3 years ago

Help please!!!!!!!!!!!!!!!!!!!!

Physics

1 answer:

Pavel [41]3 years ago

8 0

Looks like you need to review through the lesson and take notes as it tells you in the lesson what each of these are.

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Calculate the true mass (in vacuum) of a piece of aluminum whose apparent mass is 4.5000 kgkg when weighed in air. The density o

spin [16.1K]

Answer:

The true weight of the aluminium is $m_{alu} =$ 4.5021 kg

Explanation:

Given data

$m_{app}$ = 4.5 kg

$\rho_{air}$ = 1.29 $\frac{kg}{m^{3} }$

$\rho_{al}$ = 2.7× $10^{3} \frac{kg}{m^{3} }$

The true mass of the aluminium is given by

$m_{alu} = \frac{\rho_{alu}m_{app}}{\rho_{alu} -\rho_{air} }$

Put all the values in above equation we get

$m_{alu} = \frac{(2700)(4.5)}{2700-1.29}$

$m_{alu} =$ 4.5021 kg

Therefore the true weight of the aluminium is $m_{alu} =$ 4.5021 kg

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2 years ago

What is impulse measured in

MrMuchimi

Impulse = Force * time

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What situation would give you a mechanical advantage? *

const2013 [10]

D. Using a fixed-pulley system

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An iron ball weighs 80 N on the surface

Degger [83]

Answer:

2√5N will be the weight.

Explanation:

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A particle that carries a net charge of -41.8 μc is held in a region of constant, uniform electric field. the electric field vec

miss Akunina [59]

The total work done by the electric field on the charge is given by the scalar product between the electric force acting on the charge and the displacement of the charge:
$W=F d cos \theta$
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2 years ago