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3 years ago

What are some of the obstacles to exploration of outer planets or even Venus right next door?

Physics

1 answer:

weqwewe [10]3 years ago

5 0

Answer:

Distance and coping with the extreme atmospheres

Explanation:

The outer planets and venus are so far away that they are difficult to visit. Even if they were closer, the extreme atmospheric pressure and weather on these planets would make them difficult to explore.

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What is the prefix notation of 0.0000738?

MrMuchimi

Answer:

7.38 × 10-5

Explanation:

All numbers in scientific notation or standard form are written in the form m × 10n, where m is a number between 1 and 10 ( 1 ≤ |m| < 10 ) and the exponent n is a positive or negative integer.

To convert 0.0000738 into scientific notation, follow these steps:

Move the decimal 5 times to the right in the number so that the resulting number, m = 7.38, is greater than or equal to 1 but less than 10

Since we moved the decimal to the right the exponent n is negative

n = -5

Write in the scientific notation form, m × 10n

= 7.38 × 10-5

Therefore, the decimal number 0.0000738 written in scientific notation is 7.38 × 10-5 and it has 3 significant figures.

Answer = 7.38 × 10-5

4 0

3 years ago

Read 2 more answers

As we learn more, _________ are often revised. scientific laws scientific theories observation experiments

DiKsa [7]

Here is the answer that would best complete the given statement above. <span>As we learn more, SCIENTIFIC THEORIES are often revised. Scientific theories are considered as the substantial explanation of some aspect of the natural world which are gathered through scientific method. Hope this is the answer that you are looking for.</span>

4 0

4 years ago

Read 2 more answers

dayshawn is traveling at 12 m/s away from the school. dayshaun's mom is looking for him because he is late coming home so she le

vivado [14]

Speed of Dayshawn travelling towards his home is 12 m/s

Speed of her mom towards his school is 5 m/s

They both starts at same time so whenever they will meet on their path the sum of the distance covered by Dayshawn and distance covered by his mom must be equal to the total distance of school and home

Now let say they both meet after "t" time when they starts motion

so we can write the total distance between school and home as

$d = v_1*t + v_2*t$

here d = 6492 m

$v_1 = 12 m/s$ = speed of dayshawn

$v_2 = 5 m/s$ = speed of his mom

now by solving the above equation

$6492 = 12t + 5 t$

$t = \frac{6492}{17}$

$t = 381.9 s$

so they will meet after 381.9 s from start which will be 3.36 minutes from there start

Also at this time the distance covered by her mom will be

$d_2 = v_2*t$

$d_2 = 5* 381.9 = 1909.4 m$

so they will meet at distance 1909.4 m from their home

4 0

4 years ago

What’s the force of a pitching machine on a baseball? Baseballs pitched by a machine have a horizontal velocity of 30 meters/sec

Leya [2.2K]

The acceleration of the baseball is:
$a= \frac{v_f-v_i}{\Delta t}$
where $v_f$ and $v_i$ are the final and initial speed of the ball, and $\Delta t$ is the time interval in which the force acted.

Replacing the numbers, we get
$a= \frac{30 m/s-0m/s}{0.5 s}=60 m/s^2$
And at this point, we can use Newton's second law F=ma to find the value of the force of the pitching machine:
$F=ma=(0.15 kg)(60 m/s^2)=9 N$

5 0

3 years ago

The electric force between two charged objects of charge +Q0 that are separated by a distance R0 is F0. The charge of one of the

vovikov84 [41]

Answer:

$F=3F_o(\frac{R_o}{R_{o2}})^2$

Explanation:

This problem is approached using Coulomb's law of electrostatic attraction which states that the force F of attraction or repulsion between two point charges, $Q_1$ and $Q_2$ is directly proportional to the product of the charges and inversely proportional to the square of their distance of separation R.

$F=\frac{kQ_1Q_2}{R^2}..................(1)$

where k is the electrostatic constant.

We can make k the subject of formula as follows;

$k=\frac{FR^2}{Q_1Q_2}...........(2)$

Since k is a constant, equation (2) implies that the ratio of the product of the of the force and the distance between two charges to the product of charges is a constant. Hence if we alter the charges or their distance of separation and take the same ratio as stated in equation(2) we will get the same result, which is k.

According to the problem, one of the two identical charges was altered from $Q_o$ to $3Q_o$ and their distance of separation from $R_o$ to $R_{o2}$ , this also made the force between them to change from $F_o$ to $F_{o2}$ . Therefore as stated by equation (2), we can write the following;