If the length of the ramp is 2m and the height of the ramp is 1m , what is the mechanical advantage?

An electric field is e = 200 n/c i for x > 0 and e = -200 n/c i for x < 0. a cylinder of length 20 cm and radius 3 cm has

lozanna [386]

543 that's what it is

8 0

3 years ago

The record height of a man to date is 8 feet 11 inches (107 inches). If all men had identical body types, their weights would

Alisiya [41]

Answer:

679.08 lbs

Explanation:

Here it is given that

$m\propto h^3$

$m_1$ = Mass of first person = 175 pounds

$h_1$ = Height of first person = 70 inches

$m_2$ = Mass of second person

$h_2$ = Height of second person = 110 inches

$\frac{m_1}{m_2}=\frac{h_1^3}{h_2^3}\\\Rightarrow m_2=\frac{m_1h_2^3}{h_1^3}\\\Rightarrow m_2=\frac{175\times 110^3}{70^3}\\\Rightarrow m_2=679.08\ lb$

The weight of the second person would be 679.08 lbs

5 0

3 years ago

A satellite revolves around a planet at an altitude equal to the radius of the planet. the force of gravitational interaction be

padilas [110]

Let
M = the mass of the planet
n = the mass of the satellite.
r = the radius of the planet

When the satellite is at a distance r from the surface of the planet, the distance between the centers of the two masses is 2r.
The gravitational force between them is
$f_{0} = \frac{GMm}{(2r)^{2}} = \frac{1}{4} ( \frac{GMm}{r^{2}} )$
where
G = the gravitational constant.

When the satellite is on the surface of the planet, the distance between the two masses is r.
The gravitational force between them is
$f_{4} = \frac{GMm}{r^{2}} =4f_{0}$

Answer: $f_{4} = 4f_{0}$

5 0

4 years ago

Read 2 more answers

A ball rolls off a desk at a speed of 2 m/s and lands 0.50 seconds later.

Hoochie [10]

Answer:

(a) The ball lands <u>1 m</u> ahead of the desk's base.

(b) The height of the desk is <u>1.225 m.</u>

Explanation:

Given:

Initial velocity of the ball is, $u_{0}=2\ m/s$

Time of flight of the ball is, $t=0.50\ s$

(a)

Let the ball fall at a distance of $x$ from the base of the desk and let the height of the desk be $y$ .

The given motion of the ball is a projectile motion that can be divided into 2 mutually perpendicular directions.

Now, considering only the horizontal motion of the ball. The ball not under the influence of any force in the horizontal direction.

Therefore, the distance covered by the ball horizontally is given as:

$x=v_0\times t\\x=2\times 0.50=1\ m$

So, the ball lands 1 m ahead of the desk's base.

(b)

Now, consider only the vertical motion of the ball. The ball is under the influence of gravity.

So, vertical distance can be obtained using Newton's equations of motion.

We use the following equation of motion:

$y-y_0=u_{oy}t+\frac{1}{2}gt^2$

Where,

$y_{0}\rightarrow \textrm{initial vertical position of ball}\\u_{oy}\rightarrow \textrm{initial vertical velocity}\\g\rightarrow \textrm{acceleration due to gravity}\\y\rightarrow \textrm{final position of ball vertically}$

Here, as the ball leaves the desk horizontally, initial velocity is only in the horizontal direction and doesn't have any component in the vertical direction. So, $u_{oy}=0\ m/s$

Plug in $y_0=h,y=0,g=-9.8,u_{oy}=0,t=0.5$ . Solve for $h$ .

$0-h=0+\frac{1}{2}(-9.8)(0.5)^2\\-h=-4.9\times 0.25\\h=1.225\ m$

Therefore, the height of the desk is 1.225 m.

6 0

3 years ago

A small block of mass M = 0.10 kg is released from rest at point 1 at a height H = 1.8 m above the bottom of a track, as shown i

Ede4ka [16]

Answer:

D

Explanation:

A) is not correct, because the gravitation potential energy will depend on the height the block is located at. It will be calculated with the formula:

U=mgh.

If we take the ground as a zero height reference, then on point 2 the potential energy will be:

$U_{2} = 0.10kg(9.81 m/s^{2})(0.6m)$

$U_{2}=0.59 J$

While on point 3, the potential energy will be greater.

$U_{3}=0.10kg(9.81 m/s^{2})(1.2m)$

$U_{3}=1.18 J$

B) is not the right answer because the kinetic energy will vary with the height the block is located at in the fact that the energy is conserved (this is if we don't take friction into account or air resistance) so in this case:

$U_{2}+K_{2}= U_{3}+K_{3}$

We already know the potential energy at point 2. We can calculate the kinetic energy at point 3 like this:

$K_{3} =\frac{1}{2}mv_{3}^{2}$

$K_{3} =\frac{1}{2}(0.10kg)(2.5 m/s)^{2}$

$K_{3} =0.31 J$

So the kinetic energy at point 2 is given by the equation:

$K_{2} =U_{3}-U_{2}+K_{3}$

so:

$K_{2} = (1.18J)-(0.59J)+0.31J$

$K_{2} =0.9J$

As you may see the kinetic energy at point 2 is greater than the kinetic energy at point 3.

C) Is not correct because according to the first law of thermodinamics, energy is not lost, only transformed. So, since we are not taking into account friction or any other kind of loss, then we can say that the amount of mechanical energy at point 1 is exactly the same as the mechanical energy at point 3.

D) Because of what we talked about on part C, this will be the true situation, because the mechanical energy of the block will be the same no matter theh point you measure it at.

7 0

3 years ago