In 60 minutes or 3600 seconds, the tip of the minute hand traverses the circumference of a circle with radius 3.00 cm, so it moves with a tangential speed of
(3.00 cm)/(3600 s) ≈ 0.00083 cm/s = 8.3 μm/s
numerical problems:
a.) convert 300K into the celsius scale.
b.) convert 220 centigrade scale into kelvin scale.
c.) convert 20 ventigrade scale into Fahrenheit scale.
d.) convert 260 Fahrenheit into centigrade. pls help me to solve this problems
For an object spinning around a central point, what will happen if its distance from the center is decreased?
A. Nothing will change.
B. Its acceleration will decrease.
C. Its acceleration will increase.
D. The centripetal force will decrease.
Answer:
Its acceleration will increase.
Explanation:
For an object spinning around a central point, Its acceleration will increase if its distance from the centre is decreased.
What is acceleration?The rate at which an item changes its velocity is known as acceleration, a vector quantity. If an object's velocity is changing, it is acceleration. The net acceleration that objects get as a result of the combined action of gravity and centrifugal force is known as the Earth's gravity
This is the acceleration of an object in a circle of radius r at a speed v. So, centripetal acceleration is greater at high speeds and in sharp curves smaller radii and for lager radii acceleration will be less.
acceleration, a = v²/r
For an object spinning around a central point, Its acceleration will increase if its distance from the centre is decreased.
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A flywheel, rotating about its axis at a rate of 4 rev/s is acted upon by a torque of 25 Nm for 10 sec. If the wheel has moment of inertia of 1.2 kgm^2, what would be the speed of the wheel at the end in rev/s?
The applied torque increases the angular speed by the application of an
angular acceleration.
The speed after 10 seconds is approximately 37.16 rev/s.Reasons:
The speed of the flywheel at the axis = 4 rev/s
The torque applied, T = 25 N·m
The time the torque is applied, t = 10 sec
Moment of inertia of the flywheel, I = 1.2 kg·m²
Required:
The speed at the end of the 10 seconds
Solution:
T = I·α
Where;
α = Angular acceleration
[tex]\displaystyle \alpha = \frac{T}{I}[/tex]
Therefore;
[tex]\displaystyle \alpha = \frac{24 \ N\cdot m}{1.2 \ kg \cdot m^2} = \mathbf{20\frac{5}{6} \ s^{-2}}[/tex]
The rotational speed, ω = ω₀ + α·t
Which gives;
[tex]\displaystyle \mathrm{The \ angular \ speed, } \ \omega = \frac{2 \cdot \pi \times 4 \ rad }{s} = \frac{8 \cdot \pi \ rad }{s}[/tex]
ω₀ = 8·π rad/s
Which gives;
[tex]\displaystyle \omega = \mathbf{8 \cdot \pi +2 \frac{5}{6} \times 10} = 233.47[/tex]
The speed of the wheel in revolution per second is therefore;
[tex]\displaystyle Speed \ in \ rev/s = \frac{8 \cdot \pi +2 \frac{5}{6} \times 10}{2\cdot \pi} \approx 37.16[/tex]
The speed after 10 seconds is approximately 37.16 rev/s.
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This is not a question
do your work in class you would know kids
A rocket ship has several engines and thrusters. While the Solid Rocket Booster (SRB) and main engines only work together during the first 2 minutes of flight, the main engines operate for a total of 8.5 minutes after the launch. Once the SRBs are released, the main engines alone accelerate the rocket from about 1341 m/s to 7600 m/s.
What is the acceleration of the SRB and main engine during the first 2.0 minutes of flight?
A. 52 m/s2
B. 13 m/s2
C. 9.8 m/s2
D. 11 m/s2
The acceleration of the SRB and main engine during the first 2.0 minutes of flight is 52.16 m/s².
The given parameters;
initial velocity of the engine, u = 1341 m/sfinal velocity of the engine, v = 7600 m/stime of motion, t = 2 minutes = 2 x 60 s = 120 sThe acceleration of the SRB and main engine is calculated as follows;
[tex]a = \frac{\Delta v}{\Delta t } \\\\a = \frac{7600 - 1341}{2 \times 60 s} \\\\a = 52.16 \ m/s^2[/tex]
Thus, the acceleration of the SRB and main engine during the first 2.0 minutes of flight is 52.16 m/s².
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6) The modern atomic model is called the _____.
