Answer:
because it has no balance
Which value would complete the last cell?
(1 point)
3.0
100.0
25.0
4.0
Answer:
4.0
Explanation:
The following data were obtained from the question:
Force (F) = 20 N
Mass (m) = 5 kg
Acceleration (a) =.?
Force is simply defined as the product of mass and acceleration. Mathematically, it is expressed as
Force (F) = mass (m) x acceleration (a)
F = ma
With the above formula, we can obtain th acceleration of the body as follow:
Force (F) = 20 N
Mass (m) = 5 kg
Acceleration (a) =.?
F = ma
20 = 5 x a
Divide both side by 5
a = 20/5
a = 4 m/s²
Therefore, the value that will complete the last cell in the question above is 4.
A physics student stands on a cliff overlooking a lake and decides to throw a softball to her friends in the water below. She throws the softball with a velocity of 23.5 m/s at an angle of 39.5∘ above the horizontal. When the softball leaves her hand, it is 11.5 m above the water. How far does the softball travel horizontally before it hits the water? Neglect any effects of air resistance when calculating the answer.
Answer:
66.86m
Explanation:
Velocity of ball thrown, u = 23.5 m/s
Initial height of the ball above the water, H = 11.5 m
Angle of projection, θ = 39.5°
Vertical components of veloclty = usinθ
Horizontal components of veloclty = ucosθ
The soft ball hits the water after time 't'
Considering the second equation of motion
S = ut + 1/2at^2........ 1
But since the ball went through motion under gravity ( free fall ) rather than linear motion, then equation 1 can be rewritten as:
H = ut +/- 1/2gt^2
H = - 11.5m
U = usinθ
θ = 39.5°
a = -g = -9.8m/s^2
- 11.5m = 23.5(sin39.5°)t + 1/2(-9.8)t^2
-11.5m = 23.5(0.6360)t - 4.9t^2
-11.5m = 14.946t - 4.9t^2
4.9t^2 -14.946t-11.5m = 0
Since the ball drifted horizontally
D = (Ucosθ)t
Where θ = 39.5°
U = 23.5m/s t=
Alternatively,
horizontal component of the velocity is 23.5 cos 39.5º = 18.1331 m/s
now how long does it take the ball to raise to a peak and fall to the water.
vertical component of velocity = 23.5 sin 39.5º = 14.947m/s
time to reach peak t = v/g = 11.947/9.8 = 1.5252 sec
peak reached above cliff top is
h = ½gt² = ½(9.8)(1.5252)²
= ½×22.797
= 11.3985m
now the ball has to fall 11.3985+ 11.5 = 22.8985m
time to fall from that height is
t = √(2h/g) = √(2• 22.8986/9.8) = 2.1617 sec
add up the two times to get time it is in the air, 2.1617 + 1.5252 = 3.6869
now haw far does the ball travel horizontally in that time
d = vt = 18.1331 ×3.6869= 66.856m
= 66.86m
A box with mass of 2 kg is pushed directly horizontally over a horizontal surface (with friction) at a constant speed of 10 m/s. The force of the push is 60 N. How much thermal energy is generated pushing the box a distance of 15 m
Answer:
E= 600 W
Explanation:
Given that
m = 2 kg
Speed , v= 10 m/s
Force , F= 60 N
Given that box is moving with constant velocity, it means that friction force will be 60 N.
f = 60 N
Therefore total energy generated
E= f x v
E= 60 x 10 = 600 W
E= 600 W
Thus the answer will be 600 W.
The marginal cost curve
(a) Lies below the ATC curve when the ATC curve slopes upward.
(b) Intersects the AFC and ATC curves at their respective minimum points.
(c) Lies above the AVC curve when the AVC curve slopes downward.
(d) Intersects the AFC and AVC curves at their respective minimum points.
(e) Intersects the AVC and ATC curves at their respective minimum points
Answer:
c
Explanation:
The marginal cost curve image has been attached from which we can clearly, indicate that
ATC = average total cost
AFC = average fixed cost
AVC = average variable cost.
From the graph we can indicate that the marginal cost curve
(c) Lies above the AVC curve when the AVC curve slopes downward.
