If an inclined plane is 6 m long and 2 m high, what is its mechanical advantage?

Not understanding this, please help out

allsm [11]

Answer:

greater than

Explanation:

the answer is greater than

3 0

3 years ago

Supercells are updrafts that may cause which of the following?

ale4655 [162]

just for anyone looking for the answer i just took the test and the answer is tornadoes

6 0

3 years ago

When are the days in the northern hemisphere the shortest?

kolezko [41]

The Winter...? I think...? Ya, it's the Winter. :)

3 0

3 years ago

Read 2 more answers

Three particles lie in the xy plane. Particle 1 has mass m1 = 6.7 kg and lies on the x-axis at x1 = 4.2 m, y1 = 0. Particle 2 ha

krek1111 [17]

Answer:

$F=18.58\times 10^{-11}\ N$

$\theta=30.276^{\circ}$

Explanation:

Given:

mass of first particle, $m_1=6.7\ kg$

mass of second particle, $m_2=5.1\ kg$

mass of third particle, $m_3=3.7\ kg$

coordinate position of first particle in meters, $(x_1,y_1)\equiv(4.2,0)$

coordinate position of second particle in meters, $(x_2,y_2)\equiv(0,2.8)$

coordinate position of third particle in meters, $(x_3,y_3)\equiv(0,0)$

Now, gravitational force on particle 3 due to particle 1:

$F_{31}=G\frac{m_1.m_3}{r_{31}^2}$

$F_{31}=6.67\times 10^{-11} \times \frac{6.7\times 3.7}{4.2^2}$

$F_{31}=9.37\times 10^{-11}\ N$

towards positive Y axis.

gravitational force on particle 3 due to particle 2:

$F_{32}=G\frac{m_2.m_3}{r_{21}^2}$

$F_{32}=6.67\times 10^{-11} \times \frac{5.1\times 3.7}{2.8^2}$

$F_{32}=16.05\times 10^{-11}\ N$

towards positive X axis.

Now the net force

$F=\sqrt{F_{31}\ ^2+F_{32}\ ^2}$

$F=\sqrt{(10^{-11})^2(9.37^2+16.05^2)}$

$F=18.58\times 10^{-11}\ N$

For angle in counterclockwise direction from the +x-axis

$tan\theta=\frac{9.37\times 10^{-11}}{16.05\times 10^{-11}}$

$\theta=30.276^{\circ}$

4 0

3 years ago

50 points !! I need help asap.......Consider a 2-kg bowling ball sits on top of a building that is 40 meters tall. It falls to t

r-ruslan [8.4K]

1) At the top of the building, the ball has more potential energy

2) When the ball is halfway through the fall, the potential energy and the kinetic energy are equal

3) Before hitting the ground, the ball has more kinetic energy

4) The potential energy at the top of the building is 784 J

5) The potential energy halfway through the fall is 392 J

6) The kinetic energy halfway through the fall is 392 J

7) The kinetic energy just before hitting the ground is 784 J

Explanation:

1)

The potential energy of an object is given by

$PE=mgh$

where

m is the mass

g is the acceleration of gravity

h is the height relative to the ground

While the kinetic energy is given by

$KE=\frac{1}{2}mv^2$

where v is the speed of the object

When the ball is sitting on the top of the building, we have

$h=40 m$ , therefore the potential energy is not zero
$v=0$ , since the ball is at rest, therefore the kinetic energy is zero

This means that the ball has more potential energy than kinetic energy.

2)

When the ball is halfway through the fall, the height is

$h=20 m$

So, half of its initial height. This also means that the potential energy is now half of the potential energy at the top (because potential energy is directly proportional to the height).

The total mechanical energy of the ball, which is conserved, is the sum of potential and kinetic energy:

$E=PE+KE=const.$

At the top of the building,

$E=PE_{top}$

While halfway through the fall,

$PE_{half}=\frac{PE_{top}}{2}=\frac{E}{2}$

And the mechanical energy is

$E=PE_{half} + KE_{half} = \frac{PE_{top}}{2}+KE_{half}=\frac{E}{2}+KE_{half}$

which means

$KE_{half}=\frac{E}{2}$

So, when the ball is halfway through the fall, the potential energy and the kinetic energy are equal, and they are both half of the total energy.

3)

Just before the ball hits the ground, the situation is the following:

The height of the ball relative to the ground is now zero: $h=0$ . This means that the potential energy of the ball is zero: $PE=0$

The kinetic energy, instead, is not zero: in fact, the ball has gained speed during the fall, so $v\neq 0$ , and therefore the kinetic energy is not zero