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3 years ago

7

A pivot at the 35cm mark supports a meter stick. It is brought into a balance by a 0.85N weight suspended at the 5.5cm mark.comp

ute the weight of the meter stick.

1 answer:

Arte-miy333 [17]3 years ago

8 0

Answer:

2.52N

Explanation:

according to law of moments ,

F1L1=F2L2 ,where the forces are in equilibrium .

0.85(44.5)= 15F2

that is , taking their distances from 50cm point

37 .825=15F2

37.825÷15= F2

2.52N = F2

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grin007 [14]

Answer: 17.68 s

Explanation:

This problem is a good example of Vertical motion, where the main equation for this situation is:

$y=y_{o}+V_{o}t-\frac{1}{2}gt^{2}$ (1)

Where:

$y=0$ is the height of the ball when it hits the ground

$y_{o}=70 m$ is the initial height of the ball

$V_{o}=82m/s$ is the initial velocity of the ball

$t$ is the time when the ball strikes the ground

$g=9.8m/s^{2}$ is the acceleration due to gravity

Having this clear, let's find $t$ from (1):

$0=70m+(82m/s)t-\frac{1}{2}(9.8m/s^{2})t^{2}$ (2)

Rewritting (2):

$-\frac{1}{2}(9.8m/s^{2})t^{2}+(82m/s)t+70m=0$ (3)

This is a quadratic equation (also called equation of the second degree) of the form $at^{2}+bt+c=0$ , which can be solved with the following formula:

$t=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$ (4)

Where:

$a=-\frac{1}{2}(9.8m/s^{2}$

$b=82m/s$

$c=70m$

Substituting the known values:

$t=\frac{-82 \pm \sqrt{82^{2}-4(-\frac{1}{2}(9.8)(70)}}{2a}$ (5)

Solving (5) we find the positive result is:

$t=17.68 s$

7 0

3 years ago

What is the metric unit for time?

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The base unit of time in the metric and SI system is the second.

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A ferry approaches shore, moving north with a speed of 6.2 m/s relative to the dock. A person on the ferry walks from one side o

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Speed of Ferry is towards North with magnitude 6.2 m/s

Here if we assume that North direction is along Y axis and East is along X axis then we can say

$\vec v_f = 6.2 \hat j$

Now a person walk on ferry with speed 1.5 m/s towards east with respect to Ferry

so it is given as

$\vec v_{pf} = 1.5 \hat i$

also by the concept of relative motion we know that

$\vec v_{pf} = \vec v_p - \vec v_f$

now plug in all values in it

$1.5 \hat i = \vec v_p - 6.2 \hat j$

$\vec v_p = 1.5 \hat i + 6.2 \hat j$

now if we need to find the speed of the person then we need to find its magnitude

so it is given as

$v = \sqrt{1.5^2 + 6.2^2}$

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In your every day life you come across a range of motion in which ,

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A)acceleration is in the direction of motion

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Assume (unrealistically) that a TV station acts as a point source broadcasting isotropically at 3.2 MW. What is the intensity of

Dahasolnce [82]

Answer:

$I=1.5\times10^{-28}W/m^2$

Explanation:

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$r=vt=(299792458m/s)(4.3\times365\times24\times60\times60s)=4.1\times10^{16}m$

So we have:

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3 years ago