We have the next variables
x = speed of the boat in still water
y = speed of the current
(x-y) = Upstream speed
(x+y) = downstream speed
So we have the next equations for the distance
[tex]\begin{gathered} 6\mleft(x-y\mright)=408 \\ \end{gathered}[/tex][tex]\begin{gathered} 9(x+y)=882 \\ \end{gathered}[/tex]We simplify each equation
[tex]\begin{gathered} x-y=68 \\ x+y=98 \end{gathered}[/tex]we sum both equations
[tex]\begin{gathered} 2x=166 \\ x=\frac{166}{2} \\ x=83 \end{gathered}[/tex]Then we calculate the y
[tex]\begin{gathered} y=98-x \\ y=15 \end{gathered}[/tex]x = speed of the boat in still water=83 km/hr
y = speed of the current 15 km/hr
If you swing an object on a string around in a circle, how can you feel the effects of the centripetal force?(1 point)
A in the mass of the object
B in the length of the string
C in the tension in the string
D in the speed of the object
The effects of the centripetal force is felt in the tension in the string.
option C is the correct answer.
What is centripetal force?
Centripetal force is the inward or radial force experienced by an object moving in a circular path.
Mathematically, the centripetal force experienced by an object moving in a circular path is given as;
F = ma
where;
m is the mass of the objecta is the centripetal accelerationF = mv²/r
where;
m is the mass of the objectv is the velocity of the objectr is the radius of the circular pathThe centripetal force on the object is measured in Newton and it is equal to the tension in the string.
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Answer:
C in the tension in the string
Explanation:
A car stereo pulls 2.89 A from a car's battery. If the battery has a voltage of 12 V, how much power does it use?
In order to calculate the power, we can multiply the voltage and the current:
[tex]P=I\cdot U[/tex]So we have:
[tex]\begin{gathered} P=2.89\cdot12\\ \\ P=34.68\text{ W} \end{gathered}[/tex]Therefore the power is 34.68 W.
The wavelength of red helium-neon laser light in air is 632.8 nm.(a) What is its frequency?Hz(b) What is its wavelength in glass that has an index of refraction of 1.63?nm(c) What is its speed in the glass?Mm/sNeed Help?Read ItWatch ItSubmit Answer
ANSWER:
(a) 4.74*10^14 Hz
(b) 388.22 nm
(c) 184 Mm/s
STEP-BY-STEP EXPLANATION:
We have the following information:
[tex]\begin{gathered} \lambda=632.8\text{ nm} \\ n=1.63 \end{gathered}[/tex](a)
To calculate the frequency we use the following formula:
[tex]\begin{gathered} f=\frac{c}{\lambda} \\ c=3\cdot10^8\text{ m/s} \\ \lambda=632.8\text{ nm }=632.8\cdot10^{-9}\text{ m} \\ \text{replacing:} \\ f=\frac{3\cdot10^8}{632.8\cdot10^{-9}} \\ f=4.74\cdot10^{14}\text{ Hz} \end{gathered}[/tex](b)
In this case, we apply the following:
[tex]\begin{gathered} \lambda_1=\frac{\lambda}{n} \\ \text{ replacing} \\ \lambda_1=\frac{632.8}{1.63} \\ \lambda_1=388.22\text{ nm} \end{gathered}[/tex](c)
To calculate the speed it would be:
[tex]\begin{gathered} v=\frac{c}{n} \\ \text{ replacing} \\ v=\frac{3\cdot10^8}{1.63} \\ v=1.84\cdot10^8 \\ v=184\text{ Mm/s} \end{gathered}[/tex]A uniform 500 N/C electric field points in the positive y-direction and acts on an electron initially at rest. After the electron has moved4.00 cm in the field, what is the energy of electron in eV?
