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3 years ago

Help ill give you brainliest !!!

Physics

1 answer:

puteri [66]3 years ago

5 0

Answer:

1. B

2. A

3. C

4. B

5. A

6. Muscular strength is different than muscular endurance because of the fact that muscular strength is the amount of force that can be exerted in one instance. Muscular endurance is how long that you can exert that force without being completely exhausted.

7. Some benefits to strength training is the increase in muscular endurance. There is also the benefit of better muscular strength.

Explanation:

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An astronaut finds herself in a predicament in which she has become untethered from her shuttle. She figures that she could get

Blizzard [7]

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

When can a high speed velocity cause damage?'

sweet-ann [11.9K]

Answer:

50 Mph.

Explanation:

According to the National Severe Storms Laboratory, winds can really begin to cause damage when they reach <em><u>50 mph</u></em>. But here’s what happens before and after they reach that threshold, according to the Beaufort Wind Scale (showing estimated wind speeds): - at 19 to 24 mph, smaller trees begin to sway.

7 0

2 years ago

Problem set 3 » (8) river swimming a swimmer heads directly across a river, swimming at her maximum speed of 1.30 m/s relative t

jasenka [17]

Velocity of swimmer across river = 1.30 m/s

Distance arrived downstream = 48 m

Width of river = 64 m

Time taken to cross river = $\frac{Width of river}{Velocity across river}$

= $\frac{64}{1.30} =49.23 s$

Speed of river current = $\frac{Distance arrived downstream}{Time taken to cross river}$

= $\frac{48}{49.23} = 0.975 m/s$

So, the river is flowing at a speed 0.975 m/s.

5 0

2 years ago

A football quarterback runs 15.0 m straight down the playing field in 3.00 s. He is then hit and pushed 3.00 m straight backward

algol [13]

Answer:

a) $v_{1}=14.29m/s\\v_{2}=9.25m/s\\v_{3}=6.36m/s$

b) $v=+9.97m/s$

Explanation:

From the exercise we know that

$x_{1} =15m, t_{1}=3s$

$x_{2} =-3m, t_{1}=1.74s$

$x_{3} =29m, t_{3}=5.20s$

From dynamics we know that the formula for average velocity is:

$v=\frac{x_{2}-x_{1} }{t_{2}-x_{1} }$

a) For the three intervals:

$v_{1}=\frac{x_{2}-x_{1} }{t_{2}-t_{1} }=\frac{(-3-15)m}{(1.74-3)s}=14.29m/s$

$v_{2}=\frac{x_{3}-x_{2} }{t_{3}-t_{2} }=\frac{(29-(-3))m}{(5.20-1.74)s}=9.25m/s$

$v_{3}=\frac{x_{3}-x_{1} }{t_{3}-t_{1} }=\frac{(29-15)m}{(5.20-3)s}=6.36m/s$

b) The average velocity for the entire motion can be calculate by the following formula:

$v=\frac{v_{1}+v_{2}+v_{3} }{n} =\frac{(14.29+9.25+6.36)m/s}{3}=+9.97m/s$

8 0

3 years ago

Some planetary scientists have suggested that the planet Mars has an electric field somewhat similar to that of the earth, produ

aalyn [17]

Answer:

324795 C

252.637820565 N/C

$2.235844712\times 10^{-9}\ C/m^2$

Explanation:

$\epsilon_0$ = Permittivity of free space = $8.85\times 10^{-12}\ F/m$

R = Radius of Mars = $3.4\times 10^6\ m$

A = Area = $4\pi R^2$

$\phi$ = Electric flux = $3.67\times 10^{16}\ Nm^2/C$

Electric flux is given by

$\phi=\dfrac{q}{\epsilon_0}\\\Rightarrow q=\phi\epsilon_0\\\Rightarrow q=3.67\times 10^{16}\times 8.85\times 10^{-12}\\\Rightarrow q=324795\ C$

The charge is 324795 C

Electric field is given by

$E=\dfrac{\phi}{A}\\\Rightarrow E=\dfrac{3.67\times 10^{16}}{4\pi (3.4\times 10^6)^2}\\\Rightarrow E=252.637820565\ N/C$

The electric field is 252.637820565 N/C

Surface charge density is given by

$\sigma=\dfrac{q}{4\pi R^2}\\\Rightarrow \sigma=\dfrac{324795}{4\pi (3.4\times 10^6)^2}\\\Rightarrow \sigma=2.235844712\times 10^{-9}\ C/m^2$

The surface charge density is $2.235844712\times 10^{-9}\ C/m^2$

6 0

3 years ago