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3 years ago

11

Un cuerpo de 1250 kg se mueve sobre una superficie horizontal por la acción de un mo tor que produce una fuerza de 24500 N. Sabi

endo el cuerpo solo aprovecha el 45% de la fuerza que genera el motor y que el coeficiente de rozamiento es 0,15, la aceleración que experimenta el cuerpo vale:
a) 80 m/s

b) 7,9 m/s?

c) 7,3 m/s

1 answer:

bezimeni [28]3 years ago

4 0

The correct answer is b) 7,9 m/s

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Your local AM radio station broadcasts at a frequency of f = 1100 kHz. The electric-field component of the signal you receive at

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Answer:

(a) 272.73 m

(b) 0.338 N/C

Explanation:

frequency, f = 1100 kHz = 1100 x 1000 Hz

E(t) = Eo Sin(2πft)

Eo = 0.62 N/C

(a) Velocity of light, c = 3 x 10^8 m/s

wavelength, λ = c / f = (3 x 10^8) / (1100000) = 272.73 m

Thus, the wavelength is 272.73 m.

(b) at t = 3.1 microsecond = 3.1 x 10^-6 s

E = Eo Sin (2 π ft)

E = 0.62 Sin (2 x 3.14 x 1100 x 10^3 x 3.1 x 10^-6)

E = 0.62 Sin (21.4148)

E = 0.62 x 0.5449 = 0.338 N/C

Thus, the electric field at t = 3.1 microsecond s 0.338 N/C.

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To measure one wavelength, you would measure the distance between

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Answer:

the answer would be A and B

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Matter that organisms require for their life processes are

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Answer:

nutrients

Explanation:

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3 years ago

What is the kinetic energy of a 7.26 kg bowling ball that is rolling at a speed of 2m/s

vlada-n [284]

Answer:

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Explanation:

The kinetic energy of an object can be found by using the formula

$k = \frac{1}{2} m {v}^{2} \\$

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We have the final answer as

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3 years ago

A 100-turn, 2.0-cm-diameter coil is at rest in a horizontal plane. A uniform magnetic field 60∘ away from vertical increases fro

bixtya [17]

Answer:

The induced emf is $|\epsilon|=0.0261 V$

Explanation:

From the question we are told that

The number of turn is $N = 100$

The diameter of the coil is $d = 2.0 cm = 2.0 *10^{-2} m$

The uniform magnetic at initial is $B_i = 0.50 T$

The uniform magnetic at initial is $B_f = 1.50 T$

The time taken is $t = 0.60s$

The angle the magnetic field makes with vertical is $\theta = 60^o$

Generally induced emf is mathematically represented as

$\epsilon = -N \frac{d \o}{dt}$

where $d \o$ is the change magnetic flux

Magnetic flux is mathematically represented as

$\O = \= B \cdot \= A$

$= BA cos \theta$

Substituting this above

$\epsilon = -N \frac{d (BA cos \theta)}{dt}$

$\epsilon = -N A \frac{d (B cos \theta)}{dt}$

Where B is the magnetic field and A is the area which is mathematically evaluated as

$A = \frac{\pi d^2}{4}$

Substituting values

$A = \frac{\pi (2.0 *10^{-2})^2}{4}$

$A= 3.142*10^{-4}m^2$

From the equation of emf

$\epsilon = -N A \frac{d (B cos \theta)}{dt}$

dB = $B_2 -B_1$

So

$|\epsilon| = N A \frac{ (B_2 -B_1 cos \theta)}{dt}$

substituting values

$|\epsilon| = 100(3.142*10^{-4}) \frac{ (1.50 -0.50 cos(60))}{0.60}$

$|\epsilon|=0.0261 V$

6 0

3 years ago