If the force acting on an object is multiplied by ____

Which of the following statements correctly describes the position of the intake and exhaust valves during most of the power sta

I am Lyosha [343]

Both valves are closed during the power stroke.

While the fuel is burning in the cylinder, you want
all the force of the expanding gases to push the
piston down ... you don't want any of the gases
or their pressure escaping.

If either of the valves was open, even just a crack,
then part of the gases would go blooey out the valve,
and some pressure would be lost that's supposed to be
pushing the piston.

5 0

3 years ago

Read 2 more answers

A submarine is 20m below the surface of the sea. the pressure due to the water at this depth is P. on another day, the submarine

satela [25.4K]

Answer:

1.7p

Explanation:

8 0

3 years ago

On a roller coaster, riders can experience a force of up to 4 g. What is the maximum acceleration of the roller coaster?

yKpoI14uk [10]

m/s^2 is 39.2266

is the answer If thats what you needed

7 0

3 years ago

Find the value of F1 + F2 + F3.

Dovator [93]

Answer:

F = 0.78[N]

Explanation:

The given values correspond to forces, we must remember or take into account that the forces are vector quantities, that is, they have magnitude and direction. Since we have two X-Y coordinate axes (two-dimensional), we are going to decompose each of the forces into the X & y components.

For F₁

$F_{y}=2[N]$

For F₂

$F_{x}=2*cos(60)\\F_{x}=1[N]\\F_{y}=-2*sin(60)\\F_{y}=-1.73[N]$

For F₃

$F_{x}=-1*sin(60)\\F_{x}=-0.866[N]\\F_{y}=1*cos(60)\\F_{y}=0.5 [N]$

Now we can sum each one of the forces in the given axes:

$F_{x}=1-0.866=0.134[N]\\F_{y}=2-1.73+0.5\\F_{y}=0.77[N]$

Now using the Pythagorean theorem we can find the total force.

$F=\sqrt{(0.134)^{2} +(0.77)^{2}}\\F= 0.78[N]$

8 0

3 years ago

Rank the following objects by their accelerations down an incline (assume each object rolls without slipping) from least to grea

Alexxx [7]

Answer:

acceleration are

hollow cylinder < hollow sphere < solid cylinder < solid sphere

Explanation:

To answer this question, let's analyze the problem. Let's use conservation of energy

Starting point. Highest point

Em₀ = U = m g h

Final point. To get off the ramp

Em_f = K = ½ mv² + ½ I w²

notice that we include the kinetic energy of translation and rotation

energy is conserved

Em₀ = Em_f

mgh = ½ m v² +1/2 I w²

angular and linear velocity are related

v = w r

w = v / r

we substitute

mg h = ½ v² (m + I / r²)

v² = 2 gh $\frac{m}{m+ \frac{I}{r^2} }$

v² = 2gh $\frac{1}{1 + \frac{I}{m r^2} }$

this is the velocity at the bottom of the plane ,, indicate that it stops from rest, so we can use the kinematics relationship to find the acceleration in the axis ax (parallel to the plane)

v² = v₀² + 2 a L

where L is the length of the plane

v² = 2 a L

a = v² / 2L

we substitute

a = $g \ \frac{h}{L} \ \frac{1}{1+ \frac{I}{m r^2 } }$

let's use trigonometry

sin θ = h / L

we substitute

a = g sin θ \ \frac{h}{L} \ \frac{1}{1+ \frac{I}{m r^2 } }

the moment of inertia of each object is tabulated, let's find the acceleration of each object

a) Hollow cylinder

I = m r²

we look for the acerleracion

a₁ = g sin θ $\frac{1}{1 + \frac{mr^2 }{m r^2 } }$ 1/1 + mr² / mr² =

a₁ = g sin θ ½

b) solid cylinder

I = ½ m r²

a₂ = g sin θ $\frac{1}{1 + \frac{1}{2} \frac{mr^2}{mr^2} }$ = g sin θ $\frac{1}{1+ \frac{1}{2} }$

a₂ = g sin θ ⅔

c) hollow sphere

I = 2/3 m r²

a₃ = g sin θ $\frac{1}{1 + \frac{2}{3} }$

a₃ = g sin θ $\frac{3}{5}$

d) solid sphere

I = 2/5 m r²

a₄ = g sin θ $\frac{1 }{1 + \frac{2}{5} }$

a₄ = g sin θ $\frac{5}{7}$

We already have all the accelerations, to facilitate the comparison let's place the fractions with the same denominator (the greatest common denominator is 210)

a) a₁ = g sin θ ½ = g sin θ $\frac{105}{210}$