Answer:
the impulse must be the same in these two cases F t = m ([tex]\sqrt{2g h_f } - \sqrt{2g h_o}[/tex])
Explanation:
For this exercise we use the relationship between momentum and momentum
I = Δp
F t = m v_f - m v₀
To know the speed we use the conservation of energy
starting point. Highest point
Em₀ = U = m g h
fincla point. Just before the crash
Em_f = K = ½ m v²
energy is conserved
Em₀ = Em_f
m g h = ½ m v²
v = [tex]\sqrt{2gh}[/tex]
we substitute in the impulse relation
F t = m ([tex]\sqrt{2g h_f } - \sqrt{2g h_o}[/tex])
therefore we can see that as in case the initial and final heights are equal, the impulse must be the same in these two cases
Can you share me your answers ❤️❤️
Answer:
Depending, on how much it's push against together.
Explanation:
Since, the two objects are getting in contact. But, if it's a type of item/thing there's a different frictions, but I know it's normal friction when two objects comes in contact. But, its depending on how much you push it against the two items.
4 of 1
Suppose that you begin on a street corner and walk one block in one minute. How would you express the rate of your motion?
O One block per minute.
O The information is insufficient.
O Fast
O One minute per block.
Answer:
The first choice, one block per minute.
The mass of a ship before launch is 55,000 metric tons. The ship is launched down a ramp and drops a total of 10 vertical meters before coming to rest in a lock containing 75,000 cubic meters of fresh water. The specific heat of water is 4200 joules per kilogram degree Celsius. Assume all energy is transferred from the ship to the water. Determine the change in temperature of the water in degrees Celsius .
Answer:
ΔT = 17.11 °C
Explanation:
In this case, we have a ship standing on a place with a given mass and it's about to be launched to a lock containing water.
At first, before launch, the ship has a potential energy, and when the ship hits the water after being launched, this potential energy is transformed into kinetic energy.
So, let's calculate first the potential energy of the ship:
E = mgh (1)
We have the mass, gravity and height, so we need to replace the given data here. Before we do that, let's remember to use the correct units. A ton is 1000 kg, so replacing and converting we have:
E = (55000 ton * 1000 kg/ton) * (9.8 m/s²) * 10 m
E = 5.39x10⁹ J
Now this energy will be the same when the ship hits the water, only that is kinetic energy that will result in the rise of temperature. To get this rise we use the following expression:
E = m * C * ΔT (2)
We have the energy, the mass of water (assuming density of water as 1 kg/m³) and the specific heat, so, replacing in (2) and solving for ΔT we have:
ΔT = E / m * C (3)
ΔT = 5.39x10⁹ / 4200 * 75000
ΔT = 17.11 °CHope this helps
EXAMPLE
• A patient lying horizontally on a hospital bed is found to have an
enlargement of his blood vessel where the walls have weakened. The
cross-sectional area of the enlargement is 2.0A where A is the cross-
section of the normal aorta. The normal speed of blood through the
person's aorta is 0.40 m/s, and the density of blood is 1,060 kgm-3
Calculate
. a) the speed of blood in the enlargement.
• b) how much higher the pressure is in the enlargement.
Answer:
Explanation:
a ) The volume of blood flowing per second throughout the vessel is constant .
a₁ v₁ = a₂ v₂
a₁ and a₂ are cross sectional area at two places of vessel and v₁ and v₂ are velocity of blood at these places .
2A x v₁ = A x .40
v₁ = .20 m /s
b )
Let normal pressure be P₁ when cross sectional area is 2A and at cross sectional area A , pressure is P₂
Applying Bernoulli's theorem
P₁ + 1/2 ρv₁² = P₂ + 1/2 ρv₂²
P₁ - P₂ = 1/2 ρ(v₂² - v₁² )
= .5 x 1060 ( .4² - .2² )
= 63.6 Pa .
Please help 25 points!
