Answer:
the Earth magnetic field in that region is about 89.09 [tex]\mu[/tex]T
Explanation:
If the magnetic field has an inclination of [tex]79^o[/tex], and its horizontal component is 17 microTesla, then we can calculate the magnitude of the full magnetic field via the cosine function. Notice that the decomposition of a vector in horizontal and perpendicular components originates a right angle triangle where the vector is the hypotenuse of the triangle.
[tex]cos(\theta)=\frac{adjacent}{hypotenuse}\\cos(79^o)=\frac{17}{B} \\B=\frac{17}{cos(79^o)}\,\mu T \\B=89.09\,\,\mu T[/tex]
The magnitude of Earth's magnetic field is 89 μT
Trigonometric ratio is used to show the relationship between the sides and the angle of a right angle triangle.
The Earth's magnetic field is in the hypotenuse side.
Applying trigonometric ratios:
cos(θ) = adjacent / hypotenuse
cos(79) = 17 μT / hypotenuse
hypotenuse = 89 μT
The magnitude of Earth's magnetic field is 89 μT
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Two 40 W (120 V) lightbulbs are wired in series, then the combination is connected to a 120 V supply. Part A How much power is dissipated by each bulb
Answer:
The power dissipated by each bulb is [tex]P = 10.0 \ W[/tex]
Explanation:
From the question we are told that
The power rating of both bulbs is [tex]P = 40 \ W[/tex]
The voltage rating of both bulb is [tex]V = 120 \ V[/tex]
The both bulbs are connected a voltage of [tex]V_C = 120 V[/tex]
The amount of power rating of each bulb is mathematically represented as
[tex]P = \frac{V^2}{R }[/tex]
=> [tex]R = \frac{V^2}{P}[/tex]
substituting values
[tex]R = \frac{ (120)^2}{40}[/tex]
[tex]R = 360 \Omega[/tex]
Now given that the bulbs are connected is series, the equivalent resistance is evaluated as
[tex]R_{eq } = R +R[/tex]
substituting values
[tex]R_{eq } = 360 + 360[/tex]
[tex]R_{eq } =720 \ \Omega[/tex]
The current flowing through the bulbs is mathematically evaluated as
[tex]I =\frac{V_C}{R_{eq}}[/tex]
substituting values
[tex]I =\frac{120}{720}[/tex]
[tex]I = 0.1667 \ A[/tex]
Now the power dissipated by both bulbs is mathematically represented as
[tex]P = I ^2 * R[/tex]
substituting values
[tex]P = 0.1668^2 * 360[/tex]
[tex]P = 10.0 \ W[/tex]
The power that should be dissipated by each bulb is P = 10.0 W.
Calculation of the power:
Since
The power rating of both bulbs is P = 40 W.
The voltage rating of both bulbs is V = 120 V.
And, both bulks that should be connected a voltage of Vc = 120V
Now the amount of power that should be rated of each bulb should be
P = V^2/R
So, R = V^2/P
= 120^2/40
= 360Ω
The equivalent resistance should be
I = Vc/Req
= 120/720
= 0.1667 A
Now the power is = 0.1668^2 * 360
= 10.0 W
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He moves his sandpaper back and forth 45 times over a rusty area, each time moving a total distance of 0.15 m. Kevin pushes the sandpaper against the surface with a normal force of 1.8 N. The coefficient of friction for the metal/sandpaper interface is 0.92. How much work is done by the kinetic frictional force during the sanding process
Answer:
W = -11.178 J
Explanation:
In this question, Work is defined by the equation:
W = F•dx•cos(θ)
Where;
F is force
x is distance over which the force acts on the object,
θ = angle between force and direction of travel.
Now, because the force is constant in this case, we don't need make the equation an integral expression. Also, since the force of friction is always precisely opposite the direction of travel (i.e θ = 180°), the equation can be rewritten like this:
W = F•x•cos(180)
W = -F•x
The force of friction will be given by the equation:
F_fric = F_norm × coefficient of friction
Also, the total work will be the sum of all 45 passes by the sandpaper. Therefore, our final equation, when F_fric is substituted into the work equation, is:
W = -(45)×(F_norm)×(coeff of friction)×(distance)
We are given;
normal force = 1.8 N
Coefficient of friction for the metal/sandpaper interface = 0.92
Distance = 0.15 m
W = -(45) × (1.8) × (0.92) × (0.15)
W = -11.178 J
Copper wire of diameter 0.289 cm is used to connect a set of appliances at 120 V, which draw 1850 W of power total. The resistivity of copper is 1.68×10−8Ω⋅m.
