Answer:
111.6 g or 0.112 kg
Explanation:
specific heat of liquid water = 4190 J/kg⋅K
specific heat of ice = 2100 J/kg⋅K
heat of fusion for water = 3.34*10^5 J/kg
You didn't state the mass of the beaker, so, I will be assuming that the mass is negligible.
Assuming that the mass of ice required is m kg
Then the heat gained by the ice to attain zero degree will be
= m * 22.3 * 2100
= 46830m J
The heat gained by the ice to melt
= m * 3.34*10⁵ J
= 334000m J
The heat gained by water at zero degree to warm up to 30° =
m * 4190 * 30 = 125700m J
Total heat gained = 506530m J
Note: You didn't state the mass of the water and it's temperature, so I will be assuming that the mass of water is 0.3 kg, and it's temperature was 75° C
The heat lost by hot water to cool up to 30°
= .3 * 4190 * (75 - 30)
= 1257 * 45
= 56565 J
Using the relation, Heat lost = heat gained
506530m = 56565
m = 56565 / 506530 kg
m = 0.112 kg or 111.6 g
Kindly vote brainliest
Explain why a golf ball has a higher density than a ping pong ball, even though they are similar in size and shape.
Answer:
The ping pong ball is hollow.
Explanation:
Since a ping pong ball, unlike the golf ball, is only filled with air, it is less dense. Air is notably LESS dense than solids, which is what comprises the inside of a golf ball.
What is the magnitude of total charge of all the electrons in 1.2 L of liquid water?
Answer:
6.416*10^ 8C
Explanation:
First we find mass of the water
Mass= density x volume
=1.2*1000= 1200g
To find number of moles we use
N= mass/ molar mass
= 1200/18.02= 66.6mol
So the total number of water molecules is
66.6* 6.022*10^23= 401.07x 10^23moles
But each h20 moles has 10 electrons so it will be 4.01*10^26electrons
Then finally total charge will be
4.01*10^26electrons x 1.6*10^-19
= 6.416*10^ 8C
The magnitude of total charge will be "6.416×10⁸ C".
Electrons:According to the question,
Density, d = 1.2 L
Volume, V = 1000
The mass will be:
→ Mass = Density × Volume
By substituting the values,
= [tex]1.2\times 1000[/tex]
= [tex]1200[/tex] g
Now,
The number of moles will be:
→ N = [tex]\frac{Mass}{Molar \ mass}[/tex]
= [tex]\frac{1200}{18.02}[/tex]
= [tex]66.6[/tex] mol
The total no. of water molecules be:
= [tex]66.6\times 6.022\times 10^{23}[/tex]
= [tex]401.07\times 10^{23}[/tex] moles
hence,
The total charge be:
= [tex]4.01\times 10^{26}\times 1.6\times 10^{-19}[/tex]
= [tex]6.416\times 10^8[/tex] C
Thus the above response is correct.
Find out more information about magnitude here:
https://brainly.com/question/83117
A marble slides along a frictionless track at a constant speed u before encountering a section of the track that is covered with a thin layer of putty. The instantaneous velocity v of the marble as it slides past the putty, as a function of time t, is given byv(t)=u-bt+ct^3where t = 0 at the moment when the marble first comes into contact with the putty. As the marble exits the region covered in putty, the marble\'s instantaneous acceleration goes to zero. For u = 2.50 m/s, b = 3.55 m/s2, and c = 6.70 m/s4, calculate the length of the track that is covered with putty._______m
Answer:
x = 78.9 cm
Explanation:
The instantaneous velocity v of the marble as it slides past the putty, as a function of time t, is given by
[tex]v(t)=u-bt+ct^3[/tex]
u = 2.50 m/s, b = 3.55 m/s², and c = 6.70 m/s⁴
As the marble exits the region covered in putty, the marble\'s instantaneous acceleration goes to zero.
[tex]a=\dfrac{dv}{dt}\\\\a=\dfrac{d(u-bt+ct^3)}{dt}\\\\a=-b+3ct^2[/tex]
or
[tex]0-3.55+3\times 6.7t^2\\\\0=-3.55+20.1t^2\\\\t=\sqrt{\dfrac{3.55}{20.1}} \\\\t=0.42\ s[/tex]
Let x is the length of the track.
