Answer:
a) 215 cm
b) 1290 mm
c) 2.81 m³
d) 265 cm
e) 1590 mm
f) 5.26 m³
Explanation:
Lower value
1 cubit = 43 cm
Length of cylinder = 5 cubits
So,
a) Length of cylinder = 5×43 = 215 cm
Diameter of cylinder = 3 cubit
1 cubit = 43 cm = 430 mm
b) Diameter of cylinder = 3×430 = 1290 mm
Radius of cylinder = 129/2 = 64.5 cm
Volume of cylinder
[tex]v=\pi r^2h\\\Rightarrow v=\pi 0.645^2\times 2.15\\\Rightarrow v=2.81 m^3[/tex]
c) Volume of cylinder = 2.81 m³
Upper value = 53 cm
Length of cylinder = 5 cubits
So,
d) Length of cylinder = 5×53 = 265 cm
Diameter of cylinder = 3 cubit
1 cubit = 53 cm = 530 mm
e) Diameter of cylinder = 3×530 = 1590 mm
Radius of cylinder = 159/2 = 79.5 cm
Volume of cylinder
[tex]v=\pi r^2h\\\Rightarrow v=\pi 0.795^2\times 2.65\\\Rightarrow v=5.26 m^3[/tex]
f) Volume of cylinder = 5.26 m³
A two stage rocket is launched moving vertically with acceleration 5.0 m/s^2. After 10.0 s, the first stage of the rocket is ejected; the second stage is now accelerating at 8.0 m/s^2. What is the distance between the first and second stages 4.0 s after separation?
Answer:264 m
Explanation:
Given
acceleration of rocket([tex]a_1[/tex])= 5 m/s^2[/tex]
velocity after 10 s
v=u+at
[tex]v=0+5\times 10[/tex]
v=50 m/s
after first stage rocket is ejected
acceleration of second stage=[tex]8 m/s^2[/tex]
distance between first and second part after 4 sec
[tex]s=u_1t+\frac{1}{2}at^2 [/tex]
here [tex]u_1=50 m/s[/tex]
[tex]s=50\times 4+\frac{1}{2}\times 8\times 4^2[/tex]
s=200+64=264 m
(5%) Problem 17: A solid aluminum sphere of radius r1 = 0.105 m is charged with q1 = +4.4 μC of electric charge. It is surrounded concentrically by a spherical copper shell of inner radius r2 = 0.15 m and outer radius r3 = 0.242 m. An electric charge of q2 = -1.1 μC is placed on the copper shell. The situation is static.
(a) Find the magnitude of the electric field, in newtons per coulomb, inside the aluminum ball.
(b) What is the magnitude of the electric field, in newtons per coulomb, inside the copper of the shell, i.e., at a radius r from the center such that r2 < r < r3?
Answer:
Electric field at radius r inside the solid sphere is
[tex]E=\dfrac{q_{1}r}{4\pi \epsilon_{o} r_{1}^3}\ N/C\\[/tex]
Electric field at radius r between inner radius and outer radius of the shell is
E=0 N/C
Explanation:
Given
The radius of the solid sphere is [tex]r_{i}=0.105 \ m\\[/tex]
The charge on the solid sphere is [tex]q_{1}=+4.4\ \mu C[/tex]
The inner radius of the shell is [tex]r_{3}=0.15\ m[/tex]
The outer radius of the shell is [tex]r_{3}=0.242\ m[/tex]
The total charge on the shell is [tex]q_{2}=-1.1\ \mu C[/tex]
PART(A)
The magnitude of electric field at radius r where [tex] r<r_{1}[/tex] \\The volumetric charge density of the solid sphere will be
[tex]\rho=\dfrac{q_{1}}{V}\\ \rho=\dfrac{q_{1}}{\dfrac{4}{3}\pi r_{1}^3}\\[/tex]
The charge enclosed by the radius r inside the solid sphere is
rho=[tex]q_{enc}=\rho\times \dfrac{4}{3}\pi r^3\\q_{enc}=\dfrac{q_{1}}{\dfrac{4}{3}\pi r_{1}^3}\times \dfrac{4}{3}\pi r^3\\q_{enc}=\dfrac{q_{1}r^3}{r_{1}^3}\\[/tex]
According to gauss law
[tex]EA=\dfrac{q_{enc}}{\epsilon_{o}}\\E=\dfrac{\dfrac{q_{1}r^3}{r_{1}^3}}{\epsilon_{o}\times 4\pi r^2}\\E=\dfrac{q_{1}r}{4\pi \epsilon_{o} r_{1}^3}\ N/C\\[/tex]
PART(B)
The electric field at radius r where [tex]r_{2}<r<r_{3}[/tex]
The shell is conducting so due to induction of charge
The charge induced on the inner surface of the shell is equal in magnitude of the total charge on the solid sphere but polarity will be changed because a conducting shell has no net electric field inside the shell
So,
The charge on the inner surface of the shell is
[tex]q_{i}=-q_{1}\\q_{i}=-4.4\ \mu C\\[/tex]
Due to conservation of the charge on the shell
The charge accumulated on the outer surface of the shell is
[tex]q_{o}=q_{2}-q_{i}\\q_{o}=-1.1\ \mu C-(-4.4\ \mu C)\\q_{o}=-1.1\ \mu C+4.4\ \mu C\\q_{o}=3.3\ \mu C\\[/tex]
The charge enclosed by the radius r where [tex]r_{1}<r<r_{2}[/tex]
[tex]q_{enc}=q_{1}+q_{i}\\q_{enc}=4.4\times 10^{-6}-4.4\times 10^{-6}\\ q_{enc}=0\\[/tex]
According to gauss law
[tex]EA=\dfrac{q_{enc}}{\epsilon_{o}}\\ E=0\ N/C\\[/tex]
(a) Inside the aluminum sphere, the electric field is 0 N/C due to electrostatic equilibrium.
(b) Inside the copper shell, for any radius between r2 and r3, the electric field is given by (3.95 x 10⁴) / r² N/C.
To solve this problem, we need to understand the behavior of the electric field in different regions around charged spherical conductors.
