Đáp án:
p=5,74atm
Giải thích các bước giải:
áp suất khí là:
pV=nRT⇒p=nRTV=1.0,082.(77+273)5=5,74atmpV=nRT⇒p=nRTV=1.0,082.(77+273)5=5,74atm
p=5,74atm
Giải thích các bước giải:
áp suất khí là:
pV=nRT⇒p=nRTV=1.0,082.(77+273)5=5,74atmpV=nRT⇒p=nRTV=1.0,082.(77+273)5=5,74atm
Identifying the factors contributing to and acting as determinant factors of health disparities during the program theory and development process is a means of culturally tailoring the program.
a) true
b) false
Answer:
a) True
Explanation:
A program-specific message provided to an individual or group with the intention of raising awareness of a health condition, motivating behavior change, removing perceived barriers to participating in a health habit, or something else relating to the program's aims and objectives. The most effective intervention messages are usually theory-based and culturally adapted.
Phát biểu nào sau đây là SAI?
A. Cường độ điện trường là đại lượng
đặc trưng cho điện trường về phương
diện tác dụng lực.
B. Điện trường tĩnh là điện trường có
cường độ E không đổi tại mọi điểm.
C. Đơn vị đo cường độ điện trường là
vôn trên mét (V/m).
D. Trong môi trường đẳng hướng,
cường độ điện trường giảm lần so với
trong chân không
Answer:
B.
Explanation:
sana makatulong sayo
If we use 1 millimeter to represent 1 light-year, how large in diameter is the Milky Way Galaxy?
Answer:
if 1 light year was one millimeter then 105,700 light years = 105,700 mm, (or 105.7 meters in case you needed to simplify or something)
You are at a furniture store and notice that a Grandfather clock has its time regulated by a physical pendulum that consists of a rod with a movable weight on it. When the weight is moved downward, the pendulum slows down; when it is moved upward, the pendulum swings faster. If the rod has a mass of 1.23 kg and a length of 1.25 m and the weight has a mass of [10] kg, where should the mass be placed to give the pendulum a period of 2.00 seconds
Answer:
The distance is 1.026 m.
Explanation:
mass of rod, M = 1.23 kg
Length, L = 1.25 m
mass, m = 10 kg
Time period, T = 2 s
Let the distance is d.
The formula of the time period is given by
[tex]T = 2\pi\sqrt\frac{\frac{1}{3}ML^2+md^2}{(M +m)g}\\\\2\times 2 = 4\pi^2\times \frac{\frac{1}{3}\times1.23\times1.25\times 1.25+ 10d^2}{(1.23 + 10)\times9.8}\\\\11.16 = 0.64 + 10d^2\\\\d= 1.026 m[/tex]
Two skaters, both of mass 75 kg, are on skates on a frictionless ice pond. One skater throws a 0.4-kg ball at 6 m/s to his friend, who catches it and throws it back at 6.0 m/s. When the first skater has caught the returned ball, what is the velocity of each of the two skaters
Answer:
v = 0.064 m/s
Explanation:
Given that,
The mass of two skaters = 75 kg
The mass of a ball = 0.4 kg
The speed of the ball = 6 m/s
The speed of skater = 6 m/s
We need to find the velocity of each of the two skaters.
Under the values given the moment with respect to the ball and which is subsequently transmitted to people it would be given by:
[tex]P=0.4(6)+0.4(6)\\\\=4.8\ kg-m/s[/tex]
We know that,
P = mv
Where
v is the velocity of each skater.
[tex]v=\dfrac{p}{m}\\\\v=\dfrac{4.8}{75}\\\\=0.064\ m/s[/tex]
So, the velocity of each of the skaters is 0.064 m/s.
The number 0.00325 × 10-8 cm can be expressed in millimeters as A) 3.25 × 10-11 mm. B) 3.25 × 10-10 mm. C) 3.25 × 10-12 mm. D) 3.25 × 10-9 mm.
Answer:
Option B. 3.25×10¯¹⁰ mm.
Explanation:
Measurement (cm) = 0.00325×10⁻⁸ cm
Measurement (mm) =?
