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3 years ago

A 10-meter long ramp has a mechanical advantage of 5. What is the height of the ramp?

1 answer:

6 0

1. If the ramp has a length of 10 and has a mechanical advantage (MA) of 5. Then we need to find the height of the ramp.
Formula:
MA = L / H
Since we already have the mechanical advantage and length, this time we need to find the height .
MA 5 = 10 / h
h = 10 / 5
h = 2 meters

Therefore, the ramp has a length of 10 meters, a height of 2 meters with a mechanical advantage of 5.

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Define average velocity

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The average speed of an object is defined as the distance traveled divided by the time elapsed. Velocity is a vector quantity, and average velocity can be defined as the displacement divided by the time.

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3 years ago

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A ball is thrown vertically upwards with an initial velocity of 20.00 m/s. Neglecting air resistance, how long is the ball in th

algol [13]

Answer:

1) The greatest height attained by the ball equals 20.387 meters.

2) The time it takes for the ball to reach 15 meters approximately equals 1 second.

Explanation:

The greatest height will be attained when the ball stop's in the air and starts falling back to the earth.

thus using third equation of kinematics we obtain the height attained as

$v^2=u^2+2as$

where

'v' is the final speed of the ball

'u' is the initial speed of the ball

'a' is the acceleration that the ball is under which in this case equals 9.81 $m/s^{2}$

's' is the distance it covers

Thus for maximum height applying the values in the equation we get

$0=20^{2}-2\times 9.81\times h\\\\\therefore h=\frac{20^{2}}{2\times 9.81}=20.387meters$

Using the same equation we can find the speed of the ball when it reaches 15 meters of height as

$v^2=20^{2}-2\times 9.81\times 15\\\\v^{2}=105.7\\\\\therefore v=10.28m/s$

the time it takes to reduce the velocity to this value can be found by first equation of kinematics as

$v=u+at\\\\t=\frac{v-u}{a}\\\\t=\frac{10.28-20}{-9.81}=0.991seconds\approx 1second$

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3 years ago

A cup of coffee is considered a Blank Space __________.

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When you brew a coffee you're creating a Solution.

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3 years ago

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3 years ago

A ship maneuvers to within 2.46×10³ m of an island’s 1.80 × 10³ m high mountain peak and fires a projectile at an enemy ship 6.1

Nesterboy [21]

Answer:

The distance close to the peak is 597.4 m.

Explanation:

Given that,

Distance of the first ship from the mountain $d=2.46\times10^{3}\ m$

Height of island $h=1.80\times10^{3}\ m$

Distance of the enemy ship from the mountain $d'=6.10\times10^{2}\ m$

Initial velocity $v=2.55\times10^{2}\ m/s$

Angle = 74.9°

We need to calculate the horizontal component of initial velocity

Using formula of horizontal component

$v_{x}=v\cos\theta$

Put the value into the formula

$v_{x}=2.55\times10^{2}\cos74.9$

$v_{x}=66.42\ m/s$

We need to calculate the vertical component of initial velocity

Using formula of vertical component

$v_{y}=v\sin\theta$

Put the value into the formula

$v_{y}=2.55\times10^{2}\sin74.9$

$v_{y}=246.19\ m/s$

We need to calculate the time

Using formula of time

$t=\dfrac{d}{v_{x}}$

$t=\dfrac{2.46\times10^{3}}{66.42}$

$t=37.03\ sec$

We need to calculate the height of the shell on reaching the mountain

Using equation of motion

$H= v_{y}t-\dfrac{1}{2}gt^2$

Put the value in the equation

$H=246.19\times37.03-\dfrac{1}{2}\times9.8\times(37.03)^2$

$H=2397.4\ m$

We need to calculate the distance close to the peak

Using formula of distance

$H'=H-h$

Put the value into the formula

$H'=2397.4-1800$

$H'=597.4\ m$

Hence, The distance close to the peak is 597.4 m.

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3 years ago