Three waves with frequencies of 1 Hertz (Hz), 3 Hz, and 9Hz travel at the same speed. Which of the following statements is correct?
A. The 1 Hz wave contains the most energy.
B. The crests of all three waves are of equal height.
C. The wavelength of the 9Hz wave is three times that of the 3 Hz wave.
D. The 1 Hz wave has the longest wavelength.
Answer:
B
Explanation:
The crest of all three waves are of equal height
I WILL MAKE THE BRAINLEST
6. The front that moved over Roberto's area the last week of the month was humid. Based on the chart and your knowledge of fronts, what kind of front would this most likely be?
Answer choices
A. cold front
B. warm front
C. occluded front
D. stationary front
Answer:
warm front .,
Explanation:
...........
What formula could be used to find distance if you know the speed an the time
Answer: d = st
Explanation:
We know that the distance is equal to the rate (speed) times the time
d = st
A bar magnet is attached solidly to a frictionless surface and its length is aligned with the x axis.To the right of the first magnet a short distance away is a second bar magnet with its center placed on the x axis and its length perpendicular to the x axis.The second magnet is free to move.Once placed in position at rest,which best describes the initial motion of the second magnet?
a. The magnet will move away from the fixed magnet.
b. The magnet will not move.
c. The magnet will move toward the fixed magnet.
d. The magnet will start to rotate.
Answer:
the correct answer is D
Explanation:
For this exercise we must remember that the poles of the same if not repel each other and the poles of different signs attract.
With this, let's analyze the situation presented.
The two magnets are perpendicular, with the second magnet to the right of the first.
We have two cases:
* first magnet with the north pole to the right
If the north of the second magnet is upwards there is a repulsion and the south pole of the second magnet there is an attraction with the north pole of the first magnet, so it would have a force that has to rotate the second magnet.
The force with the south pole of the first magnet that is at a greater distance is less, so the resultant of force is determined by the nearest poles.
If the poles of the second magnet are reversed, that is, the South pole up and the North pole down, the same result is obtained, but with a twist in the opposite direction.
* first magnet with the south pole to the right
Repels with the south pole of the second magnet and attracts with the north of the second magnet
therefore in both possibilities the second magnet acquires a rotational movement
Consequently the correct answer is D
Work is done on an object when
A) The displacement is not zero
B) The force and the displacement are perpendicular
C) The displacement is zero
D) The force is zero
Answer:
Incomplete question but the closest ans is
A) The displacement is not zero
Explanation:
For work to be done, the direction of force has to be parallel to or in the same direction as the displacement, this can then be calculated by
[tex]W=Fs \ cos[/tex]θ
Where θ is the angle between direction of force and direction of displacement
By this equation,
if Force is zero, W is zero
if displacement is zero, W is zero
if force and displacement are perpendicular θ=90, cosθ = 0
So A is the cloest choices
A 6 kg box with initial speed 8 m/s slides across the floor and comes to a stop after 2.4 s. A) What is the coefficient of kinetic friction?B) How far does the box move? C) You put a 5 kg block in the box, so the total mass is now 11 kg, and you launch this heavier box with an initial speed of 7 m/s. How long does it take to stop?
Answer:
A. Coefficient of kinetic friction, μ = 0.34
B. The box moves a distance of 9.64 m before coming to a stop
C. The heavier box will stop after 2.1 seconds
Explanation:
a. The coefficient of kinetic or sliding friction is given as: μ = F/R
where applied force, F = m × (∆v)/t
∆v = v - u
∆v = 0 - 8 m/s = -8 m/s; t = 2.4 s
R = normal reaction = m×g
where g = 9.8 m/s²
Substituting in the kinetic friction formula; μ = m∆v/t ÷ 1/m×g
μ = ∆v/g×t
μ = 8 / 9.8 × 2.4
μ = 0.34
b. Using the equation v² = u² + 2as to calculate the distance travelled by the box
where v = 0 m/s; u = 8.0 m/s; a = ? s = ?