A) central atom model
B) plum pudding model
C) nuclear atomic model
D) electron growth model
The modern atomic model is called the nuclear atomic model.
Therefore the correct answer is option C.
What are atomic models?There are many types of atomic models proposed in past based on their individual assumptions and the experimentations
John Dalton's atomic model.
The Plum Pudding Model, developed by J.J. Thomson,
Rutherford's model provided an explanation for the presence of a nucleus inside the atom.
In accordance with the current atomic model, atoms have a nucleus made up of protons and neutrons and an ethereal gradient or cloud surrounding them that houses electrons;
The modern atomic model is called the nuclear atomic model.
Thus, the correct answer is option C.
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Goal statement write a positive statement that includes your desired outcome and your imagine preferred future after development
A goal statement may involve how to develop physical skills in a given outdoor activity (e.g., soccer) in order to reach my maximum potential. It involves a statement of purpose.
A goal statement is a statement of purpose that describes long-term professional and/or motivational objectives.
A goal statement involves writing a statement about how to reach personal objectives, i.e., how to accomplish a career, job title, sports skills, educational training, etc.
An excellent statement of purpose should ideally be clear, specific, acceptable, time-bound and realistic.
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n which case will a beat frequency most likely occur?(1 point) two different instruments playing notes at the same frequency two different instruments playing notes at the same frequency two of the same instrument playing notes at the same frequency two of the same instrument playing notes at the same frequency two of the same instrument playing notes at slightly different frequencies two of the same instrument playing notes at slightly different frequencies two different instruments playing notes at very different frequencies
Answer:
two of the same instrument playing notes at slightly different frequencies
Explanation:
The correct answer is two of the same instrument playing notes at slightly different frequencies.
What is the frequency?Frequency is the number of occurrences of a periodic event per unit of time. In the context of sound waves, frequency refers to the number of complete cycles of a sound wave that occur in one second and is measured in units of Hertz (Hz).
In this question,
A beat frequency occurs when two sound waves with slightly different frequencies interfere with each other, resulting in a periodic variation in the amplitude of the resulting wave. The beat frequency is equal to the difference in frequency between the two original sound waves.
When two of the same instrument play notes at slightly different frequencies, the sound waves produced by each instrument will have slightly different frequencies due to differences in tuning or other factors. When the two sound waves interfere with each other, they will produce a beat frequency equal to the difference between their frequencies. This beat frequency will be audible as a periodic variation in the loudness or intensity of the sound.
In the other statements, There will not be any beat frequency because the sound waves being produced have either the same frequency or very different frequencies, which do not interfere with each other in a way that produces beats.
Therefore, The statement two of the same instrument playing notes at slightly different frequencies is correct.
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1- a car speeds up to get onto the freeway. it goes from 21 m/s to 39 m/s in 4.1 seconds. How far did it travel??
2- a boulder fell off a cliff and fell for 4.1 seconds. How tall was the cliff?
Answer:
Explanation:
1) average velocity is
v = (21 + 39)/2 = 30m m/s
d = vt 30(4.1) = 123 = 120 m
2) d = ½gt²
d = ½(9.8)(4.1²)
d = 82.369 = 82 m
when rounding to the two significant digits of the question numerals.
A 5 kg box is sitting on a rough wooden surface. The coefficient of static friction between the box and surface is 0.6. If the normal force on the box is 50 N, calculate the force of friction which must be overcome to move the box. Round your answer to the nearest whole number.
The force of friction needed to overcome to move the box is 29.4N
According to Newton's second law;
[tex]\sum F_x = ma_x\\[/tex]
Taking the sum of force along the plane;
[tex]F_m -F_f = ma\\F_m -F_f = 0\\F_m=F_f = \mu R[/tex]
This shows that the moving force is equal to the frictional force
Given that
[tex]\mu = 0.6\\R = mg = 49N[/tex]
Get the frictional force;
Since
[tex]F_f = \mu R\\F_f = 0.6 \times 49\\F_f = 29.4N[/tex]
Hence the force of friction needed to overcome to move the box is 29.4N
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A friend has suggested that you go swimming in a pool having water of temperature 350 K. What would this temperature be on the Fahrenheit scale?
109°F
123°F
170°F
202°F
This temperature would be 170° F on the Fahrenheit scale. Hence, option (C) is correct.
