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3 years ago

9

A4 kg bowling ball begins rolling down a at bowling alloy at 6 m/s . When it strikes the pins, it is estimated to be moving at 5

m/s . How much onergy is lost due to friction?

1 answer:

Paul [167]3 years ago

8 0

Answer:

Energy lost due to friction is 22 J

Explanation:

Mass of the ball m = 4 kg

Initially velocity of ball v = 6 m/sec

So kinetic energy of the ball $KE=\frac{1}{2}mv^2$

$KE=\frac{1}{2}\times 4\times 6^2=72J$

Now due to friction velocity decreases to 5 m/sec

Kinetic energy become

$KE=\frac{1}{2}\times 4\times 5^2=50J$

Therefore energy lost due to friction = 72 -50 = 22 J

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Is it possible to produce a continuous and oriented aramid fiber–epoxy matrix composite having longitudinal and transverse modul

gregori [183]

Answer:

It is not possible to produce continued and oriented fiber.

Explanation:

To solve the problem it is necessary to take into account the concepts related to Fiber volume ratio. The amount of fiber in a fiber reinforced compound corresponds directly to the mechanical properties of the compound. Given the fiber volume fraction, the theoretical elastic properties of a compound can be determined. The elastic modulus of a compound in the fiber direction of a unidirectional compound can be calculated using the following equation:

$E = (1-V_f)E_m+V_fE_f$

Where,

E is the longitudinal modulus of Elasticity

$V_f$ is the fiber volume ratio

$E_m$ is the elastic modulus of the matrix

$E_f$ is the elastic modulus of the fibers

We need to consult the table of characteristics of Fibers and Reinforcements of Materials, in which they specify that the modulus of elasticity of the aramid fiber-epoxy is

$E_f = 131Gpa$

Moreover from the statement,

$E = 35Gpa$

$E_m = 3.4Gpa$

Replacing in the previous equation,

$35 = 3.4 (1-V_f)+131V_f$

$V_f = 0.25 \rightarrow longitudinal$

To make the comparison we now calculate the Fiber volume ratio through the transverse elastic modulus,

$E = \frac{E_mE_f}{(1-V_f)E_f+V_fE_f}$

Our values are given in this case as:

$E = 5.17Gpa\\E_m = 3.4Gpa \\E_f = 131Gpa$

Replacing,

$5.17 = \frac{3.4*131}{(1-V_f)(131)+V_f*3.4}$

$V_f = 0.351 \rightarrow transversal$

From both cases it is possible to conclude that it is not possible to produce a fiber of the specified material in a continuous and oriented manner, as long as the volume fraction is different in the different cases.

5 0

3 years ago

For study, the Earth can be divided into three parts:

erica [24]

The three parts of the Earth are Atmosphere, Hydrosphere and Lithosphere.

Atmosphere is the blanket of air that surrounds the earth. It is densest close to the surface and thins out as one moves higher. Atmosphere of Earth contains mainly Nitrogen, followed by Oxygen and small amounts of water vapor, Carbondioxide and other gases.

Lithosphere is the outer most part of the earth's surface. The Earth's crust and the mantle form Lithosphere.

Hydrosphere is the part of the Earth that has water. The Oceans, seas, rivers, lakes and other water bodies constitute the Hydrosphere.

Stratosphere, Mesosphere and Ionosphere are different layers of atmosphere.

Hence, for the study of the Earth, one needs to consider earth to be made of three parts- atmosphere, Lithosphere and Hydrosphere.

5 0

2 years ago

Read 2 more answers

An object is formed by attaching a uniform, thin rod with a mass of mr = 6.9 kg and length L = 4.88 m to a uniform sphere with m

Bumek [7]

Answer:

a) total moment of inertia is 1359.05 kg m^2

b) angular acceleratio is 0.854rad/sec^2

Explanation:

Given data:

m1=6.9 kg

L=4.88 m

m2=34.5 kg

R=1.22 m

we klnow that moment of inertia for rod is given as

J1=(1/12) ×m×L^2

$J1 = (1/12) \times 6.9 \times 4.88^2 = 13.693 kg m^2$

moment of inertia for sphere is given as

J1=(2/5) ×m×r^2

$J1 = (2/5) \times 34.5 \times 1.22^2 = 20.539 kg m^2$

As object rotates around free end of rod then for sphere the axis around what it rotates is at a distance of d2=L+R

For rod distance is d1=0.5*L

By Steiner theorem

for the rod we get $J_1'=J_1 + m_1\times d_1^2$

$J_1' = 13.693 + 2.44^2\times 6.9 = 54.77 kg m^2$

for the sphere we get $J_2' = J_2 + m_2\times d_2^2$

$J2' = 20.539 + 34.5\times 6.1^2 = 1304.28 kg^m2$

And the total moment of inertia for the first case is

$J_{t1} = J_1'+J_2' = 54.77 + 1304.28 = 1359.05kg.m^2$

b) F=476 N

The torque for system is given as

$M = F\times d\times sin(a)$

where a is angle between Force and distance d

and where d represent distance from rotating axis.

In this case a = 90 degree

$M = F\times L/2$

M=476*2.44 = 1161.44 Nm

The acceleration is calculated as

$a_1 = \frac{M}{J_{t1}}$

$= \frac{1161.44}{1359.05}$

= 0.854 rad/sec^2

4 0

3 years ago

Please help right now!!! MARK BRAINLIEST

hram777 [196]

For #2.
(A) The resultant velocity of the boat is the hypotenuse of a right triangle with the sides being the river and boat velocities.
The Pythagorean theorem: h^2 = a^2 + b^2
then you can use: soh cah toa to find any angles.
(B) The river velocity is not opposing the direction of travel, it increases the boats velocity from 2m/s to 2.5m/s

5 0

3 years ago

Which is the BEST example of the kind of mechanics that are studied in sports biometrics?

ikadub [295]

Answer:

-A.

Explanation:

: Hope it's Help:

[correct me if I'm not correct]

5 0

1 year ago