Given data:
* The electric field in the y-direction is,
[tex]E=500\text{ N/C}[/tex]* The distance traveled by the electron is,
[tex]\begin{gathered} d=4\text{ cm} \\ d=0.04\text{ m} \end{gathered}[/tex]Solution:
The work done in terms of electric field is,
[tex]W=\text{Edq}[/tex]where q is the charge on an electron,
Substituting the known values,
[tex]\begin{gathered} W=500\times0.04\times1.6\times10^{-19}\text{ J} \\ W=\frac{500\times0.04\times1.6\times10^{-19}}{1.6\times10^{-19}}\text{ eV} \\ W=500\times0.04\text{ eV} \\ W=20\text{ eV} \end{gathered}[/tex]This work done is stored in the charge in form of energy.
Thus, the energy of the electron in eV is 20 eV.
What is the pH of a pure WATER solution if the concentration of H+ in water is 10-7 ?
Given data:
Comcentration of H+ ion in water,
[tex]\lbrack H^+\rbrack=10^{-7}\text{ }\frac{mol}{l}[/tex]The pH value is given as,
[tex]pH=-\log _{10}\lbrack H^+\rbrack_{}[/tex]Substituting all known values,
[tex]pH=-\log _{10}\lbrack10^{-7}\rbrack[/tex]Since,
[tex]\log _{10}(a^b)=b\times\log _{10}(a)[/tex]Therefore,
[tex]\begin{gathered} pH=-7\times(-\log _{10}\lbrack10\rbrack) \\ =7\log _{10}\lbrack10\rbrack \end{gathered}[/tex]Since,
[tex]\log _{10}\lbrack10\rbrack=1[/tex]Therefore,
[tex]\begin{gathered} pH=7\times1 \\ =7 \end{gathered}[/tex]Therefore, pH of a pure water solution is 7.
if the fundamental frequency of an 80 cm long guitar string is 450 Hz, what is the speed of the traveling waves?
Given that the frequency of the wave is, f = 450 Hz
and the length of the guitar string is, L = 80 cm = 0.8 m
We have to find the speed of the wave, v.
As the frequency is the fundamental frequency, so it will be the first harmonic.
The formula to find the wavelength is
[tex]\lambda=2L[/tex]Substituting the values, we get
[tex]\begin{gathered} \lambda=2\times0.8 \\ =1.6\text{ m} \end{gathered}[/tex]The formula to find the speed of the traveling wave is
[tex]v=\lambda\times f[/tex]Substituting the values, we get
[tex]\begin{gathered} v=1.6\times450 \\ =720\text{ m/s} \end{gathered}[/tex]Thus, the speed of the traveling wave is 720 m/s
. A person pushes on a hockey puck with their stick at an angle so the vertical force is 22 N[down] and the horizontal force is 45 N [forward]. Assume the ice is frictionless.a) What is the actual force the hockey player transmits to the puck?b) What is the work done by the person pushing the hockey stick if they push the puck for 3.0s as it moves with a constant velocity of 22 m/s [forward]?c) What is the significance of the fact that both the horizontal force and motion are bothforwards?
50.08 Newton is the actual force the hockey player transmits to the puck and 3305.93 Joule is the work done by the person pushing the hockey stick if they push for 3.0s as it moves with a constant velocity of 22 m/s
(a)
Given, horizontal force Fx = 45 N, vertical force Fy = 22 N
Thus, the total force acting on the is F = [(Fx)^2 + (Fy)2]^1/2
Therefore, F = [(45N)^2 + (22 N)2]^1/2 = 50.08 N
(b)
Constant velocity v = 22 m/s and time interval t = 3.0 s
The horizontal distance travelled by the is x = v t = (22 m/s) (3.0 s) = 66 m
The work done by the person pushing the hockey stick is W = F. x = (50.08 N) (66 m) = 3305.93 J
(c) The magnitude of the horizontal force is greater than (almost double) the vertical force, so the motion of is in the same direction as the horizontal force. The horizontal force dominates here.
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If a convex lens (converging) has a focal length of 12 cm, where would you place an object in order to produce an image that has the same height, but inverted?Less than 12 cm6 cm12 cm24 cm
Given that the focal length is F = 12 cm.