A. What power is wasted in 26.0 m of this wire?
B. What is your answer if wire of diameter 0.417 cm is used?
Answer:
(a) The power wasted for 0.289 cm wire diameter is 15.93 W
(b) The power wasted for 0.417 cm wire diameter is 7.61 W
Explanation:
Given;
diameter of the wire, d = 0.289 cm = 0.00289 m
voltage of the wire, V = 120 V
Power drawn, P = 1850 W
The resistivity of the wire, ρ = 1.68 x 10⁻⁸ Ω⋅m
Area of the wire;
A = πd²/4
A = (π x 0.00289²) / 4
A = 6.561 x 10⁻⁶ m²
(a) At 26 m of this wire, the resistance of the is
R = ρL / A
R = (1.68 x 10⁻⁸ x 26) / 6.561 x 10⁻⁶
R = 0.067 Ω
Current in the wire is calculated as;
P = IV
I = P / V
I = 1850 / 120
I = 15.417 A
Power wasted = I²R
Power wasted = (15.417²)(0.067)
Power wasted = 15.93 W
(b) when a diameter of 0.417 cm is used instead;
d = 0.417 cm = 0.00417 m
A = πd²/4
A = (π x 0.00417²) / 4
A = 1.366 x 10⁻⁵ m²
Resistance of the wire at 26 m length of wire and 1.366 x 10⁻⁵ m² area;
R = ρL / A
R = (1.68 x 10⁻⁸ x 26) / 1.366 x 10⁻⁵
R = 0.032 Ω
Power wasted = I²R
Power wasted = (15.417²)(0.032)
Power wasted = 7.61 W
A straightforward method of finding the density of an object is to measure its mass and then measure its volume by submerging it in a graduated cylinder. What is the density of a 344-g rock that displaces 216 cm3 of water
Answer:
density d = 1.59 g/cm^3
The density of the rock is 1.59 g/cm^3
Explanation:
The density of an object can be derived by measuring its mass and then measuring its volume by submerging it in a graduated cylinder.
Density = mass/volume of water displaced
d = m/v ........1
Given;
mass m = 344 g
Volume of water displaced v = 216 cm^3
from equation 1, we can calculate the value of the density;
Substituting the given values;
d = 344/216 g/cm^3
d = 1.592592592592 g/cm^3
d = 1.59 g/cm^3
The density of the rock is 1.59 g/cm^3
An empty parallel plate capacitor is connected between the terminals of a 9.0-V battery and charged up. The capacitor is then disconnected from the battery, and the spacing between the capacitor plates is doubled. As a result of this change, what is the new voltage between the plates of the capacitor
Answer:
The new voltage between the plates of the capacitor is 18 V
Explanation:
The charge on parallel plate capacitor is calculated as;
q = CV
Where;
V is the battery voltage
C is the capacitance of the capacitor, calculated as;
[tex]C = \frac{\epsilon _0A}{d} \\\\q =CV = (\frac{\epsilon _0A}{d})V = \frac{\epsilon _0A V}{d}[/tex]
[tex]q = \frac{\epsilon _0A V}{d}[/tex]
where;
ε₀ is permittivity of free space
A is the area of the capacitor
d is the space between the parallel plate capacitors
If only the space between the capacitors is doubled and every other parameter is kept constant, the new voltage will be calculated as;
[tex]q = \frac{\epsilon _0A V}{d} \\\\\frac{\epsilon _0A V}{d} = \frac{\epsilon _0A V}{d} \\\\\frac{V_1}{d_1} = \frac{V_2}{d_2} \\\\V_2 = \frac{V_1d_2}{d_1} \\\\(d_2 = 2d_1)\\\\V_2 = \frac{V_1*2d_1}{d_1} \\\\(V_1 = 9V)\\\\V_2 = \frac{9*2d_1}{d_1} \\\\V_2 = 9*2\\\\V_2 = 18 \ V[/tex]
Therefore, the new voltage between the plates of the capacitor is 18 V
A 750 kg car is moving at 20.0 m/s at a height of 5.0 m above the bottom of a hill when it runs out of gas. From there, the car coasts. a. Ignoring frictional forces and air resistance, what is the car’s kinetic energy and velocity at the bottom of the hill
Answer:
Explanation:
Kinetic energy at the height = 1/2 m v²
= 1/2 x 750 x 20²
= 150000 J
Its potential energy = mgh
= 750 x 9.8 x 5
=36750 J
Total energy = 186750 J
Its total kinetic energy will be equal to 186750 J , according to conservation of mechanical energy
If v be its velocity at the bottom
1/2 m v² = 186750
v = √498
= 22.31 m /s
A solenoid with 500 turns, 0.10 m long, carrying a current of 4.0 A and with a radius of 10-2 m will have what strength magnetic field at its center
Answer:
B = 0.025T
Explanation:
In order to calculate the strength of the magnetic field at the center of the solenoid, you use the following formula:
[tex]B=\frac{\mu N i}{L}[/tex] (1)
μ: magnetic permeability of vacuum = 4π*10^-7 T/A
N: turns of the solenoid = 500
i: current = 4.0A
L: length of the solenoid = 0.10m
You replace the values of the parameters in the equation (1):
[tex]B=\frac{(4\pi*10^{-7}T/A)(500)(4.0A)}{0.10m}=0.025T[/tex]
The strength of the magnetic field at the center of the solenoid = 0.025T
Answer:
Magnetic field strength at the center is 2.51x10^-2T
Explanation:
Pls see attached file for step by step calculation
Which of the following properties of a sound wave determines its loudness (or intensity): wavelength, speed, amplitude, or frequency?
