Answer:
The lightbulbs were exploding after several mins of use
Explanation:
Answer:
the lightbulbs were exploding after several minutes of use
Explanation:
a p e x :))
SPEAR is a storage ring at the Stanford Linear Accelerator which has a circulating beam of electrons that are moving at nearly the speed of light (2.998 108 m/s). If a similar ring is about 80.0 m in diameter and has a 0.59 A beam, how many electrons are in the beam
Answer:
n = 3.1x10¹²
Explanation:
To find the number of electrons we need to find first the charge (q):
[tex] I = \frac{q}{\Delta t} \rightarrow q = I*\Delta t [/tex] (1)
Where:
I: is the electric current = 0.59 A
t: is the time
The time t is equal to:
[tex]v = \frac{\Delta x}{\Delta t} \rightarrow \Delta t = \frac{\Delta x}{v}[/tex] (2)
Where:
x: is the displacement
v: is the average speed = 2.998x10⁸ m/s
The displacement is equal to the perimeter of the circumference:
[tex] \Delta x = 2\pi*r = \pi*d [/tex] (3)
Where d is the diameter = 80.0 m
By entering equations (2) and (3) into (1) we have:
[tex]q = I*\Delta t = I*\frac{\Delta x}{v} = \frac{I\pi d}{v} = \frac{0.59 A*\pi*80.0 m}{2.99 \cdot 10^{8} m/s} = 4.96 \cdot 10^{-7} C[/tex]
Now, the number of electrons (n) is given by:
[tex] n = \frac{q}{e} [/tex]
Where e is the electron's charge = 1.6x10⁻¹⁹ C
[tex] n = \frac{q}{e} = \frac{4.96 \cdot 10^{-7} C}{1.6 \cdot 10^{-19} C} = 3.1 \cdot 10^{12} [/tex]
Therefore, the number of electrons in the beam is 3.1x10¹².
I hope it helps you!
Distinguish between homogeneous and heterogeneous mixture with 2 example each. Do not use the examples give above
It has been suggested, and not facetiously, that life might have originated on Mars and been carried to Earth when a meteor hit Mars and blasted pieces of rock (perhaps containing primitive life) free of the surface. Astronomers know that many Martian rocks have come to Earth this way. One objection to this idea is that microbes would have to undergo an enormous, lethal acceleration during the impact. Let us investigate how large such an acceleration might be. To escape Mars, rock fragments would have to reach its escape velocity of 5.0 km/s, and this would most likely happen over a distance of about 4.0m during the impact.
1) What would be the acceleration, in m/s, of such a rock fragment?
2) What would be the acceleration, in g's, of such a rock fragment?
3) How long would this acceleration last?
4) In tests, scientists have found that over 40% of Bacillius subtilis bacteria survived after an acceleration of 450000g. In light of your answer to part A, can we rule out the hypothesis that life might have been blasted from Mars to Earth?
Answer and Explanation:
Given that
v_f = 5 km/s = 5,000 m/s
d = 4 m
v_i = 0 m/s
The computation is shown below:
1. The acceleration in m/s is
Here we use the motio third equation which is
[tex]v_f^2 = v_i^2 + 2ad[/tex]
5000^2 = 0^2 + 2 (a) (4)
So
[tex]a = 3.125 \times 10^{6} m/s^2[/tex]
2. Now acceleration in g is
[tex]= \frac{3.125 \times 10^{6} m/s^2}{9.81}[/tex]
[tex]= 3.18 \times 10^{5}g[/tex]
3. The long of acceleration last is
[tex]t = \frac{v-u}{a}[/tex]
[tex]= \frac{5000 - 0}{3.125 \times 10^{6}}[/tex]
[tex]= 1.6 \times 10^{-3}s[/tex]
4.As we can see that
[tex]3.18 \times 10^{5}[/tex] is smaller than the [tex]4.5 \times 10^{5}g[/tex]
So, it should not be ruled out
6. While standing on an ice surface (assume no friction), a 60 kg skater pushes against
a wall with an average force of 240 N east for 0.360 s. Calculate the velocity
(magnitude and direction) of this student at 0.360 s.
