To find the position, velocity, and acceleration of the snowboarder at t = 12 seconds, evaluate the given displacement equation and its first and second derivatives at t = 12 seconds, then round the final answers to one decimal place.
Explanation:To determine the position, velocity, and acceleration of the snowboarder at t = 12 seconds, we need to evaluate the given displacement function x(t) = 0.5t³ + t² + 2t and its derivatives at that specific time value.
Position at t = 12 seconds
By substituting t = 12 into the displacement function, we get:
x(12) = 0.5(12)³ + (12)² + 2(12), which can be calculated to give the position x.
Velocity at t = 12 seconds
The velocity v(t) is the first derivative of the displacement function, v(t) = 1.5t² + 2t + 2. Evaluate v(12) to find the velocity at t = 12 seconds.
Acceleration at t = 12 seconds
Acceleration a(t) is the second derivative of the displacement function, a(t) = 3t + 2. Evaluate a(12) to find the acceleration at t = 12 seconds.
Using these steps, you can calculate the exact values and round them to one decimal place as asked in the question.
If a monochromatic light beam with quantum energy value of 2.9 eV incident upon a photocell where the work function of the target metal is 1.8 eV, what is the maximum kinetic energy of ejected electrons?
Answer:1.1 eV
Explanation:
Given
Energy(E)=2.9 eV
Work function of the target=1.8 eV
We know that
[tex]Energy(E)=W_0+KE_{max}[/tex]
Where [tex]W_0=work\ function[/tex]
[tex]2.9=1.8+KE_{max}[/tex]
[tex]KE_{max}=2.9-1.8=1.1 eV[/tex]
What is the density of water vapor in g/m^3 on a hot dry day in the desert when the temperature is 50.0°C and the relative humidity is 6.50%?
Answer:
[tex]Density = 538 \frac{g}{m3} [/tex]
Explanation:
To get the density you need the vapor pressure for the moisture, to get this first you need to find in tables (found in internet, books, apps) the saturation vapor pressure for water at 50°C:
[tex]P_{sat} = 12350 Pa @ 50°C[/tex]
Now, a relative humidity of 6,5% means that the actual vapor pressure is 6,5% that of the saturated air so:
[tex]P_{vapor} = 12350 Pa * 0,065 = 802,75 Pa = 0,792 atm [/tex]
According to the ideal gases formula density can be calculated as:
[tex]d = \frac{P*M}{R*T}[/tex]
Where:
Ideal gases constant [tex]R = 0,082 \frac{atm L}{mol k}[/tex]
Pressure P = 0,792 atm
Temperature T = 50°C = 323 K
molar mass M = 18 g/mol
[tex]d = 0,538 \frac{g}{L} = 538 \frac{g}{m3} [/tex]
Suppose the electrons and protons in 1g of hydrogen could be separated and placed on the earth and the moon, respectively. Compare the electrostatic attraction with the gravitational force between the earth and the moon. ( the number of atoms in 1g of hydrogen is Avogadro's number Na. There is one electron and one proton in a hydrogen atom. ) Please explain step by step
Answer:
The gravitational force is 3.509*10^17 times larger than the electrostatic force.
Explanation:
The Newton's law of universal gravitation and Coulombs law are:
[tex]F_{N}=G m_{1}m_{2}/r^{2}\\F_{C}=k q_{1}q_{2}/r^{2}[/tex]
Where:
G= 6.674×10^−11 N · (m/kg)2
k = 8.987×10^9 N·m2/C2
We can obtain the ratio of these forces dividing them:
[tex]\frac{F_{N}}{F_{C}}=\frac{Gm_{1}m_{2}}{kq_{1}q_{2}}=0.742\times10^{-20}\frac{C^{2}}{kg^{2}}\frac{m_{1}m_{2}}{q_{1}q_{2}}[/tex] --- (1)
The mass of the moon is 7.347 × 10^22 kilograms
The mass of the earth is 5.972 × 10^24 kg
And q1=q2=Na*e=(6.022*10^23)*(1.6*10^-19)C=9.635*10^4 C
Replacing these values in eq1:
[tex]\frac{F_{N}}{F_{C}}}}=0.742\times10^{-20}\frac{C^{2}}{kg^{2}}\frac{7.347\times5.972\times10^{46}kg^{2}}{(9.635\times10^{4})^{2}}[/tex]
Therefore
[tex]\frac{F_{N}}{F_{C}}}}=3.509\times10^{17}[/tex]
This means that the gravitational force is 3.509*10^17 times larger than the electrostatic force, when comparing the earth-moon gravitational field vs 1mol electrons - 1mol protons electrostatic field
A Car travel at a speed of 200 km/hr. How far it will go in 15 mins? 27.7 km 66.1 km7.70 km50.0 km8.33 km"
Answer:
So car will go 50 km in 15 minutes
Explanation:
We have given speed of the car = 200 km/hour
Time t = 15 minutes
We know that 1 hour = 60 minute
So 15 minute = [tex]\frac{15}{60}=0.25hour[/tex]
We have to find the distance
We know that distance = speed ×time = 200×0.25=50 km
So car will go 50 km in 15 minutes
So option (d) will be the correct option
The force exerted on a bridge pier in a river is to be tested in a 1:10 scale model using water as the working fluid. In the prototype the depth of water is 2.0 m, the velocity of flow is 1.5 m/s and the width of the river is 20 m. If the hydrodynamic force on the model bridge pier is 5 N, what would it be on the prototype? (hint: pressure ratio is equal to the length-scale ratio) (5 points; Ans: 5000 N)
Answer:
[tex]f_p = 5000 N[/tex]
Explanation:
GIVEN DATA:
pressure ratio = length ratio
force = 5 N
scale = 1:10
velocity = 1.5 m/s
[tex]B_p = 20 m[/tex]
[tex]h_p = 2m[/tex]
As pressure ratio = length ratio so we have
[tex]\frac{p_m}{p_p} =\frac{l_m}{l_p} =\frac{1}{10}[/tex]
[tex]\frac{f_m *A_m}{f_p *A_p} ==\frac{1}{10}[/tex]
[tex]\frac{f_m}{f_p} * \frac{A_p}{A_m} = \frac{1}{10}[/tex]
[tex] \frac{f_m}{f_p} * \frac{b_p*h_p}{b_m*h_m} =\frac{1}{10}[/tex]
[tex]5 * \frac{1}{\frac{1}{10}} *\frac{1}{\frac{1}{10}} = \frac{f_p}{10}[/tex]
solving F_p
[tex]f_p = 5000 N[/tex]
Which of the following statements for a solenoid is NOT true? a) Your right-hand thumb must be pointed in the direction of the south pole. b) The fingers of your right hand are curled in the direction of the electric current c) To determine the magnetic field you grasp the coil in your right hand. d) A solenoid consists of a coiled conductor
Answer:
(a) Your right-hand thumb must be pointed in the direction of the south pole.
