Answer:
The time of flight is [tex]T = \frac{2 u sin A}{g}[/tex].
Explanation:
Let the initial velocity is u and the angle of projection is A.
Use first equation of motion for vertical motion
Let the time to reach the maximum height is t.
[tex]v = u - gt\\\\0 = u sin A - gt \\\\t = \frac{ u sin A}{g}[/tex]
Total time of flight is
T =2 t
[tex]T = \frac{2 u sin A}{g}[/tex]
The human ear can respond to an extremely large range of intensities - the quietest sound the ear can hear is smaller than 10-20 times the threshold which causes damage after brief exposure. If you could measure distances over the same range with a single instrument, and the smallest distance you could measure was 1 mm, what would the largest be, in kilometers?
Answer:
the largest distance we can measure is 10¹⁴ km
Explanation:
Given the data in the question;
Threshold hearing = 10⁻²⁰
smallest distance measured = 1 mm
Largest distance measured will be;
⇒ ( threshold hearing )⁻¹ × smallest distance
= ( 1 / 10⁻²⁰ ) × 1 mm
= 10²⁰ × 1mm
= 10²⁰ mm
we know that; 1000 mm = 10⁶ km
Largest distance = ( 10²⁰ / 10⁶ ) km
= 10¹⁴ km
Therefore, the largest distance we can measure is 10¹⁴ km
What is air
A. A Buchner substance
B. A compound
C. An element
D. A mixture
Air is classified as a mixture. Option D is the correct answer.
Air is a combination of different gases, primarily nitrogen (about 78%), oxygen (about 21%), and small amounts of other gases such as carbon dioxide, argon, and trace elements. These gases are not chemically bonded to each other, but rather exist together in the same space. Option D is the correct answer.
In a mixture, the substances involved retain their individual properties and can be separated by physical means. This is true for air as well. The gases in air can be separated through processes like fractional distillation or filtration. It's important to note that air also contains other components such as water vapor, dust particles, and pollutants, which can vary in concentration depending on the location and environmental conditions. These components further contribute to the complex nature of air as a mixture.
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While riding his bike through a neighborhood, Joe notices a red sign out of
the corner of his eye. He quickly comes to a stop. Which active reading
strategy is he using?
A. Asking questions
B. Making mental images
C. Summarizing
D. Using visual cues
SUBMIT
Answer:
B
Explanation:
because it take to the thought of a situation
Answer:
d i think
Explanation:
and are u using IXL???
What is the "best" explanation for why the universe is the way it is?
A) god created the universe
B) there is a multiverse and this one happens to be perfect for life.
C) this is the only universe and it happens to be perfect for life.
D) It is all in illusion and none of it exists.
E) none of the above, they are all just guesses.
I know the answer I just wanna see what you guys think.
i will give brainly if you get it right.
Determine the minimum horizontal force P required to hold the crate from sliding down the plane. The crate has a mass of 50 kg and the coefficient of static friction between the crate and the plane is . ms
Answer: hello some data related to your question is missing attached below is the missing data and diagram related to the solution
answer:
P = 141.21 N
Explanation:
Given data:
Mass of crate = 50 kg
coefficient of static friction ( μ ) = 0.25
Calculate minimum horizontal force ( P ) that holds the crate from sliding
∑fx = 0
= P + Fcos θ - N*sinθ = 0
= P + 0.25N cos 30° - Nsin30° = 0
∴ P = 0.2835 N = 0
P - 0.2853 N = 0 ------- ( 1 )
∑fy = 0
- 50g + Ncosθ + Fsinθ
- 50*9.81 + Ncos30° + 0.25Nsin30°
∴ N = 494.942 N ----- ( 2 )
input 2 into 1
P - 0.2853 ( 494.942 ) = 0
P = 141.21 N
a soap bubble was slowly enlarged from radius 4cm to 6cm and amount of work necessary for enlargement is 1.5 *10 calculate the surface tension of soap bubble joules
Answer:
The surface tension is 190.2 N/m.
Explanation:
Initial radius, r = 4 cm
final radius, r' = 6 cm
Work doen, W = 15 J
Let the surface tension is T.
