Answer:
kinetic energy will be equal to 0
Explanation:
this is because at final position velocity of body will become zero.
kinetic energy eill be 8 times
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
Forensic toxicologist analyze and identify drugs that are confiscated from criminals
True
False
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].
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
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.
A 50.5-turn circular coil of radius 4.75 cm can be oriented in any direction in a uniform magnetic field having a magnitude of 0.455 T. If the coil carries a current of 22.5 mA, find the magnitude of the maximum possible torque exerted on the coil.
Answer:
The maximum torque, τ = 3.67 × 10⁻³ Nm
Explanation:
The torque τ = NiABsinθ where N = number of turns of circular coil = 50.5, i = current in circular coil = 22.5 mA = 0.0225 A, A = area of circular coil = πr² where r = radius of circular coil = 4.75 cm = 0.0475 m, B = magnetic field strength = 0.455 T and θ = 90° for maximum torque.
So, τ = NiABsinθ
τ = Niπr²Bsinθ
τ = 50.5 × 0.0225 A × π × (0.0475 m)² × 0.455 T × sin90°
τ = 0.003665 Nm
τ = 3.665 × 10⁻³ Nm
τ ≅ 3.67 × 10⁻³ Nm
So the maximum torque, τ = 3.67 × 10⁻³ Nm
A bat is flitting about in a cave, navigating via ultrasonic bleeps. Assume that the sound emission frequency of the bat is 38.9 kHz. During one fast swoop directly toward a flat wall surface, the bat is moving at 0.015 times the speed of sound in air. What frequency does the bat hear reflected off the wall?
Answer:
40085 Hz
Explanation:
We are given; Sound frequency emmision of bat;f = 38.9 kHz = 38900 Hz
Bat is moving at 0.015 times the speed of sound in air.
Speed of sound in air = 343 m/s
The formula for waves reflected off the wall is calculated from Doppler equation as:
f' = f(v + v_d)/(v - v_s)
Where;
f is the frequency = 38900 Hz
f' is the detected frequency,
v_d is the velocity of the detector = 0.015 × 343 = 5.145
v_s is the velocity of the source = 0.015 × 343 = 5.145 m/s
v is the velocity of the sound = 343 m/s
Thus;
f' = 38900(343 + 5.145)/(343 - 5.145)
f' ≈ 40085 Hz
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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Answer:
Explanation:
a) From the diagram, the load will be expressed in newton. The load will be the weight of the box on the inclined plane.
Load = mass * acceleration due to gravity.
Given the mass of the object = 100kg
acceleration due to gravity = 9.8m/s²
Load (in Newton) = 100*9.8
Load (in Newton) = 980N
b) The formula for calculating the velocity ratio of an inclined plane is expressed as VR = 1/sinθ where θ is the angle of inclination.
Given θ = 30°,
VR = 1/sin30°
VR = 1/0.5
VR = 1/(1/2)
VR = 1* 2/1
VR = 2
The velocity ratio is 2.
c) Length of the inclined plane can be calculated using the SOH, CAH, TOA trigonometry identity.
According to SOH, sinθ = opposite/hypotenuse
sin30° = 1/2 = opp/hyp
This shows that the opposite side of the triangle is 1 and the hypotenuse is 2. The length if the inclined is the length of the longest side i.e the hypotenuse. Hence the length of the inclined plane is 2m
d) Mechanical Advantage is the ratio of the load to the effort applied on an object.
Given the Load = 980N and the effort applied to the load on the incline plane = 400N
MA = Load/Effort
MA = 980/400
MA = 2.45
e) Efficiency = MA/VR * 100
Efficiency = 2.45/2 * 100
Efficiency = 122.5%
A 26-foot ladder is placed against a wall. If the top of the ladder is sliding down the wall at -2 feet per second (note that the rate is negative because the height is decreasing). At what rate is the bottom of the ladder moving away from the wall when the bottom of the ladder is 10 feet away from the wall?
Answer:
Dx/dt = 4,8 f/s
Explanation:
The ladder placed against a wall, and the ground formed a right triangle
with x and h the legs and L the hypothenuse
Then
L² = x² + h² (1)
L = 26 f
Taking differentials on both sides of the equation we get
0 = 2x Dx/dt + 2h Dh/dt (1)
In this equation
x = 10 distance between the bottom of the ladder and the when we need to find, the rate of the ladder moving away from the wall
Dx/dt is the rate we are looking for
h = ? The height of the ladder when x = 10
As L² = x² + h²
h² = L² - x²
h² = (26)² - (10)²
h² = 676 - 100
h² = 576
h = 24 f
Then equation (1)
0 = 2x Dx/dt + 2h Dh/dt
2xDx/dt = - 2h Dh/dt
10 Dx/dt = - 24 ( -2 ) ( Note the movement of the ladder is downwards)
Dx/dt = 48/10
Dx/dt = 4,8 f/s
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:
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.