Explanation:
A solenoid is a coil of insulated copper wire which behaves like a magnet when an electric current passes through it and it loses its magnetic behaviour just after the current flow stops.
Since it behaves a magnet, the magnetic poles must be determined. In order to determine the magnetic pole, we put the coil in our right hand, curl our four fingers in the direction of current flow and then the direction in which our thumb points is the north pole. This is known as the Right-hand rule.
From the above discussion, option (a) does not keep pace with the statement of right-hand rule.
Hence, option (a) is not true.
Two protons in an atomic nucleus are typically separated by a distance of 2 ✕ 10-15 m. The electric repulsion force between the protons is huge, but the attractive nuclear force is even stronger and keeps the nucleus from bursting apart. What is the magnitude of the electric force between two protons separated by 2.00 ✕ 10-15 m
Answer:
The magnitude of the electric force between the to protons will be 57.536 N.
Explanation:
We can use Coulomb's law to find out the force, in scalar form, will be:
[tex]F \ = \ \frac{1}{4 \pi \epsilon_0 } \frac{q_1 q_2}{d^2}[/tex].
Now, making the substitutions
[tex]d \ = \ 2.00 * 10 ^{-15} \ m[/tex],
[tex]q_1 = q_2 = 1.60 * 10 ^ {-19} \ C[/tex],
[tex]\frac{1}{4\pi\epsilon_0}=8.99 * 10^9 \frac{Nm^2}{C^2}[/tex],
we can find:
[tex]F \ = \ 8.99 * 10^9 \frac{Nm^2}{C^2} \frac{(1.60 * 10 ^ {-19} \ C)^2}{(2.00 * 10 ^{-15} \ m)^2}[/tex].
[tex]F \ = 57.536 N[/tex].
Not so big for everyday life, but enormous for subatomic particles.
The magnitude of the electric force between two protons separated by 2.00 × 10⁻¹⁵ m is about 57.6 Newton
[tex]\texttt{ }[/tex]
Further explanationElectric charge consists of two types i.e. positively electric charge and negatively electric charge.
[tex]\texttt{ }[/tex]
There was a famous scientist who investigated about this charges. His name is Coulomb and succeeded in formulating the force of attraction or repulsion between two charges i.e. :
[tex]\large {\boxed {F = k \frac{Q_1Q_2}{R^2} } }[/tex]
F = electric force (N)
k = electric constant (N m² / C²)
q = electric charge (C)
r = distance between charges (m)
The value of k in a vacuum = 9 x 10⁹ (N m² / C²)
Let's tackle the problem now !
[tex]\texttt{ }[/tex]
Given:
distance between protons = d = 2 × 10⁻¹⁵ m
charge of proton = q = 1.6 × 10⁻¹⁹ C
Unknown:
electric force = F = ?
Solution:
[tex]F = k \frac{q_1 q_2}{(d)^2}[/tex]
[tex]F = k \frac{q^2}{(d)^2}[/tex]
[tex]F = 9 \times 10^9 \times \frac{(1.6 \times 10^{-19})^2}{(2 \times 10^{-15})^2}[/tex]
[tex]F = 57.6 \texttt{ Newton}[/tex]
[tex]\texttt{ }[/tex]
Learn moreThe three resistors : https://brainly.com/question/9503202A series circuit : https://brainly.com/question/1518810Compare and contrast a series and parallel circuit : https://brainly.com/question/539204[tex]\texttt{ }[/tex]
Answer detailsGrade: High School
Subject: Physics
Chapter: Static Electricity
What is the longest wavelength of light that will
emitelectrons from a metal whose work function is 3.10 eV?
Answer:
The longest wavelength equals [tex]0.4\times 10^{-6}m[/tex]
Explanation:
According to Einstein's photoelectric equation we have
[tex]E_{incident}\geq \phi [/tex]
where
[tex]E_{incident}[/tex] is the energy of the incident light
[tex]\phi [/tex] is the work function of the metal
The incident energy of the light with wavelength [tex]\lambda [/tex] is given by
[tex]E_{incident}=h\cdot \frac{c}{\lambda}[/tex]
Thus the photoelectric equation reduces to
[tex]h\cdot \frac{c}{\lambda}\geq \phi\\\\h\cdot c\geq \lambda \times \phi\\\\\therefore \lambda\leq \frac{h\cdot c}{\phi}[/tex]
Thus applying values we get
[tex]\lambda\leq \frac{6.62\times 10^{-34}\times 3\times 10^{8}}{3.10\times 1.602\times 10^{-19}}\\\\\therefore \lambda\leq 0.4\times 10^{-6}m[/tex]
Hence The longest wavelength equals [tex]0.4\times 10^{-6}m[/tex]
To practice Problem-Solving Strategy 16.2 Doppler effect. The sound source of a ship’s sonar system operates at a frequency of 22.0 kHz . The speed of sound in water (assumed to be at a uniform 20∘C) is 1482 m/s . What is the difference in frequency between the directly radiated waves and the waves reflected from a whale traveling straight toward the ship at 4.95 m/s ? Assume that the ship is at rest in the water.
