Answer:
1
Explanation:
Projectile motion challenge problem. You are at war. You are stationed with a cannon that can only fire shells at 100 m/s. You may adjust the angle but it takes five seconds to do so. You are positioned on a strategically important bridge 70m high with a mission to protect it from the enemy. Spotters have alerted you to a remote controlled speedboat 3000 m away heading West to your position. It is carrying a bomb to blow up the bridge. The spotter tells you that its initial velocity is 26 m/s and its accelerating at 11 m/s2. You must fire the cannon and destroy the boat as fast as possible before it reaches the bridge. When the speedboat is 3000 m away set that as time = 0s. A. What time do you fire and what is the angle when you do fire? B. Right before you fire a 10m/s wind starts blowing from behind you towards the East. How do you adjust your fire?
Answer:
A. the time of fire is 15 seconds, and the angle of fire is approximately 23.794° above the horizontal
B. The angle of fire is increased to approximately 35.4126° above the horizontal
Explanation:
A. The height of the bridge, h = 70 m
The speed with of the shell, v₀ = 100 m/s
The location of the speedboat = 3000 m
The direction of the speedboat = 26 m/s
The acceleration of the speedboat, a = 11 m/s²
Let t represent the time of firing the shells, and let x represent the distance of the speedboat from the bridge, and let θ, represent the angle to fire with, we have;
For the speedboat, t = x/(100 × cos(θ))
We note that the time the shell can travel the 3,000 m = 30 seconds
Therefore an adequate time to fire is, t = 15 seconds
The distance the speedboat covers in 15 seconds is given as follows;
s = u·t + 1/2·a·t²
s = 26 × 15 + 1/2 × 11 × 15² = 1627.5 m
At the point the speedboat had traveled 1,627.5 m, the distance of the speedboat, x is then 3000 - 1,627.5 = 1,372.5 m from the bridge, the angle of fire is given from the following formula;
t = x/(100 × cos(θ))
15 = 1,372.5/(100 × cos(θ))
cos(θ) = 1,372.5/(15 × 100) = 0.915
θ = cos⁻¹(0.915) ≈ 23.794°
The angle of fire, θ ≈ 23.794° above the horizontal in the direction of the speedboat
B. Given that a 10 m/s wind is blowing towards East, we have;
The horizontal velocity towards East = 10 + v₀ × cos(θ)
The angle of firing is therefore, given as follows;
15 = 1,3725.5/(10 + 100 × cos(θ))
(10 + 100 × cos(θ)) = 1,3725.5/15
100 × cos(θ) = 1,372.5/15 - 10
cos(θ) = (1,372.5/15 - 10)/100 = 0.815
θ = cos⁻¹(0.815) ≈ 35.4126°
Therefore, the angle of fire, θ, will be increased to approximately 35.4126° above the horizontal in the remote controlled speedboat direction.
please help asap!!!!
Answer:
B
Explanation:
PPL how many types of turtles are there
Explanation:
356 species of diff turtles
Heat transfer occurs only in 1 case from body II to body I, 2. Heat transfer occurs only in 2 cases from body I to body II, 3. Heat transfer occurs only in 3 cases from body II to body I, 4. Heat transfer occurs in all cases from body II to body I.
Answer:
whats the question???
You are walking in Paris alongside the Eiffel Tower and suddenly a croissant smacks you on the head and knocks you to the ground. From your handy dandy tourist guidebook you find that the height of the Eiffel Tower is 144.3 m. If you neglect air resistance, calculate how many seconds the croissant dropped before it tagged you on the head assuming it started with a velocity of 0 m/s. (TWO decimal places!!)
The time taken by the croissant before it strike on the head is 5.4 s
Lets assume the height of the man is 1.3 m
Given height of the Eiffel Tower is 144.3 m
So the distance croissant travel before it hits the man on the head
= 144.3 - 1.3 m
= 143 m
According to the Second Equation of Motion which relates distance with the acceleration and time, we can say that
[tex]S = ut + \frac{1}{2} a t^{2}[/tex]
here u = 0
a = g = 9.81 m/s²
S = 143 m
therefore [tex]S = \frac{1}{2} gt^{2}[/tex]
[tex]t = \sqrt{\frac{2S}{g} }[/tex]
[tex]t = \sqrt{\frac{2 * 143}{9.81} }[/tex]
[tex]t = \sqrt{29.154}[/tex]
t = 5.4 s
Therefore, if the air resistance is neglected the croissant will drop 5.4 s before it tagged the man on his head.