Because the neutron has no charge, its mass must be found in some way other than by using a mass spectrometer. When a neutron and a proton meet (assume both to be almost stationary), they combine and form a deuteron, emitting a gamma ray whose energy is 2.2233 MeV. The masses of the proton and the deuteron are 1.007 276 467 u and 2.013 553 212 u, respectively. Based on this data, what is the mass of the neutron
Answer:
Explanation:
Energy of gamma ray = 2.2233 MeV
Let mass of neutron be n amu
mass defect of deuteron = 2.013553212 - ( 1.007 276 467 + n ) u .
in terms of energy this mass defect will be equal to energy of gamma ray
1 amu = 931 MeV
931 [ 2.013553212 - ( 1.007 276 467 + n ) ] = 2.2233
( 1.007 276 467 + n ) - 2.013553212 = .00238807733
n = 1.008664822 amu
so mass of neutron = 1.008664822 amu
(b) A cylinder of cross-sectional area 0.65m2 and
height 0.32m has a mass of 2. Ikg. If there is a
cavity inside, find the volume of the cavity.
(Density of cylinder = 11.053 kg/m^3)
Answer:
The volume of the cavity is 0.013m^3
Explanation:
To find the volume of the cavity, the major parameter missing is the diameter of the cavity itself. we can obtain this using the following steps:
Step one:
Obtain the volume of the cylinder by dividing the mass of the cylinder by the density.
Volume of the cylinder = 2.1 / 11.053 =0.19[tex]m^{3}[/tex]
Step two:
From the volume of the cylinder, we can get the radius of the cylinder.
[tex]radius = \sqrt{\frac{V}{\pi \times h}} = \sqrt{\frac{0.19}{\pi \times 0.32}} =0.44m[/tex]
Step three:
From the cross-sectional area, we can obtain the radius of the cavity.
Let the radius of the cavity be = r, while the radius of the cylinder be = R
CSA of cavity =
[tex]\pi({R^2}-r^2) = CSA\\0.65 = \pi (0.32^2-r^2)\\r= 0.115m[/tex]
Step Four:
calculate the volume of the cavity using volume =[tex]\pi r^2 \times h[/tex]
Recall that the cavity has the same height as the original cylinder
[tex]volume = \pi \times 0.115^2\times 0.32= 0.013m^3[/tex]
Will mark as BRAINLIEST.......
The Displacement x of particle moving in one dimension under the action of constant force is related to the time by equation 4x³+3x²-5x+2 , where x is in meters and t is in sec.
a)Find velocity of particle at i) t=2 sec ii) t=4 sec.
b) Find the acceleration of the particle at t=3 sec.
Explanation:
It is given that,
The Displacement x of particle moving in one dimension under the action of constant force is related to the time by equation as:
[tex]x=4t^3+3t^2-5t+2[/tex]
Where,
x is in meters and t is in sec
We know that,
Velocity,
[tex]v=\dfrac{dx}{dt}\\\\v=\dfrac{d(4t^3+3t^2-5t+2)}{dt}\\\\v=12t^2+6t-5[/tex]
(a) i. t = 2 s
[tex]v=12(2)^2+6(2)-5=55\ m/s[/tex]
At t = 4 s
[tex]v=12(4)^2+6(4)-5=211\ m/s[/tex]
(b) Acceleration,
[tex]a=\dfrac{dv}{dt}\\\\a=\dfrac{d(12t^2+6t-5)}{dt}\\\\a=24t+6[/tex]
Pu t = 3 s in above equation
So,
[tex]a=24(3)+6\\\\a=78\ m/s^2[/tex]
Hence, (a) (i) v = 55 m/s (ii) v = 211 m/s and (b) 78 m/s²
A student is planning an investigation on the properties of different types of matter. What would be the best method to find the volume of an irregularly shaped object, such as a rock?
Explanation:
Volume is the amount of space an object takes up, while mass is the amount of matter in an object. ... To find the volume of an irregular sized object, one would use the displacement method for measuring volume and place the object in water and measure the amount of water that is displaced.
Answer:
To measure the volume of an irregularly shaped object, pour some water in a measuring cylinder. Then suspend the irregularly shaped object with a thread. After that , move the object gradually downwards and immerse it in water. The volume of the irregularly shaped object is the difference between the volume of the liquid before and after. After measuring the difference, we come to know about the volume of the irregularly shaped object.