Answer:
Δf=73Hz
Explanation:
From the question we know that:
C = 1482 m/s
Vs = 0 m/s
Vr = -4.95 m/s (it's negative because it is in the opposite direction to the waves)
f0 = 22000 Hz
Applying the formula for the doppler effect:
[tex]f=(\frac{C-Vr}{C-Vs} )*fo[/tex]
f = 22073 Hz. So the difference is only 73Hz
An engineer is undertaking some reconnaissance by pacing on a construction site. The engineer takes 26 steps per 20m, the ground he is walking on has a vertical slope angle of 6. The engineer takes 152 steps. How far has he travelled horizontally (in metres)?(only enter numeric answer to 2 decimal places, no alpha characters)
Answer:
Thus the distance traveled horizontally is 116.28 m
Solution:
As per the question:
Distance covered in 26 steps = 20 m
Thus
Distance covered in one step, step size = [tex]\frac{20}{26}[/tex]
Total steps taken = 152
Now, the distance covered in 152 steps = [tex]one step size\times 152[/tex]
The distance covered in 152 steps, D = [tex]\frac{20}{26}\times 152 = 116.92 m[/tex]
The above distances are on a slope of [tex]6^{\circ}[/tex] above horizontal.
Thus the horizontal component of this distance is given by:
[tex]d_{H} = Dcos6^{\circ}[/tex]
where
[tex]d_{H}[/tex] = horizontal component of distance, D
[tex]d_{H} = 116.92cos6^{\circ} = 116.28 m[/tex]
In a compression test, a steel test specimen (modulus of elasticity = 30 x 106lb/in2) has a starting height= 2.0 in and diameter = 1.5 in. The metal yields (0.2% offset) at a load = 140,000 lb. At a load of 260,000 lb, the height has been reduced to 1.6 in. Determine (a)yield strength and (b) flow curve parameters (strength coefficient and strain-hardening exponent). Assume that the cross-sectional area increases uniformly during the test.
Answer:
Explanation:
A) we know that volume is given as V
[tex]V =\frac{\pi}{4} D^2 h[/tex]
where D = 1.5 in , h = 2.0 in
so [tex]V = \frac{\pi}{4} 1.5^2\times 2 = 3.53in^3[/tex]
[tex]Area =\frac{\pi}{4} D^2 = \frac{\pi}{4} \times 1.5^2 = 1.76 in^4[/tex]
yield strenth is given as[tex] \sigma_y = \frac{force}{area} = \frac{140,000}{1.76}[/tex]
[tex]\sigma_y = 79.224 ksi[/tex]
b)
elastic strain[tex] \epsilon = \frac{\sima_y}{E} = \frac{79.224}{30\times 10^3} = 0.00264[/tex]
strain offsets = 0.00264 + 0.002 = 0.00464 [where 0.002 is offset given]
[tex]\frac{\delta}{h} = 0.00464[/tex]
[tex]\frac{h_i -h_o}{h_o} = 0.00464[/tex]
[tex]h_i = 2\times(1-0.00464) = 1.99 inch[/tex]
area [tex]A = \frac{volume}{height} = \frac{3.534}{1.9907} = 1.775 in^2[/tex]
True strain[tex] \sigma = \frac{force}{area} = \frac{140,000}{1.775 in^2} = 78,862 psi[/tex]
At P= 260,000 lb ,[tex] A = \frac{3.534}{1.6} = 2.209 inc^2[/tex]
true stress [tex]\sigma = \frac{260,000}{2.209} = 117,714 psi[/tex]
true strain [tex]\epsilon = ln\frac{2}{1.6} = 0.223[/tex]
flow curve is given as \sigma = k\epsilon^n
[tex]\sigma_1 = 78,862 psi[/tex]
[tex]\epsilon_1 = 0.00464[/tex]
[tex]\sigma_2 = 117,714 psi[/tex]
[tex]\epsilon_2 = 0.223[/tex]
so flow curve is
[tex]78,868 = K 0.00464^n[/tex] .........1
[tex]117,714 = K 0.223^n[/tex] .........2
Solving 1 and 2
we get
n = 0.103
and K =137,389 psi
Strength coffecient = K = 137.389ksi
strain hardening exponent = n = 0.103
A driver enters a one-lane tunnel at 34.4 m/s. The driver then observes a slow-moving van 154 m ahead travelling (in the same direction as the car) at a constant 5.65 m/s. The driver applies the brakes (ignore reaction time)of the car but can only accelerate at -2.00 m/s2 because the road is wet. How fast are you moving when you hit the rear of the van?
(A) 16.3 m/s
(B) 22 m/sec
(С) 4 m/sec
(D) 0 m/s
Answer:
Ans. B) 22 m/s (the closest to what I have which was 20.16 m/s)
Explanation:
Hi, well, first, we have to find the equations for both, the driver and the van. The first one is moving with constant acceleration (a=-2m/s^2) and the van has no acceletation. Let´s write down both formulas so we can solve this problem.
[tex]X(van)=5.65t+154[/tex]
[tex]X(driver)=34.4t+\frac{(-2)t^{2} }{2}[/tex]
or by rearanging the drivers equation.
[tex]X(driver)=34.4t+t^{2}[/tex]
Now that we have this, let´s equal both equations so we can tell the moment in which both cars crashed.
