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NNADVOKAT [17]
2 years ago
8

A weather forecast for a certain day in an area shows two measurements, 18 degrees Celsius and 25 degrees Celsius. What do these

measurements indicate?
The change in air pressure as you go higher
The change in air temperature as you go higher
The low and high air pressures
The low and high air temperatures
Physics
1 answer:
slamgirl [31]2 years ago
4 0
The low and high air temperatures
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Setler79 [48]
I would say the third... force. 

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___________ force can be thought of as either a push or a pull, as long as it makes the object move in a circular path.
fomenos
I believe it would be the Centripetal force.
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A hammer exerts 49.8 N of force on the head (r=0.00510 m) of a nail. How much pressure does it exert on the nail?
Kisachek [45]

Answer:

609547.12 Pa ≈ 6.10×10^5 Pa

Explanation:

Step 1:

Data obtained from the question. This include the following:

Force (F) = 49.8 N

Radius (r) = 0.00510 m

Pressure (P) =..?

Step 2:

Determination of the area of the head of the nail.

The head of a nail is circular in nature. Therefore, the area is given by:

Area (A) = πr²

With the above formula we can obtain the area as follow:

Radius (r) = 0.00510 m

Area (A) =?

A = πr²

A = π x (0.00510)²

A = 8.17×10^-5 m²

Therefore the area of the head of the nail is 8.17×10^-5 m²

Step 3:

Determination of the pressure exerted by the hammer.

This is illustrated below:

Force (F) = 49.8 N

Area (A) = 8.17×10^-5 m²

Pressure (P) =..?

Pressure (P) = Force (F) /Area (A)

P = F/A

P = 49.8/8.17×10^-5

P = 609547.12 N/m²

Now, we shall convert 609547.12 N/m² to Pa.

1 N/m² = 1 Pa

Therefore, 609547.12 N/m² = 609547.12 Pa.

Therefore, the pressure exerted by the hammer on the nail is 609547.12 Pa or 6.10×10^5 Pa

8 0
4 years ago
The work-energy theorem states that a force acting on a particle as it moves over a ______ changes the ______ energy of the part
svetoff [14.1K]

Answer:

B. distance/potential

Explanation:

Quizlet

5 0
3 years ago
A car is rounding a 100-m-radius curve at 25 m/s.What is the minimum possible coefficient of static friction between the tires a
Crazy boy [7]

Answer:

The minimum possible coefficient of static friction between the tires and the ground is 0.64.

Explanation:

if the μ is the coefficient of static friction and R is radius of the curve and v is the speed of the car then, one thing we know is that along the curve, the frictional force, f will be equal to the centripedal force, Fc and this relation is :

Fc = f

m×(v^2)/(R) = μ×m×g

    (v^2)/(R) = g×μ

               μ = (v^2)/(R×g)

                  =  ((25)^2)/((100)×(9.8))

                  = 0.64

Therefore, the minimum possible coefficient of static friction between the tires and the ground is 0.64.

4 0
3 years ago
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