Answers
Related Questions
7 A 0.25 kg block oscillates on the end of a spring with a spring constant of 100 N/m. If the oscillation is started by elongating the spring 0.1m and giving the block a speed of 3 mis then the amplitude of the oscillation is:
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i
=(K+U)
f
0+
2
1
kA
2
=
2
1
mv
2
+
2
1
kx
2
2
1
(6.50 N/m)(0.100 m)
2
=
2
1
m(0.300 m/s)
2
+
2
1
(6.50 N/m)(5.00×10
−2
m)
2
3.25×10
−2
J=
2
1
m(0.300 m/s)
2
+8.12×10
−3
J
giving m=
9.0×10
−2
m
2
/s
2
2(2.44×10
−2
J)
=
0.542 kg
(b) ω=
m
k
=
0.542 kg
6.50 N/m
=3.46 rad/s
Then, T=
ω
2π
=
3.46 rad/s
2π rad
=
1.81 s
(c) a
max
=Aω
2
=(0.100 m)(3.46 rad/s)
2
=
1.20 m/s
2
The use of non-renewable energy resources in the UK has changed in the last 30 years.
Explain how the use of energy resources has changed in the last 30 years.
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Answer: Varies
Explanation: It became more relied upon, technology is responsible for this.
Do you expect the normal dicrystalline resistance in the forward bias to be pure ohmic resistance or what ?!
Answers
Answer:
yes
Explanation:
because is pure ohmic risistance
which of the following is used to transport sound waves
A.medium
B.vacuum
C.mass
D.light
Answers
I think the answer is d.right?
A bear slides down an incline which is oriented at 17 degrees above horizontal. What is the coefficient of friction between the bear and the incline if the bear moves with a constant velocity?
Answers
[tex]\mu = 0.306[/tex]
Explanation:
Assume that the direction down the incline is the +x-direction. We can apply Newton's 2nd law along the x-axis as
[tex]x:\:\:\:\:\:mg\sin17° - f_s = ma[/tex]
[tex]\Rightarrow mg\sin17° - \mu N = 0[/tex]
where m is the mass of the bear and g is the acceleration due to gravity and acceleration a is zero since the bear is moving with a constant velocity. Along the y-axis, we can write Newton's 2nd law as
[tex]y:\:\:\:\:\:N - mg\cos17° = 0 \Rightarrow N = mg\cos17°[/tex]
Combining these two equations together, we get
[tex]mg\sin17° = \mu(mg\cos17°)[/tex]
Solving for [tex]\mu, [/tex]
[tex]\mu = \dfrac{\sin17°}{\cos17°} = \tan17° = 0.306[/tex]
Please help if u can thanksss
Answers
Answer:
yes its okay i think so it is the correct answer
Describe how you can determine:
a) Volume of an irregular body
b) Density of a liquid
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Density of liquid try thank you so much
Answer:
a) measure the change in volume when the object is immersed; compute from range data
b) find the ratio of mass to volume for a measured mass and volume
Explanation:
a) The volume of a small enough irregular body can be found by measuring the difference in volume of the (semi-)fluid in which it is immersed, before and after immersion.
For irregular bodies for which that approach does not work, various 3D scanners are available for measuring volume and surface area. They may rely on optical (laser or camera), sonic, or radar measurements, and generally involve computation from distances to various points.
__
b) Density is the ratio of mass to volume. So, measurements of mass and volume of a liquid sample are sufficient to provide the basis for determining density.
Other methods include measuring buoyancy forces, and/or the depth of submersion of something that floats in the liquid. For specific liquids, hydrometers are available for measuring their density relative to that of water.
The refractive index of water with respect to air is 5/3 and that of glass is 3/2.What is the refractive index glass with respect to water?
Answers
Answer:
9/8
Explanation:
A photographer wants to determine the color of light he can use in the darkroom that will not expose the films he is processing. In one trial, he used a blue incandescent bulb. Which bulb can he use for another trial?
