Answer:
The average angular acceleration of the Earth is; α = 6.152 X 10⁻²⁰ rad/s²
Explanation:
We are given;
The period of 365 revolutions of Earth in 2006, T₁ = 365 days, 0.840 sec
Converting to seconds, we have;
T₁ = (365 × 24 × 60 × 60) + 0.84
T₁ = (3.1536 x 10⁷) + 0.840
T₁ = 31536000.84 s
Now, the period of 365 rotation of Earth in 2006 is; T₀ = 365 days
Converting to seconds, we have;
T₀ = 31536000 s
Hence, time period of one rotation in the year 2006 is;
Tₐ = 31536000.84/365
Tₐ = 86400.0023 s
The time period of rotation is given by the formula;
Tₐ = 2π/ωₐ
Making ωₐ the subject;
ωₐ = 2π/Tₐ
Plugging in the relevant values;
ωₐ = 2π/ 365.046306
ωₐ = 7.272205023 x 10⁻⁵ rad/s
Therefore, the time period of one rotation in the year 1906 is;
Tₓ = 31536000/365
Tₓ = 86400 s
Time period of rotation,
Tₓ = 2π /ωₓ
ωₓ = 2π / T
Plugging in the relevant values;
ωₓ = 2π/86400
ωₓ = 7.272205217 x 10⁻⁵ rad/s
The average angular acceleration is given by;
α = (ωₓ - ωₐ) / T₁
α = ((7.272205217 × 10⁻⁵) - (7.272205023 × 10⁻⁵)) / 31536000.84
α = 6.152 X 10⁻²⁰ rad/s²
Thus, the average angular acceleration of the Earth is; α = 6.152 X 10⁻²⁰ rad/s²
A 60.0-kg boy is surfing and catches a wave which gives him an initial speed of 1.60 m/s. He then drops through a height of 1.57 m, and ends with a speed of 8.50 m/s. How much nonconservative work was done on the boy
Answer:
Work = 1167.54 J
Explanation:
The amount of non-conservative work here can be given by the difference in kinetic energy and the potential energy. From Law of conservation of energy, we can write that:
Gain in K.E = Loss in P.E + Work
(0.5)(m)(Vf² - Vi²) - mgh = Work
where,
m = mass of boy = 60 kg
Vf = Final Speed = 8.5 m/s
Vi = Initial Speed = 1.6 m/s
g = 9.8 m/s²
h = height drop = 1.57 m
Therefore,
(0.5)(60 kg)[(8.5 m/s)² - (1.6 m/s)²] - (60 kg)(9.8 m/s²)(1.57 m) = Work
Work = 2090.7 J - 923.16 J
Work = 1167.54 J
A solenoidal coil with 23 turns of wire is wound tightly around another coil with 310 turns. The inner solenoid is 20.0 cm long and has a diameter of 2.20 cm. At a certain time, the current in the inner solenoid is 0.130 A and is increasing at a rate of 1800 A/s. For this time, calculate:
(a) the average magnetic flux through each turn of the inner solenoid;
(b) the mutual inductance of the two solenoids;
(c) the emf induced in the outer solenoid by the changing current in the inner solenoid.
Answer:
Explanation:
From the given information:
(a)
the average magnetic flux through each turn of the inner solenoid can be calculated by the formula:
[tex]\phi _ 1 = B_1 A[/tex]
[tex]\phi _ 1 = ( \mu_o \dfrac{N_i}{l} i_1)(\pi ( \dfrac{d}{2})^2)[/tex]
[tex]\phi _ 1 = ( 4 \pi *10^{-7} \ T. m/A ) ( \dfrac{310}{20*10^{-2} \ m }) (0.130 \ A) ( \pi ( \dfrac{2.20*10^{-2} \ m }{2})^ 2[/tex]
[tex]\phi_1 = 9.625 * 10^{-8} \ Wb[/tex]
(b)
The mutual inductance of the two solenoids is calculated by the formula:
[tex]M = 23 *\dfrac{9.625*10^{-8} \ Wb}{0.130 \ A}[/tex]
M = [tex]1.703 *10^{-5}[/tex] H
(c)
the emf induced in the outer solenoid by the changing current in the inner solenoid can be calculate by using the formula:
[tex]\varepsilon = -N_o \dfrac{d \phi_1}{dt}[/tex]
[tex]\varepsilon = -M \dfrac{d i_1}{dt}[/tex]
[tex]\varepsilon = -(1.703*10^{-5} \ H) * (1800 \ A/s)[/tex]
[tex]\varepsilon = -0.030654 \ V[/tex]
[tex]\varepsilon = -30.65 \ V[/tex]
Set the battery to a value between 0.0 V and 1.5 V. Now drag the voltage meter toward the capacitor and move the red and black leads to measure the voltage. Determine the potential difference between the two plates and whether the top plate is at higher or lower voltage than the bottom plate
Answer:
the positive terminal has higher potential(voltage) than the negative. Any terminal at the positive terminal has higher potential
Explanation:
ΔV =Vtop - Vbottom
The temperature difference between the inside and the outside of an automobile engine is 434 C°. Express this temperature difference on:
a. The Fahrenheit scale.
b. The Kelvin scale.