(a) Electric Field Inside the Aluminum Sphere
Inside a conductor in electrostatic equilibrium, the electric field is zero. Since the aluminum sphere is a conductor, the electric field inside it (for any radius r < r1) is zero.
Result: The magnitude of the electric field inside the aluminum sphere is 0 N/C.
(b) Electric Field Inside the Copper Shell
For the region inside the copper shell but outside the aluminum sphere (r2 < r < r3), we can apply Gauss's Law. According to Gauss's Law, the electric field at a distance r from the center due to a symmetric charge distribution can be calculated as follows:
Define the Gaussian surface: Here, it will be a spherical surface of radius r where r2 < r < r3.
Calculate the enclosed charge: The total charge enclosed within the Gaussian surface is only the charge of the aluminum sphere, q1 = +4.4 μC, because the copper shell's net charge is enclosed separately.
Apply Gauss's Law: Gauss's law states that the electric flux through a surface is proportional to the enclosed charge:
Equation: E * 4πr² = q1/ε₀
Solve for E:
E = q1 / (4πε₀ r²)
Substitute the values: q1 = 4.4 x 10⁻⁶ C, ε₀ = 8.854 x 10⁻¹² C²/(N·m²), and r is the distance within the copper shell.
[tex]E = \frac{4.4 \times 10^{-6} \, \text{C}}{4\pi \cdot (8.854 \times 10^{-12} \, \text{C}^2/\text{N}\cdot\text{m}^2) \cdot r^2}[/tex]
Simplifying gives:
E = (4.4 x 10⁻⁶) / (1.112 x 10⁻¹⁰ * r²)
E = (3.95 x 10⁴) / r² N/C
Hence, the magnitude of the electric field inside the copper shell, at any point where r2 < r < r3, is (3.95 x 10⁴) / r² N/C.
A river with a flow rate of 7 m3/s discharges into a lake with a volume of 9,000,000 m3. The concentration of a particular VOC in the river is 3 mg/L. The lake has a decay coefficient for the VOC of 0.3 per day. What is the concentration of the VOC in the river downstream?
Answer:
The concentration downstream reduces to 0.03463mg/L
Explanation:
Initially let us calculate the time it is required to fill the lake, since for that period of time the pollutant shall remain in the lake before being flushed out.
Thus the detention period is calculated as
[tex]t=\frac{V}{Q}\\\\t=\frac{9000000}{7}=1.285\times 10^{6}seconds\\\\\therefore t = 14.872days[/tex]
Now the concentration of the pollutant after 14.872 days is calculated as
[tex]N_{t}=N_{0}e^{-kt}[/tex]
where
[tex]N_{o}[/tex] is the initial concentration
't' is the time elapsed after which the remaining concentration is calculated
k is the dissociation constant.
Applying values we get
[tex]N_{t}=3\times e^{-0.3\times 14.872}[/tex]
[tex]N_{t}=0.03463mg/L[/tex]
A placekicker must kick a football from a point 36.0 m (about 40 yards) from the goal. Half the crowd hopes the ball will clear the crossbar, which is 3.05 m high. When kicked, the ball leaves the ground with a speed of 22.8 m/s at an angle of 51.0° to the horizontal. (a) By how much does the ball clear or fall short of clearing the crossbar? (Enter a negative answer if it falls short.)
(b) Does the ball approach the crossbar while still rising or while falling?
Answer:
a) The ball clears the crossbar by 10.6 m
b) The ball approaches the crossbar while falling
Explanation:
The position of the ball is described by the vector position r (see attached figure):
r = (x0 + v0 t cos α ; y0 + v0 t sin α + 1/2 g t²)
where:
x0 = initial horizontal position
v0 = magnitude of the initial velocity
t = time
α = launching angle
y0 = initial vertical position
g = acceleration due to gravity (-9.8 m/s²)
The vector r is composed by rx and ry (see figure):
r = (rx ; ry)
a) Let´s find the time at which the ball flies a distance of 36.0 m. If at that time the vertical component of the vector r, ry, is equal or greater than 3.05 m, then the ball will clear the crossbar.
rx = x0 + v0 t cos α = 36.0 m
Since the origin of the system of reference is located where the kicker is, x0 = 0.
36.0 m = v0 t cos α
36.0 m /(v0 cos α) = t
36.0 m / (22.8 m/s * cos 51.0°) = t
t = 2.51 s
Now let´s calculate the height of the ball at that time:
ry = y0 + v0 t sin α + 1/2 g t²
Since the kicker is on the ground, y0 = 0
ry = 22.8 m/s * 2.51 s * sin 51.0° - 1/2 * 9.8 m/s² * (2.51 s)² = 13.6 m
Since the crossbar is 3.05 m high, the ball clears it by (13.6 m - 3.05 m) 10.6 m
b) Please see the figure to figure this out ;)
If the ball approaches the crossbar while still rising, the vertical component (vy) of the velocity vector will be positive. In change, if the ball approaches the crossbar while falling the vertical component of the velocity will be negative. See the figure.
The velocity vector is given by this equation:
v = (vx ; vy)
v = ( v0 cos α ; v0 sin α + g t)
Let´s see the vertical component at time t = 2.51
vy = v0 sin α + g t
vy = 22.8 m/s * sin 51.0° - 9.8 m/s² * 2.51 s
vy = -6.88 m/s
Then, the ball approaches the crossbar while falling.
Express the following angles in radians: (a) 16.8°, (b) 53.7°, (c) 94.2°. Convert the following angles to degrees: (d) 0.258 rad, (e) 1.01 rad, (f) 7.51 rad.
