Answer:
9.42 m/s
Explanation:
a) Using Newton's law of motion formula:
[tex]\theta=\frac{(\omega+\omega_o)}{2}t\\\\where \ \theta=angular\ displacement=1\ rev =2\pi, w_o=initial\ velocity\ of\ discus\\=0\ rad/s, \omega=angular\ speed\ of\ discus\ at\ release,t=time\ = 1.2\ s.\\\\Hence:\\\\2\pi=\frac{(0+\omega)}{2}(1.2)\\\\\omega=\frac{2*2\pi}{1.2} \\\\\omega=10.47\ rad/s\\[/tex]
The speed of the discus at release (v) is:
v = ωr; where r = radius of discus
diameter = 1.8 m, r = diameter / 2= 1.6 / 2 = 0.9 m
v = ωr = 10.47 * 0.9
v = 9.42 m/s
A 7600 kg rocket blasts off vertically from the launch pad with a constant upward acceleration of 2.35 m/s2 and feels no appreciable air resistance. When it has reached a height of 595 m, its engines suddenly fail so that the only force acting on it is now gravity.
A) What is the maximum height this rocket will reach above the launch pad?
B) How much time after engine failure will elapse before the rocket comes crashing down to the launch pad?
Answer:
a) The rocket reaches a maximum height of 737.577 meters.
b) The rocket will come crashing down approximately 17.655 seconds after engine failure.
Explanation:
a) Let suppose that rocket accelerates uniformly in the two stages. First, rocket is accelerates due to engine and second, it is decelerated by gravity.
1st Stage - Engine
Given that initial velocity, acceleration and travelled distance are known, we determine final velocity ([tex]v[/tex]), measured in meters per second, by using this kinematic equation:
[tex]v = \sqrt{v_{o}^{2} +2\cdot a\cdot \Delta s}[/tex] (1)
Where:
[tex]a[/tex] - Acceleration, measured in meters per square second.
[tex]\Delta s[/tex] - Travelled distance, measured in meters.
[tex]v_{o}[/tex] - Initial velocity, measured in meters per second.
If we know that [tex]v_{o} = 0\,\frac{m}{s}[/tex], [tex]a = 2.35\,\frac{m}{s^{2}}[/tex] and [tex]\Delta s = 595\,m[/tex], the final velocity of the rocket is:
[tex]v = \sqrt{\left(0\,\frac{m}{s} \right)^{2}+2\cdot \left(2.35\,\frac{m}{s^{2}} \right)\cdot (595\,m)}[/tex]
[tex]v\approx 52.882\,\frac{m}{s}[/tex]
The time associated with this launch ([tex]t[/tex]), measured in seconds, is:
[tex]t = \frac{v-v_{o}}{a}[/tex]
[tex]t = \frac{52.882\,\frac{m}{s}-0\,\frac{m}{s}}{2.35\,\frac{m}{s} }[/tex]
[tex]t = 22.503\,s[/tex]
2nd Stage - Gravity
The rocket reaches its maximum height when final velocity is zero:
[tex]v^{2} = v_{o}^{2} + 2\cdot a\cdot (s-s_{o})[/tex] (2)
Where:
[tex]v_{o}[/tex] - Initial speed, measured in meters per second.
[tex]v[/tex] - Final speed, measured in meters per second.
[tex]a[/tex] - Gravitational acceleration, measured in meters per square second.
[tex]s_{o}[/tex] - Initial height, measured in meters.
[tex]s[/tex] - Final height, measured in meters.
If we know that [tex]v_{o} = 52.882\,\frac{m}{s}[/tex], [tex]v = 0\,\frac{m}{s}[/tex], [tex]a = -9.807\,\frac{m}{s^{2}}[/tex] and [tex]s_{o} = 595\,m[/tex], then the maximum height reached by the rocket is:
[tex]v^{2} -v_{o}^{2} = 2\cdot a\cdot (s-s_{o})[/tex]
[tex]s-s_{o} = \frac{v^{2}-v_{o}^{2}}{2\cdot a}[/tex]
[tex]s = s_{o} + \frac{v^{2}-v_{o}^{2}}{2\cdot a}[/tex]
[tex]s = 595\,m + \frac{\left(0\,\frac{m}{s} \right)^{2}-\left(52.882\,\frac{m}{s} \right)^{2}}{2\cdot \left(-9.807\,\frac{m}{s^{2}} \right)}[/tex]
[tex]s = 737.577\,m[/tex]
The rocket reaches a maximum height of 737.577 meters.
b) The time needed for the rocket to crash down to the launch pad is determined by the following kinematic equation:
[tex]s = s_{o} + v_{o}\cdot t +\frac{1}{2}\cdot a \cdot t^{2}[/tex] (2)
Where:
[tex]s_{o}[/tex] - Initial height, measured in meters.
