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3 years ago

What is 3,278,000 as scientific notation?

Physics

1 answer:

jonny [76]3 years ago

4 0

3.278*10^6 I think. Sorry if it’s wrong.

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A motorcycle is following a car that is traveling at constant speed on a straight highway. Initially, the car and the motorcycle

MAVERICK [17]

Answer:

t₂ = 3.89 s

Explanation:

given,

speed of car = 23 m/s

speed of motorcycle = 23 m/s

after time of 4 s distance between them is equal to = 53 m

motorcycle accelerates at = 7 m/s

time taken to catch up with car = ?

let t₂ be the time in which motorcycle catches car.

distance traveled by car in t₂ s

d = 23 t₂ + 53

distance traveled by motorcycle

using equation of motion

$s = ut + \dfrac{1}{2}at^2$

$s = 23 t_2 + \dfrac{1}{2}\times 7 \times t_2^2$

now, equating both the distances

$23t_2 + 53= 23 t_2 + \dfrac{1}{2}\times 7 \times t_2^2$

$t_2^2 = 15.143$

t₂ = 3.89 s

time taken by the motorcycle to catch the car is equal to 3.89 s

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3 years ago

The equation for the chemical reaction shown is not balanced. What number should replace the question mark to balance this equat

Fudgin [204]

The equation is balanced as it is.

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3 years ago

A constant force of 45 N directed at angle θ to the horizontal pulls a crate of weight 100 N from one end of a room to another a

Rasek [7]

Answer:

$W=173.48J$

Explanation:

information we know:

Total force: $F=45N$

Weight: $w=100N$

distance: $4m$

vertical component of the force: $F_{y}=12N$

-------------

In this case we need the formulas to calculate the components of the force (because to calculate the work we need the horizontal component of the force).

horizontal component: $F_{x}=Fcos\theta$

vertical component: $F_{y}=Fsen\theta$

but from the given information we know that $F_{y}=12N$

so, equation these two $F_{y}=Fsen\theta$ and $F_{y}=12N$

$Fsen\theta =12N$

and we know the force $F=45N$ , thus:

$45sen\theta=12$

now we clear for $\theta$

$sen\theta =12/45\\\theta=sin^{-1}(12/45)\\\theta =15.466$

the angle to the horizontal is 15.466°, with this information we can calculate the horizontal component of the force:

$F_{x}=Fcos\theta$

$F_{x}=45cos(15.466)\\F_{x}=43.37N$

whith this horizontal component we calculate the work to move the crate a distance of 4 m:

$W=F_{x}*D\\W=(43.37N)(4m)\\W=173.48J$

the work done is W=173.48J

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That's wave 'diffraction'.

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stepan [7]

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