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2 years ago

A pulley with a mechanical advantage of 5 will require you to pull _____ times the amount of rope.

Physics

1 answer:

Marat540 [252]2 years ago

6 0

Answer:

It will require a force of 1/5, answer A.

Explanation:

In the attached image we can see an example of an array of pulleys that will lift a 100 kg-f load.

If we analyze pulleys A,B, and C as in the image we can check a force in the cable of 20kg-f.

In the pulley D we have three forces of 20 kg-f each and those forces plus the forces in the pulley B, sum a total of 100 kg-f (60+40). This matches the mechanical advantage (100/5) = 20 kg-f

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You need to determine the density of a ceramic statue. If you suspend it from a spring scale, the scale reads 28.4 NN . If you t

shtirl [24]

Answer:

2491.23 kg/m³

Explanation:

From Archimedes principle,

R.d = weight of object in air/ upthrust in water = density of the object/density of water

⇒ W/U = D/D' ....................... Equation 1

Where W = weight of the ceramic statue, U = upthrust of the ceramic statue in water, D = density of the ceramic statue, D' = density of water.

Making D the subject of the equation,

D = D'(W/U).................... Equation 2

Given: W = 28.4 N, U = lost in weight = weight in air- weight in water

U = 28.4 - 17.0 = 11.4 N,

Constant: D' = 1000 kg/m³.

Substitute into equation 2,

D = 100(28.4/11.4)

D = 2491.23 kg/m³

Hence the density of the ceramic statue = 2491.23 kg/m³

7 0

2 years ago

You launch a cannonball at an angle of 35° and an initial velocity of 36 m/s (assume y = y₁=

velikii [3]

Answer:

Approximately $4.2\; {\rm s}$ (assuming that the projectile was launched at angle of $35^{\circ}$ above the horizon.)

Explanation:

Initial vertical component of velocity:

$\begin{aligned}v_{y} &= v\, \sin(35^{\circ}) \\ &= (36\; {\rm m\cdot s^{-1}})\, (\sin(35^{\circ})) \\ &\approx 20.6\; {\rm m\cdot s^{-1}}\end{aligned}$ .

The question assumed that there is no drag on this projectile. Additionally, the altitude of this projectile just before landing $y_{1}$ is the same as the altitude $y_{0}$ at which this projectile was launched: $y_{0} = y_{1}$ .

Hence, the initial vertical velocity of this projectile would be the exact opposite of the vertical velocity of this projectile right before landing. Since the initial vertical velocity is $20.6\; {\rm m\cdot s^{-1}}$ (upwards,) the vertical velocity right before landing would be $(-20.6\; {\rm m\cdot s^{-1}})$ (downwards.) The change in vertical velocity is:

$\begin{aligned}\Delta v_{y} &= (-20.6\; {\rm m\cdot s^{-1}}) - (20.6\; {\rm m\cdot s^{-1}}) \\ &= -41.2\; {\rm m\cdot s^{-1}}\end{aligned}$ .

Since there is no drag on this projectile, the vertical acceleration of this projectile would be $g$ . In other words, $a = g = -9.81\; {\rm m\cdot s^{-2}}$ .

Hence, the time it takes to achieve a (vertical) velocity change of $\Delta v_{y}$ would be:

$\begin{aligned} t &= \frac{\Delta v_{y}}{a_{y}} \\ &= \frac{-41.2\; {\rm m\cdot s^{-1}}}{-9.81\; {\rm m\cdot s^{-2}}} \\ &\approx 4.2\; {\rm s} \end{aligned}$ .

Hence, this projectile would be in the air for approximately $4.2\; {\rm s}$ .

8 0

11 months ago

Read 2 more answers

Spacecrafts have landed on the surface of which planet?

ki77a [65]

I'm pretty sure its Venus!!!

7 0

2 years ago

A worker pushes a crate horizontally across a warehouse floor with a force of 245 N at an angle of 55 degrees below the horizont

aev [14]

Answer:

option A

Explanation:

given,

For exerted by the worker = 245 N

angle made with horizontal = 55°

we need to calculate Force which is not used to move the crate = ?

Movement of crate is due to the horizontal component of the force.

Crate will not move due to vertical force acting on the it.

$F_y = F sin \theta$

$F_y = 245\times sin 55^0$

$F_y =200.69$

hence, worker's force not used to move the crate is equal to 200.69

The correct answer is option A

6 0

2 years ago

Read 2 more answers

A rocket has total mass Mi=360 kg , including Mf=330kg of fuel and oxidizer. In interstellar space, it starts from rest at the p

steposvetlana [31]

let the parameters are:

M=instantaneous mass of the rocket

v=velocity of the rocket

t=time

initial mass=Mi=360 kg

fuel mass=Mf=330 kg

relative speed=Ve=1500 m/s

rate of mass decay=k=2.5 kg/s

writing newton’s second law of motion:

M*du/dt=-Ve*dM/dt

==>du=-Ve*dM/M

Integration of both sides,

u=-Ve*ln(M)

using the limits with at t=0, u=0 and velocity being v(t),

at t=0, mass=Mi, at any time t, mass=Mi-k*t

v(t)-0=-Ve*ln((Mi-k*t)/Mi)

v(t)=-Ve*ln(1-(k*t/Mi))

<h3>What is needed for rocket propulsion?</h3>

Rocket propulsion is the strategy utilized to make the basic thrust to lift a rocket into the air. The force that the rocket employments to lift off from the soil is known as rocket propulsion. The third law of movement by Newton serves as the foundation for rocket propulsion. Here, the fuel is shot out from the exit mightily in arrange to cause an equal and converse reaction. The some sorts of rocket propulsion are fuelled by liquid, strong, cold gas, and particles. The rocket's mass, fuel burn rate, and weaken speed all impact speeding up.

To learn more about Rocket propulsion, visit;

brainly.com/question/15363207

#SPJ4

6 0

7 months ago