An object is dropped from a​ tower, 576 ft above the ground. The​ object's height above ground t seconds after the fall is ​s(t)=576−16t2. Determine the velocity and acceleration of the object the moment it reaches the ground.

Answers

Answer 1

Answer:

Explanation:

We are looking to solve for final velocity in this problem. We know the initial velocity is 0 since someone was holding the object before it was dropped, the displacement is -576 feet since the object had to fall that far below the point from which it was dropped (which is why it is negative), and the acceleration AT ANY POINT DURING THE FALL IS -32 ft/s/s. The equation we use to find final velocity is

v² = v₀² + 2aΔx and filling in:

v² = 0² + 2(-32)(-576) so

[tex]v=\sqrt{2(-32)(-576)}[/tex] so

v = 192 ft/s


Related Questions


A car starts from rest. If its acceleration is 1.5 m/s2 in 1.5 seconds, then calculate the
distance travelled by it.

Answers

Answer:

1.6875 m

Explanation:

Here, we are given that,

Initial velocity (u) = 0 (as it starts from rest)Acceleration (a) = 1.5 m/s²Time taken (t) = 1.5 s

Then,

Distance (s) = ?

By using the second equation of motion,

s = ut + ½at²

s is distanceu is initial velocityt is timea is acceleration

⇒ s = 0(1.5) + ½ × 1.5 × (1.5)²

⇒ s = ½ × 1.5 × 2.25

⇒ s = ½ × 1.5 × 2.25

⇒ s = ½ × 3.375

s = 1.6875 m

Distance travelled by it is 1.6875 m.

The weather map below shows air pressure reading given in millibars. Points A,B,C and D are locations of earths surface at which location is surface wind speed the lowest

Answers

Answer:

c

Explanation:

Two circular, concentric and coplanar turns of
radii R1 = 30 cm and R2 = 20 cm are traversed by currents
i1 = 5 A and i2 = 2 A, indicated in the figure. being the constant of
magnetic permeability in vacuum μ0 = 4^.10-7 Tm/A,
characterize the magnetic induction vector originating at center O.

Answers

Answer:

(check the pic)

Explanation:

hope it helps

True or False. A person who is nearsighted cannot see objects that are close to them clearly.

Answers

false, farsightedness is when you cant see close

Answer:

false

Explanation:

hope it works

Calculate the rms speed of helium atoms near the surface of the Sun at a temperature of about 5100 K. Express your answer to two significant figures and include the appropriate units.

Answers

Answer:

[tex]V_{rms}=5.6*10^3m/s[/tex]

Explanation:

From the question we are told that:

Temperature [tex]T=5100K[/tex]

Generally the equation for RMS Speed is mathematically given by

 [tex]V_{rms}=\sqrt{\frac{3kT}{m}}[/tex]

Where

 [tex]K=Boltzman's constant[/tex]

 [tex]K=1.38*10^{-23}[/tex]

And

 [tex]M=molecular mass[/tex]

 [tex]M=4*1.67*10^{-27}[/tex]

 [tex]V_{rms}=\sqrt{\frac{3(1.38*10^{-23})5100}{4*1.67*10^{-27}}}[/tex]

 [tex]V_{rms}=5.6*10^3m/s[/tex]

Write down the chemical formula and the ratio in which the four elements combine with each other in the compound Sodium- bicarbonate

Answers

Answer: The chemical formula is [tex]NaHCO_{3}[/tex] and the ratio in which the four elements combine with each other in the compound Sodium- bicarbonate is 1 : 1 : 1 : 3.

Explanation:

The chemical formula of sodium bicarbonate is [tex]NaHCO_{3}[/tex].

In this compound, there are 1 sodium atom, 1 hydrogen atom, 1 carbon atom and three oxygen atoms present.

Therefore, the ratio of atoms is 1 : 1 : 1 : 3

Thus, we can conclude that the chemical formula is [tex]NaHCO_{3}[/tex] and the ratio in which the four elements combine with each other in the compound Sodium- bicarbonate is 1 : 1 : 1 : 3.

Hai điện tích điểm Q1 = 8 C, Q2 = –6
C đặt tại hai điểm A, B cách nhau 0,1
m trong không khí. Tính cường độ điện
trường do hai điện tích này gây ra tại
điểm M, biết MA = 0,2 m

Answers

Answer:

English please

Explanation:

I don't understand the question

An object travels 4 km from A to B towards north and then 3 km from B to C towards west. Find its displacement.​

Answers

QUËSTIONS:-

An object travels 4 km from A to B towards north and then 3 km from B to C towards west. Find its displacement.