The work done is given by
W = Surface Tension x change in surface area
[tex]15 = T \times 4\pi^2(r'^2 - r^2)\\\\15 = T \times 4 \times 3.14\times 3.14 (0.06^2- 0.04^2)\\\\15 = T\times 0.0788\\\\T = 190.2 N/m[/tex]
In which states of matter will a substance have a fixed volume?
O A. Liquid and solid
O B. Solid and gas
O C. Plasma and gas
O D. Liquid and gas
Answer:
A. liquid and solid
Explanation:
A single loop of wire with an area of 0.0900 m^2 is in a uniform magnetic field that has an initial value of 3.80 T, is perpendicular to the plane of the loop, and is decreasing at a constant rate of 0.160 T/s.
Reqiured:
a. What emf is induced in this loop?
b. If the loop has a resistance of 0.600Ω, find the current induced in the loop.
Answer:
a) [tex]E=0.0144[/tex]
b) [tex]I=0.024A[/tex]
Explanation:
From the question we are told that:
Area [tex]A=0.09m^2[/tex]
Magnetic Field [tex]B=3.80T[/tex]
Rate [tex]\frac{dB}{dt}=0.160T/s[/tex]
Generally the equation for EmF E is mathematically given by
[tex]E=-A\frac{dB}{dt}[/tex]
[tex]E=-(0.0900*0.160)[/tex]
[tex]E=0.0144[/tex]
b)
at Resistance R=0.60
Generally the equation for Current I is mathematically given by
[tex]E=IR[/tex]
[tex]I=\frac{0.0144}{0.600}[/tex]
[tex]I=0.024A[/tex]
Elapsed Time
(s)
Cart Speed
(Low fan speed)
(cm/s)
Cart Speed
(Medium fan speed)
(cm/s)
Cart Speed
(High fan speed)
(cm/s)
0
1
16.4
23.0
31.7
2
31.5
64.0 cm/s
3
54.0 cm/s
36.0
89.8
4
118,81
5
120.0 cm/s
6
128.2
96.0
7
145.8
Table B
Answer:
cm/s
6
128.2
96.0
7
145.8
Table B
In the mirror diagram shown, which is the normal?
А
В
С
D
Answer:
C
Explanation:
The normal is the line which divides the angle between the incident ray (which is the ray of an object which strikes the mirror) and the reflected ray(the ray which is thrown back as the object hits the mirror surface) into two equal parts. The normal is always perpendicular to the surface. In the description agram Given , the Noa which is the line C, divides the reflected ray (line D) and the incident ray (line A) into two equal parts. The plane surface is line B and the other incident ray (line C) is perpendicular to B
Planets closer to a star will have what type of average temperature
Answer:
Mercury - 800°F (430°C) during the day, -290°F (-180°C) at night. Venus - 880°F (471°C) Earth - 61°F (16°C) Mars - minus 20°F (-28°C)30-Jan-2018
A car traveling at 14 m/s accelerates at 3.5 m/s² for 5 seconds. How much distance does it travel during that time?
Answer: 113.75
Explanation:
You know
acceleration = a = 3.5 m/s²
time = t = 5 seconds
initial velocity = u = 14 m/s
Unknown is distance = s = ?
Use equation: s = ut + [tex]\frac{1}{2}[/tex] at²
Substitute all the known values inside the equation:
s = (14*5) + 0.5 * 3.5 * 5²
s = 70 + 43.75 = 113.75 m
The car travels 113.75 metres.
The lamp has a resistance of 10 ohms each resistor has a resistance of 10 ohms what is the total resistance of the circuit
Explain how the Laws of planetary motion and Newton’s laws allow the hotel to keep moving in space.
Answer:
Explanation:
i am sorry i needed points
Hooke's law describes a certain light spring of unstretched length 38.0 cm. When one end is attached to the top of a door frame and a 6.00-kg object is hung from the other end, the length of the spring is 42.0 cm.
a. Find its spring constant.
b. The load and the spring are taken down. Two people pull in opposite directions on the ends of the spring, each with a force of 150 N. Find the length of the spring in this situation.
