Answer:
want
Explanation:
people always want the new thing especially if its technology
Suppose a wheel with a tire mounted on it is rotating at the constant rate of 3.273.27 times a second. A tack is stuck in the tire at a distance of 0.365 m0.365 m from the rotation axis. Noting that for every rotation the tack travels one circumference, find the tack's tangential speed. tangential speed: m/sm/s What is the tack's centripetal acceleration?
Answer:
The tangential speed is [tex]v = 7.5 m/s [/tex]
The centripetal acceleration is [tex]a = 154 \ m/s^2[/tex]
Explanation
Generally the angular velocity is mathematically represented as
[tex]w = e * \frac{ 2 \ pi \ rad }{s}[/tex]
substituting 3.27 rev/s for e we have that
[tex]w = 3.27 * \frac{ 2 \ pi \ rad }{s}[/tex]
=> [tex]w = 20.55 \ rad /s[/tex]
The tangential speed is mathematically represented as
[tex]v = w * r[/tex]
substituting 0.365 m for r we have that
[tex]v = 20.55 * 0.365 [/tex]
=> [tex]v = 7.5 m/s [/tex]
Generally the centripetal acceleration is mathematically represented as
[tex]a = \frac{v^2}{r}[/tex]
=> [tex]a = \frac{7.5^2}{ 0.365}[/tex]
=> [tex]a = 154 \ m/s^2[/tex]
how to find range with height and velocity
Answer:
[tex]R= \frac{ v_ i\: ^2\:\times sin\:20}{g}[/tex]
Explanation:
A cheetah runs at a constant velocity of 7 m/s. What is it’s acceleration in m/s/s
PLEASE HELP
Answer:
0 m/s²
Explanation:
Acceleration is the change in velocity over change in time. If the velocity is constant, then the acceleration is 0.
A car has an initial velocity of 50 m/s and a constant
acceleration of 5 m/s2. What is the car's velocity after 3
seconds?
A Van de Graaff generator is charged so that a proton at its surface accelerates radially outward at 1.35 ✕ 1012 m/s2. Find the following. (a) the magnitude of the electric force on the proton at that instant magnitude N direction ---Select--- (b) the magnitude and direction of the electric field at the surface of the generator magnitude N/C direction ---Select---
Answer:
(a).
We know that force is
F = m a
So
F = (1.67 x 10^(-27) x (1.38 x10^12)
F = 2.3 x 10^-15 N facing the radially outward direction
(b).
Similarly Force for charge is
F = q E = m a
So relating
E = F/q = 2.3x 10^-15 /(1.6 10^ -19
E = 144.75 N/C facing the radially outward direction
The answer to the question is
(a) The force on the proton is [tex]2.254\times10^{-15}\,N[/tex] directed radially outwards.
(b) The electric field at the surface of the generator is [tex]14031.25\, N/C[/tex] directed radially outwards.
The answer can be explained as shown below.
Given that a proton accelerates radially outward at [tex]1.35\times 10^{12}\,m/s^2[/tex].
ie; [tex]a=1.35\times 10^{12}\,m/s^2[/tex]We know the mass of a proton, [tex]m_p = 1.67\times10^{-27}\,kg[/tex].From Newton's second law we have,
[tex]F=mg[/tex]But here, [tex]m=m_p[/tex]
So, the electric force on the proton is;
[tex]F = m_p\, a= (1.67\times10^{-27}\,kg)\, \times(1.35\times 10^{12}\, m/s^2)=2.254\times10^{-15}\,N[/tex]The force on the proton is [tex]2.254\times10^{-15}\,N[/tex] directed radially outwards.Also, we know that, in electrostatics,
[tex]F=Eq\\ \\\implies E=\frac{F}{q}[/tex]The charge of a proton is, [tex]q=1.6\times 10^{-19}\,C[/tex]Therefore, the electric field is given by,
[tex]E=\frac{2.254\times 10^{-15}N}{1.6\times 10^{-19}\,C} = 14031.25\, N/C[/tex]The electric field at the surface of the generator is [tex]14031.25\, N/C[/tex] directed radially outwards.Learn more about the electric force here:
https://brainly.com/question/8960054
If a 500kg elephant is sliding across a frictionless patch of ice, how much force is needed to keep the elephant from slowing down?