The measurement in mm can be obtained as follow:
1 cm = 10 mm
Therefore,
0.00325×10⁻⁸ cm = 0.00325×10⁻⁸ cm × 10 mm / 1 cm
0.00325×10⁻⁸ cm = 3.25×10¯¹⁰ mm
Thus, 0.00325×10⁻⁸ cm is equivalent to 3.25×10¯¹⁰ mm.
The conversion from centimeter to millimeter of the number 0.00325*10^-8cm is 3.25*10^-10mm
The number given is in standard form and can be written as 3.25*10^-11 cm.
To convert this from centimeter to millimeter, we have to multiply this value by 10.
Conversion Units1 cm - 10mm100cm = 1m1000m = 1kmSo, let's 3.25*10^-11 by 10 and get our value in mm
[tex]3.25*10^-^1^1 * 10 = 3.25*10^-^1^0[/tex]
From the calculation above, we can see that option B is the right answer since it carries [tex]3.25*10^-^1^0mm[/tex]
Learn more about conversion of units here;
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A 97.6-kg baseball player slides into second base. The coefficient of kinetic friction between the player and the ground is μk = 0.555.
(a) What is the magnitude of the frictional force?
(b) If the player comes to rest after 1.22 s, what is his initial speed?
Answer:
[tex]v=6.65m/sec[/tex]
Explanation:
From the Question we are told that:
Mass [tex]m=97.6[/tex]
Coefficient of kinetic friction [tex]\mu k=0.555[/tex]
Generally the equation for Frictional force is mathematically given by
[tex]F=\mu mg[/tex]
[tex]F=0.555*97.6*9.8[/tex]
[tex]F=531.388N[/tex]
Generally the Newton's equation for Acceleration due to Friction force is mathematically given by
[tex]a_f=-\mu g[/tex]
[tex]a_f=-0.555 *9.81[/tex]
[tex]a_f=-54455m/sec^2[/tex]
Therefore
[tex]v=u-at[/tex]
[tex]v=0+5.45*1.22[/tex]
[tex]v=6.65m/sec[/tex]
Part A
Calculate the work done when a force of 6.0 N moves a book 2.0 m
Express your answer with the appropriate units.
Answer:
Work done applied = 12 newton-meter
Explanation:
Given examples:
Force applied = 6 newton
Distance of book = 2 meter
Find from the given data:
Work done
Computation:
The equation can be used to compute work.
Work done applied = Force applied x Distance of book
Work done applied = Force x Distance
Work done applied = 6 x 2
Work done applied = 12 newton-meter
Two identical ambulances with loud sirens are driving directly towards you at a speed of 40 mph. One ambulance is 2 blocks away and the other is 10 blocks away. Which of the following is true? [Note that pitch = frequency.]a) The siren from the closer ambulance sounds higher pitched to you.b) The siren from the farther ambulance sounds higher pitched to you.c) The pitch of the two sirens sounds the same to you.d) The siren from the farther ambulance sounds higher pitched, until the closer ambulance passes you.
Answer:
c) The pitch of the two sirens sounds the same to you
Explanation:
The pitch does not depend on the distance of the object from the observer.
As per the given data
pitch = frequency
Frequency = [tex]f_{0}[/tex] [tex]\frac{V +- V_{0}}{V +- V_{s}}[/tex]
[tex]f^{'}[/tex] = [tex]f_{0}[/tex] [tex]\frac{V }{V - V_{s}}[/tex]
Hence, the pitch of the two sirens remains the same for the observer.
Answer:
c) The pitch of the two sirens sounds the same to you
Explanation:
How do solar panels work with conduction, convection and radiation?
Answer:
In the case of a solar thermal panel we are trying to heat above the ambient temperature so conduction and convection will work against us by taking heat from the panel to the out- side world. ... The sun (at 6000 C surface temperature) is hotter than the solar panel so the panel will get hot due to the solar radiation.
Explanation:
The coefficent of static friction between the floor of a truck and a box resting on it is 0.38. The truck is traveling at 87.9 km/hr. What is the least distance in which the truck can stop and ensure that the box does not slide?
