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3 years ago

The foot of a ladder is 6m away from a wall. if the top of the ladder rest 8 feet up on the wall,how long is the ladder?

1 answer:

5 0

To solve this we can use Pythagorean's theorem for right triangles. The formula describes the longest side of a right triangle ( a triangle with a 90degree ). The side opposite to this angle equals the square root of the sum of the squares of the other two shorter sides. Given this equation, we have the length of the ladder as the longest side (hypotenuse) which is opposite the angle between the plane and the wall, given the height it makes with the wall 8 feet and distance of the other end of the ladder at 6m away we have below

Length of ladder = sqrt ((6 m)^2+(2.4384m)^2)

Note that 8 feet was converted to meter = 2.4384

Length of ladder is = sqrt (36 + 5.94579456) = 6.5 meters approximately

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Why are frogs said to have two lives?

Stells [14]

They live half their lives in water and most of their life on land

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3 years ago

A speed skater moving across frictionless ice at 8.4 m/s hits a 5.7 m -wide patch of rough ice. She slows steadily, then continu

malfutka [58]

Answer:

Acceleration, $a=-2.48\ m/s^2$

Explanation:

Initial speed of the skater, u = 8.4 m/s

Final speed of the skater, v = 6.5 m/s

It hits a 5.7 m wide patch of rough ice, s = 5.7 m

We need to find the acceleration on the rough ice. The third equation of motion gives the relationship between the speed and the distance covered. Mathematically, it is given by :

$v^2-u^2=2as$

$a=\dfrac{v^2-u^2}{2s}$

$a=\dfrac{(6.5)^2-(8.4)^2}{2\times 5.7}$

$a=-2.48\ m/s^2$

So, the acceleration on the rough ice $-2.48\ m/s^2$ and negative sign shows deceleration.

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3 years ago

An inkjet printer's speed is measured in ____. select one:

Vladimir [108]

<span>Back in the day, one measured a printer's speed in CPM, which stands for characters per minute. Most of the modern printers that exist today, including the inkjet printer measure their speed in PPM, which is also known as pages per minute.</span>

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3 years ago

An airplane is seller rates down a runway at 3.10 M/S squared 441.5 seconds from rest until finally left off the ground ,what is

sineoko [7]

Answer:

302129m

Explanation:

Using s= ut + 1/2at²

But u= 0

S= 1/2(3.1)(441.5)²

=302129m

6 0

3 years ago

A loaded ore car has a mass of 950 kg. and rolls on rails ofnegligible friction. It starts from rest ans is pulled up a mineshaf

stiks02 [169]

(a) 10241 W

In this situation, the car is moving at constant speed: this means that its acceleration along the direction parallel to the slope is zero, and so the net force along this direction is also zero.

The equation of the forces along the parallel direction is:

$F - mg sin \theta = 0$

where

F is the force applied to pull the car

m = 950 kg is the mass of the car

$g=9.8 m/s^2$ is the acceleration of gravity

$\theta=30.0^{\circ}$ is the angle of the incline

Solving for F,

$F=mg sin \theta = (950)(9.8)(sin 30.0^{\circ})=4655 N$

Now we know that the car is moving at constant velocity of

v = 2.20 m/s

So we can find the power done by the motor during the constant speed phase as

$P=Fv = (4655)(2.20)=10241 W$

(b) 10624 W

The maximum power is provided during the phase of acceleration, because during this phase the force applied is maximum. The acceleration of the car can be found with the equation

$v=u+at$

where

v = 2.20 m/s is the final velocity

a is the acceleration

u = 0 is the initial velocity

t = 12.0 s is the time

Solving for a,

$a=\frac{v-u}{t}=\frac{2.20-0}{12.0}=0.183 m/s^2$

So now the equation of the forces along the direction parallel to the incline is

$F - mg sin \theta = ma$

And solving for F, we find the maximum force applied by the motor:

$F=ma+mgsin \theta =(950)(0.183)+(950)(9.8)(sin 30^{\circ})=4829 N$

The maximum power will be applied when the velocity is maximum, v = 2.20 m/s, and so it is:

$P=Fv=(4829)(2.20)=10624 W$

(c) $5.82\cdot 10^6 J$

Due to the law of conservation of energy, the total energy transferred out of the motor by work must be equal to the gravitational potential energy gained by the car.

The change in potential energy of the car is:

$\Delta U = mg \Delta h$