Suppose a 118kg watermelon is held 5m above the ground before sliding along a frictionless ramp to the ground. How high above the ground is the watermelon at the moment it's kinetic energy is 4,610J

Suppose A 118kg Watermelon Is Held 5m Above The Ground Before Sliding Along A Frictionless Ramp To The

Answers

Answer 1

This question involves the concepts of kinetic energy, potential energy, and the law of conservation of energy.

The watermelon is "1.02 m" above the ground.

The total energy of the watermelon can be found by its potential energy at the highest point:

Total Energy = mgh

where,

m = mass = 118 kgg = 9.81 m/s²h = height = 5 m

Therefore,

Total Energy = (118 kg)(9.81 m/s²)(5 m)

Total Energy = 5787.9 J

LAW OF CONSERVATION OF ENERGY

Now, according to the law of conservation of energy, at the given point:

Total Energy = Kinetic Energy + Potential Energy

5787.9 J = 4610 J + mgh'

[tex]h'=\frac{5787.9\ J-4610\ J}{(118\ kg)(9.81\ m/s^2)}[/tex]

h' = 1.02 m

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Related Questions

4. When Tina went for a walk, she covered the first 300 m in 100 s and the next 200 m in 150
s. What was Tina's average speed?
A 1 m/s
B 2 m/s
C 3 m/s
D 5 m/s

Answers

2m/s

300m+200m=500m

100s+150s=250s

500m/250s= 2m/s

The average speed of tina will be 2m/s.

We have Tina who went for a walk.

We have to determine her average speed.

What is Average Speed ?

Average speed is the rate of change of displacement with respect to time.

v(avg) = Δx/Δt

According to the question -

Total Distance covered = d = 500 m

Total time taken = t = 250 s

Average speed = 500/250 = 2 m/s

Hence, the average speed of tina will be 2m/s.

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A weight lifter lifts a 1470N barbell above his head from the floor the a height of 2m. he holds the barbell there for 5 sec. how much work does he do?

Answers

Answer:

Workdone = 2940 Nm

Explanation:

Given the following data;

Force = 1470 N

Distance = 2m

Time = 5 seconds

To find the work done;

Mathematically, workdone is given by the formula;

Workdone = force * distance

Substituting into the formula, we have;

Workdone = 1470 * 2

Workdone = 2940 Nm

When an object is fully converted into energy the amount of energy liberated is
3.6×1017 J, what is the mass of the substance?

Answers

Answer:

Mass, m = 4 kg

Explanation:

Given the following data;

Energy = 3.6 * 10^17 Joules

We know that the speed of light is equal to 3 * 10⁸ m/s.

To find the mass of the substance;

The theory of special relativity by Albert Einstein gave birth to one of the most famous equation in science.

The equation illustrates, energy equals mass multiplied by the square of the speed of light.

Mathematically, the theory of special relativity is given by the formula;

[tex] E = mc^{2} [/tex]

Where;

E is the energy possessed by a substance.m is the mass.c is the speed of light.

Substituting into the formula, we have;

[tex] 3.6 * 10^{17} = m * 300000000^{2} [/tex]

[tex] 3.6 * 10^{17} = m * 9*10^{16} [/tex]

[tex] m = \frac {3.6 * 10^{17}}{9*10^{16}} [/tex]

Mass, m = 4 kg

Two creatures sit on a horizontal frictional rotating platform. The platform rotates at a constant speed. The creatures do not slip off as it rotates.


ASSUME:

Red has a mass of 5 kg

Red is 1.5 m from the center

Red has a speed of 9 m/s

Blue has a mass of 25 kg

Blue has a speed of 1.8 m/s

The force of friction on Red is EQUAL to the force of friction on Blue





DETERMINE:

How far from the center is Blue

Answers

Answer:

M v^2 / R = centripetal force

For Red: M v^2 / R = 5 * 9^2 / 1.5 = 270

For Blue M v^2 / R = 270 = 25 * 1.8^2 / Rb

So Rb = 25 * 1.8^2 / 270 = .3 m

An inventor claims to have developed a power cycle capable of delivering a net work output of 400 kJ for an energy input by heat transfer of 1200 kJ. The system undergoing the cycle receives heat from a source of 550K and rejects heat to a sink of 350K. Determine if this is a valid claim.