From F = ma = μR
a = μR/m = (μ × m × g)/m
a = μg
a = 0.34 × 9.8
a = 3.32 m/s²
This is negative acceleration or deceleration
Substituting in the equation of motion
8² + 2 × -3.32 × s = 0
-6.64s = -64
s = 9.64 m
Therefore, the box moves a distance of 9.64 m before coming to a stop
c. The coefficient of friction is independent of mass.
Using the formula in (a): μ = ∆v/g×t
t = ∆v/μg
t = 7/0.34 × 9.8
t = 2.10 s
Therefore, the heavier box will stop after 2.1 seconds
The coefficient of kinetic friction of the box is 0.34.
The distance traveled by the box is 9.6 m.
The time taken for the heavier box to stop is 2.1 s.
The coefficient of kinetic frictionThe coefficient of kinetic friction of the box is calculated as follows;
[tex]\mu mg = ma\\\\\mu g = a\\\\\mu g = \frac{v}{t} \\\\\mu = \frac{v}{gt} \\\\\mu = \frac{8}{9.8 \times 2.4} \\\\\mu = 0.34[/tex]
The distance traveled by the boxThe distance traveled by the box is calculated as follows;
[tex]v^2 = u^2 + 2as\\\\2as = -u^2\\\\s = \frac{-u^2}{2a} \\\\s = \frac{-u^2}{2\mu g} \\\\s = \frac{-(8)^2}{2\times 0.34 \times 9.8} \\\\s = -9.6 \ m\\\\|s| = 9.6 \ m[/tex]
The time taken for the heavier box to stop is calculated as;
[tex]\mu = \frac{v}{gt} \\\\t = \frac{v}{\mu g} \\\\t = \frac{7}{0.34 \times 9.8}\\\\ t = 2.1 \ s[/tex]
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1.A boy runs at a speed of 3.3 m/s straight off the end of a diving board that is 3 meters above the water
2.What is the horizontal distance the boy in # 1 travels while in the air ?
If a boy runs at a speed of 3.3 m/s straight off the end of a diving board that is 3 meters above the water, then the horizontal distance traveled by the boy would be 2.58 meters.
What are the three equations of motion?There are three equations of motion given by Newton,
v = u + at
S = ut + 1/2 × a × t²
v² - u² = 2 × a × s
As given in the problem if a boy runs at a speed of 3.3 m/s straight off the end of a diving board that is 3 meters above the water,
3 = ut + 1/2 × a × t²
3 = 0 + 0.5 × 9.8 × t²
t = 3 / 4.9
t = 0.7824
The horizontal distance traveled by the boy = 3.3 × 0.7824
= 2.58 meters
Thus, the horizontal distance traveled by the boy would be 2.58 meters.
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A student is conducting an experiment to compare the resistivity of two unknown materials by using two wires, each made of one of the materials and each connected in a circuit. The student measures the potential difference across and current in the wires. What must be the same to be able to compare the resistivities using just the potential difference and current measurements?
Answer:
is there a. b. c or d?
Explanation:
A high-voltage transmission line carries 1,000A at 700,000 V. What is the power carried by the line?
If the resistance in the wire is 1ohm/mile and the line is 110 miles long, what is the power loss due to resistive losses in this wire?
a. 11kW
b. 110kW
c. 11 MW
d. 110 MW
Answer:
d. 110 MW
Explanation:
The computation of the power loss is given below:
= I^2 × R
= 1000^2 × 110
= 110 × 10^6 W
= 110 MW
Here I represent the current through the resistance
ANd, R represent the resistance
So, the correct option is d
A 2,000 kg rocket is launched 12 km straight up at a constant acceleration into the sky at which point the rocket is travelling at 750 m/s. a) How much work was done by the rocket? b) What is the magnitude of the acceleration of the rocket? c) How long did the flight take?
