Answer:
48 m
Explanation:
As she travels at the rate of 5m/s due north, the amount of time it would take for her to cross the 80m wide river would be
t = 80 / 5 = 16 seconds
This is also the time it takes for the river to push her to the east side at the rate of 3m/s. So after 16 seconds, she would reach the opposite point at a horizontal distance from her starting of
s = 16*3 = 48 m
Two light bulbs, A and B, are connected to a 120-V outlet (a constant voltage source). Light bulb A is rated at 60 W and light bulb B is rated at 100 W. Which light bulb has a greater filament resistance?
Answer:
Bulb A has a greater resistance.
Explanation:
Electric power (P) = V²/R
P = V²/R................ Equation 1
Where P = power, V = Voltage, R = Resistance.
Make R the subject of the equation
R = V²/P ................ Equation 2
For Bulb A,
Given: V = 120 V, P = 60 W.
Substitute into equation 2
R = 120²/60
R = 240 Ω
For bulb B
Given: V =120 V, P = 100 W.
Substitute into equation 2
R = 120²/100
R = 14400/100
R = 144 Ω
Hence Bulb A has a greater resistance.
bulb A has the greater filament resistance.
What is resistance?This can be defined as the ability of a conducting material to oppose the flow of electric current.
To know the bulb with the greatest filament resistance, we use the formula below.
Formula:
P = V²/R................. Equation 1making R the subject of the equation
R = V²/P............... Equation 2Where:
P = power of the bulbV = Outlet voltageR = Resistance of the filament.For Bulb A,
Given:
P = 60 WV = 120 VSubstitute these values into equation 1
R = (120²)/60R = 240 ohms.For Bulb B,
Given:
P = 100 W.V = 60 VSubstitute these values into equation 1
R = (120²)/100R = 144 ohms.Hence, From the above, it can be seen that bulb A has the greater filament resistance.
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Electrons are ejected from a metallic surface with speeds of up to 4.60 3 105 m/s when light with a wavelength of 625 nm is used. (a) What is the work function of the surface? (b) What is the cutoff f
Complete question:
Electrons are ejected from a metallic surface with speeds ranging up to 4.60 x10⁵ m/s when light with a wavelength of 625nm is used. (a) What is the work function of the surface? (b) What is the cutoff frequency for this surface?
Answer:
Part(a) The work function of the surface is 22.177 x 10⁻²⁰ J = 1.384 eV
Part(b) The cutoff frequency for this surface is 3.347 x 10¹⁴ Hz
Explanation:
The kinetic energy (KE) of the emitted photon:
KE = 0.5mv²
m is mass of electron = 9.1 X 10⁻³¹ kg
KE = 0.5 * 9.1 X 10⁻³¹ * (460000)² = 9.628 X 10⁻²⁰ J
in eV = 9.628 X 10⁻²⁰ J x 6.242 X 10¹⁸ ev = 0.601 eV
The photon energy of the incoming radiation:
E = hf = hc/λ
c is speed of light (photon) = 3 x 10⁸
h is Planck's constant = 6.626 × 10⁻³⁴ J.s
E = (6.626 × 10⁻³⁴ *3 x 10⁸) /(625 X 10⁻⁹)
E = 31.805 X 10⁻²⁰ J
in eV = 31.805 X 10⁻²⁰ J x 6.242 X 10¹⁸ ev = 1.985 eV
Part (a) the work function of the surface
KE = hf - W
where;
W is work function
W = hf - KE
W = 31.805 X 10⁻²⁰ J - 9.628 X 10⁻²⁰ J = 22.177 x 10⁻²⁰ J
in eV = 22.177 x 10⁻²⁰ J x 6.242 X 10¹⁸ ev = 1.384 eV
Part(b) the cutoff frequency for this surface
W =hf
f = W/h
f = (22.177 x 10⁻²⁰ J)/(6.626 × 10⁻³⁴ J.s)
f = 3.347 x 10¹⁴ Hz
The work function is the minimum energy required to eject an electron from a metallic surface, which can be calculated using the maximum kinetic energy of ejected electrons and the energy of the incident light. The cutoff frequency is the minimum frequency of light required to eject an electron.
Explanation:The question is asking to calculate the work function of the surface and the cutoff frequency in a context related to the photoelectric effect. The photoelectric effect is a phenomenon in which electrons are ejected from a metallic surface when it is exposed to light of a certain frequency, in this case, light with a wavelength of 625 nm.
The energy of the incident light is used to expel electrons from the surface of the metal. Any remaining energy contributes toward the ejected electrons' kinetic energy. This can be modeled by the equation KE_maximum = hf - Φ, where KE_maximum is the electron's maximum kinetic energy, h is Planck's constant, f is the frequency of the light, and Φ is the work function of the material. The work function Φ is the minimum amount of energy required to eject an electron from the material's surface.
The cutoff frequency or threshold frequency is the minimum frequency of the incident light required to eject electrons. If the frequency of the light is less than this value, no electrons are ejected
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Light does not move infinitely fast but has a finite speed. We normally use ""c"" to indicate the speed of light in science. Write down the value for the speed of light in metric units: c = _________.
Answer:
6.71 × 10^8 mi/hr
Explanation:
Light is usually defined as an electromagnetic wave that is comprised of a definite wavelength. It is of both types, visible and invisible. The light emitted from a source usually travels at a speed of about 3 × 10^8 meter/sec. This speed of light is commonly represented by the letter 'C'.
To write it in the metric system, it has to be converted into miles/hour.
We know that,
1 minute = 60 seconds
60 minutes = 1 hour
1 kilometer = 1000 meter
1 miles = 1.6 kilometer
Now,
= [tex]\frac{3 \times\ 10^8 meter \times\ 60 sec \times\ 60 min}{1 sec \times\ 1 min \times\ 1 hr}[/tex]
= 1.08 × 10^12 m/ hr (meter/hour)
= [tex]\frac{1.08 \times\ 10^{12} meter \times\ 1 km \times\ 1 miles}{1 hr \times\ 1000 meter \times\ 1.6 km}[/tex]
= 6.71 × 10^8 mi/hr (miles/hour)
Thus, the value for speed of light (C) in metric unit is 6.71 × 10^8 mi/hr.
