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3 years ago

5

a racing car traveling initially at 8.0 m/s accelerates uniformly at 10.0 m/s^2 for 5 seconds. How far does it travel in this ti

me interval?

1 answer:

Pepsi [2]3 years ago

7 0

The car travels a distance of

(8.0 m/s) (5 s) + 1/2 (10.0 m/s²) (5 s)² = 165 m

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For a given velocity of projection in a projectile motion, the maximum horizontal distance is possible only at ө = 45°. Substant

myrzilka [38]

First let us imagine the projectile launched at initial velocity V and at angle θ relative to the horizontal. (ignore wind resistance)

Vertical component y:

The initial vertical velocity is given as Vsinθ
The moment the projectile reaches the maximum height of h, the vertical velocity will be 0, therefore the time t taken to attain this maximum height is:

h = Vsinθ - gt
0 = Vsinθ - gt
t = (Vsinθ)/g

where g is acceleration due to gravity

Horizontal component x:
The initial horizontal velocity is given as Vcosθ. However unlike the vertical component, this horizontal velocity remains constant because this is unaffected by gravity. The time to travel the horizontal distance D is twice the value of t times the horizontal velocity.
D = Vcosθ*[(2Vsinθ)/g]
D = (2V²sinθ cosθ)/g
D = (V²sin2θ)/g

In order for D (horizontal distance) to be maximum, dD/dθ = 0
That is,

2V^2 cos2θ / g = 0
And since 2V^2/g must not be equal to zero, therefore cos(2θ) = 0
This is true when 2θ = π/2 or θ = π/4

Therefore it is now<span> shown that the maximum horizontal travelled is attained when the launch angle is π/4 radians, or 45°.</span>

6 0

3 years ago

Shrinking Loop. A circular loop of flexible iron wire has an initial circumference of 168 cm , but its circumference is decreasi

DedPeter [7]

Answer:

103.1 V

Explanation:

We are given that

Initial circumference=C=168 cm

$\frac{dC}{dt}=-15cm/s$

Magnetic field,B=0.9 T

We have to find the magnitude of the emf induced in the loop after exactly time 8 s has passed since the circumference of the loop started to decrease.

Magnetic flux= $\phi=BA=B(\pi r^2)$

Circumference,C= $2\pi r$

$r=\frac{C}{2\pi}$

$r=\frac{168}{2\pi}$ cm

$\frac{dr}{dt}=\frac{1}{2\pi}\frac{dC}{dt}=\frac{1}{2\pi}(-15)=-\frac{15}{2\pi} cm/s$

$\int dr=-\int \frac{15}{2\pi}dt$

$r=-\frac{15}{2\pi}t+C$

When t=0

$r=\frac{168}{2\pi}$

$\frac{168}{2\pi}=C$

$r=-\frac{15}{2\pi}t+\frac{168}{2\pi}$

E= $-\frac{d\phi}{dt}=-\frac{d(B\pi r^2)}{dt}=-2\pi rB\frac{dr}{dt}$

$E=-2\pi(-\frac{5}{2\pi}t+\frac{168}{2\pi})B\times -\frac{15}{2\pi}$

t=8 s

B=0.9

$E=2\pi\times \frac{15}{2\pi}\times 0.9(-\frac{15}{2\pi}(8)+\frac{168}{2\pi})$

$E=103.1 V$

6 0

3 years ago

Can anyone check if my answer is correct ?

ohaa [14]

I believe your answer is correct, because 8.7*10^-7 is equal to 0.00000085347.

Hope you do well!

4 0

3 years ago

Which is present when iodine changes from brown to blue or purple?

vladimir1956 [14]

If iodine is added to a starch solution, they react with each other and the iodine darkens to an almost pitch black.

however, if iodine is added to a solution containing no starch, it will show up only as an extremely pale brown. almost colorless and hardly visible.

when following the changes in some inorganic oxidation reduction reactions, iodine may be used as an indicator to follow the changes of iodide ion and iodine element. soluble starch solution is added. only iodine element in the presence of iodide ion will give the characteristic blue black color. neither iodine element alone nor iodide ions alone will give the color result.

hope this answer really helps your question :)

5 0

3 years ago

A stoplight with weight 100 N is suspended at the midpoint of a cable strung between two posts 200 m apart. The attach points fo

Tasya [4]

There are 3 forces acting on the stoplight:

• its weight <em>W</em>, with magnitude <em>W</em> = 100 N, pointing directly downward

• two tension forces <em>T</em>₁ and <em>T</em>₂ with equal magnitude <em>T</em>₁ = <em>T</em>₂ = <em>T</em> = 1000 N, both making an angle of <em>θ</em> with the horizontal, but one points left and the other points right

The stoplight is in equilibrium, so by Newton's second law, the net vertical force acting on it is 0, such that

∑ <em>F</em> = <em>T</em>₁ sin(<em>θ</em>) + <em>T</em>₂ sin(180° - <em>θ</em>) - <em>W</em> = 0

We have sin(180° - <em>θ</em>) = sin(<em>θ</em>) for all <em>θ</em>, so the above reduces to

2<em>T</em> sin(<em>θ</em>) = <em>W</em>

2 (1000 N) sin(<em>θ</em>) = 100 N

sin(<em>θ</em>) = 0.05

<em>θ</em> ≈ 2.87°

If <em>y</em> is the vertical distance between the stoplight and the ground, then

tan(<em>θ</em>) = (15 m - <em>y</em>) / (100 m)

Solve for <em>y</em> :

tan(2.87°) = (15 m - <em>y</em>) / (100 m)

<em>y</em> = 15 m - (100 m) tan(2.87°)

<em>y</em> ≈ 9.99 m

3 0

3 years ago