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bagirrra123 [75]
3 years ago
15

One pot has a water height of 8cm. Calculate the water pressure at the bottom of the pot. They give g = 10N / kg p = 1000kg / m3

Physics
1 answer:
natima [27]3 years ago
5 0

Answer:

P = 800 Pa

Explanation:

The pressure of water at the bottom of the pot can be given by the following formula:

P = \rho g h

where,

P = Pressure at the bottom of the pot = ?

ρ = density = 1000 kg/m³

h = height of water = 8 cm = 0.08 m

Therefore,

P = (1000\ kg/m^3)(10\ m/s^2)(0.08\ m)

<u>P = 800 Pa </u>

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bogdanovich [222]

Answer:

The speed is 0.97 c.

Explanation:

Given that,

Dilated time t'= 50.0 years

Rest time t = 13.0 years

We need to calculate the speed

Using formula of time dilation

t'=\dfrac{t}{\sqrt{1-\dfrac{v^2}{c^2}}}

Where, t' = change in time

t = rest time

v = velocity

c = speed of light

Put the value into the formula

50.0=\dfrac{13.0}{\sqrt{1-\dfrac{v^2}{(3\times10^{8})^2}}}

v^2=\dfrac{(13)^2\times(c)^2-(c)^2\times50^2}{50^2}

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A balloon is rising vertically upwards at a velocity of 10m/s. When it is at a height of 45m from the ground, a parachute bails
harina [27]

(a) 30.9 m

Let's analyze the motion of the parachutist. Its vertical position above the ground is given by

y=h+ut+\frac{1}{2}gt^2

where

h = 45 m is the initial height

u = 10 m/s is the initial velocity (upward)

t is the time

g = -9.8 m/s^2 is the acceleration of gravity (downward)

Substituting t=3 s , we find the height of the parachutist when it opens the parachute:

y=45 m+(10 m/s)(3 s)+\frac{1}{2}(-9.8 m/s^2)(3 s)^2=30.9 m

(b) 44.1 m

Here we have to find first the height of the balloon 3 seconds after the parachutist has jumped off from it. The vertical position of the balloon is given by

y = h + ut

where

h = 45 m is the initial height

u = 10 m/s is the initial velocity (upward)

t is the time

Substituting t = 3 s, we find

y = 45 m + (10 m/s)(3 s) = 75 m

So the distance between the balloon and the parachutist after 3 s is

d = 75 m - 30.9 m = 44.1 m

(c) 8.2 m/s downward

The velocity of the parachutist at the moment he opens the parachute is:

v = u +gt

where

u = 10 m/s is the initial velocity (upward)

t is the time

g = -9.8 m/s^2 is the acceleration of gravity (downward)

Substituting t = 3 s,

v = 10 m/s + (-9.8 m/s^2)(3 s)= -19.4 m/s

where the negative sign means it is downward

After t=3 s, the parachutist open the parachute and it starts moving with a deceleration of

a =+5 m/s^2

where we put a positive sign since this time the acceleration is upward.

The total distance he still has to cover till the ground is

d = 30.9 m

So we can find the final velocity by using

v^2-u^2 = 2ad

where this time we have u = 19.4 m/s as initial velocity. Taking the downward direction as positive, the deceleration must be considered as negative:

a = -5 m/s^2

Solving for v,

v=\sqrt{u^2 +2ad}=\sqrt{(19.4 m/s)^2+2(-5 m/s^2)(30.9 m)}=8.2 m/s

(d) 5.24 s

We can find the duration of the second part of the motion of the parachutist (after he has opened the parachute) by using

a=\frac{v-u}{t}

where

a = -5 m/s^2 is the deceleration

v = 8.2 m/s is the final velocity

u = 19.4 m/s is the initial velocity

t is the time

Solving for t, we find

t=\frac{v-u}{a}=\frac{8.2 m/s-19.4 m/s}{-5 m/s^2}=2.24 s

And added to the 3 seconds between the instant of the jump and the moment he opens the parachute, the total time is

t = 3 s + 2.24 s = 5.24 s

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