Explanation:
(a) We need to convert following angles in radians. The conversion formula from degrees to radian is given by :
[tex]Radians=(\dfrac{\pi}{180})\times degrees[/tex]
1. If angle is 16.8°
[tex]Radians=(\dfrac{\pi}{180})\times 16.8[/tex]
16.8 degrees = 0.29 radians
2. If angle is 53.7°
[tex]Radians=(\dfrac{\pi}{180})\times 53.7[/tex]
53.7 degrees = 0.937 radians
3. If angle is 94.2°
[tex]Radians=(\dfrac{\pi}{180})\times 94.2[/tex]
53.7 degrees = 1.64 radians
(b) We need to convert following angles to degrees. The conversion formula from radian degrees to is given by :
[tex]Degrees=(\dfrac{180}{\pi})\times radians[/tex]
1. If angle is 0.258 radian
[tex]Degrees=(\dfrac{180}{\pi})\times 0.258[/tex]
0.258 radian = 14.78 degrees
2. If angle is 1.01 radian
[tex]Degrees=(\dfrac{180}{\pi})\times 1.01[/tex]
0.258 radian = 57.86 degrees
3. If angle is 7.51 radian
[tex]Degrees=(\dfrac{180}{\pi})\times 7.51[/tex]
0.258 radian = 430.29 degrees
Hence, this is the required solution.
Aaron finds a pirate’s treasure map. The treasure map says to start at an oak tree and walks 111 feet East and 234 feet South from the tree. How far did the pirate walk? What is the displacement vector for the Pirate? What is the distance and direction relative to East?
Answer:
pirate walk 259 feet
displacement vector = 111 i - 234 j
direction is along south of east at angle 64.62° , anticlockwise
total travel 345 feet
Explanation:
given data
walk east = 111 feet
walk south = 234 feet
to find out
How far did the pirate walk and displacement vector and distance and direction relative to east
solution
we consider here distance AB is 111 feet and than he turn right i.e south distance BC is 234 feet so
so angle BAC will be
tan θ = [tex]\frac{234}{111}[/tex]
θ = 64.62
and AC distance will be
AC = [tex]\sqrt{234^{2} + 111^{2}}[/tex]
AC = 259 feet
so pirate walk 259 feet
and
displacement vector is express as
displacement vector = AC ( cosθ i + sinθ j )
displacement vector = 259 ( cos64.62 i + sin64.62 j )
displacement vector = 111 i - 234 j
and
so direction is along south of east at angle 64.62° , anticlockwise
A tortoise and a hare are competing in a 1200-meter race. The arrogant hare decides to let the tortoise have a 580-meter head start. When the start gun is fired the hare begins running at a constant speed of 9 meters per second and the tortoise begins crawling at a constant speed of 5 meters per second. Let t t represent the number of seconds that have elapsed since the start gun was fired. Write an expression in terms of t t that represents the hare's distance from the starting line (in meters).
Answer:
Distance traveled by the hare = d = 9 [tex]t_{t}[/tex]
Explanation:
Tortoise travels an additional distance 580 m .
Since there is no acceleration, the distance traveled by any object is d = v t , where v is the total distance and t is the total time taken.
Speed of hare = [tex]v_{h}[/tex] = 9 m/s
Speed of tortoise = [tex]v_{t}[/tex] =5 m/s
Distance traveled by the hare = d = 9 [tex]t_{t}[/tex]
Distance traveled by tortoise = d +580 m = 5 [tex]t_{t}[/tex]
Final answer:
The distance that the hare travels from the starting line after time t is represented by the expression 9t meters.
Explanation:
To write an expression that represents the hare's distance from the starting line in meters as a function of time, we use the formula for distance traveled at constant speed d = vt, where d is distance, v is speed, and t is time. Since the hare starts running at 9 meters per second, the distance the hare travels from the starting line after t seconds can be represented by the expression 9t meters. Note that we do not include the 580-meter head start given to the tortoise in the hare's expression since we are only considering the hare's distance from the starting line and not its position relative to the tortoise.
Water leaks out of a 3,200-gallon storage tank (initially full) at the rate V '(t) = 80 -t, where t is measured in hours and V in gallons. a. How much water leaked out between 10 and 20 hours? b. How long will it take the tank to drain completely?
Answer:
water leak is 650 gallons
time required to full drain is 80 hrs
Explanation:
given data
volume V = 3200 gallon
rate = V(t) = 80 - t
to find out
how much water leak between 10 and 20 hour and drain complete
solution
we know here rate is 80 - t
so here rate will be
[tex]\frac{dV(t)}{dt}[/tex] = 80 - t
and for time 10 and 20 hour
take integrate between 10 and 20
so water leak = [tex]\int\limits^ {20}_ {10} {(80-t)} \, dt[/tex] .....................1
water leak = ( 80t - [tex]\frac{t^{2} }{2} )^{20}_{10}[/tex]
water leak = 650
so water leak is 650 gallons
and
we know here for full tank drain condition
water leak full = 80 t - [tex]\frac{t^{2} }{2} [/tex]
3200 = 80 t - [tex]\frac{t^{2} }{2} [/tex]
6400 = t² - 160 t
t = 80
so time required to full drain is 80 hrs
Ordinary glasses are worn in front of the eye and usually 2.00 cm in front of the eyeball. A certain person can see distant objects well, but his near point is 50.0 cm from his eyes instead of the usual 25.0 cm . Suppose that this person needs ordinary glasses What focal length lenses are needed to correct his vision ?What is their power in diopters?
Answer:
Focal length: 44.16cm, Power: 2.2645 diopters
Explanation:
The object and image locations are the distance from the point to the lens.
object location = 25 - 2 = 23cm
image location = 50 - 2 = -48cm
the image location is negative since the image is virtual (the light rays do not pass through the image)
We can then use the lens equation to find the focal length.
1/object location + 1/image location = 1/focal length
(1/23)-(1/48)=1/f
Focal length = 44.16cm
To find the power we use the equation
D=1/f where f is the focal length in meters
44.16cm = 0.4416m
power = 2.2645 diopters
The prescription for glasses needed to correct the person's farsightedness should have lenses with a power of approximately 4.35 diopters.
Explanation:To correct the vision of a person with a near point of 50.0 cm, we must find a lens that will allow them to clearly see objects at the normal near point of 25.0 cm. First, we calculate the lens's focal length (f), which is the distance at which it would bring parallel rays to focus. The formula for lens power (P) in diopters (D) is given by P = 1/f (in meters). The needed focal length is the new near point (25.0 cm, or 0.25 m) minus the distance the glasses are from the eye (2.00 cm, or 0.02 m). Thus, the focal length is f = 0.25 m - 0.02 m = 0.23 m. After conversion to meters, calculating the power gives P = 1/0.23 m ≈ 4.35 D.
