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2 years ago

A. In the building there are 133 steps from the ground floor to the sixth floor.

Physics

1 answer:

faust18 [17]2 years ago

7 0

(a) The work done by the person is 18.83465 kJ.

(b) The average power performed by the person during the walk is 51.4 W.

<h3>Work done by the person</h3>

The work done by the person is calculated as follows;

W = Fd

W = mgh

W = (89.2 x 9.8) x (0.162 x 133)

W = 18,834.65 J

W = 18.83465 kJ

<h3>Average power of the person</h3>

P = Fv

where;

v is velocity

v = (d)/t

v = (133 x 0.162)/(6 x 60 + 6)

v = (133 x 0.162)/(366)

v = 0.0588 m/s

P = (89.2 x 9.8) x 0.0588

P = 51.4 W

<h3>Amount of food calories burnt</h3>

4.1868 kJ = 1 Cal

18.83465 kJ = ?

= 4.5 Cal

Learn more about work done here: brainly.com/question/8119756

#SPJ1

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A wooden block of mass M resting on a frictionless, horizontal surface is attached to a rigid rod of length ℓ and of negligible

mr Goodwill [35]

Answer:

$mv l$

$\frac{M }{(M + m)}$

Explanation:

Complete question statement is as follows :

A wooden block of mass M resting on a friction less, horizontal surface is attached to a rigid rod of length ℓ and of negligible mass. The rod is pivoted at the other end. A bullet of mass m traveling parallel to the horizontal surface and perpendicular to the rod with speed v hits the block and becomes embedded in it.

(a) What is the angular momentum of the bullet–block system about a vertical axis through the pivot? (Use any variable or symbol stated above as necessary.)

(b) What fraction of the original kinetic energy of the bullet is converted into internal energy in the system during the collision? (Use any variable or symbol stated above as necessary.)

$m$ = mass of the bullet

$v$ = velocity of the bullet before collision

$r$ = distance of the line of motion of bullet from pivot = $l$

$L$ = Angular momentum of the bullet-block system

Angular momentum of the bullet-block system is given as

$L = m v r$

$L = mv l$

$V$ = final velocity of bullet block combination

Using conservation of momentum

Angular momentum of bullet block combination = Angular momentum of bullet

$(M + m) V l = m v l\\V =\frac{mv}{(M + m)}$

$K_{o}$ = Initial kinetic energy of the bullet

Initial kinetic energy of the bullet is given as

$K_{o} = (0.5) m v^{2}$

$K_{f}$ = Final kinetic energy of bullet block combination

Final kinetic energy of bullet block combination is given as

$K_{f} = (0.5) (M + m) V^{2}$

Fraction of original kinetic energylost is given as

Fraction = $\frac{(K_{o} - K_{f})}{K_{o}} = \frac{((0.5) m v^{2} - (0.5) (M + m) V^{2})}{(0.5) m v^{2}}$

Fraction = $\frac{(m v^{2} - (M + m) (\frac{mv}{(M + m)})^{2})}{m v^{2}} = \frac{(Mm v^{2} + m^{2} v^{2} - m^{2} v^{2})}{(M + m) m v^{2}}$

Fraction = $\frac{(Mm v^{2} + m^{2} v^{2} - m^{2} v^{2})}{(M + m) m v^{2}}\\ \frac{M }{(M + m)}$

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3 years ago

The __________ of exercise determines the health and fitness benefit of the exercise. A. frequency B. intensity C. time D. type

MariettaO [177]

Answer:

c- Time

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3 years ago

Read 2 more answers

A solid cylinder of mass M = 45 kg, radius R = 0.44 m and uniform density is pivoted on a frictionless axle coaxial with its sym

user100 [1]

Answer:

w_f = 1.0345 rad/s

Explanation:

Given:

- The mass of the solid cylinder M = 45 kg

- Radius of the cylinder R = 0.44 m

- The mass of the particle m = 3.6 kg

- The initial speed of cylinder w_i = 0 rad/s

- The initial speed of particle V_pi = 3.3 m/s

- Mass moment of inertia of cylinder I_c = 0.5*M*R^2

- Mass moment of inertia of a particle around an axis I_p = mR^2

Find:

- What is the magnitude of its angular velocity after the collision?

Solution:

- Consider the mass and the cylinder as a system. We will apply the conservation of angular momentum on the system.

L_i = L_f

- Initially, the particle is at edge at a distance R from center of cylinder axis with a velocity V_pi = 3.3 m/s contributing to the initial angular momentum of the system by:

L_(p,i) = m*V_pi*R

L_(p,i) = 3.6*3.3*0.44

L_(p,i) = 5.2272 kgm^2 /s

- While the cylinder was initially stationary w_i = 0:

L_(c,i) = I*w_i

L_(c,i) = 0.5*M*R^2*0

L_(c,i) = 0 kgm^2 /s

The initial momentum of the system is L_i:

L_i = L_(p,i) + L_(c,i)

L_i = 5.2272 + 0

L_i = 5.2272 kg-m^2/s

- After, the particle attaches itself to the cylinder, the mass and its distribution around the axis has been disturbed - requires an equivalent Inertia for the entire one body I_equivalent. The final angular momentum of the particle is as follows:

L_(p,f) = I_p*w_f

- Similarly, for the cylinder:

L_(c,f) = I_c*w_f

- Note, the final angular velocity w_f are same for both particle and cylinder. Every particle on a singular incompressible (rigid) body rotates at the same angular velocity around a fixed axis.

L_f = L_(p,f) + L_(c,f)

L_f = I_p*w_f + I_c*w_f

L_f = w_f*(I_p + I_c)

-Where, I_p + I_c is the new inertia for the entire body = I_equivalent that we discussed above. This could have been determined by the superposition principle as long as the axis of rotations are same for individual bodies or parallel axis theorem would have been applied for dissimilar axes.

L_i = L_f

5.2272 = w_f*(I_p + I_c)

w_f = 5.2272/ R^2*(m + 0.5M)

Plug in values:

w_f = 5.2272/ 0.44^2*(3.6 + 0.5*45)

w_f = 5.2272/ 5.05296

w_f = 1.0345 rad/s

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2 years ago

A satellite at a particular point along an elliptical orbit has a gravitational potential energy of 5100 MJ with respect to Eart

serious [3.7K]

To solve this problem we will apply the theorem given in the conservation of energy, by which we have that it is conserved and that in terms of potential and kinetic energy, in their initial moment they must be equal to the final potential and kinetic energy. This is,

$E_{initial} = E_{final}$

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Replacing the 5100MJ for satellite as initial potential energy, 4200MJ for initial kinetic energy and 5700MJ for final potential energy we have that

$KE_{final} = (PE_{initial}+KE_{initial} )-PE_{final}$

$KE_{final} = (5100+4200)-5700$

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