The energy change in an endothermic reaction is: A. Internal B. External C. Negative D. Positive
Answer:
Positive
Explanation:
In an endothermic reaction, the products are at a higher energy than the reactants. This means that the enthalpy change of the reaction (∆H) is positive
an object weights 0.250 kgf in air 0.150 in water and 0.125 in an oil.find out the density of the object and the oil
Answer:
1) The density of the object = 2500 kg/m³
2) The density of the oil = 1250 kg/m³
Explanation:
1) The information relating to the question are;
The mass of the object in air = 0.250 kgf
The mass of the object in water = 0.150 kgf
The mass of the object in the oil = 0.125 kgf
By Archimedes's principle, we have;
The upthrust on the object in water = Mass in air - mass in water = The weight of the water displaced
The upthrust on the object in water = 0.250 - 0.150 = 0.1 kgf
∴ The weight of the water displaced = 0.1 kgf
Given that the object is completely immersed in the water, we have;
The volume of the water displaced = The volume of the object
The volume of 0.1 kg of water water displaced = Mass of the water/(Density of water)
The volume of 0.1 kg of water = 0.1/1000 = 0.0001 m³
The density of the object = (Mass in air)/ volume = 0.250/0.0001 = 2500 kg/m³
The density of the object = 2500 kg/m³
2) Whereby the mass of the object in the oil = 0.125 kgf
The upthrust of the oil = The weight of the oil displaced
The upthrust of the oil on the object = Mass of the object in air - mass of the object in the oil
The upthrust of the oil on the object = 0.250 - 0.125 = 0.125 kgf
The weight of the oil displaced = The upthrust of the oil
Given that the volume of the oil displaced = The volume of the oil, we have;
The volume of the oil displaced = 0.0001 m³
The mass of the 0.0001 m³ = 0.125 kg
Therefore the density of the oil = 0.125/0.0001 = 1250 kg/m³.
The density of the oil = 1250 kg/m³.
you walk 6 block east, 2 blocks north, 3 blocks west and then 2 blocks north. the total distance you travel is blocks
Answer:
The answerI travel 13 blocksHi please, I Have An attachment on Waves, Just two Objective Questions Whoever Answers Will be Marked Brainliest thank you.
Answer:
The first answer is W and Z, since they appear to be a period apart. Dont know the second question. I did what I could, hope someone can answer the second.
Can someone explain how the weight of the block is 10.26N, with reference to an appropriate law of motion?
this process is called parellelogram method of resolving vectors.
what are some factors that affect the frequency of sound
Answer:
1. direction of propagation of sound
2.medium through which sound trsnsmitted
element X has two isotopes: X-27 and x-29. x-27 has an atomic mass of 26.975 and a relative abundance of 82.33%, and X-29 has an atomic mass of 29.018 and a relative abundance of 17.67%. calculate the atomic mass of element X. show your work
Answer:
27.34 (no unit)
Explanation:
26.975*82.33%+29.018*17.67%
=27.34
In a Young's double-slit experiment, a set of parallel slits with a separation of 0.102 mm is illuminated by light having a wavelength of 575 nm and the interference pattern observed on a screen 3.50 m from the slits.(a) What is the difference in path lengths from the two slits to the location of a second order bright fringe on the screen?(b) What is the difference in path lengths from the two slits to the location of the second dark fringe on the screen, away from the center of the pattern?
Answer:
Rounded to three significant figures:
(a) [tex]2 \times 575\; \rm nm = 1150\; \rm nm = 1.15\times 10^{-6}\; \rm m[/tex].
(b) [tex]\displaystyle \left(1 + \frac{1}{2}\right) \times (575\;\rm nm) \approx 863\; \rm nm = 8.63\times 10^{-7}\; \rm m[/tex].
Explanation:
Consider a double-slit experiment where a wide beam of monochromatic light arrives at a filter with a double slit. On the other side of the filter, the two slits will appear like two point light sources that are in phase with each other. For each point on the screen, "path" refers to the length of the segment joining that point and each of the two slits. "Path difference" will thus refer to the difference between these two lengths.
