what change do you need to make when comparing objects going upward against gravity versus downward with gravity?

A. Different amount of time in motion
B. Different length of displacement
C. Different sign + or - on the acceleration
D. Different changes to velocity

Answers

Answer 1

If an object is working against gravity, that means gravity must have the opposite sign of the object (this is why we see objects slow down at their peak height and return to us if we throw them in the air!). If the object is working with gravity, though, they must both be moving in the same direction, therefore gravitational acceleration would be the same sign as the velocity (positive or negative depending on how you define your axes).

Therefore, your answer should be C. Different sign + or - on the acceleration.

I hope this helps!

Answer 2
The answer here is c

Related Questions

A 30kg metal ball is dropped from a height of 12.5 m.  
a.  Find the final velocity when the ball  hits the ground.
b.  Find the time it takes for the ball to hit the ground.

Answers

From conservation of linear momentum, the final velocity is 15.7 m/s and the time taken is 1.6 s

What is Velocity ?

Velocity can be defined as a distance travel in a specific direction per time taken. It is a vector quantity.

Given that 30kg metal ball is dropped from a height of 12.5 m.  

a.  Find the final velocity when the ball  hits the ground.

The maximum K.E of the ball at it final velocity will be equal to its maximum P.E at height 12.5 m. That is,

K.E = P.E

1/2mv² = mgh

Where

m = 30 Kgg = 9.8 m/s²h = 12.5 mv = ?

Substitute all the parameters

1/2 × 30 × v² = 30 × 9.8 × 12.5

v² = 245

v = √245

v = 15.65 m/s

b. The time it takes for the ball to hit the ground can be found through

h = ut + 1/2gt²

but u = 0

h = 1/2gt²

Substitute all the necessary parameters

12.5 = 1/2 × 9.8 × t²

12.5 = 4.9t²

t² = 12.5/4.9

t² = 2.55

t = √2.55

t = 1.6 s

Therefore, the final velocity when the ball hits the ground is 15.65 m/s and the time it takes for the ball to hit the ground is 1.6 s

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please answer fast it doesn't have to be long please :)

Juan drives his car around a curve at a speed of 17.0 m/s. Assume the curve can be approximated by an arc of a circle of radius 122.0 m. If a force of 7380 N is required to maintain the car’s circular motion, what is the car’s mass?

Answers

The mass of the car moving with a velocity of 17 m/s is 3115.4 kg.

What is mass?

Mass is the quantity of matter a body contains. The S.I unit of mass is kilogram

To calculate the mass of the car, we use the formula below

Formula:

m = Fr/v²...................... Equation 1

Where:

m = Mass of the carF = Forcer = Radiusv = Velocity

From the question,

Given:

F = 7380 Nr = 122 mv = 17 m/s

Substitite these values into equation 1

m = (7380×122)/17²m = 3115.4 kg

Hence, the mass of the car is 3115.4 kg.

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i understand how something about the pod could have affected its velocity change.
yes or not yet.
EXPLAIN your answer choice.

Answers

A stronger force can cause a greater change in an object's velocity. The correct option is yes

What caused the pod to change direction?

The force exerted by the thrusters caused the pod to change direction. The thrusters on the ACM pod exerted the same strength force as thrusters on other pods on other missions. If the same strength force is exerted on two objects but the objects have different masses the object with less mass will have a greater change in velocity.

Therefore pod would have had to have a lighter mass meaning a smaller amount of force would have had to be exerted onto the pod making it move in the opposite direction.

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An astronaut in a satellite 1600km above the Earth experience a gravitational force magnitude 700N on earth.The Earth's radius is 6400km calculate the magnitude of the gravitational force which the astronaut experience in satellite

Answers

Answer:

The magnitude of the gravitational force which the astronaut experience in satellite [tex]= 448[/tex] Newton

Explanation:

As we know

Gravitational force  F = [tex]\frac{GMm}{r^2}[/tex]

Where G is the gravitational constant [tex]6.67 * 10^{-11}[/tex]

M is the mass of earth [tex]= 6* 10^{24}[/tex]

m is not given

r is the radius of earth [tex]6 400*10^3[/tex]

Substituting the given values, we get -

[tex]F = \frac{6.67*10^{−11}*6*10^{24}* m}{(6400*10^3)^2}[/tex]

[tex]m = 71.6[/tex] Kg

Gravitational force in the satellite

[tex]F = \frac{6.67*10^{−11}*6*10^{24}* m}{(6400*10^3 + 1600 *10^3)^2}\\F = 448[/tex]N

A student conducts an experiment to test how the temperature affects the amount of salt that can dissolve in water. In the experiment, she uses 150 milliliters of water in each trial and stirs for five minutes each time.