Answer: 1.44m/s West
Explanation:
Given the following :
Mass of skater = 60kg
Average force = 240N east
Time of = 0.360 seconds
The Velocity after 0.360seconds
Recall the relation:.
Force(f) * time(t) = mass (m) * (final Velocity(V) - initial velocity(u))
Where ; final Velocity(v) - initial velocity(u) = Magnitude of velocity(V)
Ft = mV
240 × 0.360 = 60 × V
60 × V = 86.4
V = 86.4 / 60
V = 1.44m/s
According to Newton's third law, to every action, there is an equal and opposite reaction.
The action force acts in the eastern direction hitting the wall with a force of 240N, therefore direction after hitting the wall will be in the opposite (western direction) direction, which is the direction of the reaction force
Therefore, velocity is 1.44ms West
NEED HELP!! WILL GIVE BRAINIEST!! When a driver presses the brake pedal, his car can stop with an acceleration of -5.4 meters per second squared. how far will the car travel while coming to a complete stop if its initial speed was 25 meters per second?
Answer:
d = 57.87 m
Explanation:
It is given that,
Acceleration of car due to applied brake is [tex]-5.4\ m/s^2[/tex]
Its initial speed was 25 m/s
Its final speed was 0 as its comes to rest.
Leat d is the distance covered by the car. It can be calculated using third equation of kinematics as :
[tex]v^2-u^2=2ad\\\\d=\dfrac{v^2-u^2}{2a}\\\\d=\dfrac{0^2-(25)^2}{2\times (-5.4)}\\\\d=57.87\ m[/tex]
So, the distance covered by the car is 57.87 m.
Answer:
d = 57.87 m
Explanation: have a nice day :)
A toaster oven draws 300.0 watts of power. If it is plugged into an outlet with a voltage of 115 volts, what current is in the toaster's circuit?
OA 3.45 10% amps
ОВ.
415 amps
Ос.
285 amps
OD
2.61 amps
Answer:
I = 2.61 A
Explanation:
It is given that,
Power of a toaster is 300 watts
It is plugged into an outlet with a voltage of 115 volts
We need to find the current in the toaster's circuit.
The electrical power is given by the formula as follows :
P = V I
I is electric current
[tex]I=\dfrac{P}{V}\\\\I=\dfrac{300}{115}\\\\I=2.61\ A[/tex]
So, 2.61 A of current is in the toaster's circuit.
Answer: D.
2.61 amps
Explanation: edmentum
Question 1 of 10
2 Points
What is the SI unit for gravitational potential energy?
A. Newton
B. Joule
C. Meter
D. Watt
Answer:
Joule
Explanation:
As any unit of energy (kinetic, potential, etc.), the gravitational potential energy has SI units of Joule. This s the work done by a force of 1 Newton to displace an object 1 meter in the direction of application of the force.
A light, rigid rod is 51.4 cm long. Its top end is pivoted on a frictionless horizontal axle. The rod hangs straight down at rest with a small, massive ball attached to its bottom end. You strike the ball, suddenly giving it a horizontal velocity so that it swings around in a full circle. What minimum speed at the bottom is required to make the ball go over the top of the circle?
Answer:
4.489 m/s
Explanation:
We are given;
Length of rod;L = 51.4 cm = 0.514 m
When the ball is at the top end, it's velocity will be zero and it's kinetic energy will be transformed to potential energy.
Thus; ½mv² = mgh
m will cancel out to give;
v² = 2gh
Where:
g is acceleration due to gravity = 9.8 m/s²
h is the maximum height of the ball which is equal to twice the length of the rod.
Thus, h = 2(0.514) = 1.028 m
v² = 2 × 9.8 × 1.028
v = √20.1488
v = 4.489 m/s
Un avión vuela a una velocidad de 900 km/h. Si tarda en viajar desde Canarias hasta la península 180 s ¿qué distancia recorre en ese tiempo?