[tex]s[/tex] - Final height, measured in meters.
[tex]v_{o}[/tex] - Initial speed, measured in meters per second.
[tex]a[/tex] - Gravitational acceleration, measured in meters per square second.
[tex]t[/tex] - Time, measured in seconds.
If we know that [tex]s_{o} = 595\,m[/tex], [tex]v_{o} = 52.882\,\frac{m}{s}[/tex], [tex]s = 0\,m[/tex] and [tex]a = -9.807\,\frac{m}{s^{2}}[/tex], then the time needed by the rocket is:
[tex]0\,m = 595\,m + \left(52.882\,\frac{m}{s} \right)\cdot t + \frac{1}{2}\cdot \left(-9.807\,\frac{m}{s^{2}} \right)\cdot t^{2}[/tex]
[tex]-4.904\cdot t^{2}+52.882\cdot t +595 = 0[/tex]
Then, we solve this polynomial by Quadratic Formula:
[tex]t_{1}\approx 17.655\,s[/tex], [tex]t_{2} \approx -6.872\,s[/tex]
Only the first root is solution that is physically reasonable. Hence, the rocket will come crashing down approximately 17.655 seconds after engine failure.
(a) What type of charge distribution is inside the surface? a positively charged plane parallel to the end faces of the cylinder a positive line charge situated on and parallel to the axis of the cylinder a collection of positive point charges arranged in a line at the center of the cylinder and perpendicular to the axis of the cylinder a collection of negative point charges arranged in a line at the center of the cylinder and perpendicular to the axis of the cylinder a negatively charged plane parallel to the end faces of the cylinder (b) If the radius of the cylinder is 0.66 m and the magnitude of the electric field is 300 N/C, what is the net electric flux through the closed surface? N · m2/C (c) What is the net charge inside the cylinder? nC
Answer:
a) The type of charge distribution inside the surface is a negatively charged plane parallel to the end faces of the cylinder.
b) the net electric flux is -821.09 N.m²/c
c) the net charge inside the cylinder is -7.2666 nC
Explanation:
Given the data in the question;
a) (a) What type of charge distribution is inside the surface?
Based on the Image of the question below;
The type of charge distribution inside the surface is a negatively charged plane parallel to the end faces of the cylinder.
b) If the radius of the cylinder is 0.66 m and the magnitude of the electric field is 300 N/C, what is the net electric flux through the closed surface?
we know that; Electric flux is;
∅ = -2 × E × Area
we substitute
∅ = -2 × 300 × π ( 0.66 m )²
= -821.09 N.m²/c
Therefore; the net electric flux is -821.09 N.m²/c
c) What is the net charge inside the cylinder? nC
from Gause's law;
∅ = q/∈₀
q = ∅ ∈₀
we know permittivity ∈₀ = 8.85 × 10⁻¹²)
we substitute
q = -821.09 N.m²/c × (8.85 × 10⁻¹²)
q = -7.2666 × 10⁻⁹ C
q = -7.2666 nC
Therefore, the net charge inside the cylinder is -7.2666 nC
How shadows are formed?
Answer:
Shadows are formed when an opaque object or material is placed in the path of rays of light. The opaque material does not let the light pass through it. The light rays that go past the edges of the material make an outline for the shadow.
Explanation:
Hope it is helpful.....
