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3 years ago

____ is rate of change of position.

2 answers:

4 0

I think the correct answer from the choices listed above is option B. Velocity is rate of change of position. Velocity is a physical vector quantity which means it has both direction and magnitude. The scalar value is called speed. Hope this answers the question.

Korvikt [17]3 years ago

3 0

The RATE of change of position is speed.

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Sleep is a difficult problem for dolphins because dolphins __________. A. need more sleep than their predators B. must be awake

Alex17521 [72]

Need more sleep than their predators

5 0

2 years ago

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En el MCU, la aceleración es: a) constante b) Varía en módulo, direccion, y sentido c) constante solo en el módulo d) constante

JulijaS [17]

If you put it into English in the comments I would be more then happy to help you! Thank you!

7 0

3 years ago

An airliner arrives at the terminal, and its engines are shut off. The rotor of one of the engines has an initial clockwise angu

Ilia_Sergeevich [38]

(a) 1200 rad/s

The angular acceleration of the rotor is given by:

$\alpha = \frac{\omega_f - \omega_i}{t}$

where we have

$\alpha = -80.0 rad/s^2$ is the angular acceleration (negative since the rotor is slowing down)

$\omega_f$ is the final angular speed

$\omega_i = 2000 rad/s$ is the initial angular speed

t = 10.0 s is the time interval

Solving for $\omega_f$ , we find the final angular speed after 10.0 s:

$\omega_f = \omega_i + \alpha t = 2000 rad/s + (-80.0 rad/s^2)(10.0 s)=1200 rad/s$

(b) 25 s

We can calculate the time needed for the rotor to come to rest, by using again the same formula:

$\alpha = \frac{\omega_f - \omega_i}{t}$

If we re-arrange it for t, we get:

$t = \frac{\omega_f - \omega_i}{\alpha}$

where here we have

$\omega_i = 2000 rad/s$ is the initial angular speed

$\omega_f=0$ is the final angular speed

$\alpha = -80.0 rad/s^2$ is the angular acceleration

Solving the equation,

$t=\frac{0-2000 rad/s}{-80.0 rad/s^2}=25 s$

6 0

3 years ago

....................

muminat

Answer:

-28.014 is the answer

3 0

2 years ago

Read 2 more answers

A body has masses of 0.013kg and 0.012kg in oil and water respectively, if the relative density of oil is 0.875, calculate the m

konstantin123 [22]

Answer:

the mass of the body is 0.02 kg.

Explanation:

Given;

relative density of the oil, $\gamma _0$ = 0.875

mass of the object in oil, $M_o$ = 0.013 kg

mass of the object in water, $M_w$ = 0.012 kg

let the mass of the object in air = $M_a$

weight of the oil, $W_0 = M_a - 0.013$

weight of the water, $W_w = M_a - 0.012$

The relative density of the oil is given as;

$\gamma_0 = \frac{density \ of \ oil }{density \ of \ water} = \frac{W_0}{W_w} = \frac{M_a -0.013}{M_a -0.012} \\\\0.875 = \frac{M_a -0.013}{M_a -0.012}\\\\0.875(M_a - 0.012) = M_a - 0.013\\\\0.875M_a - 0.0105 = M_a -0.013\\\\0.875M_a - M_a = 0.0105 - 0.013\\\\-0.125 M_a = -0.0025\\\\M_a = \frac{0.0025}{0.125} \\\\M_a = 0.02 \ kg$

Therefore, the mass of the body is 0.02 kg.

6 0

2 years ago