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pshichka [43]
3 years ago
9

A fisherman casts his lure at an angle of 33 degrees above the horizontal. The lure reaches a maximum height of 2.3 m. Assuming

no frictional forces, what was the initial velocity the fisherman gave the lure when he cast it?
12.34 m/s
28.45 m/s
34.91 m/s
21.29 m/s
Physics
1 answer:
larisa86 [58]3 years ago
5 0

Answer:

did you figure out the answer

Explanation:

??

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Be sure to answer all parts. Compare the wavelengths of an electron (mass = 9.11 × 10−31 kg) and a proton (mass = 1.67 × 10−27 k
Harrizon [31]

Explanation:

Given that,

(a) Speed, v=6.66\times 10^6\ m/s

Mass of the electron, m_e=9.11\times 10^{-31}\ kg

Mass of the proton, m_p=1.67\times 10^{-27}\ kg

The wavelength of the electron is given by :

\lambda_e=\dfrac{h}{m_ev}

\lambda_e=\dfrac{6.63\times 10^{-34}}{9.11\times 10^{-31}\times 6.66\times 10^6}

\lambda_e=1.09\times 10^{-10}\ m

The wavelength of the proton is given by :

\lambda_p=\dfrac{h}{m_p v}

\lambda_p=\dfrac{6.63\times 10^{-34}}{1.67\times 10^{-27}\times 6.66\times 10^6}

\lambda_p=5.96\times 10^{-14}\ m

(b) Kinetic energy, K=1.71\times 10^{-15}\ J

The relation between the kinetic energy and the wavelength is given by :

\lambda_e=\dfrac{h}{\sqrt{2m_eK}}

\lambda_e=\dfrac{6.63\times 10^{-34}}{\sqrt{2\times 9.11\times 10^{-31}\times 1.71\times 10^{-15}}}

\lambda_e=1.18\times 10^{-11}\ m

\lambda_p=\dfrac{h}{\sqrt{2m_pK}}

\lambda_p=\dfrac{6.63\times 10^{-34}}{\sqrt{2\times 1.67\times 10^{-27}\times 1.71\times 10^{-15}}}

\lambda_p=2.77\times 10^{-13}\ m

Hence, this is the required solution.

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During which time does Jamal have the greatest momentum?
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4 0
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If the mass of an object is 44 kilograms and its velocity is 10 meters per second east, how much Kinetic Energy does it have?
olya-2409 [2.1K]
Given: Mass m = 44 Kg;      Velocity v = 10 m/s

Required: Kinetic energy K.E = ?

Formula: K.E = 1/2 mv²

               K.E 1/2 (44 Kg)(10 m/s)²

               K.E = 2,200 Kg.m²/s²

               K.E = 2,200 J Answer is A
8 0
3 years ago
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