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The tops of the towers of the Golden Gate Bridge in San franciscoo , are 227 m above the water. Suppose a worker drops a 655 g w

yaroslaw [1]

Answer:

Kinetic energy is 1425.11 J.

Explanation:

Given:

Mass of the wrench is, $m=655\ g=0.655\ kg$

Height of fall is, $h=227\ m$

Force of resistance is, $F=0.141\ N$

Now, the total energy at the top is equal to the potential energy of the wrench at the top since the kinetic energy at the top is 0.

Now, potential energy at the top is given as:

$U=mgh\\U=0.655\times 9.8\times 227\\U=1457.113\ J$

Now, the potential energy at the top is converted to kinetic energy at the bottom and some energy is wasted in overcoming the resistance force by air.

Potential Energy = Kinetic energy + Energy to overcome resistance.

⇒ Kinetic energy = Potential Energy - Energy to overcome resistance.

Energy to overcome resistance force is the work done by the wrench against the resistance force and is given as:

$W_{res}=Fh=0.141\times 227=32.007\ J$

Therefore, Kinetic energy at the bottom is given as:

$K=U-W_{res}\\\\K=1457.113\ J - 32.007\ J\\\\K=1425.106\approx1425.11\ J$

Hence, the kinetic energy of the wrench be when it hits the water is 1425.11 J.

7 0

3 years ago

The world's most powerful laser is the LFEX laser in Japan. It can produce a 2 petawatt(2×10^15W) laser pulse that last for 1 ps

kogti [31]

Answer:

$2.82942\times 10^{24}\ W\m^2$

Explanation:

d = Diameter of spot = 30 μm

r = Radius of spot = $\frac{d}{2}=\frac{30}{2}=15\ mu m$

P = Power of the laser = $2\times 10^{15}\ W$

A = Area = $\pi r^2$

Intensity is given by

$I=\frac{P}{A}\\\Rightarrow I=\frac{2\times 10^{15}}{\pi\times (15\times 10^{-6})^2}\\\Rightarrow I=2.82942\times 10^{24}\ W/m^2$

The light intensity within this spot is $2.82942\times 10^{24}\ W/m^2$

8 0

3 years ago

A long, thin straight wire with linear charge density λ runs down the center of a thin, hollow metal cylinder of radius R. The c

Delvig [45]

Answer:

$E=\dfrac{\lambda }{2\pi \varepsilon _or}$

Explanation:

Given that

For straight wire

Charge density= λ

For hollow metal cylinder

Charge density=2 λ

We know that electric filed for wire given as

$E_w=\dfrac{\lambda_{wire} }{2\pi \varepsilon _or}$

$E_w=\dfrac{\lambda }{2\pi \varepsilon _or}$

Now the electric filed due to hollow metal cylinder

$E_c=\dfrac{\lambda_{cylinder} }{2\pi \varepsilon _or}$

$E_c=\dfrac{2\lambda }{2\pi \varepsilon _or}$

Now by considering the Gaussian surface r<R then only electric fild due to wire will present.So

At r<R

$E=\dfrac{\lambda }{2\pi \varepsilon _or}$

5 0

3 years ago

A propeller is modeled as five identical uniform rods extending radially from its axis. The length and mass of each rod are 0.77

Vikki [24]

The formula for the rotational kinetic energy is

$KE_{rot} = \frac{1}{2}(number \ of\ propellers)( I)( omega)^{2}$

where I is the moment of inertia. This is just mass times the square of the perpendicular distance to the axis of rotation. In other words, the radius of the propeller or this is equivalent to the length of the rod. ω is the angular velocity. We determine I and ω first.

$I=m L^{2}=(2.67 \ kg) (0.777 \ m)^{2} =2.07459 \ kgm^{2}$

ω = 573 rev/min * (2π rad/rev) * (1 min/60 s) = 60 rad/s

Then,

$KE_{rot} =( \frac{1}{2} )(5)(2.07459 \ kgm^{2}) (60\ rad/s)^{2}$

$KE_{rot} =18,671.31 \ J$

6 0

3 years ago

The rounded atomic mass of K is _______<br> K stands for _____ and has _____ valence electron(s).

netineya [11]

Answer:

Potassium

k = Potassium

Potassium Has 19 Electrons

the atomic mass is 39.0983

4 0

2 years ago

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