If a 62 kg panther sits in a tree 1.3 meters above the ground, how much gravitational potential energy does it have?

If the air resistance cause the panther to lose 200 J of energy as it falls, what is the KE of the panther just before it hits the ground?

What was the velocity of the panther just before it hit the ground?

Answers

Answer 1

Gravitational Potential Energy , Kinetic Energy and velocity of panther just before hitting the ground will be 790.41599 J , 990.41599 J , 5.6523m/s respectively.

Mass is 62 kg with a height 1.3m .

Gravitational Potential Energy means product of mass , gravitational acceleration and height . Mathematically, GPE= mgh

GPE= 62 x 1.3 x 9.8

= 790.41599 J

Energy lost = Final energy- GPE

Final Energy Δ =  790.41599 J + 200 J

=990.41599 J

To find Velocity of panther just before hitting the ground V:-

Δ⇒mV²/2

990.41599 J= 62 V²/2

V²= {990.41599 x 2}/62

= 31.9489029

V=5.6523m/s

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Related Questions

A car travels at a constant speed of 408 km/h, how long does it take to travel 80 m?​

Answers

We first apply the data to the problem.

Data:

[tex] \bold{\: \: \: \: \: \: \: \: \: \: \:V = 408km/h}[/tex]

[tex] \bold{ \: \: \: \: \: \: \: \: \: \: \: D = 80m} \\ \bold{T = ?}[/tex]

Now, we convert km/h to m/s.

Conversion:

[tex] \bold{408km/h * (1000m/1km) * (1h/3600s) }[/tex]

[tex] \bold{V = 113.3m/s}[/tex]

Then, we apply the formula that is.

Formula:

[tex] \bold{T = D/V}[/tex]

Finally we develop the problem.

Developing:

[tex] \bold{T = 80m / 113.3m/s}[/tex]

[tex] \bold{T = 0.706s}[/tex]

In a time of 0.706 seconds.

Question 3 (2 points)
Of the following exercise stimuli, ordered from the lightest to the heaviest, select the
intensity that has been shown to significantly improve affect for most people (albeit
perhaps for a relatively short period of time)
a 10-minute walk at a self-selected intensity
a 20-minute session of a cycle ergometer at 60% VO2max
a 30-minute treadmill run at 75% VO2max
a 60-minute session of aerobics at 75% VO2max
multiple 30 second intervals at 150% VO2max

Answers

A 20-minute cycle ergometer exercise at 60% VO2max has been demonstrated to dramatically improve impact for the majority of persons.

Intensity and an example are what?

the property of being strongly felt or having a powerful impact: The explosion was so loud that it could be heard from five kilometres distant. Measures of luminosity. [C or U] the strength of being something that may be measured, such as sunlight, sound, etc.

The measurement of intensity:

A measure that is produced from a randomised measure is known as an intensity measure in probability theory. Since the randomised measure of a set's expected value is the intensity measure, which is a non-random measure, it equates to the average volume that the random measure assigns.

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how can an object have the same speed and mass but change its momentum?​

Answers

Answer: It can change its direction

Explanation:

Momentum is a vector, which has magnitude and direction.  An object can have the same speed and mass (magnitude) but if it changes direction, it just changed its momentum

What is curved space?

Answers

Curved space often refers to a spatial geometry which is not "flat", where a flat space is described by Euclidean geometry. Curved spaces can generally be described by Riemannian geometry though some simple cases can be described in other ways.

A rocket blasts off from rest and attains a speed of 33.9 m/s in 12.7 s. An astronaut has a mass of 65.1 kg. What is the astronaut's
apparent weight during takeoff?

Answers

The apparent weight of the astronaut of mass 65.1 kg moving with a speed of 33.9 m/s in 2.7 s is 811.797 N.

What is weight?

Weight is the force with which a body is attracted toward the earth or a celestial body by gravitation and which is equal to the product of the mass and the local gravitational acceleration.

To calculate the astronaut's apparent weight, we use the formula below.

