Answers
Force Meter > used to measure force
Milli > Prefix that means 1/1,000
Centi > Prefix that means 1/100
Kilo > Prefixes that means 1,000
Thermometer > used to measure Temperature
Explanation: uh just trust
Related Questions
What do you call the height of a wave?
a. wavelength
b. frequency
c. amplitude
d. resonance
Answers
Answer:
amplitude is the answer
Postural deviations can result in
Answers
Answer:
Postural deviations can cause poor balance, muscle pain and skeletal stress.
Explanation:
The mass percent of hydrogen in CH₄O is 12.5%.
What is the mass percent?Mass percent is the mass of the element divided by the mass of the compound or solute.
Step 1: Calculate the mass of the compound.
mCH₄O = 1 mC + 4 mH + 1 mO = 1 (12.01 amu) + 4 (1.00 amu) + 1 (16.00 amu) = 32.01 amu
Step 2: Calculate the mass of hydrogen in the compound.
mH in mCH₄O = 4 mH = 4 (1.00 amu) = 4.00 amu
Step 3: Calculate the mass percent of hydrogen in the compound.
%H = (mH in mCH₄O / mCH₄O) × 100%
%H = 4.00 amu / 32.01 amu × 100% = 12.5%
The mass percent of hydrogen in CH₄O is 12.5%.
CO2 = 1.580 grams H2O = 0.592 grams Lookup the molar mass of each element in the compound Carbon = 12.0107 Hydrogen = 1.00794 Oxygen = 15.999 Calculate the molar mass of CH4O by adding the total masses of each element used. 12.0107 + 4 * 1.00794 + 15.999 = 32.04146 Now calculate how many moles of CH4O you have by dividing by the molar mass. m = 1.15 g / 32.04146 g/mole = 0.035891 mole Now figure out how many moles of carbon and hydrogen you have. Carbon = 0.035891 moles Hydrogen = 0.035891 moles *
Therefore, The mass percent of hydrogen in CH₄O is 12.5%.
Learn more about mass percent here:
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A 15 m uniform ladder weighing 500 N rests against a frictionless wall. The ladder makes a 60° angle with horizontal. (a) Find the horizontal and vertical forces that the earth exerts on the base of the ladder when an 800 N firefighter is 4 m from the bottom. (b) If the ladder is just on the verge of slipping when the firefighter is 9 m up, what is the coefficient of static friction between ladder and ground?
Answers
Answer:
a) F₁ = 267.3 N, N₁ = 1300 N, b) μ = 0.324
Explanation:
For this exercise we use the rotational equilibrium condition, we have a reference system is the floor and the anticlockwise rotations as positive, in the adjoint we can see a diagram of the forces
let's use subscript 1 for the ladder and 2 for the firefighter
∑ τ = 0
-W₁ x₁ - W₂ x₂ + N₁ y = 0
N₁ = [tex]\frac{W_1 x_1 + W_2 x_2}{y}[/tex] (1)
the center of mass of the ladder is at its geometric center,
d = L / 2 = 15/2 = 7.5 m
cos 60 = x₁ / d₁
x₁ = d₁ cos 60
x₁ = 7.5 cos 60
x₁ = 3.75 m
for the firefighter d₂ = 4 m
cos 60 = x₂ / d₂
x₂ = d₂ cos 60
x₂ = 4 cos 60 = 2 m
for the fulcrum d₃ = 15 m
sin 60 = y / d₃
y = d₃ sin 60
y = 15 sin 60
y = 13 m
we look for the Normal by substituting in equation 1
N₂ = [tex]\frac{500 \ 3.75 \ + 800 \ 2}{13}[/tex]
N₂ = 267.3 N
now let's use the translational equilibrium relations
X axis
F₁ - N₂ = 0
F₁ = N₂
F₁ = 267.3 N
Axis y
N₁ - W₁ -W₂ = 0
N₁ = W₁ + W₂
N₁ = 500 + 800
N₁ = 1300 N
b) for this case change the firefighter's distance d₂ = 9 m
x₂ = 9 cos 60
x₂ = 4.5 m
we substitute in 1
N₂ = \frac{500 \ 3.75 \ + 800 \ 4.5}{13}
N₂ = 421.15 N
of the translational equilibrium equation on the x-axis
fr = F₁ = N₂
fr = 421.15 N
friction force has the expression
fr = μ N
in this case the reaction of the Earth to the support of the ladder is N1 = 1300N
μ = fr / N₁
μ = 421.15 / 1300
μ = 0.324
What quantity measures the amount of space an object occupies?
