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evablogger [386]
3 years ago
11

If it takes Ashley 3 seconds to run from the batters box to first base at an average speed of 6.5 meters per second, what is the

distance she covers in that time?
Physics
2 answers:
san4es73 [151]3 years ago
7 0

Answer:

d = 19.5 m

Explanation:

Conceptual analysis

The average speed is defined as the relationship between the displacement that a body made and the total time it took to perform it.

The formula that will allow you to calculate the average speed is:

v= d/t Formula (1)

v= average speed measured in m/s

d=  displacement measured in m.

t= time interval measured in s.

Known data

v=  6.5 m/s

t= 3 s

Problem development

We replace data in the formula (1) to calculate the distance (d) :

v= d/t

6.5 m/s = d / 3 s

(6.5 m/s)*( 3 s ) = d

d = 19.5 m

Sidana [21]3 years ago
5 0
19.5 meters because 6.5*3 is 19.5
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Block B (of mass m) is initially at rest. Block A (of mass 3m) travels toward B with an initial speed v0 (vee nought) and collid
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Answer:

The maximum height reached by the two blocks is approximately 0.1147959 × v₀²

Explanation:

The mass of block B = m

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The initial velocity of block B, v₂ = 0 m/s

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By the Law of conservation of linear momentum, we have;

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Plugging in the values for the velocities gives;

3·m × v₀ + m × 0 = (3·m + m)·v₃ = 4·m·v₃

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\therefore v_3 = \dfrac{3}{4} \times v_0 = 0.75 \times v_0

The kinetic energy, K.E. of the combined blocks after the collision is given as follows;

K.E. = 1/2 × mass × v²

\therefore K.E. = \dfrac{1}{2} \times 4\cdot m \times \left (\dfrac{3}{4} \cdot v_0 \right )^2 = \dfrac{9}{8} \cdot m\cdot v_0^2

The potential energy, P.E., gained by the two blocks at maximum height = The kinetic energy, K.E., of the two blocks before moving vertically upwards

The potential energy, P.E. = m·g·h

Where;

m = The mass of the object at the given height

g = The acceleration due to gravity

h = The height at which the object of mass, 'm', is located

Therefore, for h = The maximum height reached by the two blocks, we have;

P.E. = K.E.

m \cdot g \cdot h =  \dfrac{9}{8} \cdot m\cdot v_0^2

h = \dfrac{\dfrac{9}{8} \cdot m\cdot v_0^2}{m \cdot g }  = \dfrac{9}{8} \cdot \dfrac{ v_0^2}{ g }  =  \dfrac{9}{8} \cdot \dfrac{ v_0^2}{ 9.8} = \dfrac{42}{392} \cdot  v_0^2 \approx 0.1147959   \cdot  v_0^2

The maximum height reached by the two blocks, h ≈ 0.1147959·v₀².

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