Formula:

W = m{[(v-u)/t]+g}............ Equation 1

Where:

W = The apparent weight of the astronautm = Mass of the astronautv = Final speedu = Initial speedt = Timeg = Acceleration due to gravity

From the question,

Given:

m = 65.1 kgv = 33.9 m/su = 0 m/s (from rest)t = 12.7 sg = 9.8 m/s²

Substitute these values into equation 1

W = 65.1{[(33.9-0)/12.7]+9.8]W = 65.1×12.47W = 811.797 N

Hence, the apparent weight of the astronaut is 811.797 N.

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A(n) _____ bond is one in which electrons are shared equally.

Answers

Answer:

Non-polar covalent bond

Explanation:

This type of covalent bond where electrons are shared equally between atoms is called a non-polar covalent bond.

URGENT!! ILL GIVE
BRAINLIEST!!!! AND 100 POINTS!!!!!

Answers

Answer: B

Explanation:

You will have air resistance as a type of friction, and since there isn't a normal force opposing it, gravity will bring the ball toward the ground.

b) Robert pulls at a steady speed for
2m/s how far will she travel in 10 minutes
10mins?
I need help someone pleaseeeee
Tysm

Answers

Robert pulls at a steady speed for 2m/s. She travelled 1200m in 10 minutes.

Distance : Path travelled by an object is the total distance.speed : distance travelled in an interval of time.Time : It is inversely proportional to the time.

Speed = 2m/s

time taken = 10 minutes

(1 minute = 60 seconds )

then,

10 mint = 10 x 60 = 600 seconds

To calculate how far she travel we use :

distance travelled = speed x time

distance = 2 x 600

distance = 1200m

Robert pulls at a steady speed for 2m/s. She travelled 1200m in 10 minutes.

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The half-life of chromium-51 is 28 days. If the sample contained 510 grams, how much chromium would remain after 56 days?

Answers

The quantity of 128 grams of chromium would be remaining.

The time required for half the reaction to change is called half life period.

This concept is useful for understanding carbon dating in Nuclear Physics. The term is used more to characterize any type of exponential decay. Nuclei that decay easily have shorter half life.

Given :- Half Life of Cr = 28 days

m= 510 grams

To find :- quantity of Cr after 56 days.

Formula : Nₙ= N₀/(2)ⁿ

Nₙ= 510/2²

Nₙ= 128g

Therefore 128 grams will be available after 56 days.

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a bolder of mass 45kg is pushed on a surface with a coefficient of kinetic friction of 0.85. what force has to be applied to produce an acceleration of 2 m/s^2

Answers

Answer: The answer is 386 N

Explanation:

The pressure acting on a diver in a take is 5 bar. The height of water column above him is nearly

A) 9 m B) 30 m C) 40 m D) 50 m E) 60 m

Answers

Answer:

Diver's dive depth 50 m

Explanation:

Given:

p = 5 bar = 5·10⁵ Pa  - Pressure

ρ = 1000 kg/m³ - Water density

g = 10 m/s²

______________

H - ?   Diver's dive depth

p = ρ·g·H

H = p/ ( ρ·g)

H = 5·10⁵ / (1000·10) = 50 m

2. What is the force on a 1 kg bal that is falling freely due to the pull of gravity?

Answers

Answer: 9.8 N

Explanation:

A 0.80-m-tall barrel is filled with water (with a weight density of 9800 N/m³). Find the water
pressure on the bottom of the barrel.
Express your answer to two significant figures and include the appropriate units.

Answers

Pressure on the bottom of the barrel = 1.78 * 10⁵Pₐ

a) The force applied perpendicular to the object's surface and dispersed over the area is referred to as the pressure.

Given,

height of barrel = 0.8mP

density = 9800N/m²

atmospheric pressure = Pₙ = 1.013 * 10⁵Pₐ

From the relation,

Pressure at the bottom

Pₓ = Pₙ + pgh

Pₓ = 1.013 * 10⁵ + 9800 * 9.8 * 0.8

Pₓ = 1.78 * 10⁵Pₙ

b)

Given,

mass = 1kg

area, a =1.2cm² =1.2 * 10⁻⁴ m²

Gravitational acceleration g = 9.8m/s²

We know that

Pressure on the finger,

Pf = force/area

Pf = (mass*g) / a

Pf = (1 * 9.8)/ (1.2 * 10⁻⁴)

Pf = 8.17 * 10⁴ N/m²

a) pressure on the bottom = 1.78 * 10⁵Pₐ

b) pressure on the finger = 8.17 * 10⁴ N/m²

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Find the quantity of heat required to melt completely 200g of lead initially at 27°C given that for lead melting point is 327°C. The specific heat capacity of lead is 0.14J/gK and specific latent heat of fusion for lead is 270J/g.

