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AlexFokin [52]
3 years ago
8

A mini rail gun made of a copper rod of mass m and radius r rests on two parallel copper rails that are a distance L apart and o

f length d. A magnetic field of magnitude B is directed perpendicular to the rod and the rails. When current is applied to the parallel copper rails, the rod rolls along the rails without slipping. If the rod starts from rest, what is the speed of the rod as it leaves the rails?
Physics
1 answer:
ira [324]3 years ago
7 0

Answer:

Explanation:

Magnetic Force on the rod  F = Bi L

Work done by this force = F X d

=  Bi Ld

This energy is converted into both rotational and linear kinetic energy

= 1/2 I ω² + 1/2 mv²

= 1/2 x 1/2 m r²ω²+ 1/2 mv²

= 1/4 m v² + 1/2 mv²

= 3/4 mv²

So according to conservation of energy

Bi Ld =  3/4 mv²

v = \sqrt{\frac{4BiLd}{3m} }

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Calculer l intensité du courant qui le traverse
MissTica

Answer:

Le calcul du courant se fait avec deux éléments : la tension et la valeur de la résistance. Courant (A) = tension (V) / résistance (Ohm) ce qui donne la formule I = U/R.

please mark me as brainalist

6 0
2 years ago
An astronaut is standing on the surface of a planetary satellite that has a radius of 1.74 × 10^6 m and a mass of 7.35 × 10^22 k
ExtremeBDS [4]

Answer:

2.87 km/s

Explanation:

radius of planet, R = 1.74 x 10^6 m

Mass of planet, M = 7.35 x 10^22 kg

height, h = 2.55 x 10^6 m

G = 6.67 x 106-11 Nm^2/kg^2

Use teh formula for acceleration due to gravity

g=\frac{GM}{R^{2}}

g=\frac{6.67\times 10^{-11}\times 7.35\times 10^{22}}{1.74^{2}\times 10^{12}}

g = 1.62 m/s^2

initial velocity, u = ?, h = 2.55 x 10^6 m , final velocity, v = 0

Use third equation of motion

v^{2}=u^{2}-2gh

0 = v² - 2 x 1.62 x 2.55 x 10^6

v² = 8262000

v = 2874.37 m/s

v = 2.87 km/s

Thus, the initial speed should be 2.87 km/s.

6 0
3 years ago
A vertical spring with spring stiffness constant 305 N/m oscillates with an amplitude of 28.0 cm when 0.235 kg hangs from it. Th
den301095 [7]

Answer:

The function that describe the motion in the time

y (t) = 0.28m * sin ( 36.025 * t)

Explanation:

The angular frequency of oscillation of the spring

w = √k/m

w = √305 N/m / 0.235 kg

w = 36.025 rad / s

To determine the function of the motion knowing as a motion oscillation in a amplitude a frequency

y(t) = A * sin (w t )

So

A = 28.0 cm * 1 m / 100 cm = 0.28 m

So replacing to determine the function of the motion in the time

y (t) = A sin (w t)

y (t) = 0.28m * sin ( 36.025 * t)

7 0
3 years ago
Cliff divers at Acapulco jump into the sea from a cliff 37.1 m high. At the level of the sea, a rock sticks out a horizontal dis
Stella [2.4K]

Answer:

v_x = 4.87 m/s

Explanation:

Height of the cliff is given as

h = 37.1 m

now the time taken by the diver to hit the surface is given as

h = \frac{1}{2}gt^2

37.1 = \frac{1}{2}(9.8)t^2

t = 2.75 s

Now in the same time it has to cover a distance of 13.39 m

so the speed in horizontal direction is given as

v_x = \frac{x}{t}

v_x = \frac{13.39}{2.75}

v_x = 4.87 m/s

3 0
3 years ago
Estimate the electric field at a point 2.40 cm perpendicular to the midpoint of a uniformly charged 2.00-m-long thin wire carryi
nadya68 [22]

Answer:

E = 1.85*10^{12}\frac{N}{C}

Explanation:

Hi!

The perpendicular distance 2.4cm, is much less than the distance to both endpoints of the wire, which is aprox 1m. Then the edge effect is negligible at this field point, and we can aproximate the wire as infinitely long.

The electric filed of an infinitely long wire is easy to calculate. Let's call z the axis along the wire. Because of its simmetry (translational and rotational), the electric field E must point in the radial direction,  and it cannot depende on coordinate z. To calculate the field Gauss law is used, as seen in the image, with a cylindrical gaussian surface. The result is:

E = \frac{\lambda}{2\pi \epsilon_0 r}\\\lambda=\text{charge per unit length}=\frac{4.95 \mu C}{2 m} = 2.475 \frac{C}{m}\\r=\text{perpendicular distance to wire}\\\epsilon_0=8.85*10^{-12}\frac{C^2}{Nm^2}

Then the electric field at the point of interest is estimated as:

E = \frac{\22.475}{2\pi*( 8.85*10^{-12})*(2.4*10^{-2})}\frac{N}{C}=1.85*10^{12}\frac{N}{C}

6 0
3 years ago
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