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Softa [21]
3 years ago
10

A red shirt appears red because?

Physics
2 answers:
Shalnov [3]3 years ago
8 0
<span>a. the shirt reflects red light</span>
Serggg [28]3 years ago
6 0
I'm pretty sure it's D

I hope you get it right:)
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The surface tension of water was determined in a laboratory by using the drop weight method. 100 drops were released from a bure
Lapatulllka [165]

Answer:

γ = 0.06563 N / m

9.78% difference

Explanation:

Solution:-

- Surface tension is the ability of any fluid to resist any external force which causes a decreases in surface area of the impact area due to inward compressive forces. These compressive forces occur due to cohesive nature of the fluid molecules.

- Mathematically, surface tension ( γ ) is defined as the force felt per unit length by the fluid.

                           γ = F / L

Where,

              F: Force imparted

              L: The length over which force is felt

- We are given the mass ( M ) of ( n = 100 ) water droplets to e 3.78 g. The mass of a single droplet ( m ) can be evaluated as follows:

                         m = M / n

                         m = 3.78 / 100

                        m = 0.0378 g        

- The force ( F ) imparted by a single drop of water from the burette can be determined from the force balance on a single droplet. Assuming the distance over which the drop falls is negligible and resistive forces are negligible. Then the only force acting on the droplet is due to gravity:

                        F = m*g

                        F = 0.0378*9.81*10^-3

                        F = 0.000370818 N      

- The length over which the force is felt can be magnified into a circular area with diameter equal to that of a single droplet ( d ). The circumferential length ( L ) of the droplet would be as follows:

                        L = π*d

                        L = π*( 0.0018 )

                        L = 0.00565 m

- Then the surface tension would be:

                        γ = F / L

                        γ = 0.000370818 / 0.00565

                        γ = 0.06563 N / m

- The tabulated value of water's surface tension is given as follows:

                       γa = 0.07275 N/m

- We will determine the percentage difference between the value evaluated  and tabulated value as follows:

                     p.diff = \frac{gamma_a - gamma}{gamma_a} * 100\\\\p.diff = \frac{0.07275- 0.06563}{0.07275} * 100 \\\\p.diff = 9.78 %

- The %difference between is within the allowable practical limits of 10%. Hence, the evaluated value ( γ = 0.06563 N / m ) can be accepted with 9.78% error.

8 0
3 years ago
Volcano can form along all of the following except
iragen [17]
They cannot form on a Transform Boundary

8 0
3 years ago
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What goes through evaporation
Olegator [25]
Water goes through evaporation.
6 0
3 years ago
Read 2 more answers
What is the Sl unit for momentum?<br> O kg• m<br> O kg<br> O kg• m<br> O kg<br> 2<br> m
Andrej [43]

Explanation:

<h2><em>The </em><em>S. </em><em>I. </em><em> </em><em>unit </em><em>of </em><em>momentum </em><em>is </em><em>Kg. </em><em>m/</em><em>s</em><em>e</em><em>c</em></h2>

<em>hope </em><em>it </em><em>helps </em><em>you </em>

8 0
3 years ago
What is the magnitude of the torque about his shoulder if he holds his arm straight out to his side, parallel to the floor
const2013 [10]

Complete Question

An athlete at the gym holds a 3.0 kg steel ball in his hand. His arm is 70 cm long and has a mass of 4.0 kg. Assume, a bit unrealistically, that the athlete's arm is uniform.

What is the magnitude of the torque about his shoulder if he holds his arm straight out to his side, parallel to the floor? Include the torque due to the steel ball, as well as the torque due to the arm's weight.

Answer:

The torque is  \tau = 34.3 \  N\cdot m

Explanation:

From the question we are told that

   The mass of the steel ball is  m  =  3.0 \  kg

    The length of arm is  l =  70 \ cm  = 0.7 \  m

    The mass of the arm is m_a  = 4.0 \  kg

Given that the arm of the athlete is uniform them the distance from the shoulder to the center of gravity of the arm is mathematically represented as

       r = \frac{l}{2}

=>    r = \frac{ 0.7}{2}  

=>    r = 0.35 \ m  

Generally the magnitude of torque about the athlete shoulder is mathematically represented as

      \tau =  m_a * g * r  + m * g *  L

=>    \tau =  4 * 9.8 * 0.35 + 3 * 9.8 *  0.70

=>    \tau = 34.3 \  N\cdot m

5 0
2 years ago
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