Answer:
El avión recorrió 45 km en los 180 s.
Explanation:
La relación entre velocidad, distancia y tiempo se da de la siguiente manera;
[tex]Velocidad= \dfrac{Distancia}{Hora}[/tex]
Por lo cual los parámetros dados son los siguientes;
Velocidad = 900 km/h = 250 m / s
Tiempo = 180 s
Estamos obligados a calcular la distancia recorrida
De la ecuación para la velocidad dada arriba, tenemos;
Distancia recorrida = Velocidad pf viaje × Tiempo de viaje
Distancia recorrida = 900 km/h × 180 s = 900
Distancia recorrida = 900 km/h × 1 h/60 min × 1 min/60 s × 180 s = 45 km
Por lo tanto, el avión viajó 45 km en 180 s.
Los primeros vuelos de prueba del transbordador espacial usaban un "planeador" (con masa de 980 kg incluyendo el piloto). Después de un lanzamiento horizontal a 480 km/h a una altitud de 3500 m, el planeador finalmente aterrizaba con una rapidez de 210 km/h ¿Cuál era la fuerza promedio que la resistencia del aire ejercía sobre él cuando entraba con un ángulo de planeo constante de 12° con respecto a la Tierra?
Answer:
La fuerza promedio que ejerce la resistencia del aire sobre el planeador cuando ingresa con un ángulo de planeo constante de 12 ° con respecto a la Tierra es 9613.8 N
Explanation:
Los parámetros dados son;
Masa del planeador = 980 kg.
Velocidad de lanzamiento horizontal = 480 km / h
Altitud de lanzamiento = 3500 m.
Velocidad de aterrizaje final = 210 km / h
Ángulo de deslizamiento, θ = 12 °
La ecuación para el equilibrio de fuerzas.
L · cos (θ) + D · sin (θ) = W
L · sin (θ) + D · cos (θ)
Dónde:
L = fuerza de elevación
D = fuerza de arrastre
W = Peso del "planeador"
W = Masa del "planeador" × Aceleración debido a la gravedad = 980 kg × 9.81 m / s²
W = 9613.8 N
Por lo tanto, tenemos;
L · cos (12) + D · sin (12) = 9613.8 ,,,,,,,,,,,,, (1)
L · sin (12) = D · cos (12) ,,,,,,,,,,,,, (2)
De la ecuación (2), tenemos;
L = D · (cos (12)/sin (12))
Sustituyendo el valor de L = D · (cos (12) / sin (12)) en la ecuación (1), se obtiene;
D · (cos (12)/sin (12)) · cos (12) + D · sin (12) = 9613.8
D = 9613.8/((cos (12)/sin (12)) · cos (12) + sin (12)) = 1998.82 N
L = D · (cos (12)/sin (12)) = 1998.82 · (cos (12)/sin (12)) = 9403.72 N
Por lo tanto, la fuerza da la fuerza promedio que la resistencia del aire ejerce sobre el planeador cuando ingresa con un ángulo de planeo constante de 12 ° con respecto a la Tierra;
F = √ (L² + D²) = √ (9403.72² + 1998.82²) = 9613.8 N.
Consider a satellite in a circular orbit around the earth. Why is it important to give a satellite a horizontal speed when placing it in orbit? What will happen if the horizontal speed is too small? What will happen if the speed is too large?
Answer:
In this case, the horizontal velocity of the rocket starts from the acceleration, so if its velocity drops to zero,
Explanation:
When a satellite is in orbit the most important force is the docking of gravity with the Earth
F = m a
where the acceleration is centripetal and F is the force of universal attraction
centripetal acceleration is
a = v² / r
F = m v² / r
In this case, the horizontal velocity of the rocket starts from the acceleration, so if its velocity drops to zero, the force also drops to serious and the satellite steels to Earth.
