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Sedaia [141]
2 years ago
9

An object moving at a constant velocity travels 274 m in 23 s. what is its velocity?

Physics
2 answers:
svetlana [45]2 years ago
6 0
V= 274 meters / 23 sec

V= 11.91 meters per sec
olga55 [171]2 years ago
4 0

Answer:

Velocity, v = 11.91 m/s

Explanation:

It is given that,

Distance covered by the object, d = 274 m

Time taken, t = 23 s

We need to find the velocity of the object. It is given by distance covered divided by total time taken i.e.

v=\dfrac{d}{t}

v=\dfrac{274\ m}{23\ s}

v = 11.91 m/s

So, the velocity of the object is 11.91 m/s. Hence, this is the required solution.

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exis [7]

Answer:

113.7

Explanation:

maximum distance (s) = 8.9 km

reference intensity (I0) = 1 x 10^{-12} W/m^{2}

power of a juvenile howler monkey (p) = 63 x 10^{-6} W

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intensity (I) = power/area

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area = 4πr^{2} =  4π x 210^{2} = 554,176.9 m^{2}

intensity (I) = \frac{63 x 10^{-6} }{554,176.9} = 113.7 x 10^{-12}

therefore the desired ratio I/I0 = \frac{113.7 x 10^{-12}}{1 x 10^{-12}} = 113.7

7 0
2 years ago
Determine the number of unpaired electrons in the octahedral coordination complex [fex6]3–, where x = any halide.
juin [17]
There will be four unpaired electrons
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3 0
2 years ago
A carmaker has designed a car that can reach a maximum acceleration of 12 meters/second2. The car’s mass is 1,515 kilograms. Ass
Vlada [557]
1) 15 / 12 = 1.25 ratio
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6 0
2 years ago
Consider the vector b⃗ with magnitude 4.00 m at an angle 23.5∘ north of east. what is the x component bx of this vector? express
BlackZzzverrR [31]
Decomposing the vector b on the x-axis and the y-axis, we get a rectangle triangle where the two sides are bx (x-axis) and by (y-axis), and b is the hypothenuse.
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6 0
2 years ago
If a current of 2.4 a is flowing in a cylindrical wire of diameter 2.0 mm, what is the average current density in this wire?
Gnom [1K]

The average current density in the wire is given by:

J=\frac{I}{A}

where I is the current intensity and A is the cross-sectional area of the wire.


The cross-sectional area of the wire is given by:

A=\pi r^2

where r is the radius of the wire. In this problem, r=\frac{d}{2}=\frac{2.0 mm}{2}=1.0 mm=0.001 m, so the cross-sectional area is

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8 0
2 years ago
Read 2 more answers
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