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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

#SPJ1

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In young’s double slit experiment, the measured fringe width is 0.5 mm for a Sodium light of 589 nm at a distance of 1.5 m. A br
Levart [38]

Answer:

(A).  The order of the bright fringe is 6.

(B). The width of the bright fringe is 3.33 μm.

Explanation:

Given that,

Fringe width d = 0.5 mm

Wavelength = 589 nm

Distance of screen and slit D = 1.5 m

Distance of bright fringe y = 1 cm

(A) We need to calculate the order of the bright fringe

Using formula of wavelength

\lambda=\dfrac{dy}{mD}

m=\dfrac{d y}{\lambda D}

Put the value into the formula

m=\dfrac{1\times10^{-2}\times0.5\times10^{-3}}{589\times10^{-9}\times1.5}

m=5.65 = 6

(B). We need to calculate the width of the bright fringe

Using formula of width of fringe

\beta=\dfrac{yd}{D}

Put the value in to the formula

\beta=\dfrac{1\times10^{-2}\times0.5\times10^{-3}}{1.5}

\beta=3.33\times10^{-6}\ m

\beta=3.33\ \mu m

Hence, (A).  The order of the bright fringe is 6.

(B). The width of the bright fringe is 3.33 μm.

3 0
3 years ago
a sydiver jumps out of a plane. She fall dowward at a very fast speed. Which force is acting upon it?
UNO [17]

Answer:

There will be two forces acting on her: Gravitational force and Air resisitence

6 0
3 years ago
Two balls of clay, with masses M1 = 0.49 kg and M2 = 0.47 kg, are thrown at each other and stick when they collide. Mass 1 has a
malfutka [58]

Answer:

a) p_i=1.568\hat{i}+0.752 \hat{j}

b) v_{fx}=1.668\ m.s^{-1}

c) v_{fy}=0.7999\ m.s^{-1}

Explanation:

Given masses:

m_1=0.49\ kg

m_2=0.47\ kg

Velocity of mass 1, v_1=3.2 \hat{i}\ m.s^{-1}

Velocity of mass 2, v_2=1.6 \hat{j}\ m.s^{-1}

a)

Initial momentum:

p_i=m_1.v_1+m_2.v_2

p_i=0.49\times 3.2 \hat{i}+0.47\times 1.6 \hat{j}

p_i=1.568\hat{i}+0.752 \hat{j}

b)

magnitude of initial momentum:

p_i=\sqrt{1.568^2+0.752 ^2}

p_i=1.739\ kg.m.s^{-1}

From the conservation of momentum:

p_f=p_i

m_f.v_f=1.739

v_f=\frac{1.739}{0.49+0.47}

v_f=1.85\ m.s^{-1} is the magnitude of final velocity.

Direction of final velocity will be in the direction of momentum:

tan\theta=\frac{0.752 }{1.568}

\theta=25.62^{\circ}

\therefore v_{fx}=1.85\ cos25.62^{\circ}

v_{fx}=1.668\ m.s^{-1}

c)

Vertical component of final velocity:

v_{fy}=1.85\ sin 25.62^{\circ}

v_{fy}=0.7999\ m.s^{-1}

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3 years ago
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Answer:

A prism is a clear crystal that refracts light

Explanation:

7 0
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calculate how much charge passes through the LED in 1 hour if the current flowing in the circuit below is 350mA
Morgarella [4.7K]

Given the the current flowing in the circuit and the elapsed time, the charge that passes through the LED is 1260 Coulombs.

<h3>What is Current?</h3>

Current is simply the rate of flow of charged particles i.e electrons caused by EMF or voltage.

If a charge passes through the cross-section of a conductor in a given time, the current I is expressed as;

I = Q/t

Where Q is the charge and t is time elapsed.

Given the data in the question;

  • Time elapsed t = 1hr = 3600s
  • Current I = 350mA = 0.35A
  • Charge Q = ?

We substitute our given values into the expression above to determine the charge.

I = Q/t

Q = I × t

Q = 0.35A × 3600s

Q = 1260C

Therefore,  given the the current flowing in the circuit and the elapsed time, the charge that passes through the LED is 1260 Coulombs.

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