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attashe74 [19]
3 years ago
10

A ship travels 84 km on a bearing of 17°, and then travels on a bearing of 107° for 135 km. Find the distance of the end of the

trip from the starting point, to the nearest kilometer.

Physics
1 answer:
Neporo4naja [7]3 years ago
6 0

Answer:

159km

Explanation:

Please view the diagram below;

The distance x of the ship from the staring point could be simply obtained by Pythagoras's theorem as follows;

x^2=135^2+84^2\\x^2=18225+7056\\x^2=25281\\x=\sqrt{25281}\\ x=159km

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Calculate the de Broglie wavelength of (a) a mass of 1.0 g traveling at 1.0 m s−1 , (b) the same, traveling at 1.00 × 105 km s−1
lesantik [10]

Answer:

a)\lambda=6.63\times10^{-31}m

b)\lambda=6.63\times10^{-39}m

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d)\lambda=4.03\times10^{-36}m

e)λ=∞

Explanation:

De Broglie discovered that an electron or other mass particles can have a wavelength associated, and that wavelength (λ) is:

\lambda=\frac{h}{P}=\frac{h}{mv}

with h the Plank's constant (6.63\times10^{-34}\frac{m^{2}kg}{s}) and P the momentum of the object that is mass (m) times velocity (v).

a)\lambda=\frac{6.63\times10^{-34}}{(1.0\times10^{-3}kg*1.0)}

\lambda=6.63\times10^{-31}m

b)\lambda=\frac{6.63\times10^{-34}}{(1.0\times10^{-3}*(1.00\times10^{8}))}

\lambda=6.63\times10^{-39}m

c)\lambda=\frac{6.63\times10^{-34}}{(6.65\times10^{-27}*1000)}

\lambda=9.97\times10^{-11}m

d)\lambda=\frac{6.63\times10^{-34}}{(74*2.22)}

\lambda=4.03\times10^{-36}m

e) \lambda=\frac{6.63\times10^{-34}}{(74*0)}

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6 0
3 years ago
Plane a travels at 900km/h and plane b travels at 250/5.which plane travels faster
vesna_86 [32]

Explanation:

We have,

Speed of plane a is 900 km/h

Plane b is moving at a rate of \dfrac{250\ km}{5\ h}=50\ km/h

It is required to find which plane is faster. To find which plane is faster, we need to compare their speeds.

Speed of a plane a is 900 km/h and that of plane b is 50 km/h. So, we can say that plane a is moving faster.

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