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Alborosie
3 years ago
13

Add vectors 7.5 m/s and the vector -2.9 m/s

Physics
1 answer:
densk [106]3 years ago
5 0

How To:

How do you add two vectors?  

To add or subtract two vectors, add or subtract the corresponding components. Let →u=⟨u1,u2⟩ and →v=⟨v1,v2⟩ be two vectors. The sum of two or more vectors is called the resultant. The resultant of two vectors can be found using either the parallelogram method or the triangle method .

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The wavelength of a water wave is 54 m. It has a frequency of 0.03 Hz. What is the velocity of the wave?
dalvyx [7]

Answer:

1.62 m/s

Explanation:

Wavelength of the water wave= 54 m

The frequency is 0.03 Hz

Therefore the velocity can be calculated as follows

Velocity= frequency × wavelength

= 0.03 × 54

= 1.62 m/s

6 0
3 years ago
What is peer review.why is it important​
madam [21]

Acc to

Publisso

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A peer review helps the publisher decide whether a work should be accepted.

6 0
3 years ago
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An object is dropped from a platform 100 feet high. Ignoring wind resistance, what will its speed be when it reaches the ground?
ICE Princess25 [194]

<span>if we assume the origin is at the dropping point and the object is merely dropped and not thrown up or down then y0 = 0 and v0 = 0. The equation reduces to </span>

<span>y = 0 + 0t + ½gt² </span>
<span>y = ½gt² </span>

<span>t = √(2y/g) </span>

<span>in the ft - lb - s system </span>

<span>y = -100 ft </span>
<span>g = -32.2 ft / s² </span>

<span>t = √(2y/g) </span>
<span>t = √(2(-100) / (-32.2)) </span>
<span>t = 2.5 s</span>
7 0
3 years ago
Read 2 more answers
3. Forces do ________ always cause an object to move.<br> need this ASAP
Finger [1]

Answer:

not

Explanation:

5 0
3 years ago
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A basketball player grabbing a rebound jumps 76.0 cm vertically. How much total time (ascent and descent) does the player spend.
MatroZZZ [7]

Answer: Part(a)=0.041 secs, Part(b)=0.041 secs

Explanation: Firstly we assume that only the gravitational acceleration is acting on the basket ball player i.e. there is no air friction

now we know that

a=-9.81 m/s^2  ( negative because it is pulling the player downwards)

we also know that

s=76 cm= 0.76 m ( maximum s)

using kinetic equation

v^2=u^2+2as

where v is final velocity which is zero at max height and u is it initial

hence

u^2=-2(-9.81)*0.76

u=3.8615 m/s\\

now we can find time in the 15 cm ascent

s=ut+0.5at^2

0.15=3.861*t+0.5*9.81t^2\\

using quadratic formula

t=\frac{-3.861+\sqrt{3.86^2-4*0.5*9.81(-0.15)} }{2*0.5*9.81}

t=0.0409 sec

the answer for the part b will be the same

To find the answer for the part b we can find the velocity at 15 cm height similarly using

v^2=u^2+2as

where s=0.76-0.15

as the player has traveled the above distance to reach 15cm to the bottom

v^2=0^2 +2*(9.81)*(0.76-0.15)

v=3.4595

when the player reaches the bottom it has the same velocity with which it started which is 3.861

hence the time required to reach the bottom 15cm is

t=\frac{3.861-3.4595}{9.81}

t=0.0409

8 0
3 years ago
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