In order to solve the problem, it is necessary to apply the concepts related to the conservation of momentum, especially when there is an impact or the throwing of an object.

The equation that defines the linear moment is given by

$mV_i = (m-m_O)V_f - m_OV_O$

where,

m=Total mass

$m_O =$ Mass of Object

$V_i =$ Velocity before throwing

$V_f =$ Final Velocity

$V_O =$ Velocity of Object

Our values are:

$m_1=5.3kgm_2=7.9kg\\m_3=10.5kg\\m_A=75kg\\m_{Total}=m=98.7Kg$

Solving to find the final speed, after throwing the object we have

$V_f=\frac{mV_0+m_TV_O}{m-m_O}$

We have three objects. For each object a launch is made so the final mass (denominator) will begin to be subtracted successively. In addition, during each new launch the initial speed will be given for each object thrown again.

That way during each section the equations should be modified depending on the previous one, let's start:

A) $5.3Kg\rightarrow 15m/s$

$V_{f1}=\frac{mV_0+m_TV_O}{m-m_O}$

$V_{f1}=\frac{(98.7)*0+5.3*15}{98.7-5.3}$

$V_{f1}=0.8511m/s$

B) $7.9Kg\rightarrow 11.2m/s$

$V_{f2}=\frac{mV_{f1}+m_TV_O}{m-m_O}$

$V_{f2}=\frac{(98.7)(0.8511)+(7.9)(11.2)}{98.7-5.3-7.9}$

$V_{f2} = 2.0173m/s$

C) $10.5Kg\rightarrow 7m/s$

$V_{f3}=\frac{mV_{f2}+m_TV_O}{m-m_O}$

$V_{f3}=\frac{(98.7)(2.0173)+(10.5)(7)}{98.7-5.3-7.9-10.5}$

$V_{f3} = 3.63478m/s$

Therefore the final velocity of astronaut is 3.63m/s

7 0

3 years ago

"There are two types of error that can occur when making measurements: systematic and random error. How you correct the error de

Murljashka [212]

Answer:

Systematic error can be corrected using calibration of the measurement instrument, while random error can be corrected using an average measurement from a set of measurements.

Explanation:

Random errors lead to fluctuations around the true value as a result of difficulty taking measurements, whereas systematic errors lead to predictable and consistent departures from the true value due to problems with the calibration of your equipment.

Systematic error can be corrected, by calibration of the measurement instrument. Calibration is simply a procedure where the result of measurement recorded by an instrument is compared with the measurement result of a standard value.

Random error can be corrected using an average measurement from a set of measurements or by Increasing sample size.

8 0

3 years ago

6. A golf ball is hit a distance of 300 yards in 10 sec. What is the speed of the golf ballo

katen-ka-za [31]

The speed of the ball is 27.4 m/s

Explanation:

The speed of an object is given by:

$speed=\frac{d}{t}$

where

d is the distance covered

t is the time taken

In this problem, we have:

d = 300 yards is the distance covered by the golf ball

t = 10 s is the time taken

Keeping in mind that

1 yard = 0.914 m

We can convert the distance from yards to meters:

$d = 300 \cdot 0.914 = 274.2 m$

And substituting into the equation, we find the speed of the ball:

$speed=\frac{274.2}{10}=27.4 m/s$

Learn more about speed:

brainly.com/question/8893949

#LearnwithBrainly

3 0

3 years ago