We need to find the position of the object such that the image obtained should be inverted, the same height as that of the object.
In a convex lens, for the image to be inverted and the same height as the object, it should be placed at 2F.
So, the object position will be
[tex]\begin{gathered} 2F=2\times12 \\ =24\text{ cm} \end{gathered}[/tex]Thus, the object position should be 24 cm.
Homework: Action-reaction forces area. equal in magnitude and point in the same directionb. equal in magnitude and point in opposite directionsc. unequal in magnitude but point in the same directiond. unequal in magnitude and point in opposite directions
We will have that they are:
Equal in magnitude and point in opposite directions. {Option B]
Your friend is flying a hot air balloon and you are recording them from below. When you start filming, the hot air balloon is at an angle of 63 degrees above you while staying 100 feet away from where it launched. After 5 seconds, the angle has increased to 67.9 degrees. How fast is your friend rising in feet per seconds and miles per hour.
Answer:
10 ft/s and 6.818 mi/hr
Explanation:
We can represent the situation with the following diagram:
When the angle is 63 degrees, the value of h can be calculated using a trigonometric function as:
[tex]\tan 63=\frac{h_i}{100}[/tex]Because h is the opposite side and 100 ft is the adjacent side. Solving for h, we get:
[tex]\begin{gathered} h_i=100\times\tan 63 \\ h_i=196.26\text{ ft} \end{gathered}[/tex]In the same way, when the angle is 67.9 degrees, we can calculate the height as follows:
[tex]\begin{gathered} \tan 67.9=\frac{h_f}{100} \\ h_f=100\times\tan 67.9 \\ h_f=246.27ft \end{gathered}[/tex]Now, we can calculate the speed in ft per second as follows:
[tex]s=\frac{h_f-h_i}{t}=\frac{246.27ft-196.24ft}{5s}=10\text{ ft/s}[/tex]Finally, 1 mile = 5280 ft and 1 hour = 3600 seconds, so we can convert to miles per hour as:
[tex]10\text{ ft/s }\times\frac{1\text{ mile}}{5280\text{ ft}}\times\frac{3600}{1\text{ hour}}=6.818\text{ mi/hr}[/tex]Therefore, the answers are:
10 ft/s and 6.818 mi/hr
A newly invented ride called Crazy Box in an amusement park has a strong magnet. The magnet accelerates the boxcar and its riders from zero to 35 m/s in 5 seconds. Suppose the mass of the boxcar and riders is 6,000 kg. What is the acceleration of the boxcar and its riders? What is the average net force exerted on the boxcar and riders by the magnets?
Answer:
Below
Explanation:
Acceleration = change in velocity / change in time = 35/5 = 7 m/s^2
F = m * a
= 6000 kg * 7 m/s^2 = 42 000 N
The average net force exerted on the boxcar and its riders by the magnets is 42,000 N.
What is the average net force?The average net force exerted on the boxcar and riders by the magnets is calculated as;
F(net) = ma
where;
m is massa is accelerationThe acceleration is calculated as;
a = Δv / Δt
a = 35 m/s / 5 s
a = 7 m/s²
The acceleration of the boxcar and its riders is 7 m/s².
The average net force exerted on the boxcar and riders by the magnets is;
F = 6,000 kg x 7 m/s²
F = 42,000 N
Thus, the average net force exerted on them is determined as 42,000 N.
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Imagine an asteroid at rest in space that cracks into two pieces. The first piece moves at 1.4 times the speed of the second piece. Calculate the ratio of the first piece’s mass to the second piece’s mass
Since the total linear momentum of the system was 0 before the asteroid cracking into two pieces, then the magntude of the linear momentum of each piece must be the same after the cracking.