A. Volume B.Temperature C. Mass D. Density
Answers
Answer:
mas
Explanation:
mass is the amount of space something occupies.
Question 18 of 25
Which type of reaction is shown in this energy diagram?
Energy
Products
Activation
Energy
Reoctants
to
ti
Time
A. Endothermic, because the products are lower in energy
B. Exothermic, because the reactants are lower in energy
C. Endothermic, because the reactants are lower in energy
D. Exothermic, because the products are lower in energy
Answers
Answer:
Endothermic, because the reactants are lower in energy (C)
Explanation:
From the graph, you can see the energy of the products is higher than the energy of the reactants. If you recall that when the enthalpy change Eproducts is gretater than Ereactants, the reaction is said to be endothermic.
The planet Mars is host to five functioning spacecraft, three in orbit about the planet and two on the surface of the planet. Thanks to those spacecraft, we know that the planet Mars has a mass that is 0.11 times that of Earth and a radius that is 0.53 times that of Earth. The acceleration of an object in free-fall near the surface of Mars is most nearly what in terms of the local value of g on Earth
Answers
Answer:
0.392
Explanation:
Mm = 0.11Me
Rm = 0.53Re
g = GM / r^2
G = 6.67 * 10^-11
gmars = (G * 0.11Me) / (0.53Re)^2
Recall:
gearth = GMe /Re^2
Hence, gmars in terms of gearth equals
gmars = gearth * (0.11 / 0.53^2)
gmars = gearth * 0.3915984
gmars = 0.392gearth
How fast would an object have to travel on the surface of Jupiter at the equator to keep up with the Sun (that is, so the Sun would appear to remain in the same position in the sky)? Use the facts that the radius of Jupiter is approximately 44,360 miles and its revolution is approximately 10 hours.
Answers
Answer:
27872.2 miles per hour
Explanation:
Given that :
Radius of Jupiter is approximately = 44,360 miles
Revolution is 10 hours ;
Jupiter makes one revolution in 10 hours :
Using the relation to obtain the velocity :
V = re
r = radius
w = 2π/T
Hence,
V = r * 2π/ T
V =44360 * 2 * π/10
V = 88720 * π/10
V = 278722.10 / 10
V = 27872.210
V = 27872.2 miles per hour
You are removing branches from your roof after a big storm. You throw a branch horizontally from your roof, which is a height 3.00 m above the ground. The branch lands a horizontal distance 8.00 m away from where you threw it (assuming you are the 0 position in x, and the branch traveled in the x direction). You can assume there is no air resistance. You can assume that the upwards direction is positive. What is the initial velocity in x of the branch (how fast did you throw the branch)
Answers
Answer:
The initial velocity in the x-direction with which the branch was thrown is approximately 10.224 m/s
Explanation:
The given parameters of the motion of the branch are;
The height from which the branch is thrown = 3.00 m
The horizontal distance the branch lands from where it was thrown, x = 8.00 m
The direction in which the branch is thrown = Horizontally
Therefore, the initial vertical velocity of the branch, [tex]u_y[/tex] = 0 m/s
The time it takes an object in free fall (zero initial downward vertical velocity) to reach the ground is given as follows;
s = [tex]u_y[/tex]·t + 1/2·g·t²
Where;
[tex]u_y[/tex] = 0 m/s
s = The initial height of the object = 3.00 m
g = The acceleration due to gravity = 9.8 m/s²
∴ s = 0·t + 1/2·g·t² = 0 × t + 1/2·g·t² = 1/2·g·t²
t = √(2·s/g) = √(2 × 3/9.8) = (√30)/7 ≈ 0.78246
The horizontal distance covered before the branch touches the ground, x = 8.00 m
Therefore, the initial velocity in the horizontal, x-direction with which the branch was thrown, 'uₓ', is given as follows;
uₓ = x/t = 8.00 m/((√30)/7 s)
Using a graphing calculator, we get;
uₓ = 8.00 m/((√30)/7 s) = (28/15)·√30 m/s ≈ 10.224 m/s
The initial velocity in the horizontal, x-direction with which the branch was thrown, uₓ ≈ 10.224 m/s.
