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2 years ago

Use the data to answer the question. Which statement best describes the data?

Physics

1 answer:

Natasha2012 [34]2 years ago

7 0

A statement which best describes the data is that: C. they are precise and reproducible.

<h3>What is an experiment?</h3>

An experiment can be defined as a scientific investigation which typically involves the process of manipulating an independent variable (the cause), so as to determine or measure the dependent variable (the effect).

Data can be defined as a representation of instructions or factual information in a formalized manner, especially as a series of binary digits (bits) or samples that are used by scientist in an experiment or research.

<h3>The types of data.</h3>

Generally, there are two (2) main types of data and these include:

Primary data

Secondary data

Based on the data provided in the table (see attachment), we can infer and logically deduce that a statement which best describes the data is that they are precise and reproducible.

Read more on experiments here: brainly.com/question/25322325

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Complete Question:

Which statement best describes the data?

A. They are accurate and reproducible.

B. They are accurate but not reproducible.

C. They are precise and reproducible.

D. They are precise but not reproducible.

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Find the sum of the vectors:11 km N ,11km E

vladimir1956 [14]

The resultant vector is 11√2 km due north east.

<h3><u>Explanation:</u></h3>

The vector is a type of quantity which has both magnitude and direction. This quantities when expressed needs to specify both magnitude and direction.

We need to calculate the magnitude and direction separately.

Here firstly for the magnitude,

The magnitudes are both 11 km and they are at right angles to each other.

So, the resultant magnitude = √(11² +11²) km

=11√2 km

Now for the direction, one vector is due north and the other is due east.

So the resultant vector is due north east.

So the final vector is 11√2 km due North-East.

6 0

3 years ago

Identify the statement that is true about mass extinctions.A. Mass-extinction events are times in the Earth's history when few s

nikklg [1K]

Answer:

A. Mass-extinction events are times in the Earth's history when few species evolve.

Explanation:

The extinction that occurred 65 million years ago wiped out some 50 percent of plants and animals. The event is so striking that it signals a major turning point in Earth's history, marking the end of the geologic period known as the Cretaceous and the beginning of the Tertiary period.

A wide range of animals and plants suddenly died out, from tiny marine organisms to large dinosaurs.

Species go extinct all the time. Scientists estimate that at least 99.9 percent of all species of plants and animals that ever lived are now extinct.

3 0

3 years ago

Please help me with this

AfilCa [17]

Answer:

20 N exerts no torque about the pivot.

14 N exerts a counterclockwise torque of 14 * .3 = .42 N-m

6 exerts a clockwise torque of 6 * .7 = .42 N-m

The meter stick will not turn because there is no net torque on the meter stick.

3 0

2 years ago

Scientists want to place a 3400.0 kg satellite in orbit around Mars. They plan to have the satellite orbit at a speed of 2480.0

BaLLatris [955]

Answer:

Part a)

$r = 6.96 \times 10^6 m$

Part b)

$F_g = 3.004 \times 10^3 N$

Part c)

$a = 0.88 m/s^2$

Part d)

$v = \frac{2480}{2} = 1240 m/s$

Explanation:

Part a)

As we know that the orbital speed of the satellite is given as

$v = 2480 m/s$

now we will have

$v = \sqrt{\frac{GM_{mars}}{r}}$

now we have

$M_{mars} = 6.4191 \times 10^{23} kg$

$R_{mars} = 3.397 \times 10^6 m$

now we have

$2480 = \sqrt{\frac{(6.67 \times 10^{-11})(6.4191 \times 10^{23})}{r}}$

$r = 6.96 \times 10^6 m$

Part b)

Here force between mars and satellite is the gravitational attraction force which is given as

$F_g = \frac{GM_{mars} m}{r^2}$

$F_g = \frac{(6.67\times 10^{-11})(6.4191 \times 10^{23})(3400)}{(6.96\times 10^6)^2}$

$F_g = 3.004 \times 10^3 N$

Part c)

Acceleration of satellite is the ratio of gravitational force and mass of the satellite

$a = \frac{F_g}{m}$

$a = \frac{3004}{3400}$

$a = 0.88 m/s^2$

Part d)

As we know by III law of kepler

$\frac{T_1^2}{T_2^2} = \frac{r_1^3}{r_2^3}$

here we know that T2 = 8 T1

$(\frac{1}{8})^2= \frac{r_1^3}{r_2^3}$

$\frac{r_1}{r_2} = (\frac{1}{2})^2$

so we have

$r_2 = 4r_1$

as we know that speed is given as

$v = \sqrt{\frac{GM}{r}}$

so we can say since radius is orbit becomes 4 times so the orbital speed must be half

$v = \frac{2480}{2} = 1240 m/s$

7 0

4 years ago

Review Conceptual Example 7 before starting this problem. A uniform plank of length 5.0 m and weight 225 N rests horizontally on

Andreas93 [3]

Answer:

0.82 m

Explanation:

We are given that

Weight, $w_1$ =225 N

Length,l=5 m

d=1.1 m

We have to find the distance x for which person weighs 385 N walk on the overhanging part of the plank before it just begins to tip.

Half length= $\frac{l}{2}=\frac{5}{2}=2.5 m$

$r=\frac{l}{2}-d=2.5-1.1=1.4 m$

$w_2=385 N$

When system in equilibrium then

$w_1r=w_2 x$

$225\times 1.4=385\times x$

$x=\frac{225\times 1.4}{385}$

$x=0.82 m$

6 0

4 years ago