5. (Serway 9th ed., 7-3) In 1990, Walter Arfeuille of Belgium lifted a 281.5-kg object through a distance of 17.1 cm using only his teeth. (a) How much work was done on the object by Arfeuille in this lift, assuming the object was lifted at constant speed? (b) What total force was exerted on Arfeuille’s teeth during the lift? (Ans. (a) 472 J; (b) 2.76 kN)

Answers

Answer 1

Answer:

Para resolver este problema es necesario aplicar los conceptos de Fuerza, dados en la segunda Ley de Newton y el concepto de Trabajo, como expresión de la fuerza necesaria para realizar una actividad en una distancia determinada.

El trabajo se define como

W = F*d

Where,

F = Force

d = Distance

At the same time we have that the Force by second's Newton law is equal to

F = mg

Where,

m = mass

g = Gravitational acceleration

PART A) Using our values and replacing we have that

$W = F*d\nW = mg*d\nW=281.5*9.8(17.1*10^(-2)\nW = 471.738 J\approx 472J$

PART B) Using Newton's Second law we have that,

$F = mg \nF= 281.5*9.8\nF= 2758.7 N \approx 2.76kN$

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Answers

Answer:

Explanation:

A 79-turn, 16.035-cm-diameter coil is at rest in a horizontal plane. A uniform magnetic field 43 degrees away from vertical increases from 0.997 T to 6.683 T in 56.691 s. Determine the emf induced in the coil.

Answers

Answer:

The induced emf is 0.0888 V.

Explanation:

Given that,

Number of turns = 79

Diameter = 16.035 cm

Angle = 43

Change in magnetic field $\Delta B=(6.683-0.997)= 5.686\ T$

Time = 56.691 s

We need to calculate the induced emf

Using formula of induced emf

$\epsilon=(NA\Delta B\cos\theta)/(\Delta T)$

Where, N = number of turns

A = area

B = magnetic field

Put the value into the formula

$\epsilon=(79*\pi*(8.0175*10^(-2))^2*5.686*\cos43)/(56.691)$

$\epsilon =0.0888\ V$

Hence, The induced emf is 0.0888 V.

If a pressure gauge measure an increase in 3×10^(5)Pa on an area of 0.7 m^2, calculate the increase in the force applied to the area?

Answers

Answer:210000N

Explanation:

Pressure=3x10^5pa

area=0.7m^2

Force = pressure x area

Force=3x10^5x0.7

Force=210000N

A person is making homemade ice cream. She exerts a force of magnitude 26 N on the free end of the crank handle on the ice-cream maker, and this end moves on a circular path of radius 0.26 m. The force is always applied parallel to the motion of the handle. If the handle is turned once every 2.0 s, what is the average power being expended?

Answers

Answer:

P = 31.83 W

Explanation:

Our data are,

Magnitude of the force F = 26 N

Radius of the circular path r = 0.26 m

The angle between force and handle $\theta = 0$ °

Time t = 2 s

We know that the formula to find the velocity is given by

Velocity $v = (2\pi r)/(t)$

$v= (2\pi r)/(t)$

$v=(2 \pi 0.26)/(2)$

$v= 0.8168m/s$

We know also that the formula to find the power is given by,

$P = F*v$

$P = (26)(0.8168)$

$P = 31.83 W$

A high diver of mass 60.0 kg steps off a board 10.0 m above the water and falls vertical to the water, starting from rest. If her downward motion is stopped 2.10 s after her feet first touch the water, what average upward force did the water exert on her

Answers

Answer:

The average upward force exerted by the water is 988.2 N

Explanation:

Given;

mass of the diver, m = 60 kg

height of the board above the water, h = 10 m

time when her feet touched the water, t = 2.10 s

The final velocity of the diver, when she is under the influence of acceleration of free fall.

V² = U² + 2gh

where;

V is the final velocity

U is the initial velocity = 0

g is acceleration due gravity

h is the height of fall

V² = U² + 2gh

V² = 0 + 2 x 9.8 x 10

V² = 196

V = √196

V = 14 m/s

Acceleration of the diver during 2.10 s before her feet touched the water.

14 m/s is her initial velocity at this sage,

her final velocity at this stage is zero (0)

V = U + at

0 = 14 + 2.1(a)

2.1a = -14

a = -14 / 2.1

a = -6.67 m/s²

The average upward force exerted by the water;

$F_(on\ diver) = mg - F_( \ water)\n\nma = mg - F_( \ water)\n\nF_( \ water) = mg - ma\n\nF_( \ water) = m(g-a)\n\nF_( \ water) = 60[9.8-(-6.67)]\n\nF_( \ water) = 60 (9.8+6.67)\n\nF_( \ water) = 60(16.47)\n\nF_( \ water) = 988.2 \ N$

Therefore, the average upward force exerted by the water is 988.2 N

A 2.0 cm thick brass plate (k_r = 105 W/K-m) is sealed to a glass sheet (kg = 0.80 W/K m), and both have the same area. The exposed face of the brass plate is at 80°C, while the exposed face of the glass is at 20 °C. How thick is the glass if the glass brass interface is at 65 C? Ans. 0.46 mm

Answers

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