The speed of the satellite is provides the speed, by local for smaller speeds in satellite, it descends in its orbits and when the speed is amate you have the energy to stop an orb to go to a higher orbit.
a Porsche 911 accelerates from rest to 27 metre per second due north in 5.8 seconds the mass of the car is 1400 kg what is the magnitude and direction of the average net force
Answer:
The average force has a magnitude 6524 N due north.
Explanation:
The average net force F = ma where m = mass of car = 1400 kg and a = acceleration.
a = (v - u)/t where u = initial velocity of car = 0 m/s (since it starts from rest)
v = final velocity of car = 27 m/s due north and t = time of motion = 5.8 s
a = (27 m/s - 0 m/s)/5.8 s = 27 m/s ÷ 5.8 s = 4.66 m/s
Since the direction of the velocity change is the direction of the acceleration, the acceleration is 4.66 m/s due north.
The average force, F = ma = 1400 kg × 4.66 m/s = 6524 N
Since the acceleration is due north, the average force takes the direction of the acceleration.
So the direction of the average force is due north
The average force has a magnitude 6524 N due north.
Which of the following is an example of the Doppler effect? A water bug on the surface of a pond is producing small ripples in the water. The bug is not moving. A police car with its siren on is driving towards you, and you perceive the pitch of the siren to increase. Scientists measure the speed of light and find that it is constant. The gravitational pull of two objects on each other
Answer:
A police car with its siren on is driving towards you, and you perceive the pitch of the siren to increase.
Explanation:
In Physics, Doppler effect can be defined as the change in frequency of a wave with respect to an observer in motion and moving relative to the source of the wave.
Simply stated, Doppler effect is the change in wave frequency as a result of the relative motion existing between a wave source and its observer.
The term "Doppler effect" was named after an Austrian mathematician and physicist known as Christian Johann Doppler while studying the starlight in relation to the movement of stars.
The phenomenon of Doppler effects is generally applicable to both sound and light.
An example of the Doppler effect is a police car with its siren on is driving towards you, and you perceive the pitch of the siren to increase. This is so because when a sound object moves towards you, its sound waves frequency increases, thereby causing a higher pitch. However, if the sound object is moving away from the observer, it's sound waves frequency decreases and thus resulting in a lower pitch.
Other fields were the Doppler effects are applied are; astronomy, flow management, vibration measurement, radars, satellite communications etc.
Answer:
A police car with its siren on is driving towards you, and you perceive the pitch of the siren to increase.
Explanation:
This should be the correct answer.
Describe the method you would use to measure the cicrumfrence of a cylinder using a thread and a meter
rule(4mk)
0)
Answer:
I take the thread and wrap it round the cylinder until it goes round the circumference of the cylinder.
Explanation:
I take the thread and wrap it round the cylinder until it goes round the circumference of the cylinder.
I then cut the thread at this length and extend it out on a table or floor.
I then place the length of the thread along the meter rule and mark of the position of its tip and bottom along the meter rule.
The difference between the bottom and top points is the circumference of the cylinder.
If the thread is longer than the meter rule, i mark off where the the length of the meter rule stops along the thread and then place the meter rule at that point and then mark off where the length of the thread stops.
The circumference of the cylinder is then 1 meter plus the extra measurement along the meter rule.
Two teams are playing tug-of-war. Team A, on the left, is pulling on the rope with an effort of 5000 N. If the rope is moving at a constant velocity, how hard and in which direction is team B pulling?
A. 2500 N to the left
B. 5000 N to the right
C. 2500 N to the right
D. 5000 N to the left
Explanation:
If Team A is on the left, B is on the right
if the force is constant, it means that the effort applied is equal.
So Team B is pulling 5000N to the right.
Mt Everest is at the height of 8850 m; its base is at an elevation of 5400m. The temperature here drops at the rate of 1 degree per 100m. If temperature at the base is -5, what will be the temperature at the top?
