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3 years ago

When a crown of mass 14.7 kg is submerged in water, an accurate scale reads only 13.4 kg. What material is the crown made of?

Physics

1 answer:

Bogdan [553]3 years ago

3 0

Answer:

ρ = 1.13 10⁴ km/m³

Explanation:

For this exercise we use Newton's equilibrium equation

B –W + W_scale = 0

Where B is the thrust and W_scale is the balance reading

The push is given by Archimedes' law

B = ρ_water g V

B = W- W_scale

B = m g - m_scale g

Let's calculate

B = 14.7 9.8 - 13.4 9.8

B = 12.74 N

ρ_water g V = 12.74

V = 12.74 / ρ_water g

V = 12.74 / 1000 9.8

V = 0.0013 m³

Let's use density

ρ = m / V

We replace

ρ = 14.7 / 0.0013

ρ = 1.13 10⁴ km/m³

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1- ¿Cuál sería la energía de un objeto de 50 newton de peso que se encuentra sobre una estantería de 3 metros de altura? ¿Qué ti

Liula [17]

Responder:

<h3>150 Nm </h3><h3>Energía potencial </h3>

Explicación:

El tipo de energía que posee el objeto se conoce como energía potencial. <u>La energía potencial es la energía que posee un objeto, mi virtud de su posición. </u>

Energía potencial = masa * aceleración debido a la gravedad * altura

Dado que Force = masa * aceleración debido a la gravedad

Energía potencial = Fuerza * altura

Fuerza dada = 50N y altura = 3 m

Energía potencial = 50 * 3

Energía potencial = 150 Nm

6 0

3 years ago

A car is traveling at a speed of 38.0 m/s on an interstate highway where the speed limit is 75.0 mi/h. Is the driver exceeding t

Tom [10]

Answer: Yes, he is exceeding the speed limit

Explanation:

Hi!

This is problem about unit conversion

1 mile = 1,609.344 m

Then the speed limit v is:

v = 75 mi/h = 120,700.8 m/h

1 hour = 60 min = 60*60 s = 3,600 s

v = (120,700.8/3,600) m/s = 33.52 m/s

38 m/s is higher than the speed limit v.

5 0

3 years ago

A spider begins to spin a web by first hanging from a ceiling by his fine, silk fiber. He has a mass of 0.025 kg and a charge of

Rasek [7]

Answer:

a) (5.59 × 10³) N/C

b) 0.226 N directed away from the spider.

Explanation:

a) Electric field, E, felt as a result of point charge, Q, at a distance, d away is given by

E = kQ/d²

So, magnitude of the electric field due to the charge on the second spider at the position of the first spider

Q = 4.2 µC = 4.2 × 10⁻⁶ C

k = Coulomb's constant = 8.99 × 10⁹ Nm²/C

d = 2.6 m

E = (8.99 × 10⁹ × 4.2 × 10⁻⁶)/2.6²

E = 5.59 × 10³ N/C

b) Tension in the silk fiber above the spider is the net force due to the weight of spider one and the force of repulsion due the two charges.

Force due to the two charges = Eq

where q now represents the charge of the first spider at the first point, feeling the electric field calculated in (a)

F = 5.59 × 10³ × 3.4 × 10⁻⁶ = 0.01901 N directed upwards. (That is, F = + 0.019 N)

Weight of the spider = mg = 0.025 × 9.8 = 0.245 N directed downwards. (That is, W = -0.245 N)

Net force, T = mg - F = 0.245 - 0.019 = 0.226 N (that is, 0.226 N, directed upwards, away from the spider).

8 0

3 years ago

An astronaut drops a hammer from 2.0 meters above the surface of the moon. if the acceleration due to gravity on the moon is 1.6

aivan3 [116]

Answer:

1.57 s

Explanation:

Since the motion of the hammer is a uniformly accelerated motion, the distance covered by the hammer in a time t is

$S=\frac{1}{2}at^2$

Where, in this case

S = 2.0 m is the distance covered

a = 1.62 m/s^2 is the acceleration due to gravity

t is the time taken

Re-arranging the equation, we can find the time the hammer takes:

$t=\sqrt{\frac{2S}{a}}=\sqrt{\frac{2(2.0 m)}{1.62 m/s^2}}=1.57 s$

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RideAnS [48]

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3 years ago

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