Answer:
797700000 J
Explanation:
From the question,
The work done by the rocket, is given as,
W = Ek+Ep............. Equation 1
Where Ek and Ep are the potential and the kinetic energy of the rocket respectively.
Ep = mgh............ Equation 2
Ek = 1/2mv²............. equation 3
Substitute equation 2 and equation 3 into equation 1
W = mgh+1/2mv².............. Equation 4
Where m = mass of the rocket, h = height, v = velocity of the rocket, g = acceleration due to gravity.
Given: m = 2000 kg, h = 12 km = 12000 m, v = 750 m/s, g = 9.8 m/s²
Substitute into equation 4
W = 2000(12000)(9.8)+1/2(2000)(750²)
W = 235200000+562500000
W = 797700000 J
A fan has four identical, symmetrically placed blades. The blades are rotating clockwise at twenty revolutions per second.
A) What is the smallest time interval between stroboscope flashes that will make the fan blades appear motionless?
B) What is the highest frequency (in flashes per second) at which a stroboscope will make the
fan blades appear to stand still? Show your calculation.
C) The same questions as (a) and (b), but someone has put a yellow dot on one blade, and now you want the yellow dot to appear to be standing still. Explain, and show your calculation.
D) Now the stroboscope is set for nineteen flashes per second, and the yellow dot appears to be slowly rotating. Which direction does it appear to rotate, clockwise or counterclockwise? Explain, and show your calculation.
E) The same as (d), but the stroboscope is set for twenty-one flashes per second. Explain, and show your calculation.
Answer:
A) t = 1.249 10⁻² s, B) f = 80 Hz, C) f = 20 Hz,
D) slowly advancing an angle of approximately Δθ = 0.05 rad each flash
E) In each flash it seems to go backward an angle of Δθ = -0.05 rad
Explanation:
A) To make it appear that the blades are immobile, it implies that every time the light turns on, a blade should be in the same position, therefore, as we have 4 blades, they must rotate an angle of 2π/4,
θ = π / 2
θ = 1.57 rad
taking the angle let's use the endowment kinematics relations
θ = w₀ t + ½ α t²
in general the fans rotate at constant speed α= 0
θ = w₀ t
t = θ / w₀
let's reduce the magnitudes to the SI system
w₀ = 20 rev / s (2π rad / 1rev) = 125.66 rad / s
let's calculate
t = 1.57 / 125.66
t = 1.249 10⁻² s
B) the fastest speed for the blades to rotate is when one blade of a complete turn , we use the relationship between the fecuance and the period
f = 1 / T
f = 1 / 1.25 10⁻²
f = 80 Hz
C) we have two possibilities:
* a yellow dot is placed on each sheet
In this case the angular velocity of the blade is the same at all points, therefore the results obtained should not change
* a yellow dot is placed on a single sheet.
Here for the point to remain fixed the angle of rotation must be
θ= 2π rad
the time is
t = 2π / 125.66
t = 5 10⁻² s
the maximum frequency is
f = 1/5 10⁻²
f = 20 Hz
D) The copy strobe rotates at f = 19 Hz, the time between each flash is
t = 1/19
t = 5.26 10⁻² s
this time is higher, so the angle turned is large
θ = w t
θ = 125.66 5.26 10⁻²
θ = 6.61 rad
the relationship between this angle and the angle of a circle is
θ = 1,052
We can see that it is this time the blade rotates 1 complete turns, for this the position of the blade changes us, for the other 0.052 rad the blade rotates a little more than the circumference therefore it seems that it is slowly advancing an angle of approximately
Δθ = 0.05 rad each flash
E) in this case changes the flash speed
t = 1/21
t = 4.76 10⁻² s
the angle rotated is
θ = 125.66 4.76 10⁻²
θ = 5.984 rad
θ / 2π = 0.95
in that case, the blade did not complete the turn, therefore in each flash it seems to go backward an angle
Δθ = -0.05 rad
The Willis Tower in Chicago has an observation deck 412 m above ground.