Answer:
434°C=813.2 °F
434°C=707.15 K
Explanation:
Fahrenheit is a temperature scale that sets the freezing point of water to 32 degrees Fahrenheit (° F) and the boiling point to 212 ° F (at normal atmospheric pressure). The conversion from degrees celsius to gradis fahrenheit is done by: ℉ = ℃ * 1.80+ 32.00
So, being 434°C: ℉ = 434℃ * 1.80+ 32.00= 813.2
Then: 434°C=813.2 °F
Kelvin is the temperature unit of the International System. It is one of the seven basic temperature units. Its symbol in the international system of units is K. Zero on the Celsius scale or degrees Celsius (0 ° C) is defined as the equivalent of 273.15 K. Then the conversion of Census degrees to Kelvin degrees is done by: K = ℃ + 273.15
Being 434 °C: K=434 °C+273.15=707.15
Then: 434°C=707.15 K
Consider two identical springs. At the start of an experiment, Spring A is already stretched out 3 cm, while Spring B remains at the zero position. Both springs are then stretched an additional three centimeters. What conclusion can you draw about the force required to stretch these springs during the experiment
Answer:
Explanation:
In this interesting exercise we have that spring A is 3 cm longer, due to previous experiments if these experiments did not reach the non-linear elongation point, the cosecant Km of the spring must remain the same, therefore when we lengthen the two springs these the longitudinal are lengthened.
As a consequence of the above according to Hockey law, the prediction of lengthening is the same, therefore the outside is the same in two two systems
F = K Δx
A charged particle q moves at constant velocity through a crossed electric and magnetic fields (E and B, which are both constant in magnitude and direction). Write the magnitude of the electric force on the particle in terms of the variables given. Do the same for the magnetic force
Answer:
The magnitude of the electric force on the particle in terms of the variables given is, F = qE
The magnitude of the magnetic force on the particle in terms of the variables given is, F = q (v x B)
Explanation:
Given;
a charged particle, q
magnitude of electric field, E
magnitude of magnetic field, B
The magnitude of the electric force on the particle in terms of the variables given;
F = qE
The magnitude of the magnetic force on the particle in terms of the variables given;
F = q (v x B)
where;
v is the constant velocity of the charged particle
Answer:
The magnitude of the electric force acting on a charged particle moving through an electric field = |qE|
The magnitude of the magnetic force of a charged particle moving at a particular velocity through a magnetic field = |qv × B|
Explanation:
The electric force acting on a charged particle, q, moving through an electric field, E, is given as a product of the charge on the particle (a scalar quantity) and the electric field (a vector quantity).
Electric force = qE
The magnitude of the electric force = |qE|
That is, magnitude of the product of the charge and the electric field vector.
The magnetic force acting on a charged particle, q, moving with a velocity, v, through a magnetic field, B is a vector product of qv [a product of the charge of the particle (a scalar quantity) and the velocity of the particle (a vector quantity)] and B (a vector quantity).
It is given mathematically as (qv × B)
The magnitude of the magnetic force is the magnitude of the vector product obtained.
Magnitude of the magnetic force = |qv × B|
Hope this Helps!!!
Potential difference of a battery is 2.2 V when it is connected
across a resistance of 5 ohm, if suddenly the potential difference
falls to 1.8V, its internal resistance will be
Answer:
1.1ohms
Explanation:
According to ohms law E = IR
If potential difference of a battery is 2.2 V when it is connected across a resistance of 5 ohm and if suddenly the voltage Falls to 1.8V then the current in the 5ohms resistor I = V/R = 1.8/5
I = 0.36A (This will be the load current).
Before we can calculate the value of the internal resistance, we need to know the voltage drop across the internal resistance.
Voltage drop = 2.2V - 1.8V = 0.4V
Then we calculate the internal resistance using ohms law.