Answer:
(a) 1470 N/m
(b) 48.2 m
Explanation:
Applying,
(a) F = ke.................... Equation 1
Where F = force applied to the spring, k = spring constant, e = extension
make k the subject of the equation
k = F/e............... Equation 2
But,
F = mg............. Equation 3
Where m = mass, g = acceleration due to gravity
Substitute equation 3 into equation 2
k = mg/e.............. Equation 4
From the question,
Given: m = 6 kg, e = 42-38 = 4 cm = 0.04 m
Constant: g = 9.8 m/s²
Substitute these values into equation 4
k = (6×9.8)/0.04
k = 1470 N/m
(b) Consider the end of the spring to the left which exert a force to the right
Then,
e = F/k............. Equation 5
Given: F = 150 N, k = 1470 N/m
Substitute these values into equation 5
e = 150/1470
e = 0.102 m
Hence the length of the spring is
L = 0.38+0.102 = 0.482 cm = 48.2 m
A scooter is accelerated from rest at the rate of 8m/s
. How long will it take to cover
a distance of 32m?
Explanation:
time=Distance/speed
t=32/8
t=4 seconds
A student wishes to construct a mass-spring system that will oscillate with the same frequency as a swinging pendulum with a period of 3.45 S. The student has a spring with a spring constant of 72.0 N/m. What mass should the student use to construct the mass-spring system?
Answer:
21.73 kg
Explanation:
Applying,
T = 2π√(m/k)............... Equation 1
Where T = period, m = mass on the spring, k = spring constant, π = pie.
make m the subject of the equation
m = T²k/4π²................. Equation 2
From the question,
Given: T = 3.45 s, k = 72.0 N/m, π = 3.14
Substitute these values into equation 2
m = (3.45²×72)/(4×3.14²)
m = 21.73 kg.
Hence the mass should be 21.73 kg
Consider the heaviest box of 150 lb that you can push at constant speed across a level floor, where the coefficient of kinetic friction is 0.45, and estimate the maximum horizontal force that you can apply to the box. A box sits on a ramp that is inclined at an angle of 60.0° above the horizontal. The coefficient of kinetic friction between the box and the ramp is 0.45.
If you apply the same magnitude force, now parallel to the ramp, that you applied to the box on the floor, what is the heaviest box (in pounds) that you can push up the ramp at constant speed? (In both cases assume you can give enough extra push to get the box started moving.)
Maximum horizontal force that can be applied on the box is 300.32 N.
Mass of the heaviest box that can be pushed on the ramp at constant speed is 105.16 pound.
What is meant by kinetic friction ?Kinetic friction is defined as the opposing force exerted by the surface on an object in contact with it, when there is relative motion between the two surfaces.
Here,
Mass of the box, m = 150 lb = 68.1 kg
Coefficient of kinetic friction, μ = 0.45
Maximum horizontal force that can be applied on the box is the kinetic frictional force. Frictional force,
F(k) = μmg
F(k) = 0.45 x 68.1 x 9.8
F(k) = 300.32 N
Now, the box sits on a ramp inclined at 60°
Coefficient of kinetic friction, μ = 0.45
The net force here acting on the box placed in the ramp is due to the kinetic frictional force and the weight of the box.
So,
Frictional force, F(k)' = μmgcosθ
F(k)' = 0.45 x M x 9.8 x cos 60
F(k)' = 2.2M
Weight of the box acting horizontally,
W = Mgsinθ
W = M x 9.8 x sin60
W = 8.5M
Therefore, net force,
Fn = W - F(k)'
Fn = 8.5M - 2.2M
Fn = 6.3M
The total force acting on the box is
F = F(k) - Fn
ma = 300.32 - 6.3M
Since, the box is moving with constant speed, the acceleration, a = 0
Therefore,
300.32 - 6.3M = 0
6.3M = 300.32
M = 300.32/6.3
M = 47.7 kg = 105.16 pound
Hence,
Maximum horizontal force that can be applied on the box is 300.32 N.
Mass of the heaviest box that can be pushed on the ramp at constant speed is 105.16 pound.
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A uniform electric field is oriented in the −x direction. The magnitude of the electric field is 6500 N/C. How will the equipotential surfaces associated with this electric field be oriented?