Violet pulls a rake horizontally on a frictionless driveway with a net force of 2.0 N for 5.0 m.
How much kinetic energy does the rake gain?
Answers
Answer:
10 J.
Explanation:
Given that,
Net force acting on the rake, F = 2 N
Distance moved by the rake, d = 5 m
We need to find the kinetic energy gained by the rake. We know that,
Kinetic energy = work done
So,
K = F×d
K = 2 N × 5 m
K = 10 J
So, 10 J of kinetic energy is gained by the rake.
Violet pulls a rake horizontally on a frictionless driveway with a net force of 2.0 N for 5.0 m.
How much kinetic energy does the rake gain?
Answer: 10 J
5.0 L/s water flows through a horizontal pipe that narrows smoothly from 10.0 cm diameter to 5.0 cm diameter. A pressure gauge in the narrow section reads 50 kPa. What is the reading of the pressure gauge in the wide section
Answers
Solution :
The volume rate of flow is given by : R = 5.0 L/s
[tex]$ = 5.0 \times 10^{-3} \ m^3/s$[/tex]
The radius of the pipe, [tex]$r_1= 5 \times 10^{-2} \ m$[/tex]
∴ [tex]$ 5.0 \times 10^{-3} = \pi (2.5 \times 10^{-2})^2 v_1$[/tex]
then, [tex]$v_1 = \frac{5.0 \times 10^{-3}}{(3.14)(5 \times 10^{-2})^2}$[/tex]
= 0.637 meter per second
Then the speed of the water at wider section,
[tex]$R=A_1v_1$[/tex]
Similarly, the speed of water at narrow pipe.
The radius of the [tex]$r_2 = 2.5 \times 10^{-2}$[/tex] m
[tex]$5.0 \times 10^{-3} = \pi (2.5 \times 10^{-2})^2 v_1$[/tex]
then, [tex]$v_2 = \frac{5.0 \times 10^{-3}}{(3.14)(2.5 \times 10^{-2})^2}$[/tex]
= 2.55 meter per sec
Now from Bernoulli's theorem,
[tex]$P_1 + \frac{1}{2} \rho v_1^2 =P_2 + \frac{1}{2} \rho v_2^2 $[/tex]
[tex]$P_1 = P_2 + \frac{1}{2} \rho (v_2^2 - v_1^2)$[/tex]
[tex]$= 50 \kPa + (0.5)(10^3)[(2.55)^2-(0.637)^2]$[/tex]
= 50 kPa + 3.05 kPa
= 53.05 kPa
or 53000 Pa
This question involves the concepts of Bernoulli's Theorem and Volumetric Flowrate.
The pressure reading in the wide section is "53.05 KPa".
First, we will use the volumetric flow rate to find the velocities of the water at wide and narrow sections.
[tex]V = A_1v_1[/tex]
where,
V = Volumetric Flow Rate = 5 L/s = 5 x 10⁻³ m³/s
r₁ = radius of narrow section = 5 cm/2 = 2.5 cm = 0.025 m
A₁ = Area of narrow section = πr₁² = π(0.025 m)²
v₁ = velocity at narrow section = ?
Therefore,
[tex]5\ x\ 10^{-3}\ m^3=[\pi(0.025\ m)^2](v_1)\\\\v_1=\frac{5\ x\ 10^{-3}\ m^3}{\pi (0.025\ m)^2}\\\\v_1=2.55\ m/s\\[/tex]
Similarly,
[tex]V = A_2v_2[/tex]
where,
V = Volumetric Flow Rate = 5 L/s = 5 x 10⁻³ m³/s
r₂ = radius of wide section = 10 cm/2 = 5 cm = 0.05 m
A₂ = Area of wide section = πr₁² = π(0.05 m)²
v₂ = velocity at wide section = ?