Final answer:
The speed of light in a vacuum is precisely known and has a fixed value of c = 2.99792458 × 10^8 m/s, roughly equal to 3.00 × 10^8 m/s. It's a fundamental constant in physics but is slower in materials, as defined by their index of refraction.
Explanation:
The speed of light in a vacuum is represented by the symbol c and has a fixed value of c = 2.99792458 × 108 m/s, which is essentially 3.00 × 108 m/s when rounded to three significant digits. This constant speed is a fundamental physical quantity and is crucial in many areas of physics including relativity and electromagnetism. It's important to note that the speed of light reduces when it passes through matter, which is characterized by the index of refraction n of the material.
The rate constant of a reaction is 7.8 × 10−3 s−1 at 25°C, and the activation energy is 33.6 kJ/mol. What is k at 75°C? Enter your answer in scientific notation.
The rate constant at 75°C is calculated using the two-point form of the Arrhenius equation. The original conditions, the new temperature, and the activation energy are substituted into the equation and solved for the new rate constant, k2. The result is k2 = 0.048, or 4.8 x 10^-2 s^-1.
Explanation:For calculating the rate constant at a different temperature, we can use the Arrhenius equation: k = Ae^(-Ea/RT) where k is the rate constant, A is the pre-exponential factor, Ea is the activation energy, R is the gas constant and T is the temperature in Kelvin.
To find the new temperature constant, we can transform the Arrhenius equation into the two-point form: ln(k2/k1) = (-Ea/R)(1/T2 - 1/T1).
Given:
k1 = 7.8 × 10−3 s−1, T1=25°C = 25 + 273 = 298K
Ea = 33.6 kJ/mol = 33,600 J/mol, R = 8.314 J/(mol·K)
T2 = 75°C = 75 + 273 = 348K
Substituting these values and solving for k2 (rate constant at 75°C), you get k2 = 0.048 or 4.8 x 10^-2 s^-1.
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What is the final volume of a balloon that was initially 500.0 mL at 25°C and was then heated to 50°C?
Answer:
V₂ =541.94 m L.
Explanation:
Given that
V₁ = 500 mL
T₁ = 25°C = 273 + 25 = 298 K
T₂ = 5°C = 273+50 =323 K
The final volume = V₂
We know that ,the ideal gas equation
If the pressure of the gas is constant ,then we can say that
[tex]\dfrac{V_2}{V_1}=\dfrac{T_2}{T_1}[/tex]
[tex]\dfrac{V_2}{V_1}=\dfrac{T_2}{T_1}[/tex]
Now by putting the values in the above equation we get
[tex]V_2=500\times \dfrac{323}{298}\ mL\\V_2=541.94\ mL[/tex]
The final volume of the balloon will be 541.94 m L.
V₂ =541.94 m L.
The final volume of the balloon will be "541.94 mL".
Given:
Volume,
[tex]V_1 = 500 \ mL[/tex]Temperature,
[tex]T_1 = 25^{\circ} C[/tex][tex]= 273+25[/tex]
[tex]= 298 \ K[/tex]
[tex]T_2 = 5^{\circ} C[/tex][tex]=273+50[/tex]
[tex]=323 \ K[/tex]
By using the Ideal gas equation, we get
→ [tex]\frac{V_2}{V_1} = \frac{T_2}{T_1}[/tex]
or,
→ [tex]V_2 = \frac{T_2\times V_1}{T_1}[/tex]
[tex]= \frac{500\times 323}{298}[/tex]
[tex]= 541.94 \ mL[/tex]
Thus the above approach is correct.
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What is the electric field strength just outside the surface of a conducting sphere carrying surface charge density 1.4 μC/m2μC/m2?
Answer:
[tex]E=158.19\frac{kN}{m}[/tex]
Explanation:
Gauss's theorem states that the flux of the electric field through a closed surface is equal to the the charge enclosed by the surface divided by the vacuum permittivity:
[tex]\int\limits{\vec{E}\cdot \vec{dS}} \,=\frac{q}{\epsilon_0}[/tex]
The direction of the electric field ([tex]\vec{E}[/tex]) just outside of a conductor is parallel to its surface ([tex]\vec{S}[/tex]):
[tex]\int\limits{\vec{E}\cdot \vec{dS}} \,=\int\limits{EdScos(0^\circ)} \,=E\int\limits{dS} \,=ES[/tex]
Recall that the surface charge density is defined as:
[tex]\sigma=\frac{q}{S}[/tex]
Now, we get the electric field strength:
[tex]ES=\frac{q}{\epsilon_0}\\E=\frac{q}{S\epsilon_0}\\E=\frac{\sigma}{\epsilon_0}\\E=\frac{1.4*10^{-6}\frac{C}{m^2}}{8.85*10^{-12\frac{C^2}{N\cdot m^2}}}\\\\E=158192.09\frac{N}{C}=158.19\frac{kN}{C}[/tex]
The electric field strength just outside the surface of a conducting sphere with surface charge density of 1.4 μC/m^2 can be calculated using Gauss' Law. This gives an electric field strength of approximately 1.58 x 10^5 N/C.
Explanation:The electric field strength just outside the surface of a conducting sphere, with a surface charge density of 1.4 μC/m2, can be calculated using Gauss' Law, which states that the electric field is equal to the surface charge density divided by the permittivity of free space, ε0.
It is given by the formula:
E = σ/ε0
Where E is the electric field strength, σ is the charge density, and ε0 is the permittivity of free space. Using the given value for σ (1.4 x 10-6 C/m2) and the known value for ε0 (8.85 x 10-12 C2/N.m2), we find that:
E = (1.4 x 10-6) / (8.85 x 10-12)
This simplifies to approximately E = 1.58 x 105 N/C.
Therefore, the electric field strength just outside the surface of the conducting sphere is approximately 1.58 x 105 N/C.
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Canada geese migrate essentially along a north–south direction for well over a thousand kilo-meters in some cases, traveling at speeds up to about 100 km/h. If one goose is flying at 100 km/h relative to the air but a 40-km/h wind is blowing from west to east, (a) at what angle relative to the north–south direction should this bird head to travel directly southward relative to the ground? (b) How long will it take the goose to cover a ground distance of 500 km from north to south? (Note: Even on cloudy nights, many birds can navigate by using the earth’s magnetic field to fix the north–south direction.)