Answer: Amplitude
Explanation: The intensity or loudness of a sound depends upon the extent to which the sounding body vibrates (i.e the amplitude of vibration).
Loudness is measured in units called decibels and amplitude in metre
The gas law for a fixed mass m of an ideal gas at absolute temperature T, pressure P, and volume V is PV=mRT, where R is the gas constant. Find the partial derivatives
Answer: Question for edmentum : Part D
Next, consider a scientific law known as the ideal gas law. This law deals with the relationship between temperature and pressure of a fixed mass of gas. The law states that the product of the pressure (P) and volume (V) is proportional to the temperature (T) of the mass of gas. The equation can be expressed this way, with k as a constant value:
PV = kT
Based on this equation, what are the two possible outcomes of heating up a certain mass of air?
Explanation: edmentum's sample Answer
The pressure of the gas or the volume of the gas will increase as the temperature increases. That’s because, according to the ideal gas law, pressure and volume are directly proportional to temperature. The value of k does not change.
Ideal gas law states the relation between pressure, volume, temperature and the no.of moles of a gas. The partial derivatives of this equation can be written as [tex](\frac{\partial P}{\partial T})_{n,V} = (\frac{\partial }{\partial T}\frac{nRT}{V})_{n,V}[/tex].
What is ideal gas law?According to ideal gas law the product of pressure and volume of an ideal gas is equal to the product of number of moles, temperature and universal gas constant R.
Thus, PV = nRT.
The partial derivatives are used to find the differentials of some variable with respect to other variable provided that some of the variables are constant.
We can differentiate the ideal gas law where n and R are constants. With respect to constant volume and no.of moles the PV can be differentiated as :[tex](\frac{\partial P}{\partial T})_{n,V}[/tex]. Where the constants are provided below the brackets.
Similarly the right side also can be differentiated with respect to change in temperature at constant volume and pressure as [tex](\frac{\partial }{\partial T}\frac{nRT}{V})_{n,V}[/tex].
Therefore, both these differentials are equal and thus can be equated as [tex](\frac{\partial P}{\partial T})_{n,V} = (\frac{\partial }{\partial T}\frac{nRT}{V})_{n,V}[/tex].
The same can be done with respect to change in pressure also.
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Approximately how many asteroids that are 0.98 km in radius would it take to make a planet which has a radius of 6420.0 km
Answer:
It'd take approximately 6543 asteroids.
Explanation:
In order to calculate the number of asteroids needed to make that planet we first need to determine the volume of each object. To do this we will consider them as spheres and apply the appropriate formula:
[tex]V_{asteroid} = \frac{4*\pi*0.98}{3} = 4.11 \text{ km}^3[/tex]
[tex]V_{planet} = \frac{4*\pi*6420}{3} = 26892.03 \text{ km}^3[/tex]
The number of asteroids needed are given by the division of planet's volume by the asteroid's volume.
[tex]n = \frac{26892.03}{4.11} = 6543.07[/tex]
It'd take approximately 6543 asteroids.
2. In the radioactive decay of 238???? → 234Th + 4He, a 238???? nucleus transforms to 234Th and an ejected 4He. (These are nuclei, not atoms, and thus electrons are not involved.) When the separation between 234Th and 4He is 9.0 × 10−15 m, what are the magnitudes of a) the electrostatic force between them and b) the acceleration of the 4He particle?
Answer:
Explanation:
a )
charge on nucleus of 234Th = 1.6 x 10⁻¹⁹ x 90 ( atomic number of Th is 90 )
charge on 4 He = 1.6 x 10⁻¹⁹ x 2 ( in Helium , no of proton is 2 )
force of repulsion between them
= 9 x 10⁹ x 1.6 x 10⁻¹⁹ x 90 x 1.6 x 10⁻¹⁹ x 2 / ( 9 x 10⁻¹⁵ )²
= 51.2 x 10
= 512 N .
b )
mass of 4 He
= 4 x 1.67 x 10⁻²⁷ kg
= 6.68 x 10⁻²⁷ kg
Acceleration
= Force / mass
= 512 / (6.68 x 10⁻²⁷)
= 76.64 x 10²⁷ m / s² .
Forces that act in pairs are _____ in size and ________ in direction.
Which of the following changes will increase the frequency of the lowest frequency standing sound wave on a stretching string?Choose all that apply.A. Replacing the string with a thicker stringB. Plucking the string harderC. Doubling the length of the string
Answer:
A, C
Explanation:
Since the frequency is inversely proportional to the length of a string, then I want to increase the frequency of the lowest
A. Replacing the string with a thicker string.
Thicker strings have more density. The more density the string has, the lower the sound.
