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Mila [183]
3 years ago
7

Help!

Physics
1 answer:
Artemon [7]3 years ago
5 0

Answer:

5 m/s² & 20 m

Explanation:

#1

Acceleration = Final velocity - Initial velocity/Time

Acceleration = 15 - 5/2

Acceleration = 10/2

Acceleration = 5 m/s²

#2

v² - u² = 2as

(15)² - (5)² = 2(5)(s)

225 - 25 = 10s

200 = 10s

200/10 = s

20 = s

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You lay a mirror flat on the floor with one edge against a wall and aim a laser at the mirror. The ray reflects from the mirror
joja [24]

Answer:

the angle of incidence θ is 45.56 º

Explanation:

Given data

strikes the mirror before wall x = 30.7 cm

reflected ray strikes the wall y =  30.1 cm

to find out

the angle of incidence θ

solution

let us consider ray is strike at angle  θ so after strike on surface ray strike to wall at angle 90 - θ

we will apply here right angle triangle rule that is

tan( 90 - θ) = y /x

tan( 90 - θ)  = 30.1 / 30.7

90 - θ = tan^-1 (30.1/30.7)

90 - θ = 44.4345

θ = 45.56 º

the angle of incidence θ is 45.56 º

4 0
4 years ago
To understand the distinction between mass and weight and to be able to calculate the weight of an object from its mass and Newt
Firdavs [7]

Answer: A.E.

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6 0
3 years ago
Read 2 more answers
Two resistors, R1 and R2, are
dlinn [17]

The reciprocal of the total resistance is equal to the sum of the reciprocals of the component resistances:

1/(120.7 Ω) = 1/<em>R₁</em> + 1/(221.0 Ω)

1/<em>R₁</em> = 1/(120.7 Ω) - 1/(221.0 Ω)

<em>R₁</em> = 1 / (1/(120.7 Ω) - 1/(221.0 Ω)) ≈ 265.9 Ω

3 0
3 years ago
Starting from zero, the electric current takes 2 seconds to reach half its maximum possible value in an RL circuit with a resist
Leno4ka [110]

Answer:

time=4s

Explanation:

we know that in a RL circuit with a resistance R, an inductance L and a battery of emf E, the current (i) will vary in following fashion

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Given that, at i(2)=\frac{imax}{2} =\frac{E}{2R}

⇒\frac{E}{2R}=\frac{E}{R}(1-e^\frac{-2}{\frac{L}{R}})

⇒\frac{1}{2}=1-e^\frac{-2}{\frac{L}{R}}

⇒\frac{1}{2}=e^\frac{-2}{\frac{L}{R}}

Applying logarithm on both sides,

⇒log(\frac{1}{2})=\frac{-2}{\frac{L}{R}}

⇒log(2)=\frac{2}{\frac{L}{R}}

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Now substitute i(t)=\frac{3}{4}imax=\frac{3E}{4R}

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Applying logarithm on both sides,

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now subs. \frac{L}{R}=\frac{2}{log2}

⇒t=log4\frac{2}{log2}

also log4=log2^{2}=2log2

⇒t=2log2\frac{2}{log2}

⇒t=4

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3 years ago
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Dahasolnce [82]
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