According to the law, V = Ir
V= voltage drop
I is the load current
r = internal resistance
0.4 = 0.36r
r = 0.4/0.36
r = 1.1 ohms
Three masses are located in the x-y plane as follows: a mass of 6 kg is located at (0 m, 0 m), a mass of 4 kg is located at (3 m, 0 m), and a mass of 2 kg is located at (0 m, 3 m). Where is the center of mass of the system?
A. (1 m, 2 m)
B. (2 m, 1 m)
C. (1 m, 1 m)
D. (1 m, 0.5 m)
E. (0.5 m, 1 m)
D. (1m, 0.5m)
Explanation:
The center of mass (or center of gravity) of a system of particles is the point where the weight acts when the individual particles are replaced by a single particle of equivalent mass. For the three masses, the coordinates of the center of mass C(x, y) is given by;
x = (m₁x₁ + m₂x₂ + m₃x₃) / M ----------------(i)
y = (m₁y₁ + m₂y₂ + m₃y₃) / M ----------------(ii)
Where;
M = sum of the masses
m₁ and x₁ = mass and position of first mass in the x direction.
m₂ and x₂ = mass and position of second mass in the x direction.
m₃ and x₃ = mass and position of third mass in the x direction.
y₁ , y₂ and y₃ = positions of the first, second and third masses respectively in the y direction.
From the question;
m₁ = 6kg
m₂ = 4kg
m₃ = 2kg
x₁ = 0m
x₂ = 3m
x₃ = 0m
y₁ = 0m
y₂ = 0m
y₃ = 3m
M = m₁ + m₂ + m₃ = 6 + 4 + 2 = 12kg
Substitute these values into equations (i) and (ii) as follows;
x = ((6x0) + (4x3) + (2x0)) / 12
x = 12 / 12
x = 1 m
y = (6x0) + (4x0) + (2x3)) / 12
y = 6 / 12
y = 0.5m
Therefore, the center of mass of the system is at (1m, 0.5m)
During the deceleration of an ascending elevator, the normal force on the feet of a passenger is _____ her weight. During the deceleration of a descending elevator, the normal force on the feet of a passenger is _____ her weight.
Answer: Smaller than ; larger than
Explanation:
When the elevator is moving in the upward direction, then the force acting on it is negative in nature because of
N= mg +ma, (g is gravity and a is acceleration)
here ma is negative so the N= mg-ma
Hence, it feels smaller than its original weight.
When the elevator is moving downward , then the force acting will be positive in nature
N= mg+ma,
here ma will be positive so it feels larger the original weight of passenger.
In 11.8 s, 151 bullets strike and embed themselves in a wall. The bullets strike the wall perpendicularly. Each bullet has a mass of 5 x 10^-3 kg and a speed of 1110 m/s.
Required:
a. What is the average change in momentum per second for the bullets?
b. Determine the average force exerted on the wall.
c. Assuming the bullets are spread out over an area of 3.0×10^−4m^2 obtain the average pressure they exert on this region of the wall.
Answer:
a. ΔP/Δt = 42.6 N
b. F = 42.6 N
c. P = 142042.4 Pa = 1.42 KPa
Explanation:
a.
First, we find the change in momentum of the bullets. For one bullet:
ΔP = m(Vf - Vi)
where,
ΔP = Change in Momentum = ?
m = mass of bullet = 5 x 10⁻³ kg
Vf = Final Speed = 1110 m/s
Vi = Initial Speed = 0 m/s (Since bullets are initially at rest)
Therefore,
ΔP = (3 x 10⁻³ kg)(1110 m/s - 0 m/s)
ΔP = 3.33 N.s
For 151 bullets:
ΔP = (151)(3.33 N.s)
ΔP = 502.83 N.s
Now, dividing this by time interval, Δt = 11.8 s
ΔP/Δt = 502.83 N.s/ 11.8 s
ΔP/Δt = 42.6 N
b.
According to Newton's Second Law, the force is equal to rate of change of linear momentum:
Average Force = F = ΔP/Δt
F = 42.6 N
c.
The pressure is given by:
Average Pressure = P = Average Force/Area
P = 42.6 N/ 3 x 10⁻⁴ m²
P = 142042.4 Pa = 1.42 KPa
The barricade at the end of a subway line has a large spring designed to compress 2.00 m when stopping a 1.10 ✕ 105 kg train moving at 0.350 m/s. (a) What is the force constant (in N/m) of the spring? N/m (b) What speed (in m/s) would the train be going if it only compressed the spring 0.600 m? m/s (c) What force (in N) does the spring exert when compressed 0.600 m? 2020 N (in the direction opposite to the train's motion)
Answer:
(a) k = 1684.38 N/m = 1.684 KN/m
(b) Vi = 0.105 m/s
(c) F = 1010.62 N = 1.01 KN
Explanation:
(a)
First, we find the deceleration of the car. For that purpose we use 3rd equation of motion:
2as = Vf² - Vi²
a = (Vf² - Vi²)/2s
where,
a = deceleration = ?