The prescription for the glasses should therefore have lenses with a power of approximately 4.35 diopters to correct the vision to a normal near point.
Given two charges, pick the answers that are true : The two charges act on one another no matter how far apart they are, all the way to infinity
If charge "A" is attracted to "B", then charge "B" is repelled from "A"
Plus and plus repel
If charge "A" is attracted to "B", then charge "B" is equally attracted to "A"
Minus and minus repel
Plus and minus attract
Answer:
All true except the second one.
Explanation:
The two charges act on one another no matter how far apart they are, all the way to infinity . True. Coulomb law doesn't have a distance limitation.
If charge "A" is attracted to "B", then charge "B" is repelled from "A" . False. This violates Newton 3rd Law (Action-Reaction)
Plus and plus repel . True . Charges of same sign repel and opposite sign atract.
If charge "A" is attracted to "B", then charge "B" is equally attracted to "A" . True. This is Newton 3rd Law (Action-Reaction)
Minus and minus repel . True . Charges of same sign repel and opposite sign atract.
Plus and minus attract . True . Charges of same sign repel and opposite sign atract.
A student places her 490 g physics book on a frictionless table. She pushes the book against a spring, compressing the spring by 7.10 cm , then releases the book. What is the book's speed as it slides away? The spring constant is 1550 N/m .
Answer:
book speed is 3.99 m/s
Explanation:
given data
mass m = 490 g = 0.490 kg
compressing x = 7.10 cm = 0.0710 m
spring constant k = 1550 N/m
to find out
book speed
solution
we know energy is conserve so
we can say
loss in spring energy is equal to gain in kinetic energy
so
[tex]\frac{1}{2}*k*x^2 = \frac{1}{2}*m*v^2[/tex] ..................1
put here value
[tex]\frac{1}{2}*1550*0.071^2 = \frac{1}{2}*0.490*v^2[/tex]
v = 3.99 m/s
so book speed is 3.99 m/s
Initially at rest a single-stage rocket is launched vertically from the ground. The rocket’s thrust overcomes gravity and provides the rocket a constant upward acceleration a. The fuel is exhausted 10 seconds after launch and then the motion of the rocket is free fall only due to gravity. If the total flight time is 30 s when the rocket strikes the ground, determine (a) the initial acceleration a, (b) the rocket’s impact speed as it hits the ground, (c) the height h from the ground the rockets reaches
Answer:
Explanation:
Let a be the acceleration of launch. In 10 seconds , Distance gone up in 10 seconds
s = 1/2at²
= .5 x a x 100
=50a
Velocity after 10 s
u = at = 10a
Now 50a distance in downward direction is travelled in 20 s with initial velocity 10a in upward direction.
s = ut + 1/2 gt²
50a = -10ax20 + .5 x 10x400
250a = 2000
a = 8 m s⁻² .
Velocity after 10 s
= at = 80 m/s
further height reached with this speed under free fall
h = v² / 2g
= 80 x 80 / 2 x 10
= 320 m
height achieved under acceleration
= 50a
= 50 x 8 = 400m
Total height
= 320 + 400= 720 m
velocity after falling from 720 m
v² = 2gh = 2 x10 x 720
v = 120 m/s
Answer:
a) The initial acceleration is 7.84 m/s²
b) The impact speed is 117.6 m/s
c) The height is 705.6 m
Explanation:
a) The speed from A to B is:
v = u + at
Where
u = initial speed = 0
t = 10 s
Replacing:
v = 10t (eq. 1)
The vertical distance between A to B is:
[tex]h=\frac{1}{2} at^{2} +ut=\frac{1}{2} a*(10)^{2}+0=50a[/tex] (eq. 2)
From B to C, the time it take is equal to 20 s, then:
[tex]h=vt+\frac{1}{2} at^{2}[/tex]
Replacing eq. 1 and 2:
[tex]-50a=(10a*20)-\frac{1}{2} *g*20^{2} \\-250a=-\frac{1}{2} *9.8*20^{2} \\a=7.84m/s^{2}[/tex]
b) The impact speed is equal:
[tex]v_{i} ^{2} =v^{2} +2gs[/tex]
Where
s = h = -50a
[tex]v^{2} _{i} =(10a)^{2} +2*(-9.8)*(-50a)\\v=\sqrt{100a^{2}+980a } \\v=\sqrt{(100*7.84^{2})+(980*7.84) } =117.6m/s[/tex]
c) The height is:
[tex]v_{i} ^{2} =v^{2} +2gs\\0=(10a)^{2} -2gs\\(10a)^{2} =2gs\\s=\frac{(10a)^{2} }{2g} \\s=\frac{(10*7.84)^{2} }{2*9.8} =313.6m[/tex]
htotal = 313.6 + 50a = 313.6 + (50*7.84) = 705.6 m
A rocket is fired at a speed of 75.0 m/s from ground level, at an angle of 60.0° above the horizontal. The rocket is fired toward an 11-0m-high wall, which is located 27.0 m away. The rocket attains its launch speed in a negligibly short period of time, after which its engines shut down and the rocket coasts. By how much does the rocket clear the top of the wall?
Answer:
ΔH = 33.17m
Explanation:
By knowing the amount of time it takes the rocket to travel the horizontal 27m, we will be able to calculate the height when x=27m. So:
[tex]X = V_{o}*cos(60)*t[/tex] where [tex]X=27m; V_{o}=75m/s[/tex]
Solving for t:
t=0.72s
Now we calculate the height of the rocket:
[tex]Y_{f}=V_{o}*sin(60)*t-\frac{g*t^{2}}{2} = 44.17m[/tex]
If the wall was 11m-high, the difference is:
ΔH = 33.17m
Using the equation of projectile motion, the rocket cleared the top of the wall by 38.305m.