[tex]X(van)=X(driver)[/tex]
[tex]5.65t+154=34.4t-t^{2}[/tex]
[tex]0=t^{2} -(34.4-5.65)t+154[/tex][tex]0=t^{2} -28.75t+154[/tex]
To solve this equation we use the following formulas
[tex]t=\frac{-b +\sqrt{b^{2}-4ac } }{2a}[/tex]
[tex]t=\frac{-b +\sqrt{b^{2}-4ac } }{2a}[/tex]
Where a=1; b=-28.75; c=154
So we get:
[tex]t=\frac{28.75 +\sqrt{(-28.75)^{2}-4(1)(154) } }{2(1)}=21.63s[/tex][tex]t=\frac{28.75 -\sqrt{(-28.75)^{2}-4(1)(154) } }{2(1)}=7.12s[/tex]
At this point, both answers could seem possible, but let´s find the speed of the driver and see if one of them seems ilogic.
[tex]V(driver)=V_{0} +at[/tex]}
[tex]V(driver)=34.4\frac{m}{s} -2\frac{m}{s^{2} } *(7.12s)=20.16\frac{m}{s}[/tex][tex]V(driver)=34.4\frac{m}{s} -2\frac{m}{s^{2} } *(21.63s)=-8.86\frac{m}{s}[/tex]
This means that 21.63s will outcome into a negative speed, for that reason we will not use the value of 21.63s, we use 7.12s and if so, the speed of the driver when he/she hits the van is 20.16m/s, which is closer to answer A).
Best of luck
A basketball player makes a bounce pass by throwing the ball with a speed of 5.3 m/s and at an angle 12° below the horizontal. It leaves her hands 1.0 m above the floor. Calculate how far omher feet the ball hits the ground.
Answer:
ball hit the ground from her feet is 1.83 m far away
Explanation:
given data
speed = 5.3 m/s
angle = 12°
height = 1 m
to find out
how far from her feet ball hit ground
solution
we consider here x is horizontal component and y is vertical component
so in vertical
velocity will be = v sin12
vertical speed u = 5.3 sin 12 = 1.1 m/s downward
and
in horizontal , velocity we know v = 5.3 m/s
so from motion of equation
s = ut + 0.5×a×t²
s is distance t is time a is 9.8
put all value
1 = 1.1 ( t) + 0.5×9.8×t²
solve it we get t
t = 0.353 s
and
horizontal distance is = vcos12 × t
so horizontal distance = 5.3×cos12 × ( 0.353)
horizontal distance = 1.83 m
so ball hit the ground from her feet is 1.83 m far away
Calculate the acceleration of a car (in m/s^2) that accelerates from 0 to 30 m/s in 6 s along a straight road.
Answer: 5 m/s^2
Explanation: In order to solve this question we have to use the kinematic equation given by:
Vf= Vo+a*t where V0 is zero.
we know that it takes Vf( 30 m/s) in 6 seconds
so
a=(30 m/s)/6 s= 5 m/s^2
A cube with sides of area 18 cm^2 contains a 6.0 nanoCoulomb charge. Find the flux of the electric field through the surface of the cube in unis of Nm^2/C. Enter a number with one digit behind the decimal point.
Answer:
The flux of the electric field is 677.6 Nm²/C
Explanation:
Given that,
Area = 18 cm²
Charge = 6.0 nC
We need to calculate the flux of the electric field
Using Gauss's law
[tex]\phi=\dfrac{q}{\epsilon_{0}}[/tex]
Where, q = charge
[tex]\epsilon_{0}[/tex] =permittivity of free space
Put the value into the formula
[tex]\phi=\dfrac{6.0\times10^{-9}}{8.854\times10^{-12}}[/tex]
[tex]\phi=677.6\ Nm^2/C[/tex]
Hence, The flux of the electric field is 677.6 Nm²/C.
The acceleration of a particle which is moving along a straight line is given by a = –kv0.5, where a is in meters per second squared, k is a constant, and v is the velocity in meters per second. Determine the velocity as a function of both time t and position s. Evaluate your expressions for t = 2.7 sec and at s = 6 m, if k = 0.3 m0.5 sec–1.5 and the initial conditions at time t = 0 are s0 = 2.6 m and v0 = 2.7 m/sec.
Answer:
[tex]v(t)=2.7*e^{0.5kt}\\\\ v(s)=\frac{k}{2}*s+2.7-1.3k\\\\[/tex]
For t=2.7s and k=0.3 m/s:
[tex]v(2.7s)=1.80m/s[/tex]
For s=6m and k=0.3 m/s:
[tex]v(6m)=6.69m/s\\\\[/tex]
Explanation:
Definition of acceleration:
[tex]a=\frac{dv}{dt} =0.5kv[/tex]
we integrate in order to find v(t):
[tex]\frac{dv}{v} =-0.5kdt[/tex]
[tex]\int\limits^v_0 { \frac{dv}{v}} \, =-0.5k\int\limits^t_0 {dt} \,[/tex]
[tex]ln(v)=-0.5kt+C\\\\ v=A*e^{-0.5kt}[/tex] A=constant
Definition of velocity:
[tex]v=\frac{ds}{dt} =A*e^{-0.5kt}[/tex]
We integrate:
[tex]v=\frac{ds}{dt} \\s=- (2/k)*A*e^{-0.5kt}+B[/tex] B=constant
But:
[tex]v=A*e^{-0.5kt}[/tex]⇒[tex]s= -(2/k)*v+B[/tex]
[tex]v(s)=-(\frac{k}{2} )(s-B)=D-\frac{k}{2}*s[/tex] D=other constant
Initial conditions:t = 0 are s0 = 2.6 m and v0 = 2.7 m/sec:
[tex]v(t)=A*e^{-0.5kt}\\ 2.7=Ae^{-0.5k*0}\\ 2.7=A\\[/tex]
[tex]v(s)=D-\frac{k}{2}*s\\2.7=D-\frac{k}{2}*2.6\\D=2.7+1.3k[/tex]
So:
[tex]v(t)=2.7*e^{-0.5kt}\\\\ v(s)=\frac{k}{2}*s+2.7+1.3k\\\\[/tex]
For t=2.7s and k=0.3 m/s:
[tex]v(2.7s)=2.7*e^{-0.5*0.3*2.7}=1.80m/s[/tex]
For s=6m and k=0.3 m/s:
[tex]v(6m)=\frac{0.3}{2}*6+2.7+1.3*0.3=6.69m/s\\\\[/tex]
What mass of steam at 100°C must be mixed with 499 g of ice at its melting point, in a thermally insulated container, to produce liquid water at 33.0°C? The specific heat of water is 4186 J/kg·K. The latent heat of fusion is 333 kJ/kg, and the latent heat of vaporization is 2256 kJ/kg.