A. Red incandescent bulb C. Red fluorescent bulb
B. Blue incandescent bulb D. Blue fluorescent bulb
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For a photographer that wishes to determine the color of light that he can use in a dark room that will not expose the films he is processing, having used a Blue Incandescent bulb, he should proceed to use a Red Incandescent bulb for the next trial.
The photographer in question is performing an experiment. For these kinds of experiments it is important to identify the variables present, which can be of three kinds:
Control variablesDependent variables Independent variablesFor this experiment, the dependent variable is the exposure of the light onto the films, given that this is what we wish to measure. The independent variable will be the color of the light being used which is what will affect the dependent variable.
The remaining variable must be the control variable. Unlike the previous variables, we can have more than one of these. The control variable is there to make sure that only the dependent variable is affecting the outcome. We do this by keeping the control variable the same through each trial, which is why the photographer should not change the type of bulb in the second experiment, changing only the color of the light.
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Which of the following correctly compares gravitational force and distance between two objects?
A. As the distance increases, the gravitational force decreases.
B. As the distance decreases, the gravitational force decreases.
C. There is no relationship between gravitational force and distance between two objects.
D. As the distance increases, the gravitational force increases.
Help pls…
Answers
Answer:
a. as the distances increases , the gravitational force decreases
An airplane has a mass of 33,000 kg and takes off under the influence of a constant net force of 44,000 N. What is the net force that acts on the plane's 70 kg pilot
Answers
Answer:
Explanation:
acceleration a = F/m = 44000/33000 = 1.3m/s²
F = ma = 70(1.3) = 93 N
PLEASE PLEASE HELP!!
An arrow is shot straight up in the air from the ground with an initial velocity of 54.0 m/s. If on striking the ground it
embeds itself 15.0 cm into the ground, what is the acceleration required to stop the arrow when it hits the ground?
Answers
Answer:
you have patience the distance.
Explanation:
the train leaves at 6.30.
Required information
Medical testing has established that the maximum acceleration a pilot can be subjected to without losing consciousness is
approximately 5.00g. A pilot can avoid "blackout" at accelerations up to approximately 9.00g by wearing special "g-suits"
that help keep blood pressure in the brain at a sufficient level.
What is the minimum safe radius of curvature for an unprotected pilot flying an F-15 in a horizontal circular loop at 729 km/h?
Answers
Answer:
hi there is that OK for the weekend of the following week as well
Explanation:
6th of March is fine for me
The minimum safe radius of curvature for an unprotected pilot flying an F-15 in a horizontal circular loop at 729 km/h is approximately 838.1 meters.
To determine the minimum safe radius of curvature for an unprotected pilot flying an F-15 in a horizontal circular loop, we need to consider the maximum acceleration the pilot can withstand without losing consciousness.
Given:
Maximum acceleration without losing consciousness = 5.00g
Acceleration with g-suits to avoid blackout = 9.00g
First, we need to convert the speed of the F-15 from km/h to m/s:
Speed = 729 km/h = (729 * 1000) m/3600 s ≈ 202.5 m/s
Next, we'll calculate the acceleration experienced by the pilot in the circular loop. In a horizontal circular motion, the centripetal acceleration is given by:
Acceleration = ([tex]\rm Velocity^2[/tex]) / Radius
We can rearrange the equation to solve for the radius:
Radius = ([tex]\rm Velocity^2[/tex]) / Acceleration
Using the maximum acceleration of 5.00g, we convert it to [tex]\rm m/s^2[/tex]:
Maximum acceleration = 5.00g ≈ (5.00 * 9.8) [tex]\rm m/s^2[/tex] = 49 m/s^2
Now, we can calculate the minimum safe radius of curvature:
Radius = ([tex]\rm 202.5^2[/tex]) / 49 ≈ 838.1 meters
Therefore, the minimum safe radius of curvature for an unprotected pilot flying an F-15 in a horizontal circular loop at 729 km/h is approximately 838.1 meters.
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PLEASE HELP!!!
If a 40cm rope with a 220g bob can hold a maximum tension of 3N
a) what are the maximum angular velocity and inclination angle it can reach before the rope break?
b) Angle of inclination
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Answer:
ω = 3.1 rad/s
θ = 36° from vertical
Explanation:
I will ASSUME that the bob and string is acting as a pendulum.