Answer:
a) 66.4 relative to the west in the south-west direction
b) 5.455 hours
Explanation:
a)If the wind is blowing east-ward at a speed of 40km/h, then the west component of the geese velocity must be 40km/h in order to counter balance it. Geese should be flying south-west at an angle of
[tex]cos(\alpha) = 40 / 100 = 0.4[/tex]
[tex]\alpha = cos^{-1}(0.4) = 1.16 rad = 180\frac{1.16}{\pi} = 66.4^0[/tex] relative to the West
b) The south-component of the geese velocity is
[tex]100sin(\alpha) = 100sin(66.4^0) = 91.65 km/h[/tex]
The time it would take for the geese to cover 500km at this rate is
t = 500 / 91.65 = 5.455 hours
A 0.0575 kg ice cube at −30.0°C is placed in 0.557 kg of 35.0°C water in a very well insulated container, like the kind we used in class. The heat of fusion of water is 3.33 x 105 J/kg, the specific heat of ice is 2090 J/(kg · K), and the specific heat of water is 4190 J/(kg · K). The system comes to equilibrium after all of the ice has melted. What is the final temperature of the system?
Answer:
t= 22.9ºC
Explanation:
Assuming no heat exchange outside the container, before reaching to a condition of thermal equilibrium, defined by a common final temperature, the body at a higher temperature (water at 35ºC) must give heat to the body at a lower temperature (the ice), as follows:
Qw = c*m*Δt = 4190 (J/kg.ºC)*0.557 kg*(35ºC-t) (1)
This heat must be the same gained by the ice, which must traverse three phases before arriving at a final common temperature t:
1) The heat needed to reach in solid state to 0º, as ice:
Qi =ci*m*(0ºC-(-30ºC) = 0.0575kg*2090(J/kg.ºC)*30ºC = 3605.25 J
2) The heat needed to melt all the ice, at 0ºC:
Qf = cfw*m = 3.33*10⁵ J/kg*0.0575 kg = 19147.5 J
3) Finally, the heat gained by the mass of ice (in liquid state) in order to climb from 0º to a final common temperature t:
Qiw = c*m*Δt = 4190 (J/kg.ºC)*0.0575 kg*(t-0ºC)
So, the total heat gained by the ice is as follows:
Qti = Qi + Qf + Qiw
⇒Qti = 3605.25 J + 19147.5 J + 240.9*t = 22753 J + 240.9*t (2)
As (1) and (2) must be equal each other, we have:
22753 J + 240.9*t = 4190 (J/kg.ºC)*0.557 kg*(35ºC-t)
⇒ 22753 J + 240.9*t = 81684 J -2334*t
⇒ 2575*t = 81684 J- 22753 J = 58931 J
⇒ [tex]t= \frac{58931J}{2575 J/C} = 22.9C[/tex]
⇒ t = 22.9º C
A boy throws a ball upward with a speed v0 = 12 m/s. The wind imparts a horizontal acceleration of 0.4 m/s2 to the left. At what angle θ must the ball be thrown so that it returns to the point of release? Assume that the wind does not affect the vertical motion.
Answer:
The angle is 2.33°.
Explanation:
Given that,
Speed of ball = 12 m/s
Acceleration = 0.4 m/s²
We need to calculate the time
Using formula of time of flight
[tex]t=\dfrac{2u}{g}[/tex]
[tex]t=\dfrac{2v\cos\theta}{g}[/tex]
Put the value into the formula
[tex]t=\dfrac{2\times12\cos\theta}{9.8}[/tex]
[tex]t=2.44\cos\theta[/tex]
We need to calculate the angle
Using equation of motion along vertical direction
[tex]s=ut-\dfrac{1}{2}at^2[/tex]
[tex]s=v\sin\theta\times t-\dfrac{1}{2}at^2[/tex]
Put the value in the equation
[tex]0=12\sin\theta\times2.44\cos\theta-\dfrac{1}{2}\times0.4\times(2.44\cos\theta)^2[/tex]
[tex]2\times12\sin\theta\times2.44=0.4\times(2.44)^2\cos\theta[/tex]
[tex]\tan\theta=\dfrac{0.4\times2.44}{2\times12}[/tex]
[tex]\theta=\tan^{-1}(0.04066)[/tex]
[tex]\theta=2.33^{\circ}[/tex]
Hence, The angle is 2.33°.
A gun is fired with angle of elevation . What is the muzzle speed if the maximum height of the shell is 500 m?
Answer:
The muzzle speed u =200 m/s
Explanation:
Let u be the muzzle speed.
maximum height s= 500 m
Assuming angle of elevation to be 30°
then initial vertical speed = usin30° = u/2
moreover, when the bullet reaches the maximum height its vertical velocity will be zero.
then using
[tex]v^2=u^2-2as[/tex]
a= 10 m/s^2
u= u/2
v= 0
s=500
we get
[tex]0^2=u^2/4-2\times10\times500[/tex]
u= 200 m/s.
Therefore, muzzle speed = 200 m/s
Answer:
The muzzle speed is 197.9 m/s.
Explanation:
Given that,
Maximum height = 500 m
Suppose the angle is 30°.
We need to calculate the muzzle speed
Using formula of maximum height
[tex]h_{max}=\dfrac{u^2}{2g}[/tex]
Where, h = maximum height
g = acceleration due to gravity
u = initial vertical velocity
Put the value into the formula
[tex]500=\dfrac{u^2\sin^{2}30}{2\times9.8}[/tex]
[tex]u^2=8\times500\times9.8[/tex]
[tex]u=\sqrt{4\times500\times9.8}[/tex]
[tex]u=197.9\ m/s[/tex]
Hence, The muzzle speed is 197.9 m/s.
A square plate of copper with 55.0 cm sides has no net charge and is placed in a region of uniform electric field of 82.0 kN/C directed perpendicularly to the plate.a. Find the charge density of each face of the plate.b. Find the total charge on each face.
