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andreyandreev [35.5K]
3 years ago
6

The weight of air measured in units of force per area is called _____.

Physics
2 answers:
laila [671]3 years ago
8 0
The weight of air measured in units of force per area is called air pressure.
ElenaW [278]3 years ago
8 0

Answer:

B. air pressure

Explanation:

The definition of pressure is that it is a force applied to an area.

P=\frac{F}{A}

Because weight is a force (a force is mass multiplied by acceleration, and weight is the product of the mass of an object multiplied by the acceleration of gravity, thus weight is a force), the measurement of the weight of the air over area is the air pressure.

The correct answe is B.

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Sharon is driving on a straight road. She is driving north, and her speed is
sdas [7]

Answer:

b

Explanation:

b is the correct answer

4 0
3 years ago
The idea that John Marshall, the first Chief Justice of the Supreme Court, singularly established the principle of judicial revi
____ [38]

Answer:

c. Was an idea created and supported by Congress.

Explanation:

The idea that John Marshall, the first Chief Justice of the Supreme Court, singularly established the principle of judicial review in Marbury v. Madison(1803) was an idea created and supported by Congress.

8 0
3 years ago
How will heat flow between objects?
alukav5142 [94]
Without an external agent doing work, heat will always flow from a hotter to a cooler object. Two objects of different temperature always interact. There are three different ways for heat to flow from one object to another. They are conduction, convection, and radiation.
5 0
4 years ago
Read 2 more answers
A charge of Q is fixed in space. A second charge of q was first placed at a distance r1 away from Q. Then it was moved along a s
topjm [15]

Answer:

\Delta U = \frac{Qq}{4\pi\epsilon_0}(\frac{1}{r_2^2}-\frac{1}{r_1^2})

Explanation:

The electrostatic potential energy is given by the following formula

U = \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}

Now, we will apply this formula to both cases:

U_1 = \frac{1}{4\pi\epsilon_0}\frac{Qq}{r_1^2}\\U_2 = \frac{1}{4\pi\epsilon_0}\frac{Qq}{r_2^2}

So, the change in the potential energy is

\Delta U = U_2 - U_1 = \frac{Qq}{4\pi\epsilon_0}(\frac{1}{r_2^2}-\frac{1}{r_1^2})

7 0
3 years ago
A 5-cm-high peg is placed in front of a concave mirror with a radius of curvature of 20 cm.
Andrei [34K]

Answer:

Explanation:

Using the magnification formula.

Magnification = Image distance(v)/object distance(u) = Image Height(H1)/Object Height(H2)

M = v/u = H1/H2

v/u = H1/H2...1

3) Given the radius of curvature of the concave lens R = 20cm

Focal length F = R/2

f = 20/2

f = 10cm

Object distance u = 5cm

Object height H2= 5cm

To get the image distance v, we will use the mirror formula

1/f = 1/u+1/v

1/v = 1/10-1/5

1/v = (1-2)/10

1/v =-1/10

v = -10cm

Using the magnification formula

(10)/5 = H1/5

10 = H1

H1 = 10cm

Image height of the peg is 10cm

4) If u = 15cm

1/v = 1/f-1/u

1/v = 1/10-1/15

1/v = 3-2/30

1/v = 1/30

v = 30cm

30/15 = H1/5

15H1 = 150

H1/= 10cm

5) if u = 20cm

1/v = 1/f-1/u

1/v = 1/10-1/20

1/v = 2-1/20

1/v = 1/20

v = 20cm

20/20 = H1/5

20H1 = 100

H1 = 5cm

6) If u = 30cm

1/v = 1/f-1/u

1/v = 1/10-1/30

1/v = 3-1/30

1/v = 2/30

v = 30/2 cm

v =>15cm

15/30 = Hi/5

30H1 = 75

H1 = 75/30

H1 = 2.5cm

4 0
3 years ago
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