Answer:
Explanation:
a ) from San Antonio to Houston let distance be d km .
Average speed = total distance / total time
time = distance / speed
Total time = (d / 2 x 75 ) +( d / 2 x 106 )
= .0067 d + .0047 d
= .0114 d
Average speed = d / .0114 d = 87.72 km /h
b ) from Houston back to San Antonio
Total time = (d / 2 x 106 ) +( d / 2 x 75 )
= .0047 d + .0067 d
= .0114 d
Average speed = d / .0114 d = 87.72 km /h
c )
For entire trip :
total distance = 2d
total time = 2 x .0114 d
Average speed = 2 d / 2 x .0114 d
= 87.72 km /h .
While standing at the edge of the roof of a building, a man throws a stone upward with an initial speed of 7.07 m/s. The stone subsequently falls to the ground, which is 19.3 m below the point were the stone leaves his hand. At what speed does the stone impact the ground? Ignore air resistance and use g = 9.81 m/s^2 for the acceleration due to gravity.
Answer:
the stone hits the gound with a speed of 20.7 m/s
Explanation:
Becuase gravity is constant we know that the initial upward velocity will be equal to the downward velocity when the stone has returned to its intal location.
PLEASE HELP ME EXPLAIN THESE QUESTIONS IT'S FOR ANATOMY AND PHYS
Let’s put the information you just learned into practice: perform a respiration check on a partner who is at least 12 years old. You may, as recommended by the sites, “sneak” the respiration check while checking their pulse as long as your partner has consented for you to check their “vital signs.” Telling someone you are checking their pulse, but instead are planning to check their respiratory rate, would be unethical. Based on your experiences, answer the following questions.
1.Pretend you are entering the respiration rate into a medical record. How would you record it?
2.Was the subject’s respiration rate in a health range?
3.What would be different about detecting an abnormal respiration rate if you were checking the respiration rate of a four year old?
4.Do you think you got an accurate respiration rate? Is there any reason to think the subject might have altered their breathing?
5.What exact steps did you take to measure the respiration rate, including any questions you asked or instructions you gave?
6.How long did you measure for?
7. Which parts of the DRABC acronym deal directly with the respiratory system?
8.Some first-aid practitioners also recommend a secondary assessment, which involves a head-to-toe inspection of the patient. Why would this assessment be important?
9.Under which conditions should CPR be performed? What checks or actions should be performed first?
Answer:
I tried
Explanation:
You have to check a 12 year olds respiration rate by Siting them down and trying to relax. It's best to take the respiratory rate while sitting up in a chair or in bed. Measure their breathing rate by counting the number of times their chest or abdomen rises over the course of one minute. Then Record this number. Now you have to answer the first few questions based on that.
Heart rate, blood pressure, respiratory rate and temperature are the big four vital signs.
8. Secondary assessments are used in order to determine the injury, how the injury occurred, how severe the injury is, and to eliminate further injury and that is why it is important.
9. It should only be performed when a person shows no signs of life or when they are unconscious, unresponsive, not breathing or not breathing normally.
In order to perform CPR, you need to check the scene and the person. Make sure the scene is safe, then tap the person on the shoulder and shout "Are you OK?" to ensure that the person needs help. Then pen the airway, Check for breathing, Push hard, push fast, deliver rescue breaths, continue CPR steps.
The normal respiration rate is in the range from 12 to 16 breaths per minute.
How to record respiration rate?The respiration rate is measured when a person is present at rest. For recording the respiration rate, we just count the number of breaths for one minute by counting how many times the chest rises.
The subject’s respiration rate will be considered in a health range if its respiration rate in the range from 12 to 16 breaths per minute.
Exact steps that you take to measure the respiration rate is to count the number of breaths for an entire minute or count for 30 seconds and multiply that number by two. We can measure the respiration rate for about one minute.
DRABC is an abbreviation that stands for Danger, Response, Airway, Breathing and Circulation. Airway, Breathing and Circulation are the parts of DRABC acronym that deal directly with the respiratory system.
