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2 years ago

Un bloque de 25 n se encuentra suspendido por un hilo al techo vitamina la tensión que aparece en el hilo

Physics

1 answer:

SVETLANKA909090 [29]2 years ago

7 0

Answer:

T = 25 N

Explanation:

The question says that "A 25 n block is suspended by a wire from the ceiling vitamin the tension that appears in the wire ?"

Weight of the block, W = 25 N

Weight of a body acts in downward direction and tension acts in upward direction. It would mean that,

Tension = weight of the block

T = mg

T = 25 N

Hence, the tension in the wire is 25 N.

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What is the frequency of an ocean wave that is traveling at a speed of 45 m/s if it has a wavelength of 3 meters.

vaieri [72.5K]

Answer:

Frequency, f = 15 Hz

Explanation:

We have,

Speed of an ocean wave is 45 m/s

Wavelength of a wave is 3 m

It is required to find the frequency of an ocean wave.

Speed of a wave, $v=f\lambda$ , f = frequency of ocean wave

$f=\dfrac{v}{\lambda}\\\\f=\dfrac{45}{3}\\\\f=15\ Hz$

So, the frequency of an ocean wave is 15 Hz.

5 0

3 years ago

What is the answer for number 10

Finger [1]

The vertical component is = vsinx m/s

If you know the angle, substitute the value of x.

If you know the velocity at which it is moving, substitute it for v

Hope it helps :)

3 0

3 years ago

A hypothetical planet has a mass of one-half that of the earth and a radius of twice that of the earth.

Fed [463]

<h2>Option A is the correct answer.</h2>

Explanation:

Acceleration due to gravity

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

G = 6.67 × 10⁻¹¹ m² kg⁻¹ s⁻²

Let mass of earth be M and radius of earth be r.

We have

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

Now

A hypothetical planet has a mass of one-half that of the earth and a radius of twice that of the earth.

Mass of hypothetical planet, M' = M/2

Radius of hypothetical planet, r' = 2r

Substituting

$g'=\frac{GM'}{r'^2}\\\\g'=\frac{G\times \frac{M}{2}}{(2r)^2}\\\\g'=\frac{\frac{GM}{r^2}}{8}\\\\g'=\frac{g}{8}$

Option A is the correct answer.

6 0

3 years ago

An object rotates about a fixed axis, and the angular position of a reference line on the object is given by θ = 0.220e3t, where

OLga [1]

Answer: $a_{t}=3.96$

$a_{c}=0.8712$

Explanation:

Given

$\theta =0.220e^{3t}$

r=2cm

Now angular velocity is given by $\omega =\frac{\mathrm{d}\theta}{\mathrm{d}t}$

$\omega =0.66e^{3t}$

Now linear velocity(v) is given = $\omega r$

$v=1.32e^{3t}$

Now tangential component of acceleration is given by

$a_{t}=\frac{\mathrm{d}\vec{v}}{\mathrm{d}t}=3.96e^{3t}$

at t=0

$a_{t}=3.96cm/s^2$

radial component of acceleration is given by

$a_{c}=\omega ^{2}r$

$a_{c}=0.4356e^{6t}\times 2$

$a_{c}=0.8712e^{6t} cm/s^{2}$

at t=0

$a_c=0.8712 cm/s^{2}$

5 0

3 years ago

Please help!!!!!!!!!

antiseptic1488 [7]

Answer: 1 / 4.283 x 10¹¹

the earth model will be 64 cm away from the tennis ball

Explanation:

0.03 / 7 x 10⁸ = 1 / 4.283 x 10¹¹

(1.5 x 10¹⁰)( 1 / 4.283 x 10¹¹) = 0.64285

4 0

2 years ago