Answer:
option B
Explanation:
given,
seed of airplane = 70 m/s
height = 300 m
we know,
[tex]s = ut + \dfrac{1}{2}at^2[/tex]
[tex]300 = 0 + \dfrac{1}{2} \times 9.81\times t^2[/tex]
t = 7.82 s
now, the range of the crate
R = V × t
= 70 × 7.82
= 547.44 ≅ 548 m
hence, the correct answer is option B
The car is traveling at a constant speed v0 = 80 km/hr on the level portion of the road. When the 4-percent (tan q = 4/100) incline is encountered, the driver does not change the throttle setting and consequently the car decelerates at the constant rate g sin q. Determine the speed of the car: (a) 28 seconds after passing point A ,and (b) when s = 620 m.
Answer:
11.27 m /s
2.98 m / s.
Explanation:
80 km / h = 22.22 m /s
Tanq = 4 / 100
Sinq = .0399
Deceleration acting on inclined plane = g sinq
= 9.8 x .0399
= .3910
Initial speed u = 22.22 m/s
acceleration = - .3910 ms⁻²
v = u - a t
= 22.22 - .3910 x 28
= 22.22 - 10.95
= 11.27 m /s
b ) v² = u² - 2 a s
v² = ( 22.22) ² - 2 x .3910 x 620
= 493.7284 - 484.84
= 8.8884
v = 2.98 m / s.
A charge Q is to be divided into two parts, labeled 'q' and 'Q-q? What is the relationship of q to Q if, at any given distance, the Coulomb force between the two is to be maximized?
Answer:
Explanation:
The two charges are q and Q - q. Let the distance between them is r
Use the formula for coulomb's law for the force between the two charges
[tex]F = \frac{Kq_{1}q_{2}}{r^{2}}[/tex]
So, the force between the charges q and Q - q is given by
[tex]F = \frac{K\left ( Q-q \right )q}}{r^{2}}[/tex]
For maxima and minima, differentiate the force with respect to q.
[tex]\frac{dF}{dq} = \frac{k}{r^{2}}\times \left ( Q - 2q \right )[/tex]
For maxima and minima, the value of dF/dq = 0
So, we get
q = Q /2
Now [tex]\frac{d^{2}F}{dq^{2}} = \frac{-2k}{r^{2}}[/tex]
the double derivate is negative, so the force is maxima when q = Q / 2 .
Consider the following statements about Newton's 2nd law in general. Select all of the statements that are true. Note, there may be more than one. Pay attention to synonyms: velocity and change in position are the same thing.
1. A net force causes velocity to change
2. If an object has a velocity, then we can conclude that there is a net force on the object
3. Accelerations are caused by the presence of a net force
Answer:
1. True
2. False
3. True
Explanation:
Newton's 2nd law states that the net force exerted on an object is equal to the product between the mass of the object and its acceleration:
[tex]\sum F = ma[/tex] (1)
where
[tex]\sum F[/tex] is the net force on the object
m is its mass
a is the acceleration
Furthermore, we know that acceleration is defined as the rate of change of velocity:
[tex]a = \frac{dv}{dt}[/tex]
So let's now analyize the three statements:
1. A net force causes velocity to change: TRUE. Net force (means non-zero) causes a non-zero acceleration, which means that the velocity of the object must change.
2. If an object has a velocity, then we can conclude that there is a net force on the object: FALSE. The fact that the object has a velocity does not imply anything about its acceleration: in fact, if its velocity is constant, then its acceleration is zero, which would mean that the net force on the object is zero. So this statement is not necessarly true.
3. Accelerations are caused by the presence of a net force: TRUE. This is directly implied by eq.(1): the presence of the net force results in the object having a non-zero acceleration.
The de Broglie wavelength of a 0.064-kg golf ball is 3.09 x 10^−34m. What is its speed? (h = 6.63 x 10^−34 J⋅s)
Answer:
The speed is 33.5 m/s.
Explanation:
Given that,
Mass = 0.064 kg
Wavelength [tex]\lambda= 3.09\times10^{-34}\ m[/tex]
We need to calculate the speed
Using formula of he de Broglie wavelength
[tex]\lambda=\dfrac{h}{mv}[/tex]
[tex]v=\dfrac{h}{m\lambda}[/tex]
Where, h = Planck constant
m = mass
[tex]\lambda[/tex] = wavelength
Put the value into the formula
[tex]v = \dfrac{6.63\times10^{-34}}{0.064\times 3.09\times10^{-34}}[/tex]
[tex]v=33.5\ m/s[/tex]
Hence, The speed is 33.5 m/s.
Specific heat depends on several factors. Pick the factor below that you suspect will not affect specific heat. composition
state of matter
mass
Answer:
COMPOSITION
Explanation:
the correct answer is COMPOSITION.
specific heat of the body can be defined as the heat required to raise the temperature of a unit mass of a body by 1 °C.
so, A composition of the material is not affected by the specific heat.
The unit of specific heat is Joule per kelvin (J / K ).
There are two vectors à and b, with an angle between them. Then the dot product of them is A. a. b = ab cos B. a b = (a + b) cos C. a. b = ab sino D.. b = (a + b) sino
Answer:
[tex]a.b=|a||b|\ cos\theta[/tex]
Explanation:
Let a and b are two vectors such that [tex]\theta[/tex] is the angle between them. Dot product is also known as scalar product. It is used to find the angle between two vectors such that,
[tex]a.b=|a||b|\ cos\theta[/tex]
[tex]\theta[/tex] is the angle between a and b. It can be calculated as :
[tex]\theta=cos^{-1}(\dfrac{a.b}{|a||b|})[/tex]
[tex]|a|\ and\ |b|[/tex] are the magnitude of vectors a and b such that :
[tex]|a|=\sqrt{x^2+y^2+z^2}[/tex] if a = xi +yj +zk
and
[tex]|b|=\sqrt{p^2+q^2+r^2}[/tex] if a = pi +qj +rk
So, the correct option is (a). Hence, this is the required solution.
