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andre [41]
1 year ago
7

A hiker walks due East for a distance of 6 km

Physics
1 answer:
solniwko [45]1 year ago
3 0

Resultant displacement between the base camp and the cave will be 7.211 km

Displacement is a vector quantity that refers to "how far out of place an object is"; it is the object's overall change in position.

Her resultant displacement can be calculated using Pythagoras theorem

H^{2} = P^{2} + B^{2}

     = 4^{2} + 6^{2}

     = 52

H = \sqrt{52} = 7.211 km

Resultant displacement between the base camp and the cave will be 7.211 km

To learn more about displacement here :

brainly.com/question/13271165

#SPJ1

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7 0
3 years ago
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A 4.00-kg particle moves along the x axis. Its position varies with time according to x 5 t 1 2.0t 3, where x is in meters and t
White raven [17]

Answer:

Explanation:

Given equation is ,

x =t + 2 t³ ,

dx/dt = velocity ( v ) = 1 + 6 t²

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b ) Acceleration = dv /dt = 12 t .

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Power = force x velocity = 48 t x ( 1 + 6 t²). = 48 t + 288 t³ )

work done = ∫ F dx =∫ 48 t x( 1 + 6t² )dt ; =  [48t²/2 + 48 x 6 x t³ /3 = 24 t² + 96 t³ )]₀² =  864 J

6 0
2 years ago
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What effort force will be required to lift the 20 N object using the pulley above?
Umnica [9.8K]

Answer:

I think it is 5N.

Explanation:

4 0
2 years ago
When light of wavelength 240 nm falls on a cobalt surface, electrons having a maximum kinetic energy of 0.17 eV are emitted. Fin
dusya [7]

Answer:

(a) 5.04 eV (B) 248.14 nm (c) 1.21\times 10^{15}Hz

Explanation:

We have given Wavelength of the light  \lambda = 240 nm

According to plank's rule ,energy of light

E = h\nu = \frac{hc}{}\lambda

E = h\nu = \frac{6.67\times 10^{-34} J.s\times 3\times 10^{8}m/s}{ 240\times 10^{-9} m\times 1.6\times 10^{-19}J/eV}= 5.21 eV

Maximum KE of emitted electron i= 0.17 eV

Part( A) Using Einstien's equation

E = KE_{max}+\Phi _{0}, here \Phi _0 is work function.

\Phi _{0}=E - KE_{max}= 5.21 eV-0.17 eV = 5.04 eV

Part( B) We have to find cutoff wavelength

\Phi _{0} = \frac{hc}{\lambda_{cuttoff}}

\lambda_{cuttoff}= \frac{hc}{\Phi _{0} }

\lambda_{cuttoff}= \frac{6.67\times 10^{-34} J.s\times 3\times 10^{8}m/s}{5.04 eV\times 1.6\times 10^{-19}J/eV }=248.14 nm

Part (C) In this part we have to find the cutoff frequency

\nu = \frac{c}{\lambda_{cuttoff}}= \frac{3\times 10^{8}m/s}{248.14 \times 10^{-19} m }= 1.21\times 10^{15} Hz

5 0
3 years ago
A box that weighs 5.00×10^2 N is sliding down a ramp at a constant speed. The angle the ramp makes with the horizontal is 25°. W
maxonik [38]

Answer:

0.466 (3 sig. fig.)

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5 0
2 years ago
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