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Consider an electric field that is uniform in direction throughout a certain volume. Can it be uniform in magnitude? Must it be uniform in magnitude? Answer these questions (a) assuming the volume is filled with an insulating material carrying charge described by a volume charge density
The electric field cannot and must not be of uniform magnitude if the volume is filled with an insulating material carrying charge described by a volume charge density
Consider a gaussian surface in the shape of a rectangular box with two faces perpendicular to the direction of the field. Also, assuming this surface has a volume charge density that fills it up, it would therefore contain some charge. The net flow out of this box is nonzero. The field must be stronger on one side than the other. The field's magnitude cannot be uniform. Because the field’s magnitude cannot be uniform, therefore it is also impossible for it to be mandatorily uniform. A uniform magnitude thus defies gaussian postulate.
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5. An object accelerates from 10km/h at a rate of 5m/s2. What distance has the object traveled when it
reaches a speed of 50km/h?
Explanation:
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Jordan tells a coworker, “I appreciate you filling in for me while I was sick and processing all those expense reports. You improved tremendously between the first report you processed and the last one. I can see your skills developing!” What is Jordan offering her coworker?
A.
encouragement
B.
empathy
C.
external training
D.
benefits
Answer:
Explanation:
A. Encouragement
21.
Figure 1 shows a person sliding down a zip wire.
Figure 1
Zip wire
Change
in vertical
height
(a) As the person slides down the zip wire, the change in the gravitational potential energy
of the person is 1.47 kJ
Eg=m
The mass of the person is 60 kg
gravitational field strength = 9.8 N/kg
Calculate the change in vertical height of the person.
Answer:
Seems like you desperately need answers
The potential energy change is 1.47 KJ or 1470 J. The weight of the person is given 60 kg. Thus vertical height of the person is 2.5 m.
What is potential energy?Potential energy is a form of energy, generated by virtue of the position of an object in the gravitational field. Potential energy depends on the mass, acceleration and height from the ground at which the object is placed/
Thus, P = Mgh. Where m is mass, h is height and g be the acceleration due to gravity, which is equal to 9.8 m/s². It is given that the mass of the person is 60 kg.
The potential energy in the gravitational field is 1.47 KJ or 1470 J, now the height h of the person can be calculated as follows:
h = p/mg.
= 1470 J/ (60 Kg × 9.8 m/s²)
= 2.5 m
Therefore, the vertical height of the hanging person is 2.5 m.
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A ball is falling at terminal velocity. Terminal velocity occurs when the ball is in equilibrium and the forces are balanced.
Which free body diagram shows the ball falling at terminal velocity?
The free body that shows the ball falling at terminal velocity is B. A free body diagram with 2 forces: the first pointing downward labeled F Subscript g Baseline 20 N and the second pointing upward labeled F Subscript air Baseline 20 N.
How to illustrate the information?It should be noted that at terminal velocity, the ball will stop accelerating and the net force on the ball will be zero.
In this case, for the net force to be zero, equal, their resultant force is zero. That is F₁ + F₂ = 0 ⇒ F₁ = -F₂
Since F₁ = 20 N, then F₂ = -F₁ = -20 N
Therefore, if F₁ points upwards since it is positive, then F₂ points downwards since it is negative.
Therefore, the correct option is B.
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A ball is falling at terminal velocity. Terminal velocity occurs when the ball is in equilibrium and the forces are balanced. Which free body diagram shows the ball falling at terminal velocity? A free body diagram with one force pointing downward labeled F Subscript g Baseline 20 N. A free body diagram with 2 forces: the first pointing downward labeled F Subscript g Baseline 20 N and the second pointing upward labeled F Subscript air Baseline 20 N. A free body diagram with 2 forces. The first pointing downward labeled F Subscript g Baseline 20 N and the second pointing upward labeled F Subscript air Baseline 5 N. A free body diagram with 2 forces: the first pointing downward labeled F Subscript g Baseline 20 N and the second pointing upward labeled F Subscript air Baseline 30 N.
A car has a speed of 12m/s. The mass of the car and its passengers is 1250 kg. What is the total momentum of the car/passengers? (assuming the car moves in the positive direction)
p= kgm/s
Answer:
15000kgm/s
Explanation:
Given parameters:
Speed of the car = 12m/s
Mass of car and passengers = 1250Kg
Unknown:
Momentum of the car = ?