Answer:
The mass of the steam is 91.2 g
Mass of the steam=91 grams
Explanation:
Given:
Mass of the ice=499 gFinal temperature of the liquid water[tex]=33^\circ \rm C[/tex]Latent heat of fusion=[tex]333\ \rm kJ\kg[/tex]Latent heat of vaporization =[tex]2256\ \rm kJ\kg[/tex]When steam is mixed with the ice then the heat loss by the steam will be gained by the ice so there will no overall heat gain or loss during the mixing
So According to question
Let M be the mass of the steam mixed with ice then we have
[tex]M\times2256\times10^3+M\times4186\times(100-33)=0.499\times222\times10^3+0.499\times4186\times(33-0)\\M\times2.58\times10^6=2.35\times10^5\\M=91.2\ \rm g[/tex]
A hiker treks 30 degrees south of east at a speed of 15 m/s for 30 min and then turns due west and hikes at a speed of 8m/s for another 20 min. What is the displacement of this explorer (magnitude and direction)?
Answer:19.3 km,[tex]\theta =44.40^{\circ}[/tex] south of east
Explanation:
Given
Hiker treks [tex]30 ^{\circ}[/tex] south of east at a speed of 15 m/s for 30 min and then turns due to west and hikes at speed of 8 m/s for another 20 min
Let position vector of Hiker at the end of 30 min
[tex]r=27000cos30\hat{i}-27000sin30\hat{j}[/tex]
after he turns west so new position vector of hiker is
[tex]r'=27000cos30\hat{i}-27000sin30\hat{j}-9600\hat{i}[/tex]
[tex]r'=13782.68\hat{i}-13500\hat{j}[/tex]
Therefore Displacement is given by |r'|
[tex]|r'|=\sqrt{13782.68^2+13500^2}[/tex]
[tex]|r'|=19,292.8035 m\ or 19.3 km[/tex]
for direction
[tex]tan\theta =\frac{13500}{13782.68}[/tex]
[tex]\theta =44.40^{\circ}[/tex] south of east
An electron is moving through a magnetic field whose magnitude is 9.21 × 10^-4 T. The electron experiences only a magnetic force and has an acceleration of magnitude 2.30 × 10^14 m/s^2. At a certain instant, it has a speed of 7.69 × 10^6 m/s. Determine the angle (less than 90°) between the electron's velocity and the magnetic field.
The electron can make angles of 118° and 62° with the magnetic field.
An electron moving at a speed of 4.00 × 10³ m/s in a 1.25-T magnetic field experiences a magnetic force of 1.40 × 10-16 N. What angle does the velocity of the electron make with the magnetic field?
One possible angle is 118°, and the other possible angle is 62° with the magnetic field.
The correct answer is [tex]\(\boxed{89.9999999999999\°}\).[/tex]
To determine the angle between the electron's velocity and the magnetic field, we can use the formula for the magnitude of the magnetic force on a moving charge, which is given by the Lorentz force law:
[tex]\[ F = qvB \sin(\theta) \][/tex]
where:
- [tex]\( F \)[/tex] is the magnitude of the magnetic force,
- [tex]\( q \)[/tex] is the charge of the electron,
- [tex]\( v \)[/tex] is the speed of the electron,
- [tex]\( B \)[/tex] is the magnitude of the magnetic field, and
- [tex]\( \theta \)[/tex] is the angle between the electron's velocity and the magnetic field.
The charge of an electron is [tex]\( 1.60 \times 10^{-19} \)[/tex] Coulombs (C), and the mass of an electron is approximately [tex]\( 9.11 \times 10^{-31} \)[/tex] kilograms (kg). Using Newton's second law, [tex]\( F = ma \)[/tex], where [tex]\( m \)[/tex] is the mass of the electron and [tex]\( a \)[/tex] is its acceleration, we can equate the magnetic force to the mass times acceleration:
[tex]\[ qvB \sin(\theta) = ma \][/tex]
Given that the electron experiences only a magnetic force, we can solve for [tex]\( \sin(\theta) \)[/tex]:
[tex]\[ \sin(\theta) = \frac{ma}{qvB} \][/tex]
Plugging in the given values:
[tex]\[ \sin(\theta) = \frac{(9.11 \times 10^{-31} \text{ kg})(2.30 \times 10^{14} \text{ m/s}^2)}{(1.60 \times 10^{-19} \text{ C})(7.69 \times 10^{6} \text{ m/s})(9.21 \times 10^{-4} \text{ T})} \][/tex]
[tex]\[ \sin(\theta) = \frac{(9.11 \times 2.30)}{(1.60 \times 7.69 \times 9.21)} \times 10^{-31 + 14 - (-19) - 6 - (-4)} \][/tex]
[tex]\[ \sin(\theta) = \frac{(20.953)}{(119.9424)} \times 10^{-31 + 14 + 19 - 6 + 4} \][/tex]
[tex]\[ \sin(\theta) = 0.1747 \times 10^{-31 + 14 + 19 - 6 + 4} \][/tex]
[tex]\[ \sin(\theta) = 0.1747 \times 10^{-10} \][/tex]
[tex]\[ \sin(\theta) = 1.747 \times 10^{-11} \][/tex]
Now, we can find the angle [tex]\( \theta \)[/tex] by taking the inverse sine (arcsin) of [tex]\( \sin(\theta) \)[/tex]:
[tex]\[ \theta = \arcsin(1.747 \times 10^{-11}) \][/tex]
Since the value of [tex]\( \sin(\theta) \)[/tex] is extremely small, the angle [tex]\( \theta \)[/tex] will be very close to 0 degrees. However, because the electron is experiencing an acceleration, [tex]\( \theta \)[/tex] must be slightly greater than 0 degrees. Using a calculator, we find:
[tex]\[ \theta \approx \boxed{89.9999999999999\°} \][/tex]
This result indicates that the electron's velocity is nearly parallel to the magnetic field, with an angle that is almost 90 degrees but infinitesimally less.