Let [tex]k[/tex] denote a natural number ([tex]k \in \left\lbrace0,\, 1,\, 2,\, \dots\right\rbrace[/tex].) In a double-split experiment of a monochromatic light:
A maximum (a bright fringe) is produced when light from the two slits arrive while they were in-phase. That happens when the path difference is an integer multiple of wavelength. That is: [tex]\text{Path difference} = k\, \lambda[/tex].Similarly, a minimum (a dark fringe) is produced when light from the two slits arrive out of phase by exactly one-half of the cycle. For example, The first wave would be at peak while the second would be at a crest when they arrive at the screen. That happens when the path difference is an integer multiple of wavelength plus one-half of the wavelength: [tex]\displaystyle \text{Path difference} = \left(k + \frac{1}{2}\right)\cdot \lambda[/tex].MaximaThe path difference is at a minimum (zero) at the center of the screen between the two slits. That's the position of the first maximum- the central maximum, a bright fringe where [tex]k = 0[/tex] in [tex]\text{Path difference} = 0[/tex].
The path difference increases while moving on the screen away from the center. The first order maximum is at [tex]k = 1[/tex] where [tex]\text{Path difference} = \lambda[/tex].
Similarly, the second order maximum is at [tex]k = 2[/tex] where [tex]\text{Path difference} = 2\, \lambda[/tex]. For the light in this question, at the second order maximum: [tex]\text{Path difference} = 2\, \lambda = 2 \times 575\; \rm nm = 1.15\times 10^{-6}\; \rm m[/tex].
Central maximum: [tex]k = 0[/tex], such that [tex]\text{Path difference} = 0[/tex].First maximum: [tex]k = 1[/tex], such that [tex]\text{Path difference} = \lambda[/tex].Second maximum: [tex]k = 2[/tex], such that [tex]\text{Path difference} = 2\, \lambda[/tex].MinimaThe dark fringe closest to the center of the screen is the first minimum. [tex]\displaystyle \text{Path difference} = \left(0 + \frac{1}{2}\right)\cdot \lambda = \frac{1}{2}\, \lambda[/tex] at that point.
Add one wavelength to that path difference gives another dark fringe- the second minimum. [tex]\displaystyle \text{Path difference} = \left(1 + \frac{1}{2}\right)\cdot \lambda[/tex] at that point.
First minimum: [tex]k =0[/tex], such that [tex]\displaystyle \text{Path difference} = \frac{1}{2}\, \lambda[/tex].Second minimum: [tex]k =1[/tex], such that [tex]\displaystyle \text{Path difference} = \left(1 + \frac{1}{2}\right)\cdot \lambda[/tex].For the light in this question, at the second order minimum: [tex]\displaystyle \text{Path difference} = \left(1 + \frac{1}{2}\right)\cdot \lambda = \left(1 + \frac{1}{2}\right)\times (575\; \rm nm) \approx 8.63\times 10^{-7}\; \rm m[/tex].
A net force of 0.7 N is applied on a body. What happens to the acceleration of the body in a second trial if half of the net force is applied?(1 point) The acceleration is double its original value. The acceleration is half of its original value. The acceleration is the square of its original value. The acceleration remains the same.
Answer:
The answer is The acceleration is double its original value.
Explanation:
It is because of the second trial of accelaration. Because of this, an object's acceleration doubles from its original value.Hope this helps....
Have a nice day!!!!
Answer:
The acceleration is half of its original value
Explanation:
Monochromatic light of wavelength 649 nm is incident on a narrow slit. On a screen 2.25 m away, the distance between the second diffraction minimum and the central maximum is 1.99 cm. (a) Calculate the angle of diffraction θ of the second minimum. (b) Find the width of the slit.
Answer:
a)0.51°
b)1.47×10^-4m
Explanation:
a)for a single slit experiment, the minima that has an angle of θ towards the centre needs to satisfy the expression below.
bsin(θ)= mλ.........................(*)
Where b= width of the slit
The distance on the screen from Central angle can be expressed as
Sin(θ)= y/d............. (**)
d and y is the horizontal distance between slit and screen
If we input eqn(**) into equation (*) we have
y= mλd/b................(z)
In order to find angle (θ) we have
(θ)= sin-(1.99×10^-2)/2.25
= 0.51°
Therefore, angle of diffraction θ of the second minimum is 0.51°
b)to find the width of the sloth using eqn(z) by substitute the values, we have
b= (2)(649×10^-9)(2.25)/1.99×10^-2
b= 1.47×10^-4m
Therefore, the width of the slit is 1.47×10^-4m
Astronomers can now report that active star formation was going on at a time when the universe was only 20% as old as it is today. When astronomers make such a statement, how can they know what was happening inside galaxies way back then
Answer:
First, as you may know, the light travels at a given velocity.