The current density inside a long, solid, cylindrical wire of radius a = 5.0 mm is in the direction of the central axis and its magnitude varies linearly with radial distance r from the axis according to J = J0r/a, where J0 = 420 A/m2. Find the magnitude of the magnetic field at a distance (a) r=0, (b) r = 2.9 mm and (c) r=5.0 mm from the center.
Explanation:
Given that,
Radius a= 5.0 mm
Radial distance r= 0, 2.9 mm, 5.0 mm
Current density at the center of the wire is given by
[tex]J_{0}=420\ A/m^2[/tex]
Given relation between current density and radial distance
[tex]J=\dfrac{J_{0}r}{a}[/tex]
We know that,
When the current passing through the wire changes with radial distance,
then the magnetic field is induced in the wire.
The induced magnetic field is
[tex]B=\dfrac{\mu_{0}i_{ind}}{2\pi r}[/tex]...(I)
We need to calculate the induced current
Using formula of induced current
[tex]i_{ind}=\int_{0}^{r}{J(r)dA}[/tex]
[tex]i_{ind}= \int_{0}^{r}{\dfrac{J_{0}r}{a}2\pi r}[/tex]
[tex]i_{ind}={\dfrac{2\pi J_{0}}{a}}\int_{0}^{r}{r^2}[/tex]
[tex]i_{ind}={\dfrac{2\pi J_{0}}{a}}{\dfrac{r^3}{3}}[/tex]
We need to calculate the magnetic field
Put the value of induced current in equation (I)
[tex]B=\dfrac{\mu_{0}{\dfrac{2\pi J_{0}}{a}}{\dfrac{r^3}{3}}}{2\pi r}[/tex]
[tex]B=\dfrac{\mu_{0}J_{0}r^2}{3a}[/tex]
(a). The magnetic field at a distance r = 0
Put the value into the formula
[tex]B=\dfrac{4\pi\times10^{-7}\times420\times0}{3\times5.0\times10^{-3}}[/tex]
[tex]B = 0[/tex]
The magnetic field at a distance 0 is zero.
(b). The magnetic field at a distance r = 2.9 mm
[tex]B=\dfrac{4\pi\times10^{-7}\times420\times(2.9\times10^{-3})^2}{3\times5.0\times10^{-3}}[/tex]
[tex]B = 2.95\times10^{-7}\ T[/tex]
The magnetic field at a distance 2.9 mm is [tex]2.95\times10^{-7}\ T[/tex]
(c). The magnetic field at a distance r = 5.0 mm
[tex]B=\dfrac{4\pi\times10^{-7}\times420\times(5.0\times10^{-3})^2}{3\times5.0\times10^{-3}}[/tex]
[tex]B = 8.79\times10^{-7}\ T[/tex]
The magnetic field at a distance 5.0 mm is [tex]8.79\times10^{-7}\ T[/tex]
Hence, This is the required solution.
A person starts to run around a square track with sides of 50 m starting in the bottom right corner and running counter-clockwise. Running at an average speed of 5 m/s, the jogger runs for 53.80 seconds. What is the difference between the magnitude of the person's average velocity and average speed in m/s? (please provide detailed explanation)
Answer:3.71 m/s
Explanation:
Given
square track with sides 50 m
average speed is 5 m/s
Total running time=53.8 s
Total distance traveled in this time[tex]=53.8\times 5=269 m[/tex]
i.e. Person has completed square track one time and another 69 m in second round
So displacement is 269-200=69 m
average velocity[tex]=\frac{Displacement}{time}[/tex]
[tex]=\frac{69}{53.8}=1.28 m/s[/tex]
Difference between average velocity and average speed is
5-1.28=3.71 m/s
Answer:
The difference in average speed and average velocity in terms of magnitude is 3.993 m/s
Solution:
As per the question:
The side of a square track, l = 50 m
Average speed of the runner, [tex]v_{avg} = 5 m/s[/tex]
Time taken, t = 53.80 s
Now,
The distance covered by the runner in this time:
s = [tex]v_{avg}t[/tex]
s = [tex]5\times 53.80[/tex]
s = 269 m
After covering a distance of 269 m, the person is at point A:
[tex]AQ^{2} = AR^{2} + QR^{2}[/tex]
where
AR = 19 m
QR = 50 m
Refer to fig 1.
As the runner starts from the bottom right, i.e., at Q and traveled 269 m.
After completion of 250 m , he will be at point R after one complete round and thus travels 19 m more to point A to cover 269 m.
Thus
[tex]AQ = \sqrt{19^{2} + 50^{2}} = 53.48 m[/tex]
where
AQ is the displacement
Hence,
Average velocity, v' = [tex]\frac{AQ}{t}[/tex]
v' = [tex]\frac{53.48}{53.80} = 1.007 m/s[/tex]
The difference in average speed and average velocity is:
[tex]v_{avg} - v' = 5 - 1.007 = 3.993 m/s[/tex]
A solid steel ball is thrown directly downward, with an initial speed of 7.95 m/s, from the top of a building at a height of 29.8 m. How much time (in s) does it take before striking the ground?
Answer:
1.78 s
Explanation:
Initial speed of the ball = u = 7.95 m/s and is vertically downwards.
Acceleration due to gravity = g = 9.8 m/s/s , vertically downwards.