Then:
[tex]m_1v_1=m_2v_2[/tex]If the speed of the first piece is 1.4 times the speed of the second, then:
[tex]\begin{gathered} v_1=1.4v_2 \\ \Rightarrow m_1\times1.4v_2=m_2v_2 \\ \Rightarrow1.4m_1=m_2 \\ \Rightarrow m_1=\frac{m_2}{1.4} \\ \Rightarrow\frac{m_1}{m_2}=\frac{1}{1.4} \end{gathered}[/tex]Then, the ratio of the first piece's mass to the second piece's mass is 1:1.4
The postal service will not ship goods over 50 lbs without a special label. Donovan wants to estimate the weight of his package so he doesn't exceed the weight limit. He has a cast iron skillet that weighs 5.67 lbs, a dictionary that weighs 8.34 lbs, and a set of dishes that weighs 37.88 lbs. What will the estimated weight of Donovan's package be if he rounds each item to the nearest pound before totaling the weight?
52 Lb
Explanation
Step 1
round each itme to the nearest pound.
in this case, Rounding a price to the nearest pound is the same as rounding a decimal to the closest whole number
so
for example, If the price is $3.80 you can round up to $4 because the number in the tenths position is 8. The closest whole number to 3.8 is 4
hence
[tex]\begin{gathered} cast\text{ iron=5.67 Lbs}\rightarrow6\text{ LB} \\ dictionary=8.34\text{ Lb}\rightarrow8\text{ Lb} \\ a\text{ set of disehes=}37.88\text{ Lb}\rightarrow38\text{ Lb} \end{gathered}[/tex]Step 2
now, add the cost of the items to find the weight of the package
[tex]\begin{gathered} Weight_{package}=Weight_{cast}+Weight_{dictironay}+Weight_{dishes}\text{ } \\ \text{replace} \\ Weight_{package}=6\text{ lb+8 lb +38 lb=52 lb} \end{gathered}[/tex]therefore, the answer is
52 Lb
I hope this helps you
The force of attraction between m1 and m2 is 76 N. What will this force become if m2 is tripled?
The new force is obtained as 228 N since the product of the masses is directly proportional to the force.
What is the new force?From the Newton law of universal gravitation, we can see that the gravitational force that exists between two masses is directly proportional to the product of the masses and inversely proportional to the distance between the masses.
Where the distance between the masses, remain the same, we can see that the new force would be three times the initial force since the the force and the mass are directly proportional.
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9. The graph is a plot of the velocity versus time for an objectmoving in a straight line. The x position of the objectat t = 0 seconds is 0 meters. At what time after t = 0 secondsdoes the object again pass through its initial position?(a) 1 second(b) Between 1 and 2 seconds(c) 2 seconds(d) 3 seconds
Velocity:
The velocity in simple word, is defined as the change in the position of the particle with respect to the time. The velocity is considered as a vector quantity as it has both the magnitude and direction in it. It can be measured or calculated in meter per second.
At t = 2 seconds, the speed of an object is -10 m/s. As it passes through its initial position. Hence the correct answer is (2)
A block with a mass m1 is hit by a force of magnitude F which causes the block to have an acceleration of magnitude a. If a second block of mass m2 is hit by the same force of magnitude F which causes the block to have an acceleration of magnitude 2a, then which of these could be the two masses? A) m1= 200kg ; m2= 100kgB) m1= 50kg ; m2= 25 kgC) m1= 100kg ; m2= 50kgD)m1= 10kg ; m2= 50kgE)Any of these
We are given that a force "F" accelerates an object of mass "m1". According to Newton's second law, this can be represented by the following equation:
[tex]F=m_1a[/tex]Now, we are given that a second object of mass "m2" is accelerated by "2a" using the same force. Using Newton's second law we get:
[tex]F=m_2(2a)[/tex]Now, we will divide both equations, we get:
[tex]\frac{F}{F}=\frac{m_1a}{m_2(2a)}[/tex]Now, we simplify by canceling put the "F" and the "a":
[tex]1=\frac{m_1}{2m_2}[/tex]Now, we multiply both sides by "2m2", we get:
[tex]2m_2=m_1[/tex]Therefore, the first mass must be twice the second mass.
