1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
zhuklara [117]
3 years ago
12

A block slides along a frictionless surface with 220 J of kinetic energy. It then goes up a frictionless ramp. When the kinetic

energy drops to 55 J, how much gravitational potential energy does the block have?
Physics
1 answer:
aev [14]3 years ago
8 0

Hello!

For the explanation of this energy conservation exercise, where we'll use <u>energy conservation law</u>, let's see what this principle proposes.

How you should know, mechanical energy conserves in every point, that is to say mechanical energy is same in A point like B point. (Mechanical energy will be represented by "Me")

Once time we know that, let's take the 220 Joules momentum like A point, and when 55 Joules momentum like B point.

Then, let's use the <u>energy conservation principle:</u>

Me(A) = Me(B)

  • We know Mechanical energy in A point, so just lets replace according to our data:

220 J = Me(B)

  • In B point, we know kinetic energy, but <u>we dont know gravitational potential energy</u>, so lets descompose Mechanical energy, into kinetic energy and gravitational potential energy:

220 J = Ke + Gpe

  • We know kinetic energy value, so lets replace it:

220 J = 55 J + Gpe

  • Finally, just clean Gpe and resolve it:

Gpe = 220 J - 55 J = 165 J

Gravitational potential energy is of One hundred sixty five Joules <u>(165 J).</u>

                                                                                  ║Sincerely, ChizuruChan║

You might be interested in
A 34-m length of wire is stretched horizontally between two vertical posts. The wire carries a current of 68 A and experiences a
il63 [147K]

Answer:

7.28×10⁻⁵ T

Explanation:

Applying,

F = BILsin∅............. Equation 1

Where F = magnetic force, B = earth's magnetic field, I = current flowing through the wire, L = Length of the wire, ∅ = angle between the field and the wire.

make B the subject of the equation

B = F/ILsin∅.................. Equation 2

From the question,

Given: F = 0.16 N, I = 68 A, L = 34 m, ∅ = 72°

Substitute these values into equation 2

B = 0.16/(68×34×sin72°)

B = 0.16/(68×34×0.95)

B = 0.16/2196.4

B = 7.28×10⁻⁵ T

7 0
2 years ago
What are the units of radiation?
Ksenya-84 [330]

The radiation dose absorbed by a person is measured using the conventional unit rad or the SI unit gray. The biological risk of exposure to radiation is measured using the conventional unit rem or the SI unit sievert.

I hope this helps!

3 0
3 years ago
A 110-g bullet is fired from a rifle having a barrel 0.636 m long. Choose the origin to be at the location where the bullet begi
astraxan [27]
<span>Data:
mass =  110-g bullet
d = 0.636 m
Force = 13500 + 11000x - 25750x^2, newtons.

a) Work, W

W = ∫( F* )(dx) =∫[13500+ 11000x - 25750x^2] (dx) =

W =  13500x + 5500x^2 - 8583.33 x^3 ] from 0 to 0.636  =

W = 8602.6 joule

b) x= 1.02 m

</span><span><span>W =  13500x + 5500x^2 - 8583.33 x^3 ] from</span> 0 to 1.02

W = 10383.5

c) %

[W in b / W in a] = 10383.5 / 8602.6 = 1.21 => W in b is 21% more than work in a.


</span>
7 0
3 years ago
A 217 Ω resistor, a 0.875 H inductor, and a 6.75 μF capacitor are connected in series across a voltage source that has voltage a
Nataly [62]

For an AC circuit:

I = V/Z

V = AC source voltage, I = total AC current, Z = total impedance

Note: We will be dealing with impedances which take on complex values where j is the square root of -1. All phasor angles are given in radians.

For a resistor R, inductor L, and capacitor C, their impedances are given by:

Z_{R} = R

R = resistance

Z_{L} = jωL

ω = voltage source angular frequency, L = inductance

Z_{C} = -j/(ωC)

ω = voltage source angular frequency, C = capacitance

Given values:

R = 217Ω, L = 0.875H, C = 6.75×10⁻⁶F, ω = 220rad/s

Plug in and calculate the impedances:

Z_{R} = 217Ω

Z_{L} = j(220)(0.875) = j192.5Ω

Z_{C} = -j/(220×6.75×10⁻⁶) = -j673.4Ω

Add up the impedances to get the total impedance Z, then convert Z to polar form:

Z = Z_{R} + Z_{L} + Z_{C}

Z = 217 + j192.5 - j673.4

Z = (217-j480.9)Ω

Z = (527.6∠-1.147)Ω

Back to I = V/Z

Given values:

V = (30.0∠0+220t)V (assume 0 initial phase, and t = time)

Z = (527.6∠-1.147)Ω (from previous computation)

Plug in and solve for I:

I = (30.0∠0+220t)/(527.6∠-1.147)

I = (0.0569∠1.147+220t)A

To get the voltages of each individual component, we'll just multiply I and each of their impedances:

v_{R} = I×Z_{R}

v_{L} = I×Z_{L}

v_{C} = I×Z_{C}

Given values:

I = (0.0569∠1.147+220t)A

Z_{R} = 217Ω = (217∠0)Ω

Z_{L} = j192.5Ω = (192.5∠π/2)Ω

Z_{C} = -j673.4Ω = (673.4∠-π/2)Ω

Plug in and calculate each component's voltage:

v_{R} = (0.0569∠1.147+220t)(217∠0) = (12.35∠1.147+220t)V

v_{L} = (0.0569∠1.147+220t)(192.5∠π/2) = (10.95∠2.718+220t)V

v_{C} = (0.0569∠1.147+220t)(673.4∠-π/2) = (38.32∠-0.4238+220t)V

Now we have the total and individual voltages as functions of time:

V = (30.0∠0+220t)V

v_{R} = (12.35∠1.147+220t)V

v_{L} = (10.95∠2.718+220t)V

v_{C} = (38.32∠-0.4238+220t)V

Plug in t = 22.0×10⁻³s into these values and take the real component (amplitude multiplied by the cosine of the phase) to determine the real voltage values at this point in time:

V = 30.0cos(0+220(22.0×10⁻³)) = 3.82V

v_{R} = 12.35cos(1.147+220(22.0×10⁻³)) = 11.8V

v_{L} = 10.95cos(2.718+220(22.0×10⁻³)) = 3.19V

v_{C} = 38.32cos(-0.4238+220(22.0×10⁻³)) = -11.2V

4 0
3 years ago
] A new coal-fired 750 MWe power plant with a thermal efficiency of 42% burns 9000 Btu/lb coal, which contains 1.1% sulfur. a. I
Valentin [98]

Answer:

The net emissions rate of sulfur is 1861 lb/hr

Explanation:

Given that:

The power or the power plant = 750 MWe

Since the power plant with a thermal efficiency of 42% (i.e. 0.42) burns 9000 Btu/lb coal, Then the energy released per one lb of the coal can be computed as:

\mathtt{=( 0.42\times 9000\times 1055.06) J}

= 3988126.8  J

= 3.99 MJ

Also, The mass of the burned coal per sec can be calculated by dividing the molecular weight of the power plant by the energy released per one lb.

i.e.

The mass of the coal that is burned per sec =\dfrac{750}{3.99}

The mass of the coal that is burned per sec = 187.97 lb/s

The mass of sulfur burned  = \dfrac{1.1}{100} \times 187.97 \  lb/s

= 2.067 lb/s

To hour; we have:

= 7444 lb/hr

However, If a scrubber with 75% removal efficiency is utilized,

Then; the net emissions rate of sulfur is (1 - 0.75) × 7444 lb/hr

= 0.25 × 7444 lb/hr

= 1861 lb/hr

Hence, the net emissions rate of sulfur is 1861 lb/hr

8 0
3 years ago
Other questions:
  • If a certain mass of mercury has a volume of 0.002 m^3 at a temperature of 20°c, what will be the volume at 50°c
    8·2 answers
  • Which of the following is not an example of Newton’s third law?
    10·1 answer
  • In what type of job might being a difficult person not be much of a liability ?
    8·1 answer
  • 5 uses of satellite​
    15·1 answer
  • A large sheet of charge has a uniform charge density of 9  μCm2. What is the electric field due to this charge at a point just
    6·1 answer
  • Two sheets of polarizing material are placed such that their polarizing axes are 90° to each other and no light passes through t
    15·1 answer
  • What type of thermodynamic system is best to use to describe an operating automobile engine? A) Open O B) Dilated O C) Isolated
    8·1 answer
  • HELP ME !!!Question 4 of 10
    13·2 answers
  • How many minutes are each half in soccer ?
    15·1 answer
  • A bus travels east for 3 km, then north for 4 km. What is its final displacement?
    11·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!