The equation of projectile motion is given by:
y= xtanθ +(1/2)gx²/(u²cos²θ )
Given information:
Initial velocity, u=75m/s
The angle of projection, θ = 60°
The horizontal distance, x=27m
Vertical distance will be calculated as:
y= xtanθ +(1/2)gx²/(u²cos²θ )
y=27 tan 60° + 0.5×9.8×27²/(75²cos²60°)
y=49.305m
So, the rocket cleared the top of the wall by
49.305-11 =38.305m
Therefore, using the equation of projectile motion the rocket cleared the top of the wall by 38.305m.
To know more about projectile motion, click here:
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An object moves in one dimensional motion with constant acceleration a = 5 m/s^2. At time t = 0 s, the object is at x0 = 2.7 m and has an initial velocity of v0 = 4.3 m/s. How far will the object move before it achieves a velocity of v = 6.4 m/s?
Your answer should be accurate to the nearest 0.1 m.
The object will move approximately 10.1 m before it achieves a velocity of 6.4 m/s with a constant acceleration of 5 m/s^2.
Explanation:To find the distance the object will move before it achieves a velocity of 6.4 m/s, we can use the equation v^2 = v0^2 + 2ax, where v is the final velocity, v0 is the initial velocity, a is the acceleration, and x is the distance. Rearranging the equation, we get x = (v^2 - v0^2) / (2a). Plugging in the values, we have x = (6.4^2 - 4.3^2) / (2 * 5) = 10.12 m. Therefore, the object will move approximately 10.1 m before it achieves a velocity of 6.4 m/s.
Final answer:
To find how far an object with constant acceleration a = 5 [tex]m/s^2[/tex] and initial velocity of 4.3 m/s moves before reaching a velocity of 6.4 m/s, use the kinematic equation [tex]v^2[/tex] = [tex]v0^2[/tex] + 2a(x - x0). After calculation, the object will move 5.0 m before reaching 6.4 m/s.
Explanation:
We need to calculate how far the object moves before it achieves a velocity of v = 6.4 m/s given constant acceleration a = 5 [tex]m/s^2[/tex], initial velocity v0 = 4.3 m/s, and starting position x0 = 2.7 m. We can use the kinematic equation which relates final velocity, initial velocity, acceleration, and displacement:
[tex]v^2[/tex] = [tex]v0^2[/tex] + 2a(x - x0)
Rearranging for displacement (x - x0):
(x - x0) = ([tex]v^2[/tex] - [tex]v0^2[/tex]) / (2a)
Substituting the given values:
(x - 2.7 m) = [tex](6.4 m/s)^2[/tex] - [tex](4.3 m/s)^2[/tex] / [tex](2 * 5 m/s^2)[/tex]
(x - 2.7 m) = (40.96 [tex]m^2/s^2[/tex] - 18.49 [tex]m^2/s^2[/tex]) / (10 [tex]m/s^2[/tex])
(x - 2.7 m) = 22.47 [tex]m^2/s^2[/tex] / (10 [tex]m/s^2[/tex])
(x - 2.7 m) = 2.247 m
x = 2.7 m + 2.247 m
x = 4.947 m
To the nearest 0.1 m, the object will move 5.0 m before it achieves a velocity of 6.4 m/s.
Your 64-cm-diameter car tire is rotating at 3.3 rev/swhen suddenly you press down hard on the accelerator. After traveling 250 m, the tire's rotation has increased to 6.4 rev/s. What was the tire's angular acceleration? Give your answer in rad/s^2.
Answer:
0.76 rad/s^2
Explanation:
First, we convert the original and final velocity from rev/s to rad/s:
[tex]v_o = 3.3\frac{rev}{s} * \frac{2\pi rad}{1rev} =20.73 rad/s[/tex]
[tex]v_f = 6.4\frac{rev}{s} * \frac{2\pi rad}{1rev}=40.21 rad/s[/tex]
Now, we need to find the number of rads that the tire rotates in the 250m path. We use the arc length formula:
[tex]D = x*r \\x = \frac{D}{r} = \frac{250m}{0.64m/2} = 781.25 rads[/tex]
Now, we just use the formula:
[tex]w_f^2-w_o^2=2\alpha*x[/tex]
[tex]\alpha =\frac{w_f^2-w_o^2}{2x} = \frac{(40.21rad/s)^2-(20.73rad/s)^2}{2*781.25rad} = 0.76 rad/s^2[/tex]
Explanation:
The given data is as follows.
Initial velocity, u = [tex]3.3 rev/s \times 2 \pi rad/rev[/tex]
= 20.724 rad/sec
Final velocity, v = [tex]6.4 rev/s \times 2 \times pi rad/rev[/tex]
= 40.192 rad/sec
Now, we will calculate the value of angular rotation (d) as follows.
d = No. of revolutions × [tex]2 \pi rad/rev[/tex]
d = [tex](\frac{250 m}{2 \pi r}) \times 2 \pi[/tex]
= [tex]\frac{250}{0.32}[/tex]
= 781.25 rad
Also, we know that,
[tex]v^{2} = u^{2} + 2ad[/tex]
or, a = [tex]\frac{(v^{2} - u^{2})}{2d}[/tex]
= [tex]\frac{(40.192)^{2} - (20.724)^{2})}{2 \times 781.25}[/tex]
= [tex]\frac{1615.39 - 429.48}{1562.5}[/tex]
= 0.7589 [tex]rad/s^{2}[/tex]
Thus, we can conclude that the tire's angular acceleration is 0.7589 [tex]rad/s^{2}[/tex].
Describe the difference between the velocity graph made walking at a steady rate and the velocity graph made at an increasing rate.
Answer:
The difference between the velocity graph made walking at a steady rate means that its the same value in time, that means there's no slope on the graph, so its acceleration is 0
On the other hand, if the velocity is increasing with time, the slope of the graph becomes positive, which means that the acceleration of the particle is positive.
An object is thrown upward with an initial speed of 18.5 m/s from a location 12.2 m above the ground. After reaching its maximum height it falls to the ground. What will its speed be in the last instant prior to its striking the ground?
