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2 years ago

5

Assume this piston has a maximum pressure rating of 2.5 pounds-force per square inch [psi]. The force applied to the cylindrical

piston is 100 newtons [N]. What is the diameter of the piston, in units of inches [in]

1 answer:

Leni [432]2 years ago

7 0

Explanation:

Always write what you know first about the problem

The surface of the piston that has the force applied is a circle

The area of a circle is pi x r^2

Pressure is Force / Area

Write this out

P = F / (pi x r^2)

solve for r

sqrt( F / (P x Pi)) = r

But dont forget that d = 2 x r

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Two blocks can collide in a one-dimensional collision. The block on the left hass a mass of 0.40 kg and is initially moving to t

Dmitrij [34]

Answer:

The velocity of the first block is 1.15m/s while of the second block 2.56m/s.

Explanation:

Momentum is only conserved in an isolated system, and because this problem requires us to find the value of the two variables, we need two equations; therefore, to conserved momentum the energy must be released in to the system only after the collision has occurred.

Therefore, from conservation of momentum

$m_1v_1= m_1v_{1f}+m_2v_{2f}$

$0.4(2.4)= 0.4v_{1f}+0.5v_{2f}$

$\boxed{\:0.96= 0.4v_{1f}+0.5v_{2f}}$

and from conservation of energy

$\frac{1}{2}m_1v_1^2+1.2 = \frac{1}{2} m_1v_{1f}^2+\frac{1}{2}m_2v_{2f}^2$

$m_1v_1^2+2.4 = m_1v_{1f}^2+m_2v_{2f}^2}$

$\boxed{\:3.804= 0.4v_{1f}^2+0.5v_{2f}^2.}$

Thus, we have two equations and two unknowns

$(1). \: 0.4v_{1f}+0.5v_{2f}=0.96$

${(2). \: 0.4v_{1f}^2+0.5v_{2f}^2 =3.804$

which has solutions

$(v_{1f}, v_{2f}) = (1.15,2.56)$

and

$(v_{1f}, v_{2f}) = (1.15,-2.56)$

Since the blocks cannot pass through each other, the 0.5kg block cannot have $v_{2f} = -2.56m/s$ (moves to the left) while the 0.4 kg block has $v_{1f} = 1.15m/s$ (moves to the right); therefore, we take the first solution for the velocities:

$(v_{1f}, v_{2f}) = (1.15,2.56)$ .

Thus , the velocity of the first block is 1.15m/s while of the second block 2.56m/s.

5 0

2 years ago

Read 2 more answers

Pls someone help with number 3

mars1129 [50]

Rocket thrust equation
= ( mass flow rate of fuel burnt ) X (Velocity of gas ejected ) + ( Exit Pressure - Outdoor Pressure ) X ( Area of exhaust )

In this case, we can assume the exit pressure = outdoor pressure and since area of exhaust is not given, it can be assumed to be negligible.

In this case, by Newton 3rd’s law,

Force exerted by gas on rocket
= Force exerted by rocket on gas
= (10kg/s) X (5 x 10^3 m/s)
= 5 x 10^4 N

7 0

2 years ago

How would the period of pendulum differ from an equivalent one on earth?<br>

Mars2501 [29]

Answer:

If you know that that free fall acceleration g on the Moon is about 6 times less than on the Earth, it gives you the answer: on the Moon the same pendulum will have a period about √6≈2.45 longer than on the Earth.

5 0

2 years ago

Two UFPD are patrolling the campus on foot. To cover more ground, they split up and begin walking in different directions. Offic

miss Akunina [59]

Answer:

0.256 hours

Explanation:

<u>Vectors in the plane </u>

We know Office A is walking at 5 mph directly south. Let $X_A$ be its distance. In t hours he has walked

$X_A=5t\ \text{miles}$

Office B is walking at 6 mph directly west. In t hours his distance is

$X_B=6t\ \text{miles}$

Since both directions are 90 degrees apart, the distance between them is the hypotenuse of a triangle which sides are the distances of each office

$D=\sqrt{X_A^2+X_B^2}$

$D=\sqrt{(5t)^2+(6t)^2}$

$D=\sqrt{61}t$

This distance is known to be 2 miles, so

$\sqrt{61}t=2$

$t =\frac{2}{\sqrt{61}}=0.256\ hours$

t is approximately 15 minutes

3 0

3 years ago

On a straight road (taken to be in the x direction) you drive for an hour at 60 km per hour, then quickly speed up to 120 km per

Luda [366]

Answer:

The average velocity is 180 km/hr

Explanation:

Given;

initial velocity, u = 60 km per hour

final velocity, v = 120 km per hour

initial time = 1 hour

final time = 2 hour

Initial position = 60 km/h x 1 hour = 60 km

final position = 120 km/h x 2 hour = 240 km

The average velocity is given by;

$V_{avg} = \frac{Final \ position\ - \ Initial \ position}{final \ time\ - \ initial \ time}\\\\V_{avg} = \frac{240km \ - \ 60km}{2hr\ - \ 1hr} \\\\V_{avg} = \frac{180 \ km}{1hr} \\\\V_{avg}= 180 \ km/hr$

Therefore, the average velocity is 180 km/hr

3 0

3 years ago