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1 month ago

Solve A and B using energy considerations.

1 answer:

7 0

<span>Utilize the kinematic equation vf^2 = vi^2 + 2ad, where;
vf =?
vi = 0 m/s
a = 9.8 m/s^2
d1 = 10 m
d2 = 25 m

The final velocity upon reaching the ground (d1) is vf = sqrt(2)(9.8)(10) = 14 m/s. The final velocity at the base of the cliff (d2) is vf = sqrt(2)(9.8)(25) = 22.14 m/s.

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A 100 cm3 block of lead weighs 11N is carefully submerged in water. One cm3 of water weighs 0.0098 N.

Keith_Richards [3271]

The volume of lead measures 100 cm^3

with a density of lead at 11.34 g/cm^3

. Thus, the mass of the lead block equals density multiplied by volume

$m = 100 * 11.34 = 1134 g$

$m = 1.134 kg$

Therefore, its weight in air is noted as

$W = mg = 1.134* 9.8 = 11.11 N$

Next, the buoyant force acting on the lead is defined as

$F_B = W - F_{net}$

$F_B = 11.11 - 11 = 0.11 N$

We know that

$F_B = \rho V g$

$0.11 = 1000* V * 9.8$

After solving, we find

V = 11.22 cm^3

(ii) This corresponding volume of water exerts the same weight as the buoyant force, resulting in 0.11 N

(iii) The buoyant force measures 0.11 N

(iv) The lead block sinks in water due to its density being greater than that of water.

The buoyant force acting on the lead block counterbalances its weight

$F_B = W$

$\rho V g = W$

$13* 10^3 * V * 9.8 = 11.11$

$V = 87.2 cm^3$

(ii) This volume of mercury corresponds to the buoyant force weight, confirming that the block floats within mercury, resulting in 11.11 N as its weight.

(iii) The buoyant force is recorded as 11.11 N

(iv) Given that lead's density is less than mercury's, the lead will float in the mercury medium.

Indeed, an object that has lesser density than a liquid will float; otherwise, it will sink in the liquid.

3 0

1 month ago

We fully submerge an irregular lump of material in a certain fluid. The fluid that would have been in the space now occupied by

Ostrovityanka [3204]

Answer:

the lump descends

Explanation:

The full articulation of the query is

Upon fully submerging a 3 kg lump of material in a certain fluid, the fluid that would have occupied the space now filled by the lump weighs 2 kg. (a) When released, does the lump float up, sink, or remain steady

(a)

$F_{b}$ = Buoyant force acting upward on the lump

$M$ = mass of irregular lump = 3 kg

$m$ = mass of fluid displaced = 2 kg

The upward buoyant force on the lump is given by the weight of the displaced fluid, thus

$F_{b} = mg = (2) (9.8) = 19.6 N$

the weight of the irregular lump of material is represented as

$W = mg\\W = (3) (9.8)\\W = 29.4 N$

Given that the weight of the lump downward exceeds the upward buoyant force, the lump will indeed descend

4 0

1 month ago

On a caterpillars map all distances are marked in kilometers . The caterpillars map shows the distance between two milkweed plan

ValentinkaMS [3465]

Answer:

The equivalent distance in kilometers is 4012 × $10^{-6}$ km.

Explanation:

It's known that 1 millimeter converts to $10^{-3}$ meters. Then, 1 meter converts to $10^{-3}$ kilometers. Therefore, the conversion for 1 millimeter to kilometers can be stated as