The options that meet this condition are:
[tex]m_1=200kg,m_2=100kg\text{ }[/tex][tex]m_1=50kg,m_2=25kg[/tex][tex]m_1=100kg,m_2=50kg[/tex]tion 14A 37 kg child holds a 5.9 kg package in a 67 kg boat at rest, and then the child throws the packagehorizontally from the boat at a velocity of +12.4 m/s. What is the velocity of the boat after this?Vunits
Answer:
-0.704 m/s
Explanation:
The conservation of momentum demands that since the child + boat + box system is isolated, the initial momentum must equal the final momentum.
Now since initially, everything is at rest, the initial momentum of the system is zero.
The final momentum of the system is
[tex](m_c+m_b)v_1+(m_{\text{box}})v_2[/tex]where
m_c = mass of the child
m_b = mass of the boat
m_box mass of the boat
v1 = belcity of child + boat
v2 = veloctiy of the box.
Equating the initial momentum to the final momentum gives
[tex](m_c+m_b)v_1+(m_{\text{box}})v_2=0[/tex]Now in our case
m_c = 37 kg
m_b = 67 kg
v1 = unknown
m_box = 5.9 kg
v2 = 12.4
Therefore, the above equation gives
[tex](37+67)v_1+(5.9)(12.4)=0[/tex]solving for v1 gives
[tex]v_1=\frac{5.9\cdot12.4}{37+67}[/tex]which evaluates to give (rounded to the nearest hundredth)
[tex]\boxed{v_1=-0.70m/s}[/tex]which is our answer!
1) The net external force on a golf cart is 390 N north. If the cart has a total mass of 270 kg, what arethe magnitude and directions of its acceleration?
Given data
*The net external force on a golf cart is F = 390 N
*The cart has a total mass is m = 270 kg
The formula for the magnitude of the acceleration of the cart is given as
[tex]a=\frac{F}{m}[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} a=\frac{390}{270} \\ =1.44m/s^2 \end{gathered}[/tex]Hence, the magnitude of the acceleration of the cart is 1.44 m/s^2. The direction of the acceleration is towards the north direction.
Please help me!A boy on a skateboard travels the first 600m of a trip at an average speed of 2 m/s. He then travels the next 800m in 200s and spends the last 100s at a speed of 5 m/s. Find the average speed of the bicyclist for this trip.
average speed = total distance / total time
speed x time = distance
part 1
D1= 600 m, s1=2m/s
Part 2
d2=800m, t2=200s
Part 3
s3=5m/s ,t3=100s
time1 = Distance1/speed1 = 600m/2m/s = 300s
Distance3 = speed3 x time 3 = 5 m/s x 100s = 500m
Total distance = d1+d2+d3 = 600 + 800 + 500 = 1900m
Total time = t1+t2+t3 =300 + 200 + 100 = 600 s
Avg speed = 1900m/600s = 3.17 m/s
A farm tractor tows a 3300 kg trailer up a 14 degree incline with a steady speed of 2.8 m/s. What force does the tractor exert on the trailer?
We are asked to determine the force that the tractors is exerting on a trailer up an incline. A free-body diagram of the situation is the following:
Where:
[tex]\begin{gathered} F=\text{ force of the tractor} \\ m=\text{ mass of the trailer} \\ g=\text{ acceleratio of gravity} \end{gathered}[/tex]Now, we add the forces in the direction of the incline:
[tex]\Sigma F_x=F-mg_x[/tex]To determine the x-component of "mg" we use the following right triangle:
Now, we use the function sine to determine the value of "mgx":
[tex]\sin14=\frac{mg_x}{mg}[/tex]Now, we multiply both sides by "mg":
[tex]mg\sin14=mg_x[/tex]Now, we substitute the values of "m" and "g":
[tex](3300kg)(9.8\frac{m}{s^2})\sin14=mg_x[/tex]Solving the operations:
[tex]7823.75N=mg_x[/tex]Now, we substitute the value in the sum of forces:
[tex]\Sigma F_x=F-7823.75N[/tex]Since the object is moving at a steady speed this means that the sum of forces is zero:
[tex]F-7823.75N=0[/tex]Now, we add 7823.75N to both sides:
[tex]F=7823.75N[/tex]Therefore, the tractor exerts a force of 7823.75N
A woman holds a makeup mirror with a radius of curvature of 120 cm a distance of 20 cm from her face. What is the magnification of the observed image?
