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3 years ago

What is a benefit of having a standard measurement system that can be used by scientists worldwide

1 answer:

4 0

The current international standard metric system, it is mks system based on metre, kilogram and second as well as the kelvin ampere candela and mole

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What type of image results when the object is located between 2f (2 focal lengths) and f (the focal point) of a convex lens?

VikaD [51]

Answer:

Explanation:

the image formed is real but inverted and magnified

you can remember it by R I M

hope this helps

4 0

3 years ago

The Brazilian rain forest is an area with significant biodiversity. As the rain forest is replaced with agricultural land, it is

valina [46]

B I think I’m not sure tho

4 0

2 years ago

Read 2 more answers

17. Which of the following is the closest weight in newtons of a 7.0-kilogram

marusya05 [52]

Weight = (mass) x (gravity)

Weight = (7.0 kg) x (gravity)

On Earth, where (gravity) is roughly 10 N/kg . . .

Weight = (7.0 kg) x (roughly 10 N/kg)

Weight = roughly 70 Newtons

That's B on Earth.

It would be some other number on other bodies.

4 0

3 years ago

You attach a meter stick to an oak tree, such that the top of the meter stick is 2.27 meters above the ground. later, an acorn f

Alexandra [31]

The acorn was at a height of 4.15 m from the ground before it drops.

The acorn takes a time t to fall through a distance h₁, which is the length of the scale. When the acorn reaches the top of the scale, its velocity is u.

Calculate the speed of the acorn at the top of the scale, using the equation of motion,

$s=ut+ \frac{1}{2} at^2$

Since the acorn falls freely under gravity, its acceleration is equal to the acceleration due to gravity g.

Substitute 2.27 m for s (=h₁), 0.301 s for t and 9.8 m/s² for a (=g).

$s=ut+ \frac{1}{2} at^2\\ (2.27 m)=u(0.301s)+\frac{1}{2}(9.8m/s^2)(0.301s)^2\\ u=\frac{1.8261m}{0.301s} =6.067m/s$

If the acorn starts from rest and reaches a speed of 6.067 m/s at the top of the scale, it would have fallen a distance h₂ to achieve this speed.

Use the equation of motion,

$v^2=u^2+2as$

Substitute 6.067 m/s for v, 0 m/s for u, 9.8 m/s² for a (=g) and h₂ for s.

$v^2=u^2+2as\\ (6.067m/s)^2=(0m/s)^2+2(9.8m/s^2)h_2\\ h_2=\frac{(6.067m/s)^2}{2(9.8m/s^2)} =1.878 m$

The height h above the ground at which the acorn was is given by,

$h=h_1+h_2=(2.27 m)+(1.878 m)=4.148 m$

The acorn was at a height 4.15m from the ground before dropping down.

3 0

2 years ago

Digitized information can be used by computers true or false

Slav-nsk [51]

It is true that digitized information is used by computers

8 0

2 years ago

Read 2 more answers