Solution:
Momentum is the quantity of motion a body posses;
Momentum = mass x velocity
Now insert the given parameters and solve;
Momentum = 12 x 1250 = 15000kgm/s
Which is the work required to move a charge?
O electric field
O electric field line
O electric potential
O electric potential energy
Answer:
O electric potential
Explanation:
Answer:
[tex]electric \: potential[/tex]
Explanation:
[tex]hope \: helps \\ and \: brainliest \: if \: helped[/tex]
QC A string fixed at both ends and having a mass of 4.80g , a length of 2.00 m , and a tension of 48.0 N vibrates in its second (n=2) normal mode. (b) What is the ratio of the wavelength in air of the sound emitted by this vibrating string and the wavelength of the wave on the string?
The ratio of the wavelength in air of the sound emitted by this vibrating string and the wavelength of the wave on the string is 2:1.
Frequency of the wavef = n/2L(√T/μ)
where;
n is number of nodesL is length of the stringT is the tension of the springμ is mass per unit lengthμ = 0.0048 kg / 2 m = 0.0024 kg/m
f = 2/(2 x 2) (√48/0.0024)
f = (¹/₂)(√20,000)
f = (¹/₂)(141.42)
f = 70.71 Hz
Speed of sound in stringv = √T/μ
v = (√48/0.0024)
v = 141.42 m/s
wavelength of the wave on the stringλ = v/f
λ = (141.42) / (70.71)
λ = 2 m
Wavelength in air of the sound emittedSpeed of sound in air, v = 332 m/s
λ = v/f
λ = 332 / 70.71
λ = 4.7 m
Ratio of the wavelengths = 4.7 m / 2m = 2.3/1 ≈ 2 : 1
Thus, the ratio of the wavelength in air of the sound emitted by this vibrating string and the wavelength of the wave on the string is 2:1.
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Una cantidad de aire se lleva del estado a al estado b siguiendo una trayectoria recta en una grafica PV como es mostrado en la figura. Suponga que la expansión es desde un volumen de 0.06 metros cúbicos a un volumen de 0.13 metros cúbicos y que la presión aumenta desde 125,432 pascales a 168,793 pascales.¿Cuánto trabajo efectúa el gas en este proceso?
Answer:
El trabajo de frontera ejercido por el gas es 10,297.875 joules.
Explanation:
Supongamos que el gas se comporta idealmente y que el proceso es isotérmico y adiabático. El trabajo efectuado por el gas en el proceso equivale al área bajo la curva de la gráfica PV. Puesto que es una línea recta y que tanto la presión como el volumen son variables positivas, tenemos que esa área es la suma del rectángulo inferior (límite inferior de la presión) y el triángulo superior (entre los límites inferior y superior de la presión). Es decir:
[tex]W = P_{min}\cdot (V_{2}-V_{1})+\frac{1}{2}\cdot (P_{max}-P_{min})\cdot (V_{2}-V_{1})[/tex] (1)
Donde:
[tex]W[/tex] - Trabajo de frontera ejercido por el gas, medida en joules.
[tex]V_{1}[/tex], [tex]V_{2}[/tex] - Volúmenes del gas antes y después de la expansión, medidas en metros cúbicos.
[tex]P_{min}[/tex] - Límite inferior de la presión del gas, medida en pascales.
[tex]P_{max}[/tex] - Límite superior de la presión del gas, medida en pascales.
Si sabemos que [tex]V_{1} = 0.06\,m^{3}[/tex], [tex]V_{2} = 0.13\,m^{3}[/tex], [tex]P_{min} = 125,432\,Pa[/tex] y [tex]P_{max} = 168,793\,Pa[/tex], entonces el trabajo de frontera ejercido por el gas es:
[tex]W = (125,432\,Pa)\cdot (0.13\,m^{3}-0.06\,m^{3})+\frac{1}{2}\cdot (168,793\,m-125,432\,m) \cdot (0.13\,m^{3}-0.06\,m^{3})[/tex]
[tex]W = 10,297.875\,J[/tex]
El trabajo de frontera ejercido por el gas es 10,297.875 joules.
question 2(multiple choice worth 5 points) (01.07 mc) distances in space are measured in light-years. the distance from earth to a star is 6.8 × 1013 kilometers. what is the distance, in light-years, from earth to the star (1 light-year
The distance in light years from earth to Star is 7.1 Light Years.