Answer:
The speed of the object in the last instant prior to hitting the ground is -24.1 m/s
Explanation:
The equation for the position and velocity of the object will be:
y = y0 + v0 · t + 1/2 · g · t²
v = v0 + g · t
Where:
y = height of the object at time t
v0 = initial velocity
y0 = initial height
g = acceleration due to gravity
t = time
v = velocity at time t
We know that at its maximum height, the velocity of the object is 0. We can obtain the time it takes the object to reach the maximum height and with that time we can calculate the maximum height:
v = v0 + g · t
0 = 18.5 m/s - 9.8 m/s² · t
-18.5 m/s / -9.8 m/s² = t
t = 1.89 s
Now,let´s find the max-height:
y = y0 + v0 · t + 1/2 · g · t²
y = 12.2 m + 18.5 m/s · 1.89 s + 1/2 ·(-9.8 m/s²) · (1.89 s)²
y = 29.7 m
Now, let´s see how much it takes the object to hit the ground:
In that instant, y = 0.
y = y0 + v0 · t + 1/2 · g · t²
0 = 29.7 m + 0 m/s · t - 1/2 · 9.8 m/s² · t² (notice that v0 = 0 because the object starts from its maximum height, where v = 0)
-29.7 m = -4.9 m/s² · t²
t² = -29.7 m / -4.9 m/s²
t = 2.46 s
Now, we can calculate the speed at t= 2.46 s, the instant prior to hitting the ground.
v = v0 + g · t
v = g · t
v = -9.8 m/s² · 2.46 s
v = -24.1 m/s
A square coil with 100 turns of wire has sides of length 5 cm. The resistance of the coil is 12.0 Ω. The coil is placed in a uniform magnetic field that is perpendicular to the face of the coil and whose magnitude is decreasing at a rate of 0.081 T/s. What is the magnitude of the current circulating through the coil?
Answer:
The magnitude of the current circulating through the coil is 1.68 mA.
Explanation:
Given that,
Number of turns = 100
Length = 5 cm
Resistance = 12.0 Ω
Rate of magnetic field = 0.081 T/s
We need to calculate the magnitude of the current circulating through the coil
Using formula of current
[tex]I=\dfrac{e}{R}[/tex]
[tex]I=\dfrac{-NA\dfrac{dB}{dt}}{R}[/tex]
Where, A = area
N = number of turns
R = resistance
Put the value into the formula
[tex]I=\dfrac{100\times(0.05)^2\times0.081}{12.0}[/tex]
[tex]I=0.0016875\ A[/tex]
[tex]I=1.68\times10^{-3}\ A[/tex]
[tex]I=1.68\ mA[/tex]
Hence, The magnitude of the current circulating through the coil is 1.68 mA.
The question involves Faraday's law, which relates changing magnetic fields to induced currents. Given the rate of change of the magnetic field and the resistance of the coil, the magnitude of the induced current is calculated to be 1.67 mA.
Explanation:This question involves the concept of electromagnetic induction in physics. According to Faraday's law, a changing magnetic field will induce an electromotive force (EMF) in a circuit. In this case, we're given that the magnetic field is decreasing at a rate of 0.081 T/s. This rate of change can be used to calculate the induced EMF, which is equal to the rate of change of the magnetic flux.
The magnetic flux (Φ) through the coil is given by Φ = NBA, where N is the number of turns in the coil, B is the magnetic field, and A is the area of the coil. So, the rate of change of magnetic flux (dΦ/dt) is N*d(BA)/dt. Because the area of the coil is constant, this simplifies to NAdB/dt. A is 0.0025 m^2, N is 100, and dB/dt is -0.081T/s, so dΦ/dt is -0.02 Wb/s.
Faraday's law tells us that the induced EMF (ε) is equal to -dΦ/dt, so ε has magnitude 0.02 V. If we define positive current as current that would generate a magnetic field opposing the decrease in the existing field (Lenz's Law), the induced EMF 'pushes' positive current in this direction. The current (I) can then be calculated using Ohm's Law (V = IR), giving I = ε/R = 0.02V / 12 Ω = 0.00167 A, or 1.67 mA.
Learn more about Electromagnetic Induction here:https://brainly.com/question/31384468
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Two blocks of masses 20 kg and 8.0 kg are connected togetherby
a light string and rest on a frictionless level surface.Attached to
the 8- kg mass is a second light string, which a personuses to pull
both blocks horiaontally. If the two- block systemaccelerates at
0.5 m/s ^2, what is the tension in the second stringattached to the
8-kg mass?
Answer:
The tension is 14 N
Explanation:
For this problem we have to use newton's law, so:
[tex]F=m*a[/tex]
The second string is connected to the mass of 8 kg, but the mass of 8 kg is connected to the mass of 20kg, so we can say that the second string is handling the two masses. so:
[tex]F=28kg*0.5\frac{m}{s^2}=14N[/tex]
The Baltimore Harbor water taxi is approaching the dock with a velocity of v0 = 5 m/s. The water taxi acceleration is limited to −1 m/s < a < 1 m/s, how far from the dock must the ferry begin slowing down if it is to avoid a collision?
Answer:
In order to avoid collision, the ferry must stop at a distance of 12.5 m from the dock.
Solution:
The initial velocity of the taxi, [tex]v_{o} = 5 m/s[/tex]
The minimum value of acceleration , [tex]a_{min} = - 1 m/s^{2}[/tex]
The maximum value of acceleration , [tex]a_{max} = 1 m/s^{2}[/tex]
Now,
When the deceleration starts the ferry slows down and at minimum deceleration of [tex]- 1 m/s^{2}[/tex], the ferry stops.
Thus, inthis case, the final velocity, v' is 0.
Now, to calculate the distance covered, 'd' in decelerated motion is given by the third eqn of motion:
[tex]v'^{2} = v_{o}^{2} + 2ad[/tex]
[tex]0^{2} = 5^{2} + 2\times (- 1)d[/tex]
[tex]d = 12.5 m[/tex]
One mol of a perfect, monatomic gas expands reversibly and isothermally at 300 K from a pressure of 10 atm to a pressure of 2 atm. Determine the value of q, w.
(b) Now assume the gas expands by the same amount again isothermally but now irre-
versibly against 1 atm pressure (instead of reversible expansion) and calculate again q, w, delta U and delta H.