Please understand that the string will break when the bob is at the lowest point of the swing where the vectors of gravity and centripetal acceleration align. It will NOT break at the angle of maximum inclination measured from vertical. This angle is only a component of the maximum potential energy that gets converted to maximum kinetic energy at the lowest point of the swing.
At the bottom of the swing, the string must support the weight of the bob plus supply the required centripetal acceleration.
F = mg + mω²R
F/m = g + ω²R
F/m - g = ω²R
ω = √((F/m - g)/R)
ω = √((3/0.220 - 9.8)/0.40)
ω = 3.09691...
ω = 3.1 rad/s
Potential energy will convert to kinetic energy
mgh = ½mv²
h = v²/2g
R - Rcosθ = v²/2g
R(1 - cosθ) = v²/2g
1 - cosθ = v²/2gR
cosθ = 1 - v²/2gR
cosθ = 1 - (Rω)²/2gR
cosθ = 1 - Rω²/2g
cosθ = 1 - 0.40(3.1²)/(2(9.8))
cosθ = 0.804267
θ = 36.46045...
θ = 36°
Change 1m2 in to cm2, mm2 and km2
Answers
Answer:
1m² = 10000cm²
1m² = 1000000mm²
1m² = 1 × 10^-6
Explanation:
When coverting from square meter to square centimetre, multiply the area value by 10.
When coverting from square meter to millimetre, multiply the area value by 10⁶ ( 1 million )
When converting from meter square to kilometre square, divide the area value by 10^-6 ( 0.000001)
A 770-kg two-stage rocket is traveling at a speed of 6.90×103 m/s away from Earth when a predesigned explosion separates the rocket into two sections of equal mass that then move with a speed of 2.60×103 m/s relative to each other along the original line of motion.What is the speed of each section (relative to Earth) after the explosion?How much energy was supplied by the explosion?
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Answer:
Explanation:
Let's just have our reference frame travel along with the original un broken mass. This way the original velocity is not relevant.
Each half will have a mass of 770/2 = 385 kg
Each half will have the same magnitude of velocity (conservation of momentum) which will be 2.6 x 10³/2 = 1.30 x 10³ m/s
Now add back the reference frame velocity to get velocity relative to earth.
Section one will have velocity 6.90 x 10³ + 1.30 x 10³ = 8.2 x 10³ m/s
Section two will have velocity 6.90 x 10³ - 1.30 x 10³ = 5.6 x 10³ m/s
In the moving reference frame, each half will have kinetic energy which could only come from the explosion
KE = ½(385)(1.3 x 10³)² = 325,325,000 J
2(325,325,000) = 650,650,000 J released in the explosion.
Rounding to the three significant figures of the problem numerals
E = 6.50 x 10⁸ J or 650 MJ released
What does it mean that " memory is organized in semantic
networks
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Answer:
Semantic memory is a category of long-term memory that involves the recollection of ideas, concepts and facts commonly regarded as general knowledge. Examples of semantic memory include factual information such as grammar and algebra.
HELP PLS! :/
A ball thrown straight up takes 1.89s to reach a height of 41.6m.
Taking g =9.81 m/s2 the balls initial speed was:
Help plsss
Answers
Explanation:
formula: Vi = Vf - (at)
Vi: intial velocity
Vf: final velocity
a: acceleration
t: time
fill in formula with the numbers you are given
Vi= 41.6m - ((9.81 m/s^2)(1.89s))
parenthesis first according to pemdas
Vi= 41.6m - 18.54m/s
Answer: 23.06m/s
disclaimer: I havent done physics in awhile so I have no idea if this is right. just an attempt to help steer you in the right direction hopefully. good luck
An astronaut travels to a star system 6.5 ly away at a speed of 0.80c . Assume that the time needed to accelerate and decelerate is negligible.How much time elapses in earth between the launch and the arrival of the first radio message from the astronaut saying that she has arrived? please 911
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Answer:
14.625 years
Explanation:
Because acceleration/deceleration are negligent we can simplify this problem into 2 basic parts:
First: how long before the astronaut arrives? Since the speed of the craft is 0.8 times the speed of light and the distance to travel is 6.5 light-years, the equation to answer this part is:
0.8t=6.5
solve for t by dividing by 0.8
t=6.5/0.8 or 8.125
so, she will arrive in 8.125 years.