A secondary assessment that involves a head-to-toe inspection of the patient is also necessary because it is done to do inspection of the whole body in order to check the physical condition of the patient. CPR should be performed when a person is unconscious, having abnormal breathing and not breathing.
Learn more about respiration here: https://brainly.com/question/22673336
The discharge of a pump is 3 m above the inlet. Water enters at a pressure of 138 kPa and leaves at a pressure of 1380 kPa. The specific volume of the water is 0.001 m3/kg. If there is no heat transfer and no change in kinetic or internal energy, what is the work per unit mass
Answer:
The answer is "[tex]1.271 \ \frac{KJ}{kg}\\[/tex]"
Explanation:
[tex]\Delta e_{mech} =\frac{P_2-P_1}{P} + \frac{v_{2}^2-v_{1}^2}{2}+g(z_2-z_1)\\\\\Delta e_{mech} =\frac{ 1380 -138 \times 1000 }{1000} + 0+g(3-0)\\\\P = \frac{1}{v}= \frac{1}{0.001} = 1000 \frac{kg}{m} \\\\ \Delta e_{mech} =1242 +9.81(3)= 1271.43 \frac{J}{kg} \\\\\text{work per unit pass}= 1.271 \ \frac{KJ}{kg}\\[/tex]
A mass of 10. kg is placed on the end of a 0.50-meter pendulum. What is the period of the pendulum?
Answer:
T = 1.41 seconds
Explanation:
Given that,
The mass placed in the pendulum, m = 10 kg
The length of the pendulum, l = 0.5 m
We need to find the period of the pendulum. The relation for the period of the pendulum is given by :
[tex]T=2\pi \sqrt{\dfrac{l}{g}} \\\\T=2\pi \sqrt{\dfrac{0.5}{9.8}} \\\\T=1.41\ s[/tex]
So, the time period of the pendulum is 1.41 seconds.
For anyone that needs the correct answer without POS trolls:
The answer is 1.4 s
Thank me later :)
Calculate the amount of potential difference in a capacitor of 0.9 MF , If the amount of charge is 1.4x10^-4 C .(Show all the work)
Answer:
1.56×10¯¹⁰ V.
Explanation:
From the question given above, the following data were obtained:
Capicitance (C) = 0.9 MF
Quantity of electricity (Q) = 1.4x10¯⁴ C
Potential difference (V) =?
Next, we shall convert 0.9 MF to farad (F). This can be obtained as follow:
1 MF = 10⁶ F
Therefore,
0.9 MF = 0.9 MF × 10⁶ F / 1 MF
0.9 MF = 9×10⁵ F
Finally, we shall determine the potential difference. This can be obtained as follow:
Capicitance (C) = 9×10⁵ F
Quantity of electricity (Q) = 1.4x10¯⁴ C
Potential difference (V) =?
Q = CV
1.4x10¯⁴ = 9×10⁵ × V
Divide both side by 9×10⁵
V = 1.4x10¯⁴ / 9×10⁵
V = 1.56×10¯¹⁰ V
Therefore, the potential difference is 1.56×10¯¹⁰ V.
Can someone help me answer please
Answer:
4=Conduction by convection by radiation.
Explanation:
Hope it will help you! It may be short but I don't know how to write it in blank aafai milayera lekha Hai blanks ma
at the vertices of a square with a side of 5 cm, there are identical positive charges q = 2 nC. determine the strength of the electrostatic field in the middle of one of the sides of the square
Answer:
Explanation:
To make this problem the easiest way possible, draw a picture and choose the side between the charges. The field will be zero at that point, and I'll prove it in just a second.
[tex]E=\frac{kq}{r^2}[/tex]
k is a constant with a value of [tex]8.99*10^9Nm^2/C^2[/tex]
q is the magnitude of the charge producing the field
r is the distance from the source charge to the test charge
So first determine the electric field from the charge on the bottom left corner, then we'll determine the electric from the bottom right corner.
Convert the centimeters to meters and nano-Coulombs to Coulombs
5cm = 0.05m
2 nC = 2 x 10^-9 C
[tex]E=(8.99*10^9)(2*10^-^9)/(0.025)^2=2.8768*10^4N/C[/tex]
This is pointing to the right because electric field lines point away from positive charges.
