The complete question is :
A hot-air balloon has a volume of 2100 m3 . The balloon fabric (the envelope) weighs 860 N . The basket with gear and full propane tanks weighs 1300 N .
If the balloon can barely lift an additional 3400 N of passengers, breakfast, and champagne when the outside air density is 1.23kg/m3, what is the average density of the heated gases in the envelope?
Solution :
Given volume of the hot air balloon [tex]$=2100 \ m^3$[/tex]
The balloon fabric weights = 860 N
The weight of the basket with the gear and propane tank = 1300 N
Density of outside air [tex]$= 1.23 \ kg/m^3$[/tex]
∴ Total pay load = Weight of the air displaced - weight of gas inside the balloon
Total pay load = 860 + 1300 + 3400
= 5560 N
Mass = density x volume
Weight [tex]$= \text{mass} \times g$[/tex]
Weight = volume x density [tex]$\times \text{ acceleration due to gravity (g)}$[/tex]
Weight of the displaced air = 2100 x 1.23 x 9.8
= 25313 N
Weight of the gas inside the balloon = density [tex]$\times \text{ acceleration due to gravity (g)}$[/tex] x volume
= density x 9.8 x 2100
= density x 20580 N
Therefore substituting the values, we get
⇒ 25313 - (density x 20580) = 5560
⇒ density [tex]$=\frac{19753}{20580}$[/tex]
[tex]$= 0.96 \ kg/m^3$[/tex]
So the density of the heated gas [tex]$= 0.96 \ kg/m^3$[/tex]
An electron remains suspended between the surface of the Earth (assumed neutral) and a fixed positive point charge, at a distance of 5.3 m from the point charge. Determine the charge required for this to happen.
This question is not complete, the complete question is;
An electron remains suspended between the surface of the Earth (assumed neutral) and a fixed positive point charge, at a distance of 5.3 m from the point charge. Determine the charge required for this to happen. The acceleration due to gravity is 9.8 m/s² and the Coulomb constant is 8.98755 x 10⁹ N.m²/C². Answer in units of C.
Answer: the charge required for this to happen is 1.7415 × 10¹⁹ C
Explanation:
Given the data in the question;
we know that gravitational force on electro = M×g
mass of electron m = 9.10938356 × 10⁻³¹ kilograms
charge of electron e = 1.60217662 × 10⁻¹⁹ coulombs
given that distance r = 5.3m and the Coulomb constant = 8.98755 x 10⁹ N.m²/C²
g = 9.8 m/s²
Now for the electron to be suspended, This will happen only if the electrostatic force is equal to weight of electron;
[ (1/4πε0) × q × e] / r² = m × g
so we substitute
8.98755 x 10⁹ × q × 1.60217662 × 10⁻¹⁹/[5.3]² = 9.10938356 × 10⁻³¹ × 9.8
q = 8.9271 × 10⁻³⁰ / 5.1262 × 10⁻¹¹
q = 1.7415 × 10¹⁹ C
Therefore the charge required for this to happen is 1.7415 × 10¹⁹ C
[1.43996 × 10⁹ × q] / 29.09 = 8.92719 x 10⁻³⁰
1.43996 × 10⁹ × q = 2.5969 × 10⁻²⁶
q = 2.5969 × 10⁻²⁶ / 1.43996 × 10⁹
q =
The gravitational attraction between two objects increases if
Answer:
they are closer!!
Explanation:
Hope this helped!! :D
John Purchased 6 Kg And 500 Kg of suger and 12kg and 400g of rice.what is the total mass of the suger and rice that he purchased?
Answer:
Total Mass = 18 kg 900 gram
Explanation:
Given:
Mass of sugar buy = 6 kg 500 gram
Mass of rice buy = 12 kg 400 gram
Find:
Total Mass
Computation:
Total Mass = Mass of sugar buy + Mass of rice buy
Total Mass = 6 kg 500 gram + 12 kg 400 gram
Total Mass = 18 kg 900 gram
(6)
A 100 W heater is connected to a 240 V mains supply. Calculate the:
current drawn;
resistance of the hheater.I want to know how to go about it
Answer:
Current drawn = 0.417 A (rounded to 3 significant figures)
Resistance = 576Ω (rounded to 3 significant figures)
Explanation:
Current = Power ÷ Voltage
Current = 100W / 240V
∴ Current = 0.416666666.... = 0.417A
Resistance = Voltage ÷ Current
Resistance = 240 ÷ 0.417
∴ Resistance = 575.539 = 576Ω
A guitar player can change the frequency of a string by "bending" it-pushing it along a fret that is perpendicular to its length. This stretches the string, increasing its tension and its frequency. The B string on a guitar is 64 cm long and has a tension of 74 N. The guitarist pushes this string down against a fret located at the center of the string, which gives it a frequency of 494 Hz. He then bends the string, pushing with a force of 4.0 N so that it moves 8.0 mm along the fret.