Vf = final velocity = 0 m/s (since, train finally stops)
Vi = Initial Velocity = 0.35 m/s
s = distance covered by train before stopping = 2 m
Therefore,
a = [(0 m/s)² - (0.35 m/s)²]/(2)(2 m)
a = 0.0306 m/s²
Now, we calculate the force applied on spring by train:
F = ma
F = (1.1 x 10⁵ kg)(0.0306 m/s²)
F = 3368.75 N
Now, for force constant, we use Hooke's Law:
F = kΔx
where,
k = Force Constant = ?
Δx = Compression = 2 m
Therefore.
3368.75 N = k(2 m)
k = (3368.75 N)/(2 m)
k = 1684.38 N/m = 1.684 KN/m
(c)
Applying Hooke's Law with:
Δx = 0.6 m
F = (1684.38 N/m)(0.6 m)
F = 1010.62 N = 1.01 KN
(b)
Now, the acceleration required for this force is:
F = ma
1010.62 N = (1.1 kg)a
a = 1010.62 N/1.1 x 10⁵ kg
a = 0.0092 m/s²
Now, we find initial velocity of train by using 3rd equation of motion:
2as = Vf² - Vi²
a = (Vf² - Vi²)/2s
where,
a = deceleration = -0.0092 m/s² (negative sign due to deceleration)
Vf = final velocity = 0 m/s (since, train finally stops)
Vi = Initial Velocity = ?
s = distance covered by train before stopping = 0.6 m
Therefore,
-0.0092 m/s² = [(0 m/s)² - Vi²]/(2)(0.6 m)
Vi = √(0.0092 m/s²)(1.2 m)
Vi = 0.105 m/s
Two narrow slits, illuminated by light consisting of two distinct wavelengths, produce two overlapping colored interference patterns on a distant screen. The center of the eighth bright fringe in one pattern coincides with the center of the third bright fringe in the other pattern. What is the ratio of the two wavelengths?
Answer:
The ration of the two wavelength is [tex]\frac{\lambda_1}{\lambda_2} = \frac{8}{3}[/tex]
Explanation:
Generally two slit constructive interference can be mathematically represented as
[tex]\frac{y}{L} = \frac{m * \lambda}{d}[/tex]
Where y is the distance between fringe
d is the distance between the two slit
L is the distance between the slit and the wall
m is the order of the fringe
given that y , L , d are constant we have that
[tex]\frac{m }{\lambda } = constant[/tex]
So
[tex]\frac{m_1 }{\lambda_1 } = \frac{m_2 }{\lambda_2 }[/tex]
So [tex]m_1 = 8[/tex]
and [tex]m_2 = 3[/tex]
=> [tex]\frac{m_2}{m_1} = \frac{\lambda_1}{\lambda_2}[/tex]
=> [tex]\frac{8}{3} = \frac{\lambda_1}{\lambda_2}[/tex]
So
[tex]\frac{\lambda_1}{\lambda_2} = \frac{8}{3}[/tex]
A 5.00 kg rock whose density is 4300 kg/m3 is suspended by a string such that half of the rock's volume is under water. You may want to review (Pages 370 - 372) . For help with math skills, you may want to review: Conversion Factors Part A What is the tension in the string
Answer:
The tension on the string is [tex]T = 43.302 \ N[/tex]
Explanation:
From the question we are told that
The mass of the rock is [tex]m_r = 5.00 \ kg = 5000 \ g[/tex]
The density of the rock is [tex]\rho = 4300 \ kg/m^3 = 4.3 g/dm^3[/tex]
Generally the volume of the rock is mathematically evaluated as
[tex]V = \frac{m_r}{\rho}[/tex]
substituting values
[tex]V = \frac{5000}{4.3}[/tex]
[tex]V = 1162.7 \ dm^3[/tex]
The volume of the rock immersed in water is
[tex]V_w = \frac{V}{2}[/tex]
substituting values
[tex]V_w = \frac{1162.7 }{2}[/tex]
[tex]V_w = 581.4 \ dm^3[/tex]
mass of water been displaced by the this volume is
[tex]m_w = V_w[/tex] According to Archimedes principle
=> [tex]m_w = 581.4 \ g[/tex]
[tex]m_w = 0.5814 \ kg[/tex]
The weight of the water displace is
[tex]W _w = m_w * g[/tex]
[tex]W _w = 0.5814 * 9.8[/tex]
[tex]W _w = 5.698 \ N[/tex]
The actual weight of the rock is
[tex]W_r = m_r * g[/tex]
[tex]W_r = 5.0 * 9.8[/tex]
[tex]W_r = 49.0 \ N[/tex]
The tension on the string is
[tex]T = W_r - W_w[/tex]
substituting values
[tex]T = 49.0 - 5.698[/tex]
[tex]T = 43.302 \ N[/tex]