Answers

The qhantity of heat required to melt completely 200 g of lead is 62400 J.

What is heat?

Heat can be defined as a form of energy that brings about the sensation of warmth.

To find the quantity of heat required, we use the formula below.

Formula:

Q = cm(t₂-t₁)+Cm.......... Equation 1

Where:

Q = Quantity of heatc = Specific heat capacity of leadm = Mass of leadt₂ = Final temperaturet₁ = Initial temperatureC = Specific latent heat of fussion for lead

From the question,

Given:

c = 0.14 J/gKC = 270 J/gm = 200 gt₁ = 27 °Ct₂ = 327 °C

Substitute these values into equation 1

Q = 0.14×200×(327-27)+270×200Q = 8400+54000Q = 62400 J

Hence, the quantity of heat required is 62400 J.

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A thin 2.50 kg box rests on a 5.50 kg board that hangs over the end of a table, as shown in (Figure 1).

How far can the center of the box be from the end of the table before the board begins to tilt?

Answers

Based on the principle of moments. the distance from the end of the table should the box be placed before the board begins to tilt is 8.6 cm.

What distance from the end of the table should the box be placed before the board begins to tilt?

The distance from the end of the table should the box be placed before the board begins to tilt is determined from the principle of moments as follows:

sum of clockwise moments = sum of anticlockwise moments

The block is 30 cm on the table and 20 cm outside it.

The downward force acting on the left-hand side of the box = 3/5 x 5.50  = 3.3kg.

This force acts at the center of gravity, 15 cm or 0.15 m away.

Therefore, anticlockwise moments o the left side = 3.3 x 0.15 = 0.495 J

Also, the clockwise moment on the right side = Force * distance

Force = 2/5 x 5.5 = 2.2 N

Distance from the center of gravity = 10 cm or 0.10 m away.

the clockwise moment on the right side due to the board = 2.2 x 0.1 = 0.22.

The moment due to the box with a weight of 2.5 kg at a distance of x meters will be:

the total clockwise moment on the right side = 0.28 + 2.5 * x.

When the board is just about to tilt:

0.495 = 0.28 + 2.5 * x

2.5x = 0.495 - 0.28

x = 0.086 m or 8.6 cm

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Questions
1. a. Explain why you need to use half the time taken for an echo to be
detected when you are doing calculations of distance.
b. Sound travels at 330 m/s in air. It takes 1.5s to detect the echo
from a building. Calculate the distance to the building.
(Hint: Remember that the time is for the journey there and back).
2. Explain why dolphins can find food at greater distances than bats.
3. A woman is having an ultrasound scan. Ultrasound travels at
1500 m/s inside the body. The ultrasound transmitter sends out a
pulse of ultrasound waves, which are reflected from a tissue surface
7.5 cm away. Calculate the length of time before the echo is received
(Hint: Be careful with units!)

Answers

1.a. The reason of using half time is  an echo refers to the Reflection of sound after hitting the target just like a basket ball bouncing back as it hits its target i.e. earth.

So parameters like total energy , distance, amplitude of waves,.etc..  refers to the both action and reaction journey.

Like for example if a rubber ball travels 5m to hit the wall and returns back . so the net distance = incident distance + reflection distance⇒ 10m

But in Physics to compute the accurate result of incoming signals or energy perceived we just have to take either of the 2 journeys i.e. incidence or reflection. Second perception is the concept of taking average of multiple data. When there is 2 times i.e. one time to take incidence and other one is reflection , so in such cases average of the time for close vicinity of result.

1.b . The net distance propagated is 247.5 m.

Given - Sound travels 330m/s taking time 1.5s

To find the distance travelled -

distance travelled in 1 sec= 330 m

distance for 0.75 sec= 330 x 0.75 = 247.5m

Since 1.5 sec is the total time of the process of echo.