* What is the new frequency?
Answer:
[tex]f'=504hz[/tex]
Explanation:
From the question we are told that
The B string on a guitar is 64 cm long
The B string tension tension of 74 N.
Frequency of 494 Hz
Pushed with a Force of 4.0 N
It moves 8.0 mm along the fret.
Generally the equation for frequency of ring under tension is mathematically given as
[tex]2Lf=\sqrt{x\frac{T}{\mu} }[/tex]
[tex]2*(64/100)*494=\sqrt{\frac{74}{m/0.64}[/tex]
[tex](632.32)^2={\frac{74}{m/0.64}[/tex]
[tex](632.32)^2=74*{\frac{0.64}{m}[/tex]
[tex](632.32)^2={\frac{47.36}{m}[/tex]
[tex]m=1.18450761*10^-^4[/tex]
Therefore finding the New frequency f'
[tex]f'=\frac{(\sqrt{\frac{74+11}{(\frac{1.18450761*10^-^4}{0.642})}})}{2*0.642}[/tex]
[tex]f'=\frac{(\sqrt{(\frac{74+11}{1})*(\frac{0.642}{1.18450761*10^-^4}})}{2*0.642}[/tex]
[tex]f'=\frac{(\sqrt{(\frac{85}{1})*(\frac{0.642}{1.18450761*10^-^4}})}{2*0.642}[/tex]
[tex]f'=\frac{(\sqrt{(\frac{54.57}{1.18450761*10^-^4}})}{2*0.642}[/tex]
[tex]f'=\frac{678.7471973}{2*0.642}[/tex]
[tex]f'=572.9111096hz[/tex]
A truck pushes a pile of dirt horizontally on a frictionless road with a net force of 20\, \text N20N20, start text, N, end text for 15.0\,\text m15.0m15, point, 0, start text, m, end text. How much kinetic energy does the dirt gain
Answer:
300 Nm ; 300 J
Explanation:
Given that:
Force (F) = 20 N
Distance (d) = 15 m
The kinetic energy (Workdone) = Force * Distance
Kinetic Energy = 20N * 15m
Kinetic Energy = 300Nm
K. E = 1/2
Answer:
77.0k/M
Explanation:
a batsman deflects a ball by an angle of 45 without changing its initial speed which is equal to 54km/h.what is the impulse imparted to the ball? mass of the ball is 0.15kg
Answer:
the impulse imparted to the ball is 4.158 kg.m/s
Explanation:
Please find attached image for diagram.
Given;
initial speed of the ball from point A, u = 54km/h = 15 m/s
angle of deflection of the ball along BOP, θ = 45⁰/2 = 22.5⁰
mass of the ball, m = 0.15 kg
the final speed of the ball along B, = u (in reverse direction)
The initial momentum of the ball, P₁ = mucosθ
The final momentum of the ball is P₂ in reverse direction to P₁.
the impulse imparted to the ball = change in momentum of the ball
J = ΔP = P₂ - P₁
J = mucosθ - (-mucosθ)
J = mucosθ + mucosθ
J = 2mucosθ
J = 2(0.15)15cos(22.5⁰)
J = 4.158 kg.m/s
Thus, the impulse imparted to the ball is 4.158 kg.m/s
state two uses of total internal reflection
Answer:
The phenomenon of total internal reflection of light is used in many optical instruments like telescopes, microscopes, binoculars, spectroscopes, periscopes
“DNA...”
1: has a structure of a cone.
2:cannot be repaired if it is mutated.
3:is made up of nucleotides.
4:is made up of amino acids.
