Answer:
The data suggest that the true mean HAZ depth is larger when the current setting is higher.
Need help with dot product
[tex]\textbf{A}\cdot\textbf{B} = 11.5[/tex]
Explanation:
The dot product between two vectors [tex]\textbf{A}[/tex] and [tex]\textbf{B}[/tex] is defined as
[tex]\textbf{A}\cdot\textbf{B} = AB\cos{\theta}[/tex]
where A and B are the magnitudes of the vectors [tex]\textbf{A}[/tex] and [tex]\textbf{B},[/tex] respectively and [tex]\theta[/tex] is the angle between the two. Since A = 3, B = 5 and [tex]\theta = 40°,[/tex] the dot product [tex]\textbf{A}\cdot\textbf{B}[/tex] is
[tex]\textbf{A}\cdot\textbf{B} = (3)(5)(0.766) = 11.5[/tex]
A car travelling at 79.3 Km/h on a highway has 4.22x10 5 J of kinetic energy.
a. What is the mass of the car?
b. If brakes are applied with a force of 2100 N, what distance will it take for the car to slow down to a speed of 56 Km/h?
Answer:
[tex]1.74\times10^3 kg; 100m[/tex]
Explanation:
Step a: mass of the car. Let's grab the definition of kinetic energy: [tex]K= \frac12 mv^2[/tex]. We have K, we have v (which we should convert in meters per second, dividing by 3.6) to get:[tex]4.22\times10^5 = \frac12m(22.03)^2 \rightarrow m= 2\times4.22 / 495.22 \times 10^5 = 1.74 \times 10^3 kg[/tex]
Point a is done.
Now for the (b)reaking part. (I'm sorry, it's an horrible joke, but I couldn't resist)
In theory we have the mass, we have the force, so we could find the acceleration, find how long it takes to slow down, and then find the distance traveled. Too long. Let's do things more easily: when the car slows down to 56 km/h it will have a different kinetic energy. The difference in kinetic energy is the work done by the breaking force ofer the slowing distance.
[tex]K_f-K_i=W[/tex] A quick note on signs: if you look carefully the final kinetic energy will be less than the initial value, thus the work will be negative: it means it's correct, since the work is against the motion, slowing it down. Let's get calculating, first by converting 56 kmh in m/s (15,56 m/s), then finding the final kinetic energy:
[tex]K_f =\frac12 (1.74\times10^3) (15.56)^2 =2.11 \times 10^5 J[/tex]
The difference will be the work done by the force, or
[tex](2.11 - 4.22) \times 10^5 = \vec F\cdot \vec x=Fx[/tex] where we are assuming that force and displacement have the same line of actions to simplify the dot product.
[tex]2.11\times 10^5 = 2100x = 1.00\times 10^2 m[/tex]
If an electron moves in a direction perpendicular to the same magnetic field with this same linear speed,
what is the radius of its circular orbit?
Answer:
An effect begins to alter movement, and the direction of moves in the circular path is known as centripetal force. Its measurable unit is Newton or Kilogram meter per square of the second. The product of mass and square of velocity divided by the radius of path travel by the body provide s the term centripetal force.
Explanation:
Answer:
An effect begins to alter movement, and the direction of moves in the circular path is known as centripetal force. Its measurable unit is Newton or Kilogram meter per square of the second. The product of mass and square of velocity divided by the radius of path travel by the body provide s the term centripetal force.
Explanation:
An object is moving with an initial velocity of 3.3 m/s. It is then subject to a constant acceleration of 3.7 m/s2 for 10 s. How far will it have traveled during the time of its acceleration?
I also need the complete Formula (Nothing left out)
Answer:
Explanation:
s = s₀ + v₀t + ½at²
ASSUMING the acceleration is in the direction of initial motion.
s = 0 + 3.3(10) = ½(3.7)(10²)
s = 218 m
calculate the mass of a block of ice having volume 5m³. (density of ice≈920 kg/m³)
Answer:
4600kg
Explanation:
Density=mass÷volume
920=m/5
m=920×5=4600kg
Which is the main gas that makes up the Earth's atmosphere?