Mathematically, we can see the proportionality (direct and inverse) by looking at those formulas for Frequency and Speed, when combined:
For:
[tex]f=\frac{v}{\lambda}[/tex]
[tex]f=\frac{v}{\lambda}*\sqrt{\frac{T}{D} }[/tex]
See above, how density (D) and [tex](\lambda)[/tex] wave length are inversely proportional.
C. Doubling the length of the string.
Because the length of the string is inversely proportional to the frequency.
The longer the string, the lower the frequency.
So, if we double string, we'll hear lower sounds in any string instrument
--
In short, for A, and C We can justify both since length and density are inversely proportional to the Frequency, we need longer or thicker string.
It is known that the gravitational force of attraction between two alpha particles is much weaker than the electrical repulsion. For two alpha particles at a distance d apart, calculate the ratio of the size of the gravitational attraction to that of the electrical repulsion. Specifically, find the magnitude of Gravitational/Electrical.
Answer:
The ratio of gravitational force to electrical force is 3.19 x 10^-36
Explanation:
mass of an alpha particle = 6.64 x [tex]10^{-27}[/tex] kg
charge on an alpha particle = +2e = +2(1.6 x [tex]10^{-19}[/tex] C) = 3.2 x [tex]10^{-19}[/tex] C
distance between particles = d
For gravitational attraction:
The force of gravitational attraction F = [tex]\frac{Gm^{2} }{r^{2} }[/tex]
where G = gravitational constant = 6.67 x [tex]10^{-11}[/tex] m^3 kg^-1 s^-2
r = the distance between the particles = d
m = the mass of each particle
therefore, gravitational force = [tex]\frac{6.67*10^{-11}*(6.64*10^{-27} )^{2} }{d^{2} }[/tex] = [tex]\frac{2.94*10^{-63} }{d^{2} }[/tex] Newton
For electrical repulsion:
Electrical force between the particles = [tex]\frac{-kQ^{2} }{r^{2} }[/tex]
where k is the Coulomb's constant = 9.0 x [tex]10^{9}[/tex] N•m^2/C^2
r = distance between the particles = d
Q = charge on each particle
therefore, electrical force = [tex]\frac{-9*10^{9}*(3.2*10^{-19} )^{2} }{d^{2} }[/tex] = [tex]\frac{-9.216*10^{-28} }{d^{2} }[/tex] Newton
the negative sign implies that there is a repulsion on the particles due to their like charges.
Ratio of the magnitude of gravitation to electrical force = [tex]\frac{2.94*10^{-63} }{9.216*10^{-28} }[/tex]
==> 3.19 x 10^-36
What is its diameter when the temperature is raised to 100 degrees Celsius? (b) What temperature change is required to increase its volume
The question is incomplete, the complete question is;
A spherical steel ball bearing has a diameter of 2.540cm at 25.00∘C.
(a) What is its diameter when it's temperature is raised to
100∘C?
(b) What temperature change is required to increase its volume by
1.000% ?
Answer:
a) 2.542 cm
b) 303.03°C
Explanation:
Given;
Diameter of the ball= 2.540cm
Initial temperature= 25.0°C
Final temperature= 100.0°C
Percentage increase in volume = 1.000%
Temperature coefficient of expansion for steel =11.0×10^−6/∘C
d2= d1[1 + α(T2-T1)]
d2= 2.540[1 + 11.0×10^−6(100-25)]
d2= 2.540[1 + 8.25×10^-4]
d2= 2.542 cm
From;
%V ×1/100 = V ×3α ×∆T/ V
Substituting values;
1.000 ×1/100= 3× 11.0×10^−6 × ∆T
∆T= 0.01/3× 11.0×10^−6
∆T= 303.03°C
4.A student with a mass of m rides a roller coaster with a loop with a radius of curvature of r. What is the minimum speed the rollercoaster can maintain and still make it all the way around the loop
Answer:
v = √gr
Explanation:
In order for the roller coaster to make it all the way around the loop, the centripetal force at the top of the rider must be equal to the weight the student. So, that the student does not fall during the motion. Therefore,
Centripetal Force = Weight
mv²/r = mg
v²/r = g
v² = gr
Taking square root on both sides, we get:
v = √gr
where,
v = speed of roller coaster
r = radius of curvature
g = acceleration due to gravity
Hence, the minimum speed the roller coaster can maintain and still make it all the way around the loop, is found to be:
v = √gr
An object attached to a horizontal spring is oscillating back and forth along a frictionless surface. The maximum speed of the object is 1.38 m/s, and its maximum acceleration is 6.83 m/s2. How much time elapses betwen an instant when the object's speed is at a maximum and the next instant when its acceleration is at a maximum
Answer:
t = 0.31s
Explanation:
In order to calculate the time that the object takes to travel from the point with its maximum speed to the point with the maximum acceleration, you first use the following formulas, for the maximum speed and the maximum acceleration:
[tex]v_{max}=\omega A\\\\a_{max}=\omega^2A[/tex]
A: amplitude
v_max = 1.38m/s
a_max = 6.83m/s^2
w: angular frequency
From the previous equations you can obtain the angular frequency w.
