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2 years ago

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Physics

1 answer:

mezya [45]2 years ago

6 0

Answer:

uh what are you saying

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What is the study of kinematics based upon ?<br><br>thankyou ~

NikAS [45]

Kinematics is the study of the motion of a system of bodies without directly considering the forces or potential fields affecting the motion. In other words, kinematics examines how momentum and energy are shared among interacting bodies.

8 0

2 years ago

A counterflow double-pipe heat exchanger is used to heat water from 20°C to 80°C at a rate of 1.2 kg/s. The heating is to be com

bixtya [17]

Answer:L=109.16 m

Explanation:

Given

initial temperature $=20^{\circ}C$

Final Temperature $=80^{\circ}C$

mass flow rate of cold fluid $\dot{m_c}=1.2 kg/s$

Initial Geothermal water temperature $T_h_i=160^{\circ}C$

Let final Temperature be T

mass flow rate of geothermal water $\dot{m_h}=2 kg/s$

diameter of inner wall $d_i=1.5 cm$

$U_{overall}=640 W/m^2K$

specific heat of water $c=4.18 kJ/kg-K$

balancing energy

Heat lost by hot fluid=heat gained by cold Fluid

$\dot{m_c}c(T_h_i-T_h_e)= \dot{m_h}c(80-20)$

$2\times (160-T)=1.2\times (80-20)$

$160-T=36$

$T=124^{\circ}C$

As heat exchanger is counter flow therefore

$\Delta T_1=160-80=80^{\circ}C$

$\Delta T_2=124-20=104^{\circ}C$

$LMTD=\frac{\Delta T_1-\Delta T_2}{\ln (\frac{\Delta T_1}{\Delta T_2})}$

$LMTD=\frac{80-104}{\ln \frac{80}{104}}$

$LMTD=91.49^{\circ}C$

heat lost or gain by Fluid is equal to heat transfer in the heat exchanger

$\dot{m_c}c(80-20)=U\cdot A\cdot$ (LMTD)

$A=\frac{1.2\times 4.184\times 1000\times 60}{640\times 91.49}=5.144 m^2$

$A=\pi DL=5.144$

$L=\frac{5.144}{\pi \times 0.015}$

$L=109.16 m$

6 0

2 years ago

With what force will a car hit a tree if the car has a mass of 3,415 kg and it is accelerating at

weeeeeb [17]

Answer:

$F= 17,075\ N$

Explanation:

When the car is under an accelerating force and hits a tree, the instant force received by the tree is the same force that is accelerating the car.

The accelerating force can be calculated using Newton's second law:

$F=m\cdot a$

Where m is the mass of the car and a is the acceleration.

$F=3,415\ kg\cdot 5\ m/s^2$

$\boxed{F= 17,075\ N}$

3 0

2 years ago

A triangular plate with height 6 ft and a base of 7 ft is submerged vertically in water so that the top is 2 ft below the surfac

xenn [34]

Answer:

$62.5\int\limits^6_0 {(\frac{7}{6}y^{2}-\frac{49}{3}y+56) } \, dy = 7875 lb$

Explanation:

For this problem to be easier to calculate, we can represent the triangle as a right triangle whose right angle is located at the origin of a coordinate system. (See picture attached).

With this disposition of the triangle, we can start finding our integral. The hydrostatic force can be set as an integral with the following shape:

$\int\limits^a_b$ γhxdy

we know that γ=62.5 lb/ $ft^{3}$

from the drawing, we can determine the height (or depth under the water) of each differential area is given by:

h=8-y

x can be found by getting the equation of the line, which we'll get by finding the slope of the line and using one of the points to complete the equation:

$m=\frac{y_{2}-y_{1}}{x_{2}-x{1}}$

when substituting the x and y-values given on the graph, we get that the slope is:

$m=\frac{0-6}{7-0}=-\frac{6}{7}$

once we got this slope, we can substitute it in the point-slope form of the equation:

$y_{2}-y_{1}=m(x_{2}-x_{1})$

which yields:

$y-6=-\frac{6}{7}(x-0)$

which simplifies to:

$y-6=-\frac{6}{7}x$

we can now solve this equation for x, so we get that:

$x=-\frac{7}{6}y+7$

with this last equation, we can substitute everything into our integral, so it will now look like this:

$\int\limits^6_0{(62.5)(8-y)(-\frac{7}{6}y+7)}\,dy$

Now that it's all written in terms of y we can now simplify it, so we get:

$62.5\int\limits^6_0 {(\frac{7}{6}y^{2}-\frac{49}{3}y+56)}dy$

we can now proceed and evaluate it.

When using the power rule on each of the terms, we get the integral to be:

$62.5[\frac{7}{18}y^{3}-\frac{49}{6}y^{2}+56y]^{6}_{0}$

By using the fundamental theorem of calculus we get:

$62.5[(\frac{7}{18}(6)^{3}-\frac{49}{6}(6)^{2}+56(6))-(\frac{7}{18}(0)^{3}-\frac{49}{6}(0)^{2}+56(0))]$

When solving we get:

$62.5\int\limits^6_0 {(\frac{7}{6}y^{2}-\frac{49}{3}y+56) } \, dy = 7875 lb$

6 0

3 years ago

Convert the number from scientific into standard notation: 5.9 x 10-2

guapka [62]

Move the decimal point to:
Left : (if the exponent of ten is a negative number -) ... OUR CASE HERE (-2)
or to
Right : (if the exponent is positive +).

You should move the point as many times as the exponent indicates.
Do not write the power of ten anymore.

So, standard form is:
Two points to the left {Exponent of Ten is Negative (-2)}
0.059 ... (without the 10)

6 0

2 years ago