Answers

Answer: Valid claim

Explanation:

Given

Work output is [tex]W=400\ kJ[/tex]

Work input is [tex]Q=1200\ kJ[/tex]

The temperature of the source [tex]T_h=550\ K[/tex]

The temperature of the sink is [tex]T_l=350\ K[/tex]

Efficiency is given by

[tex]\eta =\dfrac{\text{W}}{\text{Q}}\times 100[/tex]

Insert the values

[tex]\Rightarrow \eta=\dfrac{400}{1200}\times 100\\\\\Rightarrow \eta=\dfrac{100}{3}\\\\\Rightarrow \eta=33.3\%[/tex]

For ideal cycle it is

[tex]\Rightarrow \eta=\dfrac{T_h-T_l}{T_l}\times 100\\\\\Rightarrow \eta=\dfrac{550-350}{550}\times 100\\\\\Rightarrow \eta=36.36\%[/tex]

The efficiency of the cycle is less than the ideal situation, therefore, it is valid claim.

The volume of a piece of rock was 18.0 cm^3. The student measured the mass of the piece of rock as 48.6 g. Calculate the density of the rock in g/cm^3.

Answers

Answer:

Density of rock piece = 2.7 g/cm³

Explanation:

Given:

Volume of rock piece = 18 cm³

Mass of rock piece = 48.6 gram

Find:

Density of rock piece

Computation:

We know relation between density, mass and volume,

Density = Mass / Volume

Density of rock piece = Mass of rock / Volume of rock

Density of rock piece = 48.6 / 18

Density of rock piece = 2.7 g/cm³

A box proceeds along the x-axis and the figure below shows a record of its velocity as a function of time. Every grid line along the vertical axis corresponds to 2.00 m/s and each gridline along the horizontal axis corresponds to 0.500 s. (Enter your answer in m/s^2. Indicate the direction with signs of your answers. Note that t=0 at the intersection of the axes.) a) Determine the average acceleration of the box in the time interval t=0 to t=2.50 s. b) Determine the average acceleration of the box in the time interval t=2.50 s to t=7.50 s. c) Determine the average acceleration of the box in the time interval t=0 to t=10.0 s

Answers

Answer:

The answer is "[tex]\bold{0, 12.8 \ \frac{m}{s^2}, and \ 6.4 \ \frac{m}{s^2}}[/tex]"

Explanation:

For point a:

The average time interval acceleration for the box [tex]t = 0\ to\ t = 250 \ s:[/tex]

[tex]\to \alpha_{0 \ to \ 2.50}=\frac{v_f - v_i}{t_f-t_i}=\frac{0-0}{2.50-0}=0[/tex]

For point b:

The average time interval acceleration for the box

[tex]t = 250 \ s\ to\ t = 750 \ s :[/tex]  

[tex]\to \alpha_{2.50 \ to \ 7.50}=\frac{v_f - v_i}{t_f-t_i}=\frac{32-(-32)}{7.50-2.50}=12.8 \ \frac{m}{s^2}[/tex]

For point c:

The average time interval acceleration for the box [tex]t = 0\ to \ t = 10 \ s :[/tex]  

[tex]\to \alpha_{2.50 \ to \ 7.50}=\frac{v_f - v_i}{t_f-t_i}=\frac{32-(-32)}{10-0}=6.4 \ \frac{m}{s^2}[/tex]

Match these items.
1. effect observed when light from an object in cool air passes through warm air
2. line perpendicular to a surface
3. splitting of light into its component colors angle of incidence
4. alignment of light into a single vibrational direction
5. larger angle as light passes from air to water
6. cause of the sky's color
7. bouncing of light rays reflection
8. bending of light between media
9. ratio of speeds of light
10. inability of light to escape a low-velocity medium due to a large angle of approach refraction
a. scattering
b.dispersion
c.angle of incidence
d.index of refraction
e.mirage
f.reflection
g.normal
h.total inertial reflection
i.polarization
j.refraction​

Answers

Answer:

1. effect observed when light from an object in cool air passes through warm air

→ h. total internal reflection

2. line perpendicular to a surface

→ g. normal.

3. splitting of light into its component colors angle of incidence

→ b. dispersion.

4. alignment of light into a single vibrational direction

→ i. polarization

5. larger angle as light passes from air to water

→ c. angle of incidence

6. cause of the sky's color

→ f. reflection

7. bouncing of light rays reflection

→ f. reflection

8. bending of light between media

→ j. refraction.