A fairgrounds ride spins its occupants inside a flying saucer-shaped container. If the horizontal circular path the riders follow has a 9.00 m radius, at how many revolutions per minute will the riders be subjected to a centripetal acceleration whose magnitude is 1.80 times that due to gravity
Answer:
13.37 rev/min
Explanation:
acceleration due to gravity (g) = 9.8 m/s², centripetal acceleration ([tex]a_c[/tex]) = 1.8 * g = 1.8 * 9.8 m/s² = 17.64 m/s².
r = 9 m
Centripetal acceleration ([tex]a_c[/tex]) is given by:
[tex]a_c=\frac{v^2}{r} \\\\v=\sqrt{a_c*r} \\\\v=\sqrt{17.64\ m/s^2*9\ m}\\\\v=12.6\ m/s[/tex]
The velocity (v) is given by:
v = ωr; where ω is the angular velocity
Hence:
ω = v/r = 12.6 / 9
ω = 1.4 rad/s
ω = 2πN
N = ω/2π = 1.4 / 2π
N = 0.2228 rev/s
N = 13.37 rev/min
A motorcyclist is traveling north on a road at a speed of 23.0 m/s. A police car, traveling south at a speed of 40.0 m/s, approaches with its siren producing sound at a frequency of 2500 Hz. (Enter your answers in Hz. Assume that the speed of sound in air is 343 m/s.) (a) What frequency does the motorcyclist observe as the police car approaches
Answer:
f ’= 3019.8 Hz
Explanation:
This is an exercise of the Doppler effect, that is, of the relative movement of the source (s) and the observer (o)
[tex]f' = f_o \frac{v +v_o}{v -v_s}[/tex]
in the case of the source and the observer approaching, if they move away the signs are exchanged
let's calculate
f ’= 2500 [tex]\frac{343 + 23}{343 - 40}[/tex]
f ’= 2500 [tex]\frac{366}{303}[/tex]
f ’= 3019.8 Hz
What impulse must be applied to a 25.0-kg cart to cause a velocity change
of 12.0 m/s?
Answer:
Impulse of force = 300Ns
Explanation:
Given the following data;
Mass = 25kg
Change in velocity = 12m/s
To find the impulse;
Impulse is given by the formula;
[tex] Impulse \; of \; force = mass * change \; in \; velocity [/tex]
Substituting into the equation, we have;
[tex] Impulse \; of \; force = 25 * 12 [/tex]
Impulse of force = 300Ns
The vertical component of the acceleration of a sailplane is zero when the air pushes up against its wings with a lift force of 5.20 kN. Assume that the only forces on the sailplane are that due to gravity and that due to the air pushing up against its wings. What will be the gravitational force on Earth due to the sailplane
Answer:
Explanation:
The vertical component of the acceleration of a sailplane is zero , that means the sailplane is experiencing net force of zero in vertical direction . Its weight is acting in downward direction . So airplane is also experiencing an upward force equal to its weight which is making net force equal to zero on it . This force is given as 5.20 k N .
So sailplane is experiencing an upward force equal to its weight . This force is generated due to air pushing up against its wings .
We know that every force generated has equal and opposite reaction force . Air is generating force on wings of sailplane , hence wings will also exert equal force on air on downward direction . This force will be transmitted to the earth by air .
Hence the gravitational force on Earth due to the sailplane will be equal to weight of sailplane . This force is 5.2 kN .