You divide vmax and amax, and solve for w:
[tex]\frac{v_{max}}{a_{max}}=\frac{\omega A}{\omega^2 A}=\frac{1}{\omega}\\\\\omega=\frac{a_{max}}{v_{max}}=\frac{6.83m/s^2}{1.38m/s^2}=4.94\frac{rad}{s}[/tex]
Next, you take into account that the maximum speed is obtained when the object passes trough the equilibrium point, and the maximum acceleration for the maximum elongation, that is, the amplitude. In such a trajectory the time is T/4 being T the period.
You calculate the period by using the information about the angular frequency:
[tex]T=\frac{2\pi}{\omega}=\frac{2\pi}{4.94rad/s}=1.26s[/tex]
Then the required time is:
[tex]t=\frac{T}{4}=\frac{1.26s}{4}=0.31s[/tex]
Consider a container of oxygen gas at a temperature of 23°C that is 1.00 m tall. Compare the gravitational potential energy of a molecule at the top of the container (assuming the potential energy is zero at the bottom) with the average kinetic energy of the molecules. Is it reasonable to neglect the potential energy?
Answer:
Yes, it is reasonable to neglect it.
Explanation:
Hello,
In this case, a single molecule of oxygen weights 32 g (diatomic oxygen) thus, the mass of kilograms is (consider Avogadro's number):
[tex]m=1molec*\frac{1mol}{6.022x10^{23}molec} *\frac{32g}{1mol}*\frac{1kg}{1000g}=5.31x10^{-26}kg[/tex]
After that, we compute the potential energy 1.00 m above the reference point:
[tex]U=mhg=5.31x10^{-26}kg*1.00m*9.8\frac{m}{s^2}=5.2x10^{-25}J[/tex]
Then, we compute the average kinetic energy at the specified temperature:
[tex]K=\frac{3}{2}\frac{R}{Na}T[/tex]
Whereas [tex]N_A[/tex] stands for the Avogadro's number for which we have:
[tex]K=\frac{3}{2} \frac{8.314\frac{J}{mol*K}}{6.022x10^{23}/mol}*(23+273)K\\ \\K=6.13x10^{-21}J[/tex]
In such a way, since the average kinetic energy energy is about 12000 times higher than the potential energy, it turns out reasonable to neglect the potential energy.
Regards.
Based on the stability classification, if, over the airport, rising air reaches the point of saturation below 3000 m, it is likely that:_______
Answer:
It is likely that vertically developed cumuliform clouds would begin to form
Explanation:
This is because since the air is conditionally unstable below 3000mand air were to be forced to rise to a point of saturation within this particular layer of the atmosphere
a block of wood is pulled by a horizontal string across a rough surface at a constant velocity with a force of 20N. the coefficient of kinetic friction between the surfaces is 0.3 the force of the friction is
Answer:
6 N
Explanation:
From the laws of friction
F = ¶R = 0.3 × 20 = 6 N
The force of friction opposing the block's motion is 6 N.
The given parameters;
force applied on the block, F = 20 Ncoefficient of kinetic friction = 0.3The force of friction which opposes the motion of the block is obtained by applying Newton's second law of motion.
F = ma
Fₓ = μF
Substitute the given parameters to calculate the frictional force on the object.
Fₓ = 0.3 x 20
Fₓ = 6 N
Thus, the force of friction opposing the block's motion is 6 N.
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An object is placed at 10.2 cm in front of a diverging lens with a focal length of -10.6 cm. What is the magnification
Answer:
M= -0.51
Explanation:
After i calculated my v to be -5.2cm from the formula 1/f=1/v+1/u
Then m=v/u which is -0.51
A loaded 375 kg toboggan is traveling on smooth horizontal snow at 4.50 m/s when it suddenly comes to a rough region. The region is 5.40 mlong and reduces the toboggan's speed by 1.20 m/s .
a) What average friction force did the rough region exert on the toboggan?
b) By what percent did the rough region reduce the toboggan's kinetic energy?
c) By what percent did the rough region reduce the toboggan's speed?
Answer:
a) The average friction force exerted on the toboggan is 653.125 newtons, b) The rough region reduced the kinetic energy of the toboggan in 92.889 %, c) The speed of the toboggan is reduced in 73.333 %.
Explanation:
a) Given the existence of non-conservative forces (friction between toboggan and ground), the motion must be modelled by means of the Principle of Energy Conservation and the Work-Energy Theorem, since toboggan decrease its speed (associated with due to the action of friction. Changes in gravitational potential energy can be neglected due to the inclination of the ground. Then:
[tex]K_{1} = K_{2} + W_{f}[/tex]
Where:
[tex]K_{1}[/tex], [tex]K_{2}[/tex] are the initial and final translational kinetic energies of the tobbogan, measured in joules.
[tex]W_{f}[/tex] - Dissipated work due to friction, measured in joules.