Answer
given,
Side of copper plate, L = 55 cm
Electric field, E = 82 kN/C
a) Charge density,σ = ?
using expression of charge density
σ = E x ε₀
ε₀ is Permittivity of free space = 8.85 x 10⁻¹² C²/Nm²
now,
σ = 82 x 10³ x 8.85 x 10⁻¹²
σ = 725.7 x 10⁻⁹ C/m²
σ = 725.7 nC/m²
change density on the plates are 725.7 nC/m² and -725.7 nC/m²
b) Total change on each faces
Q = σ A
Q = 725.7 x 10⁻⁹ x 0.55²
Q = 219.52 nC
Hence, charges on the faces of the plate are 219.52 nC and -219.52 nC
The magnitude of the velocity vector of the car is ∣∣v→∣∣ = 78 ft/s. If the vector v→ forms an angle θ = 0.09 rad with the horizontal direction, determine the Cartesian representation of v→ relative to the (iˆ, jˆ) component system.
Answer:
[tex]\vec{v} = (77.68~{\rm ft/s})\^i + (7.01~{\rm ft/s})\^j[/tex]
Explanation:
The x- and y- components of the velocity vector can be written as following:
[tex]\vec{v}_x = ||\vec{v}||\cos(\theta)\^i[/tex]
[tex]\vec{v}_y = ||\vec{v}||\sin(\theta)\^j[/tex]
Since the angle θ and the magnitude of the velocity is given, the vector representation can be written as follows:
[tex]\vec{v} = 78\cos(0.09)\^i + 78\sin(0.09)\^j\\\vec{v} = (77.68~{\rm ft/s})\^i + (7.01~{\rm ft/s})\^j[/tex]
A particle has a velocity of v→(t)=5.0ti^+t2j^−2.0t3k^m/s.
(a) What is the acceleration function?
(b) What is the acceleration vector at t = 2.0 s? Find its magnitude and direction.
Answer:
a)[tex]a=5 i+2t j - 6\ t^2k[/tex]
b)[tex]a=\dfrac{1}{24.83}(5i+4j-24k)\ m/s^2[/tex]
Explanation:
Given that
v(t) = 5 t i + t² j - 2 t³ k
We know that acceleration a is given as
[tex]a=\dfrac{dv}{dt}[/tex]
[tex]\dfrac{dv}{dt}=5 i+2t j - 6\ t^2k[/tex]
[tex]a=5 i+2t j - 6\ t^2k[/tex]
Therefore the acceleration function a will be
[tex]a=5 i+2t j - 6\ t^2k[/tex]
The acceleration at t = 2 s
a= 5 i + 2 x 2 j - 6 x 2² k m/s²
a=5 i + 4 j -24 k m/s²
The magnitude of the acceleration will be
[tex]a=\sqrt{5^2+4^2+24^2}\ m/s^2[/tex]
a= 24.83 m/s²
The direction of the acceleration a is given as
[tex]a=\dfrac{1}{24.83}(5i+4j-24k)\ m/s^2[/tex]
a)[tex]a=5 i+2t j - 6\ t^2k[/tex]
b)[tex]a=\dfrac{1}{24.83}(5i+4j-24k)\ m/s^2[/tex]
Temperature change time base problem. Suppose V = 24 V, I = 0.1 A, for water: mw = 51 gm, cw = 4.18 J/gm ∘K-1, for resistor: mr = 8 gm, and cr = 3.7 J/gm ∘K-1. If the water is initially at room temperature, how long will it take for the water to heat up 5∘K? (Hint: dT/dt is approximately equal to Δ T / Δt .)
Answer:
[tex]t=5057.9167\ s[/tex]
Explanation:
Given:
Voltage supply to the resistor, [tex]V=24\ V[/tex]current supply to the resistor, [tex]I=0.1\ A[/tex]mass of water, [tex]m_w= 51\ g[/tex]specific heat of water, [tex]c_w=4180\ J.kg^{-1}.K^{-1}[/tex]specific heat of resistor, [tex]c_r=3700\ J.kg^{-1}.K^{-1}[/tex]mass of resistor, [tex]m_r=0.008\ kg[/tex]change in temperature, [tex]\Delta T=50\ K[/tex]Now the amount of heat required to heat the water by 50 K:
[tex]Q_w=m_w.c_w.\Delta T[/tex]
[tex]Q_w=0.051\times 4180\times 50[/tex]
[tex]Q_w=10659\ J[/tex]
Now the amount of heat required to heat the resistor by 50 K:
[tex]Q_r=m_r.c_r.\Delta T[/tex]
[tex]Q_r=0.008\times 3700\times 50[/tex]
[tex]Q_r=1480\ J[/tex]
Now the total heat to converted from the electrical energy:
[tex]Q=Q_w+Q_r[/tex]
[tex]Q=12139\ J[/tex]
Now Using Joule's law of heating:
[tex]Q=V.I.t[/tex]
[tex]12139=24\times 0.1\times t[/tex]
[tex]t=5057.9167\ s[/tex]
Final answer:
To determine the time needed to heat water by 5°K, calculate the total energy needed using the specific heat capacity of water, then use the power equation P = V × I to find the time. It will take approximately 444.125 seconds for the water to reach the desired temperature.
Explanation:
To calculate the time it will take for the water to heat up by 5°K using the provided values, first determine the total energy required to raise the temperature of the water by using the equation E = mw × cw × ΔT. Here, E is the energy in joules (J), mw is the mass of water, cw is the specific heat capacity of water, and ΔT is the change in temperature.
Using the given figures:
E = 51 g × 4.18 J/g°K × 5°K = 1065.9 J
Since power P is the rate at which energy is used and P = V × I where V is the voltage and I is the current, we can rearrange the equation to find time: t = E / P.
Substitute the power using the given voltage and current:
P = 24 V × 0.1 A = 2.4 W
The time in seconds to heat the water 5°K is thus:
t = 1065.9 J / 2.4 W = 444.125 seconds
Therefore, it will take approximately 444.125 seconds to heat up the water by 5°K.