Height of the building h = 29.8 m (traversed downwards by the steel ball).
h = u t + 1/2 g t²
29.8 = 7.95 t + 0.5 (9.8) t²
⇒ 4.9 t² +7.95 t - 29.8 = 0
Using the quadratic formula , solve for t.
t= [tex]= \frac{-b\pm \sqrt{b^2-4\times a \times c}}{2\times a}[/tex]
t = [tex]\frac{-7.95 \pm \sqrt{7.95^2-4\times 4.9 \times (-29.8)}}{2\times 9.8}[/tex] = 1.78 s, -3.4 s
Since time does not have a negative value, time taken by the stone to reach the ground = t = 1.78 s
Final answer:
To calculate the time taken for a solid steel ball to strike the ground when thrown downward from a building, use the equation of motion and plug in the given values to find the time, which is about 1.7 seconds.
Explanation:
The time taken for the ball to strike the ground can be calculated using the equation of motion:
h = (1/2) * g * t2
Where h is the initial height of the ball, g is the acceleration due to gravity, and t is the time taken. Substituting the given values, we find that it takes approximately 1.7 seconds for the ball to strike the ground.
1.48 Assume that the air volume in a small automobile tire is constant and equal to the volume between two concentric cylinders 13 cm high with diameters of 33 cm and 52 cm. The air in the tire is initially at 25 °C and 202 kPa. Immediately after air is pumped into the tire, the temperature is 30 °C and the pressure is 303 kPa. What mass of air was added to the tire? What would be the air pressure after the air has cooled to a temperature of 0 °C?
Answer:
a) 0.018 kg
b) 262 kPa
Explanation:
The volume of the concentric cylinders would be:
V = π/4 * h * (D^2 - d^2)
V = π/4 * 13 * (52^2 - 33^2) = 16500 cm^3 = 0.0165 m^3
The state equation of gases:
p * V = m * R * T
Rearranging:
m = (p * V) / (R * T)
R is 287 J/(kg * K) for air
25 C = 298 K
m0 = 202000 * 0.0165 / (287 * 298) = 0.039 kg
After pumping more air the volume remains about the same, but temperature and pressure change.
30 C = 303 K
m1 = 303000 * 0.0165 / (287 * 303) = 0.057 kg
The mass that was added is
m1 - m0 = 0.057 - 0.039 = 0.018 kg
If that air is cooled to 0 C
0 C is 273 K
p = m * R * T / V
p = 0.057 * 278 * 273 / 0.0165 = 262000 Pa = 262 kPa
Keisha looks out the window from a tall building at her friend Monique standing on the ground, 8.3 m away from the side of the building, as shown. If Keisha's line of sight makes a 30° angle with the side of the building, what is Keisha's height above the ground? Assume Monique is 1.5 m tall. A. 14 m B. 15 m C. 16 m D. 17 m
Answer:
Explanation:
GIVEN DATA:
Distance between keisha and her friend 8.3 m
angle made by keisha toside building 30 degree
height of her friend monique is 1.5 m
from the figure
[tex]\Delta ACB[/tex]
[tex]tan 30 = \frac{8.3}{h}[/tex]
[tex]h= \frac{8.3}{tan 30} = 14.376 m[/tex]
therefore
height of keisha is [tex]= h + 1.5 m[/tex]
= 14.376 + 1.5
[tex]= 15.876 \simeq 16 m[/tex]
therefore option c is correct
A secret agent skis off a slope inclined at θ = 30.2 degrees below horizontal at a speed of v0 = 20.4 m/s. He must clear a gorge, and the slope on the other side of the gorge is h = 11.7 m below the edge of the upper slope. Does he make it?
Answer:
He will make it if the gorge is no wider than 14.4 m
Explanation:
The secret agent follows a parabolic motion. We have the following data:
[tex]v_0 = 20.4 m/s[/tex] is the initial speed
[tex]\theta=30.2^{\circ}[/tex] below the horizontal is the initial angle
h = 11.7 m is the vertical distance covered by the agent before landing on the other side
Let's start by analyzing the vertical motion. The initial vertical velocity is
[tex]u_y = v_0 sin \theta = (20.4) (sin 30.2^{\circ})=10.3 m/s[/tex]
Where we have chosen downward as positive direction. Now we use the following equation:
[tex]h=u_y t + \frac{1}{2}gt^2[/tex]
where [tex]g=9.8 m/s^2[/tex] (acceleration of gravity) to find the time t at which the agent lands. Substituting the numbers:
[tex]11.7 = 10.3 t + 4.9 t^2\\4.9t^2 + 10.3t -11.7 = 0[/tex]
Which has two solutions: t = -2.92 s and t = 0.82 s. Since the negative solution is meaningless, we discard it, so the agent reaches the other side of the gorge after 0.82 s.
Now we want to find what is the maximum width of the gorge that allows the agent to safely land on the other side. For that, we need to calculate the horizontal velocity of the agent, which is constant during the motion:
[tex]u_x = u_0 cos \theta = (20.4)(cos 30.2^{\circ})=17.6 m/s[/tex]
So, the horizontal distance covered by the agent is
[tex]d = u_x t = (17.6)(0.82)=14.4 m[/tex]
So, the agent will land safely if the gorge is at most 14.4 m wide.
The secret agent will make it if the width of the gorge is less than 14.4 m.
The given parameters;
inclination of the slope, θ = 30.2⁰initial velocity, v = 20.4 m/sheight of the gorge, h = 11.7 mThe time to travel the vertical distance is calculated as follows;
[tex]h = v_0_yt + \frac{1}{2} gt^2\\\\11.7 = (20.4\times sin(30.4))t + (0.5\times 9.8)t^2\\\\11.7 = 10.32t + 4.9t^2\\\\4.9t^2 + 10.32t - 11.7 = 0\\\\solve \ the \ quadratic \ equation, \ using \ formula \ method;\\\\a = 4.9, \ b = 10.32, \ c = - 11.7\\\\t = \frac{-b \ \ + /- \ \ \sqrt{b^2 - 4ac} }{2a} \\\\t = 0.82 \ s[/tex]
The horizontal distance traveled by the secret agent is calculated as;
[tex]X = V_0_x t\\\\X = (20.4 \times cos (30.4) \times 0.82\\\\X = 14.4 \ m[/tex]
Thus, we can conclude that the secret agent will make it if the width of the gorge is less than 14.4 m.
