Answer:
11.86°
Explanation:
Projectile motion is a form of motion where an object moves in parabolic path (trajectory). Projectile motion only occurs when there is one force applied at the beginning on the trajectory, after which the only interference is from gravity.
The range of an object experiencing projectile motion is given by:
R = u²sin(2θ) / g
where u is the initial velocity, θ is the angle with horizontal and g is the acceleration due to gravity
Given that R = 6.8 m, g = 10 m/s², u = 12 m/s.
R = u²sin(2θ) / g
substituting:
6.8 = 13²sin(2θ) / 10
13²sin(2θ) = 68
sin(2θ) = 0.4024
2θ = sin⁻¹(0.4024)
2θ = 23.73
θ = 11.86°
convert 700kg into grams
Answer:
700,000g
Explanation:
1kg = 1000 Grams (or any kg x 1000)
Answer:
700 kilograms is equal to 700,000 grams
Explanation:
One kilogram has 1000 grams, or if you want the other way around, one gram is 1/1000th of one kilogram. If you follow this logic, we want to use the 'one kilogram has 1000 grams' "rule", so we would do some math here.
700 kg x 1000 g = ______ g
700*1000 = 700,000
700 kg = 700,000 g
Please let me know if I misunderstood and I hope this helped!
CLASS ENVIRONMENTAL SCIENCE
The____domain incorporates cultural norms that directly impact sustainable development. The___ domain focuses on the protection and preservation of natural and artificial resources
Blank 1: Blank 2:
A: social. A: public policy
B: economic. B: technological
C: sustainable. C: environmental
A box has a mass of 4kg and surface area 4m². Calculate the
pressure exerted by the box on the floor.
Answer:
10 pa
Explanation:
4kg* 10 (or 9.8m/s2) = 40
40N /4m2 =10
An inquisitive physics student and mountian climber climbs a 43.6 m cliff that overhangs a calm pool of water. He throws two stones vertically downward, 1.35 s apart and observes that they cause a single splash. The first stone has an initial velocity of 1.8 m/s. How long after release of the first stone do the two stones hit the water
Answer:
Explanation:
What we are basically looking for here is how long it takes the first stone to hit the water. We have everything we need to figure that out. We will use the equation
Δx = . Filling in, we will solve for t, the time is takes the first stone to hit the water (which is the same for both since they both hit the water at the same time):
which is a quadratic that we will have to factor. Get it into standard form, setting it equal to 0:
and factor to get that
t = 3.2 s and t = -2.8 s
Since time can't ever be negative, it takes 3.2 s for the stones to hit the water.
[tex]what \: is \: mirror \: {?}[/tex]
The force that the left team pulls with is 1000 N. If the right team's total mass is 300 kg and they accelerate by 1.2 m/s2, what is the force of resistance on the right team
Answer:
the force of resistance on the right team is 360 N
Explanation:
Given;
force of the left team, = 1000 N
total mass of the right team, m = 300 kg
acceleration of the right team, a = 1.2 m/s²
The force of resistance of the right team is calculated as;
Force = mass x acceleration
Force, F = 300 x 1.2
Force = 360 N
Therefore, the force of resistance on the right team is 360 N
Which statement or question is a good hypothesis
To overcome the problems that blur images and don't provide the best resolution from Earth, astronomers have started using flexible mirrors that change shape many times each second. This technique is called:
Answer:
adaptive optics
Explanation:
simple
how to solve for resistors
9514 1404 393
Answer:
A1 = 3A, A2 = 1.5A
Effective resistance = 2Ω
Explanation:
When the switch is closed, the voltage across each resistor is 6V, so the current through it (A2) is ...
A2 = 6V/(4Ω) = 1.5A
There are two parallel paths, each with that current, so the current from the battery is ...
A1 = A2 +A2 = 1.5A +1.5A = 3.0A
Then the effective resistance is ...
Reff = 6V/(3.0A) = 2.0Ω
The solution to the circuit is ...
A1 = 3A, A2 = 1.5A
Effective resistance = 2Ω
How long will it take a car to acceleration from 15.2ms to 23.Ms if the car has an average acceleration of 3.2m\s
Answer: 2.43 s
Explanation:
Initial velocity is [tex]u=15.2\ m/s[/tex]
Final velocity [tex]v=23\ m/s[/tex]
Average acceleration is [tex]a_{avg}=3.2\ m/s[/tex]
Average acceleration is change in velocity in the given amount of time
[tex]\therefore a_{avg}=\dfrac{v-u}{t}\\\\\Rightarrow 3.2=\dfrac{23-15.2}{t}\\\\\Rightarrow t=\dfrac{7.8}{3.2}\\\\\Rightarrow t=2.43\ s[/tex]
Thus, 2.43 s is required to acquire that average acceleration with 23 m/s velocity .