Identify the reactants in the combustion of methane: CH4 + O2 CO2 + O°H
A local FM radio station broadcasts at a frequency of 100.8 MHz. Calculate the energy of the frequency at which it is broadcasting. Energy
Answer:
[tex]E=6.68\times 10^{-26}\ J[/tex]
Explanation:
Given that,
The frequency of FM ratio station, f = 100.8 MHz = 100.8 × 10⁶ Hz
We need to find the energy of the wave. We know that,
Energy, E = hf
Put all the values,
[tex]E=6.63\times 10^{-34}\times 100.8\times 10^6\\\\=6.68\times 10^{-26}\ J[/tex]
So, the energy of the wave is equal to [tex]6.68\times 10^{-26}\ J[/tex].
a 50kg skater on level ice, has built up her speed to 30km/h. how far will she coast before sliding friction dissipates her energy?
Answer:
belpw
Explanation:
The distance prior to the sliding friction dispersing her energy would be:
- The distance will remain unaffected by the sliding friction i.e. 354m
As we know, When Sliding friction dissolves her energy, leading her Kinetic Energy to turn 0 on coming to the state of rest. So,
[tex]1/2 mv^2 - 1/2 mu^2 = -W[/tex] (∵ Work in -ve denotes it is done opposite to friction)
Given that,
m(mass) [tex]= 50 kg[/tex]
v(velocity) [tex]= 30 km/hr[/tex] or [tex]8.33 m/s[/tex]
The coefficient of Kinetic Friction [tex]= 0.01[/tex]
g(gravitational force) [tex]= 9.8 m/s^2[/tex]
Initial Velocity(u) [tex]= 30[/tex] × [tex]1000/3600 m/s[/tex]
[tex]= 8.33 m/s[/tex]
Now by employing the provided values,
[tex]F =[/tex] μ[tex]mg[/tex]
[tex]= (0.01) (50) (9.8)[/tex]
[tex]= 4.9[/tex]
∵ [tex]F = 4.9 N[/tex]
By using the above expression, we will find the distance;
[tex]1/2 mv^2 - 1/2 mu^2 = -W[/tex]
⇒ [tex]1/2 (50) (0)^2 - 1/2 (50) (8.33)^2 = -4.9(S)[/tex]
⇒ [tex]1734.7225 = 4.9S[/tex]
⇒ [tex]S = 1734.7225/4.9[/tex]
∵ [tex]S = 354 m[/tex]
Because [tex]1/2 mv^2 - 1/2 mu^2 = -W[/tex] [tex]= -[/tex] μmgS
⇒ [tex]S = (u^2 - v^2)[/tex]/2μ[tex]g[/tex]
Thus, the distance will remain unaffected by the sliding friction i.e. 354m
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PAY ATTENTION MY QUESTION ASK FOR RADIATION!!!
You sit with friends around a campfire, roasting marshmallows. Which
transfer of thermal energy involved in this system is an example of radiation
Answer:
The answer is c
Thermal energy moves within the air from the flames to the marshmallow.
Explanation:
Hope it helps
You sit with friends around a campfire, roasting marshmallows. then the transfer of thermal energy involved in this system is an example of radiation Thermal energy moves within the air from the flames to the marshmallow. Hence option C is correct.
What is thermal Energy ?In physics and engineering, the phrase "thermal energy" is thrown around in a lot of different situations. It can relate to a variety of distinct physical notions. Included in this are the internal energy, or enthalpy, of a body of matter and radiation; heat, which is a form of energy transfer (as is thermodynamic work); and the characteristic energy of a degree of freedom in a system described in terms of its microscopic particulate constituents (where T denotes temperature and k denotes the Boltzmann constant.
Hence option C is correct.
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How fast much an 816kg Volkswagen travel to have the same momentum as (a) a 2650kg Cadillac going 16.0 km/h? (b) a 9080-kg truck also going 16.0 km/hr?
Answer:
Explanation:
From the given information:
the car's momentum = momentum of the truck
∴
(a) 816 kg × v = 2650 kg × 16.0 km/h
v = (2650 kg × 16.0 km/h) / 816 kg
v = 51.96 km/hr
(b) 816 kg × v = 9080 kg × 16.0 km/h
v = (9080 kg × 16.0 km/h) / 816 kg
v = 178.04 km/hr
A student solving a physics problem to find the unknown has applied physics principles and obtained the expression: μkmgcosθ=mgsinθ−ma, where g=9.80meter/second2, a=3.60meter/second2, θ=27.0∘, and m is not given. Which of the following represents a simplified expression for μk?A student solving a physics problem to find the unknown has applied physics principles and obtained the expression: , where , , , and is not given. Which of the following represents a simplified expression for ?tanθ− agTo avoid making mistakes, the expression should not be simplified until the numerical values are substituted.gsinθ−agcosθThe single equation has two unknowns and cannot be solved with the information given.
