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melomori [17]
2 years ago
10

The normal boiling point of a certain liquid is , but when of urea () are dissolved in of the solution boils at instead. Use thi

s information to calculate the molal boiling point elevation constant of . Be sure your answer is rounded to the correct number of significiant digits.
Physics
1 answer:
Kruka [31]2 years ago
5 0

Answer:

100 Degrees is boiling point.

Explanation:

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Answer:

d.

Explanation:

the arrow is starts at 0,0 and ends at 2,2

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2 years ago
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5. The condition of the road surface affects total stopping distance.<br>A. true<br>B. false​
Tom [10]

Answer:

A true

Explanation:

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7 0
3 years ago
A boy exerts a 36-N horizontal force as he pulls a 52-N sled across a cement sidewalk at a constant speed. What is the coefficie
Vesna [10]

Answer:

0.69

Explanation:

Sum of the forces on the sled in the y direction:

∑F = ma

Fn - 52 N = 0

Fn = 52 N

Sum of the forces on the sled in the x direction:

∑F = ma

36 N - Fn μ = 0

Fn μ = 36 N

μ = 36 N / Fn

Substitute:

μ = 36 N / 52 N

μ = 0.69

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3 years ago
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A horizontal 745 N merry-go-round of radius
Arturiano [62]

Answer:

The kinetic energy of the merry-goround after 3.62 s is  544J

Explanation:

Given :

Weight w = 745 N

Radius r =  1.45 m

Force =  56.3 N

To Find:

The kinetic energy of the merry-go round after 3.62  = ?

Solution:

Step 1:  Finding the Mass of merry-go-round

m = \frac{ weight}{g}

m = \frac{745}{9.81 }

m = 76.02 kg

Step 2: Finding the Moment of Inertia of solid cylinder

Moment of Inertia of solid cylinder I =0.5 \times m \times r^2

Substituting the values

Moment of Inertia of solid cylinder I  

=>0.5 \times 76.02 \times (1.45)^2

=> 0.5 \times 76.02\times 2.1025

=> 79.91 kg.m^2

Step 3: Finding the Torque applied T

Torque applied T = F \times r

Substituting the values

T = 56.3  \times 1.45

T = 81.635 N.m

 Step 4: Finding the Angular acceleration

Angular acceleration ,\alpha  = \frac{Torque}{Inertia}

Substituting the values,

\alpha  = \frac{81.635}{79.91}

\alpha = 1.021 rad/s^2

 Step 4: Finding the Final angular velocity

Final angular velocity ,\omega = \alpha \times  t

Substituting the values,

\omega = 1.021 \times  3.62

\omega = 3.69 rad/s

Now KE (100% rotational) after 3.62s is:

KE = 0.5 \times I \times \omega^2

KE =0.5 \times 79.91 \times 3.69^2

KE = 544J

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