Does displacement must be straight and must include direction

1 answer:

7 0

Displacement is the total space between point a and b. For example, If I walk 3 times North, 6 times East, and 7 times North, what is the displacement? The answer is 10 because when I walked East there was on object blocking my way.

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Animal studies indicate that

notka56 [123]

Answer: mammals may be relatively better at solving problems than birds.

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3 years ago

I have a question, it concerns hydrostatic buoyancy and Archimedes' law.

saw5 [17]

Answer: the lvl wud remain the same

Explanation: as per Archimedes Principle, the weight of the water displaced by the object is equal to the weight of the object. When the ship initially went into the pool, it wud hv displaced some water. When the anchor is dropped, the level does not change coz the anchor was already in the ship and no extra weight has been added, so the weight of the anchor has already been accounted for in the first place when the ship was first placed in the pool

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3 years ago

Why are frogs said to have two lives?

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They live half their lives in water and most of their life on land

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3 years ago

A silver wire has a cross sectional area a = 2.0 mm2. a total of 9.4 × 1018 electrons pass through the wire in 3.0 s. the conduc

marta [7]

This problem uses the relationships among current I, current density J, and drift speed vd. We are given the total of electrons that pass through the wire in t = 3s and the area A, so we use the following equation to to find vd, from J and the known electron density n, so:

$v_{d} = \frac{J}{n\left | q \right |}$

<span>The current I is any motion of charge from one region to another, so this is given by:

</span> $I = \frac{\Delta Q}{\Delta t} = \frac{9.4x1018electrons}{3s} = 3189.73(A)$

The magnitude of the current density is:

$J = \frac{I}{A} = \frac{3189.73}{2x10^{-6}} = 1594.86(A/m^{2})$

Being:

$A=2mm^{2} = 2x10^{-6}m^{2}$
<span>
Finally, for the drift velocity magnitude vd, we find:

</span> $v_{d} = \frac{1594.86}{5.8x1028\left |1.60x10^{-19}|\right } = 1.67x10^{18}(m/s)$

Notice: The current I is very high for this wire. The given values of the variables are a little bit odd

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3 years ago

A 50 kg astronaut floating in space throws her 2 kg wrench to the left at 10

Hoochie [10]

The astronaut will move at 0.4 m/s in the opposite direction to the wrench

Explanation:

We can solve this problem by using the law of conservation of momentum. In fact, the total momentum of the astronaut-wrench system must be conserved before and after the launch.

Before the launch, the total momentum is zero, since the astronaut is at rest:

p = 0 (1)

After the launch, the total momentum is:

$p=mv+MV$ (2)