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Veronika [31]
3 years ago
15

If a man walks 17m east and then 17m south , the magnitude of the man's displacement is

Physics
2 answers:
alexandr402 [8]3 years ago
5 0

c^2= a^2+b^2

c^2 =(17)^2+(17)^2

c^2=289+289

c^2=578

c=24.04m


In-s [12.5K]3 years ago
4 0
Hey here is your answer u r looking for
24.041m
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Need help ASAP!! Thank you..
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3 years ago
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The boundary between the crust and mantle is marked by a seismic-velocity discontinuity called
Minchanka [31]

The boundary between the crust and mantle is marked by a seismic-velocity discontinuity is called Mohorovicic discontinuity.

Mohorovicic discontinuity was discovered by Andrija Mohorovicic in 1909 who was a Croatian seismologist. He realized that the velocity of a seismic wave is related to the material's density where it is moving through. He decoded that the acceleration of the seismic waves that are observed within outer shell of the earth is a compositional change. Thus, the acceleration should be caused by a material of higher density.

7 0
3 years ago
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Heat in a substance is an example of _____ energy?
Natali5045456 [20]
<span>Heat in a substance is still an example of heat or thermal energy. Heat or thermal energy is further defined as the amount of available molecules of kinetic energy transferred in a specific amount of time. Heat is nothing if it is not transferred from one object to another. There are many uses of heat energy. When we are cooking using our stoves, there is heat energy. When we pour hot water onto our cup of coffee, there is heat energy.</span>
3 0
3 years ago
Please help!!
kolezko [41]

Answer:

0.0469

Explanation:

v=1.50

r=32.0

i=?

equation

i=v/r

i=1.50/32.0

i=0.046875

round to 3 sig figs

i=0.0469

5 0
3 years ago
A block-spring system consists of a spring with constant k = 475 N/m attached to a 2.50 kg block on a frictionless surface. The
vovikov84 [41]

Explanation:

It is given that,

Spring constant of the spring, k = 475 N/m

Mass of the block, m = 2.5 kg

Elongation in the spring from equilibrium, x = 5.5 cm

(a) We know that the maximum elongation in the spring is called its amplitude. So, the amplitude for the resulting oscillation is 5.5 cm.

(b) Let \omega is the angular frequency. It is given by :

\omega=\sqrt{\dfrac{k}{m}}

\omega=\sqrt{\dfrac{475}{2.5}}

\omega=13.78\ rad/s

(c) Let T is the period. It is given by :

T=\dfrac{2\pi}{\omega}

T=\dfrac{2\pi}{13.78}

T = 0.45 s

(d) Frequency,

f=\dfrac{1}{T}

f=\dfrac{1}{0.45}

f = 2.23 Hz

(e) Let v is the maximum value of the block's velocity. It is given by :

v_{max}=\omega\times A

v_{max}=13.78\times 5.5\times 10^{-2}

v_{max}=0.757\ m/s

The value of acceleration is given by :

a=\omega^2A

a=(13.78)^2\times 5.5\times 10^{-2}

a=10.44\ m/s^2

Hence, this is the required solution.

7 0
3 years ago
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