[tex]v=\dfrac{dx}{dt}\\\\\text{or}\\\\\int\limits_0^x {dx} =\int\limits^{4.2}_0 {vdt} \\\\x=\int\limits^{4.2}_0 {u-bt+ct^3} \, dt\\\\x=[ut-\dfrac{bt^2}{2}+\dfrac{ct^4}{4}]_0^ {0.42}\\\\\text{Putting limits}\\\\x=0.42u-\dfrac{b(0.42)^2}{2}+\dfrac{c(0.42)^4}{4}\\\\\text{Now, put values of u,b and c}\\\\x=0.42(2.5)-\dfrac{3.55\times (0.42)^2}{2}+\dfrac{6.7\times (0.42)^4}{4}\\\\x=0.789\ m\\\\x=78.9\ cm[/tex]
So, the length of the track that is covered with putty is 78.9 cm.
The length of the track is covered with putty is 0.789 m.
How do you calculate the length of the track?The instantaneous velocity v of the marble as a function of time t is given below.
[tex]v(t)=u-bt+ct^3[/tex]
Where t = 0 at the moment when the marble first comes into contact with the putty. Also given that, u = 2.50 m/s, b = 3.55 m/s², and c = 6.70 m/s⁴.
Given that the acceleration = 0 as the marble exits the region covered in putty. The acceleration is the rate change in the velocity with respect to time.
[tex]a = \dfrac {dv}{dt}[/tex]
[tex]a = \dfrac {d(u-bt+ct^3)}{dt}[/tex]
[tex]a = -b + 3ct^2[/tex]
Substituting the values in the above equation, we get the time.
[tex]0 = -3.55 + 3 \times 6.70 t^2[/tex]
[tex]t^2 = 0.1766[/tex]
[tex]t = 0.42 \;\rm s[/tex]
Hence the time interval is t =0 to t = 0.42 seconds.
Let us consider that the length of the track is l. Then the velocity can be written as the function of length is given below.
[tex]v = \dfrac {dl}{dt}[/tex]
[tex]dl = vdt[/tex]
[tex]\int_{0}^{t}dl = \int_{0}^{0.42} vdt[/tex]
[tex]l = \int_{0}^{0.42} (u-bt+ct^3) dt[/tex]
[tex]l =[ ut-\dfrac {bt^2}{2} + \dfrac {ct^4}{4} ]_0^{0.42}[/tex]
Putting the value t = 0.42,
[tex]l = 2.50 \times 0.42 - \dfrac {3.55 \times (0.42)^2}{2} + \dfrac { 6.70 \times (0.42)^4}{4}[/tex]
[tex]l = 0.789 \;\rm m[/tex]
Hence the length of the track is covered with putty is 0.789 m.
To know more about velocity and acceleration, follow the link given below.
https://brainly.com/question/2437624.
You walk out of pace of 1.38 m/s for 7 minutes, how far do you travel in meters
Answer:
7 × 60 (minute to seconds) = 420 seconds
1.38 × 420 = 579.6m
If an X-ray tube is operating at a current of 30.0 mA:a) How many electrons are striking the target per second? b) If the potential difference between the anode and cathode of this tube is 100.0 kV, how much power is expended
Answer:
Explanation:
A. Q= It
30E-3A=q
Ne = q
30E-3/1.6*10-19= N
N= 1.8*10^16 electrons
B. Power= I x v
= 30*10^-3A x 100*10^3v
= 3000watts
Which of the following best describes how to calculate the average acceleration of any object?
a. Average acceleration is always halfway between the initial acceleration of an object and its final acceleration.
b. Average acceleration is always equal to the displacement of an object divided by the time interval.
c. Average acceleration is always equal to the change in velocity of an object divided by the time interval.
d. Average acceleration is always equal to the change in speed of an object divided by the time interval.
Answer:
c)Average acceleration is always equal to the change in velocity of an object divided by the time interval.
Explanation:
Because average acceleration= a2+a1/2
Which is equal to change in velocity which is a time rate of change of velocity v-u/t which actually explains average acceleration
A person with a black belt in taekwondo has a foot with a mass of 0.90 kg. Starting from rest, this foot attains a velocity of 8.3 m/s in 0.13 s. What is the magnitude of the average net force applied to the foot to obtain that acceleration?
Answer:
57.42 N
Explanation:
We first find the acceleration knowing that it goes from rest to 8.3 m/s in 0.13 seconds,
Therefore, the acceleration is:
[tex]a=\frac{v_f-v_i}{time} =\frac{8.3-0}{0.13 } =63.8\,\,m/s^2[/tex]
Now we use this to calculate the average net force by multiplying mass times acceleration:
F = m * a = 0.9 (63.8) = 57.42 N
Maddie Cat can ride her tricycle at an amazing speed of 3.75 meters per second.
