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Please show steps as to how to solve this problem
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Please show steps as to how to solve this problem - 1

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Answer 1
Answer:

Answer:

Torques must balance

F1 * X1 = F2 * Y2

or M1 g X1 = M2 g X2

X2 = M1 / M2 * X1 = 130.4 / 62.3 * 10.7

X2 = 22.4 cm

Torque = F1  * X2 =

62.3 gm* 980 cm/sec^2  * 22.4 cm = 137,000 gm cm^2 / sec^2

Normally x cross y   will be out of the page

r X F  for F1 will be into the page so the torque must be negative

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A 60.0-kg boy is surfing and catches a wave which gives him an initial speed of 1.60 m/s. He then drops through a height of 1.57 m, and ends with a speed of 8.50 m/s. How much nonconservative work was done on the boy

Answers

Answer:

Work = 1167.54 J

Explanation:

The amount of non-conservative work here can be given by the difference in kinetic energy and the potential energy. From Law of conservation of energy, we can write that:

Gain in K.E = Loss in P.E + Work

(0.5)(m)(Vf² - Vi²) - mgh = Work

where,

m = mass of boy = 60 kg

Vf = Final Speed = 8.5 m/s

Vi = Initial Speed = 1.6 m/s

g = 9.8 m/s²

h = height drop = 1.57 m

Therefore,

(0.5)(60 kg)[(8.5 m/s)² - (1.6 m/s)²] - (60 kg)(9.8 m/s²)(1.57 m) = Work

Work = 2090.7 J - 923.16 J

Work = 1167.54 J

What best describes the bromide ion that forms
​

Answers

Answer:

it A

Explanation:

Its a negative ion that hss one less valence electron than a netural bromine atom

A charge of uniform volume density (40 nC/m3) fills a cube with 8.0-cm edges. What is the total electric flux through the surface of this cube?

Answers

Answer:

The flux through the surface of the cube is 2.314\ Nm^(2)/C

Solution:

As per the question:

Edge of the cube, a = 8.0 cm = 8.0* 10^(- 2)\ m

Volume Charge density, \rho_(v) = 40 nC/m^(3) = 40* {- 9}\ C/m^(3)

Now,

To calculate the electric flux:

\phi = (q)/(\epsilon_(o))                                                      (1)

where

\phi = electric flux

\epsilon_(o) = 8.85* 10^(- 12)\ F/m = permittivity of free space  

Volume Charge density for the given case is given by the formula:

\rho_(v) = (Total\ charge, q)/(Volume of cube, V)                  (2)

Volume of cube, V = a^(3)

Thus

V = (8.0* 10^(- 2))^(3) = 5.12* 10^(- 4)\ m^(3)

Thus from eqn (2), the total charge is given by:

q = \rho_(v)V = 40* {- 9}* 5.12* 10^(- 4)

q = 2.048* 10^(-11)\ F = 20.48\ pF

Now, substitute the value of 'q' in eqn (1):

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Professional baseball pitchers deliver pitches that can reach the blazing speed of 100 mph (miles per hour). A local team has drafted an up-and-coming, left-handed pitcher who can consistently pitch at 42.91 m/s (96.00 mph) . Assuming a pitched ball has a mass of 0.1434 kg and has this speed just before a batter makes contact with it, how much kinetic energy does the ball have?

Answers

Answer: 132.02 J

Explanation:

By definition, the kinetic energy is written as follows:

KE = 1/2 m v²

In our question, we know from the question, the following information:

m = 0.1434 Kg

v= 42.91 m/s

Replacing in the equation for KE, we have:

KE = 1/2 . 0.1434 Kg. (42.91)² m²/s² ⇒ KE = 132.02 N. m = 132.02 J

A 5-kg moving at 6 m/s collided with a 1-kg ball at rest. The ball bounce off each other and the second ball moves in the same direction as the first ball at 10 m/sec. What is the velocity of the first ball after the collision?

Answers

Given :

A 5-kg moving at 6 m/s collided with a 1-kg ball at rest.

The ball bounce off each other and the second ball moves in the same direction as the first ball at 10 m/sec.

To Find :

The velocity of the first ball after the collision.

Solution :

We know, by conservation of momentum :

m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2

Putting all given values with directions ( one side +ve and other side -ve ).

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Hence, this is the required solution.

When water freezes, its volume increases by 9.05% (that is, ΔV / V0 = 9.05 × 10-2). What force per unit area is water capable of exerting on a container when it freezes? (It is acceptable to use the bulk modulus of water, B = 2.2 × 109 N/m2, for this problem.) Give your answer in N/cm2.

Answers

Answer:

P = 1.99 10⁸ Pa

Explanation:

The definition of the bulk module is

      B = - P / (ΔV / V)

The negative sign is included for which balk module is positive, P is the pressure and V that volume

They tell us that the variation in volume is 9.05%, that is

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    P = - B DV / V

    P = 2.2 10⁹ (0.0905)

    P = 1.99 10⁸ Pa
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