Learn more here:https://brainly.com/question/2411455
Light travels faster in warmer air. On a sunny day, the sun can heat a road and create a layer of hot air above it. Let's model this layer as a uniform one with a sharp boundary separating it from the cooler air above. Use this model to explain the formation of a mirage appearing like the shiny surface of a pool of water.
Answer:
Explanation:
If we assume there is a sharp boundary between the two masses of air, there will be a refraction. The refractive index of each medium will depend on the relative speeds of light.
n = c / v
If light travels faster in warmer air, it will have a lower refractive index
nh < nc
Snell's law of refraction relates angles of incidence and refracted with the indexes of refraction:
n1 * sin(θ1) = n2 * sin(θ2)
sin(θ2) = sin(θ1) * n1/n2
If blue light from the sky passing through the hot air will cross to the cold air, then
n1 = nh
n2 = nc
Then:
n1 < n2
So:
n1/n2 < 1
The refracted light will come into the cold air at angle θ2 wich will be smaller than θ1, so the light is bent upwards, creating the appearance of water in the distance, which is actually a mirror image of the sky.
Three negative point charges q1 =-5 nC, q2 = -2 nC and q3 = -5 nC lie along a vertical line. The charge q2 lies exactly between charge q1 and q3 which are 16 cm apart. Find the magnitude and direction of the electric field this combination of charges produces at a point P at a distance of 6 cm from the q2
Answer:
Ep= 5450N/C in direction (-x)
Explanation:
Conceptual analysis
The electric field at a point P due to a point charge is calculated as follows:
E = k*q/d²
E: Electric field in N/C
q: charge in Newtons (N)
k: electric constant in N*m²/C²
d: distance from charge q to point P in meters (m)
Equivalence
1nC= 10⁻⁹C
1cm= 10⁻²m
Graphic attached
The attached graph shows the field due to the charges q₁, q₂ y q₃ in the P (x=6, y=0):
As the charge q₁ ,q₂ , and q₃ are negative the field enter the charges.
E₁: Electric Field due to charge q₁.
E₂: Electric Field due to charge q₂.
E₃: Electric Field due to charge q₃.
Field calculation due to q₁ and q₃
Because q₁ = q₃ and d₁ = d₃, then, the magnitude of E₁ is equal to the magnitude of E₃
d₁=d₃=d
[tex]d=\sqrt{6^{2}+8^{2} } =10cm=0.1m[/tex]
q₁=q₃= -5nC= 5*10⁻⁹C
E₁ = E₃= k*q/d² = 9*10⁹*5*10⁻⁹/0.1² = 4500 N/C
E₁x = E₃x= - 4500*(6/10)= -2700 N/C
E₁y = -4500*(8/10)= -3600 N/C
E₃y= +4500*(8/10)=+3600N/C
Field calculation due to q₂
E₂x= k*q₂/d₂²= -9*10⁹*2*10⁻⁹/0.6²= -50 N/C
Magnitude and direction of the electric field in the point P (Ep)
Epx= E₁x+ E₂x+E₃x = -2700 N/C-50 N/C-2700 N/C= -5450N/C
Epy= E₁y+ E₃y= -3600 N/C+3600N = 0
Ep= 5450N/C in direction (-x)
A mass attached to a horizontal spring is stretched by 10 cm and released. It takes 0.2 sec for the mass to reach the equilibrium position. The mass is then stretched to 30 cm and released. How long does it take the mass to reach the equilibrium position?
Answer:
No change
Explanation:
The time period of an oscillating body depends on the mass of the body attached and the spring constant of the spring.
The time taken by the oscillating body to complete one vibration is called the time period of teh body.
As the time period does not depend on the amplitude of the oscillations so the time period does not change as the amplitude changes.
Thus, the time taken by the mass to reach to the equilibrium position remains same.
A long, nonconducting cylinder (radius = 6.0 mm) has a nonuniform volume charge density given by αr2, where α = 6.2 mC/m5 and r is the distance from the axis of the cylinder. What is the magnitude of the electric field at a point 2.0 mm from the axis?
Answer: 2.80 N/C
Explanation: In order to calculate the electric firld inside the solid cylinder
non conductor we have to use the Gaussian law,
∫E.ds=Q inside/ε0
E*2πrL=ρ Volume of the Gaussian surface/ε0
E*2πrL= a*r^2 π* r^2* L/ε0
E=a*r^3/(2*ε0)
E=6.2 * (0.002)^3/ (2*8.85*10^-12)= 2.80 N/C
Three point charges are arranged along the x-axis. Charge q1 = +3.00 μC is at the origin, and charge q2 = -5.00 μC is at x = 0.200 m. Charge q3 = -8.00 μC. Where is q3 located if the net force on q1 is 7.00 N in the -x-direction?