ANSWER:- displacement

[tex] {x}^{2} = {3 }^{2} + {4}^{2} \\ {x}^{2} = 9 + 16 \\ {x}^{2} = 25 \\ x = \sqrt{25} \\ x = 5[/tex]

displacement = 5 km

Explanation:

We nee to use triangle law

As displacement is the shortest path travelled, we take A-C

so, D=√4^2+3^2

D=5km

A cement block accidentally falls from rest from the ledge of a 81.5-m-high building. When the block is 14.0 m above the ground, a man, 2.00 m tall, looks up and notices that the block is directly above him. How much time, at most, does the man have to get out of the way

Answers

Answer:

0.41s

Explanation:

solve for t and y

,........

A lens with a focal length of 15 cm is placed 45 cm in front of a lens with a focal length of 5.0 cm .

Required:
How far from the second lens is the final image of an object infinitely far from the first lens?

Answers

Answer:

the required distance is 6 cm

Explanation:

Given the data in the question;

f₁ = 15 cm

f₂ = 5.0 cm

d = 45 cm

Now, for first lens object distance s = ∝

1/f = 1/s + 1/s' ⇒ 1/5 = 1/∝ + 1/s'

Now, image distance of first lens s' = 15cm  

object distance of second lens s₂ will be;

s₂ = 45 - 15 = 30 cm

so

1/f₂ = 1/s₂ + 1/s'₂

1/5 = 1/30 + 1/s'₂

1/s'₂ = 1/5 - 1/30  

1/s'₂ = 1 / 6

s'₂ = 6 cm

Hence, the required distance is 6 cm

 

The distance of the final image from the first lens will be is 6 cm.

What is mirror equation?

The mirror equation expresses the quantitative connection between object distance (do), image distance (di), and focal length (fl).

The given data in the problem is;

f₁ is the focal length of lens 1= 15 cm

f₂ s the focal length of lens 2= 5.0 cm

d is the distance between the lenses = 45 cm

From the mirror equation;

[tex]\frac{1}{f} = \frac{1}{s} +\frac{1}{s'} \\\\ \frac{1}{5} = \frac{1}{\alpha} +\frac{1}{s'} \\\\[/tex]

If f₁ is the focal length of lens 1 is 15 cm then;

[tex]s'=15 cm[/tex]

f₂ s the focal length of lens 2= 5.0 cm

s₂ = 45 - 15 = 30 cm

From the mirror equation;

[tex]\frac{1}{f_2} = \frac{1}{s_1} +\frac{1}{s_2'} \\\\ \frac{1}{5} = \frac{1}{30} +\frac{1}{s_2'} \\\\ \frac{1}{s_2'}= \frac{1}{5} -\frac{1}{30} \\\\ \frac{1}{s_2'}= \frac{1}{6} \\\\ \rm s_2'= 6 cm[/tex]

Hence the distance of the final image from the first lens will be is 6 cm.

To learn more about the mirror equation refer to the link;

https://brainly.com/question/3229491

A jogger moves from a position x =
4.0 m to a position of x = 16.0 m in
4.0 s. What was her average velocity?
(Unit = m/s)
Don't forget: velocities and displacements to
the right are +, to the left are -,

Please help me!

Answers

Answer:

3 m/s

Explanation:

We'll begin by calculating the change in displacement of the jogger. This can be obtained as follow:

Initial displacement (d₁) = 4 m

Final displacement (d₂) = 16 m

Change in displacement (Δd) =?

