Answer:
b, it's a pretty easy explanation
3. What is Newton's 1st Law?
A For every action there is an equal and opposite reaction
B. Acceleration depends on two variables, the mass of the object and the amount of
force
C. An object at rest will stay at rest, an object in motion will stay in motion, unless an
unbalanced force acts upon it.
O D. The amount of matter in an object
Answer:
A
Explanation:
for every action there is an equal and opposite reaction
Condyloid joints are, logically enough, created by joints at condyle bone markings.
True or False
Answer: False
Explanation:
It's received in elliptical cavities.
False. The statement is not entirely accurate. Condyloid joints are a type of synovial joint that allow movement in two planes, similar to an ellipsoid joint.
These joints are characterized by an oval-shaped projection of one bone fitting into an elliptical cavity of another bone. The term "condyle" does refer to a rounded projection or bump on a bone, but not all condyle bone markings create condyloid joints.
Condyloid joints are found in various parts of the body, such as the wrist (between the radius and carpal bones) and the knuckles (between the metacarpal bones and phalanges). They allow for movements such as flexion, extension, abduction, adduction, and circumduction.
It's important to note that the formation of a condyloid joint involves not only the presence of condyle bone markings but also the specific structural arrangement of the bones and the presence of articular cartilage, joint capsule, and synovial fluid. Therefore, while the term "condyle" is related to these joints, it does not accurately capture the complexity of their formation and function.
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Please write a paragraph explaining the bible verse below in your own words.
Exodus 16:19-20
Answer:
Moses had told them to not keep the food till morning but some kept some anyways because they probably thought they were going to starve or not have food the next morning but what I think it means is that you have to trust in God that he will provide for you and so when the people kept the food cuz he thought they were probably going to start of the next day it got maggots
The tires of a car make 77 revolutions as the car reduces its speed uniformly from 95.0 km/h to 65.0 km/h. The tires have a diameter of 0.90 m.
If the car continues to decelerate at this rate, how far does it go? Find the total distance.
Answer:
Explanation:
95.0 km/hr = 26.39 m/s
65 km/hr = 18.06 m/s
Circumference of a tire is 0.9π m
77 revolutions is a distance of
77(0.9π) = 69.3π m
v² = u² + 2as
a = (v² - u²) / 2s
a = (18.06² - 26.39²) / (2(69.3π))
a = -0.85 m/s²
s = (v² - u²) / 2a
s = (0² - 26.39²) / 2(-0.85)
s = 409 m
A car is traveling at 8 m/s accelerates at 3.1 m/s^2 to reach a final top speed of 56 m/s. How much time will pass before the car reaches its top speed
Please find attached photograph for your answer.
Hope it helps.
Do comment if you have any query.
How much force must be applied to push a 1.35 kg book across the desk at constant speed if the coefficient of sliding friction is 0.30?
The magnitude of the force that must be applied to push the book across the desk is 3.97 N.
The given parameters;
mass, m = 1.35 kgcoefficient of friction, μ = 0.3The acceleration of the book across the desk is calculated as follows;
a = μg
where;
g is acceleration due to gravitya = 0.3 x 9.8
a = 2.94 m/s²
The magnitude of the force that must be applied is calculated as follows;
F = ma
F = 1.35 x 2.94
F = 3.97 N
Thus, the magnitude of the force that must be applied to push the book across the desk is 3.97 N.
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Describe all the ways a bicyclist can accelerate
A tow truck pulls a car 5.00 km along a horizontal roadway using a cable having a tension of 850 N. (a.) How much work does the cable do on the car if it pulls horizontally? If it pulls at 35.0° above the horizontal? (b.) How much work does the cable do on the tow truck in both cases of part (a)? (c.) How much work does gravity do on the car in part (a)?
I think 1980is the answer because you add???
Identify:
In each case the forces are constant and the displacement is along a straight line, so
[tex]$$W=F s \cos \phi \text {. }$$[/tex]
Set-Up:
In part (a), when the cable pulls horizontally [tex]$\phi=0^{\circ}$[/tex] and when it pulls [tex]$35.0^{\circ}$[/tex] above the horizontal [tex]$\phi=35.0^{\circ}$[/tex]. In part (b), if the cable pulls horizontally[tex]$\phi=180^{\circ}$[/tex]. If the cable pulls on the car [tex]$35.0^{\circ}$[/tex] above the horizontal it pulls on the truck at below the horizontal and [tex]$\phi=145.0^{\circ}$[/tex]. For the gravity force [tex]$\phi=90^{\circ}$[/tex], since the force is vertical and the displacement is horizontal.