We have the distance from earth to a star is 6.8 × 10¹³ kilometers.
We have to convert it into light years.
What is a Light Year?Light year is a unit of astronomical distance equivalent to the distance that light travels in one year.
1 Light year = 9.4607 × 10¹³ km
According to the question, we have -
Distance from Earth to Star = 6.8 × 10¹³ km.
Now -
1 Light year = 9.4607 × 10¹² km
or
9.4607 × 10¹²km = 1 Light years
1 km = [tex]$\frac{1}{9.4607\times 10^{12} }[/tex] Light years
Therefore -
6.8 × [tex]10^{13}[/tex] km = [tex]$\frac{6.8\times 10^{13} }{9.4607\times 10^{12} }[/tex] = 0.71 x 10 = 7. 1 Light Years.
Therefore, the distance in light years from earth to Star is 7.1 Light Years.
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Is question II correct?
Answer:
it most likly right I'm not 100% sure
Your weight as measured on your bathroom scale is
a. equal to your mass
b. the force due to gravity on you
a property of mechanical equilibrium
d. all of the above
none of the above
C.
The weight as measured on the bathroom scale is the force due to gravity. The correct option is b.
What is weight?The weight of an object is the force acting on it due to gravity in science and engineering.
Weight is defined as a vector quantity in some standard textbooks, as the gravitational force acting on the object. Others define weight as a scalar quantity, the magnitude of gravity.
This is frequently expressed in the formula W = mg, where W is the weight, m is the object's mass, and g is the gravitational acceleration.
Bathroom scales for home use show your weight on a dial or a digital screen. These scales weigh you in one of two ways: mechanically, with springs, or electronically, with circuits that bend under weight, changing the current flowing through them.
Thus, the correct option is b.
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A ball free falls from the top of the roof for 5 seconds. How far did it fall?
Answer:
Taking the value of acceleration due to gravity as 9.8 m/s^2, the object fell for about 122.5 metres.
A 18000 Hz wave has wavelength of 0.03 meters. How fast is this wave
traveling?
Answer:
540m/s
Explanation:
Given parameters:
Frequency of the wave = 18000Hz
Wavelength of the wave = 0.03m
Unknown:
How fast is the wave traveling = ?
Solution:
How fast the wave is traveling is a measure of the speed of the wave;
Speed of wave = frequency x wavelength
Now insert the given parameters and solve;
Speed of wave = 18000 x 0.03 = 540m/s
A toy car accelerates at 3 m/s2 for 2 seconds. Its final velocity is 15 m/s, What is its initial velocity?
Answer:
can i have the toy car after your done asking questions with it
Explanation:
Height of cannon 5 m, initial speed of projectile 15m/s, angle of launch 0 degrees. What is the range and time in the air? Please show work!
Answer:
The range is 15.15 m and the time in the air is 1.01 s
Explanation:
Horizontal Motion
When an object is thrown horizontally (with angle 0°) with a speed v from a height h, it follows a curved path ruled exclusively by gravity until it eventually hits the ground.
The range or maximum horizontal distance traveled by the object can be calculated as follows:
[tex]\displaystyle d=v\cdot\sqrt{\frac {2h}{g}}[/tex]
To calculate the time the object takes to hit the ground, we use the equation below:
[tex]\displaystyle t=\sqrt{\frac{2h}{g}}[/tex]
The cannon is shot from a height of h=5 m with an initial speed of v=15 m/s. The range is calculated below:
[tex]\displaystyle d=15\cdot\sqrt{\frac {2*5}{9.8}}=15*1.01[/tex]
d = 15.15 m
The time in the air is:
[tex]\displaystyle t=\sqrt{\frac{2*5}{9.8}}[/tex]
t = 1.01 s
The range is 15.15 m and the time in the air is 1.01 s
An electron has a speed of 0.750 c .(a) Find the speed of a proton that has the same kinetic energy as the electron.
By relativistic energy, the proton speed is 0.024c.
We need to know about relativistic energy to solve this problem. The rest energy of the object can be determined by
Eo = m₀ . c²
where Eo is rest energy, m₀ is rest mass and c is speed of light (3 x 10⁸ m/s).
The total energy of object can be described as
E = Eo / √(1 - v²/c²)
where E is total energy, v is the object speed.
