All categories

3 years ago

What do repeated trails in an experiment allow scientists to do?

Physics

1 answer:

fredd [130]3 years ago

3 0

Answer:

Explanation:

When we do multiple trials of the same experiment, we can make sure that our results are consistent and not altered by random events. Multiple trials can be done at one time. If we were testing a new fertilizer, we could test it on lots of individual plants at the same time.

You might be interested in

At a certain distance from the center of the Earth, a 0.4-kg object has a weight of 2.0 N. (a) Find this distance. (b) If the ob

Alika [10]

Answer:

a) The distance of the object from the center of the Earth is 8.92x10⁶ m.

b) The initial acceleration of the object is 5 m/s².

Explanation:

a) The distance can be found using the equation of gravitational force:

$F = \frac{GMm}{r^{2}}$

Where:

G: is the gravitational constant = 6.67x10⁻¹¹ Nm²/kg²

M: is the Earth's mass = 5.97x10²⁴ kg

m: is the object's mass = 0.4 kg

F: is the force or the weight = 2.0 N

r: is the distance =?

The distance is:

$r = \sqrt{\frac{GMm}{F}} = \sqrt{\frac{6.67 \cdot 10^{-11} Nm^{2}/kg^{2}*5.97 \cdot 10^{24} kg*0.4 kg}{2.0 N}} = 8.92 \cdot 10^{6} m$

Hence, the distance of the object from the center of the Earth is 8.92x10⁶ m.

b) The initial acceleration of the object can be calculated knowing the weight:

$W = ma$

Where:

W: is the weight = 2 N

a: is the initial acceleration =?

$a = \frac{W}{m} = \frac{2 N}{0.4 kg} = 5 m/s^{2}$

Therefore, the initial acceleration of the object is 5 m/s².

I hope it helps you!

4 0

3 years ago

A 200. kg object is pushed 12.0 m to the top of an incline to a height of 6.0 m. If the force applied along the incline is 3000.

Nataliya [291]

Answer:

Approximately $1.2 \times 10^{4}\; {\rm J}$ (assuming that $g = 9.81\; {\rm N \cdot kg^{-1}}$ .)

Explanation:

The strength of the gravitational field near the surface of the earth is approximately constant: $g = 9.81\; {\rm N \cdot kg^{-1}}$ .

The change in the gravitational potential energy ( ${\rm GPE}$ ) of an object near the surface of the earth is proportional to the change in the height of this object. If the height of an object of mass $m$ increased by $\Delta h$ , the ${\rm GPE}$ of that object would have increased by $m\, g\, \Delta h$ .

In this question, the height of this object increased by $\Delta h = 6.0\; {\rm m}$ . The mass of this object is $m = 200\; {\rm kg}$ . Thus, the ${\rm GPE}$ of this object would have increased by:

$\begin{aligned}& (\text{Change in GPE}) \\ =\; & m\, g\, \Delta h \\ =\; & 200\; {\rm kg} \times 9.81\; {\rm N \cdot kg^{-1}} \times 6.0\; {\rm m} \\ \approx\; & 1.2 \times 10^{4}\; {\rm J}\end{aligned}$ .

(Note that $1\; {\rm N \cdot m} = 1\; {\rm J}$ .)

3 0

2 years ago

Which option is the most accurate description of speed?

puteri [66]

B I think is the answer

5 0

3 years ago

What is the smallest or most specific level of organization you and your neighbors are have in common

kow [346]

We have the same house

3 0

3 years ago

Why does an object traveling in a circle at a constant speed always accelerate ?

LUCKY_DIMON [66]

Because "acceleration" doesn't mean "speeding up". It means any change
in the speed or direction of motion.

An airplane speeding up, a bicycle slowing down, and a car going around a curve, are all doing accelerated motion.

A circle has no straight parts, so an object traveling in a circle is always changing
its direction. That means accelerated motion, even if its speed doesn't change..

6 0

3 years ago

Read 2 more answers