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Leona [35]
3 years ago
11

A 65.0-kg runner has a speed of 5.20 m/s at one instant dur- ing a long-distance event. (a) What is the runner’s kinetic energy

at this instant? (b) How much net work is required to double his speed?
Physics
1 answer:
vladimir2022 [97]3 years ago
8 0

Answer:

a)KE=878.8 J

b)W=2636.4 J      

Explanation:

Given that

mass ,m = 65 kg

Initial speed ,u = 5.2 m/s

a)

We know that kinetic energy KE is given as follows

KE=\dfrac{1}{2}mu^2

m=mass

u=velocity

Now by putting the values in the above equation we get

KE=\dfrac{1}{2}\times 65\times 5.2^2\ J

KE=878.8 J

b)

We know that

Work done by all forces = Change in the kinetic energy

The final velocity , v= 2 u = 2 x 5.2 m/s

v= 10.4 m/s

W=\dfrac{1}{2}mv^2-\dfrac{1}{2}mu^2

Now by putting the values in the above equation we get

W=\dfrac{1}{2}\times 65\times 10.4^2-\dfrac{1}{2}\times 65\times 5.2^2\ J

W=2636.4 J

a)KE=878.8 J

b)W=2636.4 J

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A horse running at 3 m/s speeds up with a constant acceleration of 5 m/s2. How fast is the
Elan Coil [88]

Answer:

The horse is going at 12.72 m/s speed.

Explanation:

The initial speed of the horse (u) = 3 m/s

The acceleration of the horse (a)= 5 m/s^{2}

The displacement( it is assumed it is moving in a straight line)(s)= 15.3 m

Applying the second equation of motion to find out the time,

s=ut+\frac{1}{2}at^{2}

15.3=3t+2.5t^{2}

2.5t^{2}+3t-15.3=0

Solving this quadratic equation, we get time(t)=1.945 s, the other negative time is neglected.

Now applying first equation of motion, to find out the final velocity,

v=u+at

v=3+1.945*5

v=3+9.72

v=12.72 m/s

The horse travels at a speed of 12.72 m/s after covering the given distance.

7 0
3 years ago
Number of vibrations (waves) in an amount of time
m_a_m_a [10]

Answer:

I think frequency not sure though

How frequently a wave or vibration occurs during a span of time, determines the waves frequency. Frequency is the number of waves per unit time. The unit for frequency if a Hertz ( 1/second). The speed a wave travels is the wavelength multiplied by this frequency. The amplitude of a wave is the maximum distance the wave is displaced.

5 0
2 years ago
An object is moving forward with a constant velocity. Which statement about this object MUST be true?
Greeley [361]

If the object's <em>velocity is constant</em> ... (it's speed isn't changing AND it's moving in a straight line) ... then the net force on the object is zero.<em> (D)</em>

Either there are no forces at all acting on the object, OR there are forces on it but they're 'balanced' ... when you add up all of their sizes and directions, they just exactly cancel each other out, and they have the SAME EFFECT on the object as if there were no forces at all.

7 0
3 years ago
Point charge μC is located at x =, y = , point charge is located at x = 0m. What are (a)the magnitude and (b)direction of the to
pishuonlain [190]

Answer: The question has some details missing. here is the complete question ; Point charge 1.5 μC is located at x = 0, y = 0.30 m, point charge -1.5 μC is located at x = 0 y = -0.30m. What are (a)the magnitude and (b)direction of the total electric force that these charges exert on a third point charge Q = 5.0 μC at x = 0.40 m, y = 0

Explanation:

  • a) First of all find the distance between the two charges;
  • x = 0, y = 0.30  and x = 0.40 m, y = 0
  • r = √( 0.4² + 0.3²)
  • = 0.5m

hence, the force F = 2Kq1q2cosθ /r²...............equation 1

but cosθ = y/r = 0.3/0.5

cosθ = 0.6

plugging back to equation 1;

F = 2 x 9 x 10^9 x 1.5 x 10^-6 x 5 x 10^-6 /0.5^2

F = 540 x 10^-3

Magnitude of Force = 0.54N

b) Direction is at angle 90

6 0
3 years ago
An archer pulls the bowstring back for a distance of 0.470 m before releasing the arrow. The bow and string act like a spring wh
never [62]

Answer:

(a) 46.94 J.

(b)  55.95 m/s

Explanation:

(a)

Potential Energy: This is the energy of a body, due to its position. The S.I unit of potential energy is Joules (J).

The formula of potential energy in a stretched spring is

Ep = 1/2ke² .......................... Equation 1

Where Ep = potential energy of the spring, k = Force constant of the spring, e = extension or compression.

Given: k = 425 N/m, e = 0.47 m.

Substitute into equation 1

Ep = 1/2(425×0.47²)

Ep = 46.94 J.

(b)

at the instant When the arrow leaves the bow, the potential energy of the arrow is converted kinetic energy of the bow.

I.e,

Ep = 1/2mv² ............. Equation 2

Where m = mass of the arrow, v = velocity of the arrow.

make v the subject of the equation

v = √(2Ep/m)............. Equation 3

Given: Ep = 46.94 J, m = 0.03 Kg.

Substitute into equation 3

v = √(2×46.96/0.03)

v = √(93.92/0.03)

v = √(3130.67)

v = 55.95 m/s

5 0
3 years ago
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