Answers

The independent variable that we have in the experiment is the temperature. Option B

What is an experiment?

We know that if we are talking about an experiment, what we have to look at is that we must consider the interplay of the variables that is taking place in the process. As we change one of the variables in the experiment, there would be a change in another variable in the experiment also.

Now, the variable that we usually beging to alter is the one that we call the independent variable. As we change the value of this particular variable, there would be a change in the independent variable.

We have been told that; a  student conducts an experiment to test how the temperature affects the amount of salt that can dissolve in water.

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Sally and Sam are in a spaceship that comes to within 15,000 km of the asteroid Ceres. Determine the force Sally experiences, in N, due to the presence of the asteroid. The mass of the asteroid is 8.7 1020 kg and the mass of Sally is 68 kg. For calculation purposes, assume the two objects to be point masses.

Answers

The gravitational force that is acting on the masses is 1.73 * 10^-2 N.

What is the force?

We know that the force of gravity is that kind of force that is going to act on any two of the objects that we have on the earth. This force as we know it is attractive force. The implication of this is that any two masses that we find on the earth are the forces that would have to be attracted to each other.

Let us note that we would have the masses that we have of the asteroid and that of sally as m1 and m2 while we have the force as;

F = G m1m2/r^2

G = gravitational constant

F = magnitude of the force

m1 and m2 = masses

r = distance of separation.

We then have;

F = 6.6 * 10^-11 * 8.7 * 10^20 * 68/(15 * 10^6)^2

F = 3.9 * 10^12/2.25 * 10^14

F = 1.73 * 10^-2 N

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In transverse waves, the movement of the particles is _________
O in a circle
O left to right
O diagonal
O up and down

Answers

Answer:

the correct answer is up and down

A resistor with a resistance of 2 MΩ is connected in series with a 5 µF capacitor and a battery with emf 10 V. The capacitor has originally a charge of 5 µC and then discharged. At what time will the charge be equal to 1µC?



A. 16 sec
B. 1.6 sec
C. 3.2 sec
D. 8 sec

Answers

C because it very simple

A force of 50N applied to the end of a lever moves that end a certain distance if the other end of the lever is moved half as far how much force does it exert?

Answers

The force exerted at other end is 100N.

Work is the product of the component of the force in the direction of the displacement and the magnitude of this displacement.Mathematically, the above statement is expressed as follows:

                              W = (F cos θ) d = F. d

Where,

W is the work done by the force.

F is the force, d is the displacement caused by force

θ is the angle between the force vector and the displacement vector

Let , F' = Force exerted on ladder at other end

       d' = distance moved = d/2 (given)

i.e.  W = F'd' and W = Fd

Using these two equations we get, F'd' = Fd

i.e. F' = Fd/d'

Given, F = 50N and d' = d/2

Putting these values in above equation we get, F' = 100N

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Which ideas did you include in your answer? Check all that apply.

Scientists use relative and absolute dating to determine age of rocks.
Scientists can organize the rocks layers based on their age on the geologic time scale.
The type of rocks and the fossils in the rock layers tell scientists about Earth’s history.

Answers

By contrasting a fossil with related rocks and fossils of known ages, relative dating can estimate the fossil's age.

By utilising radiometric dating to track the isotope decay inside the fossil or, more frequently, the rocks it is linked with, absolute dating may pinpoint an object's exact age.

What are the benefits of using relative and absolute dating of rocks?

It is possible to identify whether one artefact, fossil, or stratigraphic layer is older than another using relative dating, which does not provide particular dates.

