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MrRa [10]
3 years ago
15

Calculate the kinetic energy in joules of a 1500 kg automobile moving at 19 m/s

Physics
1 answer:
Svetlanka [38]3 years ago
5 0
Using
KE = ½mv² = ½×1500×19×19 = 270750 joules
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Two boxes on a horizontal plane with coefficient of friction µ are connected by a massless string. The left-hand box has mass m
pochemuha

Answer:

T = (3μmg - Fcosθ)/2

Explanation:

Since the boxes move to the right, the net force on box of mass 2m is

Fcosθ - 2μmg + T = ma (1)where Fcosθ = horizontal component of applied force, 2μmg = frictional force and T = tension in string and a = acceleration of box

The net force on the box with mass m is

μmg - T = ma (2) where μmg = frictional force, T = tension and a = acceleration of boxes.

Equating (1) and (2) we have

Fcosθ - 2μmg + T = μmg - T

collecting like terms

Fcosθ - 2μmg - μmg = -T - T

Fcosθ - 3μmg = -2T

(3μmg - Fcosθ)/2 = T

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3 years ago
Why did the model spacecraft go so much faster than expected on Wednesday?
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Answer:

The Wednesday test launch stored more potential energy, and launched the spacecraft at a faster speed because the stronger magnetic field closer to the magnet resulted in a greater increase in potential energy.

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2 years ago
The thinnest layer of the earth is the:
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crust

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A negative ion is (larger/smaller) than its parent atom
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3 years ago
A kayaker needs to paddle north across a 100-m-wide harbor. The tide is going out, creating a tidal current that flows to the ea
Savatey [412]

Answer:

41.81^{\circ}

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The tidal current flows to the east at 2.0 m/s and the speed of the kayaker is 3.0 m/s.

Let Vector \overrightarrow{OA} is the tidal current velocity as shown in the diagram.

In order to travel straight across the harbor, the vector addition of both the velocities (i.e the resultant velocity, \vec {R} must be in the north direction.

Let \overrightarrow{AB} is the speed of the kayaker having angle \theta measured north of east as shown in the figure.

For the resultant velocity in the north direction, the tail of the vector \overrightarrow {OA} and head of the vector \overrightarrow{AB} must lie on the north-south line.

Now, for this condition, from the triangle OAB

|\overrightarrow{AB}|\sin \theta=|\overrightarrow{OA}|

\Rightarrow \sin\theta=\frac{|\overrightarrow{OA}|}{|\overrightarrow{AB}|}=\frac 2 3

\Rightarrow \theta=\sin^{-1}\frac23

\Rightarrow \theta=41.81^{\circ}

Hence, the kayaker must paddle in the direction of 41.81^{\circ}  in the north of east direction.

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