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2 years ago

5

What can Lisa do to increase the strength of the electromagnet?

1 answer:

drek231 [11]2 years ago

4 0

Answer:

move the wire loops closer

Explanation:

because the closer t they are the more concentrated the energy is in that specific area

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Weight is the direct result of :<br><br> A. Distance <br><br> B. The amount of gravity being pulled

zheka24 [161]

B i think ........................

4 0

2 years ago

A 100-kg spacecraft is in a circular orbit about Earth at a height h = 2RE .

maria [59]

To solve this problem it is necessary to apply the concepts related to the conservation of the Gravitational Force and the centripetal force by equilibrium,

$F_g = F_c$

$\frac{GmM}{r^2} = \frac{mv^2}{r}$

Where,

m = Mass of spacecraft

M = Mass of Earth

r = Radius (Orbit)

G = Gravitational Universal Music

v = Velocity

Re-arrange to find the velocity

$\frac{GM}{r^2} = \frac{v^2}{r}$

$\frac{GM}{r} = v^2$

$v = \sqrt{\frac{GM}{r}}$

PART A ) The radius of the spacecraft's orbit is 2 times the radius of the earth, that is, considering the center of the earth, the spacecraft is 3 times at that distance. Replacing then,

$v = \sqrt{\frac{(6.67*10^{-11})(5.97*10^{24})}{3*(6.371*10^6)}}$

$v = 4564.42m/s$

From the speed it is possible to use find the formula, so

$T = \frac{2\pi r}{v}$

$T = \frac{2\pi (6.371*10^6)}{4564.42}$

$T = 8770.05s\approx 146min\approx 2.4hour$

Therefore the orbital period of the spacecraft is 2 hours and 24 minutes.

PART B) To find the kinetic energy we simply apply the definition of kinetic energy on the ship, which is

$KE = \frac{1}{2} mv^2$

$KE = \frac{1}{2} (100)(4564.42)^2$

$KE = 1.0416*10^9J$

Therefore the kinetic energy of the Spacecraft is 1.04 Gigajules.

8 0

2 years ago

A weight of 30.0 N is suspended from a spring that has a force constant of 220 N/m. The system is undamped and is subjected to a

Nimfa-mama [501]

Answer:

$F_0 = 393 N$

Explanation:

As we know that amplitude of forced oscillation is given as

$A = \frac{F_0}{ m(\omega^2 - \omega_0^2)}$

here we know that natural frequency of the oscillation is given as

$\omega_0 = \sqrt{\frac{k}{m}}$

here mass of the object is given as

$m = \frac{W}{g}$

$\omega_0 = \sqrt{\frac{220}{\frac{30}{9.81}}}$

$\omega_0 = 8.48 rad/s$

angular frequency of applied force is given as

$\omega = 2\pi f$

$\omega = 2\pi(10.5) = 65.97 rad/s$

now we have

$0.03 = \frac{F_0}{3.06(65.97^2 - 8.48^2)}$

$F_0 = 393 N$

6 0

3 years ago

4. A student measures a temperature several times. The readings lie between 29.6 and 30.2 K. This

Alona [7]

Answer:

(C) apparently written incorrectly - it should be 29.9 +- .3 K

(read 29.9 plus or minus .3 K)

8 0

2 years ago

You have an initial velocity of -3.0 m/s. You then experience an acceleration of 2.5 m/s2 for 9.0s; what

Tatiana [17]

27.9 idkkkk look it up on photomath

8 0

3 years ago