Absolute dating techniques offer more precise dates and times of origin, such as a range in age in years.

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Write words with a similar meaning to the given words for old, stubble and room help me?pls ​

Answers

Answer:

Old = senior. geriatric. senescent. unyoung. over-the-hill.

Stubble= bristles

whiskersdesigner stubblehairfacial hairbeardfive o'clock shadowFeedback

How long does it take the Earth to spin on its axis once?​

Answers

Explanation:

it takes the earth 24 hours to spin on it's axis once

Р.
Use the information to answer the following question.
The information shows the masses after the compounds are balanced.
2 LiCl + Na,o
-
Li,O +2 NaCl
84
+
62
=
X +
118
If the following materials are used during the reaction of Lithium, Oxygen, Sodium, and Chlorine, what mass of Lithium Oxide would we
expect to see after the reaction and why?
O A. 28g of Li,O, because all mass must be conserved, it would be half of the compound LiCl in the products
O
B. 44g of Li,O, because all mass must be conserved, it would be half of the compound LiCl in the products
o
C. 28g of Li_0, because all mass must be conserved
D. 44g of Li2O, because all mass must be conserved

Answers

Answer:

C. 28g of Li2O, because all mass must be conserved

Explanation:

my mom is a 8th grade science teacher lol.

Help yea I need help

Answers

Answer is D I think....

The figure below shows a cylinder filled with an ideal gas, which has a moveable piston resting on it. The cylinder's volume is initially 5.50 L, when a force on the piston of F = 14.5 kN pushes the piston downward a distance d = 0.140 m, until the volume of the cylinder is 3.00 L. The process occurs while the cylinder is in thermal contact with a large energy reservoir at a temperature of 295 K.
(A) How much work (in kJ) is done on the gas by the piston during the process?
(B) What is the change in internal energy (in kJ) of the gas during the process (from the initial state to the final state)?
(C) What is the energy transfer (in kJ) by the gas as heat during the process? (Treat the gas as the system and let the sign of your answer indicate the direction of energy flow.)
(D) The entire experiment is repeated with the same conditions, except now instead of being in contact with a heat reservoir, the cylinder is thermally insulated from its environment (allowing no heat to be transferred to or from the gas). In this case, what happens to the temperature of the gas during the process?

Answers

I uploaded the answer to[tex]^{}[/tex] a file hosting. Here's link:

bit.[tex]^{}[/tex]ly/3gVQKw3

Maritza's school provides a laptop to each student to use in the classroom and for
homework. What is the BEST way for her to use the laptop in her science class?
A.to look up bands who are coming to town?
B.to take notes as her teacher talks?
C.to complete her English assignment?
D.to email her friend in another class?

Answers

The best way for her to use the laptop in her science class  is to take notes as her teacher talks

How can a laptop be used in a classroom?Fostering online collaboration with other students.Providing curriculum support and additional information to students.Promoting better organization: Laptops help students keep track of their assignments and utilize an online school calendar.Laptops allow students to collaborate with their classmates inside and outside the classroom. They can ask questions, compare notes and share what they have learned more readily. They can also work together on group projects even if they are not in the same locationIn the last several years, schools across the country have experimented with providing each student with a laptop to facilitate learning. In 2016, Michigan State University studied the results from a number of these programs to determine if providing laptops influenced academic performance. The researchers, led by Michigan State University’s Assistant Professor Binbin Zheng, looked at nearly 100 studies of the use of laptops in the classroom, known as “one-to-one computing environments.”

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A group of students wants to determine the internal resistance of a battery. They connect the battery to a variable resistor. The students measure the potenial differnce across the battery as a function of the current throught the battery as they vary the resistance. Which of the following analyses of the data could be used to determine the internal resistance of the battery?
A. Divide the potential difference across the battery by the current through it for each data point. The average of these calculations gives the internal resistance of the battery.
B. Graph the potential difference across the battery as a function of the current through it. Extrapolate to find the y- intercept and divide this by the average of the current measurements to find the internal resistance of the battery.
C. Find the best-fit straight line for a graph of potential difference across the battery as a function of the current through it. The absolute value of the slope represents the internal resistance of the battery.
D. This data cannot be analyzed to give the internal resistance of the battery, because the potential difference across the battery does not depend on the current.