< Previous
Answer:
3. is made of nucleotides
why can the atmosphere hold on to heat
Answer:
A. Air
Explanation:
A laser beam, shining from the earth's surface, is directed at the moon whose distance from the earth on this day is 370,000 km. If the beam diverges at an angle of only 1.65 x 10-5 rad, what diameter circle will it make on the moon
Answer:
d = 6105 m
Explanation:
For this exercise we must use trigonometry to find the diameter of the has. The most common definition in optics of the angle of divergence is measured between the two ends of the beam, therefore to use this angle in trigonometry where the angle is measured with respect to the normal we must take half of this angle
θ = 1.65 10⁻⁵ / 2 = 0.825 10⁻⁵ rad
let's use the tangent
tan θ = y / L
y = L tan θ
y = 370000 103 tan (0.825 10⁻⁵)
let's be careful since the angles are in radians
y = 3025.5 m
This is the distance from the normal that corresponds to the radius of the circle, the diameter is twice the radius
d = 2 y
d = 2 3025.5
d = 6105 m
. A falling rubber ball of mass 0.025 kg strikes the ground traveling straight down at 4.0 m/s. Find the magnitude of the impulse that the ground gives to the ball if
a) the ground is soft and the ball stops dead
b) the ground is hard and the ball bounces straight back at 2.0 m/s.
Given that,
Mass of the ball, m = 0.025 kg
Initial speed, u = 4 m/s
To find,
Impulse when (a) the ground is soft and the ball stops dead
(b) the ground is hard and the ball bounces straight back at 2.0 m/s.
Solution,
Impulse = change in momentum
J = m(v-u)
(a) u = 0 (as it stops)
J = 0.025(4-0)
J = 0.1 N-m
(b) v = 2 m/s
J = 0.025(4-2)
= 0.05 N-m
Therefore, this is the required solution.
T/F: Stars die.
True
False
Answer:
The answer is true
Explanation:
could you brainliest again they said I plagarized when I didn't
12x+13y=12.01
x+y=2.323 times 10 to the 27 power
x= # of carbon - 12 atoms
y= # of carbon- 13 atoms
solve for x and y
Please help!!!!!
Answer:
[tex]y = - 2.7876 * 10^{28}[/tex]
[tex]x= 3.0199 * 10^{28}[/tex]
Explanation:
Given
[tex]12x+13y=12.01[/tex]
[tex]x+y=2.323 * 10^{27}[/tex]
Required:
Solve
Make x the subject in [tex]x+y=2.323 * 10^{27}[/tex]
[tex]x=2.323 * 10^{27} - y[/tex]
Substitute the above expression for y in [tex]12x+13y=12.01[/tex]
[tex]12(2.323 * 10^{27} - y) + 13y = 12.01[/tex]
Open brackets
[tex]12*2.323 * 10^{27} - 12y + 13y = 12.01[/tex]
[tex]12*2.323 * 10^{27} + y = 12.01[/tex]
Collect Like Terms
[tex]y = 12.01 - 12*2.323 * 10^{27}[/tex]
[tex]y = 12.01 - 2.7876 * 10^{28}[/tex]
12.01 is negligible compared to [tex]2.7876 * 10^{28}[/tex]
So:
[tex]y = - 2.7876 * 10^{28}[/tex]
Substitute the above expression for y in [tex]x=2.323 * 10^{27} - y[/tex]
[tex]x=2.323 * 10^{27} - (- 2.7876 * 10^{28})[/tex]
[tex]x=2.323 * 10^{27} + 2.7876 * 10^{28}[/tex]
[tex]x= 0.2323 * 10^{28} + 2.7876 * 10^{28}[/tex]
[tex]x= 3.0199 * 10^{28}[/tex]
At a certain distance from the center of the Earth, a 0.4-kg object has a weight of 2.0 N. (a) Find this distance. (b) If the object is released at this location and allowed to fall toward the Earth, what is its initial acceleration
Answer:
a) The distance of the object from the center of the Earth is 8.92x10⁶ m.
b) The initial acceleration of the object is 5 m/s².
Explanation:
a) The distance can be found using the equation of gravitational force:
[tex]F = \frac{GMm}{r^{2}}[/tex]
Where:
G: is the gravitational constant = 6.67x10⁻¹¹ Nm²/kg²
M: is the Earth's mass = 5.97x10²⁴ kg
m: is the object's mass = 0.4 kg
F: is the force or the weight = 2.0 N
r: is the distance =?
