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3 years ago

How much do a steel BB weight?

Physics

1 answer:

pychu [463]3 years ago

6 0

Each BB 0.349 grams and 5.386 grains. 349 mg & 5.386 grains each steel BBs.

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under normal conditions describe how increasing the temperature affects the solubility of a typical salt

Eduardwww [97]

For many solids dissolved in liquid water, the solubility increases with temperature. The increase in kinetic energy that comes with higher temperatures allows the solvent molecules to more effectively break apart the solute molecules that are held together by intermolecular attractions.

8 0

3 years ago

What engine thrust is required for a rocket of mass 35 kg to leave the launching pad? 3.5 N. 35 N. 351 N. 3,500 N. 35,000 N.

soldi70 [24.7K]

In order for the object to move upward, it needs an upward force
that's at least equal to its own weight.

Weight = (mass) x (gravity) = (35 kg) x (9.8 m/s²) = 343 N.

The engine thrust has to be more than 343 N.

7 0

3 years ago

How far from Earth must a space probe be along a line toward the Sun so that the Sun's gravitational pull on the probe balances

Tatiana [17]

Answer:

258774.9441 m

Explanation:

x = Distance of probe from Earth

y = Distance of probe from Sun

Distance between Earth and Sun = $x+y=149.6\times 10^6\ m$

G = Gravitational constant

$M_s$ = Mass of Sun = $1.989\times 10^{30}$

$M_e$ = Mass of Earth = $5.972\times 10^{24}\ kg$

According to the question

$\frac{GM_sm}{x^2}=\frac{GM_em}{y^2}\\\Rightarrow \frac{M_s}{x^2}=\frac{M_e}{y^2}\\\Rightarrow x=\sqrt{\frac{M_s\times y^2}{M_e}}\\\Rightarrow x=\sqrt{\frac{1.989\times 10^{30}\times y^2}{5.972\times 10^{24}}}\\\Rightarrow x=577.10852y$

$x+y=149.6\times 10^6\\\Rightarrow 577.10852y+y=149.6\times 10^6\\\Rightarrow 578.10852y=149.6\times 10^6\\\Rightarrow y=\frac{149.6\times 10^6}{578.10852}\\\Rightarrow y=258774.9441\ m$

The probe should be 258774.9441 m from Earth

7 0

3 years ago

In a 200-turn automobile alternator, the magnetic flux in each turn is ΦB = 2.50 10-4 cos ωt, where ΦB is in webers, ω is the an

n200080 [17]

Answer:

$10.63952sin(209.43951t)$

10.63952 V

Explanation:

$N_t$ = Number of turns = 200

$\phi_B$ = Magnetic flux = $2.5\times 10^{-4}cos(\omega t)$

$\omega_e$ = Engine angular speed = $1\times 10^{3}\ rpm$

Alternator angular speed is given by

$N_a=2\times \omega_e\\\Rightarrow N_a=2\times 1\times 10^{3}\\\Rightarrow N_a=2\times 10^{3}\ rpm$

$\omega=N_a\dfrac{2\pi}{60}\\\Rightarrow \omega=2\times 10^{3}\dfrac{2\pi}{60}\\\Rightarrow \omega=209.43951\ rad/s$

Induced emf is given by

$\epsilon=-N_t\dfrac{d\phi_B}{dt}\\\Rightarrow \epsilon=-200\dfrac{d}{dt}2.54\times 10^{-4}cos(209.43951 t)\\\Rightarrow \epsilon=200\times 2.54\times 10^{-4}\times 209.43951 sin(209.43951t)\\\Rightarrow \epsilon=10.63952sin(209.43951t)$

The function is $10.63952sin(209.43951t)$

The induced maximum emf is 10.63952 V

5 0

3 years ago

Helphpphphphpphphphpphphpph

adell [148]

Airplane with nose up: The plane's speed through the air is the square root of (80 m/s squared) plus (120 m/s squared. The whole picture is a right triangle, and the plane's speed is the hypotenuse. The angle is the angle whose tangent is (80/120). You can get it from a calculator, a book, a slide rule, or online from the site that rhymes with floogle. The man pulling the load is also a right triangle. The horizontal component is (hypotenuse) times (cosine of the angle). The vertical component is (hypotenuse) times (sine of the same angle). Fill in what you know, look up the sin and cos of 25 degrees and write those in too, and then you can solve for what you have to find.

7 0

3 years ago