Solution :
Given expression :
[tex]$\mu_k$[/tex]mgcosθ = mgsinθ − ma
Here, g = 9.8 [tex]m/s^2[/tex] , a = 3.60 [tex]m/s^2[/tex] , θ = 27°
Therefore,
[tex]$\mu_k mg \cos \theta = mg \sin \theta - ma$[/tex]
[tex]$\mu_k mg \cos \theta = m(g \sin \theta - a)$[/tex]
[tex]$\mu_k g \cos \theta = (g \sin \theta - a)$[/tex]
[tex]$\mu_k =\frac{(g \sin \theta-a)}{g \cos \theta}$[/tex]
Mow calculating the coefficient of kinetic friction as follows :
[tex]$\mu_k=\frac{g \sin \theta-a}{g \cos \theta}$[/tex]
[tex]$\mu_k=\frac{9.8 \times \sin 27^\circ-3.60}{9.8 \times \cos 27^\circ}$[/tex]
[tex]$\mu_k=0.097$[/tex]
Please help, only answer if your 1000% correct im in summer school and need to pass this class
Action reaction forces never cancel each other out because .............................? *
1 point
they do not act on objects.
they act on the same objects
they act on different objects.
none of the above
A force can be described as? *
1 point
push or pull action
pulling direction
direction of change
push action only
Using EquatIO calculate the resultant force for the horizontal (x-axis) component. Show your calculation step. *
2 points
Captionless Image
A player hits a baseball with a bat. The action force is the impact of the bat against the ball. The reaction force is .......................? *
1 point
Captionless Image
the grip of the player's hands on the ball
the weight of the ball
the air resistance of the ball
the force of the ball against the bat
Which law is this? "An object at rest tends to stay at rest and an object in motion tends to stay in motion unless acted upon by an unbalanced force." *
1 point
Newtons third law
Newtons second law
Newtons first law
None of the above
Which law is this? "If an object A exerts a force on object B, then object B must exert a force of equal magnitude and opposite direction back on object A." *
1 point
None of the above
Newtons second law
Newtons first law
Newtons third law
What can be said about a force on an object that is not stationary? *
1 point
Captionless Image
there are no forces acting on the object
there are unequal forces acting on the object
there is only one force acting on the object
there are equal and opposite acting on the object
Using EquatIO calculate the resultant force for the vertical (y-axis) component. Show your calculation step. *
2 points
Captionless Image
Identify the different force acting on a moving vehicle shown in the mage below. *
5 points
Captionless Image
A B C D E
Reaction force
Weight
Friction
Air resistance
Thrust
Reaction force
Weight
Friction
Air resistance
Thrust
Which law is this? "The acceleration of an object is directly related to the net force and inversely related to its mass." *
1 point
None of the above
Newtons third law
Newtons first law
Newtons second law
If forces acting on an object are unbalanced, what can occur to the object? *
1 point
acceleration
all of the above
change of shape
deceleration
change in direction
If a player hits a baseball with a force of 870N Calculate the acceleration (state the units) of the ball . Show your calculation step using only EquatIO. *
2 points
Captionless Image
Action reaction forces never cancel each other out because .............................? *
1 point
they act on different objects.
none of the above
they act on the same objects
they do not act on objects.
because they work on the same object
Explanation:
A push and pull factor
How many atoms of carbon, C, are in 0.020 g of carbon?
Answer:
9.6352× 10²⁰ C atoms
Explanation:
From the given information,
The molar mass of Carbon = 12 g/mol
number of moles = 0.020g/ 12 g/mol
number of moles = 0.0016 mol
If 1 mole of C = 6.022 × 10²³ C atoms
∴
0.0016 mol of C = (6.022 × 10²³ C atoms/ 1 mol of C)×0.0016 mol of C
= 9.6352× 10²⁰ C atoms
Hence, the number of carbon atoms present in 0.020 g of carbon = 9.6352× 10²⁰ C atoms
You decide to impress Grandpa by showing him how fast sound travels. You have a piece of plastic pipe with an adjustable closed end, and a 312 Hz tuning fork. The piece of pipe resonates in the 2nd resonant length when it is adjusted to a length of 81.0 cm. What is the speed of sound on that day?
