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JulijaS [17]
3 years ago
10

While following the directions on a treasure map, a pirate walks 45.0 m north and then turns and walks 7.5 m east. What single s

traight-line displacement could the pirate have taken to reach the treasure?
Physics
1 answer:
Contact [7]3 years ago
7 0

Answer:

Displacement = 45.62 m

Explanation:

Given that,

A pirate walks 45.0 m north and then turns and walks 7.5 m east.

We need to find the straight-line displacement taken to reach the treasure. We know that the displacement covered by an object is equal to the shortest path covered by it. It can be given by :

d=\sqrt{45^2+7.5^2} \\\\=45.62\ m

So, the displacement taken to reach the treasure is 45.62 m.

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Murljashka [212]

According to motion in straight line  t1≠t2


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An object is said to be in motion if its position in relation to its surroundings changes over time. It is a shift in an object's position over time. The only type of motion that exists is motion in a straight line.


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4 0
2 years ago
Describa con sus palabras que fue lo que descubrio galileo en su legendario experimento en la torre inclinada de pisa
german

Answer:

En 1589 Galileo realizó un experimento lanzando dos bolas de diferentes masas desde la famosa Torre Inclinada de Pisa para demostrar que el tiempo de caída es independiente de la masa de la bola. A través de este experimento, Galileo descubrió que los cuerpos caían casi simultáneamente, refutando la teoría de Aristóteles de que la tasa de caída era proporcional a la masa del cuerpo.  

Debido a la imperfección de los equipo de medición de esa época, la caída libre de los cuerpos era casi imposible de estudiar. En busca de una forma de reducir la velocidad de movimiento, Galileo reemplazó la caída libre por rodar sobre una superficie inclinada, donde había velocidades y resistencia del aire significativamente más bajas. Se notó que con el tiempo, la velocidad del movimiento aumenta: los cuerpos se mueven con aceleración. Se concluyó que la velocidad y la aceleración no dependen ni de la masa ni del material de la pelota.

 

5 0
3 years ago
A 2.40 cm × 2.40 cm square loop of wire with resistance 1.20×10−2 Ω has one edge parallel to a long straight wire. The near edge
Norma-Jean [14]

Answer:

current in loops is 52.73 μA

Explanation:

given data

side of square a = b  = 2.40 cm = 0.024 m

resistance R = 1.20×10^−2 Ω

edge of the loop c  = 1.20 cm = 0.012 m

rate of current = 120 A/s

to find out

current in the loop

solution

we know current formula that is

current = voltage / resistance    .................a

so current = 1/R × d∅/dt

and we know here that

flux ∅ = ( μ×I×b / 2π ) × ln (a+c/c)    ...............b

so

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so from equation a we get here current

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solve it and we get current that is

current = 4 ×10^{-7}× 1.09861 × 120

current = 52.73 ×10^{-6}  A

so here current in loops is 52.73 μA

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Answer:

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