Answers

Answer:

The answer is "Option C"

Explanation:

Ohm's Rule we remember [tex]V = I\times R[/tex]

It looks like the sample is drawn y=mx while m is curve slope

The slope of the graph is its resistance if they draw a graph v vs I.

However, the internal battery presence resistance can be modified as the straight line equation

[tex]y = mx -c[/tex]

where c is y-intercept:

[tex]E = I(R+r) \\\\ E = IR+ Ir\\\\E = V + Ir\\\\V = E- Ir\\\\V = -Ir+E[/tex]

R is really the internal battery resistance to draw graph V vs I, the slope is internal resistance to the negative slope cure.

A 7.600 kg box is resting on a horizontal surface and attached to a 2.400 kg box by a thin, light wire that passes over a frictionless pulley. The coefficient of kinetic friction between the box and the surface is 0.1500. The pulley has the shape of a hollow sphere of mass 1.200 kg and diameter 0.2800 m. The system is released from rest and allowed to move. The hanging box falls 0.6100 m before it hits the ground. What is the speed of the hanging block just before it hits the ground

Answers

Answer:

v = 1.98 m / s

Explanation:

Let's analyze this exercise a bit, initially the hanging box is at a height h, which is why it has gravitational power energy, as the system is removing the kinetic energy is zero and just when the box reaches the floor its potential energy has dropped to zero and the three bodies have kinetic energy, also between the box and the horizontal surface there is friction, so there is work. Let's use the relationship between work and energy

starting point. Before starting the movement

         Em₀ = U = m g h

final point. Just before the block hit the floor

         Em_f = K = ½ M v² + ½ I w² + ½ m v²

the speed of the two blocks must be the same to maintain the tension of the rope

The work of the friction force

the friction force opposes the movement so its work is negative

          W = - fr x

the law of equilibrium is the largest block

          N-W = 0

          N = W = Mg

we substitute

         W = - μ M g x

the relationship between the work of the non-conservative force (friction) and the energy is

         W = Em_f - Em₀

        - μ Mg x = ½ M v² + ½ I w² + ½ m v² - mg h

the moment of inertia of a hollow sphere is

         I = ⅔ m_s r²

angular and linear velocity are related

         v = w r

         w = v / r

the distance the horizontal block travels must be the same as the distance the vertical block travels

          x = h

let's substitute

         - μ M g h = ½ (M + m) v² + ½ (⅔ m_s r²) (v/r) ² - m g h

          (- μ M + m) g h = ½ (M + m + ⅔ m_s) v²

          v² = [tex]\frac{2(\mu \ M + m ) \ g h }{ M +m + \frac{2}{3} m_s}[/tex]

let's calculate

          v² = 2 (-0.15 7.6 +2.4) 9.8 0.61 / (7.6 + 2.4 + 2/3 1.2)

          v = [tex]\sqrt{\frac{42.32}{ 10.8} }[/tex]

          v = 1.98 m / s

During a neighborhood baseball game in a vacant lot, a particularly wild hit sends a 0.144 kg baseball crashing through the pane of a second-floor window in a nearby building. The ball strikes the glass at 14.9 m/s , shatters the glass as it passes through, and leaves the window at 10.5 m/s with no change of direction.

Requied:
a. What is the direction of the impulse that the glass imparts to the baseball?
b. Calculate the magnitude of this impulse (a positive number).