How far can you see out over Lake Michigan from the observation deck?
Answer:
Check Newton's Rings:
d = height of air film
s = distance from center to ring being considered
R = radius of circle considered
The approximate formula is:
d = s^2 / (2 R) or s = (2 R d)^1/2
If we just use 4000 mi for R and 1/4 mi for d the height
we get s = (2 * 4000 * 1/4)^1/2 = 2000^1/2 mi = 45 mi
The time required for one complete cycle of a mass oscillating at the end of a spring is 0.40 s. What is the frequency of oscillation?
Answer:
the frequency of the oscillation is 2.5 Hz.
Explanation:
Given;
time to complete the oscillation, t = 0.4 s
number of oscillations, n = 1
The frequency of the oscillation is calculated as;
[tex]F = \frac{n}{t} \\\\F = \frac{1}{0.4} \\\\F = 2.5 \ Hz[/tex]
Therefore, the frequency of the oscillation is 2.5 Hz.
If a biker rides west for 50 miles from his starting position, then turns and bikes back east for 80 miles. What is his displacement?
Answer:
Displacement = 30 miles due east.
Explanation:
Let the distance due west be A
Let the distance due east be B
Given the following data;
A = 50 miles
B = 80 miles
To find the displacement;
Displacement can be defined as the change in the position of a body or an object. It is a vector quantity because it has both magnitude and direction.
Thus, the displacement would be calculated by subtracting distance A from distance B because the rider rode in opposite directions.
Displacement = B - A
Displacement = 80 - 50
Displacement = 30 miles due east.
A child and a sled with a combined mass of 50.0 kg slide down a frictionless slope. If the sled starts from rest and has a speed of 3.00 m/s at the bottom, what is the height of the hill?
Answer:
The height of the hill is 0.46 m.
Explanation:
Given;
mass of the child and sled, m = 50 kg
initial velocity of the sled, u = 0
final velocity of the sled, v = 3 m/s
The height of the high is calculated from the law of conservation of energy;
P.E at top = K.E at bottom
mgh = ¹/₂mv²
gh = ¹/₂v²
[tex]h = \frac{v^2}{2g} \\\\h = \frac{3^2}{2\times 9.8} \\\\h = 0.46 \ m[/tex]
Therefore, the height of the hill is 0.46 m.
You are conducting an experiment inside an elevator that can move in a vertical shaft. A load is hung vertically from the ceiling on a string, and is stationary with respect to you. The tension in the string is measured to be exactly equal to the force due to gravity on the load. No other forces are acting on the load. Which of the following statements about the elevator are correct?
A. The elevator is an inertial frame of reference.
B. The elevator is not an inertial frame of reference.
C. The elevator may be at rest for the duration of the entire experiment.
D. The elevator may be moving at a constant velocity upward.
E. The elevator may be moving at a constant velocity downward.
F. The elevator must be accelerating
Answer: B. The elevator is not an inertial frame of reference.
F. The elevator must be accelerating.
Explanation:
Since we've been given the information that the tension in the string is measured to be exactly equal to the force due to gravity on the load and that no other forces are acting on the load, the statements about the elevator that are correct include:
• The elevator is not an inertial frame of reference.
• The elevator must be accelerating.
If the elevator isn't accelerating, the tension in the string can't be equal to the force of gravity.
QUCIK!! SOMEONE PLEASE HELP! I’LL MARK BRAINLIEST!!
Answer:
A. v = √2gh
B. No! The final velocity does not depend on the mass of the car.
C. Yes! the final velocity depends on the steepness of the hill
D. 3.28 m/s
Explanation:
A. Determination of the final velocity.
½mv² = mgh
Cancel out m
½v² = gh
Cross multiply
v² = 2gh
Take the square root of both side
v = √2gh
B. Considering the formula obtained for the final velocity i.e
v = √2gh
We can see that there is no mass (m) in the formula.
