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3 years ago

10. At point A, a runner is jogging 5 m/s. Forty seconds later, at point B

Physics

1 answer:

kondaur [170]3 years ago

5 0

Answer:

Deceleration of 0.075m/s²

Explanation:

Given parameters:

Initial velocity = 5m/s

Time = 40s

Final velocity = 2m/s

Unknown:

The jogger's acceleration = ?

Solution:

To solve this problem, we use;

a = $\frac{v - u}{t}$

a is the acceleration

v is the final velocity

u is the initial velocity

t is the time taken

a = $\frac{2 - 5}{40}$ = - 0.075m/s²

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3 years ago

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NNADVOKAT [17]

Here's what you need to know about waves:

Wavelength = (speed) / (frequency)

Now ... The question gives you the speed and the frequency,
but they're stated in unusual ways, with complicated numbers.

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Frequency = (47) / (46.9 s) = 1.747... per second. (1.747... Hz)

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The crest of one wave travels 4.16 meters in 13.7 seconds.
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3 years ago

If two objects A and B have the same kinetic energy but A has three times the momentum of B, what is the ratio of their inertias

Thepotemich [5.8K]

Answer:

$\frac{inertia_B}{inertia_A}=9$

Explanation:

First of all, let's remind that:

- The kinetic energy of an object is given by $K=\frac{1}{2}mv^2$ , where m is the mass and v is the speed

- The momentum of an object is given by $p=mv$

- The inertia of an object is proportional to its mass, so we can write $I=km$ , where k just indicates a constant of proportionality

In this problem, we have:

- $K_A = K_B$ (the two objects have same kinetic energy)

- $p_A = 3 p_B$ (A has three times the momentum of B)

Re-writing both equation we have:

$\frac{1}{2}m_A v_A^2 = \frac{1}{2}m_B v_B^2\\m_A v_A = 3 m_B v_B$

If we divide first equation by second one we get

$v_A = 3 v_B$

And if we substitute it into the first equation we get

$m_A (3 v_B)^2 = m_B v_B^2\\9 m_A v_B^2 = m_B v_B^2\\m_B = 9 m_A$

So, B has 9 times more mass than A, and so B has 9 times more inertia than A, and their ratio is:

$\frac{I_B}{I_A}=\frac{km_B}{km_A}=\frac{9m_A}{m_A}=9$

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3 years ago

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3 years ago

Light of frequency 2.5 x 1015 Hz illuminates a

Nikitich [7]

Answer:

h f = W + KE

Input energy equals work function plus KE of emitted electron

W = 6.63E-34 * 2.5E15 - 6.3 * 1.6E-19

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3 years ago