Answers

Answer:

a)The direction of the impulse that the glass imparts to the baseball Is opposite the direction of the balls motion

b)  [tex]I=0.6336Ns[/tex]

Explanation:

From the Question we are told that:

Mass [tex]m=0.144kg[/tex]

Initial Speed [tex]v_1=14.9m/s[/tex]

Final speed  [tex]v_2=10.5[/tex]

a)The direction of the impulse that the glass imparts to the baseball Is opposite the direction of the balls motion

b)

Generally the equation for impluse magnitude is mathematically given by

 [tex]I=m(v_1-v_2)[/tex]

Therefore

 [tex]I=0.144(14.9-10.5)[/tex]

 [tex]I=0.6336Ns[/tex]

A student attempts to spin a rubber stopper(m=0.5kg) in a horizontal circle with a radius of 0.75m.If the stopper completes 2.5 revolutions every second,determine the following; a)The centripetal acceleration. b)The centripetal force.​

Answers

Explanation:

a) ac = r(omega)^2

where omega = 2.5 rev/s

r = 0.75 m

We need to convert rev/s first into rad/s:

2.5 rev/s × (2pi rad/rev) = 5pi rad/s = omega

Therefore, the centripetal acceleration ac is

ac = (0.75 m)×(5pi rad/s)^2

= 185 m/s^2

b) Fc = mac

= (0.5 kg)(185 m/s^2)

= 92.5 N

Most stars are main-sequence stars, a group of stars for which size, mass, surface temperature, and radiated power are closely related. The sun, for instance, is a yellow main sequence star with a surface temperature of 5800 K. For a main-sequence star whose mass M is more than twice that of the sun, the total radiated power, relative to the sun, is approximately P/Psun = 1.5(M/Msun)3.5. The star Regulus A is a bluish main-sequence star with mass 3.8Msun and radius 3.1Rsun. What is the surface temperature of Regulus A?

Answers

Answer:

the surface temperature of Regulus A is 11724.13 K

Explanation:

Given the data in the question;

Sun's surface temperature T = 5800 K

total radiated power, relative to the sun is; P/P[tex]_{sun[/tex] = [tex]([/tex] 1.5(M/M[tex]_{sun[/tex] [tex])^{3.5[/tex]

The star Regulus A is a bluish main-sequence star with mass 3.8M[tex]_{sun[/tex]  and radius 3.1R[tex]_{sun[/tex] .

First, we determine the value power emitted by the sun or sun as follows;

P = eσAT⁴

where P is the power, e is surface emissivity, σ is Stefan Boltzmann, A is area and T is temperature.

so, lets assume emissivity of star and sun is same;

let p be power related to star and p[tex]_{sun[/tex] be power related sun.

Ratio of power radiated by star and power radiated by sun;

P/P[tex]_{sun[/tex] = eσAT⁴ / eσA[tex]_{sun[/tex]T[tex]_{sun[/tex]⁴

we know that AREA A = πR²

we input the formula for area

P/P[tex]_{sun[/tex] = eσ(πR²)T⁴ / eσ(π(R[tex]_{sun[/tex])²)T[tex]_{sun[/tex]⁴

such that we now have;

P/P[tex]_{sun[/tex] = R²T⁴ / R[tex]_{sun[/tex]²T[tex]_{sun[/tex]⁴

given that P/P[tex]_{sun[/tex] = [tex]([/tex] 1.5(M/M[tex]_{sun[/tex] [tex])^{3.5[/tex], we substitute

[tex]([/tex] 1.5(M/M[tex]_{sun[/tex] [tex])^{3.5[/tex] = R²T⁴ / R[tex]_{sun[/tex]²T[tex]_{sun[/tex]⁴

we find temperature of the star T

T = 5800  × [tex][[/tex] 1.5[tex]([/tex]M/M[tex]_{sun[/tex][tex])^{3.5[/tex] (R[tex]_{sun[/tex]²/R²)[tex]]^{1/4[/tex]

Given that;  mass M is 3.8M[tex]_{sun[/tex] and radius R is 3.1R[tex]_{sun[/tex] .

we substitute

T = 5800  × [tex][[/tex] 1.5[tex]([/tex]3.8M[tex]_{sun[/tex]/M[tex]_{sun[/tex][tex])^{3.5[/tex] (R[tex]_{sun[/tex]²/( 3.1R[tex]_{sun[/tex] )²)[tex]]^{1/4[/tex]

T = 5800  × [tex][[/tex] 1.5[tex]([/tex]3.8[tex])^{3.5[/tex] ( 1/( 3.1)²)[tex]]^{1/4[/tex]

