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Bond [772]
3 years ago
13

How much kinetic energy does a person of mass 50kg running at 3m/s have? Provide

Physics
1 answer:
ddd [48]3 years ago
4 0

Answer:

0.225 KJ kinetic energy is used for a person of mass 50 Kg running at speed of 3 m/s.

Explanation:

Given:

mass = 50 kg

velocity = v = 3 m/s

To Find :

Kinetic energy = ?

Solution:

Kinetic energy of an object is the energy that it possesses due to its motion.

The formula for kinetic energy is given as

\textrm{kinetic energy} = \frac{1}{2}\times mass\times velocity^{2}\\= \frac{1}{2}\times m\times v^{2}\\= \frac{1}{2}\times 50\times 3^{2}\\= 25\times 9\\= 225\ Joules.

But we required kinetic energy in KiloJoules i.e KJ

1 K = 1000 =10^{3}

\textrm{kinetic energy} = 225\ J\\= 0.225\times 1000\\= 0.225 \times 10^{3}\\= 0.225\ KJ

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A 10-foot ladder is leaning straight up against a wall when a person begins pulling the base of the ladder away from the wall at
docker41 [41]

Answer:

y = 4.36

Explanation:

Let the height of the ladder be L

L = 10

Also:

  • Let x = distance\ from\ the\ base\ of\ the\ ladder
  • Let y = height\ of\ the\ base\ of\ the\ ladder

When the ladder leans against the wall, it forms a triangle and the length of the ladder forms the hypotenuse.

So, we have:

L^2 = x^2 + y^2 --- Pythagoras Theorem

When the base is 9ft from the wall, this means that:

x = 9

Substitute 9 for x and 10 for L in L^2 = x^2 + y^2

10^2 = 9^2 + y^2

100 = 81 + y^2

Make y^2 the subject

y^2 = 100 - 81

y^2 = 19

Make y the subject

y = \sqrt{19

y = 4.36

<em>Hence, the true distance at that point is approximately 4.36ft</em>

8 0
3 years ago
A guitar string has a linear density of 8.30 ✕ 10−4 kg/m and a length of 0.660 m. the tension in the string is 56.7 n. when the
Sedbober [7]
Ans: Beat Frequency = 1.97Hz

Explanation:
The fundamental frequency on a vibrating string is 

f =   \sqrt{ \frac{T}{4mL} }<span>  -- (A)</span>

<span>here, T=Tension in the string=56.7N,
L=Length of the string=0.66m,
m= mass = 8.3x10^-4kg/m * 0.66m = 5.48x10^-4kg </span>


Plug in the values in Equation (A)

<span>so </span>f = \sqrt{ \frac{56.7}{4*5.48*10^{-4}*0.66} }<span> = 197.97Hz </span>

<span>the beat frequency is the difference between these two frequencies, therefore:
Beat frequency = 197.97 - 196.0 = 1.97Hz
-i</span>
3 0
3 years ago
Read 2 more answers
3. What is the main difference between a physical change and a chemical
Lana71 [14]

Answer:

B and D could both be right as they are quit similar.

Consider two rods of the same length and diameter,

Increasing the diameter of one would change the expansion qualities of that rod even though there would be no chemical changes,

However, leaving the physical appearance of both rods the same while applying a reactive substance (acid or something) to one of the rods would not necessarily change the physical appearance of that rod but could make a considerable change in the  physical properties of that rod.

3 0
3 years ago
7. 13 a turbine receives steam at 6 mpa, 600°c with an exit pressure of 600 kpa. Assume the turbine is adiabatic and neglect kin
AnnyKZ [126]

The work done by the turbine will be 708.2 kJ/kg. The work done by the turbine is the difference of the enthalpy at inlet and exit.

<h3 /><h3>What is temperature?</h3>

Temperature directs the hotness or coldness of a body. In clear terms, it is the method of finding the kinetic energy of particles within an entity. Faster the motion of particles, more the temperature.

If the given turbine is assumed to be reversible;

\rm P_I(Initial pressure)=60 mpa = 60 bar

\rm  T_i(Initial temperature)=600° C

\rm P_e (Exit pressure)=600 kpa=6 bar

The heat balance equation is;

\rm q-W_t=h_e-h_i\\\\ q=0\\\\\ W_t=h_i-h_e

The change in the entropy is;

\rm S_2-S_1=\frac{\delta q}{dt}

The work done by the turbine is;

\rm W_t = h_i-h_e\\\\ W_t =3658.4 -29.5019\\\\  W_t =708.2 \ kJ/kg

Hence,the work done by the turbine will be 708.2 kJ/kg.

To learn more about the temperature, refer to the link;

brainly.com/question/7510619

#SPJ4

6 0
2 years ago
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