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3 years ago

10

Mutations provide a basis for...

1 answer:

Sladkaya [172]3 years ago

5 0

Answer:

A.

Explanation:

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10m= (5.0) + (.5)(9.8)(5.0)

Norma-Jean [14]

M=2.45 because you multiply out the equation on the right and divide by 10

8 0

3 years ago

An ideal solenoid having a coil density of 5000 turns per meter is 10 cm long and carries a current of 4.0

Diano4ka-milaya [45]

The rule that is used to get the strength of magnetic field at the center of solenoid (B) is:
B = <span>µ x n x I where:
</span>µ is the permeability of the medium where the solenoid is based. In this problem, the medium is air which means that µ = <span>µ </span><span>o = 4 pi x 10^-7 Tm/A
</span>I is the current passing (I = 4 amperes)
n is the number of turns per unit length (5000 turns)

Substituting in the mentioned equation, we find that:
B = 4 x 3.14 x 10^-7 x 5000 x 4 = 25.132 mT

5 0

3 years ago

An astronaut lands on a new, recently discovered planet in a different star system. The astronaut measures the acceleration due

Nezavi [6.7K]

Answer:

The radius of the new planet is ~2.04 * 10⁶ m, or 2,041,752 m.

Explanation:

We can use Newton's Law of Universal Gravitation:

$\displaystyle F_g=G\frac{Mm}{r^2}$

Let's look at Newton's 2nd Law:

$F=ma$

We can set these equations equal to each other:

$\displaystyle G\frac{Mm}{r^2} =ma$

The mass of the second mass (astronaut) cancels out. We are left with:

$\displaystyle G\frac{M}{r^2} =a$

We are solving for the radius of the new planet, so we can rearrange the equation:

$\displaystyle r=\sqrt{\frac{GM}{a} }$

Substitute in our known values given in the problem (<u><em>G = 6.67 * 10⁻¹¹ </em></u><em> ; </em><u><em>M = 7.5 * 10²³</em></u><em> ; </em><u><em>a = 12</em></u>).

$\displaystyle r =\sqrt{\frac{(6.67\times 10^{-11})(7.5 \times 10^{23}}{12} }$
$r=2.04 \times 10^6$

The radius of the new planet is ~2.04 * 10⁶ m.

3 0

2 years ago

Exercise can help prevent?

Sedaia [141]

health conditions and diseases

8 0

3 years ago

Read 2 more answers

A ball of radius r rolls on the inside of a track of radius R. If the ball starts from rest at the vertical edge of the track, f

meriva

Answer:

$v=\sqrt{\dfrac{10g(R-r)}{7}}$

Explanation:

Given that

Radius of track = R

Radius of ball = r

The ball can be treated as solid sphere, so

The moment of inertia of ball

$I=\dfrac{2}{5}mr^2$

When the ball reach at the lowest position then it will have both angular and linear speed.

Condition for rolling without slipping v= ωr

Form energy conservation

$mgR=mgr+\dfrac{1}{2}mv^2+\dfrac{1}{2}I\omega^2$

v= ωr

$I=\dfrac{2}{5}mr^2$

$mgR=mgr+\dfrac{1}{2}mv^2+\dfrac{1}{2}\times \dfrac{2}{5}mr^2\omega^2$

$mg(R-r)=\dfrac{1}{2}mv^2+\dfrac{1}{2}\times \dfrac{2}{5}mv^2$

$2mg(R-r)=mv^2+\dfrac{2}{5}mv^2$

$2g(R-r)=\dfrac{7}{5}v^2$

$v=\sqrt{\dfrac{10g(R-r)}{7}}$

3 0

3 years ago