The distance is:
[tex]r = \sqrt{\frac{GMm}{F}} = \sqrt{\frac{6.67 \cdot 10^{-11} Nm^{2}/kg^{2}*5.97 \cdot 10^{24} kg*0.4 kg}{2.0 N}} = 8.92 \cdot 10^{6} m[/tex]
Hence, the distance of the object from the center of the Earth is 8.92x10⁶ m.
b) The initial acceleration of the object can be calculated knowing the weight:
[tex] W = ma [/tex]
Where:
W: is the weight = 2 N
a: is the initial acceleration =?
[tex] a = \frac{W}{m} = \frac{2 N}{0.4 kg} = 5 m/s^{2} [/tex]
Therefore, the initial acceleration of the object is 5 m/s².
I hope it helps you!
Paola conducts an investigation to determine how temperature affects plant growth. Each plant started at the same height, received the same amount of water and was planted in the same type of soil She
recorded her results every seven days for three weeks.
Pls help
Answer:
HE IS WON
Explanation:
JESUS
200. newton·meters of work is put into a machine over a distance of 20. meters. The machine does 150. newton·meters of work as it lifts a load 10. meters high. What is the mechanical advantage of the machine?
Answer:
Answer:
Mechanical advantage of the machine is 1.5
Explanation:
The formula for mechanical advantage is:
MA = OutputForce/InputForce
To calculate this, we need the force of input and the force of output.
The formula to calculate the force given the work (W) and the distance (d) is:
F = W/d
Calculating the input force Fi:
Fi = 200Nm/20m = 10 N
Calculating the output force Fo:
Fo = 150Nm/10m = 15 N
Thus, the mechanical advantage MA is:
MA = 15N / 10N = 1.5
The graph shows the amplitude of a passing wave over time in seconds (s). What is the approximate frequency of the wave shown?
A.6 Hz
B.0.3 Hz
C.0.7 Hz
D.3 Hz
Answer:
B) 0.3Hz
Explanation:
I just took the test i hope i helped and i hope you pass the test
Answer:.5
Explanation:
I took the quiz
What is her velocity?
O 1.5 m/s
O 2 m/s
O 2.5 m/s
O 5 m/s
Answer:
C. 2.5
Explanation:
It said it was right, so that's cool.
In Requiem for a Dream, the DP uses a device that straps the camera to the actor's chest to achieve a very subjective shot. What is this apparatus called
Answer:
A snorricam
Explanation:
The apparatus used here is a snorricam. A snorricam can be described to be a camera device which is used during the making of films. The snorricam is rigged to the actor. It faces him or her directly in such a way that as they walk, they would not appear to make any movements but the things around them would. It is also called the bodymount camera.
A bowling ball hits two standing bowling pins at the same time. Which of the following is true? Assume that all collisions are elastic. A. The sum of the momentum of both pins after the collision will equal the momentum of the ball before the collision. B. The momentum of the ball after the collision will equal the momentum of the ball before the collision. C. The sum of the momentum gained by both pins will equal the amount of momentum lost by the ball. D. The momentum of each pin after the collision will equal the momentum of the ball before the collision.
Answer:
C. The sum of the momentum gained by both pins will equal the amount of momentum lost by the ball.
Explanation:
The law of conservation of momentum states that total momentum before collision must be equal to the total momentum after collision.
Momentum could be transferred from one object to another during collision. When the ball hits the stationary pins, momentum is transferred from the ball to the pins.
Since the collision is elastic, momentum is conserved hence total momentum gained by the pins equals the total momentum lost by the ball.
You throw a ball into the air. Which two forces cause the ball to gradually stop moving upward and then fall back to Earth?
A.
Balanced forces
B.
Friction
C.
Normal force
D.
Gravitational force
will mark brainliest
Answer:
I'm pretty sure the answer is D
Explanation:
Honestly it's just a guess so let me know if it's right :3
Answer:
it should be A and B
Explanation:
because obviously gravity causes it to fall down and the second force acted upon it is Friction from the air and ball coming in contact with eachother. I hope this helps
Assuming the density of water is 62.4 lbm/ft3, and using standard gravity, how much does 1 gallon of water weigh?
Answer:
The weight of the water is 37.14 N.
Explanation:
Given;
density of water, ρ = 62.4 lbm/ft³
volume of water, V = 1 gallon = 0.1338 ft³
The mass of the water is calculated as;
m = ρV
[tex]m = 62.4 \ \frac{lbm}{ft^3} \ \times \ 0.1338 \ ft^3\\\\m = 8.349 \ lbm[/tex]
1 lbm = 0.454 kg
8.349 lbm = ?