Then, once she sends her message, it will travel towards earth at 1 times the speed of light or:
1.0t=6.5 which is just t=6.5
so, her message will arrive back to Earth (8.125+6.5) 14.625 years after takeoff.
The time elapsed in earth will be 14.625 years.
To understand this we need to find how much time she took to travel 6.5 light years and then how much time the signal took to reach the earth.
How much time she took to travel 6.5 light year?0.8t = 6.5 ( because 1 light year = distance travelled by light in a year)
t = 6.5/0.8
t = 8.125 years
So, she travelled for 8.125 year.
How much time it took for the signal?1.0 t = 6.5 ( because radio signal travells with the same speed of light)
t = 6.5/1.0
t = 6.5 years
Sum of these two gives us how much time elapsed on earth between two mentioned events.So, 8.125 + 6.5 = 14.625 years
So, the time elapsed on earth between the launch and the arrivel of the first radio signal is 14.625 years.
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5. A quarterback throws the football to a stationary receiver who is 31.5 m
down the field. If the football is thrown at an initial angle of 40.0° to the
ground, at what initial speed must the quarterback throw the ball for it
to reach the receiver? What is the ball's highest point during its flight?
Answers
The projectile launch equations allow to find the results for the questions about the movement of the ball are:
The initial velocity is: v₀ = 17.7 m / s. The maximum height is: y = 16 m.
Given parameters
Horizontal distance x = 31.5 m Launch angle tea = 40ºTo find
The initial speed. Maximum height.
Projectile launching is an application of kinematics, where on the x-axis there is no acceleration and on the y-axis is the gravity acceleration.
The range is the distance traveled for the same departure height, see attached.
.
R =[tex]\frac{v_o^2 \ sin 2\theta}{g}[/tex]
[tex]v_o^2 = \frac{ g R}{sin 2 \theta }[/tex]
Let's calculate.
v₀² = [tex]\frac{9.8 \ 31.5}{sin \ (2 \ 40)}[/tex]9.8 31.5 / sin (2 40.0)
[tex]v_o = \sqrt{313.46}[/tex]o = ra 313.46
v₀ = 17.7 m / s
At the point of maximum height the vertical speed is zero.
v² = v₀² - 2 g y
0 = v₀² - 2g y
y = [tex]\frac{v_o^2}{2g}[/tex]
Let's calculate.
y = [tex]\frac{17.7^2}{2 \ 9.8}[/tex]
y = 16 m
In conclusion, using the projectile launch equations we can find the results for the questions about the movement of the ball are:
The initial velocity is: v₀ = 17.7 m / s The maximum height is: y = 16 m.
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An object travels 120 m in 6 seconds. How fast is it moving?
Answers
Answer:
I actually don't understand what you are asking for....but if you denote second then...
Explanation:
Just divide 120/6=20
Hence, it's traveling 20m per second.
Answer:
20 m/s
Explanation:
Speed = Distance/Time
Speed = 120m/6s
Speed = 20 m/s
How can gravity's role in tectonic plate motion be described?
Answers
Answer: In ridge push, the mantle wells upward because of the convection and elevates the edges of spreading oceanic plates. Because these plates are higher at the spreading center, they are forced downhill due to gravity and eventually flatten out to the ocean floor.
the answer I need points for my math test
If a car travels 10 min 2 seconds, what is its speed?
A. 5 m/s
B. 0.2 m/s
C. 20 m/s
D. 2 m/s
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Does the stone remain at a constant speed? Or does it speed up?