Therefore,
[tex]5\ x\ 10^{-3}\ m^3=[\pi(0.05\ m)^2](v_2)\\\\v_2=\frac{5\ x\ 10^{-3}\ m^3}{\pi (0.05\ m)^2}\\\\v_2=0.64\ m/s\\[/tex]
Now, we will use Bernoulli's Theorem to find out the pressure wide section.
[tex]P_1 + \frac{1}{2}\rho v_1^2=P_2 + \frac{1}{2}\rho v_2^2[/tex]
where,
[tex]\rho[/tex] = density of water = 1000 kg/m³
P₁ = pressure in narrow section = 50 KPa = 50000 Pa
P₂ = pressure in wide section = ?
Therefore,
[tex]50000\ Pa + \frac{1}{2}(1000\ kg/m^3)(2.55\ m/s)^2=P_2 + \frac{1}{2}(1000\ kg/m^3)(0.64\ m/s)^2[/tex]
P₂ = 50000 Pa + 3251.25 Pa - 204.8 Pa
P₂ = 53046.45 Pa = 53.05 KPa
Learn more about Bernoulli's Theorem here:
https://brainly.com/question/13098748?referrer=searchResults
The attached picture shows Bernoulli's Theorem.
Which is a valid velocity reading for an object?
45 m/s
45 m/s north
O 0 m/s south
O 0 m/s
Answers
Answer:45 m/s north
Explanation:
a. Use the graph and the element made in question 2 to determine the mass of the star.
Answers
A ball is rolling with a constant acceleration of 13 m/s starting from rest. How long will it take to increase its velocity to a final speed of 62 m/s?
Answers
Time taken : 4.77 s
Further explanationGiven
a = 13 m/s²
vf = final velocity = 62 m/s
Required
time taken
Solution
An equation of uniformly accelerated motion
[tex]\large {\boxed {\bold {x=xo+vo.t+\frac {1} {2} at ^ 2}}}[/tex]
vf = vo + at
vf² = vo² + 2a (x-xo)
x = distance on t
vo / vi = initial speed
vt / vf = speed on t / final speed
a = acceleration
Input the value :
vo =0 ⇒from rest
vf=at
62 m/s = 13 m/s².t
t = 62 : 13
t = 4.77 s
a 90 kilogram dog runs across the dog park at a speed of 6.5 meters per second. what is the magnitude and direction of the average force required to stop the dog in .85 seconds?
Answers
Answer:
am not sure about the answer
Explanation:
you need to find out the amount of force it's going in for example 10n or 100n then you need to times it the distance then devide by the time
Two ice skaters approach each other at right angles. Skater A has a mass of 68.2 kg and travels in the x direction at 2.48 m/s. Skater B has a mass of 34.4 kg and is moving in the y direction at 1.18 m/s. They collide and cling together. Find the final speed of the couple. Answer in units of m/s.
Answers
Answer:
1.70 m/s
Explanation:
The computation of the final speed of the couple is shown below:
initial momentum of A is
= mv
= 68.2 × 2.48
= 169.136 kg
And, the initial momentum of B is
= mv
= 34.4 × 1.18
= 40.592 kg
Now magnitude is
= sqrt( A^2 + B^2)
= sqrt( 28,606.99 + 1,647.71)
= 173.94
Now the final speed is
= 173.94 ÷ (68.2 + 34.4)
= 1.70 m/s
A pendulum has a period of 6.98s. Calculate the length of the pendulum. Use
9.8m/s^2 for gravity. *
Answers
Answer:
Length, l = 0.126 meters.
Explanation:
Given the following data;
Period = 6.98s
Acceleration due to gravity, g = 9.8m/s²
To find the length, l;
[tex] Period, T = 2 \pi \sqrt {lg} [/tex]
Substituting into the equation, we have;
[tex] 6.98 = 2*3.142 \sqrt {l*9.8} [/tex]
[tex] 6.98 = 6.284 \sqrt {9.8l} [/tex]
[tex] \frac {6.98}{6.284} = \sqrt {9.8l} [/tex]
[tex] 1.1108 = \sqrt {9.8l} [/tex]
Taking the square of both sides
[tex] 1.1108^{2} = 9.8l [/tex]
[tex] 1.2339 = 9.8l [/tex]
[tex] l = \frac {1.2339}{9.8} [/tex]
Length, l = 0.126m.