The nose of an ultralight plane is pointed south, and its airspeed indicator shows 39 m/s m/s . The plane is in a 17 m/s m/s wind blowing toward the southwest relative to the earth. For help with math skills, you may want to review: Vector Addition Resolving Vector Components For general problem-solving tips and strategies for this topic, you may want to view a Video Tutor Solution of Flying in a crosswind. Part A Letting x be east and y be north, find the components of v ⃗ P/E v→P/E (the velocity of the plane relative to the earth). Express your answers in meters per second separated by a comma. View Available Hint(s) v x vx , v y vy = nothing m/s m/s Submit Part B Find the magnitude of v ⃗ P/E v→P/E . Express your answer in meters per second. View Available Hint(s) v P/E vP/E = nothing m/s m/s Submit Part C Find the direction of v ⃗ P/E v→P/E . Express your answer in degrees. View Available Hint(s) ϕ ϕ = nothing ∘ ∘ south of west Submit Provide Feedback Next
Answer:
A. [tex]|\vec v_t|=52.42\ m/s[/tex]
B. [tex]\theta=256.74^o[/tex]
Explanation:
Velocity Vector
The velocity vector has two components. Depending on the reference system they could be magnitude and direction in the polar coordinates or x-component and y-component in the rectangular coordinates system.
We are given two velocities in the form of magnitude-direction. The plane's velocity goes south at 39 m/s. The zero reference for angles is pointed East, so the south direction has a 270° angle respect to the reference. If the polar coordinates are known, the rectangular coordinates are computed as
[tex]v_{xp}=|v_p|cos\alpha_p[/tex]
[tex]v_{yp}=|v_p|sin\alpha_p[/tex]
[tex]Since\ |v_p|=39 m/s,\ \alpha_p=270^o,[/tex]
[tex]v_{xp}=39cos\ 270^o=0[/tex]
[tex]v_{yp}=39sin\ 270^o=-39[/tex]
Thus, the velocity of the plane is
[tex]\vec v_p=0\hat i-39\hat j[/tex]
The wind is blowing toward the southwest. It means its angle is 225° (3rd quadrant):
[tex]v_{xw}=|v_w|cos\alpha_w[/tex]
[tex]v_{yw}=|v_w|sin\alpha_w[/tex]
[tex]v_{xw}=17cos\ 215^o=-12.02[/tex]
[tex]v_{yw}=17sin\ 215^o=-12.02[/tex]
Thus, the velocity of the wind is
[tex]\vec v_w=-12.02\hat i-12.02\hat j[/tex]
Now we perform the vector addition to compute the plane's final speed
[tex]\vec v_t=\vec v_p+\vec v_w[/tex]
[tex]\vec v_t=0\hat i-39\hat j-12.02\hat i-12.02\hat j[/tex]
[tex]\vec v_t=-12.02\hat i-51.02\hat j[/tex]
A) The magnitude of the total velocity is
[tex]|\vec v_t|=\sqrt{(-12.02)^2+(-51.02)^2}[/tex]
[tex]\boxed{|\vec v_t|=52.42\ m/s}[/tex]
B) The direction angle is given by
[tex]\displaystyle \tan\theta=\frac{-51.02}{-12.02}=4.24[/tex]
[tex]\theta=arctan\ 4.24[/tex]
[tex]\theta=76.74^o[/tex]
This angle is west of south, we must add 180° to express it in due reference, thus
[tex]\boxed{\theta=256.74^o}[/tex]
How much taller (in m) does the Eiffel Tower become at the end of a day when the temperature has increased by 17°C? Its original height is 324 m and you can assume it is made of steel.
Answer:
324.066096 m.
Explanation:
Given that
height of the tower ,h= 324 m
The increase in temperature ,ΔT = 17°C
We know that coefficient of thermal expansion for steel ,α= 12 x 10⁻⁶ C⁻¹
The increase in height is given as
Δ h = α h ΔT
Now by putting the values in the above equation we get
Δ h= 12 x 10⁻⁶ x 324 x 17 m
Δ h=66096 x 10⁻⁶ m
Δ h=0.066096 m
Therefore the height of the tower become 324.066096 m.
Due to thermal expansion, the Eiffel Tower, made of steel, would increase in height by approximately 0.066 meters or 6.6 cm over a day when the temperature increases by 17°C.
Explanation:The height of the Eiffel Tower increases due to the phenomenon of thermal expansion, which is an increase in volume, including height, in response to an increase in temperature. The amount of expansion can be calculated using this formula: ΔL = α * L_original * ΔT. Let's apply the given values:
'α' the coefficient of linear expansion for steel is approximately 0.000012 per degree Celsius. 'L_original' is the original length in meters, which is 324 m. 'ΔT' is the change in temperature, which is 17°C.
So ΔL = 0.000012 * 324 * 17, which equates to approximately 0.066048 m. Therefore, the Eiffel Tower would increase in height by about 0.066 meters (or 6.6 cm) over the course of a day when the temperature increases by 17°C.
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At the point of fission, a nucleus of 235U that has 92 protons is divided into two smaller spheres, each of which has 46 protons and a radius of 5.9 × 10−15 m. What is the magnitude of the repulsive force pushing these two spheres apart? The value of the Coulomb constant is 8.98755 × 109 N · m2 /C 2 .
Answer:
Force = 3481.1 N.
Explanation:
Below is an attachment containing the solution.
You push a shopping cart filled with groceries (total mass = 20 kg) by applying a force to the cart 30° from the horizontal. If the force you apply has a magnitude of 86 N, what is the cart’s acceleration? Assume negligible friction.You push a shopping cart filled with groceries (total mass = 20 kg) by applying a force to the cart 30° from the horizontal. If the force you apply has a magnitude of 86 N, what is the cart’s acceleration? Assume negligible friction.2.2 m/s25.5 m/s23.7 m/s24.3 m/s2
Answer:
3.72 m/s²
Explanation:
Horizontal component of the force applied = Fcosθ = 86 cos 30 = 74.478 N
Force = mass × acceleration = 74.478 N
ma = 74.478 N
a = 74.478 / 20 since friction can be neglected = 3.72 m/s²
Answer:
[tex]3.7m/s^{2}[/tex]
Explanation:
The forces acting on the cart is displayed in the attached file below.
