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3 years ago

If the load is 13 cm from the fulcrum, how much effort is needed to lifth the load ? Help me.

1 answer:

7 0

That depends on a few things that you haven't told us about the setup.
So I'm going to assume one of them, and then give you the answer
in terms of another one:

-- Assume a Class-I lever . . . the fulcrum is between the load and the effort.

-- Then the effort needed to lift the load is

(the weight of the load) x (13 / the distance between the fulcrum and the effort)

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A truck with 0.420-m-radius tires travels at 32.0 m/s. what is the angular velocity of the rotating tires in radians per second?

andrey2020 [161]

Angular velocity of the rotating tires can be calculated using the formula,

v=ω×r

Here, v is the velocity of the tires = 32 m/s

r is the radius of the tires= 0.42 m

ω is the angular velocity

Substituting the values we get,

32=ω×0.42

ω= 32/0.42 = 76.19 rad/s

= 76.19× $\frac{1}{2\pi} *60$ revolution per min

=728 rpm

Angular velocity of the rotating tires is 76.19 rad/s or 728 rpm.

4 0

3 years ago

Calculate the kinetic energy of Julie, a 60 kg biker, traveling at a velocity of 8 m/s to the right

Vadim26 [7]

Answer:

1,920 Joules

Explanation:

K.E. = 1/2 mv2

so K.E. = 1/2 (60)(8x8) = 1,920 Joules

8 0

3 years ago

uppose the weights of Farmer Carl's potatoes are normally distributed with a mean of 9 ounces and a standard deviation of 0.8 ou

Brrunno [24]

Answer:

x= 9.53 ounces

Explanation:

Given that

Mean ,μ= 9 ounces

Standard deviation ,σ=0.8 ounces

He wants to sell only those potatoes that are among the heaviest 25%.

P=25% = 0.25

When P= 0.25 then Z=0.674

Lest take x is the the minimum weight required to be brought to the farmer's market.

We know that

x = Z . σ + μ

x= 0.674 ₓ 0.8 + 9 ounces

x= 9.53 ounces

3 0

3 years ago

A distant large asteroid is detected that might pose a threat to Earth. If it were to continue moving in a straight line at cons

Vlada [557]

Answer:

The minimum speed required is 5.7395km/s.

Explanation:

To escape earth, the kinetic energy of the asteroid must be greater or equal to its gravitational potential energy:

$K.E\geq P.E$

$\dfrac{1}{2}mv^2 \geq G\dfrac{Mm}{R}$

where $m$ is the mass of the asteroid, $R= 24,000,000\:m$ is its distance form earth's center, $M = 5.9*10^{24}kg$ is the mass of the earth, and $G = 6.7*10^{-11}m^3/kg\: s^2$ is the gravitational constant.

Solving for $v$ we get:

$v \geq \sqrt{\dfrac{2GM}{R} }$

putting in numerical values gives

$v \geq \sqrt{\dfrac{2(6.7*10^{-11})(5.9*10^{24})}{(24,000,000)} }$

$\boxed{v\geq 5739.5m/s}$

in kilometers this is

$v\geq5.7395m/s.$

Hence, the minimum speed required is 5.7395km/s.

5 0

3 years ago

a car whose mass is 1000kg is traveling at a constant speed of 10m/s. Neglecting any friction how much force will the engine hav

AURORKA [14]

This next statement is a big deal. It should be up on a board, surrounded
by flashing red and yellow lights, and hung on the wall of every Science
classroom. Although we never see it in our daily lives, it's fundamental to
the workings of the universe, and it's also Newton's first law of motion:

Without friction, it doesn't take ANY force to keep a moving object
moving. Force is only required to change the object's speed, or to
change the direction in which it's moving.

The answer to the question is: On a level road, and neglecting any friction,
the engine doesn't have to supply ANY force to keep the car going at the
same speed.

7 0

3 years ago

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