Hearing rattles from a snake, you make two rapid displacements of magnitude 1.8 m and 2.4m. Draw sketches, roughly to scale, to show how your two displacements might add to give the following resultant of magnitudes.
a. 4.2 m
b. 0.6 m
c. 3.2 m
Answers
Answer:
The answer is a 4.2m
Explanation:
Given data
Please see attached the rough drawing for your reference.
From the drawing, you ran 18m west and 2.4m south
The displacement is
= 1.8+2.4
=4.2m
A body is dropped from the roof of a 20 m high building by how much:
Does it take to reach the ground? How fast does it hit the ground?
Answers
Answer:
t = 2.01 s
Vf = 19.7 m/s
Explanation:
It's know through the International System that the earth's gravity is 9.8 m/s², then we have;
Data:
Height (h) = 20 mGravity (g) = 9.8 m/s²Time (t) = ?Final Velocity (Vf) = ?==================================================================
Time
Use formula:
[tex]\boxed{t=\sqrt{\frac{2*h}{g}}}[/tex]Replace:
[tex]\boxed{t=\sqrt{\frac{2*20m}{9.8\frac{m}{s^{2}}}}}[/tex]Everything inside the root is solved first. So, we solve the multiplication of the numerator:
[tex]\boxed{t=\sqrt{\frac{40m}{9.8\frac{m}{s^{2}}}}}[/tex]It divides:
[tex]\boxed{t=\sqrt{4.08s}}[/tex]The square root is performed:
[tex]\boxed{t=2.01s}[/tex]==================================================================
Final Velocity
use formula:
Vf = g * tReplace:
Vf = 9.8 m/s² * 2.01 sMultiply:
Vf = 19.7 m/s==================================================================
How long does it take to reach the ground?
Takes time to reach the ground in 2.01 seconds.
How fast does it hit the ground?
Hits the ground with a speed of 19.7 meters per seconds.
You are walking on a moving walkway in the airport. The length of the walkway is 59.1 m. If your velocity relative to the walkway is 2.35 m/s, and the walkway moves with a velocity of 1.85 m/s, how long will it take you to reach the other end of the walkway
Answers
Answer:
14.1seconds approx
Explanation:
Given data
Distance= 59.1m
Your velocity= 2.35m/s
Walkway velocity= 1.85m/s
Total velocity= 2.35+1.85= 4.2m/s
We know that
Velocity= distance/time
time= distance/velocity
substitute
time= 59.1/4.2
time= 14.07
time=14.1seconds approx
Hence the time is 14.1seconds approx
You are on the Pirates of the Caribbean attraction in the Magic Kingdom at Disney World. Your boat rides through a pirate battle, in which cannons on a ship and in a fort are firing at each other. While you are aware that the splashes in the water do not represent actual cannonballs, you begin to wonder about such battles in the days of the pirates. Sup-pose the fort and the ship are separated by 75.0 m. You see that the cannons in the fort are aimed so that their cannon-balls would be fired horizontally from a height of 7.00 m above the water.
(a) You wonder at what speed they must be fired in order to hit the ship before falling in the water.
(b) Then, you think about the sludge that must build up inside the barrel of a cannon. This sludge should slow down the cannonballs. A question occurs in your mind: if the can-nonballs can be fired at only 50.0% of the speed found ear-lier, is it possible to fire them upward at some angle to the horizontal so that they would reach the ship?