Given:
The radius of curvature is R = -120 cm
The object distance is u = -20 cm
Required: Magnification of the observed image.
Explanation:
The focal length can be calculated using the concave mirror as
[tex]\begin{gathered} f=\frac{R}{2} \\ =\frac{-120}{2} \\ =-60\text{ cm} \end{gathered}[/tex]The image distance can be calculated using the mirror formula as
[tex]\begin{gathered} \frac{1}{v}+\frac{1}{u}=\frac{1}{f} \\ \frac{1}{v}=\frac{1}{f}-\frac{1}{u} \\ =\frac{1}{-60}-\frac{1}{-20} \\ v=30\text{ cm} \end{gathered}[/tex]The
What is the period of the wave?
2.5 seconds
0.4 cycles/second (hertz)
3.9 cm
1.6 cm/s
Answer:
2.5 seconds
Explanation:
Perios is time divided by number of oscilliations. From the diagram the number of oscilliations is two. This can be seen by counting successive crests of the wave from the 0s mark. The 2 oscilliatoions ends in 5s. 5 divided by 2 is 2.5s.
SORRY IF ITS IN THE WRONG SUBJECT. I NEED SOMEONE ANSWER ASAP. text to answer the question. Most games that are played with a standard deck of playing cards are called trick games. In such a game, each player will take a turn playing a card, and whoever plays the winning card takes them all. These cards make up a trick, which the winner puts face-side down in a stack before playing the first card for the next round. Would the information in the text be an effective source to help answer the research question "What are the most popular magic tricks? (1 point) O No, because magic tricks that use playing cards are unpopular. O No, because the source is about card games instead of magic tricks. OYes, because the source is written by an expert on the subject. OYes, because the source explains a type of trick.
Answer: the answer should be "No, because the source is about card games instead of magic tricks."
Explanation: sorry for answering 2 weeks later
Answer:
B
Explanation:
Please helpppppppppp - the diagram shows the trajectory of a ball that is thrown horizontally from a top of a building. The ball’s vertical and horizontal velocity vectors along with the resultant factors are also indicated if the ball takes three seconds to reach the ground how fast is it moving by the time it reached the ground
A park ranger driving on a back country road suddenly sees a deer in his headlights 20m ahead. The ranger, who is driving at 11.4 m/s, immediately applies the brakes andslows down with an acceleration of 3.80 m/s2. How much distance is required for theranger's vehicle to come to rest? Only enter the number, not the units,
Given data
*The speed of the ranger who is driving at 11.4 m/s
*The given acceleration is a = -3.80 m/s^2
The formula for the distance covered by the ranger's vehicle is given by the kinematic equation of motion as
[tex]\Delta x=\frac{v^2-v^2_0}{2a}[/tex]*Here v = 0 m/s is the initial speed of the ranger's vehicle
Substitute the values in the above expression as
[tex]\begin{gathered} \Delta x=\frac{0^2-(11.4)^2}{2\times(3.80)} \\ =17.1\text{ m} \end{gathered}[/tex]The distance is required for the ranger's vehicle to come to rest is calculated as
[tex]\begin{gathered} D=20-17.1 \\ =2.9\text{ m} \end{gathered}[/tex]The stopping time is calculated as
[tex]v=v_0+at[/tex]Substitute the values in the above expression as
[tex]\begin{gathered} 0=11.4+(-3.80)(t) \\ t=3.00\text{ s} \end{gathered}[/tex]A 100-N ball suspended by a rope A is pulled to one side horizontally by another rope B and supported so that the rope A makes an angle of 30º with the vertical wall (see figure). Find the tensions of the ropes A and B. You solve it for me by fi uwu
