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3 years ago

A zebra runs across a field at a constant speed of 14m/s how far does the zebra go in 8 seconds?

Physics

2 answers:

Ratling [72]3 years ago

3 0

Answer:

112m/s

Explanation:

14x8=112 therefore meaning the zebra would run 112m/s

RoseWind [281]3 years ago

3 0

A zebra runs across a field at a constant speed of 14 m/s. The zebra would run 112 m in 8 seconds.

<u>Explanation: </u>

According to Kinematics, the distance travelled is given by the product of speed and time. Here, the speed given as 14 m/s, and time is 8 seconds. Now, the equation is as follows,

$\text {velocity} = \frac{\text {distance}(s)}{\text {time}(t)}$

The required term to find as per the question is distance, so rearranging the above equation, we get,

$s = v e l o c i t y \times t$

By substituting the given values, we get, how much distance the zebra can run in time period of 8 seconds as,

Distance, $s = 14 \times 8 = 112 m$

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Describe the energy change in the particles of a substance during melting. A) The kinetic energy of the particles remains unchan

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This causes the ice to melt. This is demonstrated in the image below.

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Read 2 more answers

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3 years ago

A horizontal pipe of diameter 1.03m has a smooth constriction to a section of diameter 0.618 m. The density of oil flowing in th

vodka [1.7K]

Velocity of the oil in the pipe: 0.76 m/s, in the constricted section: 5.87 m/s

Explanation:

We can solve this problem by using Bernoulli's equation:

$p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho v_2^2$ (1)

where

$p_1 = 7340 N/m^2$ is the pressure in the pipe

$p_2 = 5505 N/m^2$ is the pressure in the constricted section

$\rho = 821 kg/m^3$ is the density of the oil

$v_1$ is the velocity of the oil in the pipe

$v_2$ is the velocity of the oil in the constricted section

Also, according to the continuity equation,

$A_1 v_1 = A_2 v_2$

where

$A_1 = \pi r_1^2$ is the cross-sectional area of the pipe, with

$r_1 = \frac{1.03}{2}=0.515 m$ is the radius

$A_2 = \pi r_2^2$ is the cross-sectional area of the constricted section, with

$r_2=\frac{0.618}{2}=0.309 m$ is the radius

So the equation becomes

$r_1^2 v_1 = r_2^2 v_2$

So we can write

$v_2=\frac{r_1^2}{r_2^2}v_1$

Substituting into eq.(1),

$p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho (\frac{r_1^2}{r_2^2}v_1)^2$

And solving the equation for $v_1$ :

$p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho \frac{r_1^4}{r_2^4}v_1^2\\v_1=\sqrt{\frac{p_2-p_1}{\frac{1}{2}\rho-\frac{1}{2}\rho \frac{r_1^4}{r_2^4}}}=0.76 m/s$

And the velocity in the constricted section is

$v_2=\frac{r_1^2}{r_2^2}v_1=5.87 m/s$

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#LearnwithBrainly

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