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3 years ago

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Physics

1 answer:

Gennadij [26K]3 years ago

6 0

The time taken for the sailfish to reach the herring if it is 55.0 m away is 2.7 s.

<h3>Time of motion of the star fish</h3>

The time taken for the star fish to travel the given distance is calculated as follows;

s = ut + ¹/₂at²

55 = 11.8t + (0.5)(6.39)t²

55 = 11.8t + 3.195t²

3.195t² + 11.8t - 55 = 0

<em>solve the quadratic equation using formula method,</em>

where;

a = 3.195, b = 11.8, c = -55

t = 2.7 s

Thus, the time taken for the sailfish to reach the herring if it is 55.0 m away is 2.7 s.

Learn more about time of motion here: brainly.com/question/2364404

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If the sprinter from the previous problem accelerates at that rate for 20 m, and then maintains that velocity for the remainder

kakasveta [241]

Question:

A 63.0 kg sprinter starts a race with an acceleration of 4.20m/s square. What is the net external force on him? If the sprinter from the previous problem accelerates at that rate for 20m, and then maintains that velocity for the remainder for the 100-m dash, what will be his time for the race?

Answer:

Time for the race will be t = 9.26 s

Explanation:

Given data:

As the sprinter starts the race so initial velocity = v₁ = 0

Distance = s₁ = 20 m

Acceleration = a = 4.20 ms⁻²

Distance = s₂ = 100 m

We first need to find the final velocity (v₂) of sprinter at the end of the first 20 meters.

Using 3rd equation of motion

(v₂)² - (v₁)² = 2as₁ = 2(4.2)(20)

v₂ = 12.96 ms⁻¹

Time for 20 m distance = t₁ = (v₂ - v ₁)/a

t₁ = 12.96/4.2 = 3.09 s

He ran the rest of the race at this velocity (12.96 m/s). Since has had already covered 20 meters, he has to cover 80 meters more to complete the 100 meter dash. So the time required to cover the 80 meters will be

Time for 100 m distance = t₂ = s₂/v₂

t₂ = 80/12.96 = 6.17 s

Total time = T = t₁ + t₂ = 3.09 + 6.17 = 9.26 s

T = 9.26 s

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The temperature of the cooling water as it leaves the hot engine of an automobile is 240 °F. After it passes through the radiato

djverab [1.8K]

Answer:571.09 kJ

Explanation:

Given

Temperature of cooling water from engine exit $=240^{\circ} F\approx 115.55^{\circ}C$

After Passing through the radiator its temperature decreases to $175^{\circ}F\approx 79.44^{\circ}F$

specific heat of water $=4.184 J/g^{\circ}C$

Volume of water $= 1 gallon\approx 3.78 L$

density of water $\rho =1 gm/mL$

Thus mass of water $=\rho \times V=3.78\times 1=3.78 kg$

Heat transferred to the surrounding is equal to heat absorbed by cooling water

$Q=m\cdot c\cdot \Delta T$

$Q=3.78\times 4.184\times 1000\times (115.55-79.44)$

$Q=3.78\times 4.184\times 1000\times (36.11)$

$Q=571.09 kJ$

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Samantha wants to determine the force she feels in her car as she's going around a corner. What would be some of the assumptions

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