Answer:
1. 22092.31 m
2. 3188.78 m
3. 25.51 s
Explanation:
Let's split this projectile's motion into its horizontal and vertical components:
Horizontal (v_x) component: 500 * cos(30)Vertical (v_y) component: 500 * sin(30)Since this projectile is in constant acceleration (9.8 m/s² vertically downwards and 0 m/s² horizontally), we can use the kinematic equations for constant acceleration.
Let's find the time it takes for the projectile to reach its greatest height first by using our knowledge that, at the top of its trajectory, the projectile will have a final velocity of 0 m/s before changing its direction and falling down.
Therefore, we can use the equation that contains the variables a_y, v_f, v_i, and t in order to solve for time t.
This equation is: v = v_0 + at, where v is the final velocity, v_0 is the vertical component of the initial velocity, a is the acceleration in the y-direction, and t is the time of flight of the projectile.
Plug known values into the equation:
0 = 500 * sin(30) + (-9.8)tSubtract 500 * sin(30) from both sides of the equation.
-500 * sin(30) = -9.8tDivide both sides by -9.8.
t = 25.51The time it takes the projectile to reach its greatest height is 25.51 seconds.
Now we can use this time t to find the greatest height of the object, which will be the displacement in the y-direction at t = 25.51 seconds.
We can use this equation that contains the variables displacement (Δx), initial velocity (v_0), acceleration (a_y), and time (t):
Δx = v_0 t + 1/2at²Plug known values into the equation.
Δx = [500 * sin(30)] * 25.51 + 1/2(-9.8)(25.51)² Δx = 6377.5 + -4.9 * 650.7601 Δx = 6377.5 + -3188.72449 Δx = 3188.77551The displacement in the y-direction is 3188.78 m at half the time t, meaning that this is the greatest height of the projectile.
Now we can find the horizontal range of the projectile by using the same equation, but this time in the x-direction.
Δx = v_0 t + 1/2at²We will use the x-component of the initial velocity and the acceleration in the x-direction, which is always 0 m/s² assuming air resistance is negligible. Note that the time we have above (25.51 sec) is only half the total time, so we will use double this time (51.02 sec).
Δx = [500 * cos(30)] * 51.02Δx = 22092.31The horizontal range of the projectile is 22092.31 m.
given a force of 88n and an acceleration of 4m/s2 what is the mass
Answer:
The mass is 22 kilograms
Explanation:
[tex]F=ma[/tex]
[tex]88N = m(4m/s/s)[/tex]
[tex]m=\frac{88N}{4m/s^{2} }[/tex]
[tex]m=22kg[/tex]
Which example best
describes Newton's 3rd Law?
A. Paddling a canoe to make the boat move
B. A large truck takes longer to
slow down than a car
C. Falling forward in a train that is slowing down quickly
Answer:
A
Explanation:
paddling a canoe to make the boat move
You’ve been in a car traveling at 80 km/hr. How many hours will it take to travel 640 km?
What element conducts heat and electricity:
Helium
Potassium
Sulfur
Copper
Arsenic
HEY CAN ANYONE PLS ANSWER DIS I RLY NEED IT!!!
Answer:
-Covalent compounds are formed when two or more nonmetal atoms bond by sharing valence electrons.
-Ionic compounds are neutral compounds made up of positively charged ions called cations and negatively charged ions called anions.
-One similarity is both ionic and covalent bonding lead to the creation of stable molecules.
Hope this helps:)
What are the 2 types of crystallization ?
Answer:
evaporative crystallization
cooling crystallization from solution or the melt
Answer:
I think this may helps you
How far does a car travel in 30.0 s while its velocity is changing from 50.0 km/h to 80.0 km/h at a uniform rate of acceleration?
Answer:
The car will travel 541.67 m
Explanation:
Uniform Acceleration
When an object changes its velocity at the same rate, the acceleration is constant.
The relation between the initial and final speeds is:
[tex]v_f=v_o+a.t[/tex]
Where:
vf = Final speed
vo = Initial speed
a = Constant acceleration
t = Elapsed time
The acceleration can be computed by solving for a:
[tex]\displaystyle a=\frac{v_f-v_o}{t}[/tex]
And the distance traveled is calculated as follows:
[tex]\displaystyle x=v_o.t+\frac{a.t^2}{2}[/tex]
The car travels during a time of t=30 s, and its speed changes from vo=50 Km/h to vf=80 Km/h.