Answer:
78 percent nitrogenExplanation:
I hope it's helpful for you
Help pls!
A 3 kg mass is raised a distance of 14 m above the earth by a vertical force of 93 N.
The final kinetic energy of the mass, to 3 significant figures, if it was originally at rest is:
[tex] \large★·.·´¯`·.·★ {Answer}★·.·´¯`·.·★[/tex]
As we know that Kinetic Energy is the Energy that is possessed by a moving object. and if the object is at rest then it doesn't have velocity therefore there is no kinetic Energy.
In the numerical terms we can express it as : -
[tex] \sf0.00 \: \: joules[/tex][tex]꧁ \: \large \frak{Eternal \: Being } \: ꧂[/tex]
Conservation of Energy Roller Coaster A roller coaster cart of mass 100kg travels on a track with one loop. Fill in blanks A-H. А. KE=OJ PE=120000J h= А. V= B B KE=___CE PE=60000J h= _D V= E KE=__F PE=40000J h=__G_ V= KE= PE= h=Om v= K D E F G H K
(a) The height of the roller coaster at 120,000 potential energy is 122.45 m.
(b) The velocity of the roller coaster at 0 J kinetic energy is 0.
(c) The height of the roller coaster at 60,000 potential energy is 61.23 m.
(d) The velocity of the roller coaster at 60,000 J kinetic energy is 34.64 m/s.
(e) The height of the roller coaster at 40,000 potential energy is 40.82 m.
(f) The velocity of the roller coaster at 80,000 J kinetic energy is 40 m/s.
The given parameters:
mass of the roller coaster, m = 100 kgWhen the kinetic energy = 0 and potential energy = 120,000 J
The height of the roller coaster is calculated as follows;
P.E = mgh
[tex]h = \frac{P.E}{mg}\\\\h = \frac{120,000}{100 \times 9.8} \\\\h = 122.45 \ m[/tex]
Since the kinetic energy = 0, the velocity of the roller coaster = 0
When the potential energy, P.E = 60,000 J, the kinetic energy, K.E is calculated as;
P.E + K.E = M.A
P.E + K.E = 120,000
60,000 + K.E = 120,000
K.E = 120,000 - 60,000
K.E = 60,000 J
The height of the roller coaster at 60,000 potential energy is calculated as follows;
[tex]h = \frac{P.E}{mg} \\\\h = \frac{60,000}{100 \times 9.8} \\\\h =61.23 \ m[/tex]
The velocity of the roller coaster at 60,000 J kinetic energy is calculated as follows;
[tex]K.E = \frac{1}{2} mv^2\\\\v^2 = \frac{2K.E}{m} \\\\v = \sqrt{ \frac{2K.E}{m}} \\\\v = \sqrt{ \frac{2\times 60,000}{100}}\\\\v = 34.64 \ m/s[/tex]
When the potential energy, P.E = 40,000 J, the kinetic energy, K.E is calculated as;
P.E + K.E = M.A
40,000 + K.E = 120,000
K.E = 120,000 - 40,000
K.E = 80,000
The height of the roller coaster at 40,000 potential energy is calculated as follows;
[tex]h = \frac{P.E}{mg} \\\\h = \frac{40,000}{100 \times 9.8} \\\\h = 40.82 \ m[/tex]
The velocity of the roller coaster at 80,000 J kinetic energy is calculated as follows;
[tex]v = \sqrt{\frac{2K.E}{m} } \\\\v = \sqrt{\frac{2\times 80,000}{100} } \\\\v = 40 \ m/s[/tex]
Learn more here:https://brainly.com/question/19969393
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
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.
please answer this as fast as you can i need it
Answer:
it says pdf only i dont knowwhat u want me to do
a box has a mass of 4 kg and surface area 4 metre square calculate the pressure exerted by the box on the floor
- BRAINLIEST answerer ❤️✌
Answer:
9.8
Explanation:
we know ,
f=m.g(g=9.8m/s^2)
now,
p=force/area
p=m.g/a
p=4×9.8/4
p=4 pascal.
