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muminat
3 years ago
15

A 6.50 × 10–3 m2 piston compresses gas in a cylinder with a surface area of 9.75 × 10–2 m2. What is the force on the cylinder wa

lls if 50.0 N are applied to the piston?
Physics
1 answer:
vodomira [7]3 years ago
8 0
Yes it is 2cm and 1 quarter
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So I'm struggling with rearranging kinematic formulas. Does anyone have any steps or something to help.
bekas [8.4K]

Rearranging formulas is all about simple algebra rules. Just like when solving for x in an equation, you're just isolating whichever variable you want. I'll work this one out for you and hopefully it'll help, but if you need more explanation, then feel free to comment!

D = ViT + 0.5at²   Subtract ViT from both sides

D - ViT = 0.5at²    Divide both sides by 0.5t²

\frac{D - ViT}{0.5t^{2} } = \frac{0.5at^{2} }{0.5t^{2} }    I wrote this step out a little more to show how your fraction will cancel

\frac{D - ViT}{0.5t^{2} }= a    I like to flip these around so the single variable is on the right

a = \frac{D - ViT}{0.5t^{2} }

7 0
3 years ago
The index of refraction of silicate flint glass for red light is 1.620 and for violet light is 1.660 . A beam of white light in
Sonja [21]

Answer:

10.16 degrees

Explanation:

Apply Snells Law for both wavelenghts

\(n_{1}sin\theta_{1} = n_{2}sin\theta_{2}\)

For red

(1.620)(sin 25.5) = (1)(sin r)

For red, the angle is 35.45degrees

For violet

(1.660)(sin 25.5) = (1)(sin v)

For violet, the angle is 45.6 degrees

The difference is 45.6- 35.45 = 10.16 degrees

3 0
3 years ago
a 1-kg discus is thrown with a velocity of 19 m/s at an angle of 35 degrees from the vertical direction. calculate the vertical
nlexa [21]

Answer:

Vx =  10.9 m/s ,  Vy = 15.6 m/s

Explanation:

Given velocity V= 19 m/s

the angle 35 ° is taken from Y-axis so the angle with x-axis will be 90°-35° = 55°

θ = 55°

to Find Vx = ? and Vy= ?

Vx = V cos θ

Vx = 19 m/s  × cos 55°

Vx =  10.9 m/s

Vx = V sin θ

Vy = 19 m/s  × sin 55°

Vy = 15.6 m/s

6 0
3 years ago
Read 2 more answers
A 20.0-N weight slides down a rough inclined plane which makes an angle of 30 degree with the horizontal. The weight starts from
Ulleksa [173]

Answer:

1270.64\ \text{J}

Explanation:

m = Mass of object = \dfrac{mg}{g}

mg = Weight of object = 20 N

g = Acceleration due to gravity = 9.81\ \text{m/s}^2

v = Final velocity = 15 m/s

u = Initial velocity = 0

d = Distance moved by the object = 150 m

\theta = Angle of slope = 30^{\circ}

f = Force of friction

fd = Work done against friction

The force balance of the system is

\dfrac{1}{2}m(v^2-u^2)=(mg\sin\theta-f)d\\\Rightarrow \dfrac{1}{2}mv^2=mg\sin\theta d-fd\\\Rightarrow fd=mg\sin\theta d-\dfrac{1}{2}mv^2\\\Rightarrow fd=20\times \sin 30^{\circ}\times 150-\dfrac{1}{2}\times \dfrac{20}{9.81}\times 15^2\\\Rightarrow fd=1270.64\ \text{J}

The work done against friction is 1270.64\ \text{J}.

8 0
3 years ago
A string that passes over a pulley has a 0.341 kg mass attached to one end and a 0.625 kg mass attached to the other end. The pu
dalvyx [7]

Answer:

The frictional torque is \tau  = 0.2505 \ N \cdot m

Explanation:

From the question we are told that

   The mass attached to one end the string is m_1 =  0.341 \ kg

   The mass attached to the other end of the string is  m_2 =  0.625 \ kg

    The radius of the disk is  r = 9.00 \ cm  = 0.09 \ m

At equilibrium the tension on the string due to the first mass is mathematically represented as

      T_1 =  m_1 *  g

substituting values

      T_1 =  0.341 * 9.8

      T_1 =  3.342 \ N

At equilibrium the tension on the string due to the  mass is mathematically represented as

      T_2 =  m_2 *  g

     T_2 = 0.625 * 9.8

      T_2 = 6.125 \ N

The  frictional torque that must be exerted is mathematically represented as

      \tau  =  (T_2 * r ) - (T_1 * r )

substituting values  

     \tau  =  ( 6.125 * 0.09 ) - (3.342  * 0.09 )

     \tau  = 0.2505 \ N \cdot m

5 0
3 years ago
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