Converting the speeds to m/s:
vo=50 Km/h * 1000/3600 = 13.89 m/s
vf=80 Km/h * 1000/3600 = 22.22 m/s
Both numbers are shown with a 2-decimal precision but they will be used with higher accuracy in further calculations.
Calculating the acceleration:
[tex]\displaystyle a=\frac{22.22-13.89}{30}[/tex]
[tex]a = 0.28\ m/s^2[/tex]
Now for the distance:
[tex]\displaystyle x=13.89*30+\frac{0.28.30^2}{2}[/tex]
[tex]x=416.67\ m+125\ m[/tex]
x = 541.67 m
The car will travel 541.67 m
A feather is falling towards the ground at a constant speed. Of the forces listed, identify which act upon the feather.
Answer:
Gravitational force
Explanation:
Gravitational force is what is making the feather go down. It's pushing the feather down, without the Gravitational force, the feather would have been going up, and not down. Just like us. Gravitational force is what keeps us humans from floating ( unless ur an alien jklol), So yeah Gravitational force is the force that acts upon the feather to fall towards the ground at a constant speed.
Btw, u never listed the forces here just ....
Hope this helped!
A feather is falling towards the ground at a constant speed. Forces act upon the feather is gravitational force.
What is force?
A force is an effect that can alter an object's motion according to physics. An object with mass can change its velocity, or accelerate, as a result of a force. An obvious way to describe force is as a push or a pull. A force is a vector quantity since it has both magnitude and direction.
When body is falling the net acceleration that objects get as a result of the combined action of gravity and centrifugal force is known as the Earth's gravity, or g. It is a vector quantity
A feather is falling towards the ground at a constant speed. Forces act upon the feather is gravitational force.
To learn more about force refer to the link:
brainly.com/question/13191643
#SPJ2
Mr Chichester is driving his dark blue camaro to a REO Speedwagon concert. He is driving 15 m/s when he realizes he is going to be late to his favorite band so he slams on the gas for 10s and reaches a final velocity of 45 m/s. How far did Mr. Chichesters travel during this 10s?
Answer:
300m
Explanation:
step one
given data
initial speed u= 15m/s
final speed v= 45m/s
time taken to attain final speed= 10seconds
Step two:
Let us first solve for the acceleration
a= Δv/t
a= 45-15/10
a=30/10
a= 3m/s
applying the equation of motion
[tex]v^2=u^2+2as[/tex]
substituting our given data
[tex]45^2+15^2+2*3*s\\\\2025=225+6s\\\\[/tex]
collect like terms
2025-225=6s
1800=6s
divide both sides by 6
s=1800/6
s=300m
Calculate the time needed to drive a car 100 kilometers if the car is traveling at a speed of 70 km/hr.
A.14 hours
B. 7 hours
C. .7 hours
D.1.4 hours
Answer:
D.1.4 hours
Explanation:
[tex]d\ =\ vt[/tex]
[tex]t\ =\ \frac{d}{v}[/tex]
[tex]t\ =\ \frac{100}{70}[/tex]
[tex]t\ =\ 1.42857142857[/tex]
Therefore, the answer is D 1.4 hours
Cart A and B are connected by a compressed spring. Cart A has a mass of 0.8 kg and Cart B has a mass of 3.00
kg. When the spring is released, Cart B moves to the right at a speed of 4 m/s. Find the speed of Cart A.
A W
B
Answer:
?????????????????????????
I WILL GIVE YOU BRAINLIEST!!
Which physical quantity does not change when a piece of copper is heated?
Answer:
Resistance, resistivity and drift velocity varies with relaxation time which is dependent on temperature.
Number of free electrons in a conductor remains invariant even if its temperature changes.Hence correct option is option D.