= 3.79 kg
The weight of the water is calculated as;
W = mg
where;
g is acceleration due to gravity = 9.8 m/s²
W = (3.79)(9.8)
W = 37.14 N.
Therefore, the weight of the water is 37.14 N.
"When the ball leaves the ramp at Point B, students measure the horizontal" distance traveled. They repeat the experiment five times, being careful to release the ball from the same starting Point A and find the average horizontal distance traveled to be 2.0 m. One student suggests they use a stopwatch to find the time the ball is in the air whereas another student suggests they use a meter stick. Given these materials, describe the procedure students should follow to minimize error and calculate the speed of the ball as it leaves the ramp.
Answer:
Speed of ball as it leaves ramp is 2 m/s
Explanation:
From the question it is given that the average distance traveled by ball from point A to B is 2.0 m and the time taken is 1sec
one student is using stopwatch to calculate the time taken by ball and another student is using meter stick to calculate the distance traveled
since speed of ball is given by v = [tex]\frac{distance}{time}[/tex] thus,
the speed = 2 m/s
now for minimizing the error of speed it is necessary to record the readings by single students at-least 5 times and take average
by doing this, the error of speed calculation will be minimum as the it decreases the error due to random error of system caused by taking the reading by different students
What is the force of gravitational attraction between an object with a mass of 100 kg and another object that has a mass of 300 kg and are at a distance of 2m apart
Answer:
5 x 10⁻⁷N
Explanation:
Given parameters:
Mass of object 1 = 100kg
Mass of object 2 = 300kg
Distance = 2m
Unknown:
Force of gravitational attraction between the objects = ?
Solution:
From Newton's law of universal gravitation we derive an expression:
Fg = [tex]\frac{G m_{1} m_{2} }{r^{2} }[/tex]
G is the universal gravitation constant = 6.67 x 10⁻¹¹
m is the mass
r is the distance between the bodies
Now insert the parameters and solve;
Fg = 6.67 x 10⁻¹¹ x [tex]\frac{100 x 300}{2^{2} }[/tex] = 5 x 10⁻⁷N
what is letaral inversion?
Answer:
Lateral inversion is a phenomenon in which left appears to be right and vice versa. It is due to direction that light follows when it strikes a reflecting surface, generally a mirror.
why should a rain gauge be raised 30cm above the ground
To avoid splashing, rain gauges are positioned 30 cm above the surface of the ground.
What is the reason behind a rain gauge be raised 30cm above the ground?The rainfall may be precisely measured by a rain gauge without any loss from evaporation. It is a tool for measuring the amount of precipitation that falls over a certain area. As a result, it gauges rainfall. A millimeter of measured precipitation is equal to one liter of rainfall per square meter.
Standard rain gauges have a 150–170 cm channeled aperture. They are made to be a straightforward passive collector. For measuring the amount of precipitation, use a graduated measuring glass. To avoid rainwater splashing while measuring rainfall, the rain gauze should be positioned 30 cm above the ground.
Learn more about rain gauge here:
https://brainly.com/question/19820977
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7. An outfielder throws a baseball to the first baseman at a speed of 19.6
m/s and an angle of 30° above the horizontal. If the ball in the described
situation is caught at the same height from which it was thrown, calculate
the amount of time the ball was in the air. (use -9.8 m/s2 for g in this
problem] [include the numerical answer and correct unit] *
The amount of time that the ball took in air is 1 second.
We must recall that the ball was thrown as a projectile and its time of flight must be obtained from the formula for the time of flight of a projectile as follows;
T = 2usinθ/g
u = initial velocity = 19.6 m/s
θ = angle of projection = 30°
g = acceleration due to gravity = 9.8 m/s2
Substituting values;
T = 2 × 19.6 m/s sin (30°)/9.8 m/s2
T = 1 second
Learn more about projectile: https://brainly.com/question/2818076
what are conductors and insulators? Give at least five example of each
Answer: conductors are substances that allow heat or electricity to pass through . It deals with only the flow of electrons.
Eg. water, copper wire, iron rod, some ceramic materials, metallic nail.
Insulators are materials that do not allow heat or electricity to pass through.
Eg. Book, plastic, rubber, glass, paper
Explanation:
please help me i have this due tommorrow!!!