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]
The vertical position of the 100-kg block is adjusted by the screw activated wedge. Calculate the moment which must be applied to the handle of the screw to raise the block. The single thread screw has square threads with a mean diameter of 30 mm and advances 10 mm each complete turn. The coefficient of friction for the screw threads is 0.24, and the coefficient of friction for all the mating surfaces of the block and the wedge is 0.40. Neglect friction at the ball joint A
We have that for the Question "" it can be said that Calculate the moment which must be applied to the handle of the screw to raise the block is
M = 7.30 N.mFrom the question we are told
The vertical position of the 100-kg block is adjusted by the screw activated wedge. Calculate the moment which must be applied to the handle of the screw to raise the block. The single thread screw has square threads with a mean diameter of 30 mm and advances 10 mm each complete turn. The coefficient of friction for the screw threads is 0.24, and the coefficient of friction for all the mating surfaces of the block and the wedge is 0.40. Neglect friction at the ball joint A
Generally the equation for the Block is mathematically given as
[tex]\sum Fy=0[/tex]
[tex]981cos21.80 = R_2cos53.6\\\\R_2=1535N[/tex]
the equation for the Wedge is mathematically given as
[tex]\sum Fx=0\\\\1535cos36.4=Pcos21.8\\\\P=1331N[/tex]
the equation for the Screw is mathematically given as
[tex]\beta = tan^{-1}*\frac{L}{2*\pi*r} \\\\\beta = tan^{-1}*\frac{10}{2*\pi*(15)} \\\\\\beta = 6.06\\\\\theta = tan^{-1}*0.25 \\\\\theta = 14.04\\\\\\Therefore\\\\\theta + \beta = 20.1\\\\[/tex]
Therefore
[tex]M = Pr tan (\theta + \beta)\\\\M = 1331(0.015) tan20.09\\\\M = 7.30 N.m[/tex]
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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:
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
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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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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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.
A 115 kg hockey player, Adam, is skating east when he tackles a stationary 133 kg player, Bob. Afterward, they move at 1.35 m/s east. What was Adam's velocity before the collision? (Unit = m/s)
Answer:
Explanation:
Conservation of momentum
115v + 133(0) = (115 + 133)1.35
v = 2.911304...
v= 2.91 m/s east
Answer:
The velocity east is 2.91
Explanation:
Fill in the box
2.91
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:
Which properties make a metal a good material to use for electrial wires
Answer:
Most importantly metals can pass an electric current without being affected and changed by the electricity. Electrical conductivity combined with ductility makes metals the most suitable materials for electrical transmission wires.
given two vector
p= 2i + 2j + 4k
q = i - 4j + 4k
find p+ q
Answer:
3i - 2j + 8k
Explanation:
p + q = (2i + i) + (2j - 4j ) + (4k + 4k )
= 3i -2j + 8k
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
A truck moves 60 km West, and then 80 km North, and then
travels in a straight line back to its starting point. The distance
travelled by the truck is ____km and its displacement is _____km
[tex]▪▪▪▪▪▪▪▪▪▪▪▪▪ {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪[/tex]
Distance travelled by the truck is ~
[tex] \boxed{240 \: \: km}[/tex]And it's displacement is ~
[tex] \boxed{0 \: \: km}[/tex][tex] \large \boxed{ \mathfrak{Step\:\: By\:\:Step\:\:Explanation}}[/tex]
See the diagram in attachment for reference ~
Let O be the initial point, It travels 60 km towards west till point B and then 80 km towards north till point P and returns to initial point O in a straight line, now as we can observe here, it forms a right angled Triangle.
The measure of two legs is 60 km and 80 km, let's find the hypotenuse ~
According to Pythagoras theorem ~
hypotenuse² = sum of squares of other two legs
that is ~
[tex]h {}^{2} = 60 {}^{2} + 80 {}^{2} [/tex][tex] {h}^{2} = 3600 + 640 0[/tex][tex]h {}^{2} = 10000[/tex][tex]h = \sqrt{10000} [/tex][tex]h = \sqrt{100 \times 100}{}[/tex][tex]h = 100 \: \: km[/tex]So, the distance between the point A and O is 100 km
Now, The total distance is equal to the distance covered through actual path that is ~
60 km + 80 km + 100 km 240 kmAnd displacement is the distance between the final point and initial point, but since the truck returns to the point from where it started the journey, so the final and initial point is same therefore displacement is equal to 0.
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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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
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
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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Check Pic please, need help immediately
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 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 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 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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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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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 object moving at a constant velocity of 5.4 m/s travels for 12 s. How far will it move during that time?
Free-fall Acceleration is -10 m/s^2
I also need the formula
Answer:
Explanation:
s = s₀ + v₀t + ½at²
s = 0 + 5.4(12) + ½(0)12²
s = 64.8 m
you are standing ata known distance from the statue of liberty describe how you could determine its height using only a meter stick
Use the meter stick to measure your height to the level of your eyes, then use trigonometry ratio formula to calculate the height from your eyes' level and above of the statue of liberty with angle of elevation, then, add the two heights.
If you are standing at a known distance from the statue of liberty, a meter stick can be used to measure your known distance away from the statue of liberty.
To determine its height using only a meter stick, the angle at which you look at the peak of statue of liberty must be measured or known. The height of the statue of liberty can be calculated if you know the angle of elevation at which you look at the peak of the statue, and the availability of the meter stick.
Use the meter stick to measure your height to the level of your eyes, then use trigonometry ratio formula to calculate the height from your eyes' level and above of the statue of liberty. That is,
Tan Ф = opposite / adjacent
Tan Ф = H/d
H = d x TanФ
Where
H = calculated height from the eyes level and above
Ф = angle of elevation
d = known distance away from the statue
Let h = Your measured height of your body to the eyes level
Then,
The height of the statue of liberty = H + h
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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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Convert :
36°C = ... °F
373 K = ... °C
Question easy
Answer:
36 C= 96.8 F
373 K= 99.85
Explanation:
C to F: (36 x 1.8) + 32
= 64.8 +32
= 96.8 F
K to C: C= K- 273.15
C= 373-273.15
C= 99.85
____
= 36°C
=( 36 × 9/5 ) + 32
=(36 ÷ 5 × 9) + 32
=(7,2 × 9) + 32
= 64,8 + 32
= 96,8°F______
______
= 373 K
= 373 - 273
= 100°C[tex] \boxed { \sf semoga \: membantu \: :v }[/tex]
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
This is not a question
do your work in class you would know kids
An object will begin moving from rest when acted upon by which forces?
A. Forces that are slightly less than the force of friction
B. Forces that result in a net force of zero
C. Forces that are equal and act in opposite directions
D. Forces that are greater in one direction than in any other direction
Answer:
D
Explanation:
Process of elimnination + it's the only one that makes sense
An object will begin moving from rest when acted upon by forces that are greater in one direction than in any other direction. Hence, Option (D) is correct.
What is force?The definition of force in physics is: The push or pull on a massed object changes its velocity. An external force is an agent that has the power to alter the resting or moving condition of a body. It has a direction and a magnitude.
When forces that are greater in one direction than in any other direction, resultant will be unbalanced forces. Unbalanced forces are those acting on a body when the net force acting on the body is greater than zero. The body alters its state of motion when unbalanced forces act on it.
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