Answer:
The temperature at the top is -39.5 degree
Explanation:
The height of Mt Everest = 8850 m
The base is at the elevation of = 5400 m
The drop in temperature = 1 degree per 100 m
At the base, the temperature = -5
Since every 100 meter rise in height results in the one degree drop of temperature. Therefore, first find the difference between the base and height. Multiply with negative one.
Difference in heights = 8850 – 5400 = 3450
The temperature at the top = (3450 / 100) + 5 = -39.5 degree
Since there is fall in temperature so we use negative sign.
The magnitude of the electric potential depends on which factors?
O A. Mass and amount of charge
O B. Mass and distance
O C. Distance and amount of charge
O D. Density and amount of charge
Answer:
C
Explanation:
It is directly propotional to the amount of charge and inversly propotional to distance.
Answer:
C. Distance and amount of charge
Explanation:
The magnitude of the electric potential depends on which factors?
Distance and amount of charge
Show all work and answer all 4 parts. Projectile motion. 20 points. Thank you.
Answer:
i took g = 9.8m/s
A. 1.16secs
B. 2.32secs
C. 6.57m
D. 57.91m
Explanation:
A. How long does the Missile take to reach ot peak?
Time taken (t) =( U²Sin (angle) )/g
u = initial velocity = 25m/s
angle given = 30°
g = acceleration due to gravity = 9.8m/s²
t = U² x Sin (angle) / g
t = 25² x Sin(30)/9.8
t = 1.61secs
B. How long is the missile in the air in total?
T = 2t
T = 2 x 1.61 = 2.32 secs
C. what maximum Height does the missile reach?
Maximum height = U²Sin²(angle) / 2gM.H =25² x Sin(30)² / 2 x 9.8M.H=6.57mMaximum height= 6.57mD. How far does the missile travel Horizontally?
Range = U²2Sin(angle)/gRange = 25² x 2 x Sin(30) / 9.8Range = 57.91mUsing the formula W = mg, how many milliliters of water with a density of 1g/mL are required to weigh 0.75 newtons and g = 9.81 m/s2? Round to the nearest tenth. (Note: The mass will be in kg in your answer, thus convert to g and then to mL.) Answer: mL
Given that,
Weight = 0.75 N
Acceleration due to gravity = 9.81 m/s²
Density of water = 1 g/ml
We need to calculate the volume of water
Using formula of weight
[tex]W=mg[/tex]
[tex]W=\rho Vg[/tex]
[tex]mg=\rho\times V\times g[/tex]
Where, V = volume
g = acceleration
m = mass
Put the value into the formula
[tex]0.076\times 9.8\times10^3=1\times V\times9.8[/tex]
[tex]V= \dfrac{0.076\times 9.8\times10^3}{1\times9.8}[/tex]
[tex]V=76\ ml[/tex]
Hence, The volume of water is 76 mL.
Answer:
76.5mL
Explanation:
w = mg
0.75N = m * 9.81
m = 0.75 / 9.81 = 0.0765 kg
The mass in grams is 0.0765 * 1000 = 76.5g
At a density of 1 g/mL,
76.5g ÷ 1g/ml = 76.5mL
A ball is moving up an inclined plane, when it's velocity changes from 15m/s to 7m/s in 3 seconds . What is the acceleration.
Explanation:
[tex]a = \frac{change \: in \: velocity}{change \: in \: time} [/tex]
[tex]a = \frac{v - u}{t} [/tex]
[tex]v = 7 {ms}^{ - 1} [/tex]
[tex]u = 15 {ms}^{ - 1} [/tex]
[tex]t = 3 \: seconds[/tex]
[tex]a = \frac{7 - 15}{3} [/tex]
[tex]a = \frac{ - 8}{3} [/tex]
[tex]a = - 2.67 \: {ms}^{ - 2} [/tex]
please help
The statement "46 m, East" refers to the
of an object.
velocity
distance
displacement
speed
acceleration
Answer:
Displacement
Explanation:
Displacement is a vector quantity i.e it has both magnitude and direction. The direction is indicated with compass bearing i.e North, South, East or West or you could have North East, North West, South West, South East.
