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attashe74 [19]
3 years ago
5

This ball is technology too! It can be rolled, kicked, or thrown. Is that a need or a want?

Physics
2 answers:
Inga [223]3 years ago
7 0

Answer:

want

Explanation:

people always want the new thing especially if its technology

Juli2301 [7.4K]3 years ago
5 0
Wantttttrtttt............
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The​ time, t, required to drive a fixed distance varies inversely as the​ speed, r. It takes 9 hr at a speed of 20 ​km/h to driv
Harlamova29_29 [7]

Answer:

Time=7.84hrs

Explanation:

This is an inverse proportionality questions

Mathematically time t varies as 1/distance d

Hence T= dk/r

But speed r = distance/t

d= r*t= 9*20=180km/hr

We're k= constant of proportionality

At t=9hrs d=180km/hr

Hence k=r*t/d=20*9/1620=0.111

Finding t at at 1620km for 23km/hr

t=1620*0.111/23=7.84hrs

6 0
3 years ago
What is the relationship between the frequency and the pitch of a sound?
jarptica [38.1K]
The pitch of a sound come from the frequency of the soundwave, soundwaves that have bigger frequencies sound with a high pitch and soundwaves with less frequencie have low pitch.
6 0
3 years ago
Light travels from the sun to the earth, a distance of 144,000,000,000 meters, in 480 seconds. Calculate the approximate velocit
mel-nik [20]

Answer:

The answer is C.) 300,000,000 m/s

Explanation:

Light travels at a constant rate of 300,000,000 m/s. this can be determined by dividing the distance (144,000,000,000 meters) by the time (480 seconds). that's pretty fast. in fact, nothing can travel any faster than this. ever.

5 0
3 years ago
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
5. A 5.0 kg object accelerates uniformly from rest for 5.0 s and reaches a final velocity of 20.0 m/s. At 3.0 s, what is the obj
irina [24]

Answer:

jack JACKKK

Explanation:

OK YAHHHHH SO DO THIS

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