Answer: Mean weight = 2.81 N and Measurement of uncertainty = 0.82 N
Explanation:
Mean = [tex]\dfrac{\text{Sum of observations}}{\text{number of observations}}[/tex]
Mean weight is [tex]($\mu)=\frac{2.8+2.6+2.9+3.1+2.4+2.9+3.2+2.5+2.7+3.0}{10}[/tex]
[tex]=2.81$[/tex]
[tex]$\sum_{i=1}^{10}\left(x_{i}-\mu\right)^{2}=0.61[/tex]
[tex]$\sigma=\sqrt{\frac{1}{N-1} \sum_{i=1}^{10}\left(x_{i}-\mu\right)^{2}}[/tex]
[tex]=\sqrt{\frac{1}{10-1} 0.61}=0.082$[/tex]
Measurement of uncertainty will be [tex]$\sigma=0.082$[/tex]
hence, Weight [tex]$W=2.81 \pm 0.082 N$[/tex]
please helpp!
convert 1N into dyne
In the given relation F=ma a stands for write there SI unit
Answer:
a. 1 Newton = 100000 Dyne
b. a represents acceleration.
Explanation:
Newton is the standard unit (S.I) of measurement of force. Converting 1 Newton to dyne we have;
1 Newton = 10⁵ Dyne
1 Newton = 100000 Dyne
Newton's Second Law of Motion states that the acceleration of a physical object is directly proportional to the net force acting on the physical object and inversely proportional to its mass.
Mathematically, it is given by the formula;
Force = mass * acceleration
[tex] F = ma[/tex]
Hence, we can deduce that a represents the acceleration of an object and it's measured in meters per seconds square.
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
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.
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HELPPP PLSS!!!!!!
What is the chemical formula for iodine trichloride?
A. 12C|
B. ICI3
C. 3ICI
D. |1C13
Answer:
B
Explanation:
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:
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].
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.
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]
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 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]
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)
Select the correct answer.
Which figure shows a correct pattern of field lines?
A. Figure A
B. Figure B
C. Figure C
D. Figure D
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
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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
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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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).
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]
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
Calculate the RMS voltage of the following waveforms with 10 V peak-to-peak:
a. Sine wave;
b. Square wave,
c. Triangle wave.
Calculate the period of a waveform with the frequency of:
a. 100 Hz,
b. 1 kHz,
c. 100 kHz.
Answer:
a) [tex]T=0.01s[/tex]
b) [tex]T=0.001s[/tex]
c) [tex]T=0.00001s[/tex]
Explanation:
From the question we are told that:
Given Frequencies
a. 100 Hz,
b. 1 kHz,
c. 100 kHz.
Generally the equation for Waveform Period is mathematically given by
[tex]T=\frac{1}{f}[/tex]
Therefore
a)
For
[tex]T=100 Hz[/tex]
[tex]T=\frac{1}{100}[/tex]
[tex]T=0.01s[/tex]
b)
For
[tex]F=1kHz[/tex]
[tex]T=\frac{1}{1000}[/tex]
[tex]T=0.001s[/tex]
c)
For
[tex]F=100kHz[/tex]
[tex]T=\frac{1}{100*100}[/tex]
[tex]T=0.00001s[/tex]
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 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]
What is needed to Run A Brushless DC motor
Two connection methods are used for brushless DC motors. One method is to connect the coils in a loop as we compared it with the rotor winding of DC motors in Fig. 2.27. This method is called a Δ (delta) connection.
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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]
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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 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
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
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
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
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
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.