The most powerful empire between the 1500s and 1600s was the __________ Empire.
A.
Ottoman
B.
Mauryan
C.
Roman
D.
Persian
Answer:
A
Explanation:
Answer:
Ottoman
Explanation:
siri told me after I asked
A 90 kg astronaut Travis is stranded in space at a point 12 m from his spaceship. In order to get back to his ship, Travis throws a 0.50 kg piece of equipment so that it moves at a speed of 4 m/ s directly away from the spaceship towards the left . How long will it take him to reach the ship? *hint find his speed after the collision and consider it a constant speed all the way back to his spaceship*
Answer:
Explanation:
This is a recoil problem, which is just another application of the Law of Momentum Conservation. The equation for us is:
[tex][m_av_a+m_ev_e]_b=[m_av_a+m_ev_e]_a[/tex] which, in words, is
The momentum of the astronaut plus the momentum of the piece of equipment before the equipment is thrown has to be equal to the momentum of all that same stuff after the equipment is thrown. Filling in:
[tex][(90.0)(0)+(.50)(0)]_b=[(90.0)(v)+(.50)(-4.0)]_a[/tex]
Obviously, on the left side of the equation, nothing is moving so the whole left side equals 0. Doing the math on the right and paying specific attention to the sig fig's here (notice, I added a 0 after the 4 in the velocity value so our sig fig's are 2 instead of just 1. 1 is useless in most applications).
0 = 90.0v - 2.0 and
2.0 = 90.0v so
v = .022 m/s This is the rate at which he is moving TOWARDS the ship (negative was moving away from the ship, as indicated by the - in the problem). Now we can use the d = rt equation to find out how long this process will take him if he wants to reach his ship before he dies.
12 = .022t and
t = 550 seconds, which is the same thing as 9.2 minutes
A cup of mass 0.4 kg with its base radius of 4 cm is kept on a table. Calculate the pressure exerted by the cup on the table.
Answer:
779.87 N/m²
Explanation:
Pressure = Force (F) / Area (A)
Recall :
Force = ma ; mass * acceleration due to gravity
a = 9.8 m/s² ; mass = 0.4kg
Force, F = 0.4 * 9.8 = 3.92 N
Area = πr² ; r = Radius = 4cm = 4 / 100 = 0.04 m
Area = π0.04² = 0.0050265 m²
Hence,
Pressure = 3.92 / 0.0050265
Pressure = 779.86670 N/m²
Pressure = 779.87 N/m²
what is the angular speed w of the system immediately after the collision in terms f the sstem parameters and I
Answer: hello some part of your question is missing attached below is the missing detail
answer :
wf = M( v cos∅ )D / I
Explanation:
The Angular speed wf of the system after collision in terms of the system parameters and I can be expressed as
considering angular momentum conservation
Li = Lf
M( v cos∅ ) D = ( ML^2 / 3 + mD^2 ) wf
where ; ( ML^2 / 3 + mD^2 ) = I ( Inertia )
In terms of system parameters and I
wf = M( v cos∅ )D / I
With what tension must a rope with length 3.00 mm and mass 0.105 kgkg be stretched for transverse waves of frequency 40.0 HzHz to have a wavelength of 0.790 mm
Answer:
the tension of the rope is 34.95 N
Explanation:
Given;
length of the rope, L = 3 m
mass of the rope, m = 0.105 kg
frequency of the wave, f = 40 Hz
wavelength of the wave, λ = 0.79 m
Let the tension of the rope = T
The speed of the wave is given as;
[tex]v = f\lambda = \sqrt{\frac{T}{\mu} } \\\\where;\\\\\mu \ is \ mass \ per \ unit \ length\\\\\mu = \frac{0.105}{3} = 0.035 \ kg/m\\\\v = f\lambda = 40 \times 0.79 = 31.6 \ m/s\\\\v = \sqrt{\frac{T}{\mu} } \\\\v^2 = \frac{T}{\mu} \\\\T = v^2 \mu\\\\T = (31.6^2)(0.035)\\\\T = 34.95 \ N[/tex]
Therefore, the tension of the rope is 34.95 N
Give reason:
Mass and volume are called physical quantities.
a.
b
Ionath is called
Mass and volume are called physical quantities because they can be measured by using physical devices i.e. mass is measured by using beam balance and volume is measured by using metre scale .