Answer:
[tex]d=79.9m[/tex]
Explanation:
From the question we are told that:
coefficient of static friction [tex]\mu=0.38[/tex]
Velocity [tex]v=87.9=>24.41667m/s[/tex]
Generally the equation for Conservation of energy is mathematically given by
[tex]\mu*mgd = 0.5 m v^2[/tex]
[tex]d=\frac{0.5*24.42^2}{0.38*9.8}[/tex]
[tex]d=79.9m[/tex]
A ceramic tile measuring 50 cm x50cm has been designed to bear a pressure of 40 N/in . Will it with stand a force of 5 N?
Answer:
The ceramic tile will stand the force of 5 N.
Explanation:
Step 1: Calculate the area (A) of the ceramic tile
We will use the formula for the area of a square.
A = 50 cm × 50 cm = 2500 cm²
Step 2: Convert "A" to in²
We will use the conversion factor 1 in² = 6.45 cm².
2500 cm² × 1 in²/6.45 cm² = 3.9 × 10² in²
Step 3: Calculate the pressure (P) exerted by a force (F) of 5 N
We will use the following expression.
P = F/A
P = 5 N / 3.9 × 10² in² = 0.013 N/in²
Since the pressure exerted would be less than the maximum pressure resisted (40 N/in²), the ceramic tile will stand the force of 5 N.
Mỗi giây có 2,1.1018 ion+
2 và 1,8.1018
electron chạy qua tiết diện đèn ống.
Đường kính tiết diện của đèn ống là 2,0
cm. Mật độ dòng điện trung bình qua
đèn là bao nhiêu
Use a variation model to solve for the unknown value. Use as the constant of variation. The stopping distance of a car is directly proportional to the square of the speed of the car. (a) If a car travelling has a stopping distance of , find the stopping distance of a car that is travelling . (b) If it takes for a car to stop, how fast was it travelling before the brakes were applied
Complete question is;
Use a variation model to solve for the unknown value.
The stopping distance of a car is directly proportional to the square of the speed of the car.
a. If a car traveling 50 mph has a stopping distance of 170 ft, find the stopping distance of a car that is traveling 70 mph.
b. If it takes 244.8 ft for a car to stop, how fast was it traveling before the brakes were applied?
Answer:
A) d = 333.2 ft
B) 60 mph
Explanation:
Let the stopping distance be d
Let the speed of the car be v
We are told that the stopping distance is directly proportional to the square of the speed of the car. Thus;
d ∝ v²
Therefore, d = kv²
Where k is constant of variation.
A) Speed is 50 mph and stopping distance of 170 ft.
v = 50 mph
d = 170 ft = 0.032197 miles
Thus,from d = kv², we have;
0.032197 = k(50²)
0.032197 = 2500k
k = 0.032197/2500
k = 0.0000128788
If the car is now travelling at 70 mph, then;
d = 0.0000128788 × 70²
d = 0.06310612 miles
Converting to ft gives;
d = 333.2 ft
B) stopping distance is now 244.8 ft
Converting to miles = 0.046363636 miles
Thus from d = kv², we have;
0.046363636 = 0.0000128788(v²)
v² = 0.046363636/0.0000128788
v² = 3599.99658
v = √3599.99658
v ≈ 60 mph
A ball rolls over the edge of a platform with only a horizontal velocity. The height of the platform is 1.6 m and the horizontal range of the ball from the base of the platform is 20 m. What is the horizontal velocity of the ball just before it touches the ground?
Explanation:
the answer is in the above image
Which of the following changes would double the force between two charged particles?
A. Doubling the amount of charge on each particle
B. Increasing the distance between the particles by a factor of 2
C. Decreasing the distance between the particles by a factor of 2
D. Doubling the amount of charge on one of the particles
Answer:
Doubling the amount of charge on one of the particles.
