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2 years ago

A 10.0-kg animal is running at a rate of 3.0 m/s. What is the kinetic energy of the animal?

1 answer:

4 0

Kinetic energy is calculated by using formula:

Ek = 1/2*m*v^2

Now we replace values of mass and speed in this formula.

Ek = 1/2*10*3^2 = 45.0J

When you have a formula and all variable values are given in the text of question just replace variable values in equation and find the final value.

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High school???
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3 years ago

A particle of ink in a ink-jet printer carries a charge of -8x 10^-13 C and is deflected onto paper force of 3.2x10^-4. Find the

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2 years ago

In June 21 are there more hours of daylight in San Francisco or the equator explain

jonny [76]

Answer:

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4 0

3 years ago

Help please!

Talja [164]

The correct expression for the maximum speed of the object during its motion is $\frac{FT}{m}$ .

<h3>Maximum speed of the object</h3>

The maximum speed of the object is determined using the following formulas;

v(max) = Aω

where;

A is the amplitude of the motion
ω is angular speed

The maximum speed of the object can also be obtained from the maximum net force on the object,

F = ma

where;

F is the maximum net force
a is the acceleration
m is mass of the object

F = m(v/t)

mv = Ft

v = Ft/m

Thus, the correct expression for the maximum speed of the object during its motion is $\frac{FT}{m}$ .

Learn more about maximum speed here: brainly.com/question/4931057

3 0

1 year ago

A radar station detects an airplane approaching directly from the east. At first observation, the range to the plane is 375.0 m

Gekata [30.6K]

Answer:

819.78 m

Explanation:

<u>Given:</u>

OA = range of initial position of the airplane from the point of observation = 375 m
OB = range of the final position of the airplane from the point of observation = 797 m
$\theta$ = angle of the initial position vector from the observation point = $43^\circ$
$\alpha$ = angle of the final position vector from the observation point = $123^\circ$
$\vec{AB}$ = displacement vector from initial position to the final position

A diagram has been attached with the solution in order to clearly show the position of the plane.

$\vec{OA} = OA\cos \theta \hat{i}+OA \sin \theta \hat{j}\\\Rightarrow \vec{OA} = 375\ m\cos 43^\circ \hat{i}+375\ m\sin 43^\circ \hat{j}\\\Rightarrow \vec{OA} = (274.26\ \hat{i}+255.75\ \hat{j})\ m\\\vec{OB} = OB\cos \alpha \hat{i}+OB \sin \alpha \hat{j}\\\Rightarrow \vec{OB} = 797\ m\cos 123^\circ \hat{i}+797\ m\sin 123^\circ \hat{j}\\\Rightarrow \vec{OB} = (-434.08\ \hat{i}+668.42\ \hat{j})\ m$

Displacement vector of the airplane will be the shortest line joining the initial position of the airplane to the final position of the airplane which is given by:

$\vec{AB}=\vec{OB}-\vec{OA}\\\Rightarrow \vec{AB} = (-434.08\ \hat{i}+668.42\ \hat{j})\ m-(274.26\ \hat{i}+255.75\ \hat{j})\ m\\\Rightarrow \vec{AB} = (-708.34\ \hat{i}+412.67\ \hat{j})\ m$

The magnitude of the displacement vector = $\sqrt{(-708.34)^2+(412.67)^2}\ m = 819.78\ m$

Hence, the magnitude of the displacement of the plane is 819.67 m during the period of observation.

8 0

2 years ago