Answer:
q₃ is located at X= - 0.144m
Explanation:
Coulomb law:
Two point charges (q1, q2) separated by a distance (d) exert a mutual force (F) whose magnitude is determined by the following formula:
F=K*q₁*q₂/d² Formula (1)
F: Electric force in Newtons (N)
K : Coulomb constant in N*m²/C²
q₁, q₂: Charges in Coulombs (C)
d: distance between the charges in meters(m)
Equivalence
1uC= 10⁻⁶C
Data
F=0.66 N
K=8.99x10⁹N*m²/C²
q₁ = +3.00 μC = +3.00 *10⁻⁶C
q₂= -5.00 μC = -5.00 *10⁻⁶C
q₃= -8.00 μC= -8.00 *10⁻⁶C
d₁₂= 0.2m: distance between q₁ and q₂
Fn=7 N: Net force on q₁
Calculation of the magnitude of the forces exerted on q₁
F₂₁ = K*q₁*q₂/d₁₂² = (8.99 * 10⁹ * 3.00 * 10⁻⁶ * 5.00 * 10⁻⁶)/(0.2)² = 3.37 N
F₃₁ = K*q₁*q₃/d₁₃² = (8.99 * 10⁹ * 3.00 * 10⁻⁶ * 8.00 * 10⁻⁶)/(d₁₃)² = 0.21576/(d₁₃)² N
Calculation of the distance d₁₃ between q₁ and q₃
In order for the net force to be negative on the x axis, the charge q₃ must be located on the x axis to the left of the charge q₁
F₂₁: It is a force of attraction and goes to the right(+)
F₃₁: It is a force of attraction and goes to the left(-)
-Fn₁ = F₂₁ - F₃₁
-7 = 3.37 - 0.21576/(d₁₃)²
3.37 + 7 = 0.21576/(d₁₃)²
10.37*(d₁₃)²=0.21576
(d₁₃)² = (0.215769 )/ (10.37)
(d₁₃)² = 20.8 * 10⁻³
[tex]d_{13} = \sqrt{20.8 * 10^{-3}} = 0.144 m[/tex]
d₁₃=0.144m
q₃ is located at X= - 0.144m
Consider the resistances of short and open circuits. Fill in the blanks: The voltage across a short circuit will always be ______________ The voltage across an open circuit will always be ______________
Answer:
a) Zero
b) power source
Explanation:
According to Ohm's law, the voltage dropped in a resistance is proportional to the current flow and the resistor opposing to it.
[tex]V=I*R[/tex]
For the case of a short circuit, the resistance tends to zero, so the voltage will tend to zero too.
In the case of the open circuit, the resitance will tend to infinity, because we cannot obtain an infite voltage, it will be limited by the power source.
The voltage across a short circuit is always zero, due to the negligible resistance allowing easy current flow, whereas in an open circuit, which has infinite resistance preventing current flow, the voltage will be equal to the supply voltage.
Explanation:The question concerns the behaviors of electrical circuits specifically relating to short and open circuits. When considering the resistances and the resulting voltages across these types of circuits, it's crucial to understand two fundamentals:
The voltage across a short circuit will always be zero.The voltage across an open circuit will always be equal to the supply voltage.In a short circuit, a pathway exists with negligible resistance, allowing current to flow easily and resulting in virtually no voltage drop across the circuit. In contrast, an open circuit has infinite resistance because the path is broken, preventing any current from flowing; thus, the whole supply voltage appears across the open circuit.
Assume that an oil slick consists of a single layer of molecules and that each molecule occupies a cube 1.4 nm on a side. Determine the area of an oil slick formed by 1.1 m^3 of oil. Answer in units of m^2.
Answer:
The area of oil slick is calculate as [tex]785.7 m^{2}[/tex]
Solution:
Volume of oil slick, [tex]V_{o} = 1.1 m^{3}[/tex]
The thickness of one molecule on a side, w = 1.4 mm = [tex]1.4\times 10^{- 3}[/tex]
Now, in order to determine the area of oil slick, [tex]A_{o}[/tex]:
Volume, V = [tex]Area\times thickness[/tex]
Thus
[tex]Area,\ A_{o} = \frac{V_{o}}{w}[/tex]
[tex]Area,\ A_{o} = \frac{1.1}{1.4\times 10^{- 3}} = 785.7 m^{2}[/tex]
A meteoroid is first observed approaching the earth when it is 402,000 km from the center of the earth with a true anomaly of 150. If the speed of the meteoroid at that time is 2.23 km/s, calculate
(a) the eccentricity of the trajectory;
(b) the altitude at closest approach; and
(c) the speed at the closest approach.
Explanation:
Given that,
Distance = 402000 km
Speed = 2.23 m/s
Angle = 150
(a). We need to calculate the eccentricity of the trajectory
Using formula of eccentricity
[tex]\epsilon=\dfrac{v^2}{2}-\dfrac{\mu}{r}[/tex]
Put the value into the formula
[tex]\epsilon=\dfrac{2.23^2}{2}-\dfrac{398600}{402000}[/tex]
[tex]\epsilon=1.4949\ km^2/s^2[/tex]
We need to calculate the angular momentum
Using formula of the angular momentum
[tex]h^2=-\dfrac{1}{2}\dfrac{\mu^2}{\epsilon}(1-e^2)[/tex]
[tex]h^2=-\dfrac{1}{2}\dfrac{(398600)^2}{1.4949}(1-e^2)[/tex]
[tex]h^2=-5.3141\times10^{10}(1-e^2)[/tex]...(I)
The orbit equation is
[tex]h^2=\mu r(1+e\cos\theta)[/tex]
[tex]h^2=398600\times402000(1-+\cos150)[/tex]
[tex]h^2=16.02372\times10^{10}(1-e0.8660)[/tex]
[tex]h^2=16.02372\times10^{10}-e13.877\times10^{10}[/tex]....(II)
Equating the value of h²
[tex]-5.3141\times10^{10}(1-e^2)=16.02372\times10^{10}-e13.877\times10^{10}[/tex]
[tex]-5.3141+5.3141e^2=13.877e+16.02372[/tex]
[tex]5.3141e^2+13.877e-21.33782=0[/tex]
[tex]e = 0, 1.086[/tex]
(b). We need to calculate the altitude at closest approach
Put the value of e in equation (I)
[tex]h^2=16.02372\times10^{10}-1.086\times13.877\times10^{10}[/tex]
[tex]h^2=9.53298\times10^{9}\ km^4/s^2[/tex]
Now, using the formula of the altitude at closest
[tex]r_{perigee}=\dfrac{h^2}{\mu}\dfrac{1}{1+e}[/tex]
[tex]r_{perigee}=\dfrac{9.53298\times10^{9}}{398600}\dfrac{1}{1+1.086}[/tex]
[tex]r_{perigee}=11465\ km[/tex]
So, The altitude is
[tex]z_{perigee}=r_{perigee}-r_{earth}[/tex]
[tex]z_{perigee}=11465-6378[/tex]
[tex]z_{perigee}=5087\ km[/tex]
(c). We need to calculate the speed at the closest approach.
