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Illusion [34]
3 years ago
7

A 2 kg object falls 3 meters how much potential energy did the object have beforw it fell

Physics
1 answer:
Brrunno [24]3 years ago
8 0
P.E = m*g*d
where m is the mass of body , d the vertical distance & g is the gravity acceleration and it equals constant = 9.81 m/s^2
so ,, P.E = 2 * 3 * 9.81
P.E = 58.86 N.m
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The amplitude of wave-c is 1 meter.

The speed of all of the waves is (12meters/2sec)= 6 m/s.

The period of wave-a is 1/2 second.

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3 years ago
What is serve in volleyball?
madreJ [45]
Underhand serve is a type of volleyball serve in which a player holds the ball with one hand and swings the other hand in an arc motion, striking under the ball with a fist to put it into play. An underhand serve is the most common serve for beginners.
3 0
3 years ago
Read 2 more answers
A wall (of thermal conductivity 0.98 W/m · ◦ C) of a building has dimensions of 3.7 m by 15 m. The average inside and outside
disa [49]

Answer:

Answer:

the amount of energy flowing is 1.008x10⁹J

Explanation:

To calculate how much heat flows, the expression is the following:

Where

K=thermal conductivity=0.81W/m°C

A=area=6.2*12=74.4m²

ΔT=30-8=22°C

L=thickness=8cm=0.08m

t=time=16.9h=60840s

Replacing:

Explanation:

4 0
2 years ago
A confined aquifer with a porosity of 0.15 is 30 m thick. The potentiometric surface elevation at two observation wells 1000 m a
AlekseyPX

Answer:

Part (a) The flow rate per unit width of the aquifer is 1.0875 m³/day

Part (b) The specific discharge of the flow is 0.0363 m/day

Part (c) The average linear velocity of the flow is 0.242 m/day

Part (d) The time taken for a tracer to travel the distance between the observation wells is 4132.23 days = 99173.52 hours

Explanation:

Part (a) the flow rate per unit width of the aquifer

From Darcy's law;

q = -Kb\frac{dh}{dl}

where;

q is the flow rate

K is the permeability or conductivity of the aquifer = 25  m/day

b is the aquifer thickness

dh is the change in th vertical hight = 50.9m - 52.35m = -1.45 m

dl is the change in the horizontal hight = 1000 m

q = -(25*30)*(-1.45/1000)

q = 1.0875 m³/day

Part (b) the specific discharge of the flow

V = \frac{Q}{A} = \frac{q}{b} = -K\frac{dh}{dl}\\\\V = -(25 m/d).(\frac{-1.45 m}{1000 m}) = 0.0363 m/day

V = 0.0363 m/day

Part (c) the average linear velocity of the flow assuming steady unidirectional flow

Va = V/Φ

Φ is the porosity = 0.15

Va = 0.0363 / 0.15

Va = 0.242 m/day

Part (d) the time taken for a tracer to travel the distance between the observation wells

The distance between the two wells = 1000 m

average linear velocity = 0.242 m/day

Time = distance / speed

Time = (1000 m) / (0.242 m/day)

Time = 4132.23 days

        = 4132.23 days *\frac{24 .hrs}{1.day} = 99173.52, hours

4 0
3 years ago
An object that is initially not rotating has a constant torque of 3.6 N⋅m applied to it. The object has a moment of inertia of 6
Korolek [52]

Answer:

0.6

Explanation:

Angular acceleration is equal to Net Torque divided by rotational inertia, which is the rotational equivalent to Newton’s 2nd Law.  Therefore, angular acceleration is equal to 3.6/6 which is 0.6. Hope this helped!

3 0
2 years ago
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