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Vilka [71]
3 years ago
11

A container is filled with water and the pressure at the container bottom is P. If the container is instead filled with a liquid

having specific gravity 1.05, what new bottom pressure will be measured
Physics
1 answer:
den301095 [7]3 years ago
7 0
Eat ahah is when s susuehs
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A house at the bottom of a hill is fed by a full tank of water 5m deep and connected to the house by a pipe that is 110m long at
yawa3891 [41]

Answer:

a) p=964178.7\ Pa

b) h=98.2853\ m

Explanation:

Given;

  • depth of the water-tank, d=5\ m
  • length of the pipe, l=110\ m
  • inclination of the pipe from the horizontal, \theta =58^{\circ}

a)

Now the effective vertical height of the water column from the free surface of the water to the bottom of the pipe at house:

h=d+l.\sin\theta

h=5+110\sin58

h=98.2853\ m

Now the pressure due to effective water head:

p=\rho.g.h

where:

\rho= density of the liquid, here water

g= acceleration due to gravity

h= height of the liquid column

p=1000\times9.81\times 98.2853

p=964178.7\ Pa

b)

Now the height of water corresponding to this pressure will be the same as the effective water head by the law of conservation of energy.

h=98.2853\ m

6 0
3 years ago
A piece of metal and a piece of brick are both sitting in the sun if both are the same mass what property of matter will make th
xenn [34]
The piece of metal will feel hotter to touch because of its heat conductance. whereas the brick does not conduct heat as well as the metal. Hence the piece of metal feeling hotter to touch. 
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3 years ago
What is dark energy?
Gnesinka [82]
Dark energy is what is causing the accelerating expansion of the universe. It can also be defined as the theoretical energy that counteracts or acts against gravity. Others also define it as a theoretical, new kind of dynamic energy field that fills space, but whose energy has an effect opposite to that of normal energy.
5 0
3 years ago
A 600 g model rocket is on a cart that is rolling to the right at a speed of 4.0 m/s. The rocket engine, when it is fired, exert
murzikaleks [220]

Answer:

The rocket should be launched when the cart is 13.48m away from a point directly below the hoop.

Explanation:

Step 1: Data given

mass of the rocket = 600 grams

speed = 4.0 m/s

Step 2: Calculate weight

Fw = mg

with Fw = the weight (in Newton)

with m = the mass (in kg)

with g = the acceleration due to gravity (9.81 m/s²).

Fw = (0.600 kg)(9.81 m/s²)  = 5.886 N

Step 3: Calculate force available to provide acceleration

The rocket engine, when it is fired, exerts a 8.0 AND vertical thrust on the rocket.

5.886 N of that force will be used to counter the rocket's weight, leaving 2.114 N of force available to provide acceleration.  

Step 4: Calculate the rocket's upward acceleration:

Fnet = m*a

With Fnet = the net force (the force that remains after the rocket's weight is compensated)

with a = the rocket's acceleration (in m/s²)

2.114 N = (0.600 kg)*a

a = 3.52 m/s²  = the rocket's upward acceleration

Step 5: Calculate how long it will take to rise 20 meters into the air.

Δy = v0*t + 1/2 at²

with v0 = 0m/s

Δy = 1/2 at²

20 m = 1/2(3.52)t²

20 m = (1.76 m/s²)t²

11.36 = t²

t = 3.37 s

This means the rocket will take 3.37 seconds to reach the hoop.  It should be launched when the cart is 3.37 seconds away from being directly beneath the hoop.  

Step 6: Calculate the distance

v = Δx / t

4.0 m/s = Δx / 3.37 s

Δx = 13.48 m

The rocket should be launched when the cart is 13.48m away from a point directly below the hoop.

8 0
3 years ago
If the normal force exerted by the track on the car when it is at the top of the track (point B) is 6.00 N , what is the normal
Eddi Din [679]

Answer:

So, the force, F is the agent which provides the basic property of motion or rest to the body.While, the car is more obviously to have a mass, m and that the angle withe road or surface is also given which is normal to the road(i.e angle =90 degree). Then we say that lets say that the car is moving with the constant velocity of 20 m/sec and its kept unchanged by the car. So, we have the mass, m as 1 kg for the car and the value of the gravity we have the g=9.8 m/sec.

Now,

We have F=ma,

and a=v/t,

so we can have another equation for it as,

Now, providing the required data  to, it;  ∴t =2 sec,

F=(1)×(20/2),

  • <u>F=10 N.</u>
  • So, the car would be acting the force,F of about 10 N  while the car is present on the lower region of the track.

4 0
3 years ago
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