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3 years ago

4. A pipe enters a water tank at a point 6.50 m below

Physics

1 answer:

murzikaleks [220]3 years ago

8 0

Answer:

165 kPa

Explanation:

Absolute pressure is:

P = Patm + ρgh

where Patm is the atmospheric pressure,

ρ is the density,

g is acceleration due to gravity,

and h is the depth.

P = 101,300 Pa + (1000 kg/m³) (9.8 m/s²) (6.50 m)

P = 165,000 Pa

P = 165 kPa

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After you pick up a spare, your bowling ball rolls without slipping back toward the ball rack with a linear speed of 2.85 m/s. T

motikmotik

Answer:

V' = 0.84 m/s

Explanation:

given,

Linear speed of the ball, v = 2.85 m/s

rise of the ball, h = 0.53 m

Linear speed of the ball, v' = ?

rotation kinetic energy of the ball

$KE_r = \dfrac{1}{2}I\omega^2$

I of the moment of inertia of the sphere

$I = \dfrac{2}{5}MR^2$

v = R ω

using conservation of energy

$KE_r = \dfrac{1}{2}( \dfrac{2}{5}MR^2)(\dfrac{V}{R})2$

$KE_r = \dfrac{1}{5}MV^2$

Applying conservation of energy

Initial Linear KE + Initial roational KE = Final Linear KE + Final roational KE + Potential energy

$\dfrac{1}{2}MV^2 + \dfrac{1}{5}MV^2 = \dfrac{1}{2}MV'^2 + \dfrac{1}{5}MV'^2 + M g h$

$0.7 V^2 = 0.7 V'^2 + gh$

$0.7\times 2.85^2 = 0.7\times V'^2 +9.8\times 0.53$

V'² = 0.7025

V' = 0.84 m/s

the linear speed of the ball at the top of ramp is equal to 0.84 m/s

6 0

3 years ago

a car with a mass of 100 kg is stopped on the side of the road after getting a flat tire. the two people that were riding in the

notsponge [240]

Answer:

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3 years ago

Anyone.... help with this two questions...

PSYCHO15rus [73]

Answer:

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Explanation:

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3 years ago

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Which statements describe the results of the effect of angle of insolation on absorption experiment? Check all that apply.

lilavasa [31]

Answer:

At 6 minutes, the soil temperature at a 45° angle of insolation is higher than at 0°.

At 15 minutes, the soil temperature at a 90° angle of insolation is higher than at 45°

Explanation:

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3 years ago

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a flag of mass 2.5 kg is supported by a single rope. A strong horizontal wind exerts a force of 12 N on the flag. Calculate the

tatuchka [14]

The free-body diagram of the forces acting on the flag is in the picture in attachment.

We have: the weight, downward, with magnitude
$W=mg = (2.5 kg)(9.81 m/s^2)=24.5 N$
the force of the wind F, acting horizontally, with intensity
$F=12 N$
and the tension T of the rope. To write the conditions of equilibrium, we must decompose T on both x- and y-axis (x-axis is taken horizontally whil y-axis is taken vertically):
$T \cos \alpha -F=0$
$T \sin \alpha -W=$
By dividing the second equation by the first one, we get
$\tan \alpha = \frac{W}{F}= \frac{24.5 N}{12 N}=2.04$
From which we find
$\alpha = 63.8 ^{\circ}$
which is the angle of the rope with respect to the horizontal.

By replacing this value into the first equation, we can also find the tension of the rope:
$T= \frac{F}{\cos \alpha}= \frac{12 N}{\cos 63.8^{\circ}}=27.2 N$

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4 years ago