A total charge Q is distributed uniformly over a large flat insulating surface of area A . If the electric field magnitude is equal to 1000 NC/ at a point located a perpendicular distance of 0.1 m away from the center of the sheet, then the electric field at a point a perpendicular distance 0.2 m from the center of the sheet is:_______
a. 1000N/C
b. 500N/C
c. Impossible to say since we are not given Q and A
d. 250 N/C
Answer:
a. 1000N/C
Explanation:
Data mentioned in the question
Electrical field magnitude = 1000 NC
Perpendicular distance = 0.1 m
Perpendicular distance = 0.2 m
Based on the above information, the electric field is
As we know that
[tex]E = \frac{\sigma}{2\times E_o}[/tex]
where,
[tex]\sigma[/tex] = surface charge density
E = distance from nearby point to sheet i.e be independent
The distance at 0.1 and 0.2, the electric field would remain the same
So,
Based on the above explanation, the first option is correct
A 5.3 kg block initially at rest is pulled to the right along a frictionless, horizontal surface by a constant horizontal force of magnitude 21 N. Find the block's speed after it has moved through a horizontal distance of 6.4 m.
A man stands on a platform that is rotating (without friction) with an angular speed of 1.2 rev/s; his arms are outstretched and he holds a brick in each hand.The rotational inertia of the system consisting of the man, bricks, and platform about the central vertical axis of the platform is 6.0 k g times m squared. If by moving the bricks the man decreases the rotational inertia of the system to 2.0 k g times m squared, what is the resulting angular speed of the platform in rad/s
Answer:
resulting angular speed = 3.6 rev/s
Explanation:
We are given;
Initial angular speed; ω_i = 1.2 rev/s
Initial moment of inertia;I_i = 6 kg/m²
Final moment of inertia;I_f = 2 kg/m²
From conservation of angular momentum;
Initial angular momentum = Final angular momentum
Thus;
I_i × ω_i = I_f × ω_f
Making ω_f the subject, we have;
ω_f = (I_i × ω_i)/I_f
Plugging in the relevant values;
ω_f = (6 × 1.2)/2
ω_f = 3.6 rev/s
The cart travels the track again and now experiences a constant tangential acceleration from point A to point C. The speeds of the cart are 4.50 m/s at point A and 5.00 m/s at point C. The cart takes 4.00 s to go from point A to point C, and the cart takes 1.60 s to go from point B to point C. What is the cart's speed at point B
Answer:
Vi = 4.8 m/s
Explanation:
First we need to find the magnitude of constant tangential acceleration. For that purpose we use the following formula between points A and C:
a = (Vf - Vi)/t
where,
a = constant tangential acceleration from A to C = ?
Vf = Final Velocity at C = 5 m/s
Vi = Initial Velocity at A = 4.5 m/s
t = time taken to move from A to C = 4 s
Therefore,
a = (5 m/s - 4.5 m/s)/4 s
a = 0.125 m/s²
Now, applying the same equation between points B and C:
a = (Vf - Vi)/t
where,
a = constant tangential acceleration from A to B = 0.125 m/s²
Vf = Final Velocity at C = 5 m/s
Vi = Initial Velocity at B = ?
t = time taken to move from B to C = 1.6 s
Therefore,
0.125 m/s² = (5 m/s - Vi)/1.6 s
Vi = 5 m/s - (0.125 m/s²)(1.6 s)
Vi = 4.8 m/s
A long cylindrical rod of diameter 200 mm with thermal conductivity of 0.5 W/m⋅K experiences uniform volumetric heat generation of 24,000 W/m3. The rod is encapsulated by a circular seeve having an outer diameter of 400 mm and a thermal conductivity of 4 W/m⋅K. The outer surface of the sleeve is exposed to cross flow air at 27∘C with a convection coefficient of 25 W/m2⋅K.
(a) Find the temperature at the interface between the rod and sleeve and on the outer surface.
(b) What is the temperature at the center of the rod?