Tension in A rope = 58.115 N
Tension in B rope = 116.23 N
What is tension?Tension is defined as the force transmitted through a rope, cord, or wire when pulled by forces acting from opposite sides. Tension is transmitted along the length of the wire, drawing energy evenly into the bodies at both ends. T = mg + ma
Where;
T = tension, (N)
m = mass (kg)
g = gravitational force, 9.8 m/s²
A = acceleration (m/s²)
As, F = mg
100 = m x 9.8
m = [tex]\frac{100}{9.8}[/tex]
m = 10.20 kg
Let the tension in A rope be T₁ and in the B rope be T₂ which is making angle of 30⁰
The vertical component of tension T₂ will balance the weight
= T₂ cos 30 = 10.20 x 9.8
T₂ = [tex]\frac{99.96}{cos 30}[/tex]
Since cos 30 = [tex]\frac{\sqrt{3} }{2}[/tex]
T₂ = [tex]\frac{99.96}{0.86}[/tex]
T₂ = 116.23 N
The horizontal component of T₂ will balance T₁
T₂ sin 30 = T₁
116.23 sin 30 = T₁
T₁ = 116.23 × [tex]\frac{1}{2}[/tex]
T₁ = 58.115 N
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a student throws a rock straight down words from a bridge into a river below if initial speed of the rock is 10.0 Ms and it takes 2.1 s to reach the river how high is the bridge
Since this is a free fall and all the motion is going down, let's make down the positive direction (that way we won't use minus signs). A free fall is an uniform accelerated motion, this means that we can use the formula:
[tex]y-y_0=v_0t+\frac{1}{2}at^2[/tex]In this case we know the initial velocity is 10 m/s, the time is 2.1 and the acceleration of gravity is 9.8 m/s^2. Plugging this values we have that:
[tex]\begin{gathered} y-y_0=10(2.1)+\frac{1}{2}(9.8)(2.1)^2 \\ y-y_0=42.609 \end{gathered}[/tex]This means that the rock fell 42.609 meters. Therefore, the bridge is 42.609 m tall.
A/An _____ is described as a device that detects current differences and then opens the circuit preventing electrocution.circuit breakerfuseground fault interruptershort circuit
ANSWER
ground fault interrupter
EXPLANATION
A ground fault interrupter is an automatic switch that measures the difference of current between the "hot" and neutral wires. If this difference is greater than a fixed values (only a few milliamperes) the switchs opens and stops the circulation or current.
If a person is in contact with a wire in the circuit the current is diverted through the person's body to the ground. Therefore the ground fault interrupter will detect a difference of current. Hence this device is used to prevent electrocution.
A dog must be at least 17 pounds to enter the dog park.Which description best represents the weight the dog needs to be?Any value less than or equal to 17Any value equal to 17Any value greater than 17Any value greater than or equal to 17
ANSWER:
4th option: Any value greater than or equal to 17
STEP-BY-STEP EXPLANATION:
According to the statement, the first must be at least 17 pounds to enter the dog park, which means that the minimum weight is 17 pounds.
Which means the best way to represent it is any value greater than or equal to 17.
what is the greatest mass of groceries that can be lifted safely with this bag given that the bag is raised with an acceleration of 1.80 m/s^2
We will have the following:
According to the image:
First, we remember that:
[tex]F=m\cdot a[/tex]Now, we will determine he maximum mass will be:
[tex]52.0N=m\cdot(1.8m/s^2)\Rightarrow m=\frac{^{}52.0N}{1.80m/s^2}[/tex][tex]\Rightarrow m=\frac{260}{9}kg\Rightarrow m\approx28.9kg[/tex]So, the maximum mass will be approximately 28.9 kg.