"A problem involves a car of mass m going down a track from a height H, and round a loop of radius r. The loop is frictionless.
It asks for the minimum cut-off speed required, at the highest point in the loop (call it point D), such that the car makes it round the loop without falling. I know the solution; I should set the centripetal accleration equal to 9.81. In other words, contact force with the track at point D is equal to zero.
But I tried solving it by conservation of energy. At point D, the car is at a height 2r from ground level. Therefore, in order for the car to reach that height at point D, it must initially have a potential energy of mg(2r). Meaning, it should be released from a height H = 2r.
I got the wrong answer and I'm confused why that happened. Isn't that how conservation of energy work? Please clarify, where's the error in my solution?"
Answer:
Explanation:
At the topmost position, the car does not have zero velocity but it has velocity of v so that
v² /r = g or centripetal acceleration should be equal to g ( 9.8 )
Considering that, the car must fall from a height of 2r + h where
mgh = 1/2 mv²
= 1/2 m gr
So h = r/2
Hence the ball must fall from a height of
2r + r /2
= 2.5 r . So that it can provide velocity of v at the top where
v² / r = g .
What is the magnitude of the electric field at a point midway between a −5.6μC and a +5.8μC charge 9.0cm apart? Assume no other charges are nearby. Express your answer using two significant figures.
Answer:
Magnitude of electric field at mid way is [tex]5.07\times 10^{7} N/C[/tex]
Solution:
As per the question:
Q = [tex]- 5.6\mu m = - 5.6\times 10^{- 6} C[/tex]
Q' = [tex]5.8\mu m = 5.8\times 10^{- 6} C[/tex]
Separation distance, d = 9.0 cm = 0.09 m
The distance between charges at mid-way, O is [tex]\farc{d}{2} = 0.045 m[/tex]
Now, the electric at point O due to charge Q is:
[tex]E = \frac{1}{4\pi \epsilon_{o}}.\frac{Q}{(\frac{d}{2})^{2}}[/tex]
[tex]E = \frac{1}{4\pi \epsilon_{o}}.\frac{- 5.6\times 10^{- 6}}{(0.045^{2}}[/tex]
[tex]E = (9\times 10^{9})\frac{- 5.6\times 10^{- 6}}{(0.045^{2}}[/tex]
[tex]E = (9\times 10^{9})\frac{- 5.6\times 10^{- 6}}{(0.045^{2}}[/tex]
[tex]E = - 2.49\times 10^{7} N/C[/tex]
Here, negative sign is indicative of the direction of electric field which is towards the point O
Now, the electric at point O due to charge Q' is:
[tex]E' = \frac{1}{4\pi \epsilon_{o}}.\frac{Q}{(\frac{d}{2})^{2}}[/tex]
[tex]E' = (9\times 10^{9})\frac{5.8\times 10^{- 6}}{(0.045^{2}}[/tex]
[tex]E' = 2.58\times 10^{7} N/C[/tex]
Refer to Fig 1.
Since, both the fields are in the same direction:
[tex]E_{net} = E + E' = 2.49\times 10^{7} + 2.58\times 10^{7} = 5.07\times 10^{7} N/C[/tex]
The magnitude of the electric field at a point midway between two charges can be calculated using the formula E = k * (Q1-Q2) / r^2. Plugging in the values -5.6μC, +5.8μC, and 9.0cm, the magnitude of the electric field is -2.03 x 10^4 N/C.
Explanation:The magnitude of the electric field at a point midway between two charges can be calculated using the formula:
E = k * (Q1-Q2) / r^2
where E is the electric field, k is the electrostatic constant (9 x 10^9 Nm^2/C^2), Q1 and Q2 are the charges, and r is the distance between the charges.
In this case, the charges are -5.6μC and +5.8μC, and the distance is 9.0cm (0.09m).
Plugging these values into the formula:
E = (9 x 10^9 Nm^2/C^2) * ((-5.6μC) - (+5.8μC)) / (0.09m)^2
Simplifying the equation:
E = -2.03 x 10^4 N/C
A rocket accelerates straight up from the ground at 12.6 m/s^2 for 11.0 s. Then the engine cuts off and the rocket enters free fall. (a) Find its velocity at the end of its upward acceleration. (b) What maximum height does it reach? (c) With what velocity does it crash to Earth? (d) What's the total time from launch to crash?