9. ratio of speeds of light

→ d. index of refraction.

10. inability of light to escape a low-velocity medium due to a large angle of approach refraction

→ scattering.

A polar bear walks 2.5 meters to the right and then back to the left to exactly where he started. His displacement would be
?
A.
5 meters
B.
2.5 meters
C.
O meters
D.
No way to determine with given information

Answers

C. 0 metres
The displacement is 2.5 m -2.5 m which is zero

A bat emits a 40 kHz chirp to locate flying insects. If the speed of sound is 340 m/s and a bat hears the echo from the moth after 0.6 seconds, then how far away is the moth?

Answers

102 m

Explanation:

The time 0.6 sec is the time it took for the sound to travel from the bat to the moth and back. So it took 0.3 sec for the sound to reach the moth. From the definition of speed, the distance of the moth d to the bat is given by

v = d/t ---> d = vt = (340 m/s)(0.3 sec) = 102 m

What is the centripetal force for a roller coaster if the mass is 10 kg and the normal force is 25 N?

Answers

Answer:

Fc = 123 Newton

Explanation:

Net force can be defined as the vector sum of all the forces acting on a body or an object i.e the sum of all forces acting simultaneously on a body or an object.

Mathematically, net force is given by the formula;

[tex] Fnet = Fapp + Fg[/tex]

Where;

Fnet is the net force.

Fapp is the applied force.

Fg is the force due to gravitation.

Given the following data;

Normal force = 25N

Mass = 10kg

To find the centripetal force;

From the net force, we have the following formula;

Fc = N + mg

Where;

Fc is the centripetal force.

N is the normal force.

mg is the the weight of the object.

Substituting into the formula, we have;

Fc = 25 + 10(9.8)

Fc = 25 + 98

Fc = 123 Newton

When monochromatic light of an unknown wavelength falls on a sample of aluminum, a minimum potential of 2.27 V is required to stop all of the ejected photoelectrons. (The work function for aluminum is 4.08 eV.) HINT (a) Determine the maximum kinetic energy (in eV) of the ejected photoelectrons. eV (b) Determine the maximum speed (in m/s) of the ejected photoelectrons. m/s (c) Determine the wavelength in nm of the incident light. nm

Answers

Answer:

a) KE max = 3.632 * 10^{-19}

b) v = 6.31 * 10^5   m/s

c) Lambda = 195 nm

Explanation:

a) Work done is given by equation  

W = V * q = change in kinetic energy = Final KE – Initial KE

Substituting the given values, we get –  

V * 1.6*10^{-19} =0 - Initial KE

KE max = 2.27 * 1.6*10^{-19} = 3.632 * 10^{-19}

b) As we know KE = 0.5 mv^{2}

Substituting the given values, we get –  

3.632 * 10^{-19} = 0.5 * (9.11 * 10^{-31}) v^2

v = 6.31 * 10^5   m/s

c) Incident energy = W + K max

Substituting the given values we get  

hc/lambda = 4.08 * 1.6 * 10^{-19} J + 3.632 * 10^{-19} J

6.626 *10^{-34} * 3*10^8/lambda = 4.08 * 1.6 * 10^{-19} J + 3.632 * 10^{-19} J

Lambda = 1.95 * 10^7

Lambda = 195 nm

arcsin 0.9331 in degrees​

Answers

Answer:

68.9233231661

Explanation:

Just put it into your calculator, shift sin should do it but it will come up like this: [tex]sin^{-1}[/tex] which is the same as arcsin

a solid cubic centimeter of platinum weighs 21.5 N. if this cube of platinum is placed under water, what volume of water is displaced? what weight of water is displaced? (Hints: 1 cm3 of water has a mass of 1 gram; 1 gram weighs 0.0098 N; density of platinum = 21.5 g/cm3.)

