Ask question

Ask question

All categories

3 years ago

5

a mars prototype carrying scientific instruments has a mass of 1060kg. a net force of 52000 n is applied to this roverat a test

site on earth . what is the acceleration of the rover? solve the acceleration. force/ mass a =F/m

1 answer:

choli [55]3 years ago

6 0

F = 52000 N

m = 1060 kg

a= F/m = 52000 N/1060 kg = 49.0566 m/s^2

You might be interested in

What happens when the object is placed at F? Explain<br> your answer.

34kurt

Answer:

Sample Response: No image will be formed because the rays will not converge to or diverge from a common point.

Explanation:

5 0

3 years ago

Calculate the number of vacancies per cubic meter in gold (au) at 900c. the energy for vacancy formation is 0.98 ev/atom. furthe

Sergeeva-Olga [200]

My mom said its f so i think its q

8 0

3 years ago

After a party the host pours the remnants of several bottles of wine into a jug. He then inserts a cork with a 2.15-cm diameter

pychu [463]

Answer:

3893.99675 N

Explanation:

$F_1$ = 125 N

$A_1$ = $\pi 1.075^2$

$A_2$ = $\pi 6^2$

From Pascal's law

$\dfrac{F_1}{A_1}=\dfrac{F_2}{A_2}\\\Rightarrow F_2=\dfrac{F_1\times A_2}{A_1}\\\Rightarrow F_2=\dfrac{125\times 6^2}{1.075^2}\\\Rightarrow F_2=3893.99675\ N$

The force that the liquid exerts on the bottom of the bottle is 3893.99675 N

5 0

3 years ago

The equation that is used to solve second law problems is # F= ma.

maw [93]

F= Force
M=Mass
A= acceleration
F=N
Mass= in grams or kilo grams (mostly kg)
A= m/s

8 0

3 years ago

Which data set has the largest standard deviation

AfilCa [17]

Answer:

<em>The data set marked as B has the largest standard deviation</em>

Explanation:

<u>Standard Deviation</u>

It's a number used to show how a set of measurements is spread out from the average value. A low standard deviation means that most of the values are close to the average. A high standard deviation means that the numbers are more spread out.

The formula for the standard deviation is

$\displaystyle \sigma=\sqrt{\frac{\sum (x_i-\mu)^2}{n}}$

Where $x_i$ is the value of each measurement, n is the number of elements in the set, and $\mu$ is the average or media of the values, defined as

$\displaystyle \mu=\frac{\sum x_i}{n}$

Let's analyze each set of data:

A.3,4,3,4,3,4,3

The average is

$\displaystyle \mu=\frac{3+4+3+4+3+4+3}{7}=3.43$

Computing the stardard deviation:

$\sigma=\sqrt{\frac{(3-3.43)^2+(4-3.43)^2+(3-3.43)^2+(4-3.43)^2+(3-3.43)^2+(4-3.43)^2+(3-3.43)^2}{7}}$

$\sigma=0.5$

B.1,6,3,15,4,12,8

The average is

$\displaystyle \mu=\frac{1+6+3+15+4+12+8}{7}=7$

Computing the stardard deviation:

$\sigma=\sqrt{\frac{(1-7)^2+(6-7)^2+(3-7)^2+(15-7)^2+(4-7)^2+(12-7)^2+(8-7)^2}{7}}$

$\sigma=4.7$

C. 20, 21,23,19,19,20,20

The average is

$\displaystyle \mu=\frac{20+21+23+19+19+20+20}{7}=20.29$

Computing the stardard deviation:

$\sigma=\sqrt{\frac{(20-20.29)^2+(21-20.29)^2+(23-20.29)^2+(19-20.29)^2+(19-20.29)^2+(20-20.29)^2+(20-20.29)^2}{7}}$

$\sigma=1.3$

D.12,14,13,14,12,13,12

The average is

$\displaystyle \mu=\frac{12+14+13+14+12+13+12}{7}=12.86$

Computing the stardard deviation:

$\sigma=\sqrt{\frac{(12-12.86)^2+(14-12.86)^2+(13-12.86)^2+(14-12.86)^2+(12-12.86)^2+(13-12.86)^2+(12-12.86)^2}{7}}$

$\sigma=0.8$

We can see the data set marked as B has the largest standard deviation

5 0

4 years ago