A 0.14 kg baseball moving at 23.00 m s is slowed to a stop by a catcher who exerts a constant force of 370 N How long does it take this force to stop the ball Incorrect Your answer is incorrect s How far does the ball travel before stopping
Answer:
(A) The time taken for the ball to stop is 8.7 x 10⁻³ s
(B) The distance traveled by the baseball before stopping is 0.3 m
Explanation:
Given;
mass of the baseball, m = 0.14 kg
velocity of the baseball, v = 23 m/s
force exerted on the baseball by the catcher, F = 370 N
(A) The time taken for the ball to stop;
[tex]F = ma = \frac{mv}{t} \\\\t = \frac{mv}{F} \\\\t = \frac{0.14 \times 23}{370} \\\\t = 8.7\times 10^{-3} \ s\\\\t = 8.7 \ ms[/tex]
(B) The distance traveled by the baseball before stopping is calculated as;
acceleration of the ball, [tex]a = \frac{v}{t} = \frac{23}{8.7\times 10^{-3}} = 2643.678 \ m/s^2[/tex]
Distance traveled, s;
s = ut + ¹/₂at²
s = (23)(8.7 x 10⁻³) + ¹/₂(2643.678)(8.7 x 10⁻³)²
s = 0.2001 + 0.1001
s = 0.3 m
In the Daytona 500 auto race, a Ford Thunderbird and a Mercedes Benz are moving side by side down a straightaway at 73.5 m/s. The driver of the Thunderbird realizes that she must make a pit stop, and she smoothly slows to a stop over a distance of 250 m. She spends 5.00 s in the pit and then accelerates out, reaching her previous speed of 73.5 m/s after a distance of 400 m. At this point how far has the Thunderbird fallen behind the Mercedes Benz, which has continued at a constant speed
Answer:
Thunderbird has fallen behind the Mercedes Benz by 1017.49 m
Explanation:
Given the data in question;
initial speed of the ford u1 = 73.5 m/s
distance d1 = 250 m
t1 = 5.00 s
d2 = 400 m
Now, let the time taken to stop be t2 and deceleration is a1
so,
a1 = u1² / (2 × d1)
a1 = (73.5)² / (2 × 250)
a1 = 10.8045 m/s²
Now , for acceleration is a2
a2 = v² / (2 × d2)
a2 = (73.5)² / (2 × 400)
a2 = 6.7528 m/s²
total time spend = 5 + u/a1 + u/a2
total time spend = 5 + (73.5/10.8045) + (73.5/6.7528)
total time spend = 22.687 sec
Now, distance Mercedes is ahead = 22.687 × 73.5 - 400 - 250
= 1667.4945 - 400 - 250
= 1017.49 m
Therefore, Thunderbird has fallen behind the Mercedes Benz by 1017.49 m
Two students, each riding bicycles, start from the same apartment building and ride to the same building on campus, but each takes a different route. The first student rides 1300 m due east and then turns due north and travels another 1430 m before arriving at the destination. The second student heads due north for 1930 m and then turns and heads directly toward the destination.
(a) At the turning point, how far is the second student from the destination? ....m
(b) During the last leg of the trip, what direction (measured relative to due east) must the second student head? (Give your answer as a positive number from 0 to 180 degrees, either north or south of due east.) .... degrees south of east
Answer:
a) d= 1393 m
b) θ= 21º S of E.
Explanation:
a)
Since the second student goes due north, and the first student goes due east along 1300m till he turns directly northward, we conclude that when the second student turns, he is at 1300m west from the destination.Since he rode 1930 m due north, we can conclude also that the second student is 500 m past the destination in the north direction.So we can find the distance from the destination at the turning point, using the Pythagorean Theorem, taking the right triangle defined by the 1300 m segment due east, the 500 m segment due south, and which hypotenuse is the distance straight to the destination, as follows:[tex]d = \sqrt{(1300m)^{2} + (500m)^{2} } = 1393 m (1)[/tex]
b)
Taking the same right triangle than in (a), we can find the angle that makes the vector along the direction taken by the second student with the due east, applying the definition of tangent of an angle, as follows:[tex]tg \theta = \frac{500m}{1300m} = 0.385 (2)[/tex]
⇒ θ= tg⁻¹ (0.385) = 21º S of E.
Which two actions would weaken an electromagnet?
O A. Replacing the steel core of the magnet with a plastic core
O B. Decreasing the number of wraps of wire
O c. Increasing the number of wraps of wire
OD. Replacing a 6 V battery with a 9 V battery
O E. Replacing the hollow core of the magnet with a steel core
The correct options are (a &b) .
The strength of the electromagnet will vary depending on the core material utilized if cores of various materials with the same physical dimensions are employed. The number of flux lines traveling through the central core is what causes this difference in magnetic intensity.a. Replacing the steel core of the magnet with a plastic coreNo, Steel is used in permanent magnets because it takes a long time to get magnetized and remains magnetized after the current is turned off.In contrast, soft iron almost instantly acquires magnetism when the current is turned on, and it quickly loses all of its magnetism when the current is turned off.B. Decreasing the number of wraps of wireAs the number of turns increases the number of paper clips held increase. This means that the strength of the magnet increases with increasing number of turns in the coil.Learn more about an electromagnet brainly.com/question/23727978
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Answer: A & B
Explanation: just took it !