if u put g as 10 then ypu will get 10 pascal
Two blocks are set in a pully system as shown in fig below. Block A sits on the frictionless table while block B hags freely. The pully is light and frictionless towards the light string that runs over it. If the Block A has mass of 3.4 kg and Block has 3.5 kg, what would be the magnitude of the acceleration (in ms-2) of the blocks? [g = 9.8 ms=2]
Answer:
Explanation:
F = ma
a = F/m
a = mBg / (mB + mA)
a = 3.5(9.8)/(3.5 + 3.4)
a = 4.971014...
a = 5.0 m/s²
If you want to use individual Free Body Diagrams
mass A will have downward weight and upward normal forces equal at mAg
and a horizontal force of string tension T
F = ma
T = mAa
mass B will have a downward force of mBg and an upward force of T
mBg - T = mBa
substitute for T
mBg - mAa = mBa
mBg = a(mB + mA)
a = mBg / (mB + mA) which is identical to the above answer.
The center of mass of a 1600 kg car is midway between the wheels and 0.7 m above the ground. The wheels are 2.6 m apart. (a) What is the minimum acceleration A of the car so that the front wheels just begin to lift off the ground
Answer:
Explanation:
I guess we are ASSUMING that this is a rear wheel drive car as a front wheel drive car will never get the front wheel normal force to zero
If we consider it as a statics problem and choose our moment center carefully...say 0.7 m above the rear wheel to ground contact point.
Call the traction force at the rear wheels F
The normal force on the front wheels will be zero, so no moment generated by the front wheels.
Summing moments about our chosen point to zero
1600(9.8)[2.6 / 2] - F[0.7] = 0
F = 291,200
this force will create an acceleration of
a = F/m
a = 291200/1600
a = 182 m/s²
which is about 18.6 times gravity acceleration
A car is driving 12m/sec, has to stop suddenly because a pedestrian dashes out in front of the car. If the coefficient of kinetic friction between the tires and parking lot is ∪=60
what is the time, after the breaks are applied, before the car comes to a stop? Sketch the velocity time graph for the car's motion from the instant the breaks are applied until the car comes to a stop.
Answer:
Approximately [tex]2\; \rm s[/tex], assuming that the floor of this parking lot is level, [tex]\mu_{\rm k} = 0.60[/tex], and [tex]g = 9.81\; \rm m\cdot s^{-2}[/tex].
Explanation:
Let [tex]m[/tex] denote the mass of this vehicle. Weight of this vehicle: [tex]m\, g[/tex].
If the floor of this parking lot is level, the normal force on this vehicle would be equal to its weight: [tex]N = m \, g[/tex].
Given that [tex]\mu_{\rm k}[/tex], the kinetic friction between this vehicle and the ground would be consistently [tex]\mu_{\rm k} \, N = \mu_{\rm k} \, m \, g[/tex] until the vehicle comes to a stop.
Assuming that all forces on this vehicle other than friction are balanced. The net force of this vehicle during braking would be [tex](-\mu_{\rm k} \, m \, g)[/tex] (negative because this force is opposite to the direction of the motion.)
By Newton's second law of motion, the acceleration of this vehicle would be:
[tex]\begin{aligned}a &= \frac{F_\text{net}}{m} \\ &= \frac{-\mu_{\rm k} \, m \, g}{m} \\ &= -\mu_{\rm k}\, g \\ &= -0.60 \times 9.81\; \rm m\cdot s^{-2} \\ &= -5.886\; \rm m\cdot s^{-2}\end{aligned}[/tex].