Answer:
a, 71.8° C, 51° C
b, 191.8° C
Explanation:
Given that
D(i) = 200 mm
D(o) = 400 mm
q' = 24000 W/m³
k(r) = 0.5 W/m.K
k(s) = 4 W/m.K
k(h) = 25 W/m².K
The expression for heat generation is given by
q = πr²Lq'
q = π . 0.1² . L . 24000
q = 754L W/m
Thermal conduction resistance, R(cond) = 0.0276/L
Thermal conduction resistance, R(conv) = 0.0318/L
Using energy balance equation,
Energy going in = Energy coming out
Which is = q, which is 754L
From the attachment, we deduce that the temperature between the rod and the sleeve is 71.8° C
At the same time, we find out that the temperature on the outer surface is 51° C
Also, from the second attachment, the temperature at the center of the rod was calculated to be, 191.8° C
A heavy, 6 m long uniform plank has a mass of 30 kg. It is positioned so that 4 m is supported on the deck of a ship and 2 m sticks out over the water. It is held in place only by its own weight. You have a mass of 70 kg and walk the plank past the edge of the ship. How far past the edge do you get before the plank starts to tip, in m
Answer:
about 1 meter
Explanation:
The distance past the edge that the man will get before the plank starts to tip is; 0.4285 m
We are given;
Mass of plank; m = 30 kg
Length of plank; L = 6m
Mass of man; M = 70 kg
Since the plank has 2 supports which are the deck of the ship, then it means that, we can take moments about the right support before the 2m stick out of the plank.
Thus;
Moment of weight of plank about the right support;
τ_p = mg((L/2) - 2)
τ_p = 30 × 9.8((6/2) - 2)
τ_p = 294 N.m
Moment of weight of man about the right support;
τ_m = Mgx
where x is the distance past the edge the man will get before the plank starts to tip.
τ_m = 70 × 9.8x
τ_m = 686x
Now, moment of the board is counterclockwise while that of the man is clockwise. Thus;
τ_m = τ_p
686x = 294
x = 294/686
x = 0.4285 m
Read more at; https://brainly.com/question/22150651
One kind of baseball pitching machine works by rotating a light and stiff rigid rod about a horizontal axis until the ball is moving toward the target. Suppose a 144 g baseball is held 82 cm from the axis of rotation and released at the major league pitching speed of 87 mph.
Required:
a. What is the ball's centripetal acceleration just before it is released?
b. What is the magnitude of the net force that is acting on the ball just before it is released?
Answer:
a. ac = 1844.66 m/s²
b. Fc = 265.63 N
Explanation:
a.
The centripetal acceleration of the ball is given as follows:
ac = v²/r
where,
ac = centripetal acceleration = ?
v = speed of ball = (87 mph)(1 h/ 3600 s)(1609.34 m / 1 mile) = 38.9 m/s
r = radius of path = 82 cm = 0.82 m
Therefore,
ac = (38.9 m/s)²/0.82 m
ac = 1844.66 m/s²
b.
The centripetal force is given as:
Fc = (m)(ac)
Fc = (0.144 kg)(1844.66 m/s²)
Fc = 265.63 N
A 2.5 m long diving board weighs 120 N. It has two supports, one at the end and another at a distance of 1.0 m from that end. A 100 N diver stands at the other end of the board. What is the magnitude and direction of the force exerted by the support at the end of the diving board
Answer:
46.67 N Upwards (with a clockwise moment)
Explanation:
length of board = 2.5 m
weight of board = 120 N
the board has two supports, say support A and support B
support A is at one end,
support B is at 100 m from the other end.
weight of diver = 100 N
diver stands on the other end of the board.
Magnitude of support A at the end of the board
To get the magnitude and force exerted by the support at the end of the board (support A, we take moment of the forces about support B
Moment of a force is the product of force and perpendicular distance of the force about a center.
The weight of the board acts at the center of the board (1.25 m from each end of the board). That is 2.5 m from the support B.
moment of board's weight about support B is 120 x 0.25 = 30 N-m
The moment due to the weight of the board acts anticlockwise.
Weight of the diver acts at the opposite side of the board, and it acts 1 m from support B.
Moment of diver about support B is 100 x 1 = 100 N-m
Th moment due to the diver acts clockwise.
The moment due to the reaction at support A acts at a distance 1.5 m from support B
If the reaction force on support A is Fa, then the reaction about support B is Fa x 1.5 = 1.5Fa.
The moment due to support A acts clockwise.
According to moment laws, the total clockwise movement must be equal to the total anticlockwise movement.