Using formula of speed
[tex]v_{perigee}=\dfrac{h}{r_{perigee}}[/tex]
[tex]v_{perigee}=\dfrac{\sqrt{9.53298\times10^{9}}}{11465}[/tex]
[tex]v_{perigee}=8.516\ km/s[/tex]
Hence, This is the required solution.
(a) The eccentricity of the meteoroid's trajectory is approximately -0.226. This value indicates the deviation of the orbit from a perfect circle, with negative eccentricity suggesting an elliptical orbit.
(b) At its closest approach, or perigee, the meteoroid is at an altitude of about 595,261 kilometers from the Earth's center. This is the point in its trajectory where it is closest to Earth.
(c) The speed of the meteoroid at its closest approach is approximately 3.69 kilometers per second. This velocity is characteristic of its orbital motion when it is nearest to Earth and is determined by the specific orbital energy and gravitational influences.
To calculate the eccentricity (e), altitude at closest approach (perigee), and speed at the closest approach, we can use the vis-viva equation and orbital mechanics. The vis-viva equation relates the specific orbital energy (ε) to the semi-major axis (a), eccentricity (e), and velocity (v) of an object in orbit:
ε = v²/2 - μ/a
Where:
ε = specific orbital energy
v = velocity of the object
μ = standard gravitational parameter of Earth (approximately 3.986 x 10^5 km³/s²)
a = semi-major axis of the trajectory
We are given the speed of the meteoroid at a distance of 402,000 km from the Earth's center, which we'll use as v. We'll also use μ for Earth's gravitational parameter.
First, let's find the semi-major axis (a):
ε = v²/2 - μ/a
Rearrange to solve for a:
a = μ / (2μ/v² - 1/v²)
a = (3.986 x 10^5 km³/s²) / (2 * (3.986 x 10^5 km³/s²) / (2.23 km/s)² - (1 / (2.23 km/s)²))
a ≈ 491,453 km
(a) The semi-major axis (a) is approximately 491,453 km.
Next, let's calculate the eccentricity (e) using the true anomaly (ν) and the semi-major axis:
e = cos(ν) - r/a
e = cos(150°) - (402,000 km) / (491,453 km)
e ≈ -0.226
(b) The eccentricity (e) is approximately -0.226.
Now, to find the altitude at closest approach (perigee), we need to calculate the perigee distance (rp) using the semi-major axis and eccentricity:
rp = a(1 - e)
rp = (491,453 km) * (1 - (-0.226))
rp ≈ 595,261 km
(c) The perigee distance is approximately 595,261 km.
To find the speed at closest approach (closest approach velocity), we can use the vis-viva equation again:
v = sqrt(μ * (2/r - 1/a))
Where r is the distance from the center of the Earth (rp at perigee).
v = sqrt((3.986 x 10^5 km³/s²) * (2 / (595,261 km) - 1 / (491,453 km)))
v ≈ 3.69 km/s
(c) The speed at closest approach is approximately 3.69 km/s.
Learn more about meteoroid's here:
https://brainly.com/question/30123423
#SPJ6
In this first example of constant accelerated motion, we will simply consider a car that is initially traveling along a straight stretch of highway at 15 m/s. At t=0 the car begins to accelerate at 2.0 m/s2 in order to pass a truck. What is the velocity of the car after 5.0 s have elapsed?
Answer:
[tex]v_{f} =25m/s[/tex]
Explanation:
Kinematics equation for constant acceleration:
[tex]v_{f} =v_{o} + at=15+2*5=25m/s[/tex]
Two infinite planes of charge lie parallel to each other and to the yz plane. One is at x = -5 m and has a surface charge density of σ = -2.6 µC/m^2. The other is at x = 3 m and has a surface charge density of σ = 5.8 µC/m^2. Find the electric field for the following locations. (a) x < -5 m
(b) -5 m < x < 3 m
(c) x > 3 m
Answer:
a) -180.7 kN/C
b) -474.3 kN/C
c) 180.7 kN/C
Explanation:
For infinite planes the electric field is constant on each side, and has a value of:
E = σ / (2 * e0) (on each side of the plate the field points in a different direction, the fields point towards positive charges and away from negative charges)
The plate at -5 m produces a field of:
E1 = 2.6*10^-6 / (2 * 8.85*10^-12) = 146.8 kN/C into the plate
The plate at 3 m:
E2 = 5.8*10^-6 / (2 * 8.85*10^-12) = 327.5 kN/C away from the plate
At x < -5 m the point is at the left of both fields
The field would be E = 146.8 - 327.5 = -180.7 kN/C
At -5 m < x < 3 m, the point is between the plates
E = -146.8 - 327.5 = -474.3 kN/C
At x > 3 m, the point is at the right of both plates
E = -146.8 + 327.5 = 180.7 kN/C
The electric fields at the given locations are calculated by using the formula E=σ/2ε₀, and considering that the fields always point toward negative charges. Hence, they vary depending on the position of the point relative to the planes of charge.