A spring of initial length 35 cm acquires a length of 55 cm when we hang from it a mass of 3.5 kg. Calculate:
a) The elastic constant of the spring.
b) The length of the spring when we hang a mass of 5 Kg.
Answer:
the elastic constant of the spring=1.715
the length of the spring=0.28
Explanation:
we know that according to hooks law
F=-k x
F= force
k= elastic constant
x= extension or compression
given
length change from 35cm to 55 cm so delta x = L2-L1= 55-35=20 cm
now to find k we need F and F =ma
M for part a is 3.5 kg
so F=3.5 kg *9.8=34.3
now k=F/x
k=34.3/20=1.715 N/cm=171.5 N/m
now to find length given mass is 5 kg so
F= ma
F=5*9.8=49 N
so x =F/k
x=49/171.5
x=0.28
An electric lamp is marked 240v, 60w
It is left to operate for 1h. How much
heat is generated by the lamp
Answer:
H = 0.06 kWh
Explanation:
Given that,
Power of an electric lamp, P = 60 W
Voltage, V = 240 V
It is operated for 1 hour
We need to find the heat generated by the lamp. Heat generated is given by :
[tex]H=P\times t\\\\H=60\ W\times 1\ h\\\\H=60\ Wh\\\\H=0.06\ kWh[/tex]
So, 0.06 kWh of the heat is generated by the lamp.
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.
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
(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]
Eli and Andy want to find out which of the two is stronger. Eli pushes a table with a force of 120 newtons while Andy pushes the table from the opposite side with a force of 125 newtons. Ignoring the masses of Eli and Andy, what is the resultant acceleration of the table if its mass is 10.0 kilograms?
Answer:
a = 0.5 m/s²
Explanation:
the type of problem is called a Newtons second law of motion.
and the equation would be the sum of F = m * a where m = mass and a = acceleration
forces are 125N and the opposite direction is 120N
Eli pushes the table with a force of 120N towards Andy
and
Andy pushes the table with a force of 125N towards Eli
mass of table given as 10 kg.
using the equation
120N - 125N = 10kg * a
a = (120-125) / 10
a = -0.5 m/s² so the acceleration is in the direction of Andy's force towards Eli.
therefore a = 0.5 m/s²
Answer:
B.
0.50 meters/second2
Explanation:
Atoms of the same element will always have the same number of Question Blank but will have different numbers of Question Blank if their mass numbers are different.
Answer:
proton and neutron respectively.
Explanation:
Atoms of the same element will always have the same number of proton but will have different numbers of neutron if their mass numbers are different.
A 310 turn solenoid with a length of 18.0 cm and a radius of 1.60 cm carries a current of 1.90 A. A second coil of four turns is wrapped tightly around this solenoid, so it can be considered to have the same radius as the solenoid. The current in the 310 turn solenoid increases steadily to 5.00 A in 0.900 s.(a) Use Ampere's law to calculate the initial magnetic field in the middle of the 310 turn solenoid.T(b) Calculate the magnetic field of the 310 turn solenoid after 0.900 s.T(c) Calculate the area of the 4-turn coil.m2(d) Calculate the change in the magnetic flux through the 4-turn coil during the same period.Wb(e) Calculate the average induced emf in the 4-turn coil.VIs it equal to the instantaneous induced emf? Explain.(f) Why could contributions to the magnetic field by the current in the 4-turn coil be neglected in this calculation?
Answer:
Given that;
Number of turns in the solenoid N = 310
Length of the solenoid L = 18 cm = 0.18 m
Radius of the solenoid r = 1.60 cm = 0.016 m
Current in the first Circuit I₁ = 1.90A
Number of turns in second coil N₂ = 4
Final Current solenoid I₂ = 5.0 A
Time interval to change the time Δt = 0.9 s
a)
According to Ampere's law, magnetic field inside a conductor is calculated as;
B₁ = ц₀N₁I₁ / L
(ц₀ = 4π × 10⁻⁷ constant)
therefore we substitute
{(4π × 10⁻⁷) × 310 × 1.9A} / 0.18m
= 0.0041 T
b)
Magnetic field inside the solenoid after t = 0.9
B₁ = ц₀N₁I₂ / L
= {(4π × 10⁻⁷) × 310 × 5.0A} / 0.18m
= 0.0108 T
c)
Area of coil is
A = πr²
A = π × ( 0.016 )²
A = 0.000804 m²
d)
Change in magnetic influx is
dФ = (B₂ - B₁) A
= ( 0.0108 T - 0.0041 T) × 0.000804 m²
= 0.0000053868 ≈ 5.39 × 10⁻⁶
e)
Average induced emf is
e = -N₂ ( dФ / dt )
e = ( -4 ) ( 5.39 × 10⁻⁶ / 0.9)
e = - 2.39 × 10⁻⁵V ( NOTE, this is not equal to the instantaneous induced emf )
f)
The induced emf is very low, so the contributions to the magnetic field in the coil is Negative.