[tex]E=(8.99*10^9)(2*10^-^9)/(0.025)^2=-2.8768*10^4N/C[/tex]
This is pointing to the left because of the same reason. Field lines point away from positive charges.
You are able to sum them up because they are both in the x-direction. Their sum will be a net field value of zero.
An isolated, irregularly shaped piece of platinum carries -8.89 × 10-9 C of charge and is in electric equilibrium. The size of this body is about 4 mm. When the electric potential at some point on the metal\'s surface has the value V, the potential at a different point on the surface (indicate one):________.a) May equal Vb) always differs from V.c) equals V
Answer:
c) equals V
Explanation:
This is because, since the isolated, irregularly shaped piece of platinum is in electric equilibrium, the electric potential at all points on its surface is V. So that, the potential difference across any point is zero. This implies that diametrically opposite sides have the same potential and thus, the potential at other points of the surface is V since it is in electric equilibrium.
A block is pushed so that it moves up a ramp at constant speed. Identify from choices (a)-(e) below the appropriate description for the work done by the specified force while the block moves from point A to point B. (a) is zero. (b) is less than zero. (c) is greater than zero. (d) could be positive or negative depending on the choice of coordinate systems. (e) cannot be determined.
Answer:
*The work of the Normal (N) y Wy are zero answer a
*The work of the applied force (F1) is positive answer c
*The work of the friction force (fr) is negative, answer b
*The work of the Wy isnegative, answer d
Explanation:
In this exercise it is asked to identify the type of work, unfortunately the diagram cannot be seen, but in the attached we can see the diagram of a body moving upward on an inclined plane, the existing forces are shown.
As the body moves at constant speed the accelerations are zero. Let's look for the job that is defined
W = F. d
W = F d cos θ
where the dot represents the dot product and the bold letters are vectors.
* The work of the Normal (N) and the y component of the weight (Wy) are zero because they are perpendicular to the motion
answer a
* The work of the applied force (F1) is positive because it is in the same direction of motion
W = F1 Δx
answer c
* The work of the friction force (fr) is negative because the force in the displacement have opposite directions
W = -fr Δx
answer b
* the work the x component of the weight (Wx) in this case is negative
answer d
How is the speed of light measured when it is light years away?
Answer:
In a vacuum, light travels at 670,616,629 mph (1,079,252,849 km/h). To find the distance of a light-year, you multiply this speed by the number of hours in a year (8,766). The result: One light-year equals 5,878,625,370,000 miles (9.5 trillion km).Students perform an experiment in which they drop two eggs with equal mass from a balcony. In the first trial, the egg hits the ground and breaks. In the second trial, the egg hits a foam cushion and does not break or bounce.
Answer: C
Explanation: Because it is just got the same question on the Impulse and Momentum quiz
Does the shadow form by the sun remain the same for the whole day?why?
Answer:
When we are outside on a sunny day, we can see how our shadows change throughout the day. The Sun's position in the sky affects the length of the shadow. When the Sun is low on the horizon, the shadows are long. Due to the Earth's rotation, our view of the Sun changes throughout the day.
I hope this helps u! :D
Answer:
No,the shadow formed by the sun doesn't remain the same for the whole day because of the position of the sun.
help would be greatly appreciated
How does an unbalanced force cause a change in direction of an object?
Answer:
Explanation:
The net force is in the same direction of the acceleration. Acceleration changes the speed of an object.
Q 19.23: A proton is initially moving at 3.0 x 105 m/s. It moves 3.5 m in the direction of a uniform electric field of magnitude 120 N/C. What is the kinetic energy of the proton at the end of the motion
Answer:
The kinetic energy of the proton at the end of the motion is 1.425 x 10⁻¹⁶ J.
Explanation:
Given;
initial velocity of proton, [tex]v_p_i[/tex] = 3 x 10⁵ m/s
distance moved by the proton, d = 3.5 m
electric field strength, E = 120 N/C
The kinetic energy of the proton at the end of the motion is calculated as follows.