Question 2 (Multiple Choice Worth 2 points)
(03.06 LC)
Which of the following is a common groundwater contaminant?
O Ozone
O Nuclear waste
O Carbon monoxide
O Fertilizers
Question 3(Multiple Choice Worth 2 points)
10 M
Answer:
The common groundwater contaminant are fertilizers.
Explanation:
Answer:
The answer would be D( Fertilizers)
Explanation:
A fertilizers or fertilizer is any material of natural or synthetic origin that is applied to soil or to plant tissues to supply one or more plant nutrients essential to the growth of plants. Many sources of fertilizer exist, both natural and industrially produced.
whats the texture of copper wire
Copper is a reddish orange, soft metal with a bright metallic luster. It is an excellent conductor of heat and electricity. Copper surfaces when exposed to air gradually turns to a dull, brownish color.
as the mass of the cart increases, the acceleration of the cart
Answer:
decrease
Explanation:
as the mass of the cart increase the acceleration of the cart decrease
A roller coaster starts with a speed of 4.6 m/s at a point 49 m above the bottom of a dip.
Neglecting friction, what will be the speed of the roller coaster at the top of the next
slope, which is 29 m above the bottom of the dip?
Answer:
20.5m/s
Explanation:
neglecting friction the amount of energy at the bottom equals to the amount of energy at top. Law of conservation of energy.
ME= P.E+k.E
ME= mgh+ 1/2mv*v
ME = (m* 10 *49)+(1/2m(4.6*4.6))
MEtop=500.98m
MEbottom=MEtop: in order to obey the law of conservation of energy
MEbottom=mgh+ 1/2mv*v
500.98m=m*10*29+1/2m v* v
500.98m= m(290+1/2v*v) ...cancel m by m
500.98-290=1/2v*v criss cross multiple
210.98*2=v*v
v=√421.96
v=20.5 m/s
Oil at 150 C flows slowly through a long, thin-walled pipe of 30-mm inner diameter. The pipe is suspended in a room for which the air temperature is 20 C and the convection coefficient at the outer tube surface is 11 W/m2 K. Estimate the heat loss per unit length of tube.
Answer:
1.01 W/m
Explanation:
diameter of the pipe d = 30 mm = 0.03 m
radius of the pipe r = d/2 = 0.015 m
external air temperature Ta = 20 °C
temperature of pipe wall Tw = 150 °C
convection coefficient at outer tube surface h = 11 W/m^2-K
From the above, we assumed that the pipe wall and the oil are in thermal equilibrium.
area of the pipe per unit length A = [tex]\pi r ^{2}[/tex] = [tex]7.069*10^{-4}[/tex] m^2/m
convectional heat loss Q = Ah(Tw - Ta)
Q = 7.069 x 10^-4 x 11 x (150 - 20)
Q = 7.069 x 10^-4 x 11 x 130 = 1.01 W/m
The heat loss per unit length of tube should be considered as the 1.01 W/m.
Calculation of the heat loss:Since
diameter of the pipe d = 30 mm = 0.03 m
radius of the pipe r = d/2 = 0.015 m
external air temperature Ta = 20 °C
temperature of pipe wall Tw = 150 °C
convection coefficient at outer tube surface h = 11 W/m^2-K
Now
area of the pipe per unit length A should be
= πr^2
= 7.069*10^-4 m^2/m
Now
convectional heat loss Q = Ah(Tw - Ta)
Q = 7.069 x 10^-4 x 11 x (150 - 20)
Q = 7.069 x 10^-4 x 11 x 130
= 1.01 W/m
hence, The heat loss per unit length of tube should be considered as the 1.01 W/m.
Learn more about heat here: https://brainly.com/question/15170783
how does enormous energy get released from the sun
Answer:
By nuclear fission
Explanation:
The sun generates enormous energy through the process of nuclear fusion.
The core or the innermost part of the sun is characterized by high temperature and pressure. These two factors cause the separation of nuclei from electrons and the fusion of hydrogen nuclei to form a helium atom.