Thus, the final velocity does not depend on the mass of the car.
C. Considering the formula obtained for the final velocity i.e
v = √2gh
We can see that there is height (h) in the formula.
Thus, the final velocity depends on the steepness of the hill
D. Determination of the final velocity.
Height (h) = 0.55 m
Acceleration due to gravity (g) = 9.8 m/s²
Velocity (v) =?
v = √2gh
v = √(2 × 9.8 × 0.55)
v = √10.78
v = 3.28 m/s
Variations in the resistivity of blood can give valuable clues to changes in the blood's viscosity and other properties. The resistivity is measured by applying a small potential difference and measuring the current. Suppose a medical device attaches electrodes into a 1.5-mm-diameter vein at two points 5.0 cm apart.
Required:
What s the blood resistivity if a 9.0 V potential difference causes a 230μA current through the blood in the vein?
Answer:
ρ= 1.378 10⁴ Ω / m
Explanation:
Let's use ohm's law
V = i R
R = V / i
let's calculate
R = 9.0 / 230 10⁻⁶
R = 3.9 10⁴ Ω
now we can use the definite of resistance
R = ρ [tex]\frac{L}{A}[/tex]
the area of a circle is
A = π r² = π (d/2)²
ρ = R A / L
ρ = π R [tex]\frac{d^2}{4L}[/tex]
let's calculate
ρ = π 3.9 10⁴ [tex]\frac{(1.5 \ 10^{-3}^2 }{4 \ 5 \ 10^{-2}}[/tex]
ρ= 1.378 10⁴ Ω / m
What two air masses creates hurricanes?
Answer:
The warm seas create a large humid air mass. The warm air rises and forms a low pressure cell, known as a tropical depression.
Explanation:
Hurricanes arise in the tropical latitudes (between 10 degrees and 25 degrees N) in summer and autumn when sea surface temperature are 28 degrees C (82 degrees F) or higher.
Answer:
air
Explanation:
Which result is more accurate: the slope or the mean value
4 Two friction disks A and B are to be brought into contact withoutslipping when the angular velocity of disk A is 240 rpm counterclockwise. Disk A starts from rest at time t = 0 and is given a constantangular acceleration with a magnitude α. Disk B starts from rest attime t = 2 s and is given a constant clockwise angular acceleration,also with a magnitude α. Determine (a) the required angular acceleration magnitude α, (b) the time at which the contact occurs
This question is incomplete, the missing image is uploaded along this answer below;
Answer:
a) the required angular acceleration magnitude α is π rad/s² or 3.14 rad/s²
b) the time at which the contact occur is 8 seconds
Explanation:
Given the data in the question;
first we convert the given angular velocity to rad/s
angular velocity = 240 rpm = ((240/60) × 2π ) = 8π rad/s
so
ωA = 8π rad/s
next we determine angular acceleration at point A; so
ωA = at
8π rad/s = at -------let this be equation
thus, angular acceleration of disk A is ωA and rotates in counter clockwise direction.
Next we determine the velocity of point C;
Vc = rA × ωA
where Vc is velocity at point C, rA is radius of A ( 150/1000)m, { from the diagram }
so we substitute
Vc = 0.15m × 8π
Vc = 1.2π m/s
for angular velocity at point B;
Vc = rB × ωB
where rB is the radius of B ( 200/1000)m
we substitute
1.2π = 0.2 × ωB
ωB = 1.2π / 0.2
ωB = 6π rad/s
Thus, the wheel B rotates with an angular velocity of 6π rad/s in clock wise direction.
