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3 years ago

12

How do you determine the density of an empty plastic jug?

2 answers:

Andrei [34K]3 years ago

7 0

Calculate it by mass and volume.

Helga [31]3 years ago

5 0

I think mass and the volume

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Different objects don't heat at the same rate because they have different

Kaylis [27]

Specific heat

Let me know if you need further explanation!

7 0

3 years ago

A 9-μC positive point charge is located at the origin and a 6 μC positive point charge is located at x = 0.00 m, y = 1.0 m. Find

sukhopar [10]

Answer:

The coordinates of the point is (0,0.55).

Explanation:

Given that,

First charge $q_{1}=9\times10^{-6}\ C$ at origin

Second charge $q_{2}=6\times10^{-6}\ C$

Second charge at point P = (0,1)

We assume that,

The net electric field between the charges is zero at mid point.

Using formula of electric field

$E=\dfrac{kq}{r^2}$

$0=\dfrac{k\times9\times10^{-6}}{d^2}+\dfrac{k\times6\times10^{-6}}{(1-d)^2}$

$\dfrac{(1-d)}{d}=\sqrt{\dfrac{6}{9}}$

$\dfrac{1}{d}=\dfrac{\sqrt{6}}{3}+1$

$\dfrac{1}{d}=1.82$

$d=\dfrac{1}{1.82}$

$d=0.55\ m$

Hence, The coordinates of the point is (0,0.55).

3 0

3 years ago

Mike and Mitchell decide to have a foot race. They mark off a stretch of 100 yards, and recruit Cindy to work the stopwatch.

kati45 [8]

D)The boys had times that matched exactly

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3 years ago

Read 2 more answers

A particle is moving along a circular path of 2-m radius such that its position as a function of time is given by u = (5t 2) rad

OverLord2011 [107]

Answer:

Explanation:

Given

radius of path $r=2\ m$

Velocity of Particle $\theta =5t^2 rad$

where t=time in seconds

angular velocity of particle is given by

$\omega =\frac{\mathrm{d} \theta }{\mathrm{d} t}$

$\omega =2\times 5t=10\cdot t$

And angular acceleration is given by

$\alpha =\frac{\mathrm{d} \omega }{\mathrm{d} t}$

$\alpha =10 rad/s^2$

tangential acceleration is $a_t=\alpha \times r$

$a_t=10\times 2=20\ m/s^2$

Centripetal acceleration $a_c=\omega ^2\times r$

$a_c=(10t)^2\times 2=200t^2$

net acceleration is sum of tangential and centripetal force at any time t is given by

$a_{net}=\sqrt{(a_c)^2+(a_t)^2}$

$a_{net}=\sqrt{(200t)^2+(20)^2}$

$a_{net}=20\sqrt{(10t)^2+1}\ m/s$

8 0

3 years ago

When one side of the Earth is at high tide, then the opposite side of the Earth is also at high tide.

Rudiy27

<span>It is true because the near side of the moon is attracted by the gravitational force due to the alignment of the planet in one Straight line.The opposite side is due to the movement of earth towards moon side during alignment.Soon both side, there will be high tide at the same time.</span>

8 0

4 years ago