) / α
t = 43.98 / 1.258
t = 34.96 s
this is the total time, the time remaining is
Δt = t-10
Δt = 24.96 s
To find the braking torque, we use Newton's law for angular motion
τ = I α
the moment of inertia of a circular ring is
I = M r²
we substitute
τ = M r² α
we calculate
τ = 0.625 0.315² 1.258
τ = 0.078 N m
The total time taken by the wheel to come to rest is 25.18 s and the magnitude of the frictional torque is 25.18 N-m.
Given data:
The initial angular speed of wheel is, [tex]\omega = 7.0 \;\rm rps[/tex] (rps means rotation per second).
The time interval is, t' = 10.0 s.
The number of rotations made by wheel is, n = 60.0.
The mass of bike wheel is, m = 0.625 kg.
The radius of wheel is, r = 0.315 m.
The problem is based on rotational kinematics. So, apply the second rotational equation of motion as,
[tex]\theta = \omega t-\dfrac{1}{2} \alpha t'^{2}[/tex]
Here, [tex]\theta[/tex] is the angular displacement, and its value is,
[tex]\theta =2\pi \times 60\\\\\theta = 376.99 \;\rm rad[/tex]
And, angular speed is,
[tex]\omega = 2\pi n\\\omega = 2\pi \times 7\\\omega = 43.98 \;\rm rad/s[/tex]
Solving as,
[tex]376.99 = 43.98 \times 10-\dfrac{1}{2} \alpha \times 10^{2}\\\\\alpha = 1.25 \;\rm rad/s^{2}[/tex]
Apply the first rotational equation of motion to obtain the value of time to reach zero final velocity.
[tex]\omega' = \omega - \alpha t\\\\0 = 43.98 - 1.25 \times t\\\\t = 35.18 \;\rm s[/tex]
Then total time is,
T = t - t'
T = 35.18 - 10
T = 25.18 s
Now, use the standard formula to obtain the value of braking torque as,
[tex]T = m r^{2} \alpha\\\\T = 0.625 \times (0.315)^{2} \times 1.25\\\\T = 0.0775 \;\rm Nm[/tex]
Thus, we can conclude that the total time taken by the wheel to come to rest is 25.18 s and the magnitude of the frictional torque is 25.18 N-m.
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Which of these cannot be a resistor in a series or parallel circuit?
A)switch
B) battery
C) light bulb
D) all of these are resistors
Answer:
it is going to D. all of these are resistors
A 2.0 m × 4.0 m flat carpet acquires a uniformly distributed charge of −10 μC after you and your friends walk across it several times. A 5.0 μg dust particle is suspended in midair just above the center of the carpet.
Required:
What is the charge on the dust particle?
Answer:
The charge on the dust particle is [tex]q_d = 6.94 *10^{-13} \ C[/tex]
Explanation:
From the question we are told that
The length is [tex]l = 2.0 \ m[/tex]
The width is [tex]w = 4.0 \ m[/tex]
The charge is [tex]q = -10\mu C= -10*10^{-6} \ C[/tex]
The mass suspended in mid-air is [tex]m_a = 5.0 \mu g = 5.0 *10^{-6} \ g = 5.0 *10^{-9} \ kg[/tex]
Generally the electric field on the carpet is mathematically represented as
[tex]E = \frac{q}{ 2 * A * \epsilon _o}[/tex]
Where [tex]\epsilon _o[/tex] is the permittivity of free space with value [tex]\epsilon_o = 8.85*10^{-12} \ \ m^{-3} \cdot kg^{-1}\cdot s^4 \cdot A^2[/tex]
substituting values
[tex]E = \frac{-10*10^{-6}}{ 2 * (2 * 4 ) * 8.85*10^{-12}}[/tex]
[tex]E = -70621.5 \ N/C[/tex]
Generally the electric force keeping the dust particle on the air equal to the force of gravity acting on the particles
[tex]F__{E}} = F__{G}}[/tex]
=> [tex]q_d * E = m * g[/tex]
=> [tex]q_d = \frac{m * g}{E}[/tex]
=> [tex]q_d = \frac{5.0 *10^{-9} * 9.8}{70621.5}[/tex]
=> [tex]q_d = 6.94 *10^{-13} \ C[/tex]
At what minimum angle will you get total internal reflection of light traveling in diamond and reflected from ethanol? °
Answer:
34°
Using the relation
θᶜ = sin^-1(n₂/n₁),
where n1= the refractive index of light is propagating from a medium
And n2 = refractive index of medium into which light is entering
So we know that
refractive index of diamond at 589nm = 2.41= n₁
refractive index of ethanol at 589nm and 20°C = 1.36= n₂
Thus. θᶜ = sin^-1(1.361/2.417) = 0.58radians = 34°
Explanation:
A wire is carrying current vertically downward. What is the direction of the force due to Earth's magnetic field on the wire?