Explanation:The field contribution due to each individual infinite plane is E=σ/2ε₀, where ε₀ is the permittivity of the free space (8.85 x10⁻ⁱ² C²/Nm²). Now, we'll calculate the electric field for each given location:
(a) For x < -5 m: Here, the point sits to the left of both the planes. Both fields point to the right (towards negative charge), so they add up.(b) -5 m < x < 3 m: The point is between the planes. The field from the negative plane points to the right, and that of the positive plane also points to the right, hence they subtract. E = -2.6 / 2ε₀ - 5.8 / 2ε₀ .(c) x > 3 m: Here, the point is to the right of both planes. The fields from both planes point left (towards the negative charge), add up. E = -2.6 / 2ε₀ + 5.8 / 2ε₀ . Learn more about Electric Field here:
https://brainly.com/question/8971780
#SPJ3
A baseball is hit with a speed of 47.24 m/s from a height of 0.42 meters. If the ball is in the air 5.73 seconds and lands 130 meters from the batters feet, (a) at what angle did the ball leave the bat? (b) with what velocity will the baseball hit the ground?
Answer:
a)the ball will leave the bat at an angle of 61.3° .
b) the velocity at which it will hit the ground will be v = 27.1 m/s
Explanation:
Given,
v = 47.24 m
h = 0.42 m
t = 5.73 s
R = 130 m
a)We know that
R = v cosθ × t
cosθ = [tex]\dfrac{R}{v t } = \dfrac{130}{47.24\times 5.73 } =0.4803[/tex]
θ = 61.3°
the ball will leave the bat at an angle of 61.3° .
b)Vx = v cos(θ) = 47.24 x cos 61.3 = 22.7 m/s
v = u + at
Vy = 47.24 x sin 61.3 - 9.81 x 5.73
= -14.8 m/s
v = [tex]\sqrt{v_x^2 + v_y^2)}[/tex]
v = [tex]\sqrt{22.7^2 + -14.8^2}[/tex]
v = 27.1 m/s
the velocity at which it will hit the ground will be v = 27.1 m/s
The horizontal bar rises at a constant rate of three hundred mm/s causing peg P to ride in the quarter circular slot. When coordinate y reaches 150mm, find the v and a vectors for P. Hint: for a circle with origin shown x2+y2=r2. The radius of the slot is 300 mm.
Answer:
Explanation:
Given
Horizontal bar rises with 300 mm/s
Let us take the horizontal component of P be
[tex]P_x=rcos\theta [/tex]
[tex]P_y=rsin\theta [/tex]
where [tex]\theta [/tex]is angle made by horizontal bar with x axis
Velocity at y=150 mm
[tex]150=300sin\theta [/tex]
thus [tex]\theta =30^{\circ}[/tex]
position of[tex]P_x=rcos\theta =300\cdot cos30=300\times \frac{\sqrt{3}}{2}[/tex]
[tex]P_x=259.80 mm[/tex]
[tex]P=259.80\hat{i}+150\hat{j}[/tex]
Velocity at this instant
[tex]u_x=-rsin\theta =300\times sin30=-150 mm/s[/tex]
[tex]u_y=rcos\theta =300\times cos30=259.80 mm/s[/tex]
A ball thrown straight up climbs for 3.0 sec before falling. Neglecting air resistance, with what velocity was the ball thrown?
Answer:
Speed, u = 29.4 m/s
Explanation:
Given that, A ball thrown straight up climbs for 3.0 sec before falling, t = 3 s
Let u is speed with which the ball is thrown up. When the ball falls, v = 0
Using first equation of motion as :
v = u + at
Here, a = -g
So, u = g × t
[tex]u=9.8\times 3[/tex]
u = 29.4 m/s
So, the speed with which the ball was thrown is 29.4 m/s. Hence, this is the required solution.
The velocity at which the ball was thrown is 29.4 m/s.
To calculate the velocity at which the ball was thrown, we use the formula below.
Formula:
v = u+gt.............. Equation 1Where:
v = Final velocity of the ballu = Initial velocity of the ballg = acceleration due to gravity of the ballt = time.From the question,
Given:
v = 0 m/s (At maximum height)g = -9.8 m/st = 3.0 sSubstitute these values into equation 1
0 = u+3(-9.8)0 = u-29.4u = 29.4 m/sHence, The velocity at which the ball was thrown is 29.4 m/s.
Learn more bout velocity here: https://brainly.com/question/25749514
Consider an equilateral triangle with sides measuring 1.0 m in length. At each point of the triangle is a +2.0 μC charge. Calculate the Coulomb force on each charge. Recall that forces are vectors and thus your answer will require a magnitude and direction for each of the three forces.
Answer:
Th magnitude of each Force will be [tex]=62.35\times10^{-3}\ \rm N[/tex]
Explanation:
Given:
Length of each side of the equilateral triangle, L=1 mMagnitude of each point charge [tex]Q=2\ \rm \mu C[/tex]Since all the charges are identical and distance between them is same so magnitude of the force between each charge is equal.
Let F be the force between the particles. According to Coulombs Law we have
[tex]F=\dfrac{kQ^2}{L^2}\\=\dfrac{9\times10^9\times (2\times10^{-6})^2}{1^2}\\F=36\times10^{-3}\ \rm N[/tex]
Now the the force on any charge by other two charges will be F and the angle between the two force is [tex]60^\circ[/tex]
Let [tex]F_{resultant}[/tex] be the force on nay charge by other two
By using vector Law of addition we have
[tex]F_{resultant}=\sqrt{(F^2+F^2+2F\times F \times cos60^\circ)}\\=\sqrt{3}F\\=\sqrt{3}\times36\times10^{-3}\ \rm N\\=62.35\times10^{-3}\ \rm N[/tex]
The angle made by the resultant vector will be
[tex]\tan\beta=\dfrac{F\sin60^\circ}{F+F\cos60^\circ}\\\beta=30^\circ[/tex]
A car and a motorcycle start from rest at the same time on a straight track, but the motorcycle is 25.0 m behind the car. The car accelerates at a uniform rate of 3.70 m/s^2 and the motorcycle at a uniform rate of 4.40 m/s^2. How much time elapses before the MC overtakes the car? How far will each have traveled during that time?