Answer:
4905NExplanation:
The force needed to keep the elephant from slowing down is expressed as shown according to Newtons second law of motion.
Force = mass * acceleration due to gravity
Given
Mass of elephant = 500kg
acceleration due to gravity = 9.81m/s²
Force = 500*9.81
Force = 4905N
Hence the force needed to keep the elephant from slowing down is 4905N
Thermal effects refers to the:
Answer:
removal of heat by cooling towers
please help on magnitude !
Answer:
Yes, the answer is 100 N
Explanation:
Car is pushed right with 150 N and obviously frictional force will act in the opposite direction of the fore provided and that is 50 N
Hence, 150 - 50 is 100 N
Magnitude is 100 N
Direction is Towards the right
Since they asked for Magnitude the answer should be 100 N
g uppose that 3 J of work is needed to stretch a spring from its natural length of 30 cm to a length of 45 cm. (a) How much work is needed to stretch the spring from 35 cm to 37 cm? (Round your answer to two decimal places.) .02 Incorrect: Your answer is incorrect. J (b) How far beyond its natural length will a force of 10 N keep the spring stretched? (Round your answer one decimal place.)
Answer:
W= 0.319992 J and distance is 3.75 cm
Explanation:
The energy needed to stretch the spring from 30 cm to 45 cm = 3 J
Now we required to find the requirement of energy when the spring is stretched from 35 cm to 37 cm.
So first find the work done to stretch the spring from 35 cm to 45 cm.
Work done, w = (1/2) kx^2
3 = (1/2)k(0.45 – 0.30)^2
k = 266.66 N/m
now, x1 = 0.35 – 0.30 = 0.05m
x2 = 0.37 – 0.30 = 0.07m
Now the amount of work done to stretch from 35cm to 37.
w = (1/2) k (x2^2 – x1^2)
w = (1/2) (266.66) (0.07^2 – 0.05^2 )
w= 0.319992 J
(b). Given F = 10 N
F = kx
x = F / k
x = 10 / 266.66
x = 0.0375m
x = 3.75 cm
Thus, distance is 3.75cm
three short square wood blocks measuring 3.5 per side support a machine weighing 29500 lbs. What is the compressive stress in the blocks
A. 798lbs
B. 421lbs
C. 1404lbs
D. 803lb
Answer:
D. 803 lbs
Explanation:
In order to find the compressive stress on all three blocks we first need to find the normal surface area of each:
Surface Area of 1 Block = 3.5 x 3.5
Surface Area of 1 Block = 12.25
Surface Area of all 3 Blocks = A = 3 x 12.25
Area = 36.75
Now, the stress is given by the following formula:
Stress = Force/Area
Stress = 29500 lbs/36.75
Stress = 802.72 lbs
Hence, the correct option will be:
D. 803 lbs
You drop a ball from a window located on an upper floor of a building. It strikes the ground with speed v. You now repeat the drop, but you ask a friend down on the ground to throw another ball upward at speed v. Your friend throws the ball upward at the same moment that you drop yours from the window. At some location, the balls pass each other. Is this location.
Answer:
y = y₀ (1 - ½ g y₀ / v²)
Explanation:
This is a free fall problem. Let's start with the ball that is released from the window, with initial velocity vo = 0 and a height of the window i
y = y₀ + v₀ t - ½ g t²
y = y₀ - ½ g t²
for the ball thrown from the ground with initial velocity v₀₂ = v
y₂ = y₀₂ + v₀₂ t - ½ g t²
in this case y₀ = 0
y₂2 = v t - ½ g t²
at the point where the two balls meet, they have the same height
y = y₂
y₀ - ½ g t² = vt - ½ g t²
y₀i = v t
t = y₀ / v
since we have the time it takes to reach the point, we can substitute in either of the two equations to find the height
y = y₀ - ½ g t²
y = y₀ - ½ g (y₀ / v)²
y = y₀ - ½ g y₀² / v²
y = y₀ (1 - ½ g y₀ / v²)
with this expression we can find the meeting point of the two balls
Suppose a radio signal (light) travels from Earth and through space at a speed of 3 × 10^8/ (this is the speed of light in vacuum). How far (in meters) into space did the signal travel during the first 10 minutes?
Answer:
Explanation:
we know that
s=vt here v is the speed and s is distance covered by the signals
given data
v=3*10^8
t=10 min we have to convert it into seconds
1 minute=60 seconds
so
10 minutes =10*60/1 =600 seconds
now putting the value of v and t we can find the value of s
s=vt
s=3*10^8*600
s=1.8*10^11m
i hope this will help you
Speed is the rate of distance over time.
The signal will travel [tex]1.8 \times 10^{11}[/tex] meters in the first 10 minutes
The given parameters are:
[tex]s = 3 \times 10^8\ ms^{-1}[/tex] ---the speed of light
[tex]t = 10\ min[/tex] -- the time of travel
The relationship between speed, distance (d) and time is:
[tex]s = \frac dt[/tex]
Make d the subject
[tex]d = s \times t[/tex]
Substitute values for t and s
[tex]d = 3 \times 10^8\ ms^{-1} \times 10\ min[/tex]
Convert minutes to seconds
[tex]d = 3 \times 10^8\ ms^{-1} \times 10 \times 60s[/tex]
So, we have:
[tex]d = 3 \times 10^8\ m \times 10 \times 60[/tex]
[tex]d = 18 \times 10^{10}m[/tex]
Rewrite as:
[tex]d = 1.8 \times 10^{11}m[/tex]
Hence, the signal will travel [tex]1.8 \times 10^{11}[/tex] meters in the first 10 minutes
Read more about speed at:
https://brainly.com/question/21791162
Where do you feel that you are traveling at the fastest speed when on the swing?
Answer:
C
Explanation:
I think it's C, because at that point, you are going fastest. Sorry if im wrong, hope this helps.
Answer:
In between and the middle one
Explanation:
What is the correct answer choice to the question above?
Answer: A. a straight line inclined to the time axis
Explanation:
Which inference about Arthur is supported by the text?
Using the equation for for Newton's Second law, F=ma, solve the following problem. You have been given an object with a mass of 6g and an acceleration of 2 m/s2, what is the force?
Group of answer choices
A. 12N
B. 3N
C. 8N
D. 120N
Answer:
F = 0.012 N
Explanation:
Given that,
Mass of the object, m = 6 g
Acceleration, a = 2 m/s²
1 kg = 1000 grams
6 g = 0.006 kg
Force, F = ma
So,
[tex]F=0.006\ kg\times 2\ m/s^2\\F=0.012\ N[/tex]
So, the force is 0.012 N.
At a time of 30 seconds a runner passes a distance marker labeled "125 meters." If the velocity of the runner is +5.0 m/s, when did the runner pass the distance marker for 75 meters?
Answer:
Explanation
He runs at 5m/s, so in 30 s he should be at 150m. So you have to do 125m - 150m and you'll get -25m, this is his initial position. They want to know the time when he hits 75m, so you would do 75 + 25, and get 100. Then do 100m / 5m/s, and you will get 25 seconds.
Question
In order for work to be done, what three things are necessary
You must exert a force of 5N on a book to slide it across a table. If you do 2.5 J of work in the process, how far has the book moved?
Answer:
0.5 m
Explanation:
The following data were obtained from the question:
Force (F) applied = 5 N
Work done (W) = 2.5 J
Distance (s) =?
Work done (W) can be defined as the product of force (F) and distance (s) moved in the direction of the force. From the above definition, work done (W) can be represented mathematically as:
Work done (W) = Force (F) × Distance (s)
W = F × s
With the above formula, we can obtain the distance moved by the book as shown below:
Force (F) applied = 5 N
Work done (W) = 2.5 J
Distance (s) =?
W = F × s
2.5 = 5 × s
Divide both side by 5
s = 2.5/5
s = 0.5 m
Therefore, the book moved a distance of 0.5 m.
A runner completes the 300-meter dash in 38 seconds. What is the speed of the runner? Round your answer the answer to the nearest tenth.
Answer:
speed = 7.9 m/s
Explanation:
speed = total distance / time taken
speed = 300 / 38
speed = 7.89473684 m/s
to the nearest tenth
speed = 7.9 m/s
Mathew has a filtration kit, which consists of a funnel, a flask, and filter papers. Which of these mixtures can he separate using filtration?
Answer:
C. Muddy Water
The bird that migrates the farthest is the Arctic tern. Each year, the Arctic tern travels
32,000 km between the Arctic Ocean and the continent of Antarctica. Most of the
migration takes place within two four-month periods each year. Assume an Arctic
tern completes the second half of its annual migration distance in 122 days. Also
assume that during this time the tern flies directly north. If the tern flies the same
distance each day, what is its velocity in kilometers per day?
Answer: 131.14km per day
Explanation: since the second half of the terns migration takes 122 days we can assume that the full migration would take 244 days. using this we can divide the total distance by the total amount of days it takes (because speed = distance/time) which is 32,000/244, which would be 131.14
I'm confused, and I can't seem to get this question right. someone please help. Only two forces act on an object (mass = 5.00 kg), as in the drawing. (F = 55.0 N.) Find the magnitude and direction (relative to the x axis) of the acceleration of the object. (Drawing: 45 degrees, Fx is 40 N)
Answer:
17.6 m/s², 26.2° above x axis
Explanation:
Apply Newton's second law.
Sum of forces in the x direction:
∑Fₓ = ma
40 N + 55.0 N cos 45° = (5.00 kg) aₓ
aₓ = 15.8 m/s²
Sum of forces in the y direction:
∑Fᵧ = ma
55.0 N sin 45° = (5.00 kg) aᵧ
aᵧ = 7.78 m/s²
Use Pythagorean theorem to find the magnitude of the resultant.
a² = aₓ² + aᵧ²
a² = (15.8 m/s²)² + (7.78 m/s²)²
a = 17.6 m/s²
Use trigonometry to find the angle.
tan θ = aᵧ / aₓ
tan θ = (7.78 m/s²) / (15.8 m/s²)
θ = 26.2°
How much power is needed to lift the 200-N object to a height of 10 m in 4 s?
Answer:
500 watts
Explanation:
Recall that the definition of power is the amount of energy delivered per unit of time.
In our case, the energy delivered is potential energy which we can estimate as the product of the weight of the object times the distance it is lifted above ground:
200 N x 10 m = 2000 Nm
then the power is the quotient of this potential energy divided the time it took to lift the object to that position:
Power = 2000 / 4 Nm/s = 500 Nm/s = 500 watts
The space shuttle fleet was designed with two booster stages. If the second stage provides a thrust of 73 kilo-newtons and the space shuttle has an acceleration of 16,000 miles per hour squared , what is the mass of the spacecraft in units of pounds-mass ?
Answer:
m = 81281.5 pounds.
Explanation:
Given that,
Force, F = 73 kN
Acceleration of the space shuttle, a = 16000 mi/h²
1 miles/h² = 0.0001241 m/s2
16000 mi/h² = 1.98 m/s²
We need to find the mass of the spacecraft.
According to Newton's second law,
F = ma
m is mass of the spacecraft
[tex]m=\dfrac{F}{a}\\\\m=\dfrac{73\times 10^3\ N}{1.98\ m/s^2}\\\\m=36868.68\ kg[/tex]
Since, 1 kg = 2.20462 pounds
m = 81281.5 pounds
Hence, the mass of the spacecraft is 81281.5 pounds.
NEED HELP FAST!!
As the air on the surface of the Earth warms, what happens to the density of the air?
A.It decreases
B.It increases
C.It remains constant
D.It decreases, then increase
Answer:
B
Explanation:
because when the air rises the density increases
We want to predict what will happen to the density of the air on the surface of the Earth when it warms up.
We will see that the correct option is D: "It decreases, then increases"
We know that the temperature of a given object (in this case a mass of air) is related to the kinetic energy of the particles that conform it.
As the temperature increases, the kinetic energy also increases, thus, the amount of motion of each particle increases, thus, the volume of the object increases.
Now remember that:
density = Mass/Volume.
So if the volume of something increases, we will see that the density decreases (as the volume is in the denominator).
Then if the temperature of the air increases, we will see that the density of the air in the surface decreases.
But it does not end there, as only the air near the surface suffers this change of density, we will have a denser mass of air (colder air) above it. And because it is denser (has more mass in the same volume) we can say that it is heavier.
Then eventually the hot air will rise, and the cold air will fall down, thus the density of the air in the surface increases again, as the colder and denser air comes near the surface.
This is one way of how wind currents are born.
Concluding we can see that the correct option is D: "It decreases, then increases"
If you want to learn more, you can read:
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What happens if we increase the value of the resistor in forward bias connection?
What are the two main types of star clusters?
Answer:
Open and globular
Explanation:
HELP!!!
An arrow is shot into the air at an angle of 30.0 above the horizontal with a speed of 20.0 m/s. What are the x and y components of the
velocity of the arrow 1.0 s after it leaves the bowstring?
Answer:
Y(1s) = [tex]10\sqrt{3}[/tex] - 10.1
X(1s) = 10m/s
Explanation:
In annex I've done the explanation for the equations that I will just present here.
Assuming that the arrow stars from the position (0 ; 0) in the Cartesian Graphic, and with Xo and Yo the initial speeds:
[tex]Yo =\frac{\sqrt{3} }{2} . Vr\\Yo = \frac{\sqrt{3} }{2} . 20\\Yo = 10\sqrt{3} m/s \\Xo = \frac{1}{2} Vr\\Xo = 10m/s[/tex]
Ignoring friction with air, Xo = Xf
So, Xo is the same during all the movement.
X(1s) = 10m/s
For Yo is different. That component is suferring reductions from gravity.
We can find Yo(1s) with one the basic functions of cinematics:
Vf = Vo + at
Vf = Final Velocity
Vo = Start Velocity
a = aceleration - gravity (g) is negative here
t = time
Yf = Yo + gt
Yf = [tex]10\sqrt{3}[/tex] - 10.1
If you prefere, can be: Yf = 10. ([tex]\sqrt{3} - 1[/tex])
A large negatively charged object is placed on a wooden table. A neutral metallic ball rolls straight towards the object but stops before it touches it. A second neutral metallic ball rolls up along the path followed by the first ball, strikes the first ball driving it a bit closer to the negatively charged object and stops. After all balls have stopped rolling, the first ball is closer to the negatively charged object than is the second ball. At no time did either ball touch the charged object. Which statement is correct concerning the final charge on each ball
Answer:
the charge of each small sphere, which is + Q / 2
Explanation:
For this exercise we must use the fact that a charged object induces charges on nearby bodies
Induced charge comes from the fact that charges of the same sign repel and charges of different signs attract.
In this case the large, fixed ball with a -Q charge induces a positive charge in the nearest part and the negative charges are repelled to the furthest point, but the net charge on the metallic sphere remains zero. It should be emphasized that since the charges are of different signs, there is an attractive force between the two spheres.
This first metallic sphere now has a negative charge on the back, this charge induces a positive charge on the second sphere, as the charges are of a different sign, they attract each other, which is why the force is attractive.
When the first sphere stops the second sphere hits it, at this moment the charge of the two spheres is equal, therefore the induced charge in the two spheres is + Q. When the two spheres are separated, the charge on each of them is half, that is, the sphere has a charge + Q / 2 and the second sphere has a charge + Q / 2.
Therefore the first sphere is subjected to two forces: an attractive force with the large sphere of charge -Q and a repulsive force with the second sphere of charge + Q / 2.
So the first sphere must approach the big ball and the second sphere must move away from the big sphere.
This is the process of the movement of this exercise, unfortunately the statements with which to compare this process do not appear, but one of the most common questions of what is the charge of each small sphere, which is + Q / 2