[tex]echo = incidence + reflection[/tex]

so taking time of one journey = 1.5 sec/ 2 = 0.75 sec

Hence the distance traversed is 247.5 m .

2. Dolphins uses echolocation as an ability or tool to perceive the incoming waves in atmosphere that includes sound waves.

They generate a special kind of sound that makes its target i.e. food to come to it.

Now talking about bats , although they also have similar ability of echolocation , but its mechanism to capture a food is time consuming comparatively and dolphins use higher frequencies compared to bats.

3.The length of time before the echo is received is 0.0001 sec.

Given - Speed = 1500m/s , distance =7.5cm ⇒0.075m

To find how long it will take before the receptance of incoming sound signal :-

[tex]2d=speed * time \\2d=vt[/tex]

[2 x 0.075]/ 1500= t

t= 0.0001 sec

so it takes 0.0001 sec to traverse before the echo is achieved.

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Units and magnitudes 1. Make the following transformations 500 cm to m:​

Answers

Answer:

5 m

500 cm to m is 5 meters

To convert one length to another, a conversion is applied, but first we will propose some data to serve us when applying the conversation.

Data:

100 cm is the same as a meter.& 1 meter is the same as 100 cm.

Now, we convert centimeters to meters.

Conversion:

500cm • (1m / 100cm)5m

Explanation:

They are 5 meters because multiplying 500 by 1 will give you the same result of 500, then we divide the 500 with the last amount that we have left, which is 100, thus giving us a result of 5 meters.

Calculate the work done by a 40 N force pushing a 0.5 kg toy 2.3 m.

Answers

The work done calculated is 92J.

What is work done and how to find it?

Work is considered completed when a force produces motion. There are numerous aspects that affect how well a force performs. One of the aspects is the object's displacement in the force's direction. The second component is force. Work is referred to as the result of an object's displacement plus a force applied in the force's direction. Work is equal to F×S, where F represents force and S represents displacement. When a body is moved by a distance while being subjected to a force, the work done can be calculated by:

W= F×S

Given that,

Force applied= 40N

Displacement= 2.3m

So, on putting the given values in above equation,

W= 40×2.3

W= 92 J

Therefore, the answer is 92 Joule.

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At which point does the bungee jumper’s kinetic energy begin to change into potential energy?

Answers

At the bottom of the “ride,” when the jumper momentarily stops

Four park rangers are chasing an elephant, and from the GPS collar, they get the following vector readings: 720.4 m, 32.0° east of north; 570.3 m, 36.0° south of west; 170.8 m straight south. Instead of following the path, they decide to calculate the vector sum, and go directly to the animal, what do they calculate?

Answers

where ever they hear heavy ahh footsteps

A
where light does not strike.
A) mirror
C) wave
is an area
B) transparent
D) shadow
q

Answers

Answer:

d

Explanation:

because when an object stand in front of light, light bends around the other and doesn't go through it



2. A runner whose initial speed is 8.06 m/s increases her speed to 8.61 m/s in order to win a race. If the
runner takes 5.00 seconds to complete this increase in speed, what is her acceleration?

Answers

Answer:

Explanation:

Given:

V₀ = 8.06 m/s

V = 8.61 m/s

t = 5.00

________

a - ?

Acceleration:

a = (V - V₀) / t

a = (8.61 - 8.06) / 5.00 ≈ 0,11 m/s²

Answer:0.11

Explanation:

a baby carriage is sitting at the top of the hill that is 21 m high. the carriage with the baby has a mass or 1.5kg. the carriage has ___ energy. calculate it.

Answers

Answer:
309 Joules

Steps:
We can use the gravitational potential energy formula.
GPE = mgh
We are given the mass, the height, and we know gravity. Let’s solve for GPE

1.5 * 9.81 * 21 = 309 Joules

A race car leaves the starting line and travels 36,000 m in the first 600 seconds of the race. They are then forced to take a pit stop and don't go anywhere for 250 seconds. After the pit stop, they finish the race, going 24,500 m in 350 seconds.

a. What is the car's average speed during the first part of the race? (before the pit stop)

b. What is the car's average speed during the pit stop?

c. What is the car's average speed after the pit stop?

d. What is the car's average speed for the whole trip?

Answers

Answer:s= A. 60 m/s

B. s=0 m/s

c. s=70 m/s

D. s=50.4 m/s

Explanation:

Sorry if its wrong

[100 points]a force of 330 newtons is required to lift a crate up a ramp a distance of 5 meters in 2 seconds. how much power is used?

Answers

Answer:

17,482.36 J

Explanation:

Ans. W = F •d cos  = 120 N • 165 m cos 28.0 = 17,482.36 J

Answer:

Explanation:

Given:

F = 330 N

h = 5 m

t = 2 s

_________

N - ?

N = A / t = F·h / t = 330·5 / 2 = 825 W

a train travels at 36 km/hr and accelerates uniformly to 108km in 10seconds. what is the acceleration?​

Answers

Answer:

Below

Explanation:

Acceleration   =   change in velocity  / change in time

(108 km/s  - 36 km/s)   / (10 s)   = 7.2 m/s^2

what is the answer? is it equilibrium or not?​

Answers

No, the system is not in equilibrium because the forces are not balanced which caused acceleration In system.

In mechanics, a force is any action that seeks to preserve, modify, or deform a body's motion. Isaac Newton's three principles of motion, which are outlined in his Principia Mathematica, are frequently used to illustrate the idea of force (1687).

Newton's first law states that unless a force is applied to a body, it will stay in either its resting or uniformly moving condition along a straight path. According to the second law, when an external force applies on a body, the body accelerates (changes velocity) in the force's direction.

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During a medieval siege of a castle, the attacking army uses a trebuchet to hurl heavy stones at the castle walls. If the trebuchet launches the stones with a velocity of +30.0m/s at an angle of 50.0°, how long does it take the stone to hit the ground? What is the maximum distance that the trebuchet can be from the castle wall to be in range? How high will the stones go? Show all your work.

Answers

(a) The time of motion of the stone is 4.69 seconds.

(b) The maximum distance the trebuchet can be from the castle wall to be in range is 90.44 m.

(c) The maximum height reached by the stone is 26.95 m.

What is the time of motion of the stone?

The time of motion of the stone is calculated as follows;

t = 2u sinθ / g

where;

u is the initial velocity of the stoneθ is the angle of projection of the stoneg is acceleration due to gravity

t = (2 x 30 x sin50) / (9.8)

t = 4.69 seconds

The maximum distance the trebuchet can be from the castle wall to be in range is calculated as;

R = u²sin(2θ) / g

R = (30² sin(2 x 50))/(9.8)

R = 90.44 m

The maximum height reached by the stone is calculated as follows;

H = u² sin²θ / 2g

H = (30²  x (sin 50)²) / (2 x 9.8)

H = 26.95 m

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A man walking through a field travelled 30 m south and then 20 m west.
Calculate the distance he travelled and his displacement

Answers

The net distance and displacement traversed is 50m and  36.0555 m respectively.

Given - A man has walked 30m and 20m south and west respectively.

To find distance and displacement .

principle- Distance means path covered and a vector quantity. Displacement means shortest distance covered and state function. It is also scalar quantity. Displacement can be zero but distance can't be zero.

Total distance traversed= 30 + 20 = 50m

Total displacement traversed (s) ⇒√(x₁² + x₂²)

⇒√(30² + 20²)    ⇒√(900 + 400)

=√1300

=36.0555 m

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A wooden block of mass m is placed on a horizontal aluminum ramp. The angle of
the ramp is slowly increased to 10.8° above the horizontal, at which point the
mass slides down the ramp with constant speed.
Determine the coefficient of kinetic friction, %, between the wooden block and
the aluminum ramp.
Write your answer to 2 significant figures.

Answers

MAIN ANSWER- [tex]5/\sqrt{21} sec[/tex]

supporting answer-When a body is moving, it experiences resistance because it interacts with its surroundings. This resistance is a friction force. Friction opposes relative motion between systems in contact but also allows us to move, a concept that becomes obvious if you try to walk on ice. Friction is a common yet complex force, and its behavior still not completely understood. Still, it is possible to understand the circumstances in which it behaves.

body of the answer- a block of mass m with 10kgs is sliding on a inclined plane with frictionless surface

final answer-5/\sqrt{21} sec

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