Answer:
The direction of the force will be towards the east
Explanation:
From the question we are told that
The direction of the downward
Generally according to Fleming's right-hand rule(
Thumb - direction of force
Middle finger - direction of current
Index finger - direction of the magnetic field
) and the fact that the earth magnetic field acts from south to north with respect to the four cardinal points then the direction of the force will be toward the east with respect to the four cardinal point on the earth
A resistance heater having 20.7 kW power is used to heat a room having 16 m X 16.5 m X 12.3 m size from 13.5 to 21 oC at sea level. The room is sealed once the heater is turned on. Calculate the amount of time needed for this heating to occur in min. (Write your answer in 3 significant digits. Assume constant specific heats at room temperature.)
Answer:
t = 23.6 min
Explanation:
First we need to find the mass of air in the room:
m = ρV
where,
m = mass of air in the room = ?
ρ = density of air at room temperature = 1.2041 kg/m³
V = Volume of room = 16 m x 16.5 m x 12.3 m = 3247.2 m³
Therefore,
m = (1.2041 kg/m³)(3247.2 m³)
m = 3909.95 kg
Now, we find the amount of energy consumed to heat the room:
E = m C ΔT
where,
E = Energy consumed = ?
C = Specific Heat of air at room temperature = 1 KJ/kg.⁰C
ΔT = Change in temperature = 21 °C - 13.5 °C = 7.5 °C
Therefore,
E = (3909.95 kg)(1 KJ/kg.°C)(7.5 °C)
E = 29324.62 KJ
Now, the time period can be calculated as:
P = E/t
t = E/P
where,
t = Time needed = ?
P = Power of heater = 20.7 KW
Therefore,
t = 29324.62 KJ/20.7 KW
t = (1416.65 s)(1 min/60 s)
t = 23.6 min
The ancient Greek Eratosthenes found that the Sun casts different lengths of shadow at different points on Earth. There were no shadows at midday in Aswan as the Sun was directly overhead. 800 kilometers north, in Alexandria, shadow lengths were found to show the Sun at 7.2 degrees from overhead at midday. Use these measurements to calculate the radius of Earth.
Answer:
The radius of the earth is [tex]r = 6365.4 \ km[/tex]
Explanation:
From the question we are told that
The distance at Alexandria is [tex]d_a = 800 \ km = 800 *10^{3} \ m[/tex]
The angle of the sun is [tex]\theta = 7.2 ^o[/tex]
So we want to first obtain the circumference of the earth
So let assume that the earth is circular ([tex]360 ^o[/tex])
Now from question we know that the sun made an angle of [tex]7.2 ^o[/tex] so with this we will obtain how many [tex](7.2 ^o)[/tex] are in [tex]360^o[/tex]
i.e [tex]N = \frac{360}{7.2}[/tex]
=> [tex]N = 50[/tex]
With this value we can evaluate the circumference as
[tex]c = 50 * 800[/tex]
[tex]c = 40000 \ km[/tex]
Generally circumference is mathematically represented as
[tex]c = 2\pi r[/tex]
[tex]40000 = 2 * 3.142 * r[/tex]
=> [tex]r = 6365.4 \ km[/tex]
Water pressurized to 3.5 x 105 Pa is flowing at 5.0 m/s in a horizontal pipe which contracts to 1/2 its former radius. a. What are the pressure and velocity of the water after the contraction
Answer:
Explanation:
Using the Continuity equation
v X A = v' xA'
so if A is 1/2of A' then A velocity must be 2 times the A'
after-contraction v = 2 x 5.0m/s = 10m/s
Using the Bernoulli equation
p₁ + ½ρv₁² + ρgh₁ = p₂ + ½ρv₂² + ρgh₂
, the "h" terms cancel
3.5 x 10^ 5Pa + ½ x 1000kg/m³x (5.0m/s)² = p₂ + ½ x 1000kg/m³ x (10m/s)²
p₂ = 342500pa
A golfer hits a 42 g ball, which comes down on a tree root and bounces straight up with an initial speed of 15.6 m/s. Determine the height the ball will rise after the bounce. Show all your work.
