A major source of water pollution comes from that washes chemicals and other pollutants from improperty managed land.

An 80kg astronaut traveled to the moon, where gravity is one-sixth (116) as

PIT_PIT [208]

Answer:

Wmoon = 131 [N]

Explanation:

We know that the weight of a body is equal to the product of mass by gravitational acceleration.

Since we are told that the gravitational acceleration of the moon is equal to one-sixth of the acceleration of Earth's gravitation. Then we must multiply the value of Earth's gravitation by one-sixth.

$w_{moon}=\frac{1}{6} *m*g\\w_{moon}=\frac{1}{6} *80*9.81\\w_{moon}=130.8 [N] = 131 [N]$

7 0

2 years ago

Which items are matter? Pick multiple

AleksAgata [21]

Answer:

maybe phone cause matter is a solid liquid or gad

6 0

2 years ago

Two uncharged metal spheres, spaced 25.0 cm apart, have a capacitance of 26.0 pF. How much work would it take to move 12.0 nC of

viktelen [127]

Answer:

$W=2.76\times 10^{-6}\ J$

Explanation:

Given that,

The distance between two spheres, r = 25 cm = 0.25 m

The capacitance, C = 26 pF = 26×10⁻¹² F

Charge, Q = 12 nC = 12 × 10⁻⁹ C

We need to find the work done in moving the charge. We know that, work done is given by :

$U=\dfrac{Q^2}{2C}$

Put all the values,

$U=\dfrac{(12\times 10^{-9})^2}{2\times 26\times 10^{-12}}\\\\U=2.76\times 10^{-6}\ J$

So, the work done is $2.76\times 10^{-6}\ J$ .

8 0

2 years ago

RANGKAIAN RLC SERI PADA TEGANGAN BOLAK BALIK 220V R=80 OHM XL= 90 OHM XC = 30 OHM MAKA TEGANGAN PADDA INDUKTOR ADALAH???

Whitepunk [10]

Okay, 90% of this is nonsense besides the numbers maybe.

8 0

3 years ago

A certain quantity of steam has a temperature of 100.0 oC. To convert this steam into ice at 0.0 oC, energy in the form of heat

KonstantinChe [14]

Answer:

2452.79432 m/s

Explanation:

m = Mass of ice

$L_s$ = Latent heat of steam

$s_w$ = Specific heat of water

$L_i$ = Latent heat of ice

v = Velocity of ice

$\Delta T$ = Change in temperature

Amount of heat required for steam

$Q_1=mL_s\\\Rightarrow Q_1=m(2.256\times 10^6)$

Heat released from water at 100 °C

$Q_2=ms_w\Delta T\\\Rightarrow Q_2=m4186\times (100-0)\\\Rightarrow Q_2=m0.4186\times 10^6$

Heat released from water at 0 °C

$Q_3=mL_i\\\Rightarrow Q_3=m(333.5\times 10^3)\\\Rightarrow Q_3=m(0.3335\times 10^6)$

Total heat released is

$Q=Q_1+Q_2+Q_3\\\Rightarrow Q=m(2.256\times 10^6)+m0.4186\times 10^6+m(0.3335\times 10^6)\\\Rightarrow Q=3008100m$

The kinetic energy of the bullet will balance the heat

$K=Q\\\Rightarrow \frac{1}{2}mv^2=3008100m\\\Rightarrow v=\sqrt{2\times 3008100}\\\Rightarrow v=2452.79432\ m/s$

The velocity of the ice would be 2452.79432 m/s

6 0

3 years ago