By applying definitions of translation kinetic energy and work, the expression described above is now expanded and simplified:
[tex]f\cdot \Delta s = \frac{1}{2}\cdot m \cdot (v_{1}^{2}-v_{2}^{2})[/tex]
Where:
[tex]f[/tex] - Friction force, measured in newtons.
[tex]\Delta s[/tex] - Distance travelled by the toboggan in the rough region, measured in meters.
[tex]m[/tex] - Mass of the toboggan, measured in kilograms.
[tex]v_{1}[/tex], [tex]v_{2}[/tex] - Initial and final speed of the toboggan, measured in meters per second.
The friction force is cleared:
[tex]f = \frac{m\cdot (v_{1}^{2}-v_{2}^{2})}{2\cdot \Delta s}[/tex]
If [tex]m = 375\,kg[/tex], [tex]v_{1} = 4.50\,\frac{m}{s}[/tex], [tex]v_{2} = 1.20\,\frac{m}{s}[/tex] and [tex]\Delta s = 5.40 \,m[/tex], then:
[tex]f = \frac{(375\,kg)\cdot \left[\left(4.50\,\frac{m}{s} \right)^{2}-\left(1.20\,\frac{m}{s}\right)^{2}\right]}{2\cdot (5.40\,m)}[/tex]
[tex]f = 653.125\,N[/tex]
The average friction force exerted on the toboggan is 653.125 newtons.
b) The percentage lost by the kinetic energy of the tobbogan due to friction is given by the following expression, which is expanded and simplified afterwards:
[tex]\% K_{loss} = \frac{K_{1}-K_{2}}{K_{1}}\times 100\,\%[/tex]
[tex]\% K_{loss} = \left(1-\frac{K_{2}}{K_{1}} \right)\times 100\,\%[/tex]
[tex]\% K_{loss} = \left(1-\frac{\frac{1}{2}\cdot m \cdot v_{2}^{2}}{\frac{1}{2}\cdot m \cdot v_{1}^{2}} \right)\times 100\,\%[/tex]
[tex]\% K_{loss} = \left(1-\frac{v_{2}^{2}}{v_{1}^{2}} \right)\times 100\,\%[/tex]
[tex]\%K_{loss} = \left[1-\left(\frac{v_{2}}{v_{1}}\right)^{2} \right]\times 100\,\%[/tex]
If [tex]v_{1} = 4.50\,\frac{m}{s}[/tex] and [tex]v_{2} = 1.20\,\frac{m}{s}[/tex], then:
[tex]\%K_{loss} = \left[1-\left(\frac{1.20\,\frac{m}{s} }{4.50\,\frac{m}{s} }\right)^{2} \right]\times 100\,\%[/tex]
[tex]\%K_{loss} = 92.889\,\%[/tex]
The rough region reduced the kinetic energy of the toboggan in 92.889 %.
c) The percentage lost by the speed of the tobbogan due to friction is given by the following expression:
[tex]\% v_{loss} = \frac{v_{1}-v_{2}}{v_{1}}\times 100\,\%[/tex]
[tex]\% v_{loss} = \left(1-\frac{v_{2}}{v_{1}} \right)\times 100\,\%[/tex]
If [tex]v_{1} = 4.50\,\frac{m}{s}[/tex] and [tex]v_{2} = 1.20\,\frac{m}{s}[/tex], then:
[tex]\% v_{loss} = \left(1-\frac{1.20\,\frac{m}{s} }{4.50\,\frac{m}{s} } \right)\times 100\,\%[/tex]
[tex]\%v_{loss} = 73.333\,\%[/tex]
The speed of the toboggan is reduced in 73.333 %.
The average frictional force exerted on the toboggan by the rough surface is 661.5 N.
The percentage of the toboggan kinetic energy reduction is 7.11%.
The percentage of the toboggan speed reduction is 26.67%.
The given parameters;
mass of the toboggan, m = 375 kginitial speed of the toboggan, u = 4.5 m/slength of the rough region, d = 5.4 mfinal speed of the toboggan, v = 1.2 m/sThe normal force on the toboggan is calculated as follows;
Fₙ = mg
Fₙ = 375 x 9.8 = 3675 N
The acceleration of the toboggan is calculated as follows;
[tex]v^2 = u^2 + 2as\\\\2as = v^2 - u^2\\\\a = \frac{v^2 - u^2 }{2s} \\\\a = \frac{(1.2)^2 - (4.5)^2 }{2(5.4)}\\\\a = -1.74 \ m/s^2[/tex]
The coefficient of friction is calculated as follows;
[tex]\mu_k = \frac{a}{g} \\\\\mu_k = \frac{1.74}{9.8} \\\\\mu_k = 0.18[/tex]
The average frictional force exerted on the toboggan by the rough surface;
[tex]F_k = \mu_k F_n\\\\F_k = 0.18 \times 3675\\\\F_k = 661.5 \ N[/tex]
The initial kinetic energy of the toboggan is calculated as follows;
[tex]K.E_i = \frac{1}{2} mu^2\\\\K.E_i = \frac{1}{2} \times 375\times 4.5^2\\\\K.E_i = 3,796.88 \ J[/tex]
The final kinetic energy of the toboggan is calculated as follows;
[tex]K.E_f = \frac{1}{2} mv^2\\\\K.E_f = \frac{1}{2} \times 375\times 1.2^2\\\\K.E_f = 270 \ J[/tex]
The percentage of the toboggan kinetic energy reduction is calculated as follows;
[tex]\frac{K.E_f}{K.E_i} \times 100\% = \frac{270}{3796.88} \times 100\% = 7.11 \%[/tex]
The percentage of the toboggan speed reduction is calculated as follows;
[tex]\frac{1.2}{4.5} \times 100\% = 26.67 \%[/tex]
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A particle moves along a straight line with the acceleration a = (12t - 3t ^ 1/2) feet / s ^ 2, where t is in seconds. Determine your speed and position as a function of time. When t = 0, v = 0 and s = 15 feet.
