All categories

3 years ago

What are the 3 major uses of water? Describe each in 3-4 sentence

Physics

1 answer:

kirza4 [7]3 years ago

5 0

Answer:

-Drinking and Household Needs.

-Recreation.

-Industry and Commerce.

-Agriculture.

-Thermoelectricity/Energy.

Explanation:

You might be interested in

A silver bar 0.125 meter long is subjected to a temperature change from 200°C to 100°C. What will be the length of the bar after

Delicious77 [7]

$\Delta L= \alpha L_0 (T_f-T_i)$

= (18 x 10^-6 /°C)(0.125 m)(100° C - 200 °C)

= -0.00225 m

New length = L + ΔL
= 1.25 m + (-0.00225 m)
= 1.248

D

5 0

3 years ago

Read 2 more answers

The surface is tilted to an angle of 37 degrees from the horizontal, as shown above in Figure 3. The blocks are each given a pus

hoa [83]

Answer:

Incomplete question: "Each block has a mass of 0.2 kg"

The speed of the two-block system's center of mass just before the blocks collide is 2.9489 m/s

Explanation:

Given data:

θ = angle of the surface = 37°

m = mass of each block = 0.2 kg

v = speed = 0.35 m/s

t = time to collision = 0.5 s

Question: What is the speed of the two-block system's center of mass just before the blocks collide, vf = ?

Change in momentum:

$delta(P)=F*delta(t)$

$P_{f} -P_{i}=F*delta(t)$

$2m(v_{f} -v_{i})=F*delta(t)$

$v_{i} =0.35-0.35=0$

It is neccesary calculate the force:

$F=(m+m)*g*sin\theta$

Here, g = gravity = 9.8 m/s²

$F=(0.2+0.2)*9.8*sin37=2.3591N$

$v_{f} =\frac{F*delta(t)}{2m} =\frac{2.3591*0.5}{2*0.2} =2.9489m/s$

6 0

3 years ago

Raul dug a hole in his yard to repair a water pipe. It took him 2 seconds to apply a force of 50 Newtons to push the shovel 0.25

Sergeu [11.5K]

Answer:

Option B. 6.25 J/S

Explanation:

Data obtained from the question include:

t (time) = 2secs

F (force) = 50N

d (distance) = 0.25m

P (power) =?

The power can be obtained by using the formula P = workdone/time.

P = workdone / time

P = (50 x 0.25)/ 2

P = 6.25J/s

3 0

3 years ago

A spring stretches 0.2 cm per newton of applied force. An object is suspended from the spring and a deflection of 3 cm is observ

siniylev [52]

Answer:

$m=1.53kg$

Explanation:

To solve this problem we use the Hooke's Law:

$F=k*\Delta x$ (1)

F is the Force needed to expand or compress the spring by a distance Δx.

The spring stretches 0.2cm per Newton, in other words:

1N=k*0.2cm ⇒ k=1N/0.2cm=5N/cm

The force applied is due to the weight

$F=mg$

We replace in (1):

$mg=k*\Delta x$

We solve the equation for m:

$m=k*\Delta x/g=5*3/9.81=1.53kg$

7 0

3 years ago

g Two radiation modes (one at the center frequency lIo and the other at lIO+?lI) are excited with 1000 photons each. Determine t

Schach [20]

Answer:

a) P=0.25x10^-7

b) R=B*N2*E

c) N=1.33x10^9 photons

Explanation:

a) the spontaneous emission rate is equal to:

1/tsp=1/3 ms

the stimulated emission rate is equal to:

pst=(N*C*o(v))/V

where

o(v)=((λ^2*A)/(8*π*u^2))g(v)

g(v)=2/(π*deltav)

o(v)=(λ^2)/(4*π*tp*deltav)

Replacing values:

o(v)=0.7^2/(4*π*3*50)=8.3x10^-19 cm^2

the probability is equal to:

P=(1000*3x10^10*8.3x10^-19)/(100)=0.25x10^-7

b) the rate of decay is equal to:

R=B*N2*E, where B is the Einstein´s coefficient and E is the energy system

c) the number of photons is equal to:

N=(1/tsp)*(V/C*o)

Replacing:

N=100/(3*3x10^10*8.3x10^-19)

N=1.33x10^9 photons

7 0

3 years ago