Total clockwise movements = 100 N-m + 1.5Fa
Total anticlockwise moment = 30 N-m
according to moment laws,
100 + 1.5Fa = 30
1.5 Fa = 30 - 100 = -70
Fa = -70/1.5 = -46.67 N
The magnitude of the force exerted at support A is equal but opposite to the reaction at support A and is equal to 46.67 N
An air bubble underwater has the same pressure as that of the surrounding water. As the air bubble rises toward the surface (and its temperature remains constant), the volume of the air bubble.
a) increase or decrease, depending on the rate it rises.
b) decreases.
c) increases.
d) remains constant
Answer:
D Remains constant
Explanation:
In a device called the ballistic pendulum, a compressed spring is used to launch a steel ball horizontally into a soft target hanging from a string. The ball embeds in the target and the two swing together from the string. Describe the energy transfers and/or transformations that take place during the use of the ballistic pendulum and at what points they occur
Answer:
When the spring in the ballistic pendulum is compressed, energy is stored up in the spring as potential energy. When the steel ball is launched by the spring, the stored up potential energy of the compressed spring is transformed and transferred into the kinetic energy of the steel ball as it flies off to hit its target. On hitting the soft target, some of the kinetic energy of the steel ball is transferred to the soft target (since they stick together), and they both start to swing together. During the process of swinging, the system's energy is transformed between kinetic and potential energy. At the maximum displacement of the ball from its point of rest, all the energy is converted to potential energy of the system. At the lowest point of travel (at the rest point), all the energy of the system is transformed into kinetic energy. In between these two points, energy the energy of the system is a combination of both kinetic and potential energy.
In the end, all the energy will be transformed and lost as heat to the surrounding; due to the air resistance around; bringing the system to a halt.
The power dissipated in each of two resistors is the same. The current across resistor A is triple that across resistor B. If the resistance of resistor B is R, what is the resistance of A?
Answer:
Explanation:
this is the answer to your question
While sitting in your car by the side of a country road, you see your friend, who happens to have an identical car with an identical horn, approaching you. You blow your horn, which has a frequency of 260 Hz; your friend begins to blow his horn as well, and you hear a beat frequency of 5.0 Hz. Part A How fast is your friend approaching you
Answer:
-6.49 m/s
Explanation:
This is doppler effect.
The equation is;
F_l = [(v + v_l)/(v + v_s)]F_s
Where;
F_l is frequency observed by the listener
v is speed of sound
v_l is speed of listener
v_s is speed of source of the sound
F_s is frequency of the source of the sound
In this question, the source of the sound is the moving vehicle.
Thus;
F_l = F_beat + F_s
We are given beat frequency (f_beat) as 5 Hz while source frequency (F_s) as 260 Hz.
So,
F_l = 5 + 260
F_l = 265 Hz
Since listener is sitting by car, thus; v_l = 0 m/s
Thus,from our doppler effect equation, let's make v_s the subject;
v_s = F_s[(v + v_l)/F_l] - v
Speed of sound has a value of v = 344 m/s
Thus;
v_s = 260[(344 + 0)/265] - 344
v_s = -6.49 m/s
This value is negative because the source is moving towards the listener
3.Cuanto Calor pierden 514 ml de agua si su temperatura desciende de 12°C a 11°C. Expresa el resultado en calorias.
514 cal
51.4 Kcal
514J/cal
5.149 Cal
Answer:
514 cal
Explanation:
In order to calculate the lost heat by the amount of water you first take into account the following formula:
[tex]Q=mc(T_2-T_1)[/tex] (1)
Q: heat lost by the amount of water = ?
m: mass of the water
c: specific heat of water = 1cal/g°C
T2: final temperature of water = 11°C
T1: initial temperature = 12°C
The amount of water is calculated by using the information about the density of water (1g/ml):
[tex]m=\rho V=(1g/ml)(514ml)=514g[/tex]
Then, you replace the values of all parameters in the equation (1):
[tex]Q=(514g)(1cal/g\°C)(11\°C-12\°C)=-514cal[/tex]
The amount of water losses a heat of 514 cal
Two metal spheres are hanging from nylon threads. When you bring the spheres close to each other, they tend to attract. Based on this information alone, discuss all the possible ways that the spheres could be charged. Is it possible that after the spheres touch, they will cling together? Explain.
Explanation:
In the given question, the two metal spheres were hanged with the nylon thread.
When these two spheres were brought close together, they attracted each other. The attraction between these spheres is the result of the opposite charges between them.
The possible ways by which these two metal spheres can be charged are by induction that is touching the metal or by rubbing them.
During induction, the same charges are transferred to each sphere. In this case, either both the spheres will be negatively charged or positively charged.