Answer:
v = 6t² − 2t^³/₂
s = 2t³ − ⅘t^⁵/₂ + 15
Explanation:
a = 12t − 3t^½
Integrate to find velocity.
v = ∫ a dt
v = ∫ (12t − 3t^½) dt
v = 6t² − 2t^³/₂ + C
Use initial condition to find C.
0 = 6(0)² − 2(0)^³/₂ + C
C = 0
v = 6t² − 2t^³/₂
Integrate to find position.
s = ∫ v dt
s = ∫ (6t² − 2t^³/₂) dt
s = 2t³ − ⅘t^⁵/₂ + C
Use initial condition to find C.
15 = 2(0)³ − ⅘(0)^⁵/₂ + C
15 = C
s = 2t³ − ⅘t^⁵/₂ + 15
How do your results from ray tracing compare to your results from using the thin-lens equation?
What is the focal length of a convex lens that produces an image 10 cm away with a magnification of -0.5? Show all calculations in your answer.
Answer:
f = 6.66 cm
Explanation:
For this exercise we will use the constructor equation
1 / f = 1 / p + 1 / q
where f is the focal length, p is the distance to the object and q is the distance to the image
the expression for magnification is
m = h '/ h = - q / p
with this we have a system of two equations with two unknowns, in the problem they give us the distance to the image q = 10 cm and a magnification of m = -0.5
-0.5 = - q / p
p = - q / 0.5
p = - 10 / 0.5
p = 20 cm
now we can with the other equation look for the focal length
1 / f = 1/20 + 1/10
1 / f = 0.15
f = 6.66 cm
The driver of a car travels at 90 km / h, observes some children playing on the road 50 m away, and applies the brakes, managing to stop 4 seconds later, if the total mass of the driver-vehicle system is 2200 kg What was the force developed by the brakes? Was there an accident? Explain with mathematical arguments, please help.
Answer:
13,750 N
Yes
Explanation:
Given:
v₀ = 90 km/h = 25 m/s
v = 0 m/s
t = 4 s
Find: a and Δx
a = Δv / Δt
a = (0 m/s − 25 m/s) / (4 s)
a = -6.25 m/s²
F = ma
F = (2200 kg) (-6.25 m/s²)
F = -13,750 N
Δx = ½ (v + v₀) t
Δx = ½ (0 m/s + 25 m/s) (4 s)
Δx = 50 m
5) With a power of 0.12HP it is possible to move a piece of furniture from one place to another, in 14 seconds, applying a force of 70N on it. Calculate: a- The work done. b- The distance traveled.
Answer:
a) 1250 J
b) 17.9 m
Explanation:
Convert horsepower to watts:
0.12 HP = 89.5 W
a) Work = power × time
W = (89.5 W) (14 s)
W = 1250 J
b) Work = force × distance
1250 J = (70 N) d
d = 17.9 m
The voltage between the cathode and the screen of a television set is 30 kV. If we assume a speed of zero for an electron as it leaves the cathode, what is its speed (m/s) just before it hits the screen
Answer:
The speed is [tex]v =10.27 *10^{7} \ m/s[/tex]
Explanation:
From the question we are told that
The voltage is [tex]V = 30 kV = 30*10^{3} V[/tex]
The initial velocity of the electron is [tex]u = 0 \ m/s[/tex]
Generally according to the law of energy conservation
Electric potential Energy = Kinetic energy of the electron
So
[tex]PE = KE[/tex]
Where
[tex]KE = \frac{1}{2} * m* v^2[/tex]
Here m is the mass of the electron with a value of [tex]m = 9.11 *10^{-31} \ kg[/tex]
and
[tex]PE = e * V[/tex]
Here e is the charge on the electron with a value [tex]e = 1.60 *10^{-19} \ C[/tex]
=> [tex]e * V = \frac{1}{2} * m * v^2[/tex]
=> [tex]v = \sqrt{ \frac{2 * e * V}{m} }[/tex]
substituting values
[tex]v = \sqrt{ \frac{2 * (1.60*10^{-19}) * 30*10^{3}}{9.11 *10^{-31}} }[/tex]
[tex]v =10.27 *10^{7} \ m/s[/tex]
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Answer: Net electrostatic force on C is 24.2×[tex]10^{-2}[/tex] Newtons.
