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3 years ago

You should be particularly careful a restraining a struggling rabbit because it can

1 answer:

6 0

You have no options here so I'll just answer. It can cause a rise in heart rate and greatly increases the risk of overheating and even death. If you grab the rabbit too hard, you risk breaking/fracturing a bone or causing other kinds of damage, whether externally or internally, to the rabbit.

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A nonconducting sphere has radius R = 2.81 cm and uniformly distributed charge q = +2.35 fC. Take the electric potential at the

Sladkaya [172]

Answer:

(a). The electric potential at 1.650 cm is $-1.219\times10^{-4}\ V$ .

(b). The electric potential at 2.81 cm is $-3.759\times10^{-4}\ V$ .

Explanation:

Given that,

Radius of sphere R=2.81 cm

Charge = +2.35 fC

Potential at center of sphere

$V = 0$

(a). We need to calculate the potential at a distance r = 1.60 cm

Using formula of potential difference

$V_(r)-V_(0)=-\int_{0}^{r}{E(r)}dr$

$V_{r}-0=-\int_{0}^{r}{\dfrac{qr}{4\pi\epsilon_{0}R^3}}dr$

$V_{r}=-(\dfrac{qr^2}{8\pi\epsilon_{0}R^3})_{0}^{1.60\times10^{-2}}$

$V_{r}=-(\dfrac{2.35\times10^{-15}\times(1.60\times10^{-2})^2}{8\times\pi\times8.85\times10^{-12}\times(2.81\times10^{-2})^3})$

$V_{r}=-0.00012190\ V$

$V_{r}=-1.219\times10^{-4}\ V$

The electric potential at 1.650 cm is $-1.219\times10^{-4}\ V$ .

(b). We need to calculate the potential at a distance r = R

Using formula of potential difference

$V_{R}=-\dfrac{2.35\times10^{-15}}{8\pi\times8.85\times10^{-12}\times2.81\times10^{-2}}$

$V_{R}=-0.0003759\ V$

$V_{R}=-3.759\times10^{-4}\ V$

The electric potential at 2.81 cm is $-3.759\times10^{-4}\ V$ .

Hence, This is the required solution.

5 0

3 years ago

An object's potential energy is set it cannot change is this true

zepelin [54]

The energy achieved I think

8 0

3 years ago

Read 2 more answers

What are two ways in which friction can be useful?

pishuonlain [190]

If something is going down a hill it can help slow it down

it can stop you from flying off a rollercoaster

6 0

3 years ago

Read 2 more answers

A 15.7-g aluminum block is warmed to 53.2 °c and plunged into an insulated beaker containing 32.5 g of water initially at 24.5 °

ipn [44]

The equilibrium temperature of aluminium and water is 33.2°C

We know that specific heat of aluminium is 0.9 J/gm-K, and that of water is 1 J/gm-K

Now we can calculate the equilibrium temperature

(mc∆T)_aluminium=(mc∆T)_water

15.7*0.9*(53.2-T)=32.5*1*(T-24.5)

T=33.2°C

6 0

3 years ago

Read 2 more answers

A bungee jumper starts with 1000 J in their GPE store. After they jump they fall and are brought to a stop with the bungee cord.

pochemuha

Answer:

energy is equal to 1000 J

Explanation:

When the jumper is in the tent, he has a given height, this height gives him a gravitational potential energy, which forms his initial mechanical energy of 1000 J. After jumping, this energy is converted into elastic energy of the rope plus a remainder of potential energy gravitational, it does not reach the ground, but as the friction is negligible the total mechanical energy is conserved, therefore its energy is equal to 1000 J

This is a case of energy transformation, but the total value of mechanical energy does not change

8 0

3 years ago