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weeeeeb [17]
3 years ago
12

Highway safety engineers want to design roadside barriers that will crumple

Physics
2 answers:
EastWind [94]3 years ago
5 0

Answer:

0.84\:\text{s}

Explanation:

The impulse-momentum theorem states that the impulse on an object (F\Delta t) is equal to the change in momentum of that object (\Delta p).

Set up the following equation:

F\Delta t=\Delta p

Solving for change in momentum:

The momentum of an object is equal to p=mv, where m is the mass of the object and v is the velocity of the object. Since the person's final velocity will be zero, their final momentum will also be zero. Therefore, the person's change in momentum is 68\cdot 27-0=1836\:\text{kgm/s}.

Solving for time:

2180\cdot\Delta t = 1836,\\\Delta t =\frac{1836}{2180},\\\Delta t =\boxed{0.84\:\text{s}}

victus00 [196]3 years ago
3 0

Answer:

If the engineers know that the

If the engineers know that themaximum force that a person can safely withstand is 2180 N, approximately, <u>0</u><u>.</u><u>8</u><u>4</u><u> </u><u>second</u><u> </u>is required to crumple the barrier to safely slow the person

is required to crumple the barrier to safely slow the personwith this force.

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A veritical brass rod of circular section is loaded by placing a 10 kg wt on top of it .if it's length is 1 m. it's radius of cr
Inga [223]

Answer:

4.37 * 10^-4 J

Explanation:

Energy stored :

mgΔl / 2

m = mass = 10kg ; g = 9.8m/s² ; r = cross sectional Radius = 1cm = 1 * 10-2 m

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Y = Youngs modulus = Y=3.5 ×10^10 ; l = Length = 1m

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mgΔl / 2

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3 years ago
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salantis [7]

\boxed{\sf \tau=rFsin\theta}

Put values

\\ \rm\hookrightarrow 40=2Fsin40

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\\ \rm\hookrightarrow 0.64F=20

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Mamont248 [21]
That's wave 'diffraction'.
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melamori03 [73]

Answer:

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Explanation:

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Answer:

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