Answer:
336.96m/s
Explanation:
answer is in photo above
The speed of sound on that day is 336.96 m/s.
What is speed?Speed is distance travelled by the object per unit time. Due to having no direction and only having magnitude, speed is a scalar quantity With SI unit meter/second.
given that:
frequency of the sound = 312 Hz.
2nd resonant length = 81.0 cm.
If the wavelength of sound is λ; the 2nd resonant length of plastic pipe with an adjustable closed end = 3λ/4
Hence, wavelength of sound is = 81×(4/3) cm = 0.81 ×(4/3) m.
So, the speed of sound on that day is = frequency × wavelength
= 312 Hz × 0.81 ×(4/3) m.
= 336.96 m/s.
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A kettle operates from a 120 V outlet. It has a heating element with a resistance of 8.0 Ω . Calculate the current going through the element.
Answer:
I = 15A
Explanation:
V = I*R
120V = I*8.0ohms
I = 120V/8.0ohms
I = 15A
Answer:
I=15A
Explanation:
you know what to do.
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Give them the brain list.
Suppose 3 mol of neon (an ideal monatomic gas) at STP are compressed slowly and isothermally to 0.19 the original volume. The gas is then allowed to expand quickly and adiabatically back to its original volume.
Required:
a. Find the highest temperature attained by the gas.
b. Find the lowest temperature attained by the gas.
c. Find the highest pressure attained by the gas.
d. Find the lowest pressure attained by the gas.
Answer:
a. 273 K b. 90.1 K c. 5.26 atm d. 0.33 atm
Explanation:
For isothermal expansion PV = constant
So, P₁V₁ = P₂V₂ where P₁ = initial pressure of gas = 1 atm (standard pressure), V₁ = initial volume of gas, P₂ = final pressure of gas and V₂ = final volume of gas,
So, P₁V₁ = P₂V₂
P₂ = P₁V₁/V₂
Since V₂/V₁ = 0.19,
P₂ = P₁V₁/V₂
P₂ = 1 atm (1/0.19)
P₂ = 5.26 atm
For an adiabatic expansion, PVⁿ = constant where n = ratio of molar heat capacities = 5/3 for monoatomic gas
So, P₂V₂ⁿ = P₃V₃ⁿ where P₂ = initial pressure of gas = 5.26 atm, V₂ = initial volume of gas, P₃ = final pressure of gas and V₃ = final volume of gas,
So, P₂V₂ⁿ = P₃V₃ⁿ
P₃ = P₂V₂ⁿ/V₃ⁿ
P₃ = P₂(V₂/V₃)ⁿ
Since V₃ = V₁ ,V₂/V₃ = V₂/V₁ = 0.19
1/0.19,
P₃ = P₂(V₂/V₃)ⁿ
P₃ = 5.26 atm (0.19)⁽⁵/³⁾
P₃ = 5.26 atm × 0.0628
P₃ = 0.33 atm
Using the ideal gas equation
P₃V₃/T₃ = P₄V₄/T₄ where P₃ = pressure after adiabatic expansion = 0.33 atm , V₃ = volume after adiabatic expansion, T₃ = temperature after adiabatic expansion P₄ = initial pressure of gas = P₁ = 1 atm , V₄ = initial volume of gas = V₁ and T₄ = initial temperature of gas = T₁ = 273 K (standard temperature)
P₃V₃/T₃ = P₄V₄/T₄
T₃ = P₃V₃T₄/P₄V₄
T₃ = (P₃/P₄)(V₃/V₄)T₂
Since V₃ = V₄ = V₁ and P₄ = P₁
V₃/V₄ = 1 and P₃/P₄ = P₃/P₁
T₃ = (P₃/P₁)(V₃/V₄)T₂
T₃ = (0.33 atm/1 atm)(1)273 K
T₃ = 90.1 K
So,
a. The highest temperature attained by the gas is T₁ = 273 K
b. The lowest temperature attained by the gas = T₃ = 90.1 K
c. The highest pressure attained by the gas is P₂ = 5.26 atm
d. The lowest pressure attained by the gas is P₃ = 0.33 atm