The same motor is used in rockets with different masses. The rockets have different accelerations. According to Newton’s second law, how is acceleration expected to change as the rocket mass increases? As rocket mass increases, acceleration decreases. There are no changes in acceleration, as it would depend on the amount of force. As rocket mass increases, acceleration increases. Acceleration cannot be predicted based on changes in mass.
Answer:
As rocket mass increases, acceleration decreases.
Explanation:
From Newton's second law of motion;
F= ma
Where;
m= mass of the object
a= acceleration of the object
Hence we can write;
a= F/m
This implies that an increase in mass (m) will lead to a decrease in acceleration if the force on the object is held constant.
Hence, if the rockets have different masses, they will have different accelerations.
Hello!
---------
As rocket mass increases, acceleration decreases.
Hope this helps! The rest are available on Quizlet at "Unit 6: Lesson 4 Force, Mass and Acceleration". Thanks and good luck!
A bus is travelling at 10m/s. It accelerates at 2m/s^2 over a distance of 20m. Calculate it's final velocity
Answer:
13.4 m/s
Explanation:
Given:
Δx = 20 m
v₀ = 10 m/s
a = 2 m/s²
Find: v
v² = v₀² + 2aΔx
v² = (10 m/s)² + 2 (2 m/s²) (20 m)
v = 13.4 m/s
At summer camp, the swimming course runs the length (L) of a small lake. To determine the length of the course, the camp counselors measure the two "dry" legs of a right triangle. What is the length in meters of the swimming course in the figure below?
Answer:
47.17 m
Explanation:
From the diagram of the question attached, The length of the legs are 25 m and 40 m . This legs form a right angle triangle with the length of the swimming course (L).
Pythagoras theorem states that for a right angle triangle with hypotenuse a and legs b and c, then:
a² = b² + c²
In the triangle, the length of the swimming course (L) is the hypotenuse and the two legs are 25 m and 40 m. Using Pythagoras:
L² = 25² + 40²
L² = 625 + 1600
L² = 2225
L = √2225
L = 47.17 m
You have a reservoir held at a constant temperature of –30°C. You add 400 J of heat to the reservoir. If you have another reservoir at 0°C, how much heat must you add to it so that both reservoirs have the same increase in entropy?
Answer:
449.38 J
Explanation:
ΔS = ΔQ/T
Where ΔS = entropy change
Q = quantity of heat
T = temperature
First reservoir :
T = –30°C = - 30 + 273 = 243K
Q = 400 J
Second reservoir :
T = 0°C = 273K
Q =?
To have same increase in entropy for both reservoirs :
Q/T of first reservoir = Q/T of second reservoir
400/243 = Q/273
243 * Q = 400 * 273
Q = (400 * 273) / 243
Q = 109,200 / 243
Q = 449.38271
Q = 449.38 J
A particular celestial body orbits at a particular speed. For every two orbits it makes, another celestial body orbits three times. This orbital resonance would correspond to which musical interval?
Answer:
Explanation:
frequency of first body f₁ = 2 / T where T is time taken by it for making two orbits
frequency of second body f₂ = 3 / T
ration of two frequency
f₁ / f₂ = 2 / 3
This ratio is called perfect fifth in musical interval .
Determine the gradient and the co-ordinates of the x and y intercept of line whose equation is 2y + 3x = 1
Answer:
The x - intercept is 1/3
The y - intercept is 1/2
The gradient is -3/2
Explanation:
To find the x - intercept of the equation 2y + 3x = 1, we find the value of x when y = 0. So,
2y + 3x = 1
2(0) + 3x = 1
0 + 3x = 1
3x = 1
x = 1/3
So, the x - intercept is 1/3
To find the y - intercept of the equation 2y + 3x = 1, we find the value of y when x = 0. So,
2y + 3x = 1
2y + 3(0) = 1
2y + 0 = 1
2y = 1
y = 1/2
So, the y - intercept is 1/2
To find the gradient of the equation 2y + 3x = 1, we re-write it in gradient intercept form by making y subject of the formula.
So, 2y + 3x = 1
2y = -3x + 1
y = -3x/2 + 1/2
The coefficient of x which equals -3/2 is the gradient.
The gradient is -3/2
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 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.
in how many ways can five basketball players be placed in three postitions?
Answer:
Well if they playing a game like that
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:
