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3 years ago

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What factor has the greatest impact on flexibility?

1 answer:

hjlf3 years ago

4 0

That would be c. :) :) :)

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Please HELP ME If the motion of an object changes, what must be true about the forces acting on that object?

atroni [7]

Answer:

If there is a net force acting on an object, the object will have an acceleration and the object's velocity will change. ... Newton's second law states that for a particular force, the acceleration of an object is proportional to the net force and inversely proportional to the mass of the object.

Explanation:

4 0

3 years ago

Read 2 more answers

umka2103 [35]

Answer:

$x=2.4365\ m$

and

$x=-1.4365\ m$

Explanation:

Given:

first charge, $q_1=5\times 10^{-3}\ C$

second charge, $q_2=3\times 10^{-3}\ C$

position of first charge, $x_1=-2\ m$

position of second charge, $x_2=-1\ m$

Now since there are only 2 charges and of the same sign so they repel each other. This repulsion will be zero at some point on the line joining the charges.

<u>Now, according to the condition, electric field will be zero where the effects of field due to both the charges is equal.</u>

$E_1=E_2$

since first charge is greater than the second charge so we may get a point to the right of the second charge and the distance between the two charges is 1 meter.

$\frac{1}{4\pi.\epsilon_0} \frac{q_1}{(r+1)^2} =\frac{1}{4\pi.\epsilon_0} \frac{q_2}{(r)^2}$

$\frac{5\times 10^{-3}}{(r+1)^2} = \frac{3\times 10^{-3}}{(r)^2}$

$3(r^2+1+2r)=5r^2$

$2r^2-6r-3=0$

$r=3.4365 \&\ r=-0.4365$

Since we have assumed that the we may get a point to the right of second charge so we calculate with respect to the origin.

$x=-1+3.4365=2.4365\ m$

and

$x=-1-0.4365=-1.4365\ m$

6 0

3 years ago

Can some one help me with this question its hard :( down below ill give brainliest

tensa zangetsu [6.8K]

I’m pretty sure it’s c sorry if I’m wrong

3 0

3 years ago

Replacing an object attached to a spring with an object having 14 the original mass will change the frequency of oscillation of

Pachacha [2.7K]

Answer:

<em>The frequency changes by a factor of 0.27.</em>

<em></em>

Explanation:

The frequency of an object with mass m attached to a spring is given as

$f$ = $\frac{1}{2\pi } \sqrt{\frac{k}{m} }$

where $f$ is the frequency

k is the spring constant of the spring

m is the mass of the substance on the spring.

If the mass of the system is increased by 14 means the new frequency becomes

$f_{n}$ = $\frac{1}{2\pi } \sqrt{\frac{k}{14m} }$

simplifying, we have

$f_{n}$ = $\frac{1}{2\pi \sqrt{14} } \sqrt{\frac{k}{m} }$

$f_{n}$ = $\frac{1}{3.742*2\pi } \sqrt{\frac{k}{m} }$

if we divide this final frequency by the original frequency, we'll have

==> $\frac{1}{3.742*2\pi } \sqrt{\frac{k}{m} }$ ÷ $\frac{1}{2\pi } \sqrt{\frac{k}{m} }$

==> $\frac{1}{3.742*2\pi } \sqrt{\frac{k}{m} }$ x $2\pi \sqrt{\frac{m}{k} }$

==> 1/3.742 = <em>0.27</em>

7 0

3 years ago

A trooper is moving due south along the freeway at a speed of 30.1 m/s when a red car passes the trooper. The red car moves with

romanna [79]

Answer:

The correct answer to the following question will be "41.87 m".

Explanation:

The given values are:

The speed of trooper = $30.1 \ m/s$

The velocity of red car = $45.4 \ m/s$

Now,

A red car goes as far as possible until the speed or velocity of the troops is the same as that of of the red car at

$t=\frac{45-30}{2.7}$ (∵ $time=\frac{distance}{time}$ )

$t=5.56 \ sec$

then,

The distance covered by trooper,

$t1=30\times 5.56+\frac{1}{2}\times 2.7\times (5.56)^2$

$=208.33 \ m$

The distance covered by red car,

= $45\times 5.56$

= $250.2 \ m$

Maximum distance = $250.2-208.33$

= $41.87 \ m$

6 0

3 years ago