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xeze [42]
3 years ago
10

I don't understand why when you use a straw the higher pressure outside of it pushes the water up the straw.

Physics
2 answers:
stepan [7]3 years ago
6 0
Pressure is force acting on an area .
in the same area, more pressure means more force.

2 person, pressure outside and one in straw push water against each other. more pressure outside, so more force it pushes ... where will the water flow?
ValentinkaMS [17]3 years ago
3 0
When material stuff is free to move, it moves away from stronger force on it to weaker force on it. OK ? Now take it to the next step: When material stuff is free to flow ... like liquid or gas ... it flows from higher pressure to lower pressure. That's what makes wind, and that's why water comes out of a garden hose when you open the nozzle. It doesn't amaze you at all when you INCREASE the pressure at one end of the straw (blow into it), and air moves DOWN and out of it. So it shouldn't bother you that when you DECREASE the pressure at your end, anything that's free to flow UP the straw will do that.
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When is the mass of an object if it exerts a force of 160 N and an acceleration of 8.15m/s^2
yan [13]

Answer:

f=ma.......m=f/a......m=20kg

4 0
3 years ago
How do I calculate speed, velocity, and acceleration? I need the formulas too
vodomira [7]
Speed is the secular unit, which means it only measures "how fast it goes."
speed= \frac{distance}{time}
Velocity is a victory unit, which means it measures "both how fast it goes AND the direction."
velocity= displacement/time
Acceleration is the change of velocity over time. 
acceleration =  \frac{change in velocity}{time}
5 0
3 years ago
The electric potential inside a charged spherical conductor of radius R is is given by V = keQ/R, and the potential outside is g
Bumek [7]

Answer:

For outer points of shell

E = \frac{k_eQ}{r^2}

Now for inner point of shell

E = 0

Explanation:

As we know that out side the shell electric potential is given as

V = \frac{K_e Q}{r}

inside the shell the electric potential is given as

V = \frac{K_e Q}{R}

now we know the relation between electric potential and electric field as

E = - \frac{dV}{dr}

so we can say for outer points of the shell

E = -\frac{dV}{dr}

E = - \frac{d}{dr}(\frac{K_eQ}{r})

E = \frac{k_eQ}{r^2}

Now for inner point again we can use the same

E = - \frac{dV}{dr}

E = - \frac{d}{dr}(\frac{K_eQ}{R})

E = 0

6 0
3 years ago
Used to measure mass
Lemur [1.5K]
Measuring mass with a balance
5 0
3 years ago
Read 2 more answers
A flat, 181 ‑turn, current‑carrying loop is immersed in a uniform magnetic field. The area of the loop is 4.97 cm2 and the angle
Sidana [21]

Answer:

B = 0.135T

Explanation:

To find the magnitude of the magnetic field you use the following formula, for the torque produced by a magnetic field B in a loop:

\tau=NIABsin\theta   (1)

τ: torque = 1.51*10^-5 Nm

I: current = 2.47mA = 2.47*10^-3 A

B: magnitude of the magnetic field

A: area of the loop = 4.97cm^2 = 4.97(10^-2m)^2=4.97*10^-4m^2

N: turns = 181

θ: angle between B and the magnetic dipole (same as the direction  of the normal to the plane)

You replace the values of the parameters in (1). Furthermore you do B the subject of the formula:

B=\frac{\tau}{NIAsin\tetha}=\frac{1.51*10^{-5}Nm}{(181)(2.47*10^{-3}A)(4.97*10^{-4}m^2)(sin30.1\°)}=0.135T

3 0
3 years ago
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