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3 years ago

10

An object moving at a constant velocity will always have a what

2 answers:

NNADVOKAT [17]3 years ago

8 0

<span>An object moving at a constant velocity will always have a zero acceleration. </span>

telo118 [61]3 years ago

4 0

<em><u>Answer:</u></em>

zero acceleration

<em><u>Explanation:</u></em>

<u>Acceleration</u> is defined as the rate change of velocity with respect to time

<u>This means that:</u>

$acceleration = \frac{delta_v_e_l_o_c_i_t_y}{delta_t_i_m_e}$

Now, since the velocity is given as a constant, this means that the value of the velocity does not change which means that the change of velocity is zero.

<u>Substitute in the formula:</u>

$acceleration = \frac{0}{delta_t_i_m_e} = 0$

<u>Therefore,</u> for constant velocity, the acceleration is zero.

Hope this helps :)

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Refer to the diagram shown below.

The force, F, is applied at 5 cm from the elbow.

For dynamic equilibrium, the sum of moments about the elbow is zero.
Take moments about the elbow.
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3 years ago

When antimatter interacts with an equal mass of ordinary matter, both matter and antimatter are converted completely into energy

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Answer:

v = 5.88 10⁷ m / s

Explanation:

For this exercise we use the relation

E = m c²

also indicate that all energy is converted into kinetic energy

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3 years ago

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3 years ago

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3 years ago

An airplane is traveling at 250 m/s in level flight. If the airplane is to make a change in direction, it must travel is a horiz

Nonamiya [84]

Answer:

The radius of curvature of the curved path of the airplane is 23784.356 meters (23.784 kilometers).

Explanation:

We assume that airplane can be represented as a particle. The free body diagram of the vehicle is presented below as attachment, whose variables are:

$W$ - Weight of the airplane, measured in newtons.

$F$ - Lift, measured in newtons.

$\theta$ - Banking angle, measured in sexagesimal degrees.

The equations of equilibrium associated with the airplane are, respectively:

$\Sigma F_{r} = F\cdot \sin \theta = m\cdot \frac{v^{2}}{R}$ (Eq. 1)

$\Sigma F_{z} = F\cdot \cos \theta - W = 0$ (Eq. 2)

From (Eq. 2):

$F = \frac{W}{\cos \theta}$

In (Eq. 1):

$W\cdot \tan \theta = m\cdot \frac{v^{2}}{R}$

By using the definition of weight, we eliminate the mass of the airplane:

$g\cdot \tan \theta = \frac{v^{2}}{R}$

Where:

$g$ - Gravitational acceleration, measured in meters per square second.

$v$ - Speed, measured in meters per second.

$R$ - Radius of curvature, measured in meters.

Lastly, we clear the radius of curvature with the expression:

$R = \frac{v^{2}}{g\cdot \tan \theta}$

If we know that $v = 250\,\frac{m}{s}$ , $g = 9.807\,\frac{m}{s^{2}}$ and $\theta = 15^{\circ}$ , the radius of curvature is:

$R = \frac{\left(250\,\frac{m}{s} \right)^{2}}{\left(9.807\,\frac{m}{s^{2}} \right)\cdot \tan 15^{\circ}}$

$R = 23784.356\,m$

The radius of curvature of the curved path of the airplane is 23784.356 meters (23.784 kilometers).

6 0

4 years ago