How much work is the adult completing 1.6j 0.6j 20j 1,500j I NEED THIS DONE FAST

Carol drops a stone in a mine shaft 122.5 metres deep.How

Verizon [17]

Answer:

T = 5.36 s

Explanation:

given,

depth of the mine shaft = 122.5 m

speed of the sound = 340 m/s

time taken = ?

time taken by the stone to reach at the bottom

using equation of motion

$s = u t + \dfrac{1}{2}gt^2$

initial speed , u = 0 m/s

$t = \sqrt{\dfrac{2s}{g}}$

$t = \sqrt{\dfrac{2\times 122.5}{9.8}}$

t = 5 s

time taken by the sound to travel

d =v x t

$t = \dfrac{d}{v}$

$t = \dfrac{122.5}{340}$

t = 0.36 s

total time taken for the sound to reach carol after dropping the stone

T = 5 + 0.36

T = 5.36 s

7 0

3 years ago

You have a 3.00-liter container filled with N₂ at 25°C and 4.45 atm pressure connected to a 2.00-liter container filled with Ar

LuckyWell [14K]

Answer : The final pressure in the two containers is, 2.62 atm

Explanation :

Boyle's Law : It is defined as the pressure of the gas is inversely proportional to the volume of the gas at constant temperature and number of moles.

$P\propto \frac{1}{V}$

Thus, the expression for final pressure in the two containers will be:

$PV=P_1V_1+P_2V_2$

$P=\frac{P_1V_1+P_2V_2}{V}$

where,

$P_1$ = pressure of N₂ gas = 4.45 atm

$P_2$ = pressure of Ar gas = 2.75 atm

$V_1$ = volume of N₂ gas = 3.00 L

$V_2$ = volume of Ar gas = 2.00 L

P = final pressure of gas = ?

V = final volume of gas = (4.45 + 2.75) L = 7.2 L

Now put all the given values in the above equation, we get:

$P=\frac{(4.45atm)\times (3.00L)+(2.75atm)\times (2.00L)}{7.2L}$

$P=2.62atm$

Thus, the final pressure in the two containers is, 2.62 atm

8 0

2 years ago

What type of energy can take up space

Arada [10]

Answer:

Yes energy does take up space.

Explanation:

Every form of energy has a defining characteristic; sound is the vibration of molecules, electricity is the movement of electrons, and mass is the thing that take up space.

8 0

2 years ago

Answer either one or both questions! Must explain and show work to receive brainliest!

MariettaO [177]

F x L = W x X whereW=weight is total load = 80, L is length from fulcrum which is the unknown and what we are solving for. x= length we know. and F equals 50 force we know. So (W*X)/F=LL equals 64

3 0

2 years ago

On a straight road, a car speeds up at a constant rate from rest to 20 m/s over a 5 second interval and a truck slows at a const

IceJOKER [234]

Answer:

a)

Explanation:

Since the car speeds up at a constant rate, we can use the kinematic equation for distance (assuming that the initial position is x=0, and choosing t₀ =0), as follows:

$x_{fc} = v_{o}*t + \frac{1}{2}*a*t^{2} (1)$

Since the car starts from rest, v₀ =0.
We know the value of t = 5 sec., but we need to find the value of a.
Applying the definition of acceleration, as the rate of change of velocity with respect to time, and remembering that v₀ = 0 and t₀ =0, we can solve for a, as follows:

$a_{c} =\frac{v_{fc}}{t} = \frac{20m/s}{5s} = 4 m/s2 (2)$

Replacing a and t in (1):

$x_{fc} = v_{o}*t + \frac{1}{2}*a*t^{2} = \frac{1}{2}*a*t^{2} = \frac{1}{2}* 4 m/s2*(5s)^{2} = 50.0 m. (3)$

Now, if the truck slows down at a constant rate also, we can use (1) again, noting that v₀ is not equal to zero anymore.
Since we have the values of vf (it's zero because the truck stops), v₀, and t, we can find the new value of a, as follows:

$a_{t} =\frac{-v_{to}}{t} = \frac{-20m/s}{10s} = -2 m/s2 (4)$

Replacing v₀, at and t in (1), we have:

$x_{ft} = 20m/s*10.0s + \frac{1}{2}*(-2 m/s2)*(10.0s)^{2} = 200m -100m = 100.0m (5)$

Therefore, as the truck travels twice as far as the car, the right answer is a).

7 0

2 years ago