How many windows should the real tree house have

A 91-kg astronaut and a 1300-kg satellite are at rest relative to the space shuttle. The astronaut pushes on the satellite, givi

WARRIOR [948]

Answer:

18.2145 meters

Explanation:

Using the conservation of momentum, we have that:

$m1v1 + m2v2 = m1'v1' + m2'v2'$

m1 = m1' is the mass of the astronaut, m2=m2' is the mass of the satellite, v1 and v2 are the inicial speed of the astronaut and the satellite (v1 = v2 = 0), and v1' and v2' are the final speed of the astronaut and the satellite. Then we have that:

$0 + 0 = 91*v1' + 1300*0.17$

$v1' = -1300*0.17/91 = -2.4286\ m/s$

The negative sign of this speed just indicates the direction the astronaut goes, which is the opposite direction of the satellite.

If the astronaut takes 7.5 seconds to come into contact with the shuttle, their initial distance is:

$distance = 2.4286 * 7.5 = 18.2145\ meters$

8 0

3 years ago

A solenoid with 35 turns per centimeter carries a current I. An electron moves within the solenoid in a circle that has a radius

castortr0y [4]

Answer:

The current of the solenoid is 0.0129 A.

Explanation:

The movement of the electron within the solenoid in a circle is produced by equaling the magnetic force and the centripetal force, as follows:

$F_{B} = F_{c}$

$e*v \mu_{0}*n*I = \frac{m*v^{2}}{r}$

$I = \frac{m*v}{e* \mu_{0}*n*r}$

Where:

I: is the current

m: is the electron's mass = 9.1x10⁺³¹ kg

v: is the electron's speed = 3.0x10⁵ m/s

μ₀: is the permeability magnetic = 4πx10⁻⁷ T.m/A

n: is the number of turns per unit length = 35/cm

r: is the radius of the circle = 3.0 cm

e: is the electron's charge = 1.6x10⁻¹⁹ C

$I = \frac{m*v}{e*\mu_{0}*n*r} = \frac{9.1 \cdot 10^{-31} kg*3.0 \cdot 10^{5} m/s}{1.6 \cdot 10^{-19} C*4\pi \cdot 10^{-7} T.m/A*3500/m*0.03 m} = 0.0129 A$

Therefore, the current of the solenoid is 0.0129 A.

I hope it helps you!

3 0

3 years ago

Two power lines run parallel for a distance of 222 m and are separated by a distance of 40.0 cm. if the current in each of the t

earnstyle [38]

1) Magnitude of the force:

The magnetic field generated by a current-carrying wire is
$B= \frac{\mu_0I}{2 \pi r}$
where
$\mu_0$ is the vacuum permeability
I is the current in the wire
r is the distance at which the field is calculated

Using I=135 A, the current flowing in each wire, we can calculate the magnetic field generated by each wire at distance
$r=40.0 cm=0.40 m$ ,
which is the distance at which the other wire is located:
$B= \frac{\mu_0 I}{2 \pi r}= \frac{(4 \pi \cdot 10^{-7} N/A^2)(135 A) }{2 \pi (0.40 m)}=6.75 \cdot 10^{-5} T$

Then we can calculate the magnitude of the force exerted on each wire by this magnetic field, which is given by:
$F=ILB=(135 A)(222 m)(6.75 \cdot 10^{-5}T)=2.03 N$

2) direction of the force:
The two currents run in opposite direction: this means that the force between them is repulsive. This can be determined by using the right hand rule. Let's apply it to one of the two wires, assuming they are in the horizontal plane, and assuming that the current in the wire on the right is directed northwards:
- the magnetic field produced by the wire on the left at the location of the wire on the right is directed upward (the thumb of the right hand is directed as the current, due south, and the other fingers give the direction of the magnetic field, upward)

Now let's apply the right-hand rule to the wire on the right:
- index finger: current --> northward
- middle finger: magnetic field --> upward
- thumb: force --> due east --> so the force is repulsive

A similar procedure can be used on the wire on the left, finding that the force exerted on it is directed westwards, so the force between the two wires is repulsive.

6 0

3 years ago

Solve the equation x=3logy2 for y.

melisa1 [442]

X = 3 · log(Y²)

X = 3 · 2·log(Y)

X/6 = log(Y)

10^(X/6) = 10^log(Y)

Y = 10^(X/6)

6 0

3 years ago

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