Explanation: Coulomb's Law is used to determine Electrostatic Force. Its formula is:
F = k.[tex]\frac{q_{0}.q_{1}}{r^{2}}[/tex]
where:
k is electrostatic constant (k = 8.987×[tex]10^{9}[/tex] Nm²/C²);
q is the charge of the object in Coulumb;
r is the distance between charges;
The net force is the sum of all the forces acting on C, so:
Force B on C:
They are both positive, so there is a relpusive force acting between them on the y-axis.
[tex]F_{BC} = 8,987.10^{9}.\frac{4.35.10^{-3}.9.67.10^{-4}}{(6.14.10^{2})^{2}}[/tex]
[tex]F_{BC} = 10.03.10^{-2}[/tex] N
Force D on C:
There is an atractive force between them on the x-axis.
[tex]F_{CD} = 8.987.10^{9}.\frac{9.67.10^{-4}.1.92.10^{-3}}{(1.42.10^{3})^{2}}[/tex]
[tex]F_{CD} = 13.64.10^{-4}[/tex] N
Force A on C:
First, find the distance between objects:
The distance is a diagonal line that divides the rectangle into a right triangle. Distance is square of the hypotenuse .
[tex]r^{2} = (6.14.10^2)^{2} + (1.42.10^{3})^{2}[/tex]
[tex]r^{2} = 37.72.10^{4}[/tex]
and hypotenuse: r = [tex]6.14.10^2[/tex]m
There is an atractive force between charges, but there are components of the force in x- and y-axis. So, because of that, force will be:
[tex]F_{CA} = F_{CA}[/tex].sinα + [tex]F_{CA}.[/tex]cosα
[tex]F_{CA} = 8.987.10^{9}.\frac{3.12.10^{-3}.9.67.10^{-4}}{37.72.10^{4}}[/tex]
[tex]F_{CA} = 7.2.10^{-2}[/tex]
The trigonometric relations is taken from the rectangle:
sinα = [tex]\frac{6.14.10^{2}}{6.14.10^{2}}[/tex]
cosα = [tex]\frac{1.42.10^{3}}{6.14.10^{2}}[/tex]
[tex]F_{CA}.[/tex]cosα = [tex]7.2.10^{-2}(\frac{1.42.10^{3}}{6.14.10^{2}})[/tex] = 0.17
[tex]F_{CA}.[/tex]sinα = [tex]7.2.10^{-2}.(\frac{6.14.10^{2}}{6.14.10^{2}} )[/tex] = 0.072
[tex]F_{CA} =[/tex] 0.17î + 0.072^j
Now, sum up all the terms in its respective axis:
X: [tex]13.64.10^{-4} + 0.17 =[/tex] 0.1714
Y: [tex]10.03.10^{-2} + 7.2.10^{-2}[/tex] = 0.1723
These forms another right triangle, whose hypotenuse is the net electrostatic force:
[tex]F_{net} = \sqrt{(0.1714)^{2} + (0.1723)^2}[/tex]
[tex]F_{net} = 24.3.10^{-2}[/tex] N
The net electrostatic force acting on C has magnitude [tex]F_{net} = 24.3.10^{-2}[/tex] N.
A ball is thrown straight up in the air at an initial speed of 30 m/s. At the same time the ball is thrown, a person standing 70 m away begins to run toward the spot where the ball will land. How fast will the person have to run to catch the ball just before it hits the gro
Answer:
11.44 m/s
Explanation:
We are given;
Initial Speed of ball;v_o = 30 m/s
Distance from person to ball;d = 70 m
Now, from projectile motion, the ball will hit the ground in time given by the formula;
t = (2v_o)/g
Wherr;
t is time taken
v_o is initial velocity
g is acceleration due to gravity
Plugging in the values with g = 9.8 m/s² as acceleration due to gravity, we have;
t = 2(30)/9.8
t = 6.12s
Now, the person will need to run 70m in 6.12s
We know that velocity/speed = distance/time
Thus;
Speed = 70/6.12
Speed = 11.44m/s
Complete the following statement: When a net torque is applied to a rigid object, it always produces a:______.
a. constant acceleration.
b. rotational equilibrium.
c. constant angular velocity.
d. constant angular momentum.
e. change in angular velocity.
When a net torque is applied to a rigid object, it always produces a change in angular velocity. (e.)
The word which best completes the given sentence, "When a net torque is applied to a rigid object, it always produces a_____" is:
E. Change in angular velocityAccording to the given question, we are asked to ask what would be produced when a net torque is applied to a rigid object based on the vector quantities involved.
As a result of this, we can see that there is a change in angular velocity when there is a net torque applied as the rigid object would change the rate at which it rotates.
Therefore, the correct answer is option E
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