Answer:
12.2 m
Explanation:
Given:
v₀ = 15.6 m/s
v = 0 m/s
a = -10 m/s²
Find: Δy
v² = v₀² + 2aΔy
(0 m/s)² = (15.6 m/s)² + 2 (-10 m/s²) Δy
Δy = 12.2 m
[tex] \LARGE{ \boxed{ \rm{ \green{Answer:}}}}[/tex]
Given,
The initial speed is 15.6 m/s The mass of the ball is 42g = 0.042kgFinding the initial kinetic energy,
[tex]\large{ \boxed{ \rm{K.E. = \frac{1}{2}m {v}^{2}}}}[/tex]
⇛ KE = (1/2)mv²
⇛ KE = (1/2)(0.042)(15.6)²
⇛ KE = 5.11 J
|| ⚡By conservation of energy, the potential energy at the highest point will also be 5.11 J, since there is no kinetic energy at the highest point because the ball is not moving (we neglect energy lost due to air resistance, heat, sound, etc.) ⚡||
So, we have:
[tex] \large{ \boxed{ \rm{P.E. = mgh}}}[/tex]
⇛ h = PE/(mg)
⇛ h = 5.11 J /(0.042 × 9.8)
⇛ h = 12.41 m
✏The ball will rise upto a height of 12.41 m
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Select from the following for the next two questions:
A virtual, inverted and smaller than the object
B real, inverted and smaller than the object
C virtual, upright and smaller than the object
D real, upright and larger than the object
E virtual, upright and larger than the object
F real, inverted and larger than the object
G virtual, inverted and larger than the object
H real, upright and smaller than the object
An object is placed 46.9 cm away from a converging lens. The lens has a focal length of 10.0 cm. Select the statement from the list above which best describes the image an objesthse place 46.9 cm away from a spherical convex mirror. The radius of curvature of the mirror is 20.0 cm. Select the statement from the An object is placed 46.9 cm away from a spherical convex mirror. The radius of curvature of the mirror is 20.0 cm. Select the statement from the list above which best describes the image.
Answer:
Explanation:
1 )
An object is placed 46.9 cm away from a converging lens. The lens has a focal length of 10.0 cm.
Since the object is placed at a distance more than twice the focal length , its image will be inverted , real and will be of the size less than the size of object . So option B is applied .
B) real, inverted and smaller than the object.
2 )
An object is placed 46.9 cm away from a spherical convex mirror. The radius of curvature of the mirror is 20.0 cm.
The object is placed at a point beyond its radius of curvature, its image will be formed at a point between f and C or between focal point and centre of curvature . Its size will be smaller than size of object and it will be real and inverted .
B) real, inverted and smaller than the object.
What is the average value of the magnitude of the Poynting vector (intensity) at 1 meter from a 100-watt light bulb radiating in all directions
Answer:
I = 7.96 W / m²
Explanation:
The light bulb emits a power of P = 100W, this power is distributed over the surface of a sphere, thus the emission is in all directions.
Intensity is defined by power per unit area
I = P / A
The area of a sphere is
A = 4π r²
we substitute
I = P / (4π r²)
in this case it tells us that the distance is r = 1 m
let's calculate
I = 100 / (4π 1²)
I = 7.96 W / m²
If the rods with diameters and lengths listed below are made of the same material, which will undergo the largest percentage length change given the same applied force along its length?a. d, 3L b. 3d, L c. 2d, 2L d. 4d, L
Answer:
The highest percentage of change corresponds to the thinnest rod, the correct answer is a
Explanation:
For this exercise we are asked to change the length of the bar by the action of a force applied along its length, in this case we focus on the expression of longitudinal elasticity
F / A = Y ΔL/L
where F / A is the force per unit length, ΔL / L is the fraction of the change in length, and Y is Young's modulus.
In this case the bars are made of the same material by which Young's modulus is the same for all
ΔL / L = (F / A) / Y
the area of the bar is the area of a circle
A = π r² = π d² / 4
A = π / 4 d²
we substitute
ΔL / L = (F / Y) 4 /πd²
changing length
ΔL = (F / Y 4 /π) L / d²
The amount between paracentesis are all constant in this exercise, let's look for the longitudinal change
a) values given d and 3L
ΔL = cte 3L / d²
ΔL = cte L /d² 3
To find the percentage, we must divide the change in magnitude by its value and multiply by 100.
ΔL/L % = [(F /Y 4/π 1/d²) 3L ] / 3L 100
ΔL/L % = cte 100%
b) 3d and L value, we repeat the same process as in part a
ΔL = cte L / 9d²
ΔL = cte L / d² 1/9
ΔL / L% = cte 100/9
ΔL / L% = cte 11%
c) 2d and 2L value
ΔL = (cte L / d ½ )/ 2L
ΔL/L% = cte 100/4
ΔL/L% = cte 25%
d) value 4d and L
ΔL = cte L / d² 1/16
ΔL/L % = cte 100/16
ΔL/L % = cte 6.25%
The highest percentage of change corresponds to the thinnest rod, the correct answer is a
Astronomers think planets formed from interstellar dust and gases that clumped together in a process called? A. stellar evolution B. nebular aggregation C. planetary accretion D. nuclear fusion
Answer:
C. planetary accretion
Explanation:
Astronomers think planets formed from interstellar dust gases that clumped together in a process called planetary accretion.
Answer:
[tex]\boxed{\sf C. \ planetary \ accretion }[/tex]
Explanation:
Astronomers think planets formed from interstellar dust and gases that clumped together in a process called planetary accretion.
Planetary accretion is a process in which huge masses of solid rock or metal clump together to produce planets.
A resistor and a capacitor are connected in series to an ideal battery of constant terminal voltage. At the moment contact is made with the battery, the voltage across the resistor is
Answer:
At the moment contact is made with the battery, the voltage across the resistor is equal to the batteries terminal voltage
Explanation;
Because at series connection the battery and resistor have equal voltage
