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How much must a woman weigh (force) if the pressure she exerts while standing on one foot has an area of 0.4 m2 exerts a pressur

aleksley [76]

Why do teachers give such hard questions? anyways hopefully your question will get answered! (btw i also need the points but yeah gl!!!)

7 0

3 years ago

Read 2 more answers

A 1 kg object can be accelerated at 10 m/s^2. If you apply this same force to a 4kg object, what will its acceleration be?

AleksAgata [21]

Answer:

$a2 = 2.5~m/s^2$

Explanation:

<u>Mechanical Force</u>

According to the second Newton's law, the net force F exerted by an external agent on an object of mass m is:

F = m.a

Where a is the acceleration of the object.

Assume we apply some given force F to an object of m1=1 Kg that produces an acceleration $a1=10 m/s^2$ , then:

F = m1.a1

The same force F is now applied to a second object m2=4 Kg that produces an acceleration a2, then:

F = m2.a2

Dividing both equations:

$\displaystyle 1=\frac{m1.a1}{m2.a2}$

Solving for a2:

$\displaystyle a2=\frac{m1.a1}{m2}$

Substituting values:

$\displaystyle a2=\frac{1*10}{4}$

$a2 = 2.5~m/s^2$

7 0

3 years ago

A small, solid cylinder with mass = 20 kg and radius = 0.10 m starts from rest and rotates without friction about a fixed axis t

Mice21 [21]

Answer:

Explanation:

angular acceleration α = 5 rad /s ²

θ = 1/2 α t² , θ is angle of rotation , t is time .

= .5 x 5 x 4²

= 40 rad .

θ = l / r , θ is angle formed , l is length unwound , r is radius o wheel .

l = θ x r

= 40 x .1

= 4 m .

4 0

3 years ago

What are some predicted affects of climate change linked to global warming

Alla [95]

Decrease in atmospheric ozone, which results in lower protection from various stellar particles

5 0

3 years ago

Guys I really need help with these 2 questions , it's for my final plz help asap

mars1129 [50]

Answer:

(1) Initial speed, $u=0$

Final speed, $v=165.76m/s$

Average speed, $v_a_v_g=82.87m/s$

(2) Force of gravity, $F_g=12.8\times10^1^5N$

Explanation:

(1)

Given,

Distance, $S=300meter$

Time, $t=3.62second$

It is given that drag racer started at rest.

So Initial speed, $u=0$

Using Newton's second equation of motion,

$S=ut+\frac{1}{2}at^2\\300=0+\frac{a\times3.62^2}{2} \\a=45.79m/s^2$

Newton's first equation of motion,

$v=u+at\\=0+45.79\times3.62\\=165.76 m/s$

So, Final speed, $v=165.76m/s$

Average speed is defined as totle distance divided by totle time.

$v_a_v_g=\frac{S}{t}\\=\frac{300}{3.62} \\=82.87m/s$

So, Average speed, $v_a_v_g=82.87m/s$

(2)

Gravitation: It is the natural phenomenon in which two different bodies attract each other by virtue of their masses.

According to Newton's law of gravitation, the force of attraction between two bodies is directly proportional to the masses of the bodies and inversely proportional to square of distance between centers of mass of the bodies.

$F_g\propto\frac{m_1m_2}{r^2} \\F_g=G\frac{m_1m_2}{r^2}$ where $G$ is constant of proportionality and known as gravitation constant.

Given,

Mass of Jupiter, $m_1=1.9\times10^2^7kg$

Mass of Ganymede, $m_2=1.48\times10^2^3kg$

Distance between their centers of mass, $r=1.21\times10^1^2meter$

$F_g=G\frac{m_1m_2}{r^2}\\=\frac{6.67\times10^-^1^1\times1.9\times10^2^7\times1.48\times10^2^3}{(1.21\times10^1^2)^2} \\=12.8\times10^1^5N$

So, Force of gravity, $F_g=12.8\times10^1^5N$

7 0

3 years ago