Execute:
(a) When the cable is horizontal, [tex]$W=(850 \mathrm{~N})\left(5.00 \times 10^{3} \mathrm{~m}\right) \cos 0^{\circ}=4.25 \times 10^{6} \mathrm{~J}$[/tex].
When the cable is[tex]$35.0^{\circ}$[/tex] above the horizontal,[tex]$W=(850 \mathrm{~N})\left(5.00 \times 10^{3} \mathrm{~m}\right) \cos 35.0^{\circ}=3.48 \times 10^{6} \mathrm{~J}$[/tex].
(b)[tex]$\cos 180^{\circ}=-\cos 0^{\circ}$[/tex] and [tex]$\cos 145.0^{\circ}=-\cos 35.0^{\circ}$[/tex],
So the answers are [tex]$-4.26 \times 10^{6} \mathrm{~J}$[/tex] and [tex]$-3.48 \times 10^{6} \mathrm{~J}$[/tex].
(c) Since [tex]$\cos \phi=\cos 90^{\circ}=0, W=0$[/tex] in both cases.
Evaluate: If the car and truck are taken together as the system, the tension in the cable does no net wnetwork
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You have a 25 W and a 100 W bulb at home. When you connect one or the other, which bulb carries the greater current
Answer:
100 w bulb has more current.
Explanation:
P=V^2/R.
when velocity is constant power is inversly proportional to resistence, so 25 W will have an hogher resistance tjan a 100 w bulb.
P=VI
when v is constant, power is directly proportional to I. hence, 100 w bulbs will carry more current.
When you connect one or the other, the bulb that's connected carries more current than the one that's not connected.
When you connect BOTH of them, the 100W bulb carries more current than the 25W one.
2) A ray of light in air is approaching the boundary with water at an
angle of 52 degrees. Determine the angle of refraction of the light
ray. (Refractive index of air = 1, water = 1.33)
=
Answer:
Explanation:
ASSUMING the 52° is the angle of incidence measured from the perpendicular to the surface
n₁sinθ₁ = n₂sinθ₂
1 sin52 = 1.33sinθ₂
θ₂ = arcsin(sin52 / 1.33)
θ₂ = 36°
as measured from the perpendicular to the surface
8.92 A 45.0 kg woman stands up in a 60.0 kg canoe 5.00 m long. She walks from a point 1.00 m from one end to a point I .00 m from the other end (Fig. P8.92). If you ignore resistance to motion Of the canoe in the water, how far does the canoe move during fis process
The distance that the canoe moves in this process is 1.29 meters.
We first have to find the center of mass
[tex]X = \frac{MsXs+McXc}{Mw+Mc} \\\\[/tex]
Where
Ms = Woman's mass = 45
Mc = Canoe's mass = 60kg
Xs = position from left= 1 cm
Xc = position from left end of canoe's mass = 2.5cm
When we put these values into the equation we have:
[tex]X=\frac{45*1+60*2.5}{45+60} \\\\= 1.857\\[/tex]
The center of gravity lies at the center of this boat. Therefore,
[tex]Xc = \frac{L}{2} \\\\L = 5 m long\\\\\frac{5}{2} =2.5[/tex]
5.00 - 1. 00 = 4 meters
[tex]\frac{45*4+60*2.5}{45+60} = 3.14meters\\\\[/tex]
To get the distance that is moved by this canoe
distance = 3.143-1.857
= 1.286
≈ 1.29 meters
The distance that the canoe moves in this process is 1.29 meters.
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A merry-go-round rotates from rest with an angular acceleration of 1.08 rad/s2. How long does it take to rotate through (a) the first 3.74 rev and (b) the next 3.74 rev
Answer:
Let ω1 be the initial angular speed and ω2 the final angular speed:
α = (ω2- α1) / t
corresponding to a = (v2 - v1) / t
S (distance corresponds to theta)
1 rev = 2 pi and 3.74 rev = 7.48 pi = 23.5 radians
S = 1/2 a t^2 linear or S = 1/2 α t^2 angular acceleration
23.5 = 1/2 * 1.08 t^2 and t = 6.60 sec for first 3.84 rev
b) ω1 = 1.08 * 6.6 = 7.13 rad/sec initial speed for second 3.74
23.5 = 7.13 t + .54 t^2 compare to S = v1 t + 1/2 a t^2
.54 t^2 + 7.13 t - 23.5 = 0
t^2 + 13.2 t - 43.5 = 0
t = 2.7 sec for next 3.74
Check:
7.13 * 2.7 + .54 * 2.7^2 = 23.2 rad = 3.7 rad
A power plant running at 31 % efficiency generates 270 MW of electric power. Part A At what rate (in MW) is heat energy exhausted to the river that cools the plant
The rate of heat energy exhausted to the river is 600.96 MW.