Since the motion is only along the horizontal, we can conclude that the force bringing about that motion is the horizontal force.And from second newton law of motion, we deduce that
F=ma,
F=force, m=mass and a=acceleration
from the given question,
F=86N, mass,m=20kg,
Hence we can write the horizontal component of the force as
[tex]F_{X}=Fcos\alpha \\F_{x}=86cos30\\F_{x}=74.48\\Hence \\a=\frac{F_{x}}{m} \\a=\frac{74.48}{20}\\ a=3.7m/s^{2}[/tex]
A cliff diver positions herself on a cliff that angles downwards towards the edge. The length of the top of the cliff is 50.0 m and the angle of the cliff is θ = 21.0° below the horizontal. The cliff diver runs towards the edge of the cliff with a constant speed, and reaches the edge of the cliff in a time of 6.10 s. After running straight off the edge of the cliff (without jumping up), the diver falls h = 30.0 m before hitting the water.
After leaving the edge of the cliff how much time does the diver take to get to the water?
How far horizontally does the diver travel from the cliff face before hitting the water?
Remember that the angle is at a downward slope to the right.
Final answer:
The diver takes approximately 2.18 seconds to reach the water after leaving the edge of the cliff. The diver travels approximately 3.21 meters horizontally from the cliff face before hitting the water.
Explanation:
To find the time it takes for the diver to reach the water after leaving the edge of the cliff, we can use the equation of motion:[tex]h = 1/2 * g * t^2,[/tex]where h is the height of the cliff and g is the acceleration due to gravity. Rearranging the equation to solve for t, we get t = sqrt(2h/g). Plugging in the values given, we have t = sqrt(2*30/9.8) ≈ 2.18 s.
To find the horizontal distance the diver travels, we can use the equation s = v * t, where s is the distance, v is the horizontal velocity, and t is the time. Rearranging the equation to solve for v, we get v = s/t. Plugging in the values given, we have v = 7/2.18 ≈ 3.21 m/s.
An object’s velocity is measured to be vx(t) = α - βt2, where α = 4.00 m/s and β = 2.00 m/s3. At t = 0 the object is at x = 0. (a) Calculate the object’s position and acceleration as functions of time. (b) What is the object’s maximum positive displacement from the origin?
Answer:
Explanation:
Given
[tex]v_x(t)=\alpha -\beta t^2[/tex]
[tex]\alpha =4\ m/s[/tex]
[tex]\beta =2\ m/s^3[/tex]
[tex]v_x(t)=4-2t^2[/tex]
[tex]v=\frac{\mathrm{d} x}{\mathrm{d} t}[/tex]
[tex]\int dx=\int \left ( 4-2t^2\right )dt[/tex]
[tex]x=4t-\frac{2}{3}t^3[/tex]
acceleration of object
[tex]a=\frac{\mathrm{d} v}{\mathrm{d} t}[/tex]
[tex]a=-4t[/tex]
(b)For maximum positive displacement velocity must be zero at that instant
i.e.[tex]v=0[/tex]
[tex]4-2t^2=0[/tex]
[tex]t=\pm \sqrt{2}[/tex]
substitute the value of t
[tex]x=4\times \sqrt{2}-\frac{2}{3}\times 2\sqrt{2}[/tex]
[tex]x=3.77\ m[/tex]
The definitions of acceleration and velocity allow to find the results for the questions about the motion of the particle are:
A) the function of the acceleration is: a = -4t
and the position function is: x = 4 t - ⅔ t³
B) The maximum displacement is: x = 3.77 m
Given parameters
The velocity of the body v = α-β t² with α = 4 m/s and β = 2 m/s²To find
a) position and acceleration as a function of time,
b) maximum displacement,
The acceleration of defined as the change in velocity with time.
a = [tex]\frac{dv}{dt}[/tex]
Let's calculate.
a = - 2βt
a = - 2 2 t
a = -4 t
The speed is defined by the variation of the position with respect to time.
v = [tex]\frac{dx}{dt}[/tex]
dx = v dt
We integrate.
∫ dx = ∫ v dt
x - x₀ = ∫ (α - β t²)
x-x₀ = αt - βt³/ 3
we substitute.
x = 4 t - ⅔ t³
B) To find the maximum displacement we use the first derivative to be zero.
[tex]\frac{dx}{dt}[/tex] = 0
4 - 2t² = 0
t² = 2
t = √2 = 1.414 s
Let's find the position for this time.
x = 4 √2 - ⅔ (√2)³
x = 3.77 m
In conclusion using the definitions of acceleration and velocity we can find the result for the questions about the motion of the particle are:
A) the function of the acceleration is: a = -4t
and the position function is: x = 4 t - ⅔ t³
B) The maximum displacement is: x = 3.77 m
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Tim and Rick both can run at speed v_r and walk at speed v_w, with v_r > v_w They set off together on a journey of distance D. Rick walks half of the distance and runs the other half. Tim walks half of the time and runs the other half.
How long does it take Rick to cover the distance D?
Express the time taken by Rick in terms of v_r, v_w, and D.
Find Rick's average speed for covering the distance D.
Express Rick's average speed in terms of v_r and v_w.
How long does it take Tim to cover the distance?
Express the time taken by Tim in terms of v_r, v_w, and D.
Who covers the distance D more quickly?
In terms of given quantities, by what amount of time, Delta t, does Tim beat Rick?
It will help you check your answer if you simplify it algebraically and check the special case v_r = v_w
Express the difference in time, Delta t in terms of v_r, v_w, and D.
In the special case that v_r = v_w, what would be Tim's margin of victory Delta t(v_r = v_w)?
A ray of light is incident on an air/water interface.
The ray makes an angle of θ1 = 32 degrees with respect to the normal of the surface. The index of the air is n1 = 1 while water is n2 = 1.33.
Choose an expression for the angle (relative to the normal to the surface) for the ray in the water, θ2.
a) θ2 = sin (θ1).n1/n2
b) θ2 = asin (n1/n2)
c) θ2 = asin (sin(θ1).n2/n1)
d) θ2 = asin (sin(θ1).n1/n2)
Answer:
[tex]\theta_2=sin^{-1}(\dfrac{n_1\ sin\theta_1}{n_2})[/tex]
Explanation:
Given that,
The ray makes an angle of 32 degrees with respect to the normal of the surface.
