Answer:
Weight of car = 36034.88 lb.ft/s²
Explanation:
We are told the pressure of the four tires of your car is; P = 35.0 psi
Also, the surface area of contact is; A = 32 in²
Thus;
Weight of tires = Pressure × Area
W_tires = 35 × 32
Weight_tires = 1120lb.
To get the weight of the car, we will multiply the tire weight by acceleration due to gravity.
It's value in ft/s² is g = 32.174 ft/s²
Thus;
Weight of car = 1120 × 32.174
Weight of car = 36034.88 lb.ft/s²
A farmer plants the same crop in a field year after year. Every year there are months during which the field is left without any plants. The farmer notices a decline in soil
quality during these months. What causes this decrease in soil quality?
Erosion
Drought
Desertification
Consumption of nutrients by the crops
Answer:
Drought
Explanation:
how is a trench and a tsunami related? 6-8 sentences
Answer: A tsunami is a very long-wavelength wave of water that is generated by sudden displacement of the seafloor or disruption of any body of standing water. Tsunami are sometimes called "seismic sea waves", although they can be generated by mechanisms other than earthquakes. Tsunami have also been called "tidal waves", but this term should not be used because they are not in any way related to the tides of the Earth. Because tsunami occur suddenly, often without warning, they are extremely dangerous to coastal communities. Ocean trenches are steep depressions in the deepest parts of the ocean [where old ocean crust from one tectonic plate is pushed beneath another plate, raising mountains, causing earthquakes, and forming volcanoes on the seafloor and on land.
Explanation:
In the figure, a 32 cm length of conducting
wire that is free to move is held in place
between two thin conducting wires. All of the
wires are in a magnetic field. When a 6.0 A
current is in the wire, as shown in the figure,
the wire segment moves upward at a constant
velocity.
The acceleration of gravity is 9.81 m/s?.
a) Assuming the wire slides without friction
on the two vertical conductors and has a mass
of 0.13 kg, find the magnitude of the minimum
magnetic field that is required to move the
wire.
Answer in units of T. b) What is the direction?
Answer:
.66354 T
Explanation:
Use F=ILB
B = [tex]\frac{F}{IL}[/tex]
B = Magnetic field
F= force due to magnetic
I= current
L= length in meters
F = mg
Final formula:
B=[tex]\frac{mg}{IL}[/tex]
B=[tex]\frac{(.13)(9.8)}{(6)(.32)}[/tex]
B= .66354
ayo btw ion know how to find direction, my b G
The minimum magnetic field required to move the wire is 66354 T.
The direction of magnetic field is normal to the page outwards.
What is magnetic field?The region surrounding a magnet that experiences the effects of magnetism is known as the magnetic field. When describing the distribution of the magnetic force within and around a magnetic object in nature, the magnetic field is a useful tool.
Given parameter:
Current passing through the wire, I = 6.0 A.
Length of the wire ,L = 32 cm = 0.32 m.
Mass of the wire, m = 0.13 Kg.
Acceleration due to the gravity, g = 9.8 m/s².
We know that, force acting on a current caring wire due to magnetic field is, F=ILB
Where,
B = Required magnetic field.
To find the minimum magnetic field that is required to move the
wire, force acting on a current caring wire due to magnetic field is equal to weight the wire, that is, mg.
Hence, we can write,
mg = ILB
⇒ B = mg/IL
= (0.13 * 9.8)/(6.0 * 0.32)
=0.66354 Tesla
Hence, the minimum magnetic field is 0.66354 Tesla.
b) By using Maxwell's right hand thumb Rule along current flow, the direction of magnetic field is determined as normal to the page pointing outwards.