How far can she go in 5.0 seconds?
Distance = (speed) x (time)
Distance = (3.75 m/s) x (5.0 s)
Distance = 18.75 meters
Answer: 3.75 x 5.0= 18.75 meters
Explanation:
To practice Problem-Solving Strategy 22.1 for electric force problems. Two charged particles, with charges q1=qq1=q and q2=4qq2=4q, are located on the x axis separated by a distance of 2.00cm2.00cm . A third charged particle, with charge q3=qq3=q, is placed on the x axis such that the magnitude of the force that charge 1 exerts on charge 3 is equal to the force that charge 2 exerts on charge 3.Find the position of charge 3 when qqq = 2.00 nCnC .
Answer:
x₂ = 0.01336
Explanation:
In this exercise we use the translational equilibrium equation
F₁₃ - F₂₃ = 0
F₁₃ = F₂₃
at the point where charge 3 is placed the two electric forces have the same magnitude
let us use the expression of Coulomb's law for the electric force
F₁₃ = ka q₁ q₃ / r₁₃²
F₂₃ = ka q₂ q₃ / r₂₃²
we substitute
k q₁ q₃ / r₁₃² = k q₂ q₃ / r₂₃²
now imprescriptibility suppose that particle 1 is at the origin of the coordinate system, particle 2 is at a distance d = 2.00cm = 2 10-2 m, therefore let's call the distance from particle 1 to particle 3 as x
r₁₃ = x
R₂₃ = d-x
In the exercise we are given the charges for the particle1 q1 = q, for the particle 2 the charge is q2 = 4q the distance between them is d = 2.00cm = 0.0200 m, the value of q = 2.00 nC = 2.00 10⁻⁹ C
let's substitute these values
q₁ / x₂ = q₂ / (d-x)²
let's clear x
(d-x)² = q₂ / q₁ x²
d² - 2dx + x₂ = q₂ /q₁ x²
x² (1-q₂ / q₁) - 2d x + d² = 0
let's substitute the values and solve the quadratic equation
x² (1 - 4q / q) - 2 0.02 x + 0.02² = 0
-3 x² - 0.04 x + 0.0004 = 0
x² + 0.0133 x - 0.0001333 = 0
x = [-0.0133 ±√(0.0133² + 4 0.00013333)] / 2
x = [-0.0133 + - 0.026664] /2
x₁ = -0.01998 m
x₂ = 0.01336 m
Since load 3 must be between charged 1 and 2 the correct answer is
x₂ = 0.01336
Simpson drives his car with an average velocity of 24 m/s toward the east. how long will it take him to drive 560 km on a perfectly straight highway?
Answer:
The time taken to drive on this high way is 6.48 hours
Explanation:
Given;
average velocity, v = 24 m/s
distance of travel, d = 560 km = 560,000 m
The distance traveled is given by;
[tex]d = (\frac{v+u}{2} )t\\\\d =( 24 \ m/s)t\\\\t = \frac{d}{24} \\\\t = \frac{560,000}{24}\\\\t = 2333.33 \ s = 6.48 \ hrs[/tex]
Therefore, the time taken to drive on this high way is 6.48 hours
A container weighs 450 lb when filled with 4 ft3 of fluid. If the empty container weighs 50 lb, determine the specific weight, mass density and specific gravity of this fluid.
Answer:
112.5 lb/ft³
405.93 kg/m³
0.406
Explanation:
given that
Weight of the container, w(c) = 450 lb
Volume of fluid, v(f) = 4 ft³
Weight of empty container, w(e) = 50 lb
Specific weight is said to be the weight per unit volume. Thus, the specific weight of this fluid is
w(s) = 450 / 4
w(s) = 112.5 lb/ft³
Mass density is the ratio of the mass of the fluid, with respect to the volume.
ρ = m/v
Since the question gives us the mass in weight, we'd have to convert it to mass from weight.
mass = weight / acceleration due to gravity
mass = 450 / 9.81
mass = 45.87 kg
Also, since the mass is going to be in kg, the volume also has to be in m³, thus needing a conversion.
