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3 years ago

a beaker has mass of 125 g. what is the mass of liquid if the beaker plus liquid have a mass of 232 g?

Physics

1 answer:

MrRissso [65]3 years ago

7 0

Answer: 107g

Explanation:

Just subtract 125 from 232, like this: 232-125=107g.

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a force that is able to act at a distance

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Calculate the potential energy of a +1.0μC point charge sitting 0.1m from a -5.0μC point charge.

AfilCa [17]

Answer:

P.E. = -0.449 J

Explanation:

Potential energy of a charge particle in any electrostatic field is defined as the amount of work done ( in negative ) to bring that charge particle from any position to a new position r.

Now Potential energy is defined by this formula,

P.E. = k q₁ q₂/ r

where P.E. is the potential energy.

k = 1/( 4πε₀) = 8.99 × 10⁹ C²/ ( Nm²)

q₁ = charge of one particle = +1.0μC

q₂ = charge of another particle = -5.0μC

r = distance = 0.1 m

Now , P.E. = 8.99 × 10⁹C²/ ( Nm²) * ( -5.0 × 10⁻⁶ C ) × ( 1 × 10⁻⁶ C ) / 0.1 m

P.E. = -0.449 J

8 0

3 years ago

400 N of load can be overcome by an effort of 50 N by using a lever. Calculate the mechanical advantage of the lever.

mars1129 [50]

Answer:

load (l)=400N

Effort(E)=50N

mechanical advantage (MA)= load ÷Effort

(ma)=400÷50

(ma)=8

Explanation:

I copy pasted from the answer from the same question. Remember to first check if ur question is there

4 0

3 years ago

A bucket of water of mass 10 kg is pulled at constant velocity up to a platform 30 meters above the ground. This takes 10 minute

Rudik [331]

Answer:

W = 2352 J

Explanation:

Given that:

mass of the bucket, M = 10 kg

velocity of pulling the bucket, v = 3 $m.min^{-1}$

height of the platform, h = 30 m

time taken, t = 10 min

rate of loss of water-mass, m = $0.4 kg.min^{-1}$

Here, according to the given situation the bucket moves at the rate,

$v=3 m.min^{-1}$

The mass varies with the time as,

$M=(10-0.4t) kg$

Consider the time interval between t and t + ∆t. During this time the bucket moves a distance

∆x = 3∆t meters

So, during this interval change in work done,

∆W = m.g∆x

<u>For work calculation:</u>

$W=\int_{0}^{10} [(10-0.4t).g\times 3] dt$

$W= 3\times 9.8\times [10t-\frac{0.4t^{2}}{2}]^{10}_{0}$

$W= 2352 J$

5 0

3 years ago