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Find the projection of vector vector A=2i-8j+k in the direction of vector B=3i-4j-12k
The certain forest moon travels in an approximately circular orbit of radius
14,441,566 m with a period of 6 days 10 hr, around its gas giant exoplanet host. Calculate the mass of the exoplanet from this
information. (Units: kilograms)
Answer:
Mass of Exoplanet = 0.58 kg
Explanation:
First, we will calculate the speed of the forest moon:
[tex]speed = v = \frac{Circumference}{time}\\[/tex]
circumference = 2πr = 2π(14441566 m) = 90739035.3 m
time = 6 days 10 hr = (6 days)(24 h/1 day)(3600 s/1 h) + (10 h)(3600 s/1 h)
time = 554400 s
Therefore,
[tex]v = \frac{90739035.3\ m}{554400\ s}\\\\v = 163.67\ m/s[/tex]
We know that the centripetal force on forest moon will be equal to the gravitational force given by Newton's Gravitational Law, as follows:
[tex]Centripetal\ Force = Gravitational\ Force\\\frac{m_{moon}v^2}{r} = \frac{Gm_{moon}m_{exoplanet}}{r^2}\\\\m_{exoplanet} = \frac{v^2r}{G}\\\\m_{exoplanet} = \frac{(163.67\ m/s)^2(14441566)}{6.67\ x\ 10^{-11}\ N.m^2/kg^2}[/tex]
Mass of Exoplanet = 0.58 kg
...........................
Answer: kooi kooi
how r u and thanks for the free points :)
Joe is carrying a load of building supplies and likely injuring his back by supporting 882 in of weight. Calculate the mass of what he is carrying
Answer:
Answer is c because in exothermic reactions, energy releases.in cellular respiration,atp+h20 (water) = adp+ pi + energyso celullar respiration is exothermic and releases energy
Explanation:
Answer is c because in exothermic reactions, energy releases.in cellular respiration,atp+h20 (water) = adp+ pi + energyso celullar respiration is exothermic and releases energy
Which of the following would fill in the table where "A" is?
Symbol
Element Name
Atomic Number Mass Number
#nº
#p
#e
Net Charge
Si
Silicon
A (#13)
27
B (#14 14 C (#15)
0
D (#16)
Potassium
18
39
19
18 E (#17)
A. 3
B. 13
C. 14
D. 28
Answer:
.lyrjhg.
Explanation:
qqlqfgtjnfh
is this correct and why
Answer:
correct ooooooooooooo
Consider two insulating balls with evenly distributed equal and opposite charges on their surfaces, held with a certain distance between the centers of the balls. Construct a problem in which you calculate the electric field (magnitude and direction) due to the balls at various points along a line running through the centers of the balls and extending to infinity on either side. Choose interesting points and comment on the meaning of the field at those points. For example, at what points might the field be just that due to one ball and where does the field become negligibly small? Among the
things to be considered are the magnitudes of the charges and the distance between the centers of the balls. Your instructor may wish for you to consider the electric field off axis or for a more complex array of charges, such as those in a water molecule.
Answer:
interest point:
1) Point on the left side
2) Point within the radius r₁ of the first sphere
3) Point between the two spheres
4) point within the radius r₂ of the second sphere
5) Right side point
Explanation:
In this case, the total electric field is the vector sum of the electric fields of each sphere, to simplify the calculation on the line that joins the two spheres
We will call the sphere on the left 1 and it has a positive charge Q with radius r1, the sphere on the right is called 2 with charge -Q with radius r2. The total field is
E_ {total} = E₁ + E₂
E_{ total} = [tex]k \frac{Q}{x_1^2} + k \frac{Q}{x_2^2}[/tex]
the bold indicate vectors, where x₁ and x₂ are the distances from the center of each sphere. If the distance that separates the two spheres is d
x₂ = x₁ -d
E total = [tex]k \frac{Q}{x_1^2} - k \frac{Q}{(x_1 - d)^2}[/tex]
Let's analyze the field for various points of interest.
1) Point on the left side
in this case
E_ {total} = [tex]k Q \ ( \frac{1}{x_1^2} - \frac{1}{(x_1 +d)2} )[/tex]
E_ {total} = [tex]k \frac{Q}{x_1^2}[/tex] [tex]( 1 - \frac{1}{(1 + \frac{d}{x_1} )^2 } )[/tex]
We have several interesting possibilities:
* We can see that as the point is further away the field is more similar to the field created by two point charges
* there is a point where the field is zero
E_ {total} = 0
x₁² = (x₁ + d)²
2) Point within the radius r₁ of the first sphere.
