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rusak2 [61]
2 years ago
6

Three sticks are arranged to form a right triangle. If the lengths of the three sticks are 0.47 m, 0.62 m and 0.78 m, what are t

he three angles of the triangle?
Physics
1 answer:
Svetradugi [14.3K]2 years ago
6 0

The angles in the triangle are 91 degrees, 53 degrees and 36 degrees respectively.

<h3>What is the cosine rule?</h3>

From the cosine rule we know that;

c^2 = a^2 + b^2 - 2abcosC

Since;

a = 0.47 m

b =  0.62 m

c =  0.78 m

Then;

(0.78)^2 = (0.47)^2 + (0.62)^2 - 2(0.47 * 0.62)cosC

0.61 = 0.22 + 0.38 - 0.58 cosC

0.61 - ( 0.22 + 0.38) = - 0.58 cosC

0.01 = - 0.58 cosC

C = cos-1(0.01/-0.58)

C = 91 degrees

Using the sine rule;

b/Sin B = c/Sin C

0.62/sinB = 0.78/sin 91

0.62/Sin B = 0.78

B = sin-1 (0.62//0.78)

B = 53 degrees

Angle A is obtained from the sum of angles in a triangle;

180 - (91 + 53)

A = 36 degrees

Learn more about triangle:brainly.com/question/2773823

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a0.155 kg arrow is shot from ground level, upward at 31.4 m/s. what is its kinetic energy (ke) when it is 30.0 m above the groun
swat32

The kinetic energy (KE) of a 0.155 kg arrow that is shot from ground level, upward at 31.4 m/s, when it is 30.0 m above the ground is 30.85 J

Assuming air friction is negligible,

a = - 9.8 m / s²

u = 31.4 m / s

s = 30 m

v² = u² + 2 a s

v² = 31.4² + ( 2 * - 9.8 * 30 )

v² = 985.96 - 588

v² = 397.96 m / s

KE = 1 / 2 m v²

KE = 1 / 2 * 0.155 * 397.96

KE = 0.0775 * 397.96

KE = 30.85 J

Therefore, the kinetic energy ( KE ) when it is 30.0 m above the ground is 30.85 J

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5 0
2 years ago
Why was Thompson's Plum pudding model a significant departure from previous models?
PSYCHO15rus [73]

Answer:

Thomson's cathode-ray tube experiments led him to develop the plum-pudding model, which stated that each atom had positively charged particles spread throughout its negatively charged matter. Reword the statement so it is true. ... More alpha particles were deflected than he expected.

Explanation:

Well.. I hope it helps you..

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5 0
3 years ago
A simple hydraulic lift is made by fitting a piston attached to a handle into a 3.0-cm diameter cylinder. The cylinder is connec
stiv31 [10]

Answer:

Approximately 3.1 \times 10^4 \; \rm N (assuming that the acceleration due to gravity is g = 9.81\; \rm kg \cdot N^{-1}.)

Explanation:

Let A_1 denote the first piston's contact area with the fluid. Let A_2 denote the second piston's contact area with the fluid.

Similarly, let F_1 and F_2 denote the size of the force on the two pistons. Since the person is placing all her weight on the first piston:

F_1 = W = m \cdot g = 50\; \rm kg \times 9.81 \; \rm kg \cdot N^{-1} =495\; \rm N.  

Since both pistons fit into cylinders, the two contact surfaces must be circles. Keep in mind that the area of a square is equal to \pi times its radius, squared:

  • \displaystyle A_1 = \pi \times \left(\frac{1}{2} \times 3.0\right)^2 = 2.25\, \pi\;\rm cm^{2}.
  • \displaystyle A_2 = \pi \times \left(\frac{1}{2} \times 24\right)^2 = 144\, \pi\;\rm cm^{2}.

By Pascal's Law, the pressure on the two pistons should be the same. Pressure is the size of normal force per unit area:

\displaystyle P = \frac{F}{A}.

For the pressures on the two pistons to match:

\displaystyle \frac{F_1}{A_1} = \frac{F_2}{A_2}.

F_1, A_1, and A_2 have all been found. The question is asking for F_2. Rearrange this equation to obtain:

\displaystyle F_2 = \frac{F_1}{A_1} \cdot A_2 = F_1 \cdot \frac{A_2}{A_1}.

Evaluate this expression to obtain the value of F_2, which represents the force on the piston with the larger diameter:

\begin{aligned}F_2 &= F_1 \cdot \frac{A_2}{A_1} \\ &= 495\; \rm N \times \frac{2.25\, \pi\; \rm cm^2}{144\, \pi \; \rm cm^2} \approx 3.1 \times 10^4\; \rm N\end{aligned}.

6 0
3 years ago
Which of the following is the best explanation of work?
Mariulka [41]
The correct option is C
7 0
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what is the vapor pressure of water if the total vapor pressure is 1.5 atm and the pressure of oxygen is 0.2 atm?
andreyandreev [35.5K]

Since the total pressure of each gas is the sum of the partial pressures of all the individual gases, the vapor pressure of oxygen is  1.3 atm.

From Dalton's law of partial pressure, the total pressure is the sum of the individual partial pressures of the gases;

PT = PA + PB + PC + --------

Given that;

Vapor pressure of oxygen = 0.2 atm

Total pressure = 1.5 atm

Vapor pressure of oxygen = ?

Partial pressure of oxygen =  1.5 atm - 0.2 atm = 1.3 atm

The vapor pressure of oxygen is  1.3 atm.

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8 0
3 years ago
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