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3 years ago

A cheetah runs at a constant velocity of 7 m/s. What is it’s acceleration in m/s/s PLEASE HELP

Physics

1 answer:

uysha [10]3 years ago

5 0

Answer:

0 m/s²

Explanation:

Acceleration is the change in velocity over change in time. If the velocity is constant, then the acceleration is 0.

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Leni [432]

Answer: 4

Explanation:

when energy is greater then frequency should increase and wavelength should decrease.

Therefore answer is 4

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3 years ago

A uniform line charge of density λ lies on the x axis between x = 0 and x = L. Its total charge is 7 nC. The electric field at x

DedPeter [7]

Answer:

The electric field at x = 3L is 166.67 N/C

Solution:

As per the question:

The uniform line charge density on the x-axis for x, 0< x< L is $\lambda$

Total charge, Q = 7 nC = $7\times 10^{- 9} C$

At x = 2L,

Electric field, $\vec{E_{2L}} = 500N/C$

Coulomb constant, K = $8.99\times 10^{9} N.m^{2}/C^{2}$

Now, we know that:

$\vec{E} = K\frac{Q}{x^{2}}$

Also the line charge density:

$\lambda = \frac{Q}{L}$

Thus

Q = $\lambda L$

Now, for small element:

$d\vec{E} = K\frac{dq}{x^{2}}$

$d\vec{E} = K\frac{\lambda }{x^{2}}dx$

Integrating both the sides from x = L to x = 2L

$\int_{0}^{E}d\vec{E_{2L}} = K\lambda \int_{L}^{2L}\frac{1}{x^{2}}dx$

$\vec{E_{2L}} = K\lambda[\frac{- 1}{x}]_{L}^{2L}] = K\frac{Q}{L}[frac{1}{2L}]$

$\vec{E_{2L}} = (9\times 10^{9})\frac{7\times 10^{- 9}}{L}[frac{1}{2L}] = \frac{63}{L^{2}}$

Similarly,

For the field in between the range 2L< x < 3L:

$\int_{0}^{E}d\vec{E} = K\lambda \int_{2L}^{3L}\frac{1}{x^{2}}dx$

$\vec{E} = K\lambda[\frac{- 1}{x}]_{2L}^{3L}] = K\frac{Q}{L}[frac{1}{6L}]$

$\vec{E} = (9\times 10^{9})\frac{7\times 10^{- 9}}{L}[frac{1}{6L}] = \frac{63}{6L^{2}}$

Now,

If at x = 2L,

$\vec{E_{2L}} = 500 N/C$

Then at x = 3L:

$\frac{\vec{E_{2L}}}{3} = \frac{500}{3} = 166.67 N/C$

4 0

4 years ago

In which way does blue light change as it travels

____ [38]

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7 0

3 years ago

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There are 8.8 liters of gas in a piston at a pressure of 1.75 atmospheres. The temperature remains constant, and the gas is comp

saveliy_v [14]

The new pressure P2 is 2.48 atmosphere.

<u>Explanation:</u>

Here, the one of the product of pressure and volume is equal to the products of pressure and volume of other.

By using Boyles's law,

pressure is inversely proportional to volume,

P1 V1 = P2 V2

where P1, V1 represents the first pressure and volume,

P2, V2 represents the second pressure and volume

P2 = (P1 V1) / V2

= (1.75 $\times$ 8.8) / 6.2

P2 = 2.48 atmosphere.

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3 years ago

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Answer:

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3 years ago

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