Answer:
a) 138.6 m/s
b) 762.3 m
c) 122.3 m/s
d) 24.47
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration
[tex]v=u+at\\\Rightarrow v=0+12.6\times 11\\\Rightarrow v=138.6 \ m/s[/tex]
Velocity at the end of its upward acceleration is 138.6 m/s
[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow s=0\times t+\frac{1}{2}\times 12.6\times 11^2\\\Rightarrow s=762.3\ m[/tex]
Maximum height the rocket reaches is 762.3 m
[tex]v^2-u^2=2as\\\Rightarrow v=\sqrt{2as-u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 762.3-0^2}\\\Rightarrow v=122.3\ m/s[/tex]
The velocity with which the rocket crashes to the Earth is 122.3 m/s
[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow 762.3=0\times t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{762.3\times 2}{9.81}}\\\Rightarrow t=12.47\ s[/tex]
Total time from launch to crash is 12.47+11 = 24.47 seconds
Final answer:
The rocket's velocity at the end of its upward acceleration is 138.6 m/s. The maximum height it reaches is 1353.2 m. The rocket crashes to Earth with a velocity of -1.0 m/s.
Explanation:
(a) To find the velocity at the end of the rocket's upward acceleration, we can use the formula:
v = u + at
Where:
u = initial velocity = 0 m/s (since the rocket starts from rest)
a = acceleration = 12.6 m/s2
t = time = 11.0 s
Substituting the values, we get:
v = 0 + 12.6 * 11.0 = 138.6 m/s
Therefore, the velocity at the end of its upward acceleration is 138.6 m/s.
(b) To find the maximum height the rocket reaches, we can use the second equation of motion:
s = ut + 0.5at2
Where:
s = distance
u = initial velocity
a = acceleration
t = time
In this case, we need to consider the time taken for both the upward acceleration and free fall.
For the upward acceleration:
u = 0 m/s (since the rocket starts from rest)
a = 12.6 m/s2
t = 11.0 s
Substituting the values, we get:
s1 = 0 * 11.0 + 0.5 * 12.6 * (11.0)2 = 388.5 m
For the free fall:
u = 138.6 m/s (velocity at the end of the upward acceleration)
a = -9.8 m/s2 (acceleration due to gravity)
t = ?
To find the time for free fall, we can use the equation:
u = at
Substituting the values, we get:
138.6 = -9.8t
Solving for t, we get:
t = -14.1 s
However, time cannot be negative in this case. So, we take the absolute value of t:
t = 14.1 s
Substituting the values in the equation for free fall distance, we get:
s2 = 138.6 * 14.1 + 0.5 * (-9.8) * (14.1)2 = 964.7 m
The maximum height reached by the rocket is s1 + s2 = 388.5 m + 964.7 m = 1353.2 m.
(c) To find the velocity at which the rocket crashes to Earth, we again consider the free fall phase. Using the equation:
v = u + at
Where:
u = 138.6 m/s (velocity at the end of the upward acceleration)
a = -9.8 m/s2 (acceleration due to gravity)
t = 14.1 s
Substituting the values, we get:
v = 138.6 - 9.8 * 14.1 = -1.0 m/s
The velocity at which the rocket crashes to Earth is -1.0 m/s. The negative sign indicates that the velocity is directed downward.
(d) The total time from launch to crash is the sum of the time for upward acceleration (11.0 s) and the absolute value of the time for free fall (14.1 s). Therefore, the total time is 11.0 s + 14.1 s = 25.1 s.
According to the reading, which of the following is NOT an obstacle to using big data for analysis and decision making? O Poor reporting process b. Limited types of data Insufficient data d. Lack of talent to evaluate and analyze data All of the above
Answer:
B .Limited types of data Insufficient data
Explanation:
To analyse and making decision from big data we have following thing
- Sufficient data
- data analyst
- Accurate data
- data Privacy
- data storage
So, limited data or data that cannot be copied is not and obstacle in data handling. hence option B Limited types of data Insufficient data is correct.
a car is travelling at 25m/s when it begins to slow down and eventually stops in a 98.5m A. What is the car’s acceleration?
B. At what time after the brakes were applied was the car traveling at 12.5m/s?
C. At what time did the car travel only half the stopping distance?
Answer:
(a) -3.173 m/s^2
(b) 3.94 s
(c) 2.47 s
Explanation:
initial velocity, u = 25 m/s
final velocity, v = 0
distance, s = 98.5 m
(a) Let a be the acceleration of the car
Use third equation of motion
[tex]v^{2}=u^{2}+2as[/tex]
0 = 625 + 2 x a x 98.5
a = -3.173 m/s^2
(b) v = 12.5 m/s
u = 25 m/s
a = - 3.173 m/s^2
Let the time is t.
Use first equation of motion
v = u + a t
12.5 = 25 - 3.173 t
t = 3.94 s
(c) s = 98.5 / 2 = 49.25 m
u = 25 m/s
a = - 3.173 m/s^2
Let the time be t.
Let v be the velocity at this distance.
Use third equation of motion
[tex]v^{2}=u^{2}+2as[/tex]
[tex]v^{2}=25^{2}+2\times {-3.173}\ times 49.25tex]
v = 17.17 m/s
Use first equation of motion
v = u + at
17.17 = 25 - 3.173 x t
t = 2.47 s
An object has a charge of-3.8 μC. How many electrons must be removed so that the charge becomes +2.6 μC?
Answer:
The answer is [tex] 3.994 \times 10^{13}\ electrons[/tex]
Explanation:
The amount of negative charge that must be removed is
[tex]\Delta = Final\ charge - initial\ Charge = 2.6 - (-3.8) = 6.4 \ \mu C = 6.4 \times 10^{-6}\ C[/tex]
and the charge of one electron is
[tex]1 e = 1.60217662\times 10^{-19} \ C[/tex]
So the amount of electrons we need to remove is
[tex]x = \frac{6.4 \times 10^{-6}}{1.60217662\times 10^{-19}} \approx 3.994 \times 10^{13}\ electrons[/tex]
Two particles, one with charge -5.45 × 10^-6 C and one with charge 4.39 × 10^-6 C, are 0.0209 meters apart. What is the magnitude of the force that one particle exerts on the other? Two new particles, which have identical positive charge q3, are placed the same 0.0209 meters apart, and the force between them is measured to be the same as that between the original particles. What is q3?
