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2 years ago

Actividad: Sitúese en el siglo III a.c. el rey de Siracusa, Hierón II, le pide

Physics

1 answer:

zvonat [6]2 years ago

7 0

Answer:

ρ_c < ρ_gold the crown is not gold

Explanation:

For this exercise we use the translational equilibrium condition

∑ F = 0

in the air

T₁ - W = 0

T₁ = W

in water

T + B - W = 0

where T is the scale reading, B the liquid thrust and W the weight

B = W -T

B = 7.84 - 6.84

B = 1 N

thrust is

B = ρ g V_water

V_water = $\frac{B}{\rho g}$

V_water = $\frac{1}{1000 \ 9.8}$

V_water = 1.02 10⁻⁴ m³

how the crown is totally submerged

V_c = V_water

V_c = 1.02 10⁻⁴ m³

now we can calculate the density of the crown

ρ_c = m_c / Vc

W = mg

m = W / g

we substitute

ρ_c = $\frac{W}{g \ V_c}$

ρ_c = $\frac{7.84}{ 9.8 \ 1.02 \ 10^{-4}}$

ρ_c = 0.784 10⁴ kg / m³

ρ_c = 7.84 10³ kg / m³

We can see that this density is less than the density of gold

ρ_c < ρ_gold

therefore the crown is not gold

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If a migrating loon flies at an average speed of 19 m/s, what is its average speed in mi/hr?

zlopas [31]

Unit conversion is a way of converting some common units into another without changing their real value. The average speed of the migrating loon flies is 45.5019 miles/hr.

<h3>What is Unit conversion?</h3>

Unit conversion is a way of converting some common units into another without changing their real value. for, example, 1 centimeter is equal to 10 mm, though the real measurement is still the same the units and numerical values have been changed.

Given that migrating loon flies at an average speed of 19 m/s. Now, since 1 meter is equal to (1/1609.34) miles and 1 second is equal to (1/3600) seconds. Therefore, we can write the speed as,

$\rm Speed = 19\ \dfrac{meter}{second}\\$

$=19 \times \dfrac{\frac{1}{1609.34}\rm\ miles}{\frac{1}{3600} \rm\ hours}$

= 19 × (3600/1609.34) miles/hr

= 42.5019 miles/ hr

Hence, the average speed of the migrating loon flies is 45.5019 miles/hr.

Learn more about Units conversion here:

brainly.com/question/4736731

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4 0

1 year ago

. One long wire carries a current of 30 A along the entire x axis. A second-long wire carries a current of 40 A perpendicular to

d1i1m1o1n [39]

Answer:

magnitude of net magnetic field at given point is

$B = 5 \times 10^{-6} T$

Explanation:

As we know that magnetic field due to a long current carrying wire is given as

$B = \frac{\mu_o i}{2\pi r}$

here we we will find the magnetic field due to wire which is along x axis is given as

$i = 30 A$

r = 2 m

now we have

$B_1 = \frac{4\pi \times 10^{-7} (30)}{2\pi (2m)}$

$B_1 = 3\times 10^{-6} T$ into the plane

Now similarly magnetic field due to another wire which is perpendicular to xy plane is given as

$i = 40 A$

r = 2 m

now we have

$B_2 = \frac{4\pi \times 10^{-7} (40)}{2\pi (2m)}$

$B_2 = 4\times 10^{-6} T$ along + x direction

Since the two magnetic field is perpendicular to each other

So here net magnetic field is given as

$B = \sqrt{B_1^2 + B_2^2}$

$B = 5 \times 10^{-6} T$

5 0

3 years ago

Singly charged uranium-238 ions are accelerated through a potential difference of 2.20 kV and enter a uniform magnetic field of

Aleonysh [2.5K]

Answer:

r = 0.0548 m

Explanation:

Given that,

Singly charged uranium-238 ions are accelerated through a potential difference of 2.20 kV and enter a uniform magnetic field of 1.90 T directed perpendicular to their velocities.

We need to find the radius of their circular path. The formula for the radius of path is given by :

$r=\dfrac{1}{B}\sqrt{\dfrac{2mV}{q}}$