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Aleks04 [339]
3 years ago
7

Which tool would you use to measure the amount of rainfall?

Physics
2 answers:
alina1380 [7]3 years ago
8 0

Answer:

A. Graduated Cylinder

Explanation:

A thermometer would just tell you the temperature.

A scale would just tell you how much it weighs.

And a stop watch would do nothing but show numbers until it got wet enough and broke.

The only logical answer is A since it is also a measuring flask WITH incriments already on the side of it to tell you how much water you collect.

Hope this helps!!

zheka24 [161]3 years ago
5 0

Answer:

a

Explanation:

You might be interested in
Pages 64 to 65 of your reading material trace an extended problem that involes lifting a bag of sugar up to a shelf at first it
nadezda [96]

Answer:

d) the amount of work is the same whether the bag is moved all at once or in two stages, provided the total height lifted is the same in either case.

Explanation:

While moving the bag to the shelf in one shot we can say that the total work done is given as

W = mg(2H)

here we know that

2H = total height raised by the bag

now when we raise the bag to first shelf and then move it to next shelf

then we will have

W = W_1 + W_2[tex][tex]W = mgH + mgH

W = 2mgH

so the correct answer will be

d) the amount of work is the same whether the bag is moved all at once or in two stages, provided the total height lifted is the same in either case.

7 0
3 years ago
The cabinet is mounted on coasters and has a mass of 45 kg. The casters are locked to prevent the tires from rotating. The coeff
stira [4]

Answer:

the force P required for impending motion is 132.3 N

the largest value of "h" allowed if the cabinet is not to tip over is 0.8 m

Explanation:

Given that:

mass of the cabinet  m = 45 kg

coefficient of static friction μ =  0.30

A free flow body diagram illustrating what the question represents is attached in the file below;

The given condition from the question let us realize that ; the casters are locked to prevent the tires from rotating.

Thus; considering the forces along the vertical axis ; we have :

\sum f_y =0

The upward force and the downward force is :

N_A+N_B = mg

where;

\mathbf { N_A  \ and  \ N_B} are the normal contact force at center point A and B respectively .

N_A+N_B = 45*9.8

N_A+N_B = 441    ------- equation (1)

Considering the forces on the horizontal axis:

\sum f_x = 0

F_A +F_B  = P

where ;

\mathbf{ F_A \ and \ F_B } are the static friction at center point A and B respectively.

which can be written also as:

\mu_s N_A + \mu_s N_B  = P

\mu_s( N_A +  N_B)  = P

replacing our value from equation (1)

P = 0.30 ( 441)    

P = 132.3 N

Thus; the force P required for impending motion is 132.3 N

b) Since the horizontal distance between the casters A and B is 480 mm; Then half the distance = 480 mm/2 = 240 mm = 0.24 cm

the largest value of "h" allowed for  the cabinet is not to tip over is calculated by determining the limiting condition  of the unbalanced torque whose effect is canceled by the normal reaction at N_A and it is shifted to N_B:  

Then:

\sum M _B = 0

P*h = mg*0.24

h =\frac{45*9.8*0.24}{132.3}

h = 0.8 m

Thus; the largest value of "h" allowed if the cabinet is not to tip over is 0.8 m

6 0
3 years ago
The wavelength of sound wave is 30 m. Frequency is 7 Hz. What is the wave speed
MaRussiya [10]

Answer:23hz

Explanation:

6 0
3 years ago
Why do you think sound travels faster in steel than it does in wood? (Hint: Think about the densities of the two materials, and
Crazy boy [7]

Answer:

If sound waves of the same energy were passed through a block of wood and a block of steel, which is more dense than the wood, the molecules of the steel would vibrate at a slower rate. Sound moves faster through denser air because the molecules are closer together in dense air and sound can be more easily passed on.

Explanation:

6 0
3 years ago
Saturated water vapor at 200 kPa is condensed into a saturated liquid via a constant-pressure process inside of a piston-cylinde
evablogger [386]

Answer:

The process is not possible

Explanation:

if we want to determine if the process is possible , we can check with the second law of thermodynamics

ΔS≥ ∫dQ/T

for a constant temperature process ( condensation)

ΔS≥ 1/T ∫dQ

and from the first law of thermodynamics

ΔH = Q - ∫VdP , but P=constant → dP=0 → ∫VdP=0

Q=ΔH

then

ΔS≥ΔH/T

from steam tables

at P= constant = 200 Kpa → T= 120°C = 393 K

at P= constant → H vapor = 2201.5 kJ/kg ,  H liquid = 1.5302 kJ/kg

, S vapor= 7.1269 kJ/kg , S liquid 1.7022 kJ/kg

therefore

ΔH = H vapor - H liquid = 2201.5 kJ/kg -  1.5302 kJ/kg = 2199.9698 kJ/kg

ΔS = S vapor - S liquid = 7.1269 kJ/kg - 1.7022 kJ/kg = 5.4247 kJ/kg

therefore since

ΔS required  = ΔH/T = 2199.9698 kJ/kg/(393 K)= 5.597 kJ/kg K

and

ΔS= 5.4247 kJ/kg  ≤ ΔS required=5.597 kJ/kg K

the process is not possible

5 0
3 years ago
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