Answers
Answer:
it would remain the same speed
Explanation: the rock isnt going down a hill or anything so therefore if gravity isnt pulling it down a slope then it would remain the same pace
50 points help
Column I Column II
______ 1. acceleration a. change in distance over time
______ 2. speed b. time interval
______ 3. velocity c. scalar
______ 4. Δt. d. change in position
______ 5. Magnitude only e. change in velocity over time
______ 6. Δx f. change in displacement over time
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[tex]\\ \sf\longmapsto Acceleration\longrightarrow Change\:in\: Velocity\:over\:time[/tex]
[tex]\\ \sf\longmapsto Speed\longrightarrow Change\:in\: Distance\:over\:Time[/tex]
[tex]\\ \sf\longmapsto Velocity\longrightarrow Change\:in\: Displacement\: over\:time[/tex]
[tex]\\ \sf\longmapsto ∆t\longrightarrow Time\: interval [/tex]
[tex]\\ \sf\longmapsto Magnitude\:only\longrightarrow Scaler[/tex]
[tex]\\ \sf\longmapsto ∆x=Change\:in\: position [/tex]
The figure shows the light intensity on a screen behind a double slit. The slit spacing is 0.22 mm and the screen is 2.0 m behind the slits (Figure 1). What is the wavelength of the light?
Answers
The wavelength of the light is 550 nm
For a double slit interference pattern with slit spacing, d we have
dsinθ = mλ where d = slit spacing = 0.22 mm = 0.22 × 10⁻³ m, m = number of maximum fringe = 2(from the picture) and λ = wavelength of light.
Thus sinθ = mλ/d
Also, tanθ = L/D where L = distance between central maximum and fringe = 2.0 cm/2 = 1.0 cm = 1 × 10⁻² m and D = distance between slit and screen = 2.0 m
Since θ is small, sinθ ≅ tanθ
So, mλ/d = L/D
Making λ subject of the formula, we have
λ = dL/mD
Substituting the values of the variables into the equation, we have
λ = dL/mD
λ = 0.22 × 10⁻³ m × 1 × 10⁻² m/(2 × 2.0 m)
λ = 0.22 × 10⁻⁵ m²/4.0 m
λ = 0.055 × 10⁻⁵ m
λ = 0.55 × 10⁻⁶ m
λ = 550 × 10⁻⁹ m
λ = 550 nm
So, the wavelength of the light is 550 nm
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A hot air balloon rising vertically is tracked by an observer located 2 miles from the lift-off point. At a certain moment, the angle between the observer's line-of-sight and the horizontal is , and it is changing at a rate of 0.1 rad/min. How fast is the balloon rising at this moment
Answers
We have that for the Question, it can be said that
the balloon rising at [tex]0.266miles/min[/tex]From the question we are told
An observer located 2 miles from the lift-off point. At a certain moment, the angle between the observer's line-of-sight and the horizontal is , and it is changing at a rate of 0.1 rad/min.From,
[tex]tan\theta = \frac{h}{2}[/tex]
differentiate with respect to h
[tex]sec^2\theta * \frac{do}{dz} = \frac{1}{2} * \frac{dh}{dz}\\\\\frac{dh}{dz} = 2 sec^\theta * \frac{d\theta}{dz}\\\\\theta = \frac{\pi}{6} and \frac{d\theta}{dz} = 0.1rad/min\\\\\frac{dh}{dz} = 2sec^2 (\frac{\pi}{6}) * (0.1)\\\\= 0.266miles/min[/tex]
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A car drives at a constant speed of 45km in 20mn. The speed of the car is..?
Answers
Answer:
83.8851 mi/h
Please write a paragraph explaining the bible verse below in your own words.
Exodus 16:1-3
Answers
Answer: Moses
Explanation: The Israelites are angry at Aaron and Moses, because they lead them into this desert where there is no food, or water. It was told that after the Israelites left Egypt they would wander the desert for forty years. The Israelites complain that they would have had plenty of food if they had just stayed enslaved in Egypt. They are not grateful that God has liberated them from slavery.
True or false. The buildings in cities are getting larger
Answers
Answer:
trueExplanation:
cuase the city is getting more people than before