Juanita ran one mile around her school track in six minutes. What is
her average speed, and what is the magnitude of her average velocity?
10 mph, 0 mph
6 mph, 0 mph
6 mph, 6 mph
10 mph, 10 mph
Answer:
The correct option is a) 10 mph, 0 mph.
Explanation:
1. The average speed (S) is a magnitude given by:
[tex] S = \frac{D}{T} [/tex]
Where:
D: is the total distance = 1 mi
T: is the total time = 6 min
[tex] S = \frac{D}{T} = \frac{1 mi}{6 min}*\frac{60 min}{1 h} = 10 mph [/tex]
Hence, the average speed is 10 mph.
2. The average velocity is a vector:
[tex] V = \frac{\Delta d}{\Delta t} = \frac{d_{f} - d_{i}}{t_{f} - t_{i}} [/tex]
Where:
[tex]d_{f}[/tex]: is the final distance
[tex]d_{i}[/tex]: is the initial distance
[tex]t_{f}[/tex]: is the final time
[tex]t_{i}[/tex]: is the initial time
Since Juanita ran one mile around her school track, the final position is the same that the initial position, so the magnitude of the average velocity is zero.
Therefore, the correct option is a) 10 mph, 0 mph.
I hope it helps you!
In the hydraulic system depicted, the cylinder on the left has a diameter of 2 inches and the cylinder on the right has a diameter of 6 inches. If 100 lbs of force was applied to the cylinder on the left, what force would be exerted on the cylinder on the right
Answer:
F2 = 900 lbs
Explanation:
From pascal principle;
F1/A1 = F2/A2
Force on cylinder at left; F1 = 100 lbs
Diameter of cylinder at left; d1 = 2 inches
Diameter of cylinder at right; d2 = 6 inches
Formula for area of top of cylinder = πr²
Thus;
Area of top of left cylinder; A = π × 2² = 4π
Area of top of right cylinder; A = π × 6² = 36π
Thus;
100/4π = F2/36π
F2 = (36π × 100)/4π
F2 = 900 lbs
The 75.0 kg hero of a movie is pulled upward at a constant velocity by a rope. What is the tension on the rope?
Answer:
750 N
Explanation:
the tension on the rope is the weight of the hero
Calculate surface tension of an enlarged radius of 4cm to 6cm and amount of work necessary for enlargement was 1.5×10^-4 joules
Answer:
[tex]T=7.5*10^{-5}[/tex]
Explanation:
From the question we are told that:
Radius [tex]r=2cm[/tex]
Work done [tex]W=1.5×10^-4 joules[/tex]
Generally the equation for Work done is mathematically given by
[tex]W=T.dA[/tex]
Therefore
[tex]T=\frac{W}{dr}[/tex]
[tex]T=\frac{1.5*10^{-4}}{2}[/tex]
[tex]T=7.5*10^{-5}[/tex]
what is measurement?
While traveling north on an expressway, a car traveling 60 mph (miles per hour) slows down to 30 mph in 12 minutes due to traffic conditions
Answer:
acceleration = - 150 m/s^2
distance = 9 miles.
Explanation:
initial speed, u = 60 mph
time, t = 12minutes = 0.2 hour
final speed, v = 30 mph
Let the acceleration is a and the distance is s.
By the first equation of motion
v = u + at
30 = 60 + a x 0.2
a = - 150 m/s^2
Let the distance is s.
Use third equation of motion is
[tex]v^2 = u^2 + 2 a s \\\\30^2 = 60^2 + 2 \times 150\times s\\\\s = 9 miles[/tex]
A glass tube in the shape of a letter J has the shorter limb sealed and the longer limb open. Mercury is poured into the tube until the levels in either limb is the same when the tube is vertical.In this position, the length of the air column in the sealed limb is 6.3cm.More mercury is then poured into the tube until the length of the trapped air column is 42cm.Calculate the difference in the levels of mercury in the limbs if a nearby mercury barometer reads 75.0cm and the reading of a nearby thermometer has not changed?
