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2 years ago

You spy on an anchor on the bottom of a lake. What is the direction and amount of force the water exerts on it: a) zero

Physics

1 answer:

vaieri [72.5K]2 years ago

5 0

Answer:

c) up, less than the weight of the anchor

Explanation:

An object floats when the weight of the water the object displaces is equal to the weight of the object. In this case the weight of the anchor and the weight of the water it displaces is not the same so it sinks. Buoyancy is the force that acts on body when it is in water.

When buoyancy is less than the force that the object is generating due to gravity the object sinks. Buoyancy always acts up.

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Why should people pay attention to scientists when making decisions?

Vesna [10]

Answer:

Explanation:

All the other answers are false scientist make frequent mistake in there hypothesis B. is a opinion so is C

And a scientist job is to study and OBSERVE a field of science

6 0

3 years ago

A generator converts blank energy into blank energy

jok3333 [9.3K]

Answer: Mechanical Energy and Electrical Energy

Explanation: A generator converts mechanical energy into electrical energy, while a motor does the opposite - it converts electrical energy into mechanical energy.

6 0

2 years ago

PLEASE HELP I NEED IT RIGHT NOW

morpeh [17]

The answer should be B :)

7 0

3 years ago

Read 2 more answers

Blood pressure is usually measured by wrapping a closed air-filled jacket equipped with a pressure gage around the upper arm of

Sever21 [200]

Answer:

a) High and low pressures are 15.999 kilopascals and 10.666 kilopascals, respectively.

b) High and low pressures are 2.320 pounds per square inch and 1.547 pounds persquare inch, respectively.

c) High and low pressures are 1.632 meters water column and 1.088 meters water column, respectively.

Explanation:

a) High and low pressures in kilopascals:

101.325 kPa equals 760 mm Hg, then, we can obtain the values by a single conversion:

$p_{high} = 120\,mm\,Hg\times \frac{101.325\,kPa}{760\,mm\,Hg}$

$p_{high} = 15.999\,kPa$

$p_{low} = 80\,mm\,Hg\times \frac{101.325\,kPa}{760\,mm\,Hg}$

$p_{low} = 10.666\,kPa$

High and low pressures are 15.999 kilopascals and 10.666 kilopascals, respectively.

b) High and low pressures in pounds per square inch:

14.696 psi equals 760 mm Hg, then, we can obtain the values by a single conversion:

$p_{high} = 120\,mm\,Hg\times \frac{14.696\,psi}{760\,mm\,Hg}$

$p_{high} = 2.320\,psi$

$p_{low} = 80\,mm\,Hg\times\frac{14.696\,psi}{760\,mm\,Hg}$

$p_{low} = 1.547\,psi$

High and low pressures are 2.320 pounds per square inch and 1.547 pounds persquare inch, respectively.

c) High and low pressures in meter water column in meters water column:

We can calculate the equivalent water column of a mercury column by the following relation:

$\frac{h_{w}}{h_{Hg}} = \frac{\rho_{Hg}}{\rho_{w}}$

$h_{w} = \frac{\rho_{Hg}}{\rho_{w}}\times h_{Hg}$ (Eq. 1)

Where:

$\rho_{w}$ , $\rho_{Hg}$ - Densities of water and mercury, measured in kilograms per cubic meter.

$h_{w}$ , $h_{Hg}$ - Heights of water and mercury columns, measured in meters.

If we know that $\rho_{w} = 1000\,\frac{kg}{m^{3}}$ , $\rho_{Hg} = 13600\,\frac{kg}{m^{3}}$ , $h_{Hg, high} = 0.120\,m$ and $h_{Hg, low} = 0.080\,m$ , then we get that:

$h_{w, high} = \frac{13600\,\frac{kg}{m^{3}} }{1000\,\frac{kg}{m^{3}} } \times 0.120\,m$

$h_{w, high} = 1.632\,m$

$h_{w, low} = \frac{13600\,\frac{kg}{m^{3}} }{1000\,\frac{kg}{m^{3}} } \times 0.080\,m$

$h_{w, low} = 1.088\,m$

High and low pressures are 1.632 meters water column and 1.088 meters water column, respectively.

4 0

2 years ago

A small object carrying a charge of -4.00 nC is acted upon by a downward force of 19.0 nN when placed at a certain point in an e

Gala2k [10]

Explanation:

Given that,

Charge acting on the object, $q=-4\ nC=-4\times 10^{-9}\ C$

Force acting on the object, $F=19\ nC=19\times 10^{-9}\ C$ (in downward direction)

(a) The electric force acting in the electric field is given by :

$F=qE$

E is the electric field

$E=\dfrac{F}{q}$

$E=\dfrac{19\times 10^{-9}\ N}{4\times 10^{-9}\ C}$

E = 4.75 N/C

The direction of electric field is as same as electric force. But it is negative charge. So, the direction of electric field is in upward direction.

(b) The charge on the proton is, $q=1.6\times 10^{-19}\ C$

The force acting on the proton is :

$F=qE$

$F=1.6\times 10^{-19}\times 4.75$

$F=7.6\times 10^{-19}\ N$

If the charge on the proton is positive, the force on the proton is in upward direction.

Hence, this is the required solution.

8 0

2 years ago