Δd = d₂ – d₁

Δd = 16 – 4

Δd = 12 m

Finally, we shall determine the determine the average velocity. This can be obtained as follow:

Change in displacement (Δd) = 12 m

Time (t) = 4 s

Velocity (v) =?

v = Δd / t

v = 12 / 4

v = 3 m/s

Thus, the average velocity of the jogger is 3 m/s

which team won the champions league in 2020 2021​

Answers

Answer:

Chelsea F.C

Explanation:

Chelsea F.C

Soccer

Suppose you walk 13.0 m straight west and then 25.0 m straight south. How far are you from your starting point (in m)

Answers

Answer:

28.2 m

Explanation:

Applying,

Pythagoras theorem,

a² = b²+c²............... Equation 1

Where a = The distance from my starting point to my current point, b = distance walked due west, c = distance walked due south

From the question,

Given: b = 13 m, c = 25 m

Substitute these values into equation 1

a² = 13²+25²

a² = 169+625

a² = 794

a = √794

a = 28.18 m

a ≈ 28.2 m

The speed of a 2.0-kg object changes from 30 m/s to 40 m/s during a 5.0-second time interval.
During this same time interval, the velocity of the object changes its direction by 90°. What is the
magnitude of the average total force acting on the object during this time interval?
a. 30 N
b. 20 N
c. 15 N
d. 40 N
e. 10 N
Which is the correct answer?

Answers

Answer:

F = 2 * 30 / 5 = 12 N to stop forward motion

F = 2 * 40 / 5 = 16 N to accelerate to 90 degrees

(12^2 + 16^2)^1.2 = 20 N   average force applied

The magnitude of the average total force acting on the object during this time interval is 20 N.

The given parameters:

Mass of the object, m = 2.0 kgInitial velocity, u = 30 m/sFinal velocity, v = 40 m/sTime of motion, t = 5.0 s

The magnitude of the average total force acting on the object during this time interval is calculated as follows;

[tex]F = \frac{mv }{t} \\\\F_1 = \frac{2(40)}{5} \\\\F_1 = 16\ N\\\\F_2= \frac{2(30)}{5} \\\\F_2 = 12 \ N\\\\F = \sqrt{F_1^2 + F_2^2} \\\\F = \sqrt{16^2 + 12^2} \\\\F = 20 \ N[/tex]

Learn more about resultant force here:  https://brainly.com/question/25239010

7. An electric train moving at 20km/hrs
. Accelerates to a speed of 30km/hrs. in
20 sec, find the distance travelled in meters during the period of
acceleration​

Answers

Answer

NB:

- speed, U is measure in m/s

- acceleration, a is measured in m/s²

-time t in seconds , s

Therefore conversation must be made

Speed U = 20km/hrs

=20km÷1hr

But 20km= 20×1000=20000m

1hr= 1×60min×60sec=3600s

U=20000÷3600=5.56m/s

a=30km/hrs

=30km÷1hr

But 30km=30×1000=30000

1hr=3600s

a=30000÷3600=8.33m/s²

From the equation of motion

S=Ut + ½ at².

Where s= distance

S = 5.56m/s × 20s + ½(8.33m/s²)(20s)²

S = 1777.3m

A girl and her bicycle have a total mass of 40.0 kg. At the top of the hill her speed is 5.0 m/s, and her speed doubles as she rides down the hill. The hill is 10.0 m high and 100 m long. How much kinetic energy and potential energy is lost to friction

Answers

Answer:

The kinetic energy and potential energy lost to friction is 2,420 J.

Explanation:

Given;

total mass, m = 40 kg

initial velocity of the girl, Vi = 5 m/s

hight of the hill, h = 10 m

length of the hill, L = 100 m

initial kinetic energy of the girl at the top hill:

[tex]K.E_{i} = \frac{1}{2} mv_i^2 = \frac{1}{2} \times 40 \times (5)^2\\\\K.E_{i} = 500 \ J[/tex]

initial potential energy of the girl at the top hill:

[tex]P.E_{i} = mgh_i = 40 \times 9.8 \times 10\\\\P.E_{i}= 3920 \ J[/tex]

Total energy at the top of the hill:

E = 500 J + 3920 J

E = 4,420 J

At the bottom of the hill:

final velocity = double of the initial velocity = 2 x 5 m/s = 10 m/s

hight of the hill = 0

final kinetic energy of the girl at the bottom of the hill:

[tex]K.E_{f} = \frac{1}{2} mv_f^2 \\\\K.E_f = \frac{1}{2} \times 40 \times (10)^2 = 200 0 \ J[/tex]

final potential energy of the girl at the bottom of the hill:

[tex]P.E_f = mgh_f = 40 \times 9.8 \times 0 = 0[/tex]

Based on the principle of conservation of energy;  

the sum of the energy at the top hill = sum of the energy at the bottom hill

The energy at the bottom hill is less due to energy lost to friction.

