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2 years ago

What is the acceleration of a rocket that speeds up from 50 m/s to 1000 m/s in 3 seconds?

Physics

1 answer:

Zepler [3.9K]2 years ago

8 0

Answer:

= 316.67 m/s^2

Explanation:

Acceleration formula =

A = $\frac{V_{f} - V_{i} }{t}$

$A = \frac{1000 - 50}{3}$

= $\frac{950}{3}$

= 316.67 m/s^2

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A soccer player runs 75m in a straight line down a soccer field in 11 seconds. What

Ratling [72]

Answer:

6.9m

Explanation:

I divide 75m by 12 seconds.

Please give BRAINLIEST

5 0

2 years ago

A scientist who studies teenage behavior was interested in determining if teenagers spend more time playing computer games then

natita [175]

Answer:

greater than 0.10

Explanation:

The null hypothesis is:

$H_{0} = 10.2$

The alternate hypotesis is:

$H_{1} > 10.2$

Our test statistic is:

$t = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

In which X is the statistic, $\mu$ is the mean, $\sigma$ is the standard deviation and n is the size of the sample.

We have that:

$t = 0.45$

We are testing if X is greater than 0.45, so our pvalue is 1 subtracted by the pvalue of z = t = 0.45.

z = 0.45 has a pvalue of 0.6736

1 - 0.6735 = 0.3264

So our pvalue is 0.3264, which is greater than 0.10.

So the correct answer is:

greater than 0.10

7 0

3 years ago

A uranium and iron atom reside a distance R = 37.50 nm apart. The uranium atom is singly ionized; the iron atom is doubly ionize

postnew [5]

Answer:

r=15.53 nm

$F=9.57\times 10^{-13}N$

Explanation:

Lets take electron is in between iron and uranium

Charge on electron $q_1= -1.602\times 10^{-19}C$

Charge on iron $q_2= 2\times 1.602\times 10^{-19}C$

Charge on uranium $q_3= 1.602\times 10^{-19}C$

We know that force between two charge

$F=K\dfrac{q_1 q_2}{r^2}$

$K=9\times 10^9\dfrac{N-m^2}{c^2}$

For equilibrium force between electron and iron should be force between electron and uranium

Lets take distance between electron and uranium is r so distance between electron and iron will be 37.5-r nm

Now by balancing the force

$K\dfrac{q_1 q_2}{r^2}=K\dfrac{q_1 q_3}{(37.5-r)^2}$

$K\dfrac{q_1q_2}{(37.5-r)^2}=K\dfrac{q_1 q_3}{r^2}$

$q_2= 2\timesq_1,q_3=q_1$

$\dfrac{q_1\times 2\timesq_1}{r^2}=\dfrac{q_1\times q_1}{(37.5-r)^2}$

So r=15.53 nm

So force

$F=9\times 10^9\dfrac{1.602\times 10^{-19}\times 1.602\times 10^{-19}}{(15.53\times 10^{-9})^2}$

$F=9.57\times 10^{-13}N$

7 0

3 years ago

In music, the note G above middle C has a frequency of about 392 hertz. If the speed of sound in the air is 340 m/s, what is the

morpeh [17]

Answer:

0.87 meters

Explanation:

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2 years ago

The National Ambient Air Quality Standards (NAAQS) are maximum allowable levels for _____ harmful pollutants. sixteen six sixty

Harman [31]

6

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3 years ago

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