Answers

Answer:

Mp =Pp g    where p = density of solid Platinum

Wp  = Mp Pp g     weight of mass M

Volume displaced is 1 cm^3

Weight of water = .0098 N   since 1 cm^3 displaced

Or Ww = 1 cm^3 * .001 kg / cm^3 * 9.8 N/kg = .0098 N

A film of oil lies on wet pavement. The refractive index of the oil exceeds that of the water. The film has the minimum nonzero thickness such that it appears dark due to destructive interference when viewed in visible light with wavelength 678 nm in vacuum. Assuming that the visible spectrum extends from 380 to 750 nm, what is the longest visible wavelength (in vacuum) for which the film will appear bright due to constructive interference

Answers

Answer:

Explanation:

For destructive interference , the condition is

2μt = nλ

2μt = n x 678

For constructive interference , the condition is

2μt = (2n+1)λ₁ /2

n x 678 = (2n+1)λ₁ /2

λ₁ = 1356 n / ( 2n + 1 )

λ₁ = 1356  / ( 2 + 1/n )

For longest wavelength , denominator should be smallest or n should be largest . The longest value of n is infinity so

λ₁ = 1356  /  2

= 678 nm .

If two speeds are in exact opposite directions, the combined speed is the difference in the speeds.

A. True
B. False

Answers

B. False if the speeds are going different directions there speed shall be different
It’s totally false ♥️

TAKE 100 POINTS!!!please help look the picture​!!!!​

Answers

Answer:

A major challenge in the drug delivery field is to enhance transport of therapeutics across biological barriers such as the blood brain barrier (BBB), the small intestine, nasal, skin and the mouth mucosa.

Answer:

An aqueous stagnant layer that overlies the apical membrane and the subepithelial blood flow are potential barriers to the absorption of drugs that readily penetrate the absorbing cell of the epithelium. The apical, basal, and basement membranes are potential barriers to the absorption of less permeable drugs

What unit is current measured in?

Answers

What unit? I don’t see anything to help you answer.

Biodiversity decline poses a problem in an ecosystem because

Answers

Answer:

Biodiversity decline continues due to a rapidly expanding human population. Habitat is damaged in order to meet growing needs for agriculture, urban development, water and materials. Fish, wildlife and plants are overharvested, despite mounting evidence that many harvesting practices are unsustainable.

Science questions!! Please help!!
Post Assessment on Investigating the Immune System
please help!! Please choose the right answers!! Dont guess if you dont know the answers!!

Answers

9: A
10: A
11: D
12: A
13: D
14: B
15: A.

Can you please mark me brainliest THANKS
hope this helped

Two organisms that can reproduce and produce fertile offspring are defined as a: *

Answers

A species is often defined as a group of organisms that can reproduce naturally with one another and create fertile offspring.

g 2. In a laboratory experiment on standing waves a string 3.0 ft long is attached to the prong of an electrically driven tuning fork which vibrates perpendicular to the length of the string at a frequency of 60 Hz. The weight (not mass) of the string is 0.096 lb. a) [5 pts] What tension must the string be under (weights are attached to the other end) if it is to vibrate in four loops

Answers

Answer:

The tension in string will be "3.62 N".

Explanation:

The given values are:

Length of string:

l = 3 ft

or,

 = 0.9144 m

frequency,

f = 60 Hz

Weight,

= 0.096 lb

or,

= 0.0435 kgm/s²

Now,

The mass will be:

= [tex]\frac{0.0435}{9.8}[/tex]

= [tex]0.0044 \ kg[/tex]

As we know,

⇒  [tex]\lambda=\frac{2L}{n}[/tex]

On substituting the values, we get

⇒     [tex]=\frac{2\times 0.9144}{4}[/tex]

⇒     [tex]=0.4572 \ m[/tex]

or,

⇒  [tex]v=f \lambda[/tex]

⇒      [tex]=0.4572\times 60[/tex]

⇒      [tex]=27.432 \ m/s[/tex]

Now,

⇒  [tex]v=\sqrt{\frac{T}{\mu} }[/tex]

or,

⇒  [tex]T=\frac{m}{l}\times v^2[/tex]

On putting the above given values, we get

⇒      [tex]=\frac{0.0044}{0.9144}\times (27.432)^2[/tex]

⇒      [tex]=\frac{752.51\times 0.0044}{0.9144}[/tex]

⇒      [tex]=3.62 \ N[/tex]

When silver nitrate and beryllium chloride react, silver chloride and beryllium nitrate form. What are the coefficients in this equation? (Note: Be sure to keep the reactants and products in the same order that they appear in the question.)
A. 2,1,1,2
B. 1,2,2,1
C. 1,2,1,2
D. 2,1,2,1 ​

Answers

Answer:

D. 2,1,2,1 ​

Explanation:

The equation of the reaction is; 2AgNO3 + BeCl2 + ---------》2AgCl + Be(NO3)2

The rule applied in balancing chemical reaction equation is that the number of atoms of each element on the reactants side must be the same as the number of atoms of the same element on the products sides.

