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3 years ago

Tony completed a 720 km journey with an average speed of

Physics

1 answer:

yan [13]3 years ago

5 0

Answer:

87 km/h

Explanation:

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_______________ appears on the ecg as having no p wave, a wide qrs complex, and t waves that deflect in the opposite direction f

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It would probably be <span>Premature Ventricular Contractions.

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3 years ago

Calculate the momentum of a 2500kg car with a velocity of 20 m/s

Sedaia [141]

Answer:

50000 kg.m/s

Explanation:

p = mv

p = 2500 × 20

p = 50000 kg. m/s

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3 years ago

Which of these would most likely be a parts of a lab procedure?

vladimir2022 [97]

C . Record the time to complete a chemical reaction

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3 years ago

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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:

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

7 0

3 years ago

A machinist turns the power on to a grinding wheel, at rest, at time t = 0 s. The wheel accelerates uniformly for 10 s and reach

Nookie1986 [14]

Answer:489 Revolutions

Explanation:

Given

Angular deceleration $(\alpha ) =1.5rad/s^2$

Given wheel angular velocity =96 rad/s when machine is turned off

time taken by machine to reach zero angular velocity

$0=\omega _0+(\alpha)t$

0=96+(-1.5)t

t=64 sec

angular displacement is given by

$\theta =\omega_0t+\frac{1}{2}\alpha t^2$

$\theta =96(64)-\frac{1}{2}(-1.5)(64^2)=3072 degree$

For revolutions = $\frac{3072}{2\cdot \pi}=488.86 \approx 489 revolution during Slowdown$

5 0

3 years ago