Answer:
t = 8.45 sec
car distance d = 132.09 m
bike distance d = 157.08 m
Explanation:
GIVEN :
motorcycle is 25 m behind the car , therefore distance need to covered by bike to overtake car is 25+ d, when car reache distance d at time t
for car
by equation of motion
[tex]d = ut + \frac{1}{2}at^2[/tex]
u = 0 starting from rest
[tex]d = \frac{1}{2}at^2[/tex]
[tex]t^2 = \frac{2d}{a}[/tex]
for bike
[tex]d+25 = 0 + \frac{1}{2}*4.40t^2[/tex]
[tex]t^2= \frac{d+25}{2.20}[/tex]
equating time of both
[tex]\frac{2d}{a} = \frac{d+25}{2.20}[/tex]
solving for d we get
d = 132 m
therefore t is[tex] = \sqrt{\frac{2d}{a}}[/tex]
[tex]t = \sqrt{\frac{2*132}{3.70}}[/tex]
t = 8.45 sec
each travelled in time 8.45 sec as
for car
[tex]d = \frac{1}{2}*3.70 *8.45^2[/tex]
d = 132.09 m
fro bike
[tex]d = \frac{1}{2}*4.40 *8.45^2[/tex]
d = 157.08 m
To find the time when the motorcycle overtakes the car, equate the distances they have traveled given their individual accelerations and solve for time. Once the time is known, calculate the distance each has traveled using the equations of motion for uniform acceleration.
Explanation:To determine when the motorcycle (MC) overtakes the car, we need to calculate the time at which both have traveled the same distance, considering the initial 25.0 m advantage of the car. We can use the equation of motion d = ut + (1/2)at2 where d is the distance traveled, u is the initial velocity, a is the acceleration, and t is the time.
The car starts from rest, so its initial velocity is 0, and it accelerates at 3.70 m/s2. The motorcycle also starts from rest, with an acceleration of 4.40 m/s2, but it needs to cover an additional 25.0 m to catch up with the car.
We can set the equations equal to each other to find the time t when the distances are equal:
Distance covered by the car: Car_d = (1/2)(3.70 m/s2)t2Distance covered by the MC: MC_d = 25.0 m + (1/2)(4.40 m/s2)t2To find the time when MC overtakes the car, we equate Car_d to MC_d and solve for t:
(1/2)(3.70)t2 = 25.0 + (1/2)(4.40)t2
After solving for t, we can calculate the distance each has traveled using the original equations of motion for uniform acceleration.
A moving van travels 10km North, then 4 km east, drops off some furniture and then drives 8 km south. (a) Sketch the path of the moving van. (2) How far is the van from its starting point? (Hint: it is the magnitude of the displacement vector).
Answer:
4.47 km
Explanation:
If we draw the path of the van then we get a shape with two exposed points A and D. If we draw a line from point D perpendicular to BA we get point E. This gives us a right angled triangle ADE.
From Pythagoras theorem
AD² = AE² + ED²
[tex]AD=\sqrt{AE^2+ED^2}\\\Rightarrow AD=\sqrt{2^2+4^2}\\\Rightarrow AD=\sqrt{20}\\\Rightarrow AD=4.47\ km[/tex]
Hence, the van is 4.47 km from its initial point
If the gravitational force between objects of equal mass is 2.30 x 10^‐8 N when the objects are 10.0 m apart, what is the mass of each object?
Answer:
Mass of each objects, m = 185.69 kg
Explanation:
The gravitational force between two equal masses is, [tex]F=2.3\times 10^{-8}\ N[/tex]
Separation between the masses, d = 10 m
Let m is the mass of each object. The gravitational force is given by :
[tex]F=G\dfrac{m^2}{d^2}[/tex]
[tex]m=\sqrt{\dfrac{F.d^2}{G}}[/tex]
[tex]m=\sqrt{\dfrac{2.3\times 10^{-8}\times (10)^2}{6.67\times 10^{-11}}}[/tex]
m = 185.69 kg
So, the mass of each objects is 185.69 kg. Hence, this is the required solution.
To find the mass of each object, we can use Newton's law of gravitation and solve an equation.
Explanation:To find the mass of each object, we can use Newton's law of gravitation, which states that the gravitational force between two objects is given by the equation: F = G * M₁ * M₂ / R². Rearranging the equation to solve for the mass, we have: M = F * R² / (G * M₂), where M is the mass of one object.
Using the given values, the gravitational force is 2.30 x 10^‐8 N and the distance between the objects is 10.0 m. Plugging these values into the equation, we have: M = (2.30 x 10^‐8 N) * (10.0 m)² / (6.67 × 10-¹¹ Nm²/kg² * M₂).
Since the masses of the two objects are equal, we can assume M₁ = M₂. Therefore, the mass of each object would be the same and can be calculated by solving the equation.
Learn more about mass of objects here:https://brainly.com/question/6240825
#SPJ11
Students in an introductory physics lab are performing an experiment with a parallel-plate capacitor made of two circular aluminum plates, each 21 cm in diameter, separated by 1.0 cm . Part A How much charge can be added to each of the plates before a spark jumps between the two plates? For such flat electrodes, assume that value of 3× 10 6 N/C of the field causes a spark. Express your answer with the appropriate units. q max q m a x = nothing nothing SubmitRequest Answer Provide Feedback Next
Answer:
0.92 μC
Explanation:
In a parallel-plate capacitor, the electric field formed is equal to the charge density divited by the vacuum permisivity e0, as there are no dielectric between the plates. e0 is equal to 8.85*10^-12 C^2/Nm^2. The charge density is the total charge of each individual plate divided by its area. Then, the maximum charge allowed will be equal to:
[tex]E = \frac{o}{e_0} = \frac{Q}{Ae_0} \\ Q = E*A*e_0 = 3*10^6 N/C * (0.25*\pi *(0.21m)^2)*8.85*10^{-12}C^2/Nm^2 = 9.196 *10^{-7} C[/tex]
or 0.92 μC