Now,
a) Determine the required angular acceleration magnitude α
we find the the angular acceleration of disk B after 2 seconds, using the expression;
ωB = at
where angular acceleration is a and t is time ( t - 2)
we substitute
ωB = at
6π = a( t - 2) -------- let this be equation 2
now, lets substract equation 1 form equation 2
(6π = a( t - 2)) - (8π = at)
(6π = at - 2a) - ( 8π = at)
-2π = 0 + -2a
2π = 2a
a = 2π/2
a = π rad/s² or 3.14 rad/s²
Therefore, the required angular acceleration magnitude α is π rad/s² or 3.14 rad/s²
b) determine the time at which the contact occurs;
from equation 1
8π = at
we substitute in the value of a
8π = π × t
t = 8π / π
t = 8 seconds
Therefore, the time at which the contact occur is 8 seconds
What is cutoff wavelength?
Answer:
The cutoff wavelength is the minimum wavelength in which a particular fiber still acts as a single mode fiber. Above the cutoff wavelength, the fiber will only allow the LP01 mode to propagate through the fiber (fiber is a single mode fiber at this wavelength).
Explanation:
A spring is stretched 5 mm by a force of 125 N. How much will the spring stretch
when 250 N force is applied?
Answer:
10 mm
Explanation:
We'll begin by calculating the spring constant of the spring. This can be obtained as follow:
Extention (e) = 5 mm
Force (F) = 125 N
Spring constant (K) =?
F = Ke
125 = K × 5
Divide both side by 5
K = 125 / 5
K = 25 N/mm
Finally, we shall determine how much the spring will stretch when a 250 N force is applied. This can be obtained as follow:
Force (F) = 250 N
Spring constant (K) = 25 N/mm
Extention (e) =?
F = Ke
250 = 25 × e
Divide both side by 25
e = 250 / 25
e = 10 mm
Thus, the spring will stretch 10 mm when a 250 N force is applied.
Calculate the amount of heat given off by 640 g of water cooling from 76 °C to 28° C. Specific heat of water = 4.816 J/g C. Show your step by step process on how you have arrived at your answer. *
Answer:
47947.52 J.
Explanation:
From the question,
Amount of heat given of (Q) = mc(t₁–t₂).................... Equation 1
Where m = mass of water, c = specific heat capacity of water, t₁ = initial temperature, t₂ = final temperature.
Given, m = 640 g = 640 g, c = 4.816 J/g°C, t₁ = 76 °C, t₂ = 28 °c.
Substitute these values into equation 1 above
Q = 640×4.816(48)
Q = 147947.52 J.
Hence the amount of heat given off is 47947.52 J.
When driving across Death Valley in the summertime, it is recommended that you release some air from your tires before making the crossing. Using the principles of Kinetic Molecular Theory (KMT), explain why it is a good idea to follow this recommendation.
According to the ideal gas law, pressure will rise as a gas's temperature rises. There is a limit to how much the tire can expand before the rubber gives in to the pressure build-up.
What the principles of Kinetic Molecular Theory?For every 10 degrees that the temperature drops, the inflation pressure in tires typically decreases by 1 to 2 psi. Moreover, as the tire pressure heats up during the first 15 to 20 minutes of driving, it will increase by one psi every five minutes.
The ideal gas law states that pressure will increase as a gas's temperature increases. Before the rubber gives in to the pressure build up, the tire can only expand so far.
Therefore, The pressure in your tires will increase due to the increased particle movement in hot air, which will cause the centre of the tread to bow out and wear out first. Increasing the demand for new tires.
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Explain how a common housecat gets “worms.”explain(science)
Answer:
Ingenstion
Explanation:
The most common tapeworms that infect cats worldwide are Dipylidium caninum and Taenia taeniaeformis. Dipylidium caninum is transmitted to cats by fleas. The immature fleas larvae ingest the eggs of the worm, but infection is then passed on to a cat when it swallows an infected flea during grooming.
Answer:
The most common worms that housecats get are tapeworms. The housecats always lick their fur to clean them. The tapeworms love where it is damp. After the housecats clean their fur, they will play in the house sometimes they will roll on the floor. Then they get tapeworm eggs, after the eggs hatch the tapeworm will live on the fur of the cats.
Explanation: