All categories

2 years ago

How long would it take for a penny to fall from the top of the empire state building? the building is 381 meters tall.

1 answer:

8 0

Here, S = ut + 1/2 at²
381 = 0*t + 1/2 * 9.8 * t²
762/9.8 = t²
t = √77.75
t = 8.82 sec

In short, Your Answer would be 8.82 s

Hope this helps!

You might be interested in

Predict the products of the combustion of methanol, CH3OH(l).

gregori [183]

Answer:

Carbon dioxide and water

Explanation:

The products of complete combustion are always carbon dioxide and water.

The balanced reaction is:

4 CH₃OH + 3 O₂ → 4 CO₂ + 2 H₂O

8 0

2 years ago

Read 2 more answers

A proton is accelerated from rest through a potential difference of 2.5 kV and then moves perpendicularly through a uniform 0.60

Bumek [7]

Answer:

1.2cm

Explanation:

V=(2ev/m)^1/2

=(2*1.6*10^19 x2500/ 1.67*10^27)^1/2

=6.2x10^5m/s

Radius of resulting path= MV/qB

= 1.67*10^-27x6.92*10^6/1.6*10^-16 x0.6

=0.012m

=1.2cm

5 0

3 years ago

Read 2 more answers

To determine the height of a tall building such as Sears Tower in Chicago, Illinois a ball was dropped from the top of the build

Darya [45]

Answer:

The height of Sears Tower is 1448.5 feet.

Explanation:

<h3>We apply the free fall formula to the ball: </h3><h3> $y=v_{o} *t+\frac{1}{2} *g*t^{2}$ </h3><h3>y: The vertical distance the ball moves at time t </h3><h3> $v_{o}$ i: Initial speed </h3><h3>g=Gravity acceleration= $9.8*(\frac{\frac{1ft}{0.305m} }{s^{2} } )$ </h3>

Known information

We know that the vertical distance (y) that the ball moves in 9,5s is equal to height of Sears Tower (h).

Too we know that the ball is released from rest, then, $v_{0}$ =0

Height of Sears Tower calculation:

We replace in the equation 1 the data following;

$y=h$

$v_{o} =0$

$g=32,1\frac{ft}{s^{2} }$

$t= 9,5s$

$h=0*9.5+\frac{1}{2} *32.1*9.5^{2}$

$h=1448.5 ft$

Answer: The height of Sears Tower is 1448.5 ft

6 0

3 years ago

How much faster does sound travel in the ocean been in the rain forest?show your work.

Natasha2012 [34]

In rainforest sound travels faster because in the ocean sound travels very slow due to the pressure

5 0

3 years ago

The World-War II battleship U.S.S Massachusetts used 16-inch guns whose barrel lengths were 15 m long. The shells each of mass 1

Vaselesa [24]

Answer:

The explosive force experienced by the shell inside the barrel is 23437500 newtons.

Explanation:

Let suppose that shells are not experiencing any effect from non-conservative forces (i.e. friction, air viscosity) and changes in gravitational potential energy are negligible. The explosive force experienced by the shell inside the barrel can be estimated by Work-Energy Theorem, represented by the following formula:

$F\cdot \Delta s = \frac{1}{2}\cdot m \cdot (v_{f}^{2}-v_{o}^{2})$ (1)

Where:

$F$ - Explosive force, measured in newtons.

$\Delta s$ - Barrel length, measured in meters.

$m$ - Mass of the shell, measured in kilograms.

$v_{o}$ , $v_{f}$ - Initial and final speeds of the shell, measured in meters per second.

If we know that $m = 1250\,kg$ , $v_{o} = 0\,\frac{m}{s}$ , $v_{f} = 750\,\frac{m}{s}$ and $\Delta s = 15\,m$ , then the explosive force experienced by the shell inside the barrel is:

$F = \frac{m\cdot (v_{f}^{2}-v_{o}^{2})}{2\cdot \Delta s}$

$F = \frac{(1250\,kg)\cdot \left[\left(750\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}\right]}{2\cdot (15\,m)}$

$F = 23437500\,N$

The explosive force experienced by the shell inside the barrel is 23437500 newtons.

6 0

2 years ago