4 ft³ = 0.113 m³
Mass density = 45.87 / 0.113
Mass density = 405.93 kg/m³
Specific gravity of the fluid is simply given as mass density / 1000. Therefore,
g(s) = 405.93 / 1000
g(s) = 0.406
please vote brainliest
The fastest land animal is the cheetah. A cheetah can run 100 meter dash in 5.95 seconds, calculate the cheetahs average speed in meters per second
Answer:
16.80672269 m/s
Explanation:
100 divide by 5.95 seconds
Answer:
16.8m/s
Explanation:
S=D/T
=
[tex] \frac{100}{5.95} [/tex]
=16.80672269
=16.8
Would a human feel more gravity on a bigger planet or less gravity
Answer: you would feel more gravity
Explanation: the more mass of an object the more pull
What is the opposite of yummy??? PLEASE I GIVE BRainlIEST NO NOOBS PLZZ
Answer:
tasteless
Explanation:
Can an objects displacement be greater than or equal to the objects distance?
if an object is not accelerating, then one knows for sure that it is_____.
Which is the most dense solid?
Answer:
Bolt
Explanation:
A baseball is thrown horizontally at 46 m/s. The ball slows down at a rate of 5 m/s2. How long is the ball in the air before coming to rest?
Answer:
9.2 s
Explanation:
Given:
v₀ = 46 m/s
v = 0 m/s
a = -5 m/s²
Find: t
v = at + v₀
0 m/s = (-5 m/s²) t + 46 m/s
t = 9.2 s
Find the height or length of these natural wonders in km, m, and cm.A. A cave system with a mapped length of 354 miles.B. A waterfall that drops 1,235.2 ft.C. A 21320 ft tall mountain.
D. A canyon with a depth of 6630 ft.
Explanation:
Some standard unit conversion are 1 mile = 1.609344 km, 1 ft.= 30.48 cm,
1 km= 1000 m or 1 m = 0.001 km and 1 m= 100 cm or 1 cm=0.01 m.
Now, use these values to convert the given lengths.
A. length of cave = 5 miles (given)
From standard value 1 mile = 1.609344 km
[tex]\Rightarrow[/tex] 5 miles = [tex]\times[/tex] 1.609344 km= 8.04672 km,
[tex]\Rightarrow[/tex] 5 miles =8.04672 [tex]\times[/tex] 1000 m= 8046.72 m [ as 1 km = 1000 m]
[tex]\Rightarrow[/tex] 5 miles = 8046.72 [tex]\times[/tex] 100 cm=804672 cm
B. Height of the waterfall = 1235 ft. (given)
1 ft.= 30.48 cm [ from standard value]
[tex]\Rightarrow[/tex] 1235.2 ft.= 1235.2[tex]\times[/tex] 30.48 cm=37648.896 cm,
[tex]\Rightarrow[/tex] 1235.2 ft.=37648.896 [tex]\times[/tex] 0.01 m =376.48896 m [ as 1 cm = 0.01 m]
[tex]\Rightarrow[/tex] 1235.2 ft.=376.48896 [tex]\times[/tex] 0.001 km =0.37648896 km [ as 1 m = 0.001 km]
C. Height of the mountain= 21320 ft. (given)
From standard value: 1 ft.= 30.48 cm
[tex]\Rightarrow[/tex] 21320 ft.= 21320 [tex]\times[/tex] 30.48 cm=649833.6 cm,
[tex]\Rightarrow[/tex] 21320 ft.=649833.6 [tex]\times[/tex] 0.01 m =6498.336 m [ as 1 cm = 0.01 m]
[tex]\Rightarrow[/tex] 21320 ft.=6498.336 [tex]\times[/tex] 0.001 km =6.498336 km [ as 1 m = 0.001 km].
D. Depth of canyon =6630 ft.
From standard value: 1 ft.= 30.48 cm
[tex]\Rightarrow[/tex] 6630 ft.= 6630 [tex]\times[/tex] 30.48 cm=202082.4 cm,
[tex]\Rightarrow[/tex] 6630 ft.=202082.4 [tex]\times[/tex] 0.01 m =2020.824 m [ as 1 cm = 0.01 m]
[tex]\Rightarrow[/tex] 6630 ft.=2020.824 [tex]\times[/tex] 0.001 km =2.020824 km [ as 1 m = 0.001 km].
assume this car is driven off a cliff. How many arrows of force need to be drawn in the free body diagram? Assum no air resistance
Answers:
five
one
four
three
Answer:
4
Explanation:
friction
weight
normal reaction
force to overcome inertia
Answer:
1
Explanation:
--->
A string has mass 5.0 grams, and is stretched with 180 N of tension. A wave travels on this string with frequency 260 Hz and wavelength 0.60 m. Determine the length of the string.
Answer:
3.68×10¯⁶ m
Explanation:
The following data were obtained from the question:
Mass (m) = 5 g
Tension (T) = 180 N
Frequency (f) = 260 Hz
Wavelength (λ) = 0.60 m.