In this case, according to Gauus' law, the charge is on the surface of the sphere at the point, there is no charge inside so this sphere has no electric field on its inner point
E_ {total} = [tex]-k \frac{Q}{x_2^2} = -k \frac{Q}{((d-x_1)^2}[/tex]
this expression holds for the points located at
-r₁ <x₁ <r₁
3) Point between the two spheres
E_ {total} = [tex]k \frac{Q}{x_1^2} + k \frac{Q}{(d+x_1)^2}[/tex]
This champ is always different from zero
4) point within the radius r₂ of the second sphere, as there is no charge inside, only the first sphere contributes
E_ {total} = [tex]+ k \frac{Q}{(d-x_1)^2}[/tex]+ k Q / (d-x1) 2
point range
-r₂ <x₂ <r₂
5) Right side point
E_ {total} = [tex]k \frac{Q}{(x_2-d)^2} - k \frac{Q}{x_2^2}[/tex]
E_ {total} = [tex]- k \frac{Q}{x_2^2} ( 1- \frac{1}{(1- \frac{d}{x_2})^2 } )[/tex]- k Q / x22 (1- 1 / (x1 + d) 2)
we have two possibilities
* as the distance increases the field looks more like the field created by two point charges
* there is a point where the field is zero
Help me please I have other ones like this too on my page please help!
Check out this app! It's millions of students helping each other get through their schoolwork. https://brainly.app.link/qpzV02MawO
Answer:
OK we appreciate your concern.
1. A person kicks a rock off a cliff horizontally with a speed of 20 m/s. It takes 7.0 seconds to hit the
ground, find:
a. height of the cliff
b. final vertical velocity
C. range
D.speed and angle of impact
This problem involved half projectile.
initial velocity, vo = 20 m/s
time of flight, t = 7 s
(a) Simply use the formula to get the height, h:
h = vo*t - (1/2)gt^2
(b) To get the final vertical velocity or terminal velocity (vf), use the formula:
(vf)^2 - (vo)^2 = 2gh
(c) Use the formula find the horizontal distance traveled, R:
R = vo * cos(θ) * t
But since the angle involved with respect to horizontal is zero, and cos(0) = 1, we have
R = vo * t
Hope this helps~ `u`
Jai
Suppose a rocket in space is accelerating at 1.5 m/s2. If, at a later time, the rocket quadruples its thrust (i.e., net propelling force), what is the new acceleration?
In medieval warfare, one of the greatest technological advancement was the trebuchet. The trebuchet was used to sling rocks into castles. You are asked to study the motion of such a projectile for a group of local enthusiast planning a medieval war reenactment. Unfortunately an actual trebuchet had not been built yet, so you decide to first look at the motion of a thrown ball as a model of rocks thrown by a trebuchet. Specifically, you are interested in how the horizontal and the vertical components of the velocity for a thrown object change with time. 1. Make a large rough sketch of the trajectory of the ball after it has been thrown. Draw the ball in at least five different positions; two when the ball is going up, two when it is going down, and one at its maximum height. Label the horizontal and vertical axes of your coordinate system.
2. On the sketch, draw and label the expected acceleration vectors of the ball (relative sizes and directions) for the five different positions. Decompose each acceleration vector into its vertical and horizontal components.
3. On the sketch, draw and label the velocity vectors of the object at the same positions you chose to draw your acceleration vectors. Decomposes each velocity vector into its vertical and horizontal components. Check to see that the changes in the velocity vector are consistent with the acceleration vectors.
4. Looking at the sketch, how does someone expect the ball's horizontal acceleration to change with time? Could you give a possible equation giving the ball's horizontal acceleration as a function of time? Graph this equation. If there are constants in your equation, what kinematic quantities do they represent? How would someone determine these constants from the graph?
5. Looking at the sketch, how does someone expect the ball's horizontal velocity to change with time? Is it consistent with the statements about the ball's acceleration from the previous question? Could you give a possible equation for the ball's horizontal velocity as a function of time? Graph this equation. If there are constants in the equation, what kinematic quantities do they represent? How would someone determine these constants from the graph?
6. Could you give a possible equation for the ball's horizontal position as a function of time? Graph this equation. If there are constants in the equation, what kinematic quantities do they represent? How would someone determine these constants from the graph? Are any of these constants related to the equations for horizontal velocity or acceleration?
7. Repeat questions 4-6 for the vertical component of the acceleration, velocity, and position. How are the constants for the acceleration, velocity and position equations related?
Answer:
2) a_y= -g 3) vₓ=constant v_y = v_{oy} - g t, 4) vₓ = v₀ₓ - ax t
5) changes the horizontal speed, should change range
7) changes the vertical speed change the maximum height
Explanation:
1) After reading your long writing, we are going to solve the exercise, in the attachment you can see the different vectors.
2) The acceleration vectors are vertical and directed downwards due to the attraction of the Earth (gravity force) this force is constant, on the x axis there is no acceleration
3) the velocity vectors on the x-axis are constant because there are no relationships and the y-axis changes value according to the expression
v_y = v_{oy} - gt
at the point of maximum height, vy = 0 is equal to the maximum height
4) For someone to change the horizontal acceleration we must assume a friction with the air, in this case they relate it would be in the opposite direction to the horizontal speed
In the graph it would be directed to the left, therefore the velocity would be
vₓ = v₀ₓ - ax t
5 and 6) If someone changes the horizontal speed, they should change the range of the shot for greater horizontal speed, the rock goes further.
the equations of motion are
x = v₀ₓ t
y = v_{oy} t - ½ g t²
7) If someone changes the vertical speed change the maximum height, but not the scope of the shot, for higher speed higher maximum height,
the equations of motion are the same.
You are comparing the beam waste for two different situations with the goal of using the smallest beam waste possible. A Nd-YAG laser system emits light at 532 nm and the beam is 8 mm in diameter. You also have a Ti-sapphire laser that emits at 855 nm and has a beam diameter of 6 mm. Compare the beam waist for both laser systems using a focusing lens with a focal length of 10 mm. Assume the light fills the lenses in each case
Answer:
comparing the beam waist for both lasers ( ratio of the beam waists )
4.536 μm / 2.117 μm = 2.14
Explanation:
Nd-YAG laser system : emits at 532 nm , beam diameter = 8 mm
Ti-sapphire laser system : emits at 855 nm , Beam diameter = 6mm
Comparing the beam waist for both lase systems using a focusing lens
Focal length = 10 mm
assumption : light fills lenses in each laser system
Beam waist radius ( W ) = [tex](\frac{2\beta }{\pi } )(\frac{F}{D} )[/tex]
β = wavelength , D = diameter illuminated , F = focal length
For
Nd-YAG laser system
β = 532 mm , D = 8 mm
hence ( Wn ) = [tex](\frac{2\beta }{\pi } )(\frac{F}{D} )[/tex] = ( 2*532 / π ) ( 10 / 8 ) = 2.117 μm
For
Ti-sapphire laser
β = 855 nm , D = 6 mm
hence ( Wt ) [tex](\frac{2\beta }{\pi } )(\frac{F}{D} )[/tex] = ( 2* 855 ) / π ) ( 50 / 6 ) = 4.536 μm
comparing the beam waist for both lasers ( ratio of the beam waists )
4.536 μm / 2.117 μm = 2.14
Need help with questions 2–3 ASAP pls ease help
Answer:
A and D
Explanation:
❤❤ plz follow have a food day
identify the types of motion in each activity.1.walking a long a hallway. 2.motion of the blades of the fan. 3.earths rotation 4.ball moving on the ground. 5.soldiers marching.
Answer:
Explanatation
1 is just walking
2 spinging
3 roatating
4 rolling
5 stomping there feet
Those should be right but if im wrong then just someone eles the brainly
SCIENCE
The growth of algae in ocean water is limited by their need for
a.
warm ocean currents.
b.
carbon dioxide and sunlight.
c.
dissolved oxygen.
d.
low salinity.
Answer:
c is trueeeeeeeeeeeeeee
The growth of algae in ocean water is limited by their need for dissolved oxygen. Thus, the correct answer is option C.
What is algae?The term "algae" refers to a large and diverse group of photosynthetic eukaryotic organisms. It is a polyphyletic grouping of species from several distinct clades. Most are aquatic and autotrophic, lacking many of the distinct cell and tissue types found in land plants, such as stomata, xylem, and phloem.
Algae, like all organisms, normally grow in balance with their ecosystems, with the amount of nutrients in the water limiting their growth. Deep ocean water contains fewer algae because algae require sunlight and carbon dioxide to thrive.
Therefore, due to need for dissolved oxygen the growth of algae in ocean water is limited.
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A satellite, moving in an elliptical orbit, is 368 km above Earth's surface at its farthest point and 164 km above at its closest point. (a) Calculate the semimajor axis of the orbit. Incorrect: Your answer is incorrect. m (b) Calculate the eccentricity of the orbit. Incorrect: Your answer is incorrect. Did you find the semimajor axis a from the greatest and smallest radii
Answer:
a) 6636 km
b) 0.0154
Explanation:
The height above the earth at its furthest point is 368 km
The height above the earth at its closest point is 164 km
Radius of the Earth is 6370 km
The distance of the satellite from the center of the earth to the furthest point is 6370 + 368 km = 6738 km
The distance of the satellite from the center of the earth to the closest point is 6370 + 164 = 6534 km
If we add together the sum of the distance of the satellite from the furthest and its closest distance, it is equal to the 2 major semi axis.
Basically,
2a = R + r
a = (R + r) / 2
a = (6738 + 6534) / 2
a = 13272 / 2
a = 6636 km
Eccentricity, e = (a - r) / a
Eccentricity, e = (6636 - 6534) / 6636
Eccentricity, e = 102 / 6636
Eccentricity, e = 0.0154
1. If airbags reduce the impact force from an accident why has there been questions over their safety?
2. Are airbags the safest option to prevent serious injury or death from a car accident?
Answer:Air bags can leave you in even more injury, From the impact they give
You could end up with a broken nose,arm
concussion
A ball is thrown so that its speed increases by 20 m/s in 10 seconds. What is the ball’s acceleration?
Answer: a= 2 m/s²
Explanation: acceleration = change of speed/ time = 20 m/s / 10 s
Experiments carried out on the television show Mythbusters determined that a magnetic field of 1000 gauss is needed to corrupt the information on a credit card's magnetic strip. (They also busted the myth that a credit card can be demagnetized by an electric eel or an eelskin wallet.) Suppose a long, straight wire carries a current of 5.0 A .
How close can a credit card be held to this wire without damaging its magnetic strip?
Answer:
his distance is too small (r = 0.01 mm), therefore the cut can be at any distance
Explanation:
For this exercise let's use the ampere law.
Let's use a cylinder as the circulating surface
∫ B. ds = μ₀ I
in this case the field is circular and ds is circular therefore the angle between them is zero and cos 0 = 1
B 2π r = μ₀ I
r = [tex]\frac{\mu_o I}{2\pi B}[/tex]
The field needed to demagnetize the card is B = 1000 gauss = 0.1 T
r = [tex]\frac{4\pi 10^{-7} 5.0 }{2\pi \ 0.1}[/tex]
r = 2 10⁻⁷ 5.0/0.1
r = 1 10⁻⁵ m
this distance is too small (r = 0.01 mm), therefore the cut can be at any distance
Please help !!!!!!!!!!!!
Answer:
152 Volts
Explanation:
First, the Resistors are in series. So, the net resistance, R = 3 + 4 +3 + 4 +5
= 19 ohms
Using, V = IR;
V (the needed P.D) = 8 x 19 = 152 Volts
will be needed to successfully transport 8 amps of current round the circuit
Answer:
4q
Explanation:
wait nvm i dont know
an ice skater is moving across a flat and level skating rink and is speeding up. which one of the following statement is true of the ice skater
a) its potential energy is constant
b) its potential energy is increasing
c) its potential energy is decreasing
Answer:
A
Explanation:
The angular momentum of a system of particles around a point in a fixed inertial reference frame is conserved if there is no net external torque around that point:
d
→
L
d
t
=
0
or
→
L
=
→
l
1
+
→
l
2
+
⋯
+
→
l
N
=
constant
.
Note that the total angular momentum
→
L
is conserved. Any of the individual angular momenta can change as long as their sum remains constant. This law is analogous to linear momentum being conserved when the external force on a system is zero.
How can a mutation result in a new trait
They are called beneficial mutations. They lead to new versions of proteins that help organisms adapt to changes in their environment. Beneficial mutations are essential for evolution to occur. They increase an organism's changes of surviving or reproducing, so they are likely to become more common over time.
Mutation is when there is a particular change in the bases of the DNA, this changes how enzymes and proteins work, and there may be a chance that this change will be helpful.
Riders in an amusement park ride shaped like a Viking ship hung from a large pivot are rotated back and forth like a rigid pendulum. At each end of the swing the ship hangs motionless for a moment before the ship swings down under the influence of gravity. Assume that this motionless point occurs when the bar connecting the pivot point and the ship is horizontal.
Required:
a. Assuming negligible friction, find the speed of the riders at the bottom of its arc, given the system's center of mass travels in an arc having a radius of 14.0 m and the riders are near the center of mass.
b. What is the centripetal acceleration at the bottom of the arc?
c. Draw a free body diagram of the forces acting on a rider at the bottom of the arc.
d. Find the force exerted by the ride on a 60.0 kg rider and compare it to her weight.
e. Discuss whether the answer seems reasonable.
Answer:
a) v = 16.57 m / s, b) a = 19.6 m / s², d) N = 1.76 10³ N, N / W = 3
Explanation:
This exercise looks interesting, but I think you have some problem with the writing, the questions seem a bit disconnected from the initial text.
Let's answer the questions.
a) For this part we can use energy considerations.
Starting point. The upper part of the trajectory indicates that the arm is horizontally
Em₀ = U = m g h
in this case h = r
Final point. For lower of the trajectory
Em_f = K = ½ m v²
as they indicate that there is no friction
Em₀ = em_f
mgh = ½ m v²
v = [tex]\sqrt{2gh}[/tex]
let's calculate
v = [tex]\sqrt{2 \ 9.8 \ 14.0}[/tex]
v = 16.57 m / s
b) the centripetal acceleration has the formula
a = v² / r
a = 16.57² / 14.0
a = 19.6 m / s²
c) see attached where the diagram is
where N is the normal and w the weight
d) let's use Newton's second law
N-W = m a
N - mg = m ar
N = m (g + a)
let's calculate
N = 60.0 (9.8 + 19.6)
N = 1.76 10³ N
the relationship with weight is
N / W = 1.76 10³/( 60 9.8)
N / W = 3
normal is three times greater than body weight
e) the answer is reasonable since by Newton's first law the body must continue in a straight line, therefore to change its trajectory a force must be applied to deflect it
What happens during heat
transfer?
A. Heat always flows from cool to warm.
B. Heat always flows from warm to cool.
C. Heat always flows from warm to hot.
Answer:
B
Explanation:
A marshmallow is fired from ground level with an initial speed of 41.5 m/s at an
angle of 33.5° above the horizontal. (a) Determine the maximum height reached
by the marshmallow. (b) Determine the horizontal range that the marshmallow
travels during its flight.
What causes coastal erosion
La erosión costera es la pérdida o desplazamiento de tierra, o la remoción a largo plazo de sedimentos y rocas a lo largo de la costa debido a la acción de olas, corrientes, mareas, agua impulsada por el viento, hielo transportado por el agua u otros impactos de marejadas ciclónicas.