During the fusion process, energy is released.
Two projectiles are thrown from the same point with the velocity of49ms-1. First is
projected making an angle with the horizontal and the second at an angle of (90- ). The
second is found to rise 22.5m higher than the first. Find the heights to which each will rise?
Answer:
Height of the first projectile = 49.98 m
Height of the second projectile = 72.52 m
Explanation:
From the given information;
Two projectiles are thrown from the same point with the velocity of49m/s
First is projected making an angle θ with the horizontal
and the second at an angle of 90 - θ.
Thus; for the first height to the horizontal; we have;
[tex]H_1 = \dfrac{v^2 sin^2 \theta}{2g}[/tex] ----- (1)
the second height in the vertical direction is :
[tex]H_2 = \dfrac{v^2 cos^2 \theta}{2g}[/tex] -----(2)
However; the second is found to rise 22.5 m higher than the first; so , we have :
[tex]\dfrac{v^2 cos^2 \theta}{2g}= 22.5 + \dfrac{v^2 sin^2 \theta}{2g}[/tex]
Let's recall that :
Cos²θ = 1 - Sin²θ
Replacing it into above equation; we have:
[tex]\dfrac{v^2}{2g} - \dfrac{v^2 sin^2 \theta}{2g}= 22.5 + \dfrac{v^2 sin^2 \theta}{2g}[/tex]
[tex]\dfrac{v^2}{2g} - 22.5 = \dfrac{v^2 sin^2 \theta}{g}[/tex]
[tex]\dfrac{1}{2 } \dfrac{v^2}{g} - 22.5 = \dfrac{v^2 sin^2 \theta}{g}[/tex]
[tex]\dfrac{1}{2 } - \dfrac {9.8 \times 22.5}{(49)^2} = sin^2 \theta[/tex]
[tex]\dfrac{1}{2 } - \dfrac {220.5}{2401} = sin^2 \theta[/tex]
[tex]sin^2 \theta= 0.408[/tex]
From (1);
[tex]H_1 = \dfrac{v^2 sin^2 \theta}{2g}[/tex]
[tex]H_1 = \dfrac{49^2 \times 0.408}{2*9.8}[/tex]
[tex]H_1 = \dfrac{979.608}{19.6}[/tex]
[tex]\mathbf{H_1 =49.98 \ m }[/tex]
Height of the first projectile = 49.98 m
Similarly;
From(2)
[tex]H_2 = \dfrac{v^2 cos^2 \theta}{2g}[/tex]
[tex]H_2 = \dfrac{v^2 (1-sin^2 \theta)}{2g}[/tex]
[tex]H_2 = \dfrac{49^2 (1-0.408 )}{2 \times 9.8}[/tex]
[tex]H_2 = \dfrac{2401 (0.592 )}{19.6}[/tex]
[tex]H_2 = \dfrac{1421.392}{19.6}[/tex]
[tex]\mathbf{H_2 = 72.52 \ m}[/tex]
Height of the second projectile = 72.52 m
An airplane on a runway has an initial speed of 15 meters/second. What is its final velocity when it travels 600 meters on the runway with a constant acceleration of 6 meters/second2?
Answer:
86.17m/s
Explanation:
Since the motion is of constant acceleration, we can use one of the equations of motion as follows;
v² = u² + 2as
Where;
v = final velocity of the airplane
u = initial velocity of the airplane
s = distance covered by the plane
a = acceleration
From the question,
u = 15m/s
s = 600m
a = 6m/s²
Substitute these values into the equation above as follows;
v² = 15² + 2(6)(600)
v² = 225 + 7200
v² = 7425
v = √7425
v = 86.17m/s
Therefore, the final velocity of the airplane is 86.17m/s
helppppp me it's urgent please
the last one is the light will move in some manner not determined by the information here.
Light bends away from the normal, because it's moving from higher to lower refractive index.
Same bend-direction as when it goes from water into air.