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3 years ago

10

Time period of a simple pendulum is measured at karachi, what change will occur in the time period ,if it is measured on mount e

verest ,explain?

1 answer:

iren [92.7K]3 years ago

3 0

The period T of a simple pendulum is given by:

$T = 2\pi \sqrt{ \frac{m}{g} }$

where m is the mass of the pendulum and g is the gravitational acceleration.

The period of the pendulum will increase when the gravitational acceleration decreases.

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Will give correct answer brainliest

sineoko [7]

$200 \times 80\%$

<h2><em>calculate</em></h2>

<em> $200 \times \frac{80}{100}$ </em>

<h2><em>reduce </em><em>the </em><em>numbers</em></h2>

<em> $2 \times 80$ </em>

<h2><em>multiply</em></h2>

<em> $= 160$ </em>

<h2><em>there </em><em>for </em><em>we </em><em>have </em><em>a </em><em>solution</em><em> to</em><em> the</em><em> </em><em>equation</em></h2>

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How often does a 16-year-old need a medical evaluation from a doctor?

a_sh-v [17]

Answer:

Explanation: A

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A 70.0-kg person throws a 0.0480-kg snowball forward with a ground speed of 33.5 m/s. A second person, with a mass of 55.0 kg, c

saw5 [17]

Answer:

The final velocity of the thrower is $\bf{3.88~m/s}$ and the final velocity of the catcher is $\bf{0.029~m/s}$ .

Explanation:

Given:

The mass of the thrower, $m_{t} = 70~Kg$ .

The mass of the catcher, $m_{c} = 55~Kg$ .

The mass of the ball, $m_{b} = 0.0480~Kg$ .

Initial velocity of the thrower, $v_{it} = 3.90~m/s$

Final velocity of the ball, $v_{fb} = 33.5~m/s$

Initial velocity of the catcher, $v_{ic} = 0~m/s$

Consider that the final velocity of the thrower is $v_{ft}$ . From the conservation of momentum,

$&& m_{t}v_{ft} + m_{b}v_{fb} = (m_{t} + m_{b})v_{it}\\&or,& v_{ft} = \dfrac{(m_{t} + m_{b})v_{it} - m_{b}v_{fb}}{m_{t}}\\&or,& v_{ft} = \dfrac{(70 + 0.0480)(3.90) - (0.0480)(33.5)}{70}\\&or,& v_{ft} = 3.88~m/s$

Consider that the final velocity of the catcher is $v_{fc}$ . From the conservation of momentum,

$&& (m_{c} + m_{b})v_{fc} = m_{b}v_{it}\\&or,& v_{fc} = \dfrac{m_{b}v_{it}}{(m_{c} + m_{b})}\\&or,& v_{fc} = \dfrac{(0.048)(33.5)}{(55.0 + 0.0480)}\\&or,& v_{fc} = 0.029~m/s$

Thus, the final velocity of thrower is $3.88~m/s$ and that for the catcher is $0.029~m/s$ .

8 0

3 years ago

Plz help..During takeoff a plane accelerates at 4m/s^2 and takes 40s to reach takeoff speed.

matrenka [14]

Answer:

The velocity of the plane at take off is 160 m/s.

The distance travel by the plane in that time is 3200 meter.

Explanation:

Given:

Acceleration, a = 4 m/s²

Time, t = 40 s

u = 0 i .e initial velocity

To Find:

velocity , v = ?

distance , s =?

Solution:

we have first Kinematic equation

v = u + at

∴ v = 0 + 4×40

∴ v = 160 m/s

Now by Third Kinematic equation

$s = ut + \frac{1}{2}at^{2}$

∴ s = 0 + 0.5 × 4× 40²

∴ s = 3200 meter

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