In vaccum, this velocity is c = 3x10^8 m/s.
And we know that:
distance = velocity*time
Now, if some object (like a star ) is really far away, the light that comes from that star may take years to reach the Earth.
This means that the images that the astronomers see today, actually happened years and years ago (So the night sky is like a picture of the "past" of the universe)
Also, for example, if an astronomer sees some particular thing, he can apply a model (a "simplification" of some phenomena that is used to simplify it an explain it) and with the model, the scientist can infer the information of the given thing some time before it was seen.
The astronomers could know what was happening inside galaxies way back then by the fact that;
they examine the spectra of galaxies (or the overall colors of galaxies) with the highest redshifts they can find
Astronomers Measure the wavelength of the light that is stretched, so the light is seen as 'shifted' towards the red part of the spectrum by using spectroscopy. This measure is also called redshift.
This invokes a ray of light through a triangular prism that splits the light into various components known as spectrum.
The way the astronomers could use this concept to know what was happening in the galaxies before is by examining the spectra of galaxies that have the highest redshifts.
Read more at; https://brainly.com/question/15995216
A small cylinder is rolled along a ruler and completes two revolutions. The circumference is the distance around the outside of a circle. What is the circumference of the cylinder? A 4.4 cm B 5.2 cm C 8.8 cm D 10.2 cm
Answer; 4.4cm
Explanation: There is a ruler upon which the cylinder start from 1.4cm and reaches 10.2cm
distance traveled =10.2-1.4=8.8
since this cylinder is small so the linear distance can be approximately taked as rotational distance(as in case of point charge) so
2x2πxr =8.8
so the circumference will be 2πr=4.4cm
What did Bohr's model of the atom include that Rutherford's model did not have?
a nucleus
energy levels
electron clouds
smaller particles
Answer:
The correct option is energy levels
Explanation:
Rutherford's model of an atom suggests that an atom has a tiny positively charged central mass (now called the nucleus) which is surrounded by electrons (negatively charged) in a cloud-like manner.
Bohr's model went a bit further than the Rutherford's model in describing an atom by suggesting that the electrons which surrounds in the nucleus travel in fixed circular orbits. This description by Bohr was able to describe the energy levels of orbitals which assumes that smallest orbitals have the lowest energy while the largest orbitals have the highest energy.
Answer:
energy levels
hope this helped!
Explanation:
observe the virual skateboarder coming down the hill and over the ramp describe how each of newton’s laws of motion can be observed in this action you can choose the dry wet or muddy conditions or some combination of these
Answer:
first part the skater goes down a constant slope ramp, initially he has Newton's second law
pply Newton's third law, the normal is the reaction to the support of the body on the surface
the ramp shoots off. axis becomes zero and therefore with Newton's first law its speed
Explanation:
It is the description of this movement let's write Newton's laws.
* The first law that a body goes at constant speed or zero if the sum of the external forces is zero
* the second law is F = m a
* The third law states that the forces act in pairs of equal magnitude and opposite direction, one applied to each body.
Let's apply these laws to our case
In the first part the skater goes down a constant slope ramp, initially he has Newton's second law when he accelerates from the initial velocity of zero to a terminal velocity.
The expression for this is
Wₓ - fr = ma
W sin θ - μ W cos θ = m a
W = mg
g (sin θ - μ cos) = a
the value of the coefficient of kinetic friction depends on the condition of the surface, dry, wet or muddy
This is Newton's second law
On the Y axis, which is perpendicular to the ramp we have
N- [tex]W_{y}[/tex] = 0
If we apply Newton's third law, the normal is the reaction to the support of the body on the surface, note that it can be different from the weight.
In the second part when he is on the ramp.
In the ramp the skater enters with a speed v, suppose that the ramp has an incline so that the skater can jump, in this case the angle is positive with respect to the axis x
In this case the analysis is similar to the previous one
Newton's second law gives the acceleration of the skater, who when he reaches the end of the ramp shoots off.