It is not possible that after the sphere touch each other they will cling together because the same charge repels each other and during touching, if one sphere is neutral, then the charged one will transfer the same charge. And as we know that same charge repel each other therefore they will repel each other.
13. Under what condition (if any) does a moving body experience no energy even though there
is a net force acting on it?
(2 marks)
Answer:
When the Net Forces are equal to 0
Explanation:
Momentum of a body can be defined as product of mass and velocity. It is in the same direction as in velocity. When the momentum of a body doesn't change, it is said to be conserved. If the momentum of a body is constant, the the net forces acting on a body becomes zero. When net forces acting on a body is zero, it means that no kinetic energy is being lost or gained, hence the kinetic energy is also conserved. If no energy is being gained or lost, it means that the body will experience no energy.
A hawk is flying horizontally at 18.0 m/s in a straight line, 230 m above the ground. A mouse it has been carrying struggles free from its grasp. The hawk continues on its path at the same speed for 2.00 s before attempting to retrieve its prey. To accomplish the retrieval, it dives in a straight line at constant speed and recaptures the mouse 3.00 m above the ground. (a) Assuming no air resistance, find the diving speed (magnitude of the total velocity vector) of the hawk. (b) What angle did the hawk make with the horizontal during its descent? (c) For how long did the mouse "enjoy" free fall?
Answer:
a) vd = 47.88 m/s
b) θ = 80.9°
c) t = 6.8 s
Explanation:
In the situation of the problem, you can assume that the trajectory of the hawk and the trajectory of the mouse form a rectangle triangle.
One side of the triangle is the horizontal trajectory of the hawk after 2.00s of flight, the other side of the triangle is the distance traveled by the mouse when it is falling down. And the hypotenuse is the trajectory of the hawk when it is trying to recover the mouse.
(a) In order to calculate the diving speed of the hawk, you first calculate the hypotenuse of the triangle.
One side of the triangle is c1 = (18.0m/s)(2.0s) = 36m
The other side of the triangle is c2 = 230m - 3m = 227 m
Then, the hypotenuse is:
[tex]h=\sqrt{(36m)^2+(227m)^2}=229.83m[/tex] (1)
Next, it is necessary to calculate the falling down time of the mouse, this can be done by using the following formula:
[tex]y=y_o+v_ot+\frac{1}{2}gt^2[/tex] (2)
yo: initial height = 230m
vo: initial vertical speed of the mouse = 0m/s
g: gravitational acceleration = -9.8m/s^2
y: final height of the mouse = 3 m
You replace the values of the parameters in (2) and solve for t:
[tex]3=230-4.9t^2\\\\t=\sqrt{\frac{227}{4.9}}=6.8s[/tex]
The hawk traveled during 2.00 second in the horizontal trajectory, hence, the hawk needed 6.8s - 2.0s = 4.8 s to travel the distance equivalent to the hypotenuse to catch the mouse.
You use the value of h and 4.8s to find the diving speed of the hawk:
[tex]v_d=\frac{229.83m}{4.8s}=47.88\frac{m}{s}[/tex]
The diving speed of the Hawk is 47.88m/s
(b) The angle is given by:
[tex]\theta=cos^{-1}(\frac{c_1}{h})=cos^{-1}(\frac{36m}{229.83m})=80.9 \°[/tex]
Then angle between the horizontal and the trajectory of the Hawk when it is descending is 80.9°
(c) The mouse is falling down during 6.8 s
You are playing ice hockey with friends on a frozen pond. The hockey puck (mass 0.2 kg) is sliding on the ice at 10 m/s when it starts to hail. A hailstone lands on the puck, causing it to slow to 8 m/s. What mass does the hailstone have
Answer:
The mass of the hailstone is [tex]m_2 = 0.05[/tex]
Explanation:
From the question we are told that
The mass of the hockey puck is [tex]m_1 = 0.2 \ kg[/tex]
The initial speed of the puck is [tex]u_1 = 10 \ m/s[/tex]
The final speed of the puck is [tex]v_f = 8 \ m/s[/tex]
According to the principles of linear momentum
[tex]m_1u_1 + m_2u_2 = (m_1 + m_2)v_f[/tex]
At initial the velocity of the hailstone was zero so
[tex]u_2 = 0 \ m/s[/tex]
So
[tex]m_1 u_1 = (m_1 + m_2 )v_f[/tex]
substituting values
[tex]0.2 * 10 = (0.2 + m_2 )* 8[/tex]
=> [tex]2 = (0.2 + m_2 )* 8[/tex]
=> [tex]0.25 = 0.2 + m_2[/tex]
=> [tex]m_2 = 0.05[/tex]