An inner city revitalization zone is a rectangle that is twice as long as it is wide. The width of the region is growing at a rate of 24 m per year at a time when the region is 300 m wide. How fast is the area changing at that point in time
Answer:
[tex]\frac{dA}{dt} = 28800 \ m^2/year[/tex]
General Formulas and Concepts:
Pre-Algebra
Order of Operations: BPEMDAS
Brackets Parenthesis Exponents Multiplication Division Addition Subtraction Left to RightEquality Properties
Geometry
Area of a Rectangle: A = lwAlgebra I
Exponential Property: [tex]w^n \cdot w^m = w^{n + m}[/tex]Calculus
Derivatives
Differentiating with respect to time
Basic Power Rule:
f(x) = cxⁿ f’(x) = c·nxⁿ⁻¹Explanation:
Step 1: Define
Area is A = lw
2w = l
w = 300 m
[tex]\frac{dw}{dt} = 24 \ m/year[/tex]
Step 2: Rewrite Equation
Substitute in l: A = (2w)wMultiply: A = 2w²Step 3: Differentiate
Differentiate the new area formula with respect to time.
Differentiate [Basic Power Rule]: [tex]\frac{dA}{dt} = 2 \cdot 2w^{2-1}\frac{dw}{dt}[/tex]Simplify: [tex]\frac{dA}{dt} = 4w\frac{dw}{dt}[/tex]Step 4: Find Rate
Use defined variables
Substitute: [tex]\frac{dA}{dt} = 4(300 \ m)(24 \ m/year)[/tex]Multiply: [tex]\frac{dA}{dt} = (1200 \ m)(24 \ m/year)[/tex]Multiply: [tex]\frac{dA}{dt} = 28800 \ m^2/year[/tex]Answer:
28,800 m²/yr
Explanation:
This rectangle has dimensions such that:
width = wlength = 2wWe are given [tex]\displaystyle \frac{dw}{dt} = \frac{24 \ m}{yr}[/tex] and want to find [tex]\displaystyle \frac{dA}{dt} \Biggr | _{w \ = \ 300 \ m} = \ ?[/tex] when w = 300 m.
The area of a rectangle is denoted by Area = length * width.
Let's multiply the width and length (with respect to w) together to have an area equation in terms of w:
[tex]A=2w^2[/tex]Differentiate this equation with respect to time t.
[tex]\displaystyle \frac{dA}{dt} =4w \cdot \frac{dw}{dt}[/tex]Let's plug known values into the equation:
[tex]\displaystyle \frac{dA}{dt} =4(300) \cdot (24)[/tex]Simplify this equation.
[tex]\displaystyle \frac{dA}{dt} =1200 \cdot 24[/tex] [tex]\displaystyle \frac{dA}{dt} =28800[/tex]The area is changing at a rate of 28,800 m²/yr at this point in time.
1. João tem um tempo de reação de 0,2 s, pedala a bicicleta com velocidade de 5 m/s. Ele se depara com uma situação de emergência e aciona os freios. Qual foi a distância percorrida antes de frear? Tempo = 0,2s Velocidade = 5m/s = ou = ⋅
Responda:
1 metro
Explicação:
Dado que :
Tempo de reação = 0,2 segundos
Velocidade de deslocamento = 5 m / s
A distância percorrida antes da frenagem pode ser obtida usando a relação:
Distância = velocidade de viagem * tempo de reação
Distância percorrida antes da frenagem = (5 m / s) * (0,2s)
Distância percorrida antes da frenagem = 1 metro
Help pleasseeee URGENT
If a 0.16 kg 8-ball at rest is hit by the 0.17 kg cue ball that is moving at a speed of 2 m/s, what is the speed of the 8-ball if the cue ball is completely stopped after the collision?
Answer:
The speed of the 8-ball is 2.125 m/s after the collision.
Explanation:
Law Of Conservation Of Linear Momentum
The total momentum of a system of masses is conserved unless an external force is applied. The momentum of a body with mass m and velocity v is calculated as follows:
P=mv
If we have a system of masses, then the total momentum is the sum of all the individual momentums:
[tex]P=m_1v_1+m_2v_2+...+m_nv_n[/tex]
When a collision occurs, the velocities change to v' and the final momentum is:
[tex]P'=m_1v'_1+m_2v'_2+...+m_nv'_n[/tex]
In a system of two masses, the law of conservation of linear momentum is simplified to:
[tex]m_1v_1+m_2v_2=m_1v'_1+m_2v'_2[/tex]
The m1=0.16 Kg 8-ball is initially at rest v1=0. It is hit by an m2=0.17 Kg cue ball that was moving at v2=2 m/s.
