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alexandr402 [8]
3 years ago
6

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

lace the other in a container full of mercury and observe that it floats. Comparing the buoyant forces in the two cases you conclude that
a.) the buoyant force in water is smaller than in mercury

b.) the buoyant force in the water is larger than that in mercury

c.) the buoyant force in the water is zero and that in mercury is non - zero

d.) the buoyant force in the water is equal to that in mercury

e.) no conclusion can be made about the respective values of the buoyant forces
Physics
1 answer:
Elis [28]3 years ago
3 0

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).  

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A 0.40-kg block is attached to the end of a horizontal ideal spring and rests on a frictionless surface. The block is pulled so
lubasha [3.4K]

Answer:

160N/m

Explanation:

According to Hooke's law which states that the extension of an elastic material is directly proportional to the applied force provided that the elastic limit is not exceeded. Mathematically,

F = ke where

F is the applied force

k is the spring constant

e is the extension

From the formula k = F/e

Since the body accelerates when the block is released, F = ma according to Newton's second law of motion.

The spring constant k = ma/e where

m is the mass of the block = 0.4kg

a is the acceleration = 8.0m/s²

e is the extension of the spring = 2.0cm = 0.02m

K = 0.4×8/0.02

K = 3.2/0.02

K = 160N/m

The spring constant of the spring is therefore 160N/m

4 0
3 years ago
For a constant force, the impulse could be increased by: A.decreasing the distance. B.keeping the time the force acts constant.
Andre45 [30]
C. Increasing the time of the force
This is because
Impulse = FΔt
5 0
3 years ago
An ice skater starts with a velocity of 2.25 m/s in a 50.0 direction. after 8.33 m/s in a 120 direction. what is the y-component
Bond [772]

The y-component of the acceleration is 0.33 m/s^2

Explanation:

The y-component of the acceleration is given by:

a_y = \frac{v_y-u_y}{t}

where

v_y is the y-component of the  final velocity

u_y is the y-component of the initial velocity

t is the time elapsed

For the ice skater in this problem, we have:

u_y = u sin \theta_1 = (2.25 m/s)(sin 50.0^{\circ})=1.72 m/s

where

u = 2.25 m/s is the initial velocity

\theta_1 = 50.0^{\circ} is the initial  direction

v_y = v sin \theta_2 = (4.65)(sin 120^{\circ})=4.03 m/s, where

v = 4.65 m/s is the final velocity

\theta_2 = 120.0^{\circ} is the final direction

The time elapsed is

t = 8.33 s

Therefore, we can find the y-component of the acceleration:

a_y=\frac{4.03-1.72}{8.33}=0.33 m/s^2

Learn more about acceleration:

brainly.com/question/9527152

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3 0
3 years ago
A light spring having a force constant of 115 N/m is used to pull a 9.00 kg sled on a horizontal frictionless ice rink. The sled
Ksivusya [100]

Answer:

Stretch in the spring = 0.1643 (Approx)

Explanation:

Given:

Mass of the sled (m) = 9 kg

Acceleration of the sled (a) = 2.10 m/s ²

Spring constant (k) = 115 N/m

Computation:

Tension force in the spring  (T) = ma

Tension force in the spring  (T) = 9 × 2.10

Tension force in the spring  (T) = 18.9 N

Tension force in the spring = Spring constant (k) × Stretch in the spring

18.9 N = 115 N  × Stretch in the spring

Stretch in the spring = 18.9 / 115

Stretch in the spring = 0.1643 (Approx)

8 0
3 years ago
A moose runs through the woods and covers 825 m North in 118 s. What is the average velocity of the moose?
weqwewe [10]
  • Total displacement=825m
  • Total Time=118s

Average Velocity=Total Displacement/Total Time

\\ \sf\longmapsto Average\: Velocity=\dfrac{825}{118}

\\ \sf\longmapsto Average\:Velocity\approx 8m/s

4 0
2 years ago
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