If this rule is properly applied to the reaction between silver nitrate and beryllium chloride to form silver chloride and beryllium nitrate , the coefficients in the equation are;  2,1,2,1

How many tons and pounds (like together, not two different numbers) is 9,920 pounds? Please help,

Answers

Answer:

About 5 tons (4.96)

Learning Goal: To understand the concept of moment of inertia and how it depends on mass, radius, and mass distribution.
In rigid-body rotational dynamics, the role analogous to the mass of a body (when one is considering translational motion) is played by the body's moment of inertia. For this reason, conceptual understanding of the motion of a rigid body requires some understanding of moments of inertia. This problem should help you develop such an understanding.
The moment of inertia of a body about some specified axis is I = cmr^2, where c is a dimensionless constant, m is the mass of the body, and r is the perpendicular distance from the axis of rotation. Therefore, if you have two similarly shaped objects of the same size but with one twice as massive as the other, the more massive object should have a moment of inertia twice that of the less massive one. Furthermore, if you have two similarly shaped objects of the same mass, but one has twice the size of the other, the larger object should have a moment of inertia that is four times that of the smaller one.
Two spherical shells have their mass uniformly distrubuted over the spherical surface. One of the shells has a diameter of 2 meters and a mass of 1 kilogram. The other shell has a diameter of 1 meter. What must the mass m of the 1-meter shell be for both shells to have the same moment of inertia about their centers of mass?

Answers

Answer:

 m₂ = 4 kg

Explanation:

The moment of inertia is defined by

         I = ∫ r² dm

for bodies with high symmetry it is tabulated, for a spherical shell

        I = 2/3 m r²

in this case the first sphere has a radius of r₁ = 2m and a mass of m₁ = 1 kg, the second sphere has a radius r₂ = 1m.

They ask what is the masses of the second spherical shell so that the moment of inertia of the two is the same.

        I₁ = ⅔ m₁ r₁²

        I₂ = ⅔ m₂ r₂²

They ask that the two moments have been equal

        I₁ = I₂

        ⅔ m₁ r₁² = ⅔ m₂ r₂²

         m₂ = (r₁ / r₂) ² m₁

let's calculate

         m₂ = (2/1) ² 1

         m₂ = 4 kg

Please help, confused on this whole lesson

Answers

Answer:

1,4,3and 2 would happen hope this helps

What does Binding Energy in the nucleus represent?

Answers

Answer:

Nuclear binding energy is the energy required to split a nucleus of an atom into its component parts: protons and neutrons, or, collectively, the nucleons. The binding energy of nuclei is always a positive number, since all nuclei require net energy to separate them into individual protons and neutrons.

I hope you find this helpful.

Please please help me please please help please please

Answers

Answer:

Refraction

The rainbow is created because the index of refraction of water droplets changes as a function of wavelength. So, when the light enters the water droplet different colors will bend at different angles thus producing a dispersive effect known as a rainbow.

Answer:

D: Refraction

Explanation:

Refraction is the spitting of the electromagnetic spectrum, or the disembling of colours. For example, white is a mixture of all the colours.

The ratio of carbon-14 to nitrogen-14 in an artifact is 1:7. Given that the half-
life of carbon-14 is 5730 years, how old is the artifact?

Definitely will be giving brainliest!! Thanks!!

Answers

Answer:

17,190 years old

Explanation:

A p e x

A cargo spacecraft was orbiting the earth. It is equipped with rocket and has an initial mass of 30 metric ton. It wants to catch up with the international space station to deliver the cargo. So its rocket engine was fired for 1 minute. The engine ejects mass at a rate of 30 kg/s with an exhaust velocity of 3.1 km/s. The pressure at the nozzle exit is 5 kPa and the diameter of exit area is 94.4 cm. What is the thrust of the engine in a vacuum

Answers

Answer:

Explanation:

A cargo spacecraft was orbiting the earth. It is equipped with rocket and has an initial mass of 30 metric ton. It wants to catch up with the international space station to deliver the cargo. So its rocket engine was fired for 1 minute. The engine ejects mass at a rate of 30 kg/s with an exhaust velocity of 3.1 km/s. The pressure at the nozzle exit is 5 kPa and the diameter of exit area is 94.4 cm. What is the thrust of the eng

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