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3 years ago

A narrow region between two air masses of different densities is a

Physics

1 answer:

Sav [38]3 years ago

5 0

A front is a narrow region between two air masses of different densities.

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What must the driver do when approaching an intersection and seeing the traffic light turn from green to yellow?

NISA [10]

The person should start to slow down but if close enough or in the intersection go threw. Otherwise come to a complete stop until the light turns green again

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3 years ago

is it possible to breed a black haired long tailed retriever? If so, what percentage is possible? PLEASE answer the corrects ans

Pachacha [2.7K]

Answer: it can be considered a genetic mutation with a history of a Golden Retriever in their blood but it is very rare. and there our some black retrievers you can buy too. i hope i helped

Explanation:

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2 years ago

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Does displacement = Δx?

Volgvan

Typically no. Displacement can be in multiple directions as a vector. of something is traveling only along x, then it would be true though this is usually not the case.

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3 years ago

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The instantaneous speed of a particle moving along one straight line is v(t) = ate−6t, where the speed v is measured in meters p

beks73 [17]

Answer:

v_max = (1/6)e^-1 a

Explanation:

You have the following equation for the instantaneous speed of a particle:

$v(t)=ate^{-6t}$ (1)

To find the expression for the maximum speed in terms of the acceleration "a", you first derivative v(t) respect to time t:

$\frac{dv(t)}{dt}=\frac{d}{dt}[ate^{-6t}]=a[(1)e^{-6t}+t(e^{-6t}(-6))]$ (2)

where you have use the derivative of a product.

Next, you equal the expression (2) to zero in order to calculate t:

$a[(1)e^{-6t}-6te^{-6t}]=0\\\\1-6t=0\\\\t=\frac{1}{6}$

For t = 1/6 you obtain the maximum speed.

Then, you replace that value of t in the expression (1):

$v_{max}=a(\frac{1}{6})e^{-6(\frac{1}{6})}=\frac{e^{-1}}{6}a$

hence, the maximum speed is v_max = ((1/6)e^-1)a

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3 years ago

Rotational dynamics about a fixed axis: A person pushes on a small doorknob with a force of 5.00 N perpendicular to the surface

FrozenT [24]

Answer:

I = 2 kgm^2

Explanation:

In order to calculate the moment of inertia of the door, about the hinges, you use the following formula:

$\tau=I\alpha$ (1)

I: moment of inertia of the door

α: angular acceleration of the door = 2.00 rad/s^2

τ: torque exerted on the door

You can calculate the torque by using the information about the Force exerted on the door, and the distance to the hinges. You use the following formula:

$\tau=Fd$ (2)

F: force = 5.00 N

d: distance to the hinges = 0.800 m

You replace the equation (2) into the equation (1), and you solve for α:

$Fd=I\alpha\\\\I=\frac{Fd}{\alpha}$

Finally, you replace the values of all parameters in the previous equation for I:

$I=\frac{(5.00N)(0.800m)}{2.00rad/s^2}=2kgm^2$

The moment of inertia of the door around the hinges is 2 kgm^2

3 0

4 years ago