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3 years ago

Manuel is holding a 5kg box. How much force is the box exerting on him?In what direction

1 answer:

3 0

The force the box is exerting on Manuel is the weight of the box, downward:
$W=mg=(5 kg)(9.81 m/s^2)=49.05 N$
and this force is perfectly balanced by the constraint reaction applied by Manuel's hand, pushing upward.

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Is this good? Its for a science test ~

Digiron [165]

Yes..............................

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3 years ago

44. A mediados de la década de 1960, la McGill University lanzó sensores metereológicos de gran altitud al dispararlos desde un

Afina-wow [57]

La cinemática permite encontrar la respuesta para la aceleracion del cuerpo en el cañón es:

a = 1,8 10⁶ m/s²

La cinemática estudia el movimiento de los cuerpos, buscando relaciones entre la posicion, la velocidad y la aceleración.

v² = v₀² + 2 a x

Donde v y v₀ son la velocidad actual e inicial, respectivamente, a es la aceleracion y x la distancia recorrida.

Indica que la longitud de cañon es x= 18 m la velocidad de salida es

v= 29000 km/h ( $\frac{1000m}{1 km}$ ) ( $\frac{1h}{3600s}$ s) = 8,055,56 m/s.

La velocidad inicial del proyectil es cero.

a = $\frac{v^2}{2x}$

a = $\frac{ 8055.56^2}{2 \ 18}$

a = 1,8 10⁶ m/s²

En conclusión usando la cinemática podemos encontrarla respuesta para la aceleracion del cuerpo en el cañón es:

a = 1,8 10⁶ m/s²

Aprender mas aquí: brainly.com/question/19793086

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2 years ago

suggest a Form of renewable energy that could be used to power an electric ticket machine that only needs to work during dayligh

Stella [2.4K]

Answer:

Solar Energy

Explanation:

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3 years ago

A circular wire loop has a radius of 7.50 cm. a sinusoidal electromagnetic plane wave traveling in air passes throughthe loop, w

ExtremeBDS [4]

Intensity of electromagnetic wave is given as

$I = 2[\frac{B_{rms}^2}{2\mu_0}\times c]$

given that

$I = 0.0275 W/m^2$

$0.0275 = \frac{B_{rms}^2}{\mu_0} \times c$

here we know that

$\mu_0 = 4\pi \times 10^{-7}$

$c = 3 \times 10^8 m/s$

now we have

$0.0275 = \frac{B_{rms}^2}{4 \pi \times 10^{-7}}(3 \times 10^8)$

$B_{rms} = 1.07 \times 10^{-8} T$

now we will have

$B_0 = \sqrt2 B_{rms}$

$B_0 = 1.52 \times 10^{-8} T$

frequency of wave is given as

$f = \frac{c}{\lambda}$

$f = \frac{3 \times 10^8}{6.90} = 4.35 \times 10^7 Hz$

now the induced EMF is given as

$EMF = B_0A2\pi f$

$EMF = 1.52 \times 10^{-8} \times \pi(0.075)^2 \times (2\pi \times 4.25 \times 10^7)$

$EMF = 0.0733 Volts$

7 0

3 years ago

Read 2 more answers

A pendulum bob executing simple harmonic motion has 2cm and 12Hz as amplitude and frequency respectively. Calculate the period o

Jobisdone [24]

Answer:

Explanation:

The amplitude is extraneous information for this question.

Period (T) is simply a measure of time (how long it takes to complete a full cycle).

The frequency (f), is the reciprocal (how many cycles it can complete per unit of time).

The frequency reported is 12Hz. Hz is a unit meaning "cycles per second"

Thus 12cycles = 1second.

To find the number of seconds per cycle, simply create the other unit fraction and simplify:

$T=\frac{1second}{12cycles}\\T=0.8\bar{3}\frac{seconds}{cycle}\\$

$T=0.83seconds$

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1 year ago