Explanation:
The force between two charges is given by :
[tex]F=\dfrac{kq_1q_2}{r^2}[/tex]
Where
r is the distance between charges
or
[tex]F\propto \dfrac{1}{r^2}[/tex]
On doubling the charge on one of the particle,
F' = 2F
So, the force gets doubled. Hence, the correct option is (d).
equations of kinematics new dimensions
Answer:
Δx Displacement
t Time interval
v0 Initial velocity
v Final velocity
a Constant acceleration
Explanation:
I dont know if this helps because these are kinda like, examples... hope it helps tho :)
A car is travelling at a speed of 30m/s on a straight road. what would be the speed of the car in km
Answer:
[tex] = \frac{30 \times {10}^{ - 3} }{1} \\ = 0.03 \: km \: per \: second[/tex]
Answer:
108 km/hr or 0.03 km/s
Explanation:
conversion factor for m/s to km/hr is 5/18
conversion factor for m/s to km/s is 1/1000
two factor of a number are 5 and 6 .what is the number show working
Answer:
30
Explanation:
since [tex]\frac{30}{5}[/tex]=6
[tex]\frac{30}{6}[/tex]=5
then both 5 and 6 are factors of 30
Have a nice day
4. Water stands 12.0 m deep in a storage tank whose top is open to the atmosphere at
1.00 atm. The density of water is given as 1000 kg/m² and some pressure conversion
are 1 Pa = 1 N/m² while 1 atm = 101 325 Pa.
a) What is the absolute pressure at the bottom of the tank?
b) What is the gauge pressure at the bottom of the tank?
[4]
[4]
Answer:
[tex]P=217600Pa[/tex]
Explanation:
From the question we are told that:
Density [tex]\rho=1000kg/m^3[/tex]
Depth of Water [tex]d=12.0m[/tex]
Generally the equation for Pressure is mathematically given by
[tex]P=\rho gh[/tex]
[tex]P=1000*9.8*12[/tex]
[tex]P=117600N/m^2[/tex]
Therefore
Absolute Pressure=P+P'
Where
P=Pressure under water
P'=Atmospheric Pressure
Therefore
[tex]P_A=P+P'[/tex]
[tex]P_A=117,600+10^5[/tex]
[tex]P=217600Pa[/tex]
A projectile of mass m is fired horizontally with an initial speed of v0 from a height of h above a flat, desert surface. Neglecting air friction, at the instant before the projectile hits the ground, find the following in terms of m, v0, h, and g : Are any of the answers changed if the initial angle is changed?
Complete question is;
A projectile of mass m is fired horizontally with an initial speed of v0 from a height of h above a flat, desert surface. Neglecting air friction, at the instant before the projectile hits the ground, find the following in terms of m, v0, h and g:
(a) the work done by the force of gravity on the projectile,
(b) the change in kinetic energy of the projectile since it was fired, and
(c) the final kinetic energy of the projectile.
(d) Are any of the answers changed if the initial angle is changed?
Answer:
A) W = mgh
B) ΔKE = mgh
C) K2 = mgh + ½mv_o²
D) No they wouldn't change
Explanation:
We are expressing in terms of m, v0, h, and g. They are;
m is mass
v0 is initial velocity
h is height of projectile fired
g is acceleration due to gravity
A) Now, the formula for workdone by force of gravity on projectile is;
W = F × h
Now, Force(F) can be expressed as mg since it is force of gravity.
Thus; W = mgh
Now, there is no mention of any angles of being fired because we are just told it was fired horizontally.
Therefore, even if the angle is changed, workdone will not change because the equation doesn't depend on the angle.
B) Change in kinetic energy is simply;
ΔKE = K2 - K1
Where K2 is final kinetic energy and K1 is initial kinetic energy.
However, from conservation of energy, we now that change in kinetic energy = change in potential energy.
Thus;
ΔKE = ΔPE
ΔPE = U2 - U1
U2 is final potential energy = mgh
U1 is initial potential energy = mg(0) = 0. 0 was used as h because at initial point no height had been covered.
Thus;
ΔKE = ΔPE = mgh
Again like a above, the change in kinetic energy will not change because the equation doesn't depend on the angle.
C) As seen in B above,
ΔKE = ΔPE
Thus;
½mv² - ½mv_o² = mgh
Where final kinetic energy, K2 = ½mv²
And initial kinetic energy = ½mv_o²
Thus;
K2 = mgh + ½mv_o²
Similar to a and B above, this will not change even if initial angle is changed
D) All of the answers wouldn't change because their equations don't depend on the angle.