After the collision, the cue ball comes to rest v2'=0. It's required to find the final speed v1' after the collision.
The above equation is solved for v1':
[tex]\displaystyle v'_1=\frac{m_1v_1+m_2v_2-m_2v'_2}{m_1}[/tex]
[tex]\displaystyle v'_1=\frac{0.16*0+0.17*2-0.17*0}{0.16}[/tex]
[tex]\displaystyle v'_1=\frac{0.34}{0.16}[/tex]
[tex]v'_1=2.125\ m/s[/tex]
The speed of the 8-ball is 2.125 m/s after the collision.
It is not possible to accurately calculate instantaneous velocity.
TRUE
FALSE
What is the mass of an object moving with 80N of force and an acceleration of 8 m/s2?
Answer:
10 Kg
Explanation:
Force is equal to mass times acceleration
therefore mass is equal to force divided by acceleration
please mark me brainliest
An object is placed 5 cm away from a lens of magnification 3.2.
How far away from the lens is the image formed?
Answer:
16 cm
Explanation:
Magnification (M)= v/u
where V = image distance from lens
u = object distance from lens
Given that the object is 5 cm away from the lens
The magnification of the lens is 3.2
image distance from lens (v) = magnification * object distance = 3.2 * 5 =16cm
1.
Asthenosphere
a. the liquid layer located between the mantle and the inner
core
2.
Continental crust
b. the inner most solid layer of Earth
3.
Convection currents
the parts of Earth's crust located beneath the ocean that
make up the seafloor
4.
Crust
5.
Density
d. a transition zone beneath the lithosphere that is semi-
fluid, sometimes referred to as the plastic layer
6.
Inner core
e. the hard, rigid outermost layer of Earth
7
Lithosphere
f. all the parts of the Earth's crust that are not beneath the
ocean
8
Mantle
9.
Mesosphere
g the layer that is directly below Earth's crust and is divided
into 3 parts: lithosphere, asthenosphere, and mesosphere
h. the amount of force on a substance
10.
Oceanic crust
1. the solid lower layer of the mantle
11.
Outer core
j. the rigid, solid upper mantle connected to Earth's crust
12
Pressure
ki instrument used to measure earthquake waves
13.
Seismograph
1. the layer of Earth that is split into larger pieces: it floats on
the asthenosphere and moves due to convection currents
14.
Tectonic plates
m. the measure of mass per volume; the number of particles
in a given space
in movement of liquid and gaseous matter in a circular
motion caused by a difference in temperature; warm rises
and cool sinks, causing a circular current
Answer:
dsfghrtykuyjfcjuktj,ilyk
Explanation:
jgbnm,g bcm
A 0.3 kg bird flies south with a speed of 5 m/s. What is its kinetic energy?
A. 1.5 J
B. 2.5 J
C. 3.8 J
D. 8.3 J
Answer:
right answer is number C that is 3.8J since KE=1/2mv²
In the data table , distance is measured in meters and time is in seconds. Calculate the mans average velocity using the equation (average velocity= total distance/total time.
Answer:
3.626 m/s
Explanation:
v=d/t
1. -0.02/0 = 0 m/s
2. 0.86/0.2 = 4.3 m/s
3. 1.71/0.4 = 4.275 m/s
4. 2.54/0.6 = 4.23 m/s
5. 3.32/0.8 = 4.15 m/s
6. 4.08/1.0 = 4.08 m/s
7. 4.79/1.2 = 3.99 m/s
8. 5.48/1.4 = 3.91 m/s
9. 6.15/1.6 = 3.84 m/s
10. 6.76/1.8 = 3.76 m/s
11. 7.37/2.0 = 3.66 m/s
12. 7.92/2.2 = 3.6 m/s
13. 8.45/2.4 = 3.52 m/s
14. 8.96/2.6 = 3.45 m/s
the mean of these numbers is 3.626
his average velocity ks 3.626 m/s
Pls help me!! Thank u sm
Answer:
D
Explanation:
No machine have 100% efficiency. Why?