Answers
Answer:
a) v₀ₓ = 62.76 m / s, b) θ₁ = 17.6º, θ₂ = 67.0º
Explanation:
We can solve this exercise using the projectile launch ratios
a) Let's find the time it takes for the bullet to reach the water level
y = y₀ + v_{oy} t - ½ g t²
when it reaches the water its height is zero y = 0, as the bullet is fired horizontally its initial vertical velocity is zero
0 = y₀ + 0 - ½ g t²
t =[tex]\sqrt{2y_o/g}[/tex]
t = [tex]\sqrt{2 \ 7 /9.8}[/tex]
t = 1,195 s
now we can calculate the speed with the horizontal movement
x = v₀ₓ t
v₀ₓ = x / t
v₀ₓ = 75.0 / 1.195
v₀ₓ = 62.76 m / s
b) if the speed of the bullets is half of that found
v₀ = 62.76 / 2 = 31.38 m / s
let's write the expressions for the distance
x = v₀ cos θ t
y = y₀ + v_{oy} sin θ t - ½ g t²
t = [tex]\frac{x}{v_o \ cos \theta}[/tex]
we substitute
[tex]0 = y_o + v_o sin \theta \ \frac{x}{v_o \cos \thetay} - 1/2 g \ (\frac{x}{v_o \ cos \theta})^2[/tex]
[tex]0 = y_o + x tan \theta - \frac{1}{2} g \ \frac{x^2}{ v_o^2 \ cos^2 \theta}[/tex]
let's use the identified trigonometry
sec² θ = 1 + tan² θ
sec θ = 1 / cos θ
we substitute
[tex]0 = y_o + x tan \theta - \frac{g x^2}{2 v_o^2} ( 1 + tan^2 \theta)[/tex]
[tex]\frac{g x^2}{2v_o^2} tan^2 \theta - x tan \theta + \frac{gx^2}{2v_o^2} - y_o = 0[/tex]
we change variable
tan θ = H
[tex]\frac{gx^2}{2 v_o^2 } H^2 - x H + \frac{gx^2}{2v_o^2}-y_o =0[/tex]
we subtitle the values
[tex]\frac{9.8 \ 75^2}{2 \ 31.38^2} H^2 - 75 H + \frac{9.8 \ 75^2}{2 \ 31.38^2}-7 =0[/tex]
27.99 H² - 75 H + 20.99 = 0
H² - 2.679 H + 0.75 = 0
we solve the quadratic equation
H = [2.679 ± [tex]\sqrt{2.679^2 - 4 0.75}[/tex]] / 2
H = [2,679 ± 2,044] / 2
H₁ = 0.3175
H₂ = 2.3615
now we can find the angles
H₁ = tan θ₁
θ₁ = tan⁻¹ H₁
θ₁ = tan⁻¹ 0.3175
θ₁ = 17.6º
θ₂ = 67.0º
for these two angles the bullet hits the boat
On a scale of 1-10 how much do you care of what people think of you?
Answers
Answer:
3
Explanation:
my family i hope thinks of me. And I don't have friends for them to think of me.
Thomas knows that many machines transform electrical energy into other forms of energy
Answers
So? Gd for him though
Answer:
Only the car transforms electrical energy into more than one form of energy.
Explanation:
The motion of the car is mechanical energy but it can also transform into another energy witch is electrical energy
How does Doppler ultrasound technology differ from ultrasound technology
that does not use the Doppler effect?
A. Doppler ultrasound collects data from moving objects.
B. Other ultrasound technology creates images, but Doppler
ultrasound does not.
C. Doppler ultrasound creates images, but other ultrasound
technology does not.
D. Doppler ultrasound is based on absorption of sound, and other
ultrasound technology is based on reflection.
Answers
Answer:
A. Doppler ultrasound collects data from moving objects
Explanation:
Did the test !!
Answer:A. Doppler ultrasound collects data from moving objects.
Explanation: just got it right
on my test
kinetic energy portfolio in part 2 the independent changes to----?
Answers
A car accelerates from zero to a speed of 110
km/hr in 15 seconds. What is the car's rate of
acceleration?
Answers
The car's rate of acceleration : a = 2.04 m/s²
Further explanationGiven
speed = 110 km/hr
time = 15 s
Required
The acceleration
Solution
110 km/hr⇒30.56 m/s
Acceleration is the change in velocity over time
a = Δv : Δt
Input the value :
a = 30.56 m/s : 15 s
a = 2.04 m/s²
A building inspector standing on the top floor of a building wishes to determine the depth of the elevator shaft. They drop a rock from rest and hear it hit bottom after 2.56 as. (a) How far (in m) is it from where they drop the rock to the bottom of the shaft
Answers
Answer:
d = 29.89 m
Explanation:
To solve this, we need to separate this problem in two parts.
One part would be the the time taken by the rock to actually hit the bottom, and the other part would be the time taken by the sound to reach the inspector.
Joining these two times we have:
t = t₁ + t₂ (1)
This time is 2.56 s.