Answer:
Explanation:
Given
1 mole of perfect, monoatomic gas
initial Temperature[tex](T_i)=300 K[/tex]
[tex]P_i=10 atm[/tex]
[tex]P_f=2 atm[/tex]
Work done in iso-thermal process[tex]=P_iV_iln\frac{P_i}{P_f}[/tex]
[tex]P_i[/tex]=initial pressure
[tex]P_f[/tex]=Final Pressure
[tex]W=10\times 2.463\times ln\frac{10}{2}=39.64 J [/tex]
Since it is a iso-thermal process therefore q=w
Therefore q=39.64 J
(b)if the gas expands by the same amount again isotherm-ally and irreversibly
work done is[tex]=P\Delta V[/tex]
[tex]V_1=\frac{RT_1}{P_1}=\frac{1\times 0.0821\times 300}{10}=2.463 L[/tex]
[tex]V_2=\frac{RT_2}{P_2}=\frac{1\times 0.0821\times 300}{2}=12.315 L[/tex]
[tex]\Delta W=1\times (12.315-2.463)=9.852 J[/tex]
[tex]\Delta q=\Delta W=9.852 J[/tex]
[tex]\Delta U=0[/tex]
A car that weighs 1.5 × 10^4 N is initially moving at a speed of 36 km/h when the brakes are applied and the car is brought to a stop in 17 m. Assuming that the force that stops the car is constant, find (a) the magnitude of that force and (b) the time required for the change in speed. If the initial speed is doubled, and the car experiences the same force during the braking, by what factors are (c) the stopping distance and (d) the stopping time multiplied? (There could be a lesson here about the danger of driving at high speeds.)
Answer:
Explanation:
Initial velocity u = 36 km/h = 10 m /s
v = 0 , accn = a , Time taken to stop = t , distance covered to stop = s
v² = u² - 2as
u² = 2as
a = u² / 2s
= 10 x 10 / 2 x 17
= 2.94 ms⁻²
Force applied = mass x acceleration
= 15000 / 9.8 x 2.94
= 4500 N
b )
v = u + at
0 = 10 - 2.94 t
t = 10 / 2.94
= 3.4 s
c )
from the relation
u² = 2as
it is clear that stopping distance is proportional to u², if acceleration a is constant .
If initial speed u is doubled , stopping distance d will become 4 times or 17 x 4 = 68 m .
d )
u = at
if a is constant time taken to stop will be proportional to initial velocity.
If initial velocity is doubled , time too will be doubled. Or time will become
3.4 x 2 = 6.8 s .
The radius of Mercury (from the centerto just above the atmosphere)
is 2440km (2440103
m),and its mass is 0.31024
kg.An object is launched straight up from just above the atmosphere
ofMercury.
(a) What initial speed is needed so that when the object is farfrom
Mercury its final speed is2000 m/s?
Answer:
u = 12962.11 m/s
Explanation:
Given that,
The radius of mercury, [tex]r=2440\ km=2440\times 10^3\ m[/tex]
Mass of Mercury, [tex]M=3\times 10^{24}\ kg[/tex]
Final speed of the object, v = 2000 m/s
Let u is its initial speed when the object is far from Mercury. It can be calculated by applying the conservation of energy as :
Initial kinetic energy + gravitational potential energy = final kinetic energy
[tex]\dfrac{1}{2}mu^2+(-\dfrac{GmM}{r})=\dfrac{1}{2}mv^2[/tex]
[tex]\dfrac{1}{2}u^2+(-\dfrac{GM}{r})=\dfrac{1}{2}v^2[/tex]
[tex]\dfrac{1}{2}u^2=\dfrac{1}{2}v^2+\dfrac{GM}{r}[/tex]
[tex]u^2=2\times (\dfrac{1}{2}v^2+\dfrac{GM}{r})[/tex]
[tex]u^2=2\times (\dfrac{1}{2}(2000)^2+\dfrac{6.67\times 10^{-11}\times 3\times 10^{24}}{2440\times 10^3})[/tex]
u = 12962.11 m/s
So, the initial speed of the object is 12962.11 kg. Hence, this is the required solution.
A 9.0 μF capacitor, a 13 μF capacitor, and a 16 μFcapacitor are connected in series. What is their equivalent capacitance? Express your answer using two significant figures.
Answer:
Equivalent capacitance, [tex]C'=3.9\ \mu F[/tex]
Explanation:
Capacitance, [tex]C_1=9\ \mu F[/tex]
Capacitance, [tex]C_2=13\ \mu F[/tex]
Capacitance, [tex]C_3=16\ \mu F[/tex]
Let C' is the equivalent capacitance of the combination of capacitors. It is given by :
[tex]\dfrac{1}{C'}=\dfrac{1}{C_1}+\dfrac{1}{C_2}+\dfrac{1}{C_3}[/tex]
[tex]\dfrac{1}{C'}=\dfrac{1}{9}+\dfrac{1}{13}+\dfrac{1}{16}[/tex]
[tex]C'=3.99\ \mu F[/tex]
or
[tex]C'=3.9\ \mu F[/tex]
So, their equivalent capacitance is [tex]3.9\ \mu F[/tex]. Hence, this is the required solution.
Final answer:
The equivalent capacitance of a 9.0 μF, 13 μF, and a 16 μF capacitors connected in series is approximately 4.0 μF when expressed with two significant figures.
Explanation:
The question concerns the calculation of the equivalent capacitance when multiple capacitors are connected in series. To find the equivalent capacitance of the 9.0 μF, 13 μF, and 16 μF capacitors connected in series, you use the formula for capacitors in series:
1/Ceq = 1/C1 + 1/C2 + 1/C3
Substituting the given values:
1/Ceq = 1/9.0 μF + 1/13 μF + 1/16 μF
Calculating the reciprocals and summing them gives:
1/Ceq = 0.111 + 0.077 + 0.063
1/Ceq = 0.251
Now, the equivalent series capacitance, Ceq, can be calculated by taking the reciprocal of the result:
Ceq = 1 / 0.251
Ceq is approximately 3.98 μF.