What is efficiency?The ratio of usable output to total input can be used to objectively measure efficiency. The efficiency of the device is defined as the ratio of energy converted to a useable form to the original amount of energy supplied.
Given parameters:
Efficiency of the power plant; η = 31 %
Output electric power; O = 270 MW.
We know that, Efficiency of the power plant;
η = (Output electric power/ input power)× 100%
⇒ input power = (Output electric power × 100)/η
⇒ input power = (270 × 100)/31 MW
= 870.96 MW.
So, the rate of heat energy exhausted to the river that cools the plant = Input power- output power
= (870.96 - 270) MW
= 600.96 MW.
Hence, heat energy exhausted to the river that cools the plant is 600.96 MW.
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what is a non-economic benefit to international trade
Answer:
non economic benefits to international trade means not being too much profit not selling too much things /items in international trade
Explain different layers of atmosphere and the pressure in each layer. Draw diagram
Answer:
Our atmosphere has five different layers. They are:
1. Troposphere: This is the most important layer of the atmosphere with an average height of 13 km from the earth. It is in this layer that we find the air that we breathe. Almost all the weather phenomena such as rainfall, fog and hailstorm occur here.
2. Stratosphere: This layer extends up to a height of 50 km. It presents the most ideal condition for flying airplanes. It contains a layer of ozone gas which protects us from the harmful effect of the sun rays.
3. Mesosphere: This layer extends up to a height of 80 km. Meteorites bum up in this layer on entering from the space.
4. Thermosphere: In this layer, the temperature rises very rapidly with increasing height. The ionosphere is a part of this layer. It extends between 80-400 km. This layer helps in radio transmission. Radio waves transmitted from the earth the reflected back to the earth by this layer.
5. Exosphere: It is the uppermost layer where there is very thin air. Light gases such as helium and hydrogen float into space from here.
A body at higher temperature contains more heat? Is this true ?
A car of mass 5000 kg was initially moving at 100 km/h and stops at a distance of 55 m. Find the
magnitude of the net force (in N) acting to stop the car.
Answer:
|F| = 35 kN
Explanation:
a = F/m
100 km/hr(1000 m/km / 3600 s/hr) = 27.8 m/s
v² = u² + 2as
a = (v² - u²) / 2s
F/m = (v² - u²) / 2s
F = m(v² - u²) / 2s
F = 5000(0² - 27.8²) / 2(55)
F = - 35,072.9517...
Two objects are being lifted by a machine. One object has a mass of 2 kg, and is lifted at a speed of 2
m/s. The other has a mass of 4 kg and is lifted at a rate of 3 m/s.
a. Which object has more kinetic energy while it is being lifted?
Answer:
Kinetic energy = (1/2) (mass) (speed²)
First object: (1/2) (2 kg) (2 m/s)² = 4 joules .
Second object: (1/2) (4 kg) (3 m/s)² = 18 joules .
The second object had more kinetic energy than the first one had.
Explanation:
Answer:
Kinetic energy = (1/2) (mass) (speed²)
First object: (1/2) (2 kg) (2 m/s)² = 4 joules .
Second object: (1/2) (4 kg) (3 m/s)² = 18 joules .
The second object had more kinetic energy than the first one had.
10. What is Newton's 3rd Law?
O A. For every action there is an equal and opposite reaction.
B. Acceleration depends on two variables, the mass of the object and the amount of
force.
C. An object at rest will stay at rest, an object in motion will stay in motion, unless an
unbalanced force acts upon it.
D. The amount of matter in an object.
7. What is Newton's 2nd Law? *
Answer:
A. For every action there is an equal and opposite reaction.
A less than youthful 82.6 kg physics professor decides to run the 26.2 mile (42.195 km) Los Angeles Marathon. During his months of training, he realizes that one important component in running a successful marathon is carbo-loading, the consumption of a sufficient quantity of carbohydrates prior to the race that the body can store as glycogen to burn during the race. The typical energy requirement for runners is 1 kcal/km per kilogram of body weight, and each mole of oxygen intake allows for the release of 120 kcal of energy by oxidizing (burning) glycogen.