Answer:
35.4 cm
Explanation:
We have that when the level of mercury on either limb is the same, the pressure of the trapped air, P₁ = Atmospheric pressure
Also the initial height of the mercury in the tube = The reading of the barometer = 75.0 cm
The initial length of the air column, l₁ = 6.3 cm
The final length of the air column, l₂ = 4.2 cm (The length is expected to decrease due to compression)
The volume, V = l × A
Where;
A = The cross sectional area of the tube
Therefore, the volume of the air column is directly proportional to the length of the air column
∴ V ∝ l
According to Boyles law, we have;
P₁·V₁ = P₂·V₂
Where;
P₁ = The initial pressure in the air column before more mercury is added
V₁ = The initial volume occupied by the air in the air column
P₂, and V₂ are the final pressure and volume of the air column respectively
Given that V = l·A, we can write;
P₁·l₁·A = P₂·l₂·A
P₂ = P₁·l₁·A/(l₂·A) = P₁·l₁/(l₂) = P₁ × 6.3/4.2 = 1.5·P₁
The pressure in the air column after more mercury is added, P₂ = 1.5 × P₁
P₁ = Atmospheric pressure, therefore;
The pressure in the air column after more mercury is added, P₂ = 1.5 × Atmospheric pressure
Pressure = h·ρ·g
Where;
ρ = The density of the substance
g = The acceleration due to gravity
h = The height of the column of the fluid
Given that the density and the gravitational force, can be taken as constant, we have that the pressure of the fluid is directly proportional to the height of the fluid column
Therefore, when the pressure doubles, the height of the fluid column doubles, and when the factor of increase is 1.5, we have;
The final level of the mercury, h₂ = 1.5·h₁ = 1.5×75 cm = 112.5 cm
The initial length of the closed end of the J tube, [tex]h_{closed1}[/tex] = 6.3 cm + 75 cm = 81.3 cm
The final length of the mercury in the closed end, [tex]h_{closed2}[/tex] = 81.3 cm - 4.2 cm = 77.1 cm
The difference in the level of mercury, Δh = h₂ - [tex]h_{closed2}[/tex]
∴ Δh = 112.5 cm - 77.1 cm = 35.4 cm
The difference in the levels of mercury in the limbs, Δh = 35.4 cm
A 4 LBfweight is attached to a spring suspended from the ceiling. When the weight comes to rest at equilibrium, the spring has been stretched 3 inches. The damping constant for the system is 2 LBf−sec ft . If the weight is raised 9 inches above equilibrium and given an initial upward velocity of 2 ft/sec, determine the equation of motion of the weight and give its damping factor, quasiperiod and quasifrequency.
Answer:
attached below
Explanation:
The initial conditions :
x(0) = - 9 inches = -3/4 ft
x'(0) = - 2 ft/sec
spring stretched 3 inches = 1/4 ft
mass = w / g = 4 Ib / 32 ft/sec^2 = 1/8 slug
the spring constant ( k ) = w / l = 4 / ( 1/4 ) = 16 Ib/ft
applying the second law of motion
m d^2x/dt^2 + b dx/dt + kx
= 1/8 d^2x/dt^2 + 2 dx/dt + 16 x
= d^2x/dt^2 + 16 dx/dt + 128 x ------- ( 1 )
we will resolve the above equation to obtain the required equation of motion x( t )
Attached below is the remaining part of the solution
What is the dependent variable in this
experiment?
DONE
Biologists designed an experiment to test
the effect of compost on the development
of root crops. They tested several different
crops, including carrots, potatoes, beets,
and onions. They grew most of the plants
in the greenhouse, but due to space issues,
they had to grow some outdoors. They gave
all the plants the same amount of compost.
They obtained the compost from a local
farmer and from the local hardware store.
They ran out of the farmer's compost, so
some of the plants received that compost
when the seeds were planted and other
plants got hardware store compost after
the plants had already started growing.
What is the independent variable in this
experiment?
DONE
Answer:
"the plants had already started growing."
Explanation:
I think this is the answer because the definition of a dependent variable is the variable that is being affected by the change. Since the plants had already started growing BECAUSE of "They ran out of the farmer's compost, so
some of the plants received that compost
when the seeds were planted and other
plants got hardware store compost after
the plants had already started growing."
Sorry if I am wrong, I am just a 4th grader, pls don't hate on me, I am just trying to help :)
Answer:
It's compost
Explanation:
In case you needed the dependent variable, its the amount of plant growth
6. An object is fired from the gound at 275 m/s at an angle of 55° N of E.
a. How far away did the object first hit the ground?
b. what is the maximum height that the object reaches?
there u go fella hope u understood
A battery of emf 5V and internal resistance 2ohm is joined to a resistor of 8ohm.Calculate the terminal potential difference.
Answer:
4V
Explanation:
First, we calculate the total resistance to the given battery cell of emf 5V. The total resistance is the sum of all the resistance in the cell i.e.