To express this answer using two significant figures, the equivalent capacitance is 4.0 μF.
An airplane is heading due south at a speed of 430 km/h . A wind begins blowing from the northwest at a speed of 85.0 km/h (average).What should the plane's course shift be so that it will fly due south?
Answer:
It should fly 8° to west of south at 430km/h
Explanation:
According to the diagram. X components for both velocities must have the same magnitude in order to get the resultant velocity due south.
[tex]V_{w}*cos(45) = V_{A}*sin(\alpha )[/tex] Solving for α:
α = 8.03°
Plot the velocity vs. time and the position vs. time for a car that travels at 20 m/s for 20 seconds, then accelerates in 10 seconds to 30 m/s, travels at this speed for 20 seconds, and the brakes and comes to rest in 10 s.
Answer:
The position-time graph and the velocity-time graph has been shown in the figure attached.
Explanation:
Given:
For t = 0 s to t = 20 s,
The car moves with a constant velocity whose position-time and the velocity time graph has been shown with a red line.
For t = 20 s to t = 30 s,
The car moves with a constant acceleration whose position-time and the velocity time graph has been shown with a green line.
For t = 30 s to t = 50 s,
The car moves with a constant velocity whose position-time and the velocity time graph has been shown with a blue line.
For t = 50 s to t = 60 s,
The car moves with a constant deceleration whose position-time and the velocity time graph has been shown with a black line.
For a constant velocity, the velocity-time graph of the particle is a straight line parallel to a time axis and the position-time graph is a straight line inclined at some positive angle with the time axis.
For a constant acceleration, the velocity-time graph is a straight line incline at a positive angle with the time axis and the position-time graph is a parabolic curved line having upward concavity.
For a constant deceleration, the velocity-time graph is a straight line incline at a negative angle with the time axis and the position-time graph is a parabolic curved line having downward concavity.
Which of the following has the greatest kinetic energy? A. a mass of 4m at velocity v. B. a mass of 3m at velocity 2v. C. a mass of 2m at velocity 3v. D. a mass of m with a velocity of 4v.
Answer:
Option C
Explanation:
Kinetic energy is the energy that the body possesses by virtue of its motion.
The formula for Kinetic energy is given by [tex]\frac{1}{2} mv^2[/tex]
Using this formula let us find kinetic energy for the bodies given and find out which is the greatest
A) KE = [tex]\frac{1}{2} (4m)(v^2) = 2mv^2[/tex]
B) KE =[tex]\frac{1}{2} (3m)(2v)^2 = 6mv^2[/tex]
C) KE = [tex]\frac{1}{2} (2m)(3v)^2 = 9mv^2[/tex]
D) KE = [tex]\frac{1}{2} (3)(4v)^2 = 8mv^2[/tex]
Comparing these we find that 9mv^2 is the highest.
Hence option C is the answer.
To which types of thermodynamic processes will the relation Tds =đQ apply? What is the correct term for the symbol T in this expression, and what exactly does dQ represent? (5) (b) Calculate the change in entropy when tkg of ice is melted at 273.16K Calculate the change in entropy when lkg of water is heated from 273 16K to 300K (15) (c) Why is the change in entropy much larger for one of the calculations in part (b) of this question
Answer:
a) Reversible processes
b) 1222.73 J/K and 392.144 J/K
c) Because of breaking bonds that are a stable state, while accelerating molecules in the other is less difficult.
Explanation:
Since it is equality, it should be applied to reversible processes, and even better to isothermal processes, since the Temperature remains constant. The T is the absolute temperature (measured in °K) and dQ is the heat absorbed by the system, which DEPENDS on the process.
b) The heat absorbed in the fusion process depends on the latent heat of fusion, L, of the water, which is 334 J/g. It says t kg, but I assume it was a mistake in the typing, so the change in entropy is calculated for 1 kg of water melting as follows:
[tex]\Delta S=\frac{\Delta Q}{T}=\frac{mass*L}{T}=\frac{1kg*334J/g*(1000 g/kg)}{273.16 K} \\\Delta S= 1222.73 J/K[/tex]
Now, to proceed with the change of entropy for water heated from 273.16K to 300K we use the specific heat of water, which is 4184 J/kg°K as follows:
[tex]\Delta S=m_{mass}c_{water}*ln(\frac{T_f}{T_i} )=1kg*4184\frac{J}{Kg*K}*ln(\frac{300}{273.16})\\ \Delta S=392.144 J/K[/tex]
c) In the solid state, water molecules have different bonds with other water molecules creating the crystals. In the liquid state, each molecule moves freely with less interaction between molecules. So, it required more energy to break these bonds and alter this ordered state than just accelerating the molecules in the liquid state.
If the fundamental frequency of a piece of conduit is 707 Hz, and the speed of sound is 343 m/s, determine the length of the piece of conduit (in m) for each of the following cases. (a) the piece of conduit is closed at one end (b) the piece of conduit is open at both ends
Answer:
i)0.1213m
ii)0.2426 m
Explanation:
The formula to apply here are;
For a closed conduit at one end the fundamental frequency produced by a pipe of length L is ;
f=v/4L where v is speed of sound and L is length of conduit
707=343/4L
707*4L=343
2828L=343
L=343/2828
L=0.1213m
For an open conduit, the fundamental frequency produced by pipe of length L is given by;
f=v/2L where v is speed of sound and L is length of conduit
707=343/2L
707*2L=343
1414L=343
L=343/1414
L=0.2426 m
If a 10kg block is at rest on a table and a 1200N force is applied in the eastward direction for 10 seconds, what is the acceleration on the block?
Answer:
a = 120 m/s²
Explanation:
We apply Newton's second law in the x direction:
∑Fₓ = m*a Formula (1)
Known data
Where:
∑Fₓ: Algebraic sum of forces in the x direction
F: Force in Newtons (N)
m: mass (kg)
a: acceleration of the block (m/s²)
F = 1200N
m = 10 kg
Problem development
We replace the known data in formula (1)
1200 = 10*a
a = 1200/10
a = 120 m/s²