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3 years ago

A triangle has sidesof length 12cm 16cm and 21 cm determine the 3 angles contained in a triangle

Physics

1 answer:

Deffense [45]3 years ago

3 0

Angles will probably be 96° 35° and 49°
I tried to explain it with the working out, just re arrange the cosine rule for angles instead of side lengths.
if you need it explaining at all just message me :)
hoped it helped

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A glass bottle of soda is sealed with a screw cap. The absolute pressure of the carbon dioxide inside the bottle is 1.50 x 105 P

MArishka [77]

Answer:

$F \approx 19.5 N$

Explanation:

From the question we are told that:

Pressure of $P_{CO_2}=1.50 * 105 Pa.$

Bottle cap area $A_b= 4.40 * 10-4 m^2$

Generally the equation for Resultant pressure $P_r$ is give as is mathematically given by

$P_r=P_{CO_2}-P_a$

Where

$P_a=atmospheric\ pressure = 1.013*10^5 pa$

$P_r=1.50 * 105 Pa-1.013*10^5 pa$

$P_r=0.487*10^5 pa$

Generally the equation for Force exerted by screw F is give as is mathematically given by

$F = P*A\\F = 0.487*10^5*4.00*10^-4\\ F = 19.48 N$

$F \approx 19.5 N$

4 0

2 years ago

Air as an ideal gas enters a diffuser operating at steady state at 5 bar, 280 K with a velocity of 510 m/s. The exit velocity is

Nataly [62]

Answer:

Explanation:

Calculating the exit temperature for K = 1.4

The value of $c_p$ is determined via the expression:

$c_p = \frac{KR}{K_1}$

where ;

R = universal gas constant = $\frac{8.314 \ J}{28.97 \ kg.K}$

k = constant = 1.4

$c_p = \frac{1.4(\frac{8.314}{28.97} )}{1.4 -1}$

$c_p= 1.004 \ kJ/kg.K$

The derived expression from mass and energy rate balances reduce for the isothermal process of ideal gas is :

$0=(h_1-h_2)+\frac{(v_1^2-v_2^2)}{2}$ ------ equation(1)

we can rewrite the above equation as :

$0 = c_p(T_1-T_2)+ \frac{(v_1^2-v_2^2)}{2}$

$T_2 =T_1+ \frac{(v_1^2-v_2^2)}{2 c_p}$

where:

$T_1 = 280 K \\ \\ v_1 = 510 m/s \\ \\ v_2 = 120 m/s \\ \\c_p = 1.0004 \ kJ/kg.K$

$T_2= 280+\frac{((510)^2-(120)^2)}{2(1.004)} *\frac{1}{10^3}$

$T_2 = 402.36 \ K$

Thus, the exit temperature = 402.36 K

The exit pressure is determined by using the relation: $\frac{T_2}{T_1} = (\frac{P_2}{P_1})^\frac{k}{k-1}$

$P_2=P_1(\frac{T_2}{T_1})^\frac{k}{k-1}$

$P_2 = 5 (\frac{402.36}{280} )^\frac{1.4}{1.4-1}$

$P_2 = 17.79 \ bar$

Therefore, the exit pressure is 17.79 bar

7 0

3 years ago

On a nice summer day,Kim takes her niece Madison for a walk in her stroller.If they start from rest and accelerate at a rate of

11111nata11111 [884]

2.5m/s

Explanation:

Given parameters:

Initial velocity = 0m/s

Acceleration = 0.5m/s²

time of travel = 5s

Solution:

Final velocity = ?

Solution:

Acceleration can be defined as the change in velocity with time:

Acceleration = $\frac{Final velocity - Initial velocity}{time}$

From the equation above, the unknown is final velocity:

Final velocity - initial velocity = Acceleration x time

since initial velocity = 0

Final velocity = 0.5 x 5 = 2.5m/s

Learn more:

Acceleration brainly.com/question/3820012

#learnwithBrainly

6 0

3 years ago

Rod AB has a diameter of 200mm and rod BC has a diameter of 150mm. Find the required temperature increase to close the gap at C.

Leni [432]

Answer:

$T&=\frac{\sigma_{AB}+\sigma_{BC}}{2\alpha E}$

Explanation:

The given data :-

Diameter of rod AB ( d₁ ) = 200 mm.

Diameter of rod BC ( d₂ ) = 150 mm.

The linear co-efficient of thermal expansion of copper ( ∝ ) = 1.6 × 10⁻⁶ /°C

The young's modulus of elasticity of copper ( E ) = 120 GPa = 120 × 10³ MPa.
Consider the required temperature increase to close the gap at C = T °C
Consider the change in length of the rod = бL

Solution :-

$\begin{aligned}\sum H& =0\\-R_A+R_C&=0\\R_A&=R_C\\R_A&=R\\R_C&=R\\R_{A}&=\text{reaction\:force\:at\:A}\\R_{C}&=\text{reaction\:force\:at\:C}\\\sigma_{AB}&=\text{axial\:stress\:at\:A}\\\sigma_{BC}&=\text{axial\:stress\:at\:B}\\\sigma_{AB}&=\frac{R}{A_{A}}\\&=\frac{R_{A}}{A_{A}}\\\sigma_{BC}&=\frac{R_{B}}{A_{B}}\\&=\frac{R}{A_{B}}\\\frac{\sigma_{AB}}{\sigma_{BC}}&=\frac{A_{B}}{A_{B}}\\&=\frac{\frac{\pi}{4}\cdot 150^{2}}{\frac{\pi}{4}\cdot 200^{2}}\\&=\frac{9}{16}\end{aligned}$

$\begin{aligned}\delta L&= (\delta L _{thermal} +\delta L_{axial})_{AB} + ( \delta L _{thermal} +\delta L_{axial})_{BC}\\0& = (\delta L _{thermal} +\delta L_{axial})_{AB} + ( \delta L _{thermal} +\delta L_{axial})_{BC}\\&=\left[\alpha\:T\:L+\left(\frac{-RL}{AE}\right)\right]_{AB}+\left[\alpha\:T\:L+\left(\frac{-RL}{AE}\right)\right]_{BC}\\&=2\:\alpha\:T\:L-\frac{L}{E}(\sigma_{AB}+\sigma_{BC})\\T&=\frac{\sigma_{AB}+\sigma_{BC}}{2\alpha E}\end{aligned}$

5 0

3 years ago

Which of the following types of particles can be categorized into different kind of elements?

KatRina [158]

The answer would be atoms

4 0

3 years ago

Read 2 more answers