(a) If the professor finishes the marathon in 4:45:00 h, what is the professor's oxygen intake rate, in liters per minute, during the race if he metabolizes all of the carbo-loaded glycogen during the race and the ambient temperature is 21.5°C? 2.28 Read the problem statement again carefully. Is the air at standard temperature and pressure during the marathon? How would this affect the volume of 1 mol of oxygen? L/min
(b) The human body has an efficiency of 25.0%. Only 25.0% of the energy released from oxidizing glycogen is used as macroscopic mechanical energy, and the remaining 75.0% is used for body processes such as pumping blood and respiration, and then leaves the body through the skin via radiation, evaporative cooling, and other processes. What is the average mechanical power (in W) generated by the professor during the run? 197.561 What is the total energy required by the professor during the run? How efficient is the human body, and how long did the race last? W
(c) What is the change in entropy (in J/K) of the professor's body if his core temperature has risen to 38.3°C during the run and his skin temperature is at 36.0°C during the marathon? J/K
(d) What is the change of the entropy (in J/K) of the air surrounding the professor during the race if the ambient temperature remains constant at 21.5°C? J/K
A 12.0 kg box resting on a horizontal, frictionless surface is attached to a 5.00 kg weight by a thin, light wire that passes over a frictionless pulley (Figure 1). The pulley has the shape of a uniform solid disk of mass 2.60 kg and diameter 0.500 m. A) After the system is released, find the tension in the horizontal segment of the wire. B) After the system is released, find the tension in the vertical segment of the wire. C) After the system is released, find the x-component of the acceleration of the box. D) After the system is released, find the x- and y-components of the force that the axle exerts on the pulley.
Answer:
Explanation:
Assuming x direction is horizontal any y direction is vertical
As the moment of inertia of the pulley is I = (½)mR²
The pulley mass will appear as ½ its actual mass in the system.
F = ma
5.00(9.81) = (5.00 + 2.60/2 + 12.0)a
a = 49.05 / 18.3
a = 2.6803278...
a = 2.68 m/s²
A) T = 12.0(2.68) = 32.16... = 32.2 N
B) T = 5.00(9.81 - 2.68) = 35.65 = 35.7 N
C) aₓ = 2.68 m/s²
D) Fx = 32.2 N
Fy = 35.7 + 2.60(9.81) = 61.2 N
A)The tension in the horizontal segment of the wire will be 32.2 N
B)The tension in the vertical segment of the wire will be 35.7 N
What is acceleration?Acceleration is defined as the rate of change of the velocity of the body. Its unit is m/sec².It is a vector quantity. It requires both magnitudes as well as direction to define.
The given data in the problem is;
m is the mass box resting on a horizontal =12.0 kg
the uniform solid disk of mass of 2.60 kg
d is the diameter = 0.500 m.
A)The tension in the horizontal segment of the wire will be 32.2 N
The acceleration is found as;
[tex]\rm F = ma \\\\ a= \frac{49.05}{18.3} \\\\ \rm a= 2.68 \ m/sec^2[/tex]
The tension in the horizontal segment of the wire is found as;
T=ma
T= 12.0 ×2.68
T=32.2 N
Hence the tension in the horizontal segment of the wire will be 32.2 N
B)The tension in the vertical segment of the wire will be 35.7 N
T=m(g-a )
T= 5.00 × (9.8-2.68)
T=35.7 N
Hence the tension in the vertical segment of the wire will be 35.7 N
C) the x-component of the acceleration of the box. will be 2.68 m/s².
aₓ = 2.68 m/s²
D)The x- and y-components of the force that the axle exerts on the pulley will be 32.2 and 61.2 N respectively.
Fx = 32.2 N
Fy = 35.7 + 2.60(9.81) = 61.2 N
Hence the x- and y-components of the force that the axle exerts on the pulley will be 32.2 and 61.2 N respectively.
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What happened when a lobster release a claw? a. The lobster attacks with the claw. b. The lobster let the claw go. C. The lobster bites off the claw. D. The lobster eats with the claw.
Answer:b
Explanation:
Answer:
The lobster let the claw go. so, B.
Explanation:
i hope this helps.
Two parallel plates, separated by 0.20 m, are connected to a 12-V battery. An electron released from rest at a location 0.10 m from the negative plate. When the electron arrives at a distance 0.050 m from the positive plate, what is the potential difference between the initial and final points
The potential difference between the initial and final point is 3.0 V.
The given parameters:
distance between the plates, d = 0.2 mvoltage across the plates, V = 12 Vposition of the electron from negative plate, x₁ = 0.1 mposition of the electron from the positive plate, x₂ = 0.0 5mThe potential difference between the initial and final point is calculated as follows;
[tex]E = \frac{V}{d} \\\\\frac{V_1}{d_1} = \frac{V_2}{d_2}[/tex]
where:
[tex]d_2[/tex] is the distance of the electron between the positive and negative plate
[tex]0.1 + d_2 + 0.05 = 0.2\\\\d_2 + 0.15 = 0.2\\\\d_2 = 0.2 - 0.15\\\\d_2 = 0.05 \ m[/tex]
[tex]V_2 = \frac{V_1d_2}{d_1} \\\\V_2 = \frac{12 \times 0.05}{0.2} \\\\V_2 = 3.0 \ V[/tex]
Thus, the potential difference between the initial and final point is 3.0 V.