What is temperature?The physical concept of temperature indicates in numerical form how hot or cold something is. A thermometer is used to determine temperature. Thermometers are calibrated using a variety of temperature scales, which historically defined distinct reference points and thermometric substances.
The most popular scales are the Celsius scale, sometimes known as centigrade, with the unit symbol °C, the Fahrenheit scale (°F), and the Kelvin scale (K), with the latter being mostly used for scientific purposes.
the relation between Kelvin scale and Fahrenheit scale is given by:
(F - 32)/180 = (K - 273)/100
F - 32 = (350 - 273)(9/5)
F = 32 + (350 - 273)(9/5)
F = 170
Hence, this temperature would be 170° F on the Fahrenheit scale.
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A branch falls from a tree. How fast is the branch moving after 0.28 seconds?
A. 2.7 m/s
B. 1.3 m/s
C. 4.4 m/s
D. 3.1 m/s
Answer:
A. 2.7 m/s
Explanation:
Answer:
[tex]\boxed {\boxed {\sf A. \ 2.7 \ m/s}}[/tex]
Explanation:
We want to find how fast a branch is falling after 0.28 seconds.
Essentially, we want to find its final velocity at exactly 0.28 seconds. We will use the following kinematic equation:
[tex]v_f= v_i+at[/tex]
The branch fell from the tree, so it initially started at rest or 0 meters per second. The branch is in free fall, so its acceleration is due to gravity, or 9.8 meters per second squared. It falls for 0.28 seconds.
[tex]v_i[/tex]= 0 m/s a= 9.8 m/s²t= 0.28 sSubstitute the values into the formula.
[tex]v_f= 0 \ m/s + (9.8 \ m/s^2)(0.28 \ s)[/tex]
Multiply the numbers in parentheses.
[tex]v_f= 0 \ m/s +(9.8 \ m/s/s * 0.28 \ s )[/tex]
[tex]v_f= 0 \ m/s +2.744 \ m/s[/tex]
Add.
[tex]v_f= 2.744 \ m/s[/tex]
The original measurement of time has 2 significant figures, so our answer must have the same. For the number we found, that is the tenth place. The 4 in the hundredth place tells us to leave the 7.
[tex]v_f \approx 2.7 \ m/s[/tex]
The branch is moving at a velocity of approximately 2.7 meters per second.
a body of mass 15 kg accelerates from rest of the rate of 4.0 ms^-2. determine the distance with the body travel in 25 seconds
The distance traveled by the body in the given time is 1,250 m.
The given parameters;
mass of the body, m = 15 kgacceleration of the body, a = 4 m/s²time of motion, t = 25 sinitial velocity, u = 0The distance traveled by the body in the given time is calculated as follows;
[tex]s =ut + \frac{1}{2} at^2\\\\s = 0 \ + \ \frac{1}{2} (4)(25^2)\\\\s =1,250 \ m[/tex]
Thus, the distance traveled by the body in the given time is 1,250 m.
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what two things make up an ionic bond?
A 10kg crate sits at rest on a rough flat surface. Astudent decides to pull the crate by attaching a rope at a 37 degree angle. Although the student pulls the rope with a force of 600 newtons, the coefficient of kinetic friction is large and has the force the student applies remains constant, how much time after he begins pulling the crate will it take before the crate has traveled a distance of 1.0 meter?
Answer:
Explanation:
Normal force of the surface on the box will be
N = mg - Fsinθ
Ν = 10(9.8) - 600sin37
N = -263
As normal force cannot be less than zero, the applied force lifts the crate off the surface.
Now it's just a matter of finding the acceleration
In the horizontal direction, the acceleration is
a = F/m
a = (600cos37) / 10
a = 47.9181... m/s²
the crate weight is mg = 10(9.8) = 98 N.
In the vertical direction the acceleration is
a = ((600sin37 - 98) / 10)
a = 26.3089... m/s²
total acceleration is
a = √(47.9181² + 26.3089²)
a = 54.6653... m/s²
s = ½at²
t = √(2s/a)
t = √(2(1.0)/54.6653)
t = 0.19127...
t = 0.19 s
A ball is thrown up onto a roof, landing 4 sec later at height of 20m above the release level. The balls path just before landing is angled at 60 degree with the roof.
a) find the horizontal distance d it travels.
b) what is the magnitude of the balls initial velocity?
c) what is the angle (relative to the horizontal) of the balls initial velocity?
Answer: 20
Explanation:
Please help, I keep trying a bunch of things but keep getting them wrong. I don't know where I am going wrong here.
1. Boyle's Law states the volume and pressure of a gas are inversely proportional.
Name the three units of the constant of proportionality between pressure and volume in alphabetical order. (**I have the first two)
2. The ideal gas law can be written as (PV/nT=R). Name the units for R.
The units of the constant of proportionality between pressure and volume in alphabetical order are
1. Celsius (°C)
2. Fahrenheit (°F)
3. Kelvin (K)
The units for R, that is, the ideal gas constant are
1. J K⁻¹ mol⁻¹
2. L atm K⁻¹ mol⁻¹
We will start by completing the Boyle's Law stated
Boyle's Law states the volume and pressure of a gas are inversely proportional, provided that the temperature remains constant.
This means temperature is the constant of proportionality.
Now, we will name the three units of the constant of proportionality, that is, temperature. The units are
1. Degree Celsius (°C)
2. Degree Fahrenheit (°F)
3. Kelvin (K)
2. In the ideal gas equation (PV/nT=R), R represents the ideal gas constant.
The units for R, that is, the ideal gas constant are
1. J K⁻¹ mol⁻¹
2. L atm K⁻¹ mol⁻¹
Hence,
The units of the constant of proportionality between pressure and volume in alphabetical order are
1. Celsius (°C)
2. Fahrenheit (°F)
3. Kelvin (K)
The units for R, that is, the ideal gas constant are
1. J K⁻¹ mol⁻¹
2. L atm K⁻¹ mol⁻¹
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3) A force of magnitude Fx acting in the x-direction on a 2.00 kg particle varies in time as shown
in FIGURE 2. Find
a) The impulse of the force
b) The final velocity of the particle if it is initially at rest
c) The final velocity of the particle if it is initially moving along the x-axis with velocity
of -2.00 ms -1
Answer:
a) Impuise of force =F∗?(t) = area of F-T graph area= impulse =triangle + rectangle + triangle = 0.5*4*2 + 4*1 + 0.5*4*2 = 12 N-s (b) impulse = change in momentum \(= mExplanation:
Question: A NEO distance from the Sun is 1.18 AU. What is its relative speed compared to Earth (round your answer to 3 decimal places)
Its relative speed compared to Earth is 0.921
The speed of the object v = 2πr/T where r = radius of orbit and T = period of orbit.
Let v = speed of earth, r = radius of earth orbit = 1 AU and T = period of earth orbit.
So, v = 2πr/T
Also, v' = speed of NEO, r' = radius of NEO orbit = distance of NEO from sun = 1.18 AU and T' = period of NEO orbit.
So, v' = 2πr'/T'
v'/v = 2πr'/T' ÷ 2πr/T
v'/v = r'/r × T/T'
From Kepler's law, T² ∝ r³
So, T'²/T² = r'³/r³
(T'/T)² = (r'/r)³
T'/T = √[(r'/r)]³
T/T' = √[(r'/r)]⁻³
So, substituting this into the equation, we have
v'/v = r'/r × T/T'
v'/v = r'/r × √[(r'/r)]⁻³
v'/v = √[(r'/r)]⁻¹
Since r' = 1.18 AU and r = 1 AU, r'/r = 1.18
So, v'/v = √[(r'/r)]⁻¹
v'/v = √[(1.18)]⁻¹
v'/v = [1.0863]⁻¹
v'/v = 0.921
So, its relative speed compared to Earth is 0.921
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An object accelerates from rest, with a constant acceleration of 6.4 m/s2, what will its velocity be after 7s?
I also need the Formula
Hi there!
The formula for velocity given acceleration:
v = at
Plug in given values:
v = 6.4(7) = 44.8 m/s
If a battery causes a wire to carry a current of 4 Amps how many coulombs of charge flow past any point in the wire in 3 seconds
[tex]▪▪▪▪▪▪▪▪▪▪▪▪▪ {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪[/tex]
According to above question ~
Current (I) = 4 AmperesTime (t) = 3 seconds Charge (q) = ?Let's find the charge (q) by using formula ~
[tex]I = \dfrac{q}{t} [/tex][tex]4 = \dfrac{q}{3} [/tex][tex]q = 4 \times 3[/tex][tex]q = 12 \: \: coulombs[/tex]Hence, 12 coulombs of charge flow past any point in the wire in 3 seconds
từ độ cao h=2m một vật bắn lên với vận tốc ban đầu V0=10(m/s), hợp với phương ngang 1 góc 30 độ.Hãy xác minh
a. Thời gian chuyển động của vật ?
b. Độ lớn vận tốc tại điểm chạm đất
Answer:
A IS THE ANSWER
Explanation:
HOPE IT HELPS AND PLEASE MARK AS BRAINLIST
Check Pic please, need help immediately
what is the velocity of this graph between points a and b? 0.0m/s 2.5m/s 5.0m/s 6.0m/s?
Answer:
Pick c is the right one
Explanation:
5.0m/s
please answer this as fast as you can i need it
Answer:
it says pdf only i dont knowwhat u want me to do
an observer sees two spaceships flying apart with speed .99c. What is the speed of one spaceship as viewed by the other
Answer:
V2 = (V1 - u) / (1 - V1 u / c^2)
V1 = speed of ship in observer frame = .99 c to right
u = speed of frame 2 = -.99 c to left relative to observer
V2 = speed of V1 relative to V2
V2 = (.99 - (-.99 ) / (1 - .99 (-.99)) c
V2 = 1.98 / (1 + .99^2) c = .99995 c
Objects 1 and 2 attract each other with a gravitational force of 178 units. If the mass of object 1 is one-fourth the original value AND the mass of object 2 is tripled AND the distance separating objects 1 and 2 is halved, then the new gravitational force will be _____ units.
Explanation:
Fgravity = G*(mass1*mass2)/D²
G is the gravitational constant, which has the same value throughout our universe.
D is the distance between the objects.
now, several numbers change.
Fgravitynew = G*((1/4)*mass1*3*mass2)/(1/2 * D)² =
= G*((3/4)*mass1*mass2)/(D²/4) =
= (3/4)* (G*(mass1*mass2)/D²) *4 =
= 4*(3/4)* (G*(mass1*mass2)/D²) =
= 3* (G*(mass1*mass2)/D²) = 3* Fgravity
the new gravitational force will be 3×178 = 534 units.
The elevation at the base of a ski hill is 350 m above sea level. A ski lift raises a skier (total mass=72 kg, including equipment) to the top of the hill. If the skier's gravitational potential energy relative to the base of the hill is now 9.2 x 105 J, what is the elevation at the top of the hill?
The elevation at the top of the hill is 1,653.85 m.
The given parameters;
initial height of the skier, h₁ = 350 mlet the final height of the skier at the hill top, = h₂total mass, m = 72 kggravitational potential energy of the skier, P.E = 9.2 x 10⁵ JThe elevation at the top of the hill is calculated as follows;
[tex]P.E = mg\Delta h\\\\P.E = mg(h_2 -h_1)\\\\h_2 -h_1 = \frac{P.E}{mg} \\\\h_2 = \frac{P.E}{mg} + h_1\\\\h_2 = \frac{9.2 \times 10^5 }{72 \times 9.8} \ + \ 350 \ m\\\\h_2 = 1,653.85 \ m[/tex]
Thus, the elevation at the top of the hill is 1,653.85 m.
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A crane is lifting a 500 lb car. If the power of the crane 1.82 hp, find the velocity of the car.
Answer:
Explanation:
550 ft•lb/s / hp•(1.82 hp) / 500 lb = 2.00 ft/s
A 0.60-kgkg ball, attached to the end of a horizontal cord, is rotated in a circle of radius 1.3 mm on a frictionless horizontal surface. Part A If the cord will break when the tension in it exceeds 60 NN , what is the maximum speed the ball can have
11.4 m/s
Explanation:
The cord will break when the centripetal force exerted on it meets or exceeds the maximum tension [tex]T_{max}[/tex] that it can handle.
[tex]T_{max} = m\dfrac{v_{max}^2}{r}[/tex]
Solving for [tex]v_{max},[/tex] we get
[tex]v_{max}^2 = \dfrac{rT_{max}}{m}[/tex]
or
[tex]v_{max} = \sqrt{\dfrac{rT_{max}}{m}} =\sqrt{\dfrac{(1.3\:\text{m})(60\:\text{N})}{(0.6\:\text{kg})}}[/tex]
[tex]\:\:\:\:\:= 11.4\:\text{m/s}[/tex]