[tex]E_{friction} \ + E_{bottom}= E_{top}\\\\E_{friction} = E_{top} - E_{bottom}\\\\E_{friction} = 4,420 \ J - 2,000 \ J\\\\E_{friction} = 2,420 \ J[/tex]

Therefore, the kinetic energy and potential energy lost to friction is 2,420 J.

7. A car is travelling along a road at 30 ms when a pedestrian steps into the road 55 m ahead. The
driver of the car applies the brakes after a reaction time of 0.5 s and the car slows down at a rate of
10 ms. What happens?

Answers

Answer: Car collide with man

Explanation:

Given

Speed of car is [tex]u=30\ m/s[/tex]

Distance of the man from the car is [tex]s=55\ m[/tex]

Reaction time [tex]t_r=0.5\ s[/tex]

Rate of deceleration [tex]a_d=-10\ m/s^2[/tex]

Distance traveled in the reaction time [tex]d_o=30\times 0.5=15\ m[/tex]

Net effective distance to cover [tex]d=55-15=40\ m[/tex]

Distance required to stop the car

[tex]\Rightarrow v^2-30^2=2(-10)(s)\\\Rightarrow 0-900=-20s\\\Rightarrow s=45\ m[/tex]

Require distance is more than that of net effective distance. Hence, car collides with the man.

How far did you travel in 10 hours if you drove at a constant speed of 5km/hr? *

Answers

Answer:

50km

Explanation:

So we know we drove 5km a hour.

We drove like this for 10 hours.

To find the total km, we must multiply the time driven by the speed:

time*speed=distance

Plug in our values:

10*5km=50km

So your answer is 50km.

Hope this helps!

The answer is 50km/hr

Two identical loudspeakers 2.0 m apart are emitting sound waves into a room where the speed of sound is 340 m/sec. John is standing 5.0m in front of one of the speakers, perpendicular to the line joining the speakers, and hears a maximum in the intensity of the sound. What is the lowest possible frequency of sound for which this is possible?

Answers

Answer: The lowest possible frequency of sound for which this is possible is 212.5 Hz.

Explanation:

It is known that formula for path difference is as follows.

[tex]\Delta L = (n + \frac{1}{2}) \times \frac{\lambda}{2}[/tex]    ... (1)

where, n = 0, 1, 2, and so on

As John is standing perpendicular to the line joining the speakers. So, the value of [tex]L_{1}[/tex] is calculated as follows.

[tex]L_{1} = \sqrt{(2)^{2} + (5)^{2}}\\= 5.4 m[/tex]

Hence, path difference is as follows.

[tex]\Delta L = (5.4 - 5) m = 0.4 m[/tex]

For lowest frequency, the value of n = 0.

[tex]\Delta L = (0 + \frac{1}{2}) \times \frac{\lambda}{2} = \frac{\lambda}{4}[/tex]

[tex]\lambda = 4 \Delta L[/tex]

where,

[tex]\lambda[/tex] = wavelength

The relation between wavelength, speed and frequency is as follows.

[tex]\lambda = \frac{\nu}{f}\\4 \Delta L = \frac{\nu}{f}\\[/tex]

where,

[tex]\nu[/tex] = speed

f = frequency

Substitute the values into above formula as follows.

[tex]f = \frac{\nu}{4 \Delta L}\\f = \frac{340}{4 \times 0.4 m}\\= 212.5 Hz[/tex]

Thus, we can conclude that the lowest possible frequency of sound for which this is possible is 212.5 Hz.

Kaseem is taking his bicycle for a ride. His bicycle is a system, and its main purpose is to provide transportation. What is the main input into this system? What is the desired output of this system?

Answers

His bicycle that operates on a system for transportation

Which of the following would produce a star with the longest lifespan

1) a star with no mass
2)a star with low mass
3) a star with medium mass 4) a star with high mass 5)these options would not all be the same

Answers

Answer:

1

Explanation:

because stars with the lowest Mass live longer because they burn out slower than a star with high mass

I need help with physics question.

Answers

(D)

Explanation:

Assuming that the charge q is moving perpendicular to the magnetic field B, the magnitude of the force experienced by the charge is

F = qvB = (2.9×10^-17 C)(4.0×10^5 m/s)(1.7T)

= 2.0×10^-11 N

1 airplane
travel due north at 300 km while another airplane travels Due South and 300 km are there speed the same or their velocities the same ​

Answers

Answer:

Explanation:

Speed is scalar and velocity is vector. Vector values imply direction as well as magnitude. Therefore, speed and velocity are not the same. The speeds of these 2 planes are the same at 300km/hr, but the velocity of the plane traveling north is +300km/hr while the velocity of the plane traveling south is -300km/hr if we define north as positive and south as negative.

E=kq/r^2 chứng minh điện thế V=kq/r từ mối liên hệ giữa điện trường E và điện thế V

Answers

Answer:

hindi ko maintindihan teh

What is the reason why many ships and plane ones traveled in this boundary got disappeared and no debris are found in the bermuda triangle?

Answers

Karl Kruszelnicki has insisted the reason why so many ships and planes vanish without trace in the area between Bermuda, Florida, Puerto Rico is nothing to do with aliens or fire-crystals from the lost city of Atlantis.

Instead, the Australian scientist ‘revealed’, the high number of disappearances is explained by nothing more supernatural than plain old human error plus bad weather and the fact that lots of planes and ships enter that area of the Atlantic Ocean in the first place.

A rocket at fired straight up from rest with a net upward acceleration of 20 m/s2 starting from the ground. After 4.0 s, the thrusters fail and the rocket continues to coast upward with insignificant air resistance. (a) What is the maximum height reached by the rocket

Answers

Answer:

The maximum height reached by the rocket is 486.53 m

Explanation:

Given;

initial velocity of the rocket, u = 0

acceleration of the rocket, a= 20 m/s²

duration of the rocket first motion, t = 4 s

The distance traveled by the rocket before its thrust failed

h₁ = ut + ¹/₂at²

h₁ = 0 + ¹/₂ x 20 x 4²

h₁ = 160 m

The second distance moved by the rocket is calculated as follows;

The velocity of the rocket before its thrust failed;

v = u + at

v = 0 +  20 x 4

v = 80 m/s

This becomes the initial velocity for the second stage

At maximum height, the final velocity = 0

[tex]v_f^0 = v_i^2 - 2gh_2\\\\0 = (80)^2 - (2 \times 9.8)h_2\\\\0 = 6400 - 19.6h_2\\\\19.6h_2 = 6400\\\\h_2 = \frac{6400}{19.6} \\\\h_2 = 326.53 \ m[/tex]

The maximum height reached by the rocket = h₁ + h₂

                                                                          = 160 + 326.53

                                                                          = 486.53 m

Which series represents the flow of thermal energy in a heat engine?
Fuel → air inside piston chamber → air outside piston chamber
Fuel → air inside piston chamber → piston movement
Fuel → piston movement
Fuel → air outside piston chamber

Answers

It’s fuel> air inside piston chamber piston movement

Answer:

B - Fuel → air inside piston chamber → piston movement

Explanation:

A 2000 kg car experiences a braking force of 10000N and skids to a 14 m stop. What was the speed of the car just before the brakes.

Answers

Answer:

V = 11.83 m/s

Explanation:

Given the following data;

Mass = 2000 kg

Force = 10000N

Distance = 14 m

To find the final velocity of the car;

First of all, we would determine the acceleration of the car;

Acceleration = force/mass

Acceleration = 10000/2000

Acceleration = 5 m/s²

Next, we would use the third equation of motion to find the final velocity;

[tex] V^{2} = U^{2} + 2aS [/tex]

Where;

V represents the final velocity measured in meter per seconds.

U represents the initial velocity measured in meter per seconds.

a represents acceleration measured in meters per seconds square.

S represents the displacement measured in meters.

Substituting into the equation, we have;

V² = 0² + 2*5*14

V² = 0 + 140

V = √140

V = 11.83 m/s

1 point
Q.29. A stone has a weight of 5.7 N.
The gravitational field strength g is 10
N/kg.What is the mass of the stone?
O A 0.57 kg​

Answers

Answer:

weight/mass = gravitational field strength

Given :

Weight of stone = 5.7 N

Gravitational field strength (g) = 10 N/kg

Taking Mass of stone x

=> 5.7/x = 10

x = 10 * 5.7

x = 57 kg

Therefore mass of stone is 57 kg

Are you aware of human rights violation happening in the community and explain​

Answers

Answer:

Individuals who commit serious violations of international human rights or humanitarian law, including crimes against humanity and war crimes, may be prosecuted by their own country or by other countries exercising what is known as “universal jurisdiction.”

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