Consider work-energy theorem;
W = ΔK.E
[tex]W =K.E_f - K.E_i[/tex]
where;
K.Ef is the final kinetic energy
W is work done in moving the proton = F x d = (EQ) x d = EQd
[tex]K.E_f =EQd + \frac{1}{2}m_pv_p_i^2[/tex]
[tex]m_p \ is \ mass \ of \ proton = 1.673 \ \times \ 10^{-27} kg \\\\Q \ is \ charge \ of \ proton = 1.6 \times 10^{-19} C[/tex]
[tex]K.E_f = 120\times 1.6 \times 10^{-19} \times 3.5 \ + \ \frac{1}{2}(1.673\times 10^{-27})(3\times 10^5)^2 \\\\[/tex]
[tex]K.E_f = 6.72\times 10^{-17} \ + \ 7.53 \times 10^{-17} \\\\K.E_f = 14.25 \times 10^{-17} J\\\\K.E_f = 1.425\times 10^{-16} \ J[/tex]
Therefore, the kinetic energy of the proton at the end of the motion is 1.425 x 10⁻¹⁶ J.
Fred's lightbulb is 45% efficient, and Fran's is 75% efficient. If they both use the same amount of electric energy, which produces more light energy?
Answer:
Frank's 75% efficient light bulb will shine brighter.
Explanation:
The brightness of a bulb is gotten from the power equation;
P = I²R
The more the power rating in watts, the more the brightness.
Now, if they both use the same amount of energy but yet have different efficiency, it means we will just multiply the efficiency by the power.
Thus, 75% efficiency will yield more power than a 45% efficient one.
Therefore, Frank's light bulb will shine brighter.
A car sits in an entrance ramp to a freeway, waiting for a break in the traffic. Then the driver accelerates with constant acceleration along the ramp and onto the freeway. The car starts from rest, moves in a straight line, and has a speed of 19 m/s when it reaches the end of the 110-m-long ramp. The traffic on the freeway is moving at a constant speed of 19 m/s. What distance does the traffic travel while the car is moving the length of the ramp
Answer:
the distance travelled by the traffic, while the car is moving the length of the ramp is 220 m
Explanation:
Given the data in the question;
first we determine acceleration using the kinematic equation below;
v² - u² = 2as
a = v² - u² / 2s
our initial velocity is zero, v is 19 m/s and distance s is 110 m
so we substitute
a = (19² - 0²) / 2×110
a = 361 / 220
a = 1.6409 m/s²
Next, the time t taken by the car to travel along the length of the ramp will be;
t = v - u / a
we substitute
t = (19 - 0) / 1.6409
t = 11.579 sec
so the distance travelled by a body moving with constant speed u is given by following expression:
s = ut + [tex]\frac{1}{2}[/tex] at²
so we substitute 19 m/s for u, 11.579 sec for t and 0 m/s² for a
s = (19 m/s × 11.579 s) + [tex]\frac{1}{2}[/tex] × 0 × (11.579)²
s = (19 m/s × 11.579 s) + 0
s = 220 m
Therefore, the distance travelled by the traffic, while the car is moving the length of the ramp is 220 m
to see if the original results are Which career field is an applied science?
geology
biotechnology
physics
chemistry
Answer:
it is chemistry
Explanation:
An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the
power consumed will be-
(a) 100 W
(b) 75 W
(c) 50 W
(d) 25 W
Answer:
50 W
Explanation:
Case 1
Power = V * I
100 = 220 * I
I = [tex]\frac{100}{220}[/tex] A
Case 2
P = V * I
P = 110 * [tex]\frac{100}{220}[/tex]
P = 50 W
I think the answer is 50 W
Hope it helps
What is the maximum height achieved if a 0.400 kg mass is thrown straight upward with an initial speed of 40.0 m⋅s−1? Ignore the effect of air resistance
The maximum height : 81.63 m
Further explanationGiven
0.4 kg mass
vo = initial speed = 40 m/s
Required
the maximum height
Solution
We can use the law of conservation energy(ME=PE+KE) or use parabolic motion
For parabolic motion :
h max = (vo²sin²θ)/2g
θ = 90°(straight upward)
Input the value :
h max = (40²sin²90°)/2 x 9.8
h max = 81.63 m
What are the two main ways in which chemical bonds are formed
Answer:
The two main types of bonds formed between atoms are ionic bonds and covalent bonds.