Total resistance = 2Ω + 8Ω = 10Ω
Using ohms law equation to calculate the current passing through the battery cell:
V = IR
Where; V = voltage, I = current, R = resistance
5V = I × 10Ω
I = 5/10
I = 0.5A
Terminal voltage is calculated by the us of the following equation:
V=emf−IR
Where; R is internal resistance
V = 5 - (0.5 × 2)
V = 5 - 1
V = 4V
Therefore, the potential difference across the terminals of the battery cell is 4V
Which of the following is defined as a force that pushes and pulls the current through the circuit? Group of answer choices D) resistance B) electricity A) current C) voltage
Answer:
C voltage
Explanation:
Voltage is the change in electric potential so basically current flows from high potential to low potential due to voltage.
Question 23 of 23
Suppose a current flows through a copper wire. Which two things occur?
O A. The field is parallel to the direction of flow of the current.
B. An electric field forms around the wire.
OC. A magnetic field forms around the wire.
U
D. The field is perpendicular to the direction of flow of the current.
SUBM
Answer:
The field is parallel to the direction of flow of the current.
5- Clasifica los siguientes cambios de la materia, anotando delante de cada uno cambio físico (F) o cambio químico (Q): • Disolver azúcar en agua • Freir una chuleta • Arrugar un papel • El proceso de la digestión • Secar la ropa al sol • Congelar una paleta de agua • Hacer un avión de papel • Oxidación del cobre • Romper un lápiz • Prender fuegos artificiales • Excavar un hoyo • Quemar basura
Answer:
1) Disolver azúcar en agua - Cambio químico - Es un caso de una solución en donde el solvente es el azúcar y el soluto es el agua.
2) Freir una chuleta - Cambio físico - Es un proceso de cocinado por un transferencia de calor y transferencia de masa.
3) Arrugar un papel - Cambio físico - Se aplica fuerzas externas para deformar el papel.
4) El proceso de la digestión - Cambio químico - Reducción de los alimentos a desechos y la absorción de nutrientes por el contacto con jugos gástricos o ambientes intestinales.
5) Secar la ropa al sol - Cambio físico - El secado es un fenómeno de transferencia de masa.
6) Congelar una paleta de agua - Cambio físico - Cambio del estado líquido al estado sólido por transferencia de calor.
7) Hacer un avión de papel - Cambio físico - Aplicación de fuerzas externas para plegar y doblar la hoja de papel.
8) Oxidación del cobre - Cambio químico - Proceso de corrosión por contacto con iones que se transportan en el ambiente.
9) Romper un lápiz - Cambio físico - Proceso de ruptura por esfuerzo normal a causa de un momento como consecuencia de una fuerza externa aplicada sobre el lápiz.
10) Prender fuegos artificiales - Cambio químico - Reacción química de reducción-oxidación.
11) Excavar un hoyo - Cambio físico - Remoción de tierra por trabajo físico.
12) Quemar basura - Cambio químico - Reacción de combustión.
Explanation:
A continuación, veremos que representa cada caso:
1) Disolver azúcar en agua - Cambio químico - Es un caso de una solución en donde el solvente es el azúcar y el soluto es el agua.
2) Freir una chuleta - Cambio físico - Es un proceso de cocinado por un transferencia de calor y transferencia de masa.
3) Arrugar un papel - Cambio físico - Se aplica fuerzas externas para deformar el papel.
4) El proceso de la digestión - Cambio químico - Reducción de los alimentos a desechos y la absorción de nutrientes por el contacto con jugos gástricos o ambientes intestinales.
5) Secar la ropa al sol - Cambio físico - El secado es un fenómeno de transferencia de masa.
6) Congelar una paleta de agua - Cambio físico - Cambio del estado líquido al estado sólido por transferencia de calor.
7) Hacer un avión de papel - Cambio físico - Aplicación de fuerzas externas para plegar y doblar la hoja de papel.
8) Oxidación del cobre - Cambio químico - Proceso de corrosión por contacto con iones que se transportan en el ambiente.
9) Romper un lápiz - Cambio físico - Proceso de ruptura por esfuerzo normal a causa de un momento como consecuencia de una fuerza externa aplicada sobre el lápiz.
10) Prender fuegos artificiales - Cambio químico - Reacción química de reducción-oxidación.
11) Excavar un hoyo - Cambio físico - Remoción de tierra por trabajo físico.
12) Quemar basura - Cambio químico - Reacción de combustión.