[tex]\huge\fcolorbox{black}{pink}{Answer}[/tex]
Because said machines have to operate in the physical world with constraints caused by the laws of physics. It’s not possible to do useful work without encountering some of these issues. The mains ones are:
Heat lost from the systemFriction (including air resistance)Electrical resistance[tex]\sf\blue{hope\:it\:helps}[/tex]
What force acts on a projectile when it is launched in the air?
A. gravity
B. friction
Hair sticking to a balloon after it
has been rubbed on a girl's head is
an example of
Answer:
Friction
Explanation:
Answer:
It causes electrostatic charging, which makes it stick to your hair,
a tiger moving with constant accleration covers the distance between two points 70 meter apart in 7 seconds . its speed as it pass the second point 15 meter per second . 1) what is the speed at the first point. 2) what us its accleration
Answer:
a = 1.428 [m/s²]
v₀ = 5 [m/s]
Explanation:
To solve this problem we must use the following equation of kinematics.
[tex]x=x_{o}+v_{o}*t+\frac{1}{2}*a*t^{2}[/tex]
where:
x = final point [m]
x₀ = initial point [m]
v₀ = initial velocity [m/s]
a = acceleration [m/s²]
t = time [s]
But we need to use this additional equation.
[tex]v_{f}=v_{o}+a*t[/tex]
where:
vf = final velocity = 15 [m/s]
Now we can use this equation, replacing it, in the first one. We must bear in mind that the difference among x - x₀ is equal to 70 [m]
[tex]x-x_{o}=v_{o}*t+\frac{1}{2}*a*t^{2} \\x-x_{o}=(v_{f}-a*t)*t+\frac{1}{2} *a*t^{2}\\70=(15-a*t)*t+\frac{1}{2}*a*t^{2}\\70=15*t-a*t^{2} +\frac{1}{2}*a*t^{2} \\70=15*t-\frac{1}{2}*a*t^{2}\\70=15*(7)-\frac{1}{2} *a*(7)^{2}\\105-70=0.5*a*49\\35=24.5*a\\a=1.428[m/s^{2} ][/tex]
Now replacing this value in the second equation, we can find the initial velocity.
[tex]15=v_{o}+1.428*7\\v_{o}=5[m/s][/tex]
What is the frequency of a wave with a wavelength of 2 and velocity of 20m/s?
Answer:
I think it is 10 Hz
Explanation:
Sorry if it's wrong
Which star is closest to our solar system (after the Sun)?
Alpha Centauri A
The two main stars are Alpha Centauri A and Alpha Centauri B, which form a binary pair. They are an average of 4.3 light-years from Earth. The third star is Proxima Centauri. It is about 4.22 light-years from Earth and is the closest star other than the sun
Have a nice Day , Hope this helped you I would appreciate it if you could mark my answer brainliest
Answer:
The answer is A
Dylan has two cubes of iron. The larger cube has twice the mass of the smaller cube. He measures the smaller cube. Its mass is 20 grams, and its density is 7.87 g/cm3. What’s the larger cube’s volume? The larger cube’s volume is about cm3.
Answer:
5.08cm³
Explanation:
Given parameters
The larger cube has twice the mass of the smaller cube
Mass of smaller cube = 20g
Mass of the larger cube = 2 x 20 = 40g
Density of the smaller cube = 7.87g/cm³
Unknown:
Volume of the larger cube = ?
Solution:
Density is the mass per unit volume of substance. For all samples of a substance, the density value is the same.
So, the density of the small and large iron is the same
Density = [tex]\frac{mass}{volume}[/tex]
Volume = [tex]\frac{mass}{density}[/tex]
So;
Volume of larger cube = [tex]\frac{40}{7.87}[/tex] = 5.08cm³
Answer:
5.08cm^3
Explanation:
Ya sabes que los materiales se utilizan dependiendo de la fun- ción que deben tener: transparentes (cristales), conductores de la electricidad (cables de tendidos eléctricos), muy resistentes (vigas), inoxidables (cuberterías), etc. Pero, ¿interesarán materiales de densidad grande o pequeña? ¿Por qué? Busca ejemplos de tres materiales que justifiquen tu respuesta (por ejemplo, el hilo conductor de las líneas de alta tensión, los colchones de espuma de alta densidad, la fibra de carbono, etc). ¿Qué características tiene la fibra de carbono en cuanto a resis- tencia y densidad? Busca objetos que se elaboren con fibra de carbono.