Explanation:
Hope this helped <3
From what does oil form?
A. marine organisms
B. terrestrial plants
C. dinosaurs
D. lava or magma
Answer: marine organisms
Explanation:
i just took the test
Carousel conveyors are used for storage and order picking for small parts. The conveyorsrotate clockwise or counterclockwise, as necessary, to position storage bins at the storageand retrieval point. The conveyors are closely spaced, such that the operators travel timebetween conveyors is negligible. The conveyor rotation time for each item equals 1 minute;the time required for the operator to retrieve an item after the conveyor stops rotatingequals 0.25 minute. How many carousel conveyors can one operator tend without creatingidle time on the part of the conveyors
Answer:
the number of carousel conveyors that an operator can operate without any idle time is 5
Explanation:
Given the data in the question;
first we express the equation for number of carousel conveyors that can be operated by an operator;
n' = [tex]\frac{(a + t)}{( a + b)}[/tex]
where a is the concurrent activity time ( 0.25 minute )
b is the independent operator activity time
t is the independent machine activity time( 1 )
Now independent activity time is zero as the operator is not performing any inspection or packaging tasks.
So time taken for the operator to retrieve the finished item at the end of the process is the concurrent activity and independent machine activity time, the conveyor rotation time of each item
so
we substitute
0.25min for a, 1 for t and 0min for b
n' = [tex]\frac{(0.25min + 1min)}{( 0.25min+ 0 min)}[/tex]
n' = 1.25 min / 0.25
n' - 5
Therefore, the number of carousel conveyors that an operator can operate without any idle time is 5
Describe how global context (scientific and technical innovation) is connected with force, friction,energy and motion in 200 words.
A car comes to a bridge during a storm and finds the bridge washed out. The driver must get to the other side, so he decides to try leaping it with his car. The side the car is on is 20.8 mm above the river, whereas the opposite side is a mere 1.3 mm above the river. The river itself is a raging torrent 53.0 mm wide.
A) How fast should the car be traveling just as it leaves the cliff in order to just clear the river and land safely on the opposite side?
B) What is the speed of the car just before it lands safely on the other side?
Answer:
A) 26.5 m/s
B) 33.0 m/s
Explanation:
A)
Once the car leaves the cliff, as no other influence than gravity acts on it, and since it causes the car an acceleration in the vertical direction only, in the horizontal direction, it keeps moving at the same speed until it reaches to the other side.So, we can apply the definition of average velocity to find this speed as follows:[tex]v_{x} = \frac{\Delta x}{\Delta t} (1)[/tex]
We know the value of Δx, which is just the wide of the river (53.0m), but we need to find also the value of Δt.This time is given by the vertical movement, whic.h is independent from the horizontal one, because both movements are perpendicular each other.Since the only influence in the vertical direction is due to gravity, the car is accelerated by gravity, with constant acceleration downward equal to g = -9.8m/s² (taking the upward direction as positive).Since the acceleration is constant, we can use the following kinematic equation, as follows:[tex]\Delta y = y_{f} - y_{o} = v_{o} * t + \frac{1}{2} * g *t^{2} (2)[/tex]
if we take the river level as our x-axis, this means that yf = 1.3 m andy₀ = 20.8 m.