Length (L) =?
Next, we shall determine the period of the wave. This can be obtained as follow:
Period is simply defined as the time taken to complete 1 oscillation. Mathematically, it is represented as:
Period (T) = 1/frequency (f)
T = 1/f
Thus, with the above formula, the period of the wave can be obtained as shown below:
Frequency (f) = 260 Hz
Period (T) =?
T = 1/f
T = 1/260
T = 3.85×10¯³ s
Finally, we shall determine the length of the string as follow:
Period (T) = 3.85×10¯³ s
Acceleration due to gravity (g) = 9.8 m/s²
Pi (π) = 3.14
Length (L) =?
T = 2π√(L/g)
3.85×10¯³ = 2 × 3.14 × √(L/9.8)
3.85×10¯³ = 6.28 × √(L/9.8)
Divide both side by 6.28
3.85×10¯³ / 6.28 = √(L/9.8)
Take the square of both side
(3.85×10¯³ / 6.28)² = L/9.8
Cross multiply
L = (3.85×10¯³ / 6.28)² × 9.8
L = 3.68×10¯⁶ m
Therefore, the length of the string is 3.68×10¯⁶ m
Please help!
A tennis ball falls off a shelf and bounces several times. Each bounce is lower than the one before. Soon the ball stops. Why doesn’t the ball keep bouncing?
A) some of the energy is changed into mass
B) The energy becomes stored in the ball
C) Energy is transferred to the air and ground
Answer:
c
Explanation:
energy isn't stored and it doesn't change to mass
The ball has some kinetic energy when it hits the floor, but some of it is changed, so it loses some of it each time it bounces. After a few bounces, the ball has so little kinetic energy remaining that it stops bouncing. Thus, option C is correct.
What doesn’t the ball keep bouncing?Since the gravitational potential energy, which can be transformed back into kinetic energy on the rebound, increases with the drop height, the consequent bounce height will likewise rise.
Because it has the most elasticity, the rubber ball will rebound the highest when all three balls are dropped from the same height. Because rubber is incredibly elastic, it squishes or compresses as it touches the ground but immediately snaps back to its original shape.
Therefore, Additionally, some energy is changed into other forms like heat and sound. The majority of these additional energy sources are lost and not recovered, causing the ball to bounce back to a lesser height.
Learn more about ball here:
https://brainly.com/question/19930452
#SPJ2
g An astronaut must journey to a distant planet, which is 189 light-years from Earth. What speed will be necessary if the astronaut wishes to age only 12 years during the round trip? (Give your answer accurate to four decimal places.)
Answer:
The value is [tex]v = 2.999 *10^{8} \ m/s[/tex]
Explanation:
From the question we are told that
The time taken to travel to the planet from earth is [tex]t = 189 \ light-years[/tex]
The time to be spent on the ship is [tex]t_{s} = 12 \ years[/tex]
Generally speed can be obtained using the mathematical relation represented below
[tex]t_s = 2 * t * \sqrt{1 - \frac{v^2}{c^2 } }[/tex]
The 2 in the equation show that the trip is a round trip i.e going and coming back
=> [tex]12 = 2 * 189 * \sqrt{1 - \frac{v^2}{(3.0*10^{8})^2 } }[/tex]
=> [tex]v = 2.999 *10^{8} \ m/s[/tex]
A particle makes 240 revolutions per minute on a circle of radius 2m. Find I. its period ii. angular velocity III. linear velocity.
Answer:
(i) The period, T is 0.25 s
(ii) The angular velocity, ω is 25.136 rad/s
(iii) The linear velocity is, v is 50.27 m/s
Explanation:
Given;
number of revolution of the particle, N = 240 revolutions per minutes
radius of the circle, R = 2m
(i) The period is given by;
[tex]T = \frac{2\pi}{\omega}[/tex]
where;
ω is angular velocity
[tex]\omega = (240 \frac{rev}{min}) *(\frac{1 \ min}{60 \ s} )*(2\pi \frac{rad}{rev})\\\\ \omega = 25.136 \ rad/s[/tex]
[tex]T = \frac{2\pi }{\omega}\\\\ T = \frac{2\pi}{25.136}\\\\T = 0.25 \ s[/tex]
(ii) angular velocity, ω = 25.136 rad/s
(iii) linear velocity is given by;
v = ωR
v = (25.136 rad/s) x (2 m)
v = 50.27 m/s
How does diet affect diabetes
Answer:
A diet high in fat, calories, and cholesterol increases your risk of diabetes. A poor diet can lead to obesity (another risk factor for diabetes) and other health problems. A healthy diet is high in fiber and low in fat, cholesterol, salt, and sugar. Also, remember to watch your portion size.