The kinetic energy is
KE = E - Eo
From the question above, we know that :
mp = 1.6 x 10¯²⁷ kg
me = 9.1 x 10¯³¹ kg
c = 3 x 10⁸ m/s
ve = 0.75c
Find the rest energy of electron
Eo = me . c²
Eo = 9.1 x 10¯³¹ . (3 x 10⁸)²
Eo = 8.19 x 10¯¹⁴ joule
Eo = 8.19 x 10¯¹⁴ / (1.6 x 10¯¹⁹) eV
Eo = 511875 eV
Determine the total energy of electron
E = Eo / √(1 - ve²/c²)
E = 511875 / √(1 - (0.75c)²/c²)
E = 511875 / 0.66
E = 773882.26 eV
Calculate the kinetic energy of electron
KE = E - Eo
KE = 773882.26 - 511875
KE = 262007.26 eV
Find the rest energy of proton
Eo = mp . c²
Eo = 1.6 x 10¯²⁷ . (3 x 10⁸)²
Eo = 1.44 x 10¯¹⁰ joule
Eo = 1.44 x 10¯¹⁰ / (1.6 x 10¯¹⁹) eV
Eo = 900000000 eV
Determine the total energy of proton
E = KE + Eo
E = 262007.26 + 900000000
E = 900262007.3 eV
Find the speed of proton
900262007.3 = 900000000 / √(1 - v²/c²)
√(1 - v²/c²) = 0.9997
1 - v²/c² = 0.9994
v²/c² = 1 - 0.9994
v²/c² = 5.82 x 10¯⁴
v² = 5.82 x 10¯⁴ c²
v = 0.024c
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NEED HELP ASAP
ONLY ANSWER IF YK THE ANSWERS
Answer:
there yah go that's the answer
hi i need help pls help me
Answer:
best answer would be a
Explanation:
d is just a law
c is just the start of a paragraph
and b would be a revision or addtion to the already amendments
Discuss why physically active job does not guarantee better physical fitness
Question 14 (3 points)
What part of volcano carries magma from the magma chamber to the vent? (3 points)
Vent
Magma Chamber
Lava Flow
Conduit
d
you may check it online
Lars is standing near the edge of a 90-meter cliff. He throws a ball upward, but does not catch it, and it falls to the bottom of the cliff face. From when he threw the ball upward to when it hit the ground below, 5.55 seconds passed. What was the initial vertical velocity of Lars' throw?
a. -43 m/s
b. -11 m/s
c. 43 m/s
d. 11 m/s
Answer:
c. 43 m/s
Explanation:
Given the following data;
Displacement, S = 90 meters
Time, t = 5.55 seconds
To find the initial velocity;
We would use the second equation of motion given by the formula;
[tex] S = ut + \frac {1}{2}at^{2}[/tex]
Where;
S represents the displacement or height measured in meters. u represents the initial velocity measured in meters per seconds. t represents the time measured in seconds. a represents acceleration measured in meters per seconds square.We know that acceleration due to gravity is -9.8m/s² because the direction is downward.
Substituting into the equation, we have;
[tex] 90 = u*5.55 + \frac {1}{2}*(-9.8)*5.55^{2}[/tex]
[tex] 90 = u5.55 - 4.9*30.8025[/tex]
[tex] 90 = u5.55 - 150.93225[/tex]
Rearranging the equation, we have;
[tex] u5.55 = 90 + 150.93225[/tex]
[tex] u5.55 = 240.93225[/tex]
[tex] u = \frac {240.93225}{5.55}[/tex]
Initial velocity, u = 43.41 ≈ 41 m/s
a ball is thrown upward at 25 m/s from the ground. what distance has the ball travelled after 5 seconds?
Answer:
125m
Explanation:
25m/s x 5 = 125
dentify the following terms associated with the water cycle.
Water changes from a gas to a liquid phase.
Water falls to the Earth in the form of a liquid or solid.
Liquid water changes into a gas.
Water evaporates from the leaves of plants.
Answer:
condensation, precipitation, evaporation, transpiration
Explanation:
condensation is when gas condenses into liquid, precipitation is when water comes from the sky like in rain, evaporation is when water is converted from liquid to gas, transpiration is when water is evaporated from leaves.
When two polarizers are overlapped, what angle relative to one another will allow for all of the incoming light to be blocked? *
Answer:
90°
Explanation:
The intensity of the transmitted light is proportional to the square of the cosine of the angle between polarizer axes. That cosine is zero when the angle is 90°.