Now, as we are asked to determine the distance from the top floor to the bottom, and we have two times taken in different ways, one by sound and the other the actual, we can say the same thing on distance, we need a distance relationed to the time taken by rock to hit the bottom, and the other distance relationet to the time taken by sound to reach the inspector.
Doing this we have that the distance traveled by the rock is:
y₁ = gt²/2
y₁ = 9.8t²/2 = 4.9t₁² (2)
Now, the distance traveled by sound would be:
y₂ = v * t₂ = 336t₂ (3)
Remember that the speed of the sound is 336 m/s
From this last expression (3), we can actually write t₂ in function of t₁, using (1):
2.56 = t₁ + t₂
t₂ = 2.56 - t₁ (4)
Replacing (4) in (3):
y₂ = 336(2.56 - t₁) (5)
Now that we have y₁ and y₂, we can equal (2) and (5), both expressions to get the value of t₁, and then, calculate the distance:
4.9t₁² = 336(2.56 - t₁)
4.9t₁² = 860.16 - 336t₁
4.9t₁² + 336t₁ - 860.16 = 0
Using the quadractic formula, we can calculate t₁:
t₁ = -336 ±√(336)² + 4*4.9*860.16 / (2*4.9)
t₁ = -336 ±√129,7555.136 / 9.8
t₁ = -336 ± 360.21 / 9.8 Using only the positive value we have:
t₁ = 2.47 s
This means that the rock hits the bottom in 2.47 s, and the remaining 0.09 s belongs to the time taken by sound. (2.47 + 0.09 = 2.56 s)
With this, we can calculate the distance of the rock using expression (2):
y₁ = 4.9 * (2.47)²
y₁ = 29.89 mHope this helps
NEED HELP WITH THIS PLEASE
Answers
Answer:
A
Explanation:
The Moon has a smaller mass than the Earth. If you were to travel to the moon your weight would....
A Increase
B Decrease
C Stay the Same
D Vary with day and night
Answers
Answer:The Moon has a smaller mass than the Earth. If you were to travel to the moon your weight would..
Explanation: It would decrease.
The Moon has a smaller mass than the Earth. If you were to travel to the moon your weight would decrease because the acceleration due by gravity on the moon is less than the acceleration due to gravity on the earth, therefore the correct answer is option C.
What is gravity?It can be defined as the force by which a body attracts another body towards its center as the result of the gravitational pull of one body and another.
In comparison to the Earth, the Moon is less massive. Your weight would drop if you traveled to the moon because the acceleration caused by gravity there is lower than that caused by gravity here on Earth.
As a result of the less gravity on the moon, the weight would decrease.
Thus, the correct answer is option C.
To learn more about gravity from here, refer to the link;
brainly.com/question/4014727
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A plane wishes to fly due north to an airport which is 205 km away. The plane can fly at a speed in still air of 220 km/h. A wind of 43 km/h blows from east to west.
a. In which direction,relative to north, should the plane head to reach it’s destination?
b. How long does this take?
Answers
Answer:
nique ta mama
Explanation:
Two loudspeakers are about 10 mm apart in the front of a large classroom. If either speaker plays a pure tone at a single frequency of 400 HzHz, the loudness seems pretty even as you wander around the room, and gradually decreases in volume as you move farther from the speaker. If both speakers then play the same tone together, what do you hear as you wander around the room
Answers
Answer:
I hear points of low volume sound and points of high volume of sound.
Explanation:
This is because, since the two sources of sound have the same frequency and are separated by a distance, d = 10 mm, there would be successive points of constructive and destructive interference.
Since their frequencies are similar, we should have beats of high and low frequency.
So, at points of low frequency, the amplitude of the wave is smallest and there is destructive interference. The frequency at this point is the difference between the frequencies from both speakers. Since the frequency from both speakers is 400 Hz, we have, f - f' = 400 Hz - 400 Hz = 0 Hz. So, the volume of the sound is low(zero) at these points.
Also, at points of high frequency, the amplitude of the wave is highest and there is constructive interference. The frequency at this point is the sum between the frequencies from both speakers. Since the frequency from both speakers is 400 Hz, we have, (f + f') = 400 Hz + 400 Hz = 800 Hz. So, the volume of the sound is high at these points.
So, as you wander around the room, I should hear points of high and low sound across the room.