T = 5800  × [tex][[/tex] 1.5( 106.9652 ) ( 1/(9.61) [tex]]^{1/4[/tex]

T = 5800  × [tex][[/tex] 16.69592 [tex]]^{1/4[/tex]

T = 5800  ×  2.02140152

T = 11724.13 K

Therefore, the surface temperature of Regulus A is 11724.13 K

a charge of 2 * 10^-9C is placed at the origin, and another charge of 4 * 10^-9C is placed at x = 1.5m. find the point between these charges where a charge of 3 * 10^-9C should be placed so that the net electric force on it is zero

Answers

Answer:

  x₁ = 0.62 m

Explanation:

In this exercise the force is electric, given by Coulomb's law

         F =[tex]k \frac{q_1q_2}{r^2}[/tex]

This force is a vector, since the three charges are in a line we can reduce the vector sum to a scalar sum.

For the sense of force let us use that charges of the same sign repel and charges of the opposite sign attract.

     ∑ F = F₁₂ - F₂₃

They ask us to find the point where the summaries of the force is zero.

      F₁₂ - F₂₃ = 0

      F₁₂ = F₂₃

let's fix a reference system located in the first charge (more to the left), the distance between the two charges is d = 1.5 m and x is the distance to the location of the second sphere

      k q₁q₂ / x² = k q₂q₃ / (d-x) ²

      q₁ (d-x) ² = q₃ x²

       

let's solve

       d² - 2 x d + x² = [tex]\frac{q_3}{q_1}[/tex]  x²

       x² (1 -  [tex]\frac{q_3}{q_1}[/tex]) - 2x d + d² = 0

we substitute the values

       x² (1- 4/2) - 2 1.5 x + 1.5² = 0

       x² (-1) - 3.0 x + 2.25 = 0

       

       x² + 3 x - 2.25 = 0

let's solve the quadratic equation

       x = [-3 ± [tex]\sqrt{ 3^2 + 4 \ 2.25}[/tex]] / 2

       x = [-3 ± 4.24] / 2

       x₁ = 0.62 m

       x₂ = 3.62 m

since it indicates that the charge q₂ e places between the spheres, the correct solution is

            x₁ = 0.62 m

A rock is thrown horizontally of a bridge at 8.00 m/s. It hits the water's surface below 3.4 seconds later. How high is the bridge? How far from the bridge does it hit the water?

I was provided with the equations:
x= v * t and y= (Vi * t)1/2 * a * t^2

Answers

Answer:

56.7m

Explanation:

horizontal velocity does not matter in this situation because it stays constant (no unbalanced force acting upon it)

a=9.81m/s^2              

t=3.4s

Vi=0m/s

d=Vi*t+1/2at^2

d=1/2at^2

d=1/2*9.81m/s^2*(3.4s)^2

d=4.905m/s^2*11.56s^2

d=56.7m

1. As air in Earth's atmosphere is heated by
the Sun, the warmer air expands and
becomes less dense than the cooler air. As a
result, the warmer air rises and the cooler
air sinks. What is the cycle of warm air
rising and cool air sinking called?

tidal current

surface current

deep current

convection current

Answers

The correct answer is D) convection current.

Convection currents are the cycle of warm air rising and cool air sinking that occurs in Earth's atmosphere (as well as in other fluids) as a result of temperature differences. When a fluid (such as air) is heated, it expands and becomes less dense, causing it to rise. When the warmer fluid cools, it contracts and becomes denser, causing it to sink. This cycle of rising and sinking creates a circular flow known as a convection current.

what are you made of

Answers

Answer:

Body stuff

Explanation:

:) <3

Body stuff... plz lol

Photo used to help with the question is below!! Please answer! Will mark BRAINLIEST!

⬇⬇⬇⬇⬇⬇⬇

-----------------------------------------------------------------------------------------------------------

Chose the correct answer:

-reflection

-refraction

-absorption

-none of the above

Answers

dookey dookey dookey dookey dookey

file:///C:/Users/Admin/Downloads/GUIA%20N.%206%20(2).pdf
porfa copia el link
y has el trabajo porfa te doy 23 puntos

Answers

1. Create a new question

2. Provide an image of your question/work so we don't have to download anything.

Batman shoots a grappling hook
34.6 m/s at an 80.2° angle. What is the magnitude only (no direction) of the velocity of the hook 1.09 s later?

