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2 years ago

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An oil has a kinematic viscosity of 1.25 × 10^-4 and a specific gravity of 0.8. What is the viscosity of this oil in SI unit of

kg/m-s? In the unit of centipoise?

1 answer:

emmainna [20.7K]2 years ago

5 0

Step 1 of 3
Calculate the density of oil as follows:

Here, SG is the specific gravity and is the density of standard fluid (water).
Substitute 0.8 for SG and for .

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A spearfisher stands in shallow water and sees a fish a few feet in front of her. She throws her spear directly toward the posit

DENIUS [597]

Answer:

the spear will end up above the fish relative to the actual position of the fish.

Explanation:

due to refraction of light coming from the fish the fish will appear slightly above from its real position

So due to this refraction the spearfisher will throw the spear directly at the image of the fish due to which it will not reach the position of fish but it will reach the position above the fish.

So here we can say that the spear will end up above the fish relative to the actual position of the fish

5 0

3 years ago

A loaded 375 kg toboggan is traveling on smooth horizontal snow at 4.50 m/s when it suddenly comes to a rough region. The region

zmey [24]

Answer:

a) The average friction force exerted on the toboggan is 653.125 newtons, b) The rough region reduced the kinetic energy of the toboggan in 92.889 %, c) The speed of the toboggan is reduced in 73.333 %.

Explanation:

a) Given the existence of non-conservative forces (friction between toboggan and ground), the motion must be modelled by means of the Principle of Energy Conservation and the Work-Energy Theorem, since toboggan decrease its speed (associated with due to the action of friction. Changes in gravitational potential energy can be neglected due to the inclination of the ground. Then:

$K_{1} = K_{2} + W_{f}$

Where:

$K_{1}$ , $K_{2}$ are the initial and final translational kinetic energies of the tobbogan, measured in joules.

$W_{f}$ - Dissipated work due to friction, measured in joules.

By applying definitions of translation kinetic energy and work, the expression described above is now expanded and simplified:

$f\cdot \Delta s = \frac{1}{2}\cdot m \cdot (v_{1}^{2}-v_{2}^{2})$

Where:

$f$ - Friction force, measured in newtons.

$\Delta s$ - Distance travelled by the toboggan in the rough region, measured in meters.

$m$ - Mass of the toboggan, measured in kilograms.

$v_{1}$ , $v_{2}$ - Initial and final speed of the toboggan, measured in meters per second.

The friction force is cleared:

$f = \frac{m\cdot (v_{1}^{2}-v_{2}^{2})}{2\cdot \Delta s}$

If $m = 375\,kg$ , $v_{1} = 4.50\,\frac{m}{s}$ , $v_{2} = 1.20\,\frac{m}{s}$ and $\Delta s = 5.40 \,m$ , then:

$f = \frac{(375\,kg)\cdot \left[\left(4.50\,\frac{m}{s} \right)^{2}-\left(1.20\,\frac{m}{s}\right)^{2}\right]}{2\cdot (5.40\,m)}$

$f = 653.125\,N$

The average friction force exerted on the toboggan is 653.125 newtons.

b) The percentage lost by the kinetic energy of the tobbogan due to friction is given by the following expression, which is expanded and simplified afterwards:

$\% K_{loss} = \frac{K_{1}-K_{2}}{K_{1}}\times 100\,\%$

$\% K_{loss} = \left(1-\frac{K_{2}}{K_{1}} \right)\times 100\,\%$

$\% K_{loss} = \left(1-\frac{\frac{1}{2}\cdot m \cdot v_{2}^{2}}{\frac{1}{2}\cdot m \cdot v_{1}^{2}} \right)\times 100\,\%$

$\% K_{loss} = \left(1-\frac{v_{2}^{2}}{v_{1}^{2}} \right)\times 100\,\%$

$\%K_{loss} = \left[1-\left(\frac{v_{2}}{v_{1}}\right)^{2} \right]\times 100\,\%$

If $v_{1} = 4.50\,\frac{m}{s}$ and $v_{2} = 1.20\,\frac{m}{s}$ , then:

$\%K_{loss} = \left[1-\left(\frac{1.20\,\frac{m}{s} }{4.50\,\frac{m}{s} }\right)^{2} \right]\times 100\,\%$

$\%K_{loss} = 92.889\,\%$

The rough region reduced the kinetic energy of the toboggan in 92.889 %.

c) The percentage lost by the speed of the tobbogan due to friction is given by the following expression:

$\% v_{loss} = \frac{v_{1}-v_{2}}{v_{1}}\times 100\,\%$

$\% v_{loss} = \left(1-\frac{v_{2}}{v_{1}} \right)\times 100\,\%$

If $v_{1} = 4.50\,\frac{m}{s}$ and $v_{2} = 1.20\,\frac{m}{s}$ , then:

$\% v_{loss} = \left(1-\frac{1.20\,\frac{m}{s} }{4.50\,\frac{m}{s} } \right)\times 100\,\%$

$\%v_{loss} = 73.333\,\%$

The speed of the toboggan is reduced in 73.333 %.

5 0

3 years ago

Consider a concave spherical mirr or that has focal length f = +19.5 cm.

lidiya [134]

The distance of an object from the mirror's vertex if the image is real and has the same height as the object is 39 cm.

<h3>What is concave mirror?</h3>

A concave mirror has a reflective surface that is curved inward and away from the light source.

Concave mirrors reflect light inward to one focal point and it usually form real and virtual images.

<h3>Object distance of the concave mirror</h3>

Apply mirrors formula as shown below;

1/f = 1/v + 1/u

where;

f is the focal length of the mirror
v is the object distance
u is the image distance

when image height = object height, magnification = 1

u/v = 1

v = u

Substitute the given parameters and solve for the distance of the object from the mirror's vertex

1/f = 1/v + 1/v

1/f = 2/v

v = 2f

v = 2(19.5 cm)

v = 39 cm

Thus, the distance of an object from the mirror's vertex if the image is real and has the same height as the object is 39 cm.

Learn more about concave mirror here: brainly.com/question/27841226

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7 0

1 year ago

Which of the following is not a vector quantity?

baherus [9]

Answer: The answer is density

3 0

3 years ago

Two birds have the same mass but one has bigger feet. How would pressure that they produce on the ground be different ?

Mama L [17]

Answer:

the pressure per square inch is greater from the smaller feet.

Explanation:

different weight distribution

3 0

3 years ago