Charlotte throws a paper airplane into the air, and it lands on the ground. Which best explains why this is an example of projectile motion? The paper airplane’s motion is due to horizontal inertia and the vertical pull of gravity. A force other than gravity is acting on the paper airplane. The paper airplane’s motion can be described using only one dimension. A push and a pull are the primary forces acting on the paper airplane.
highschool physics, not college physics
Answer:
Answer:
A). The paper airplane’s motion is due to horizontal inertia and the vertical pull of gravity.
Explanation:
Edge.
Answer:
The motion of the paper airplane is best explained by horizontal inertia and vertical pull of gravity.
Explanation:
What is horizontal inertia and vertical pull of gravity?Inertia is the property by which the body wants to remain in its position unless any external for is applied. Here horizontal inertia is inertia of motion which is acting horizontally .
While vertical pull is due to the earth .
In a paper airplane , four forces act .these forces provide it flight.These forces are horizontal inertia , vertical pull downwards , lift by air and drag.Hence horizontal inertia and vertical pull best explain the projectile motion of paper airplane.
Also read it;
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A train is moving at a constant
speed of 55.0 m/s. After 5.00
seconds, how far has the train
gone?
cara
(Units = m)
Answer:
Distance = speed * time
55*5
275 meters.
The train would have covered a distance of 275 m
What is distance ?
We can define distance as to how much ground an object has covered despite its starting or ending point.
Distance = speed * time
given
speed= 55 m/s
time = 5 sec
Distance = 55 * 5 = 275 m
The train would have covered a distance of 275 m
learn more about distance
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A 17-kg sled is being pulled along the horizontal snow-covered ground by a horizontal force of 33 N. Starting from rest, the sled attains a speed of 1.6 m/s in 9.8 m. Find the coefficient of kinetic friction between the runners of the sled and the snow.
Answer:
[tex]\mu=0.185[/tex]
Explanation:
From the question we are told that:
Mass [tex]m=17kg[/tex]
Force [tex]F=33N[/tex]
Velocity [tex]v=1.6m/s[/tex]
Distance [tex]d= 9.8m[/tex]
Generally the equation for Work done is mathematically given by
[tex]W=\triangle K.E+\triangle P.E[/tex]
Where
[tex]\triangle K.E=(F-F_f)*2[/tex]
[tex]F_f=F+\frac{\triangle K.E}{d}[/tex]
[tex]F_f=33+\frac{0.5*17*1.6^2}{9.8}[/tex]
[tex]F_f=30.8N[/tex]
Since
[tex]f = \mu*m*g[/tex]
[tex]\mu= 30.8/(m*g)[/tex]
[tex]\mu= 30.8/(17*9.81)[/tex]
[tex]\mu=0.185[/tex]
How does the magnitude of the normal force exerted by the ramp in the figure compare to the weight of the static block? The normal force is:______ a. greater than the weight of the block. b. possibly greater than or less than the weight of the block, depending on whether or not the ramp surface is smooth. c. equal to the weight of the block. d. possibly greater than or equal to the weight of the block, depending on whether or not the ramp surface is smooth. less than the weight of the block.
Answer:
less than the weight of the block.
Explanation:
From the free body diagram, we get.
The normal force is N = Mg cosθ
The tension in the string is T = Mg sinθ
Wight of the block when the block is static, W = Mg
Now since the magnitude of cosθ is in the range of : 0 < cosθ < 1,
therefore, the normal force is less than the weight of the static block.
Can someone write this question clearly and send it to me? Don't just say the answer. Draw and write clearly please
Explanation:
acceleration is weight*gravity
tension is the weight In Newtons
A major artery with a 1.3 cm^2 cross-sectional area branches into 18 smaller arteries, each with an average cross-sectional area of 0.6 cm^2. By what factor is the average velocity of the blood reduced when it passes into these branches?
