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Nina [5.8K]
3 years ago
9

Use appropriate units and significant figures.  USE THE LAW OF COSINES AND LAW OF SINES.

Physics
1 answer:
N76 [4]3 years ago
3 0

Answer:

The resultant velocity is 86.1 mi/h.    

Explanation:

The law of cosines is given by:

c^{2} = a^{2} + b^{2} - 2abcos(\theta)

Where:

c: is the resultant velocity =?

a: is the velocity of the plane = 75.0 mi/h

b: is the velocity of the wind = 15.0 mi/h  

θ: is the angle between "a" and "b"                          

The angle between "a" and "b" can be found as follows:

\theta = 180.0 - 46.0 = 134.0 ^{\circ}

Now, by using the law of cosines we have:

c^{2} = (75.0)^{2} + (15.0)^{2} - 2*75.0*15.0*cos(134.0) = 7413.0

c = 86.1 mi/h    

Therefore, the resultant velocity is 86.1 mi/h.    

The law of sines is:

\frac{a}{sin(\gamma)} = \frac{b}{sin(\alpha)} = \frac{c}{sin(\theta)}

Where:

γ: is the angle between "b" and "c"

α: is the angle between "a" and "c"

So, if we want to find "c" by using the law of sines, we need to know another angle besides θ (γ or α), and the statement does not give us.

I hope it helps you!        

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Answer:

Explanation:

plate separation = 2.3 x 10⁻³ m

capacity C₁ = ε A / d

= ε A / 2.3 x 10⁻³

C₂ = ε A / 1.15 x 10⁻³

\frac{C_2}{C_1} = \frac{2.3}{1.15}

a ) when charge remains constant

energy = \frac{q^2}{2C}

q is charge and C is capacity

energy stored initially E₁= \frac{q^2}{2C_1}

energy stored finally E₂ = \frac{q^2}{2C_2}

\frac{E_1}{E_2} = \frac{C_2}{C_1} = \frac{2.3}{1.15}

E_2 = \frac{1.15}{2.3 } \times E_1

= \frac{1.15}{2.3 } \times 8.38

= 4.19 J

b )

In this case potential diff remains constant

energy of capacitor = 1/2 C V²

energy is proportional to capacity as V is constant .

\frac{E_2}{E_1} = \frac{C_2}{C_1}

\frac{E_2}{8.38} = \frac{2.3}{1.15}

E_2 = 16.76 .

8 0
3 years ago
A ball bearing of radius of 1.5 mm made of iron of density
Serjik [45]

Answer:

\boxed{\sf Viscosity \ of \ glycerine \ (\eta) = 14.382 \ poise}

Given:

Radius of ball bearing (r) = 1.5 mm = 0.15 cm

Density of iron (ρ) = 7.85 g/cm³

Density of glycerine (σ) = 1.25 g/cm³

Terminal velocity (v) = 2.25 cm/s

Acceleration due to gravity (g) = 980.6 cm/s²

To Find:

Viscosity of glycerine (\sf \eta)

Explanation:

\boxed{ \bold{v =  \frac{2}{9}  \frac{( {r}^{2} ( \rho -  \sigma)g)}{ \eta} }}

\sf \implies \eta =  \frac{2}{9}  \frac{( {r}^{2}( \rho -  \sigma)g )}{v}

Substituting values of r, ρ, σ, v & g in the equation:

\sf \implies \eta =  \frac{2}{9}  \frac{( {(0.15)}^{2}  \times  (7.85 - 1.25) \times 980.6)}{2.25}

\sf \implies \eta =  \frac{2}{9}  \frac{(0.0225 \times 6.6 \times 980.6)}{2.25}

\sf \implies \eta =  \frac{2}{9}  \times  \frac{145.6191}{2.25}

\sf \implies \eta =  \frac{2}{9}  \times 64.7196

\sf \implies \eta =  2 \times 7.191

\sf \implies \eta =  14.382 \: poise

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3 years ago
A 1500-kg car accelerates from 0 to 25 m/s in 7.0s with negligible friction and air resistance. What is the average power delive
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Answer:

90 hp

Explanation:

Power = work / time

P = ½ (1500 kg) (25 m/s)² / 7.0 s

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P = 90 hp

4 0
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A 300 g glass thermometer initially at 23 ◦C is put into 236 cm3 of hot water at 87 ◦C. Find the final temperature of the thermo
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Answer:

74^{\circ} C

Explanation:

We are given that

Mass of glass,m=300 g

T_1=23^{\circ}

Volume,V=236cm^3

Mass of water=density\times volume=1\times 236=236 g

Density of water=1g/cm^3

Temperature of hot water,T=87^{\circ}

Specific heat of glass,C_g=0.2cal/g^{\circ}C

Specific heat of water,C_w=1 cal/g^{\circ}C

Q_{glass}=m_gC_g(T_f-T_1)=300\times 0.2(T_f-23)

Q_{water}=m_wC_w(T_f-T)=236\times 1(T_f-87)

Q_{glass}+Q_{water}=0

300\times 0.2(T_f-23)+236\times 1(T_f-87)

60T_f-1380+236T_f-20532=0

296T_f=20532+1380=21912

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