The refractive index of air, [tex]n_1=1[/tex]
The refractive index of water, [tex]n_2=1.33[/tex]
Snell's law is given by :
[tex]n_1\ sin\theta_1=n_2\ sin\theta_2[/tex]
[tex]sin\theta_2=\dfrac{n_1\ sin\theta_1}{n_2}[/tex]
[tex]\theta_2=sin^{-1}(\dfrac{n_1\ sin\theta_1}{n_2})[/tex]
So, option (4) is correct. Hence, this is the required solution.
The answer is option d.
The correct expression for the angle (relative to the normal to the surface) for the ray in the water is d) θ2 = asin (sin(θ1).n1/n2), based on Snell's law of refraction.
Explanation:The question addresses the refraction of light, specifically the change in angle as light moves from air to water. According to Snell's law, which is used to calculate the angle of refraction, the correct expression in your options is d) θ2 = asin (sin(θ1).n1/n2). Here's a step by step process:
First, it's important to understand that light changes direction when it moves from one medium to another, a process known as refraction.Snell's law mathematically expresses this change and is written as n1*sin(θ1) = n2*sin(θ2). In your case, you want to find the angle θ2. So, rearranging Snell's law to solve for θ2 gives you θ2 = asin(n1*sin(θ1)/n2).Learn more about Refraction:https://brainly.com/question/2459833?referrer=searchResults
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A hockey pick sliding along a frictional surface strikes a box at rest, after the collision the two objects stick together and move at the same final speed. Which of the following describes the change in momentum and energy of the puck during the collision?
a. puck loses some but not all of its original momentum.
b. one cannot determine
c. puck conserves original momentum, but loses all mechanical energy
d. puck loses some momentum but conserves mechanical energy
e. puck loses conserves all momentum and mechanical energy
f. conserves momentum but loses some mechanical energy
Answer:
Explanation:
Option a is correct
If puck and pick constitute a system then the momentum of the system is conserved but not this may not be valid for the puck .
Option e is correct
If puck and pick is the system then momentum is conserved but because of the presence of friction, mechanical energy is not conserved.
Friction will cause the energy to dissipate in heat.
A trapezoidal channel with 6.0 ft bed width, 3 ft water depth and 1:1 side slope, carries a discharge of 250 ft3/s. Determine whether the flow is supercritical or subcritical.
Answer
given,
width of trapezoidal channel, b = 6 ft
depth of water, d = 3 ft
discharge,Q = 250 ft³/s
now, we have to calculate Froude number
[tex]F_r = \dfrac{Q}{A\sqrt{gD}}[/tex]
Where D is the hydraulic radius
[tex]D = \dfrac{A}{P}[/tex]
[tex]F_r = \dfrac{Q}{A\sqrt{g\times \dfrac{A}{P}}}[/tex]
P is the width of the channel
P = b + 2 z d
P = 6 + 2 x 1 x 3
P = 13 ft
A = d(b + z d) = 3 (6 + 3) = 27 ft²
g = 32.2 ft/s²
now,
[tex]F_r = \dfrac{Q}{A\sqrt{g\times \dfrac{A}{P}}}[/tex]
[tex]F_r = \dfrac{250}{27\sqrt{32.2\times \dfrac{27}{13}}}[/tex]
F_r = 1.087
F_r > 1
Froude number is greater than 1 so, the flow is Super critical flow.
An object starts from rest and accelerates at a rate of 2 rad/s^2 until it reaches an angular speed of 24 rad/s. The object then accelerates at a rate of -3 rad/s2 until it stops. Through what angular displacement (in rad) does the object move between when it starts moving and when it stops?
Answer:
Total angular displacement will be 240 radian
Explanation:
In first case object starts from rest so initial angular speed [tex]\omega _i=0rad/sec[/tex]
Angular acceleration is given [tex]\alpha =2rad/sec^2[/tex]
Final angular speed[tex]\omega _f=24rad/sec[/tex]
From third equation of motion [tex]\omega _f^2=\omega _i^2+2\alpha \Theta[/tex]
So [tex]24^2=0^2+2\times 2\times \Theta[/tex]
[tex]\Theta =144[/tex] radian
Now in second case as the objects finally stops
So final velocity [tex]\omega _f=0rad/sec[/tex]
Angular acceleration [tex]\alpha =-3rad/sec^2[/tex]
So [tex]0^2=24^2-2\times 3\times \Theta[/tex]
[tex]\Theta =96[/tex] radian
So total angular displacement will be 96+144 = 240 radian
To find the angular displacement, we can analyze the motion of the object in two phases: the first phase of acceleration and the second phase of deceleration. We can calculate the time and angular displacement in each phase using the formulas for angular speed and angular displacement. Finally, we can add the angular displacements from both phases to find the total angular displacement.
Explanation:To find the angular displacement, we need to analyze the motion of the object in two phases: the first phase of acceleration and the second phase of deceleration. In the first phase, the object starts from rest and accelerates at a rate of 2 rad/s^2 until it reaches an angular speed of 24 rad/s. We can use the formula:
Final Angular Speed = Initial Angular Speed + Angular Acceleration * Time
Using this formula, we can find the time taken in the first phase. Then, we can calculate the angular displacement during the first phase using the formula:
Angular Displacement = Initial Angular Speed * Time + 0.5 * Angular Acceleration * Time^2
In the second phase, the object decelerates at a rate of -3 rad/s^2 until it stops. We can use the same formulas to find the time and angular displacement in the second phase. Finally, we can add the angular displacements from both phases to get the total angular displacement.
The total angular displacement will be the sum of the angular displacements in the two phases.
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A driver has a reaction time of 0.50s , and the maximum deceleration of her car is 6.0m/s2 . She is driving at 20m/s when suddenly she sees an obstacle in the road 50m in front of her.
Can she stop the car in time to avoid the collision?
Answer:
given,
reaction time.t_r = 0.50 s
deceleration of the car = 6 m/s²
initial speed,v = 20 m/s
distance at which the car stop = ?
distance travel by the car in reaction time
d= v x t_r
d = 20 x 0.5 = 10 m
using equation of motion
distance travel to stop the car
v² = u² + 2 a s
0² = 20² - 2 x 6 x s
12 s = 400
s = 33.33 m
Total distance travel by the car
D = 10 + 33.33
D = 43.33 m
Hence, the car stops before to avoid collision.
