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2 years ago

A car accelerates at a constant rate of 3 m/s2 for 5 seconds. If it reaches a velocity of 27 m/s, what was its initial velocity?

Physics

1 answer:

Kaylis [27]2 years ago

4 0

The initial velocity of a car that accelerates at a constant rate of 3m/s² for 5 seconds is 12m/s.

CALCULATE INITIAL VELOCITY:

The initial velocity of the car can be calculated by using one of the equation of motion as follows:

V = u + at

Where;

V = final velocity (m/s)
u = initial velocity (m/s)
a = acceleration due to gravity (m/s²)
t = time (s)

According to this question, a car accelerates at a constant rate of 3 m/s² for 5 seconds. If it reaches a velocity of 27 m/s, its initial velocity is calculated as follows:

u = v - at

u = 27 - 3(5)

u = 27 - 15

u = 12m/s.

Therefore, the initial velocity of a car that accelerates at a constant rate of 3m/s² for 5 seconds is 12m/s.

Learn more about motion at: brainly.com/question/974124

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Answer:

Explanation:

We shall first calculate the velocity at height h = 575 m .

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v² = u² + 2 a s

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v² = 0 + 2 x 2.2 x 575

v = 50.3 m /s

After 575 m , rocket moves under free fall so g will act on it downwards

If it travels further by height H

from the relation

v² = u² - 2 g H

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2 years ago

A Hall probe, consisting of a rectangular slab of current-carrying material, is calibrated by placing it in a known magnetic fie

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Answer:

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(b) 0.389 T

Solution:

As per the question:

Hall emf, $V_{Hall} = 20\ mV = 0.02\ V$

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Hall emf, $V'_{Hall} = 69\ mV = 0.069\ V$

Now,

Drift velocity, $v_{d} = \frac{V_{Hall}}{B}$

$v_{d} = \frac{0.02}{0.10} = 0.2\ m/s$

Now, the expression for the electric field is given by:

$E_{Hall} = Bv_{d}sin\theta$ (1)

And

$E_{Hall} = V_{Hall}d$

Thus eqn (1) becomes

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where

d = distance

$B = \frac{V_{Hall}}{v_{d}sin\theta}$ (2)

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$B = \frac{0.069}{0.2\times sin90} = 0.345\ T$

(b) When $\theta = 60^{\circ}$

$B = \frac{0.069}{0.2\times sin60} = 0.398\ T$

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3 years ago

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maksim [4K]

Answer:

The answer is the option a.

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