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2 years ago

Is the meter, centimeter, or millimeter more useful for measuring the length and width of your classroom? why?

Physics

1 answer:

Verizon [17]2 years ago

8 0

A meter because centimeters and millimeters are way smaller than meters

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What would MOST LIKELY happen if the amount of incoming solar energy decreased without a change in the amount of reflection or o

777dan777 [17]

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8 0

2 years ago

Read 2 more answers

What pressure will 14. 0 g of co exert in a 3. 5 l container at 75°c?

Ronch [10]

The pressure will 14. 0 g of co exert in a 3. 5 l container at 75°c is 4.1atm.

Therefore, option A is correct option.

Given,

Mass m = 14g

Volume= 3.5L

Temperature T= 75+273 = 348 K

Molar mass of CO = 28g/mol

Universal gas constant R= 0.082057L

Number of moles in 14 g of CO is

n= mass/ molar mass

= 14/28

= 0.5 mol

As we know that

PV= nRT

P × 3.5 = 0.5 × 0.082057 × 348

P × 3.5 = 14.277

P = 14.277/3.5

P = 4.0794 atm

P = 4.1 atm.

Thus we concluded that the pressure will 14. 0 g of co exert in a 3. 5 l container at 75°c is 4.1atm.

learn more about pressure:

brainly.com/question/22613963

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5 0

11 months ago

During which two time intervals does the particle undergo equal displacement?

san4es73 [151]

Answer:

BC and DE

Explanation:

In the given figure, the velocity time graph is shown. We know that the area under v-t curve gives the displacement of the particle.

Area under AB, $d_1=\dfrac{1}{2}\times 2\times 10=10\ m$

Area under BC, $d_2=\dfrac{1}{2}\times 2\times 4=4\ m$

Area under CD, $d_3=\dfrac{1}{2}\times 2\times 7=7\ m$

Area under DE, $d_4=\dfrac{1}{2}\times 2\times 4=4\ m$

Area under EF, $d_5=\dfrac{1}{2}\times 2\times 3=3\ m$

So, form above calculations it is clear that, during BC and DE undergo equal displacement. Hence, the correct option is (c) "BC and DE = 4 meters".

4 0

3 years ago

A 1.50 cm high diamond ring is placed 20.0 cm from a concave mirror with radius of curvature 30.00 cm. The magnification is ____

Rama09 [41]

Answer:

Magnification, m = -0.42

Explanation:

It is given that,

Height of diamond ring, h = 1.5 cm

Object distance, u = -20 cm

Radius of curvature of concave mirror, R = 30 cm

Focal length of mirror, f = R/2 = -15 cm (focal length is negative for concave mirror)