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jek_recluse [69]

2 years ago

14

If you are pushing a box toward your friend with a force of 20 N, and your friend is pushing the box toward you with a force of

30 N, what will happen to the box?

1 answer:

yan [13]2 years ago

7 0

Answer : the box will move backwards with a force 10 N

Explanation :

F1 = 20 N

F2 = 30 N

F = F2 - F1 = 30 - 20 = 10 N

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A scientist is conducting research to determine how much pressure a dam can withstand. Why are good observations important to hi

Natalija [7]

The things that a scientist should consider while observing the force is the environmental conditions,the force that is expected to act on the dam, the means to contain that force, and compare different types of designs in accordance with the location of the dam

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2 years ago

What is the wavelength, in nm, of the line in the hydrogen spectrum when one n value is 3 and the other n value is 6?

iren [92.7K]

Answer:

$\lambda=1090nm$

Explanation:

Rydberg formula is used to calculate the wavelengths of the spectral lines of many chemical elements. For the hydrogen, is defined as:

$\frac{1}{\lambda}=R_H(\frac{1}{n_1^2}-\frac{1}{n_2^2})$

Where $R_H$ is the Rydberg constant for hydrogen and $n_1$ , $n_2$ are the lower energy state and the higher energy state, respectively.

$\frac{1}{\lambda}=1.10*10^{7}m^{-1}(\frac{1}{3^2}-\frac{1}{6^2})\\\frac{1}{\lambda}=9.17*10^{5}m^{-1}\\\lambda=\frac{1}{1.09*10^{6}m^{-1}}\\\lambda=1.09*10^{-6}m*\frac{10^{9}nm}{1m}\\\lambda=1090nm$

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3 years ago

When Jim and Rob ride bicycles, Jim can only accelerate at three-quarters the acceleration of Rob. Both startfrom rest at the bo

Natali5045456 [20]

Answer:

46.4 s

Explanation:

5 minutes = 60 * 5 = 300 seconds

Let g = 9.8 m/s2. And $\theta$ be the slope of the road, s be the distance of the road, a be the acceleration generated by Rob, 3a/4 is the acceleration generated by Jim . Both of their motions are subjected to parallel component of the gravitational acceleration $gsin\theta$

Rob equation of motion can be modeled as s = a_Rt_R^2/2 = a300^2/2 = 45000a[/tex]

Jim equation of motion is $s = a_Jt_J^2/2 = (3a/4)t_J^2/2 = 3at_J^2/8$

As both of them cover the same distance

$45000a = 3at_J^2/8$

$t_J^2 = 45000*8/3 = 120000$

$t_J = \sqrt{120000} = 346.4 s$

So Jim should start 346.4 – 300 = 46.4 seconds earlier than Rob in other to reach the end at the same time

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3 years ago

Where did the missouri compromise imaginary line run

lara31 [8.8K]

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3 years ago

Brainliest please help<br><br>tell me if am right <br>if not correct me <br><br><br>please

REY [17]

Answer:

See the answers below

Explanation:

To solve this problem we must use the following equation of kinematics.

$v_{f}=v_{o}+a*t$

where:

Vf = final velocity [m/s]

Vo = initial velocity [m/s]

a = acceleration [m/s²]

t = time [s]

<u>First case</u>

Vf = 6 [m/s]

Vo = 2 [m/s]

t = 2 [s]

$6=2+a*2\\4=2*a\\a=2[m/s^{2} ]$

<u>Second case</u>

Vf = 25 [m/s]

Vo = 5 [m/s]

a = 2 [m/s²]

$25=5+2*t\\t = 10 [s]$

<u>Third case</u>

Vo =4 [m/s]

a = 10 [m/s²]

t = 2 [s]

$v_{f}=4+10*2\\v_{f}=24 [m/s]$

<u>Fourth Case</u>

Vf = final velocity [m/s]

Vo = initial velocity [m/s]

a = acceleration [m/s²]

t = time [s]

$v_{f}=5+8*10\\v_{f}=85 [m/s]$

<u>Fifth case</u>

Vf = final velocity [m/s]

Vo = initial velocity [m/s]

a = acceleration [m/s²]

t = time [s]

$8=v_{o}+4*2\\v_{0}=8-8\\v_{o}=0$

8 0

3 years ago