(Unit = m/s)

Answers

Answer:

24.132 m/s

Explanation:

Note

[tex]U[/tex] = Initial Velocity

[tex]U_x[/tex] = Initial Horizontal Velocity

[tex]U_y[/tex] = Initial Vertical Velocity

[tex]V[/tex] = Final Velocity

[tex]V_x[/tex] = Final Horizontal Velocity

[tex]V_y[/tex] = Final Vertical Velocity

[tex]B[/tex] = launch angle

[tex]g[/tex] = gravity

[tex]t[/tex] = time

[tex]U_x=U*cos(B)[/tex]

[tex]U_y=U*sin(B)[/tex]

The horizontal component of the velocity is constant throughout the flight. So [tex]U_x=V_x[/tex] It can be defined as

[tex]V_x=U*cos(B)[/tex]

We can use the kinematics equation

[tex]V=U+at[/tex]

Gravity is acting downwards; gravity would be negative

[tex]V_y=U_y-gt[/tex]

The magnitude of the velocity can be defined as

[tex]V=\sqrt{V_x^2+V_y^2}[/tex]

Inserting some of the other equations gives us an equation at a given time (t).

[tex]V=\sqrt{(U*cos(B))^2+(U*sin(B)-gt)^2}[/tex]

[tex]V=\sqrt{(UcosB)^2+(UsinB)^2+(gt)^2-2gtU*sinB[/tex]

[tex]V=\sqrt{U^2+g^2*t^2-2*t*g*U*sinB}[/tex]

[tex]V(t)=\sqrt{U^2+g^2t^2-2tgUsinB}[/tex]

We are given

[tex]U=34.6[/tex]

[tex]B=80.2[/tex]

[tex]t=1.09[/tex]

[tex]g=9.81[/tex]

[tex]V(1.09)=\sqrt{34.6^2+9.81^2*1.09^2-2*1.09*9.81*34.6*sin80.2}[/tex]

[tex]V(1.09)=\sqrt{34.6^2+114.338-2*1.09*9.81*34.6*sin80.2}[/tex]

[tex]V(1.09)=\sqrt{34.6^2+114.338-739.94868*sin80.2}[/tex]

[tex]V(1.09)=\sqrt{34.6^2+114.338-729.151}[/tex]

[tex]V(1.09)=\sqrt{1197.16+114.338-729.151}[/tex]

[tex]V(1.09)=\sqrt{582.347}[/tex]

[tex]V(1.09)=24.132[/tex]

A rating/review would be much appreciated. Hope this helps!

Let me know if you have any questions about my work

The rectangular coordinates of a point are (5.00, y) and the polar coordinates of
this point are (r, 67.4°).
a. What is the value of the polar coordinate r in this case?
b. In what quadrant are both the sine and tangent negative?

Answers

Answer:

Explanation:

Polar coordinates formula

Summary. To convert from Polar Coordinates (r,θ) to Cartesian Coordinates (x,y) : x = r × cos( θ ) y = r × sin( θ )

The leaning tower of Pisa is about 56 meters tall. A ball released from the top takes 3.4 seconds to reach the ground. The final velocity of the ball before it hits the ground is 33 meters/second. Assuming that the ball experienced a constant acceleration throughout this descent, calculate the magnitude of the acceleration.

Answers

The magnitude of the constant acceleration of the ball is 9.7 m/s².

What is acceleration?

Acceleration is the rate of change of velocity.

To calculate the constant acceleration of the ball, we use the formula below.

Formula:

a = (v-u)/t................. Equation 1

From the question,

Given:

a = Acceleration,v = Final velocityu = Initial velocityt = Time

From the question,

Given:

v = 33 m/su = 0 m/s t = 3.4 seconds

Substitute these values into equation 1

a = (33-0)/3.4a = 9.7 m/s²

Hence, the acceleration is 9.7 m/s².

Learn more about acceleration here:  https://brainly.com/question/26216029

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