A construction worker drops a hammer from a 15 m high building. How fast is it going before it hits the ground?

There are many environmental benefits to using nuclear power. One is not an environmental benefit of nuclear power. That is:

FinnZ [79.3K]

The Answer is here
Nuclear energy has no place in a safe, clean, sustainable future. Nuclear energy is both expensive and dangerous, and just because nuclear pollution is invisible doesn't mean it's clean

6 0

3 years ago

Two children are riding on the edge of a merry-go-round that has a mass of 100.kg and radius of 1.60m and is rotating at 20.0rpm

Gre4nikov [31]

Here since both children and merry go round is our system and there is no torque acting on this system

So we will use angular momentum conservation in this

$I_1\omega_1 = I_2\omega_2$

now here we have

$I_1 = \frac{MR^2}{2} + m_1R^2 + m_2R^2$

$I_1 = \frac{100(1.60)^2}{2} + (22 + 28)(1.60)^2$

$I_1 = 256$

Now when children come to the position of half radius

then we will have

$I_2 = \frac{MR^2}{2} + m_1(\frac{R}{2})^2 + m_2(\frac{R}{2})^2$

$I_2 = \frac{100(1.6)^2}{2} + (28 + 22)(0.8)^2$

$I_2 = 160$

now from above equation we have

$256 (20.0 rpm) = 160(\omega_2)$

$\omega_2 = 32 rpm$

8 0

3 years ago

n isolated charged soap bubble of radius R0=7.45 cmR0=7.45 cm is at a potential of V0=307.0 volts.V0=307.0 volts. If the bubble

Gnesinka [82]

Complete Question

An isolated charged soap bubble of radius R0 = 7.45 cm is at a potential of V0=307.0 volts. V0=307.0 volts. If the bubble shrinks to a radius that is 19.0%19.0% of the initial radius, by how much does its electrostatic potential energy ????U change? Assume that the charge on the bubble is spread evenly over the surface, and that the total charge on the bubble r

Answer:

The difference is $U_f -U_i = 16 *10^{-7} J$

Explanation:

From the question we are told that

The radius of the soap bubble is $R_o = 7.45 \ cm = \frac{7.45}{100} = 0.0745 \ m$

The potential of the soap bubble is $V_1 =307.0 V$

The new radius of the soap bubble is $R_1 = 0.19 * 7.45=1.4155\ cm = 0.014155 \ m$

The initial electric potential is mathematically represented as

$U_i = \frac{V_1^2 R_o }{2k }$

The final electric potential is mathematically represented as

$U_f = \frac{V_2^2 R_1 }{2k }$

The initial potential is mathematically represented as

$V_1 = \frac{kQ}{R_o}$

The final potential is mathematically represented as

$V_2 = \frac{kQ}{R_1}$

Now

$\frac{V_2}{V_1} = \frac{R_o}{R_1}$

substituting values

$\frac{V_2}{V_1} = \frac{7.45}{1.4155} = \frac{1}{0.19}$

=> $V_2 = \frac{V_1}{0.19}$

So

$U_f = \frac{V_1^2 R_2 }{2k * 0.19^2}$

Therefore

$U_f -U_i = \frac{V_1^2 R_2 }{2k * 0.19^2} - \frac{V_1^2 R_o }{2k }$

$U_f -U_i = \frac{V_1^2}{2k} [\frac{ R_1 }{ * 0.19^2} - R_o]$

where k is the coulomb's constant with value $9*10^{9} \ kg\cdot m^3\cdot s^{-4}\cdot A^2.$

substituting values

$U_f -U_i = \frac{307^2}{9 * 10^{9}} [\frac{ 0.014155 }{ 0.19^2} - 0.0745]$

$U_f -U_i = 16 *10^{-7} J$

8 0

3 years ago

Why doesn't an object thrown in an upward direction fall the same distance in each time interval as it descends toward Earth? (A

Stels [109]

I'm not that smart but I think it is c I really hope It helps

4 0

3 years ago

Find the work done in pumping gasoline that weighs 6600 newtons per cubic meter. A cylindrical gasoline tank 3 meters in diamete

almond37 [142]

Answer:

work done in pumping the entire fuel is 1399761 J

Explanation:

weight per volume of the gasoline = 6600 N/m^3

diameter of the tank = 3 m

length of the tank = 6 m

The height of the tractor tank above the top of the tank = 5 m

The total volume of the fuel is gotten below

we know that the tank is cylindrical.

we assume that the fuel completely fills the tank.

therefore, the volume of a cylinder =

where r = radius = diameter ÷ 2 = 3/2 = 1.5 m

volume of the cylinder = 3.142 x x 6 = 42.417 m^3

we then proceed to find the total weight of the fuel in Newton

total weight = (weight per volume) x volume

total weight = 6600 x 42.417 = 279952.2 N

therefore,

the work done to pump the fuel through to the 5 m height = (total weight of the fuel) x (height through which the fuel is pumped)

work done in pumping = 279952.2 x 5 = 1399761 J

5 0

3 years ago