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2 years ago

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Question #2

1 answer:

Contact [7]2 years ago

8 0

Answer:

Distance 20 km and Displacement 0 km

His displaceent is 0 km because he ends his walk where he started. The total distance of his walk is 20 km because he walks 10 km to the store + 10km back home.

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You have two identical pure silver ingots. You place one of them in a glass of water and observe it to sink to the bottom. You p

Elis [28]

Answer: a)

Explanation:

The buoyant force, as stated by Archimedes’ principle, is equal to the weight of the liquid that occupies the same volumen as the submerged object, as follows:

Fb = δ.V.g

If this force is larger than the weight of the object (that means that the fluid is denser than the solid), the object floats, which is the case for silver and mercury.

Instead, silver density is larger than water density, which explains why the pure silver ingot sinks.

Finally, as mercury is denser than water, we conclude that for a same object, the buoyant force in mercury is larger than in water (exactly 13.6 times greater).

3 0

2 years ago

One component of a metal sculpture consists of a solid cube with an edge of length 38.9 cm. The alloy used to make the cube has

vovangra [49]

Answer:

The mass of the cube is 420.8 kg.

Explanation:

Given that,

Length of edge = 38.9 cm

Density $\rho= 7.15 \times10^{3}\ kg/m^3$

We need to calculate the volume of cube

Using formula of volume

$V = 38.9^3$

$V=0.058863\ m^3$

We need to calculate the mass of the cube

Using formula of density

$\rho = \dfrac{m}{V}$

$m = V\times\rho$

$m =0.058863\times7.15 \times10^{3}$

$m=420.8\ kg$

Hence, The mass of the cube is 420.8 kg.

7 0

2 years ago

The drawing shows three identical springs hanging from the ceiling. Nothing is attached to the first spring, whereas a 4.50-N bl

Pavlova-9 [17]

Answer:

a. 30 N / m

b. 9.0 N

Explanation:

Given that

Unstretched length of the spring, $L_o$ = 20.0cm = 0.2m

a) When the mass of 4.5N is hanging from the second spring, then extended length Is

$L_1$ = 35.0cm = 0.35m

So, the change in spring length when mass hangs is

$x = L_1 - L_o$

= (0.35 - 0.20) m

= 0.15m

As spring are identical

Let us assume that the spring constant be "k", so at equilibrium

Restoring Force on spring = Block weightage

kx = W = 4.50

$k= \frac{4.50}{x} = \frac{4.50}{0.15}$

= 30 N / m

b) Now for the third spring, stretched the length of spring is

$L_2$ = 50cm = 0.5m

So, the change in spring length is

$x'= L_2 - L_o$

= (0.5-0.20)m

= 0.30m

At equilibrium,

Restoring Force on spring = Block weightage

Now using all mentioned and computed values in above,

$W'= kx'$

= 30(0.3)

= 9.0 N

4 0

2 years ago

The solar energy strikes the deck at the rate of 1400 W on every square metre.

Elena-2011 [213]

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8 0

1 year ago

A 0.3 mm long invertebrate larva moves through 20oC water at 1.0 mm/s. You are creating an enlarged physical model of this larva

AleksandrR [38]

Answer:

Explanation:

For the problem, we should have same reynolds number

ρvd/mu = constant

1000×1×10⁻³×0.3×10⁻³/1.002×10⁻³ = 1400×0.5×d/600

d = 25.66 cm

5 0

2 years ago