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2 years ago

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You exert a 138 N push the leftmost of two identical blocks of mass 244 g connected by a spring of stiffness 605 kg/s2. After pu

shing the block a distance 15 cm, you release it; by this time the rightmost block has moved a distance 5 cm. (a) What is the energy in the oscillations between the blocks

1 answer:

Reil [10]2 years ago

6 0

Answer:

the energy in the oscillations between the blocks is 3.025 J

Explanation:

Given the data in the question;

Force f = 138 N

stiffness of spring k = 605 kg/s²

mass of block = 202 g = 0.202 kg

pushing the block a distance 15 cm, the rightmost block has moved a distance 5 cm

i.e

x₁ = 15 cm

x₂ = 5cm

the energy in the oscillations between the blocks will be;

E $_A$ = E $_B$ = $\frac{1}{2}$ k( Δx )²

we substitute

= $\frac{1}{2}$ × k( 15 - 5 )² × 10⁻⁴

= $\frac{1}{2}$ × 605 × ( 10 )² × 10⁻⁴

= $\frac{1}{2}$ × 605 × 100 × 10⁻⁴

= 3.025 J

Therefore, the energy in the oscillations between the blocks is 3.025 J

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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

tankabanditka [31]

Answer:

I wanna say 4 I'm not sure though

4 0

2 years ago

Now, consider whether the following statements are true or false:_____________________________________.

Igoryamba

Answer:

A)F

B)F

C)T

Explanation:

A)

This is not always true that the product of the two vector's always non-negative .It may be positive or negative.

That is why this statement is false.

B)

The unit vectors in three perpendicular direction is i ,j and k.

We know that for unit vectors

i.i=j.j=k.k= 1

i .j= j.k=k.i=0

Therefore only parallel component of the force will contribute in the integral part instead of perpendicular component of the force.

That is why this statement is false.

C)

This statement is True.

A)F

B)F

C)T

7 0

3 years ago

During the contraction of the heart, 65 cm3 blood is ejected from the left ventricle into the aorta with a velocity of approxima

vova2212 [387]

Answer:

The value $F = 0.1396 \ N$

Explanation:

From the question we are told that

The volume blood ejected is $V = 65 \ cm^3 = 65*10^{-6} \ m^3$

The velocity of the blood ejected is $v = 103 \ cm/s = \frac{103}{100} = 1.03 \ m/s$

The density of blood is $\rho = 1060 \ kg/m^3$

The heart beat is $R = 59 \ bpm(beats \ per \ minute) = \frac{59}{60}= 0.9833\ bps$

The average force exerted by the blood on the wall of the aorta is mathematically represented as

$F = 2 * \rho * V * R * v$

=> $F = 2 * 1060 * 65*10^{-6} * 0.9833 * 1.03$

=> $F = 0.1396 \ N$

8 0

3 years ago

Suppose that the coefficient of kinetic friction between Zak's feet and the floor, while wearing socks, is 0.250. Knowing this,

insens350 [35]

Answer:

The distance travel before stopping is 1.84 m

Explanation:

Given :

coefficient of kinetic friction $\mu_{k} = 0.250$

Zak's speed $v = 3 \frac{m}{s}$

Gravitational acceleration $g = 9.8$ $\frac{m}{s^{2} }$

Work done by frictional force is given by,

$W = \Delta K$

$\mu _{k} mg d = \frac{1}{2} m v^{2}$

$d = \frac{v^{2} }{2 g \mu _{k} }$

$d = \frac{9}{2 \times 9.8 \times 0.250}$

$d = 1.84$ m

Therefore, the distance travel before stopping is 1.84 m

3 0

3 years ago