At the same time, due to in the vertical direction the car has no initial velocity, this means that v₀ = 0.Replacing by the values in (2) , and solving for t:[tex]t = \sqrt{\frac{2* \Delta y}{g} } = \sqrt{\frac{2*19.5m}{9.8m/s2} } = 2 s (3)[/tex]
If we choose t₀ =0 ⇒ Δt = t = 2 sReplacing Δx and Δt in (1):[tex]v_{x} = \frac{\Delta x}{\Delta t} = \frac{53.0m}{2s} = 26.5 m/s (4)[/tex]
B)
When the car is just landing in the other side, the velocity of the car has two components, the horizontal one that we just found in A) and a vertical one.Due to the initial velocity in the vertical direction was just zero, we can find the final velocity just applying the definition of acceleration, with a =g, as follows:[tex]v_{fy} = g*t = -9.8m/s2*2 s = -19.6 m/s (5)[/tex]
Since both components are perpendicular each other, we can find the magnitude of the velocity vector (the speed) using the Pythagorean Theorem, as follows:[tex]v = \sqrt{v_{x}^{2} + v_{fy}^{2} } } = \sqrt{(26.5m/s)^{2} + (-19.6m/s)^{2}} = 33.0 m/s (6)[/tex]
How long would it take for a car to travel a distance of 220 kilometers if it is traveling at a velocity of 55 km/hr South?
Your answer:
220 hours
12,100 hours
4 hours
0.25 hours
Convert an acceleration of 12m/s^2 to km/h^2
Answer:
43.2
because to convert from m/sec to kmph we need to multiply by 3600/1000
Choose the words that make each statement correct.
(i) After being released from rest in a uniform electric field, a pro- ton will move [(a) in the same direction as; (b) opposite the direction of] the electric field to regions of [(c) higher; (d) lower] electric potential.
(ii) After being released from rest in a uniform electric field, an electron will move [(e) in the same direction as; (f) opposite the direction of] the electric field to regions of [(g) higher; (h) lower] electric potential.
Answer:
i). (a) in the same direction as , (d) lower
ii). (f) opposite the direction of, (g) higher
Explanation:
An proton may be defined as a sub atomic particle and it has a positive electrical charge. Its mass is slightly less than that of a neutron. When a proton is placed in an electrical field that is uniformly charged, it is at rest. When the proton first moves out from rest from the uniform electric field, it will move in a direction which is same as that of the electric field and it will move to a region of higher potential.
An electron is defined as the subatomic particle having negative electric charge. When an electron is released form rest from an uniform electric field, it will move in the opposite direction of the uniform electric field and will move to the region of lower electric potential.
Block A, mass 250 g , sits on top of block B, mass 2.0 kg . The coefficients of static and kinetic friction between blocks A and B are 0.34 and 0.23, respectively. Block B sits on a frictionless surface. What is the maximum horizontal force that can be applied to block B, without block A slipping
Answer:
F = 69.3 N
Explanation:
For this exercise we use Newton's second law, remembering that the static friction force increases up to a maximum value given by
fr = μ N
We define a reference system parallel to the floor
block B ( lower)
Y axis
N - W₁-W₂ = 0
N = W₂ + W₂
N = (M + m) g
X axis
F -fr = M a
for block A (upper)
X axis
fr = m a (2)
so that the blocks do not slide, the acceleration in both must be the same.
Let's solve the system by adding the two equations
F = (M + m) a (3)
a =[tex]\frac{F}{ M+m}[/tex]
the friction force has the formula
fr = μ N
fr = μ (M + m) g
let's calculate
fr = 0.34 (2.0 + 0.250) 9.8
fr = 7.7 N
we substitute in equation 2
fr = m a
a = fr / m
a = 7.7 / 0.250
a = 30.8 m / s²
we substitute in equation 3
F = (2.0 + 0.250) 30.8
F = 69.3 N
How much time does it take for a car traveling 19.1 m/s to cover a distance of 1200 m?
Answer:
63
Explanation:
it would take around 63 if ur asking for 1200/19.1
A person in a laboratory is asked to taste and describe her conscious experience of two different drinks. The person talks about how sweet each drink tastes to her, what the drinks remind her of, and which one she finds most pleasant.
Answer:
The laboratory testing is done on metrics. The research will help find the preference of a person among various drinks.
Explanation:
The laboratory testing is for research purposes. When a person is introduced with two different drinks he may have different experiences. The taste of two different drinks is sweet. A persons behavior towards the taste is psychological also. If he is already informed that the drinks will be sweet, he might start feeling sweeter when he tastes it even if the drink has no taste.