In other words, braking would reduce the velocity of this vehicle by a constant [tex]5.886\; \rm m\cdot s^{-1}[/tex] every second until the vehicle comes to a stop. Calculate the time it would take to reduce the velocity of this vehicle from [tex]v_{0} = 12\; \rm m\cdot s^{-1}[/tex] to [tex]v_{1} = 0\; \rm m\cdot s^{-1}[/tex]:
[tex]\begin{aligned}t &= \frac{v_{1} - v_{0}}{a} \\ &= \frac{0\; \rm m\cdot s^{-1} - 12\; \rm m\cdot s^{-1}}{-5.886\; \rm m \cdot s^{-2}} \\ &\approx 2.0\; \rm s \end{aligned}[/tex].
Acceleration is the slope of the velocity-time graph. Since the acceleration here is constant, the velocity-time graph of this vehicle would be a line with a negative slope.
A racing car on the straight accelerates from 100 km/h to 316 km/h in three seconds.
What is its acceleration?
40m/s2
30m/s2
20m/s2
72m/s2
Answer:
[tex]20m/s^2[/tex]
Explanation:
Solution is attached. I apologize if it is a little messy.
6) An object is released from rest at the top of a ramp inclined at 30. degrees up from the horizontal. Due to friction, the ramp is only 20. % efficient. What is the object's speed after it slides down ALONG the ramp for 2.0 m? *
Answer:
Explanation:
I've been doing these types of problems for many years and I don't think I've ever seen an "efficiency" rating on a ramp.
I'm going to ASSUME that 20% efficient means that 80% of the Potential energy that gets converted becomes system internal heat energy.
Potential energy at the start of a 2.0 m slide
PE = mgh = mg2sin30 = mg2(½) = mg J
0.8mg J gets converted to heat and 0.2mg converts to kinetic energy
0.2mg = ½mv²
v² = 0.4g
v = √(0.4(9.8)) = 1.979898... ≈ 2.0 m/s
A tennis player strikes the tennis ball with an initial velocity of 44.7 m/s horizontally. The ball is initially 1.28 m above the ground and 12.9 m from the 0.914 m tall net. Does the tennis ball make it over the net?
Hi there!
We can begin by finding the total time taken for the ball to reach the net using the equation:
dₓ = vₓt
12.9 = 44.7t
12.9/44.7 = t = 0.289 s
Now, we can use the following equation to solve for displacement in the Y direction:
d = y₀ + vit + 1/2at²
There is no initial vertical velocity, so:
d = y₀ + 1/2at²
Plug in known values:
d = 1.28 + 1/2(-9.8)(0.289²)
d = 0.87m
Thus, since 0.87 m < 0.914 m, the tennis ball does NOT make it over the net.
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
How much did the pressure drop in the storm's center from November 9, 1200z, until November 11, 0000z
The pressure dropped by 24 MB in the storm's center from November 9, 1200z, until November 11, 0000z.
Using the attached map below, in the morning of November 9, the pressure situated at the storm's center = 1000 MB isobar located at the center.
Meanwhile, on November 11, 0000z the pressure situated at the storm's center = 976 MB isobar
The difference in this pressure is regarded as the pressure drop in the storm's center and it is determined as follows;
= 1000 MB - 976 MB
= 24 MB
Therefore, we can conclude that the pressure drop by 24 MB in the storm's center from November 9, 1200z, until November 11, 0000z.
Learn more about pressure drop in an isobaric process here:
https://brainly.com/question/13089696?referrer=searchResults
Is electrical energy the same or different from energy it takes to play a soccer game
Answer:
Diffrent
Explanation:
Electrical energy is electrical charges moving. When you play soccer kinetic energy is used. Kinetic energy is the movement of atoms, objects, and electrons.
a boat's engine can give it a velocity of 25m/s. if the boat heads east across a river which of the following due south with a velocity of 8.5m/s; what is the resultant velocity of the boat? (remember you must find both a magnitude and direction!)
Answer:
Explanation:
v = √(25² + 8.5²) = 26.40549... = 26 m/s
θ = arctan(8.5/25) = 18.77803... = 19° S of E
If the penny is thrown horizontally at 25 m/s from the 170 meter building, how long will it take for the penny to hit the ground?