Answer:
When the blood passes into the smaller branches, its average velocity reduces by a factor of 0.12
Explanation:
Given;
initial area of the artery, A₁ = 1.3 cm²
Area of each smaller 18 arteries, a₂ = 0.6 cm²
Total area of the smaller 18 arteries, A₂ = 18 x 0.6 cm²
Apply flow rate equation;
Q = AV
where;
Q is the flow rate of the blood
V is the average velocity of the blood
If the flow rate is constant, then;
A₁V₁ = A₂V₂
[tex]V_2 = \frac{A_1V_1}{A_2} = \frac{1.3\times V_1}{18\times 0.6} \\\\V_2 = 0.12 \ V_1[/tex]
When the blood passes into the smaller branches, its average velocity reduces by a factor of 0.12
The first and second coils have the same length, and the third and fourth coils have the same length. They differ only in the cross-sectional area. According to theory, what should be the ratio of the resistance of the second coil to the first coil and the fourth coil to the third
Answer:
The ratio of the resistances of second coil to the first coil is the ratio of square of radius of the first coil to the square of radius of second coil.
And
The ratio of the resistances of fourth coil to the third coil is the ratio of square of radius of the third coil to the square of radius of fourth coil.
Explanation:
The resistance of the coil is directly proportional to the length of the coil and inversely proportional to the area of coil and hence inversely proportional to the square of radius of the coil.
So, the ratio of the resistances of second coil to the first coil is the ratio of square of radius of the first coil to the square of radius of second coil.
And
The ratio of the resistances of fourth coil to the third coil is the ratio of square of radius of the third coil to the square of radius of fourth coil.
A solid aluminum sphere of radius R has moment of
inertia I about an axis through its center. What is the
moment of inertia about a central axis of a solid
aluminum sphere of radius 2R?
1. 21
2. 41
3. 87
4. 161
5. 321
Answer:
5. 32I
Explanation:
The moment of inertia of a solid sphere about its central axis is given by
I = [tex]\frac{2}{5} MR^2[/tex] ------------------(i)
Where;
M = mass of the sphere
R = radius of the sphere.
From the question;
Case 1: The aluminum sphere has a radius R and moment of inertia I.
This means that we can substitute these values of R and I into equation (i) and get;
I = [tex]\frac{2}{5} MR^2[/tex] --------------(ii)
M is the mass of the aluminum sphere and is given by;
M = pV
Where;
p = density of aluminum
V = Volume of the sphere = [tex]\frac{4}{3} \pi R^3[/tex]
=> M = p([tex]\frac{4}{3} \pi R^3[/tex]) --------------------(*)
Case 2: An aluminum sphere with a radius of 2R instead.
Let the moment of inertia in this case be I' and mass be M'
Substituting R = 2R, M = M' and I = I' into equation (i) gives
I' = [tex]\frac{2}{5} M'(2R)^2[/tex] ------------------(iii)
Where;
M' = pV'
p = density of aluminum
V' = volume of the sphere = [tex]\frac{4}{3} \pi (2R)^3[/tex]
=> M' = p([tex]\frac{4}{3} \pi (2R)^3[/tex])
Rewriting gives;
M' = p([tex]\frac{4}{3} \pi (2)^3(R)^3[/tex])
M' = p([tex]\frac{4}{3} \pi8(R)^3[/tex])
M' = 8p([tex]\frac{4}{3} \pi R^3[/tex])
From equation (*), this can be written as
M' = 8M
Now substitute all necessary values into equation (ii)
I' = [tex]\frac{2}{5} M'(2R)^2[/tex]
I' = [tex]\frac{2}{5} (8M)(2R)^2[/tex]
I' = [tex]\frac{2}{5} (8M)(2)^2(R)^2[/tex]
I' = [tex]\frac{2}{5} (8M)(4)(R)^2[/tex]
I' = [tex]\frac{2}{5} (32M)(R)^2[/tex]
I' = [tex]32[\frac{2}{5}MR^2][/tex]
Comparing with equation (ii)
I' = [tex]32[I][/tex]
Therefore, the moment of inertia about a central axis of a solid
aluminum sphere of radius 2R is 32I
don't answer for points you will be reported
Explanation:
Glasses or Contacts. You might not realize it, but if you wear glasses or contact lenses, this is light refraction at play. ...
Human Eyes. Human eyes have a lens. ...
Prism. ...
Pickle Jar. ...
Ice Crystals. ...
Glass. ...
Twinkling Stars. ...
Microscope or Telescope.