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Name two environmental factors, one natural and one human-made, that could account for the trend or pattern in bird (quail, wren) and rodent (mice, rabbits) populations before 1997. Help!
Factors such as more crops, fewer predators, invasive species, and changes in the environment explain a decrease or increase in the birds and rodent populations.
Before 1997, both rodents and birds populations increased due to different factors. Moreover, these factors can be classified as natural or human-made factors. Here are the factors that mainly affected these animals:
Bird population:
Increase:
Decrease in predators that increase birds chances to survive (natural factor)
Increase in crops that improve food access for birds (human factor)
Decrease:
Invasive species or increase of predators (natural factor)
Destruction of natural habitats due to an increase in industry and extraction (human factor)
Rodent population:
Increase:
More adaptability that increased rodents chances to survive (natural factor)Increase in crops that improve food access for rodents (human factor)Decrease:
Invasive species or increase of predators (natural factor)Increase in rabbit consumption and use of rodent control products (human factor)Learn more in: https://brainly.com/question/18123593
An airplane accelerates from a speed of 33 m/s at the constant rate of 3.0 m/s2 over a distance of 500 m. What the final velocity?
Answer:
Explanation:
v² = u² + 2as
v² = 33² + 2(3.0)(500)
v² = 4089
v = 63.9452...
v = 64 m/s
There is a uniform magnetic field of magnitude B, pervading all space, perpendicular to the plane of rod and rails. The rod is released from rest, and it is observed that it accelerates to the left. In what direction does the magnetic field point?
The right hand rule to find the direction of the magnetic field for a falling bar is:
The charge is positive the magnetic field is outgoing, horizontally and towards us. The charge of the bar is negative, the magnetic field is incoming, that is horizontal away from us.
The magnetic force is given by the vector product of the velocity and the magnetic field.
F = q v x B
Where the bolds indicate vectors, F is the force, q the charge on the particle, v the velocity and B the magnetic field.
In the vector product, the vectors are perpendicular, which is why the right-hand rule has been established, see attached:
The thumb points in the direction of speed. Fingers extended in the direction of the magnetic field. The palm is in the direction of the force if the charge is positive and in the opposite direction if the charge is negative.
They indicate that the bar is dropped, therefore its speed is vertical and downwards, it moves to the left therefore this is the direction of the force, we use the right hand rule, the magnetic field must be horizontal, we have two possibilities:
If the charge is positive the magnetic field is outgoing, horizontally and towards us. If the charge of the bar is negative, the magnetic field is incoming, that is, horizontal away from us
In conclusion using the right hand rule we can find the direction of the magnetic field for a falling bar is:
The charge of the bar is negative, the magnetic field is incoming, that is horizontal away from us.The charge is positive the magnetic field is outgoing, horizontally and towards us.
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Paano ka makaka tulong sa upang maiwasan ang suliranin ukol sa climate change
Answer:
Climate Change
Ito ang pagbabago sa normal na panahon sa isang lugar. Maaaring abnormal na pagbabago sa ulan na nattanggap ng isang lugar sa isang taon o ang abnormal na pagbabago sa init ng panahon.
Paano masusulusyunan ito?
Tanggalin sa pagkakasaksak ang mga gadgets kapag puno na
Huwag pumutol ng mga puno
Iwasan ang pagsakay sa transportasyon kung malapit lang ang pupuntahan
Explanation:
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what is translator disadvantage?
Machine translators:
Level of accuracy can be very low.
Accuracy is also very inconsistent across different languages.
Machines can't translate context.
Mistakes are sometimes costly.
Sometimes translation simply doesn't work.
Human translators:
Turnaround time is longer.
Translators rarely work for free.
Unless you use a translation agency, with access to thousands of translators, you're limited to the languages any one translator can work with.
Easiest way to Find fahrenhiet to celsius please i need necessary 20 for the fastest correct answer
Answer:
thx for the points
Explanation:
no need brainliest
Is carbon a source or a sink of energy? Why?
Carbon-Based Fuels Dominate Global Energy Use
Crude oil, coal, and natural gas supply about 85% of the energy used in the world. Fossil fuels are valuable as sources of energy because they contain hydrocarbons and other carbon-based materials.