A lens of focal length 10.0 cm

PLEASE PLEASE HELP!!!!!!!

Marta_Voda [28]

Answer:

the object will move

Explanation:

well if you push an object it will move.

4 0

3 years ago

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A platinum sphere with radius 0.0199 m is totally immersed in mercury. Find the weight of the sphere, the buoyant force acting o

zepelin [54]

Answer:

Weight = 6.919 n

Buoyant force = 4.397 N

Apparent weight = 2.522 N

Explanation:

radius, r = 0.0199 m

density of platinum = 2.14 x 1064 kg/m^3

density of mercury = 1.36 x 1064 kg/m^3

Weight of sphere= volume of sphere x density of platinum x g

= 4/3 x 3.14 x (0.0199)^3 x 2.14 x 10^4 x 9.8 = 6.919 N

Buoyant force acting on sphere = Volume of sphere x density of mercury x g

= 4/3 x 3.14 x (0.0199)^3 x 1.36 x 10^4 x 9.8 = 4.397 N

Apparent weight = True weight - Buoyant force = 6.919 - 4.397 = 2.522 N

8 0

4 years ago

The continental crust is:

vlada-n [284]

the relatively thick part of the earth's crust that forms the large landmasses. It is generally older and more complex than the oceanic crust.

3 0

3 years ago

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these four trucks are identical each box loaded on the truck has the same mass choose the truck that has the greatest force of g

likoan [24]

Answer:

the green truck

Explanation:

the truck has 6 boxes which weighs more than 4, 3, or 2

4 0

4 years ago

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For no apparent reason, a poodle is running at a constant speed of 5.00 m/s in a circle with radius 2.9 m . Let v⃗ 1 be the velo

Nostrana [21]

At a constant speed of 5.00 m/s, the speed at which the poodle completes a full revolution is

$\left(5.00\,\dfrac{\mathrm m}{\mathrm s}\right)\left(\dfrac{1\,\mathrm{rev}}{2\pi(2.9\,\mathrm m)}\right)\approx0.2744\,\dfrac{\mathrm{rev}}{\mathrm s}$

so that its period is $T=3.644\,\frac{\mathrm s}{\mathrm{rev}}$ (where 1 revolution corresponds exactly to 360 degrees). We use this to determine how much of the circular path the poodle traverses in each given time interval with duration $\Delta t$ . Denote by $\theta$ the angle between the velocity vectors (same as the angle subtended by the arc the poodle traverses), then

$\Delta t=0.4\,\mathrm s\implies\dfrac{3.644\,\mathrm s}{360^\circ}=\dfrac{0.4\,\mathrm s}\theta\implies\theta\approx39.56^\circ$

$\Delta t=0.2\,\mathrm s\implies\dfrac{3.644\,\mathrm s}{360^\circ}=\dfrac{0.2\,\mathrm s}\theta\implies\theta\approx19.78^\circ$

$\Delta t=7\times10^{-2}\,\mathrm s\implies\dfrac{3.644\,\mathrm s}{360^\circ}=\dfrac{7\times10^{-2}\,\mathrm s}\theta\implies\theta\approx6.923^\circ$

We can then compute the magnitude of the velocity vector differences $\Delta\vec v$ for each time interval by using the law of cosines:

$|\Delta\vec v|^2=|\vec v_1|^2+|\vec v_2|^2-2|\vec v_1||\vec v_2|\cos\theta$

$\implies|\Delta\vec v|=\begin{cases}3.384\,\frac{\mathrm m}{\mathrm s}&\text{for }\Delta t=0.4\,\mathrm s\\1.718\,\frac{\mathrm m}{\mathrm s}&\text{for }\Delta t=0.2\,\mathrm s\\0.6038\,\frac{\mathrm m}{\mathrm s}&\text{for }\Delta t=7\times10^{-2}\,\mathrm s\end{cases}$

and in turn we find the magnitude of the average acceleration vectors to be

$\implies|\vec a|=\begin{cases}8.460\,\frac{\mathrm m}{\mathrm s^2}&\text{for }\Delta t=0.4\,\mathrm s\\8.588\,\frac{\mathrm m}{\mathrm s^2}&\text{for }\Delta t=0.2\,\mathrm s\\8.625\,\frac{\mathrm m}{\mathrm s^2}&\text{for }\Delta t=7\times10^{-2}\,\mathrm s\end{cases}$

So that takes care of parts A, C, and E. Unfortunately, without knowing the poodle's starting position, it's impossible to tell precisely in what directions each average acceleration vector points.

5 0

3 years ago