At this point the force in the x (horizontal) axis becomes zero and therefore with Newton's first law its speed this axis remains constant and the force in the y axis is the force of gravity and has an acceleration that changes if velocity according to Newton's second law
Answer:look at explanations
Explanation:
in how many ways can five basketball players be placed in three postitions?
Answer:
Well if they playing a game like that
Is there a way for us to control motion
Answer:
They are:
1) change position
2) distract yourself
3) Get fresh air
4) Face the direction you are going.
5) Drink water.
6) Play music.
7) Put your eyes on horizon.
Explanation:
Hope it helps.
a 2-n force is applied to a spring, and there is displacement of 0.4 m. how much would the spring be displaced if a 5-n force was applied?
Answer:1m
Explanation:
2n=0.4m
5n=?
5n×0.4/2n=1m
A missile is moving 1350 m/s at a 25° angle it needs to hit a target 23,500 m away in a 55° direction in 10.2 seconds what is the magnitude of its final velocity
Answer:
3504 m/s
Explanation:
Let x be the horizontal component of distance
y - vertical component of distance
t-time
ax- horizontal component of acceleration
ay-Vertical component of acceleration
Vx-horizontal component of velocity
Vy-Vertical component of velocity
horizontally: x = V_x ×t + ½×a_x×t²
plugging the values we get
23500× cos 55º = 1350×cos25.0º × 10.20 + ½×a_x× (10.20)²
⇒ax = 19.2 m/s²
Moreover,
V'x = V_x + a_x×t = 1350×cos25.0º + 19.2×10.20= 1419 m/s
similarly in vertical direction:
y = V_y×t + ½×a_y×t²
23500×sin55º = 1350×sin25.0º×10.20s + ½×a_y×(10.20)²
⇒a_y = 258 m/s²
Also,
V'y = V_y + a_y×t = 1350×sin25.0º + 258×10.20 = 3204 m/s
Therefore
V = √(V'x² + V'y²) = 3504 m/s
therefore, magnitude of final velocity of missile=3504 m/s
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Suppose an experiment is designed to test the durability of batteries in different conditions. All of the batteries tested are double-A (AA) Brand X. All sets of batteries are preconditioned in different environmental conditions for exactly 168 hours (1 week).
Set 1: 0°C (freezing point of water)
Set 2: 24°C (approximately room temperature)
Set 3: 37°C (approximately body temperature)
The batteries are then continuously used to power identical mechanical drummer toys. As long as the toy keeps drumming the battery is considered functional. The drumming time for each toy is measured as an indication of battery durability. In this experiment, which condition is not controlled?
A.) temperature
B.) brand of batteries
C.) test for durability
D.) type of battery (battery size)
Answer:
I assume its c. Since its talking about testing.
Explanation:
Answer:
The answer is test of durability
Explanation:
Can you solve this question please help me with this
Answer:
Explanation:
The velocity ratio of a wheel and axle is the ratio of the radius (R) of the wheel to the radius (r) of the axle. It is expressed as;
VR = R/r
Since radius = diameter/2
VR = (D/2)/(d/2)
VR = D/d
D is the diameter of the wheel and 'd' is the diameter of the axle.
Given VR = 3 and d = 5cm
3 = D/5
D = 15 cm
If the diameter of the wheel is 15cm, the radius of the wheel will be 15/2 = 7.5cm.
b) Workdone by the load = Load * distance moved by load
Given load = 60kg
Distance moved by load = 2π*radius of axle
Distance moved by load = 2π(0.025) = 0.157
workdone by load = 60* 0.157 = 9.42J
Effort = Workdone by load/distance moved by the wheel
Effort = 9.42/2π(0.075)
Effort = 9.42/0.471
Effort = 20kg
Hence the effort applied is 20kg
c) MA = Load/Effort
MA = 60/20
MA = 3
d) Efficiency = MA/VR * 100%
Efficiency = 3/3 * 100%
Efficiency = 100%
Forensic toxicologist analyze and identify drugs that are confiscated from criminals
True
False
Describe the motion of water waves.
Answer:
Water waves are an example of waves that involve a combination of both longitudinal and transverse motions. As a wave travels through the waver, the particles travel in clockwise circles. The radius of the circles decreases as the depth into the water increases.
