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Vera_Pavlovna [14]
3 years ago
9

Can someone please help meee!!

Physics
1 answer:
Paraphin [41]3 years ago
5 0

The answer is 9800 J/S

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During a tornado in 2008 the Peachtree Plaza Westin Hotel in downtown Atlanta suffered damage. Suppose a piece of glass dropped
Darina [25.2K]

Answer:

Time = t = 6.62 s

Explanation:

Given data:

Height = h = 215 m

Initial velocity = v_{i} = 0 m/s

gravitational acceleration = g = 9.8 m/s²

Time = t = ?

According to second equation of motion

                            h = v_{i}t + \frac{1}{2}gt^{2}

As initial velocity is zero, So the first term of right hand side of above equation equal to zero.

                                 h = \frac{1}{2}gt^{2}

                                        t² = \frac{2h}{g}

                                         t =\sqrt{\frac{2h}{g} }

                                         t = \sqrt{\frac{(2)(215) }{9.8} }

                                         t = 6.62 s

3 0
3 years ago
he tune-up specifications of a car call for the spark plugs to be tightened to a torque of 47 N⋅m . You plan to tighten the plug
S_A_V [24]

Answer:

207.4 N

Explanation:

The torque \tau  on a body is

\tau = r* F  where r is the radius vector from the point of rotation to the point at which force F is applied.

The product of r and F is equal to the product of magnitude of r and F multiplied by the sine of angle between both vectors.

Therefore, torque is also given by

\tau = rF\sin \theta

Where \theta is the angle between r and F.

Use the expression of torque.

Substitute L for r in the equation \tau = rF\sin \theta

\tau = LF\sin \theta

Where L is the length of the wrench.

Making F the subject

F = \frac{\tau }{{L\sin \theta }}

Force required to pull the wrench is given as,

F = \frac{\tau }{{L\sin \theta }}

Substitute 47{\rm{ N}} \cdot {\rm{m}}  for \tau, 25 cm for L, and 115o for \theta  

\begin{array}{c}\\F = \frac{{47{\rm{ N}} \cdot {\rm{m}}}}{{\left( {25{\rm{ cm}}} \right)\sin {{115}^{\rm{o}}}}}\left( {\frac{{1{\rm{ cm}}}}{{{{10}^{ - 2}}{\rm{ m}}}}} \right)\\\\ = 207.435{\rm{ N}}\\\\ \approx 207.4{\rm{ N}}\\\end{array}  

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