A swimmer bounces straight up from a diving board and falls feet first into a pool. She starts with a velocity of 3.00 m/s, and her takeoff point is 1.80 m above the pool.(a) How long are her feet in the air?(b) What is the highest point above the board her feet reach? (c) What is her velocity when her feet hit the water?
Answer:
A) 0.9844 s
B) x2 = 0.4587 m
C) v = 6.657 m/s
Explanation:
We are given;
Height of take off point above pool; x1 = 1.8 m
Initial take off velocity; u = 3 m/s
Final velocity at highest point before free fall; v = 0 m/s
B) To find the highest point above the board her feet reaches means the distance from take off to the top of the motion just before free fall.
Thus, we will be using equation of motion and we have;
v² = u² + 2gs
Now, let s = x2 which will be the distance between take off and the top before free fall.
So;
v² = u² + 2g(x2)
Now,since the motion is against gravity, g will be negative.
Thus;
v² = u² + 2(-9.81)(x2)
Plugging in the relevant values to give;
0² = 3² - (19.62x2)
19.62(x2) = 9
x2 = 9/19.62
x2 = 0.4587 m
A) We want to find how long her feet is in air.. It means we want to find out the time to get to a distance of x1 and also the time to achieve the distance (x1 + x2) on free-fall.
Thus, using equation of motion;
v = u + gt
Again, g = -9.81
Thus;
0 = 3 - 9.81t1
9.81t1 = 3
t1 = 3/9.81
t1 = 0.3058 s
Now, for the time taken to achieve the distance (x1 + x2) on free-fall, we will use the formula;
s = ut + ½gt²
Where s = (x1 + x2) = 1.8 + 0.4587 = 2.2587 m
And now, u = 0 m/s because the start of the free fall is from maximum height with velocity of 0 m/s. Again, g = - 9.81 m/s²
Thus;
2.2587 = 0 - ½(-9.81)(t2)²
2.2587 = 4.905(t2)²
(t2)² = 2.2587/4.905
(t2)² = 0.4605
t2 = √0.4605
t2 = 0.6786 s
Thus, total time of feet in air = t1 + t2 = 0.3058 + 0.6786 = 0.9844 s
C) Velocity when feet hit the water would be given by;
v = u + gt
Where u = 0 m/s and t = t2 = 0.6786
Since it's in direction of gravity, g = 9.81 m/s
v = 0 + (0.6786 × 9.81)
v = 6.657 m/s
A flag pole is attached to the side of a building 5.00 ft off the ground and it hangs down 30 degrees from the vertical. If the pole is 6.00 ft in length, how high in feet is the top of the pole at its highest point (sticking out from building)? (Hint: draw a picture) E
Answer:
h ’= 0.6963 m
this is the height of the part of the post it supports.
Explanation:
For this exercise we can use the trigonometric relationships
Let's start by finding the vertical component of the height of the pole
cos 30 = x / L
x = L cos 30
x = 6 cps 30
x = 5.196 m
with this length, the post has the same height as the building, so there are
Aₓ = L -x
Δx = 6 -5.196
Δx = 0.804 ft
the height of this section is
cos 30 = h ’/ Dx
h ’= 0.804 cos30
h ’= 0.6963 m
this is the height of the part of the post it supports.
What is the equation of the line on the graph?
Trevor and Nick are taxi drivers. Trevor drives a taxi using diesel oil while Nick drives a taxi using LPG. Whose taxi will cause higher levels of air pollution and why?
Answer:
Trevor's taxi would cause higher levels of air pollution
Explanation:
Trevor's taxi use diesel oil.
Diesel is less cleaner than LPG.
Compared to automotive pollution from petrol and diesel, pollutants from LPG-driven cars include lower amounts of petroleum hydrocarbons, nitrogen oxides , sulphur oxides, ozone contamination and particulate matter.
Use the mass and density data to calculate the volume of corn syrup to the nearest tenth. Mass of corn syrup = 57.3 g Density of corn syrup = 1.38 g/mL
Density = (mass) / (volume) (definition)
(Density) x (volume) = (mass)
(volume) = (mass) / (density)
Volume = (57.3 g) / (1.38 g/mL)
Volume = 41.5 mL
Answer:
41.5
Explanation:
Got answer correct on EDG