Which phenomenon related to light allows scientists to measure the distance between a star and Earth? A. Dark patches appear in the electromagnetic spectrum emitted by stars. B. Light waves spread out as they move farther from their source. C. Stars have different colors based on their temperature. D. Stars have unique elements that make up their atmospheres.
Answer:
The Answer Is B not C
Explanation:
Answer:b
Explanation:
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.
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
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
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 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.
3. A block of mass m is suspended by strings as shown in the figure. The tension in the horizontal string
is 36 N. The angle 0 is 60°. Find the mass of the block and the tension forces in string A and string
B
I am having trouble trying to answer this problem. Any help?
Answer:
mass of the block is 7.2Kg
Tension on string A = 72N
Tension on the string with an angle is = 108N
Explanation:
The Horizontal string will have the cos component of the force...
Hence, mg cos60 = 36
by keeping acceleration due to gravity = 10 m/s^2, we get,
m = 7.2Kg
The tension in String A = mg = 7.2*10 = 72N
since the whole string holding the block is pulled by another string with 36 N
the resultant force will be the tension of the string with an angle.
Hence, 72 +36 = 108 N
Please let me know whether I got this right or not...
A statement about what happens in nature that seems to be true all the time ES
Answer:
their is always a animal or bug in nature
Explanation:
What is the correct answer choice to the question above?
Answer: A. a straight line inclined to the time axis
Explanation:
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]
A squirrel is running a race where she is on track for her average velocity to be 6.0 m/s. She is distracted by a dummy of an attractive male squirrel and pauses for 3.0 s. As a result her average velocity ends up being 5.0 m/s instead. What is the length of the race? [HINT: construct two equations with the same two unknowns in them and you can solve the system of equations]
Answer:
90 m
Explanation:
There are two unknowns: the amount of time the squirrel spent running, and the length of the race. Let's call these t and x, respectively.
The average velocity is the total distance divided by the total time.
5.0 m/s = x / (t + 3.0)
The total distance is the time she spent running times the speed she ran at.
x = (6.0 m/s) t
Substitute and solve:
5 = 6t / (t + 3)
5 (t + 3) = 6t
5t + 15 = 6t
t = 15
She ran for 15 seconds (not including the 3 seconds she stopped). So the length of the race is:
x = (6.0 m/s) (15 s)
x = 90 m
Thermal effects refers to the:
Answer:
removal of heat by cooling towers
If horizontal velocity is 5 m/s, and vertical velocity is 8 m/s, what is the magnitude of the resultant velocity?
Answer:
40 m/s
Explanation:
A model airplane is flying north with a velocity of of 15m/s. A strong wind blowing east at 12m/s
What is the airplanes resultant speed (magnitude of vector)
What is the airplanes heading (direction of velocity vector
Answer:
1.) 19.21 m/s
2.) 57 degree
Explanation: Given that a model airplane is flying north with a velocity of of 15m/s. A strong wind blowing east at 12m/s
The airplanes resultant speed can be calculated by using pythagorean theorem
R = sqrt ( 15^2 + 12^2 )
R = sqrt( 225 + 144 )
R = sqrt( 369 )
R = 19.21 m/s
The magnitude of the resultant vector is 19.21 m/s
The direction of velocity vector will be:
Tan Ø = 15 /12
Tan Ø = 1.25
Ø = tan^-1(1.25)
Ø = 57.04
Ø = 57 degree.
Therefore, the airplanes is heading 57 degree in the horizontal direction of velocity vector
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°
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
which is an example of a vector quantity
A.time
B.speed
C. acceleration
D. distance
Answer:
The answer to your question is C
Explanation:
acceleration is a vector quantity because it has both magnitude and direction
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?
What are the two main types of star clusters?
Answer:
Open and globular
Explanation:
When a wave goes through a substance, _______________ has occurred.
When a wave gets trapped or stuck in a substance, ______________ has occurred.
When a wave bounces off of a surface, ______________ has occurred.
The ____________________ states that the angle of an incoming wave is equal to the angle of the reflected wave.
Answer:
A) When a wave goes through a substance, _______________ has occurred.
This is refraction or transmission.
Refraction refers to the changes in the wave that occur when it enters a new medium, in this case, the substance.
Transmission refers to the transfer of energy that is now in the medium.
Usually, we have both of those together.
B) When a wave gets trapped or stuck in a substance, ______________ has occurred.
This is total internal reflection. This happens when, by the Snel Law, the angle of the transmited wave is 90° (or larger) which is traduced in no transmitted wave, this is why the wave is "trapped" in the substance.
C) When a wave bounces off of a surface, ______________ has occurred.
This is reflection.
D) The ____________________ states that the angle of an incoming wave is equal to the angle of the reflected wave.
This is the Law of Reflection, that says "when light is reflected from a surface, the angle of reflection is the same as the angle of incidence"
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
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.
How many elements in oxygen gas? PLEASE ANSWER!
Answer:
8
Explanation:
Answer:
Chemical Properties of Oxygen
At standard temperature and pressure (STP), two atoms of the element bind to form dioxygen, a colorless, odorless, tasteless diatomic gas with the formula O2. Oxygen is a member of the chalcogen group on the periodic table and is a highly reactive nonmetallic element.
Explanation:
What happens if we increase the value of the resistor in forward bias connection?
4. What is the momentum of a 78.0 N bowling ball with a velocity of 8.00 m/s?
Momentum = (mass) x (speed)
We don't know the mass of the bowling ball, but we know how much it weighs. So if we knew the acceleration of gravity in the place where the ball is, we could calculate its mass. The question doesn't tell us what planet the bowling ball is on. So if we only use the information given in the question, there's no way to find the answer, and we're stuck here.
But I urgently need the points, so I'm going to go out on a limb here and make a big assumption: I'll assume that the alley where this ball is rolling and bowling is located on planet Earth, where the acceleration of gravity is around 9.8 m/s² everywhere on the planet's surface. NOW I can go ahead and answer the question that I just invented.
weight = m g
Mass = (weight) / (g)
Mass = (78.0 N) / (9.8 m/s²)
Mass = 7.96 kilograms
Momentum = (mass) x (speed)
Momentum = (7.96 kg) x (8.00 m/s)
Momentum = 63.7 kg-m/s
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.
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:
https://brainly.com/question/14482731
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