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Oxana [17]
3 years ago
13

An object is dropped at a height of 12 m from the ground. How fast is it moving just before it hits the ground?

Physics
1 answer:
sp2606 [1]3 years ago
6 0

The final speed of the object is 15.3 m/s

Explanation:

The object in the problem is moving by uniformly accelerated motion (it is in free fall), so we can use the following suvat equation:

v^2-u^2=2as

where

v is the final velocity

u is the initial velocity

a=g=9.8 m/s^2 is the acceleration of gravity (taking downward as positive direction)

s is the vertical displacement

For the object in this problem:

u = 0 (it starts from rest)

s = 12 m is the displacement of the object

Solving for v,

v=\sqrt{u^2+2as}=\sqrt{0+2(9.8)(12)}=15.3 m/s

Learn more about free fall motion here:

brainly.com/question/1748290

brainly.com/question/11042118

brainly.com/question/2455974

brainly.com/question/2607086

#LearnwithBrainly

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Consider a uniformly charged sphere of radius Rand total charge Q. The electric field Eout outsidethe sphere (r≥R) is simply tha
AlexFokin [52]

1) Electric potential inside the sphere: \frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})

2) Ratio Vcenter/Vsurface: 3/2

3) Find graph in attachment

Explanation:

1)

The electric field inside the sphere is given by

E=\frac{1}{4\pi \epsilon_0}\frac{Qr}{R^3}

where

\epsilon_0=8.85\cdot 10^{-12}F/m is the vacuum permittivity

Q is the charge on the sphere

R is the radius of the sphere

r is the distance from the centre at which we compute the field

For a radial field,

E(r)=-\frac{dV(r)}{dr}

Therefore, we can find the potential at distance r by integrating the expression for the electric field. Calculating the difference between the potential at r and the potential at R,

V(R)-V(r)=-\int\limits^R_r  E(r)dr=-\frac{Q}{4\pi \epsilon_0 R^3}\int r dr = \frac{-Q}{8\pi \epsilon_0 R^3}(R^2-r^2)

The potential at the surface, V(R), is that of a point charge, so

V(R)=\frac{Q}{4\pi \epsilon_0 R}

Therefore we can find the potential inside the sphere, V(r):

V(r)=V(R)+\Delta V=\frac{Q}{4\pi \epsilon_0 R}+\frac{-Q}{8\pi \epsilon_0 R^3}(R^2-r^2)=\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})

2)

At the center,

r = 0

Therefore the potential at the center of the sphere is:

V(r)=\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})\\V(0)=\frac{3Q}{8\pi \epsilon_0 R}

On the other hand, the potential at the surface is

V(R)=\frac{Q}{4\pi \epsilon_0 R}

Therefore, the ratio V(center)/V(surface) is:

\frac{V(0)}{V(R)}=\frac{\frac{3Q}{8\pi \epsilon_0 R}}{\frac{Q}{4\pi \epsilon_0 R}}=\frac{3}{2}

3)

The graph of V versus r can be found in attachment.

We observe the following:

- At r = 0, the value of the potential is \frac{3}{2}V(R), as found in part b) (where V(R)=\frac{Q}{4\pi \epsilon_0 R})

- Between r and R, the potential decreases as -\frac{r^2}{R^2}

- Then at r = R, the potential is V(R)

- Between r = R and r = 3R, the potential decreases as \frac{1}{R}, therefore when the distance is tripled (r=3R), the potential as decreased to 1/3 (\frac{1}{3}V(R))

Learn more about electric fields and potential:

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7 0
3 years ago
A silver bar 0.125 meter long is subjected to a temperature change from 200°C to 100°C. What will be the length of the bar after
Delicious77 [7]
\Delta L=  \alpha L_0 (T_f-T_i)

= (18 x 10^-6 /°C)(0.125 m)(100° C - 200 °C)

= -0.00225 m

New length = L + ΔL
= 1.25 m + (-0.00225 m)
= 1.248

D
5 0
3 years ago
Read 2 more answers
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lana66690 [7]

Answer:

Very sorry if this is late

Explanation:

Currently, in Olympia, Washington, it is raining because a cold front with high pressure has met an area with low pressure. This occurs due to the fact that storms are caused by sudden differences in air pressure.

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If you're going 80 mph how long does it take to go 80 miles
Elza [17]
1 mile. Is this a joke lol
6 0
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A car is moving at 30.0 km/h when it accelerates at 2.0 m/s for 3.6 s. what is the car final speed?
Lana71 [14]

Answer:

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Explanation:

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3 0
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