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3 years ago

The surface tension of isopropanol in air has a value of 23.00 units and the

Physics

2 answers:

Y_Kistochka [10]3 years ago

7 0

Answer:

It's A & C

Explanation:

hichkok12 [17]3 years ago

4 0

Answer:

A and C

Explanation:

A.P.E.X

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A runner of mass 51.8 kg starts from rest and accelerates with a constant acceleration of 1.31 m/s^2 until she reaches a velocit

Stells [14]

Answer:32.24 s

Explanation:

Given

mass of runner (m)=51.8 kg

Constant acceleration(a)= $1.31 m/s^2$

Final velocity (v)=5.47 m/s

Time taken taken to reach 5.47 m/s

v=u+at

$5.47=0+1.31\times t$

$t=\frac{5.47}{1.31}=4.17 s$

Distance traveled during this time is

$s=ut+\frac{1}{2}at^2$

$s=\frac{1}{2}\times 1.31\times 4.17^2=11.42 m$

So remaining distance left to travel with constant velocity=153.57 m

thus time $=\frac{distance}{speed}$

$t_2=\frac{153.57}{5.47}=28.07 s$

Total time=28.07+4.17=32.24 s

6 0

3 years ago

Please help me.

vesna_86 [32]

A) the tension in the string.

4 0

3 years ago

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Two identical point charges of +Q coul are separated by a distance of 10 cm.

tatiyna

Answer:

IM NOT SMART

Explanation:

5 0

3 years ago

At time t=0, a particle is located at the point (3,6,9). It travels in a straight line to the point (5,2,7), has speed 8 at (3,6

Elis [28]

The particle has constant acceleration according to

$\vec a(t)=2\,\vec\imath-4\,\vec\jmath-2\,\vec k$

Its velocity at time $t$ is

$\displaystyle\vec v(t)=\vec v(0)+\int_0^t\vec a(u)\,\mathrm du$

$\vec v(t)=\vec v(0)+(2\,\vec\imath-4\,\vec\jmath-2\,\vec k)t$

$\vec v(t)=(v_{0x}+2t)\,\vec\imath+(v_{0y}-4t)\,\vec\jmath+(v_{0z}-2t)\,\vec k$

Then the particle has position at time $t$ according to

$\displaystyle\vec r(t)=\vec r(0)+\int_0^t\vec v(u)\,\mathrm du$

$\vec r(t)=(3+v_{0x}t+t^2)\,\vec\imath+(6+v_{0y}t-2t^2)\,\vec\jmath+(9+v_{0z}t-t^2)\,\vec k$

At at the point (3, 6, 9), i.e. when $t=0$ , it has speed 8, so that

$\|\vec v(0)\|=8\iff{v_{0x}}^2+{v_{0y}}^2+{v_{0z}}^2=64$

We know that at some time $t=T$ , the particle is at the point (5, 2, 7), which tells us

$\begin{cases}3+v_{0x}T+T^2=5\\6+v_{0y}T-2T^2=2\\9+v_{0z}T-T^2=7\end{cases}\implies\begin{cases}v_{0x}=\dfrac{2-T^2}T\\\\v_{0y}=\dfrac{2T^2-4}T\\\\v_{0z}=\dfrac{T^2-2}T\end{cases}$

and in particular we see that

$v_{0y}=-2v_{0x}$

and

$v_{0z}=-v_{0x}$

Then

${v_{0x}}^2+(-2v_{0x})^2+(-v_{0x})^2=6{v_{0x}}^2=64\implies v_{0x}=\pm\dfrac{4\sqrt6}3$

$\implies v_{0y}=\mp\dfrac{8\sqrt6}3$

$\implies v_{0z}=\mp\dfrac{4\sqrt6}3$

That is, there are two possible initial velocities for which the particle can travel between (3, 6, 9) and (5, 2, 7) with the given acceleration vector and given that it starts with a speed of 8. Then there are two possible solutions for its position vector; one of them is

$\vec r(t)=\left(3+\dfrac{4\sqrt6}3t+t^2\right)\,\vec\imath+\left(6-\dfrac{8\sqrt6}3t-2t^2\right)\,\vec\jmath+\left(9-\dfrac{4\sqrt6}3t-t^2\right)\,\vec k$

4 0

3 years ago

You design toys for a toy company. Your boss wants you to hook up the lights in the toy car you are working on in the cheapest w

bezimeni [28]

Answer:

put the car on fire

Explanation:

if you put it on fire you would have a lot of light now

7 0

3 years ago

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