9514 1404 393
Answer:
about 5.89 seconds
Explanation:
The penny will hit the ground at the same time it would if it were simply dropped. The equation for the vertical motion is ...
h(t) = -4.9t^2 +170 . . . . . where 170 is the initial height in meters
h(t) = 0 when ...
4.9t^2 = 170
t = √(170/4.9) ≈ 5.89
The penny will hit the ground in about 5.89 seconds.
plz answer the question.
Answer:
a
Explanation:
sana po makatulong <3♡♡
3) A 60. kg person is in an elevator. The elevator starts from rest and then accelerates upwards at 2.0 m/s^2 for 4.0 seconds. Calculate the work done by the normal force on the person. *
Answer:
WD = 960 J
Explanation:
WD = work done (J)
F = force (N)
s = displacement (m)
m = mass (kg) = 60
a = acceleration (m/s²) = 2
t = time (s) = 4
u = initial velocity (m/s) = 0
The formulas or equations that are relevant ate:
WD = F × s
F = m × a
s = u + at
We want to find WD, so we need to now the force and the displacement (or distance);
We calculate force, in Newtons, with the formula F = ma:
F = 60 × 2
F = 120 N
We also need displacement, which get with the formula s = u + at:
s = 0 + 2(4)
s = 8 m
Now we have F and s, we can calculate WD:
WD = 120 × 8
WD = 960 J
Methodology:
Starting with what you want to find, in this case WD, list the formula/s you could use;
Then, identify the information you need for the formula and whether or not you are given that information;
Next, list the formulas for the information you don't have and once again, identify whether the information you are given is sufficient to use those formulas;
Once you can calculate all necessary information, then proceed to calculate the values and finally, the answer;
I suggest also keeping a list of all the variables as I've done at the top of my working so it is clear for you to see and use.
An airplane accelerates down a runway at 3.20 m/s2 for 32.8 s until is finally lifts off the ground. Determine the distance traveled before takeoff.
Answer:
The distance traveled before takeoff is 1720 m
Explanation:
Given:a = + 3.2 m/s²
t = 32.8 s
Vᵢ = 0 m/s
To Find:d = ?
Now,
d = Vᵢ × t + 0.5 × a × t²
d = (0 m/s) × (32.8 s) + 0.5 × (3.20 m/s²) × (32.8 s)²
d = 1720 m
Thus, The distance traveled before takeoff is 1720 m
-TheUnknownScientist 72
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
2. Which of the following contributions did Louie De Broglie do for electronic structure of matter? A. determined the speed of electron of hydrogen atom B. proposed a theory that electrons showed characteristics similar to light C. provided mathematical operation for the characteristics of light D. recorded the movement of proton in the nucleus of an atom
❤️
Answer:
In 1924 Louis de Broglie introduced the idea that particles, such as electrons, could be described not only as particles but also as waves. This was substantiated by the way streams of electrons were reflected against crystals and spread through thin metal foils.
Explanation:
I know I probably didn't answer your question, I just used all of my knowledge that I learned about Louie De Broglie. Hope it helps!
Why is it important for the community to take action
Answer:
Why is community action important? ... Involving communities in the design and delivery of services can help to achieve a number of objectives, including: Building community and social capacity – helping the community to share knowledge, skills and ideas. Community resilience – helping the community to support itself
Two trucks leave at different times (from the same place) headed for the same city. Both trucks arrive at the same time. Based on this information, which of the following sentences is true? Select one:
a. The trucks travelled the same distance in the same amount of time.
b. The trucks were traveling at the same average speed.
c. The trucks travelled different distances.
d. The truck that left later was travelling faster.
Answer:
d. The truck that left later was travelling faster
Explanation:
Both trucks travelled from the same place to the same place, meaning they both travelled the same distance;
They both arrive at the same time, but the second truck left later so it took less time to travel the distance than the first truck;
The only variable that can account for this difference is speed;
The one that left later, therefore, must have been going faster.