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Romashka-Z-Leto [24]
3 years ago
11

Explain in terms of impulse how padding reduces forces in a collision. State this in terms of a real example, such as the advant

ages of a carpeted vs. tile floor for a day care cente
Physics
1 answer:
Lena [83]3 years ago
3 0

Answer:

Impulse = Average force x time of contact

Explanation:

Impulsive force is a force which is very large but applied on a body for a very small duration of time.

Impulse is given by the change in momentum of the body.

Impulse = Average force x small time interval

When padding is there, the time interval of contact is large and thus, the force exerted by the body is small.

So, when a person falls on the tile floor, there is no compression and thus, the time of contact is very small and thus the impulsive force is very large, due to  which the body may damage.

So, when a person falls on the carpeted floor, there is a compression and thus, the time of contact is comparatively large and thus the impulsive force is small, due to  which the body may safe.

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si tienes 2 circuitos uno en serie y otro en paralelo con el mismo valor en sus resistencias, para cuál circuito la resistencia
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circuito paralelo

Explanation:

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A long solenoid with 1.65 103 turns per meter and radius 2.00 cm carries an oscillating current I = 6.00 sin 90πt, where I is in
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Answer:

The  electric field  is 35\cos(90\pi t)\ mV/m

Explanation:

Given that,

Radius = 2.00 cm

Number of turns per unit length n= 1.65\times10^{3}

Current I = 6.00\sin 90\pi t

We need to calculate the induced emf

\epsilon =\mu_{0}nA\dfrac{dI}{dt}

Where, n = number of turns per unit length

A = area of cross section

\dfrac{dI}{dt}=rate of current

Formula of electric field is defined as,

E=\dfrac{\epsilon}{2\pi r}

Where, r = radius

Put the value of emf in equation (I)

E=\dfrac{\mu_{0}nA\dfrac{dI}{dt}}{2\pi r}....(II)

We need to calculate the rate of current

I=6.00\sin 90\pi t....(III)

On differentiating equation (III)

\dfrac{dI}{dt}=90\pi\times6.00\cos(90\pi t)

Now, put the value of rate of current in equation (II)

E=\dfrac{4\pi\times10^{-7}\times1.65\times10^{3}\times\pi\times(2.00\times10^{-2})^2\times90\pi\times6.00\cos(90\pi t)}{2\pi\times 2.00\times10^{-2}}

E=35\cos(90\pi t)\ mV/m

Hence, The  electric field  is 35\cos(90\pi t)\ mV/m

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