Explanation:
Given that,
Charge 1, [tex]q_1=-5.45\times 10^{-6}\ C[/tex]
Charge 2, [tex]q_2=4.39\times 10^{-6}\ C[/tex]
Distance between charges, r = 0.0209 m
1. The electric force is given by :
[tex]F=k\dfrac{q_1q_2}{r^2}[/tex]
[tex]F=9\times 10^9\times \dfrac{-5.45\times 10^{-6}\times 4.39\times 10^{-6}}{(0.0209)^2}[/tex]
F = -492.95 N
2. Distance between two identical charges, [tex]r=0.0209\ m[/tex]
Electric force is given by :
[tex]F=\dfrac{kq_3^2}{r^2}[/tex]
[tex]q_3=\sqrt{\dfrac{Fr^2}{k}}[/tex]
[tex]q_3=\sqrt{\dfrac{492.95\times (0.0209)^2}{9\times 10^9}}[/tex]
[tex]q_3=4.89\times 10^{-6}\ C[/tex]
Hence, this is the required solution.
A 1.8 kg hammer moving at 6.0 m/s drives a nail 30mm into a board. Compute the average resistance of the board on the nail using theWork-Energy theorem.
Answer:
F = 1080 N
Explanation:
given,
mass of the hammer = 1.8 kg
velocity of the hammer = 6 m/s
distance into board = 30 mm = 0.03 m
to calculate average resistance force = F
kinetic energy of the hammer is equal to the work done by the hammer
[tex]\dfrac{1}{2}mv^2 = Force\times displacement[/tex]
[tex]\dfrac{1}{2}mv^2 = F\times d[/tex]
[tex]\dfrac{1}{2}\times 1.8\times 6^2 = F\times 0.03[/tex]
F = 1080 N
hence, the average resistance force is equal to F = 1080 N
Two cars drive on a straight highway. At time t=0, car 1 passes mile marker 0 traveling due east with a speed of 20.0 m/s. At the same time, car 2 is 1.2 km east of mile marker 0 traveling at 30.0 m/s due west. Car 1 is speeding up with an acceleration of magnitude 0.10 m/s^2 , and car 2 is slowing down with an acceleration of magnitude 0.30 m/s^2. At what time do the cars pass next to one another?
Answer:
The cars pass next to one another after 25.28 s.
Explanation:
When the cars pass next to one another, the position of both cars is the same relative to the center of the system of reference (marker 0 in this case). Then:
Position of car 1 = position of car 2
The position of an accelerating object moving in a straight line is given by this equation:
x = x0 +v0 t +1/2 a t²
where
x = position at time t
x0 = initial position
v0 = initial speed
t = time
a = acceleration
If the position of car 1 = position of car 2 then:
0 km + 20.0 m/s * t + 1/2 * 0.10 m/s² * t² = 1.2 km - 30.0 m/s * t + 1/2 * 0.30 m/s² * t²
Note that the acceleration of car 2 has to be positive because the car is slowing down and, in consequence, the acceleration has to be opposite to the velocity. The velocity is negative because the direction of car 2 is towards the origin of our system of reference. Let´s continue:
0 km + 20.0 m/s * t + 1/2 * 0.10 m/s² * t² = 1.2 km - 30.0 m/s * t + 1/2 * 0.30 m/s² * t²
1200 m - 50.0 m/s * t + 0.10 m/s² * t² = 0
Solving the quadratic equation:
t = 25.28 s
t = 474. 72 s We discard this value because, if we replace it in the equation of the position of car 2, we will get a position of 20762 m, which is impossible because the position of car 2 can´t be greater than 1200 m.
Then, the cars pass next to one another after 25.28 s
A point of charge 4.9 μC is placed at the origin (x1 = 0) of a coordinate system, and another charge of -1.1 μC is placed on the x-axis at x2 = 0.28 m. a) Where on the x-axis can a third charge be placed in meters so that the net force on it is zero?
b) What if both charges are positive; that is, what if the second charge is 1.9 μC?
To find the location on the x-axis where a third charge can be placed so that the net force on it is zero, we need to consider the forces exerted by the two charges. If both charges are positive, the net force on a third charge placed on the x-axis will never be zero.
Explanation:a) To find the location on the x-axis where a third charge can be placed so that the net force on it is zero, we need to consider the forces exerted by the two charges. The net force will be zero when the two forces are equal in magnitude but opposite in direction. Using Coulomb's law, we can calculate the force between the first charge and the third charge at an unknown position x3. We can then set that force equal to the force between the second charge and the third charge at the same position x3, and solve for x3.
b) If both charges are positive, the net force on a third charge placed on the x-axis will never be zero. Positive charges repel each other, so the forces will always be in the same direction.
The area of a rectangular park is 4 mi^2. The park has a width that is equal to "w", and a length that is 3 mi longer than the width of the park. Find the dimensions of the park.
Answer:
l= 4 mi : width of the park
w= 1 mi : length of the park
Explanation:
Formula to find the area of the rectangle:
A= w*l Formula(1)
Where,
A is the area of the rectangle in mi²
w is the width of the rectangle in mi
l is the width of the rectangle in mi
Known data
A = 4 mi²
l = (w+3)mi Equation (1)
Problem development
We replace the data in the formula (1)
A= w*l
4 = w* (w+3)
4= w²+3w
w²+3w-4= 0
We factor the equation:
We look for two numbers whose sum is 3 and whose multiplication is -4
(w-1)(w+4) = 0 Equation (2)
The values of w for which the equation (2) is zero are:
w = 1 and w = -4
We take the positive value w = 1 because w is a dimension and cannot be negative.
w = 1 mi :width of the park
We replace w = 1 mi in the equation (1) to calculate the length of the park:
l= (w+3) mi
l= ( 1+3) mi
l= 4 mi