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4 years ago

Ladybug walks 10 cm forward and 5 cm backwards in 20 seconds what is the average speed of ladybug what is the average velocity

Physics

2 answers:

Tamiku [17]4 years ago

4 0

Hey!

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Formula: Distance / Time = Speed

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Solution:

Speed = 15/20 = 0.75cm/sec

Velocity = 5/20 = 0.25cm/sec

10 + 5 = 15 (distance traveled)

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Answer:

Speed = 0.75cm/sec

Velocity = 0.25cm/sec going forward.

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Hope This Helped! Good Luck!

lbvjy [14]4 years ago

3 0

Answer:

Average speed = 0.0075 m/s

Average velocity = 0.0025 m/s along forward direction

Explanation:

Speed is the ratio of distance and time and velocity is the ratio of displacement and time.

Distance traveled = 10 + 5 = 15 cm = 0.15 m

Displacement = 10 - 5 = 5 cm = 0.05 m

Time = 20 seconds

$\texttt{Average speed = }\frac{\texttt{Distance}}{\texttt{Time}}\\\\\texttt{Average speed = }\frac{0.15}{20}=7.5\times 10^{-3}m/s\\\\\texttt{Average velocity = }\frac{\texttt{Displacement}}{\texttt{Time}}\\\\\texttt{Average velocity = }\frac{0.05}{20}=2.5\times 10^{-3}m/s$

Average speed = 0.0075 m/s

Average velocity = 0.0025 m/s along forward direction

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A uniform rod is hung at one end and is partially submerged in water. If the density of the rod is 5/9 that of water, find the f

VashaNatasha [74]

Answer:

$\frac{y}{L}$ = 0.66

Hence, the fraction of the length of the rod above water = $\frac{y}{L}$ = 0.66

and fraction of the length of the rod submerged in water = 1 - $\frac{y}{L}$ = 1 - 0.66 = 0.34

Explanation:

Data given:

Density of the rod = 5/9 of the density of the water.

Let's denote density of Water with w

And density of rod with r

So,

r = 5/9 x w

Required:

Fraction of the length of the rod above water.

Let's denote total length of the rod with L

and length of the rod above with = y

Let's denote the density of rod = r

And density of water = w

So, the required is:

Fraction of the length of the rod above water = y/L

y/L = ?

In order to find this, we first need to find out the all type of forces acting upon the rod.

We know that, a body will come to equilibrium if the net torque acting upon a body is zero.

As, we know

F = ma

Density = m/v

m = Density x volume

Volume = Area x length = X ( L-y)

So, let's say X is the area of the cross section of the rod, so the forces acting upon it are:

F = mg

F = (Density x volume) x g

g = gravitational acceleration

F1 = X(L-y) x w x g (Force on the length of the rod submerged in water)

where,

X (L-y) = volume

w = density of water.

Another force acting upon it is:

F = mg

F2 = X x L x r x g

Now, the torques acting upon the body:

T1 + T2 = 0

F1 ( y + $(\frac{L-y}{2})$ ) g sinФ - F2 x ( $\frac{L}{2}$ ) x gsinФ = 0

plug in the equations of F1 and F2 into the above equation and after simplification, we get:

$(L^{2} - y^{2} ) . w$ = $L^{2}$ . r

where, w is the density of water and r is the density of rod.

As we know that,

r = 5/9 x w

So,

$(L^{2} - y^{2} ) . w$ = $L^{2}$ . 5/9 x w

Hence,

$(L^{2} - y^{2} )$ = $\frac{5L^{2} }{9}$

$\frac{L^{2} - y^{2} }{L^{2} }$ = $\frac{5}{9}$

Taking $L^{2}$ common and solving for $\frac{y}{L}$ , we will get

$\frac{y}{L}$ = 0.66

Hence, the fraction of the length of the rod above water = $\frac{y}{L}$ = 0.66

and fraction of the length of the rod submerged in water = 1 - $\frac{y}{L}$ = 1 - 0.66 = 0.34

8 0

3 years ago

Is the classification for an instrument that produces sound whne a string or strings stretched between two points is plucked?

Yuki888 [10]

The correct answer for the question is Chordophone
Chordophone is an instrument in which a stretched, vibrating string produces the initial sound. Strings instruments produce sound through the vibration of strings. The length, tightedness, and thickness determines the sound produced by the strings.

5 0

3 years ago

How do I solve such problem???

pashok25 [27]

As far as I'm concerned, this is a bogus question, or at least a severely corrupted one.

The three numbers given can NOT all be true on Earth.

-- It rolled off the table at 7.6 m/s . By golly, there you are! Its initial horizontal velocity is 7.6 m/s, and it has no vertical velocity until it leaves the table.

-- There are no horizontal forces that we're aware of acting on the object. So it maintains the same horizontal velocity for the rest of the story. It's 10.5m away from the table in (10.5 m) / (7.8 m/s) = 1.35 second .

-- Vertically, it's just an object dropped from 17.6m off the floor. Shockingly, the distance it falls in time 'T' is (1/2 g) T². In 1.35 second, that's 8.88 meters ! . . . only about halfway to the floor !

-- In order to fall 17.6 m to the floor, it would need 1.89 seconds. In <u>that</u> length of time, however, it would travel (7.8 m/s) x (1.89 s) = 14.78 m away from the base of the table.

So you see, either . . .

-- the table is NOT 17.6m tall, or

-- the object does NOT roll off of the table at 7.8 m/s, or

-- it does NOT land 10.5 m away from the base of the table.

OR . . .

-- the table is not on Earth, and gravity is not 9.8 m/s² !

We often see questions posted on Brainly with not enough given information, OR with some information given that's not needed because it's not involved the answer.

THIS one is different, and it's unusual. In this one, we have<em> too much</em> given information, we can't ignore any of it because it's all related, but it's inconsistent and it CAN't all be true.

(Unless the whole story takes place on a mystery planet that is not Earth. Which I'm not going to take the time and effort right now to figure out what the acceleration of gravity has to be in order to make all of the given information compatible.)

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3 years ago

Where is the frequency of ultrasound in relation to the range of human ability to hear

kykrilka [37]

Answer:

ultra sounds have frequency higher than the upper audible limit of human hearing, for healthy, young adults.

Explanation:

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3 years ago

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