Raindrops hitting the side windows of a car in motion often leave diagonal streaks even if there is no wind. Why? Is the explanation the same or different for diagonal streaks on the windshield?
Answer:
because of the raindrop velocity relative of the car has a vertical and horizontal component
Explanation:
The car moves in a horizontal direction relative to the ground. The raindrops fall in the vertical direction relative to the ground. Their velocity relative to the moving car has both vertical and horizontal components and this is the reason for the diagonal streaks on the side window. The diagonal streaks on the windshield arise from a different reason. The drops are pushed off to one side of the windshield because of air resistance.By what order of magnitude is something that runs in nanoseconds faster than something that runs in milliseconds?
To solve this problem we will define the order of magnitude of both points, then we will obtain the radius and obtain the conclusion of the order of magnitude.
A nanosecond is one billionth of a second while and a millisecond is one millionth of a second
[tex]\frac{\text{millisec}}{\text{nanosec}} = \frac{10^{-3}}{10^{-9}} = 10^6[/tex]
Therefore something that runs in nanoseconds is six times faster than something that runs in milliseconds
Word magnitude means the extent of something. It is the property that determines the object is larger or smaller than the other. The object's magnitude can be arranged into class.
Nano second is the SI unit of time that is equal to bn of a second. That is 10⁻⁹ seconds. The nanosecond is a one billion th of the second, whereas an millisecond is 1000th of a second. The nanosecond process is thus 1,000,000 times faster.The correct answer is 100,00,00.
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The driver of a car wishes to pass a truck that is traveling at a constant speed of 20.0 m/s (about 45 mi/h). Initially, the car is also traveling at 20.0 m/s, and its front bumper is 24.0 m behind the truck’s rear bumper. The car accelerates at a constant 0.600 m/s2, then pulls back into the truck’s lane when the rear of the car is 26.0 m ahead of the front of the truck. The car is 4.5 m long, and the truck is 21.0 m long. (a) How much time is required for the car to pass the truck? (b) What distance does the car travel during this time? (c) What is the final speed of the car?
Answer:
a) 15.864s
b) 392.78m
c) 29.52 m/s
Explanation:
The total distance (relative to the truck) that the (front bumper of the) car travels from 24m behind the truck's rear bumper to in front of the car is
distance from car's front bumper to the truck's rear bumper + distance from truck's rear bumper to truck's front bumper (truck's length) + distance from truck's front bumper to car's rear bumper's + distance from the car's rear bumper to the car's front bumper (car's length)
= 24 + 21 + 26 + 4.5 = 75.5 m
As they start at the same speed, we can draw the following equation of motion for the car distance relative to the truck
[tex]s = at^2/2[/tex]
[tex]75.5 = 0.6t^2/2[/tex]
[tex]t^2 = 251.67[/tex]
[tex]t = \sqrt{251.67} = 15.864s[/tex]
b) The actual distance relative to Earth that the car has traveled during this time is the distance car traveled relative to the truck plus distance truck traveled relative to Earth within this time
= 75.5 + 20*15.864 = 392.78 m
c) final speed of the car is the initial speed plus the change in speed
[tex]v = v_0 + \Delta v = v_0 + at = 20 + 15.864*0.6 = 29.52 m/s[/tex]
To pass the truck, it takes the car 33.3 seconds to accelerate and overtake the truck. The car travels a distance of 333 meters during this time. The final speed of the car is 1.80 m/s.
Explanation:u is the initial velocity of the car and a is the acceleration. Since the car is initially traveling at the same speed as the truck (20.0 m/s) and accelerates at a constant rate of 0.600 m/s², the equation becomes: t = (0 - 20) / -0.600. Solving for t gives us t = 33.3 seconds. To find the distance traveled by the car during this time, we can use the equation: s = ut + (1/2)at², where s is the distance, u is the initial velocity, t is the time, and a is the acceleration. Plugging in the values, we get: s = 20(33.3) + (1/2)(-0.600)(33.3)². Solving for s gives us s = 333 meters. To find the final speed of the car, we can use the equation: v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time. Plugging in the values, we get: v = 20 + (-0.600)(33.3). Solving for v gives us v = 1.80 m/s.
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Two identical loudspeakers 2.0 m apart are emitting 1800 Hz sound waves into a room where the speed of sound is 340 m/s.
Is the point 4.0 m directly in front of one of the speakers, perpendicular to the plane of the speakers, a point of maximum constructive interference, perfect destructive interference, or something in between?
Answer:
a point of destructive interference.
Explanation:
the wavelength of the sound:
λ= v/f
v= velocity of sound =340 m/s
f= frequency of sound wave= 1800 Hz
L_1 = 4 m
then speaker is at the distance of
[tex]L_2 = sqrt(4^2+2^2)[/tex]
= 2√5 m
ΔL = L_2-L_1
x = ΔL/λ
Now, if this result is an integer, the waves will add up at the point. If it is nearly an integer + 0.5, the waves will have a destructive interference at the point. If it is neither of them , then point is "something in between".
[tex]x= \frac{2\sqrt{5}-4 }{\frac{340}{1800} } =2.4995[/tex]
which is close to 2.5, an integer + 0.5. So it's a point of destructive interference.
The result is within an integer value of +0.5, thus its a point of destructive interference.
The given parameters;
distance between the speakers, d = 2.0 mfrequency, f = 1800 Hzspeed of the sound, v = 340 m/sdistance below the speakers, c = 4 mThe resultant distance between the speakers is calculated as follows;
[tex]L = \sqrt{2^2 + 4^2} \\\\L = 4.47 \ m[/tex]
The wavelength of the sound wave is calculated as;
[tex]v = f\lambda\\\\\lambda = \frac{v}{f} \\\\\lambda = \frac{340}{1800} \\\\\lambda = 0.188 \ m[/tex]
Now, determine if the point is constructive interference, perfect destructive interference, or something in between?
[tex]x = \frac{\Delta L}{\lambda} \\\\x = \frac{4.47 - 4}{0.188} \\\\x = 2.5[/tex]
The result is within an integer value of +0.5, thus its a point of destructive interference.
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