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Aneli [31]
2 years ago
10

What is the density of an object if it has a volume of 162 ml with a mass(273) that is shown in the picture below?

Physics
2 answers:
creativ13 [48]2 years ago
7 0

Answer:

1.685185185185185 (you can round to whatever your teacher asks)

Explanation:

Formula for Density

D=m/v (Density equals mass over volume)

273/162

(/) means division.

e-lub [12.9K]2 years ago
7 0
It’s 1.65 I just rounded it real quick but that the short answer
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Answer:

The phenomenon known as "tunneling" is one of the best-known predictions of quantum physics, because it so dramatically confounds our classical intuition for how objects ought to behave. If you create a narrow region of space that a particle would have to have a relatively high energy to enter, classical reasoning tells us that low-energy particles heading toward that region should reflect off the boundary with 100% probability. Instead, there is a tiny chance of finding those particles on the far side of the region, with no loss of energy. It's as if they simply evaded the "barrier" region by making a "tunnel" through it.

Explanation:

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2 years ago
A horizontal force of 92.7 N is applied to a 40.5 kg crate on a rough, level surface. If the crate accelerates at 1.13 m/s2, wha
lord [1]

Answer:

The value is F_f =  46.935 \  N

Explanation:

From the question we are told that

    The  magnitude of the horizontal force is F  =  92.7 \  N

     The mass of the crate is  m  =  40.5 \  kg

     The acceleration of the crate is  a =  1.13 \ m/s

Generally the net force acting on the crate is mathematically represented as

       F_{net} =  F -  F_f =  ma

Here F_f is force of kinetic friction (in N) acting on the crate

      So  

            92.7  -  F_f =  40.5 * 1.13

=>         F_f =  46.935 \  N

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3 years ago
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How far from the castle wall does the launched rock hit the ground?
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King Arthur's knights use a catapult to launch a rock from their vantage point on top of the castle wall, 14 m above the moat. The rock is launched at a speed of 27 m/s and an angle of 32degrees above the horizontal.
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7nadin3 [17]

Explanation:

The power P dissipated by a heater is defined as

P = VI

where V is the voltage and I is the current.

a) The current running through a 130-W heater is

I = \dfrac{P}{V} = \dfrac{130\:\text{W}}{120\:\text{V}} = 1.08\:\text{A}

b) The resistance <em>R</em><em> </em>of the heater is

P = VI = (IR)I = I^2R

where V= IR is our familiar Ohm's Law.

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R = 110.8\:Ω

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2 years ago
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