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STatiana
3 months ago
13

This is a problem about a child pushing a stack of two blocks along a horizontal floor. The masses of the blocks, and the coeffi

cients of friction between the two blocks and between the lower block and the floor will be given. In order to do the pushing, the child will only be touching one of the two blocks. The mass of the upper block in the stack is 0.760 kg . The mass of the lower block in the stack is 1.630 kg . The coefficients of friction between the two blocks are: static 0.790, and kinetic 0.660. The child's mother, who likes to encourage his experiments, has oiled a small strip of the horizontal floor so that it is very slick; the coefficient of kinetic friction between the oiled section of floor and the lower block is only 0.080 and the coefficient of static friction is insignificantly different. Before the pushing starts, here is a question about the vertical forces acting on the two blocks.
Required:
What is the vertical component of the contact force on the lower block by the floor?
Physics
1 answer:
Sav [3.1K]3 months ago
7 0

Answer:

 N = 23.4 N

Explanation:

Let's tackle this problem after reading through the lengthy explanation.

The normal force, which we refer to as the contact force in this scenario, can be determined by applying the equation for translational equilibrium along the y-axis.

            N - w₁ -w₂ =

            N = m₁ g + m₂ g

            N = g (m₁ + m₂)

Now, perform the calculations:

            N = 9.8 (0.760 + 1.630)

            N = 23.4 N

This represents the upward force exerted by the two blocks resting on the surface.

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A cylinder has 500 cm3 of water. After a mass of 100 grams of sand is poured into the cylinder and all air bubbles are removed b
Keith_Richards [3271]

Answer: SG = 2.67

The specific gravity for the sand is 2.67

Explanation:

Specific gravity is determined by the formula: density of the substance/density of water

Provided information;

Mass of sand m = 100g

The volume of sand equals the volume of water it displaces

Vs = 537.5cm^3 - 500 cm^3

Vs = 37.5cm^3

Calculating density of sand = m/Vs = 100g/37.5 cm^3

Ds = 2.67g/cm^3

Density of water Dw = 1.00 g/cm^3

Thus, the specific gravity of the sand can be expressed as

SG = Ds/Dw

SG = (2.67g/cm^3)/(1.00g/cm^3)

SG = 2.67

The specific gravity of the sand stands at 2.67

3 0
2 months ago
A 10-turn conducting loop with a radius of 3.0 cm spins at 60 revolutions per second in a magnetic field of 0.50T. The maximum e
kicyunya [3294]

Answer:

Maximum emf = 5.32 V

Explanation:

Provided data includes:

Number of turns, N = 10

Radius of loop, r = 3 cm = 0.03 m

It made 60 revolutions each second

Magnetic field, B = 0.5 T

We are tasked to determine the maximum emf produced in the loop, which is founded on Faraday's law. The induced emf can be calculated by:

\epsilon=\dfrac{d(NBA\cos\theta)}{dt}\\\\\epsilon=NBA\dfrac{d(\cos\theta)}{dt}\\\\\epsilon=NBA\omega \sin\omega t\\\\\epsilon=NB\pi r^2\omega \sin\omega t

For the maximum emf, \sin\omega t=1

Therefore,

\epsilon=NB\pi r^2\omega \\\\\epsilon=NB\pi r^2\times 2\pi f\\\\\epsilon=10\times 0.5\times \pi (0.03)^2\times 2\pi \times 60\\\\\epsilon=5.32\ V

Hence, the maximum emf generated in the loop is 5.32 V.

3 0
3 months ago
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