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dalvyx [7]
3 years ago
6

A 3,000-N force gives an object an acceleration of 15 m/s^2. The mass of the object is

Physics
2 answers:
Alex787 [66]3 years ago
8 0
The correct answer is m= 200 kg

Explanation: The solution for solving the mass kg the object using Newton’s Law of Accerelation is F= ma.
Bingel [31]3 years ago
7 0
<h2>Answer:</h2>

<em>Hello, </em>

<h3><u>QUESTION)</u></h3>

According to the second Newton's Law,

<em>✔ We have : F = m x a ⇔ m = F/a </em>

  • m = 3000/15
  • m = 200 kg

The mass of the object is therefore 200 kg.

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Given a particle that has the velocity v(t) = 3 cos(mt) = 3 cos (0.5t) meters, a. Find the acceleration at 3 seconds. b. Find th
dalvyx [7]

Answer:

Explanation:

a )  V = 3 cos(0.5t)

differentiating with respect to t

dv /dt = -3 x .5 sin0.5t

= -1.5 sin0.5t.

acceleration = - 1.5 sin 0.5t

when t = 3 s

acceleration = - 1.5 sin 1.5

= - 1.496 ms⁻²

v = 3 cos.5t

b )  dx/dt = 3 cos 0.5 t

dx = 3 cos 0.5 t dt

integrating on both sides

x = 3 sin .5t / .5

x = 6 sin0.5t

At t = 2 s

x = 6 sin 1

x = 5.05 m

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3 years ago
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3 years ago
In some circumstances, it is useful to look at the linear velocity of a point on the blade. The linear velocity of a point in un
mihalych1998 [28]

Answer:

v=wr

Explanation:

<u>Tangent and Angular Velocities</u>

In the uniform circular motion, an object describes the same angles in the same times. If \theta is the angle formed by the trajectory of the object in a time t, then its angular velocity is

\displaystyle w=\frac{\theta}{t}

if \theta is expressed in radians and t in seconds the units of w is rad/s. If the circular motion is uniform, the object forms an angle 2\theta in 2t, or 3\theta in 3t, etc. Thus the angular velocity is constant.

The magnitude of the tangent or linear velocity is computed as the ratio between the arc length and the time taken to travel that distance:

\displaystyle v=\frac{\theta r}{t}

Replacing the formula for w, we have

\boxed{ v=wr}

4 0
3 years ago
The near point of an eye is 48.5 cm. A corrective lens is to be used to allow this eye to focus clearly on objects at the distan
Masteriza [31]

Answer:

The focal length of the lens should be -51.5 cm (a concave lens).

Explanation:

The purpose of the lens is to make objects at 48.5 cm appear at the healthy near point. The healthy near point is 25.0 cm.

We use the lens formula

\dfrac{1}{f} = \dfrac{1}{u} + \dfrac{1}{v}

where <em>f</em> = focal length, <em>u</em> = object distance and <em>v</em> = image distance.

In this case, <em>u</em> = 48.5 cm and <em>v</em> = -25.0 cm.

<em>v</em> is negative because the image is virtual an not real. (Here, we are using the real-is-positive sign convention)

\dfrac{1}{f} = \dfrac{1}{48.5} + \dfrac{1}{-25.0} = -\dfrac{23.5}{1212.5}

f = -51.5

The negative sign indicates the lens is concave.

3 0
2 years ago
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