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Irina-Kira [14]
4 years ago
7

Two farmers are trying to move a large rock that sits in a field. They use a long stick as a lever and a brick as the fulcrum, a

s shown in the diagram. They apply a force on the stick at the location shown by the arrow. However, the rock does not move.
Which change would BEST allow the farmers to move the rock?
A) Moving the brick closer to the rock.
B) Moving the brick to the other side of the rock.
C) Applying a smaller force at the same location on the stick.
D) Applying an equal force at a location farther away from the rock.

Physics
2 answers:
kumpel [21]4 years ago
4 0

D is the answer for usatestprep

diamong [38]4 years ago
4 0

Answer:

D) Applying an equal force at a location farther away from the rock.

Explanation:

Here big stone is moved by applying the technique of torque

Here the applied force on the rod will give torque due to which we will have

\tau = \vec r \times \vec F

so it is product of force and its distance from fulcrum

So if the the product will be more then we will have more torque in output and that will easily shift the heavy rock

So here the torque can be increased by increasing the product of force and its distance from fulcrum

So here correct answer will be

D) Applying an equal force at a location farther away from the rock.

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An object starts from rest at time t = 0.00 s and moves in the +x direction with constant acceleration. The object travels 14.0
Sergio039 [100]

Answer:

The acceleration of the object is 9.3 m/s²

Explanation:

For a straight movement with constant acceleration, this equation for the position applies:

x = x0 + v0 t + 1/2 a t²

where

x = position at time t

x0 = initial position

v0 = initial velocity

a = acceleration

t = time

we have two positions: one at time t = 1 s and one at time t = 2 s. We know that the difference between these positions is 14.0 m. These are the equations we can use to obtain the acceleration:

x₁ = x0 + v0 t + 1/2 a (1 s)²

x₂ = x0 + v0 t + 1/2 a (2 s)²

x₂ - x₁ = 14 m

we know that the object starts from rest, so v0 = 0

substracting both equations of position we will get:

x₂ - x₁ = 14

x0 + v0 t + 1/2 a (2 s)² - (x0 + v0 t + 1/2 a (1 s)²) = 14 m

x0 + v0 t + 2 a s² - x0 -v0 t - 1/2 a s² = 14 m

2 a s² - 1/2 a s² = 14 m

3/2 a s² = 14 m  

a = 14 m / (3/2 s²) = <u>9.3 m/s² </u>

5 0
4 years ago
On which of saturn’s moons did the cassini-huygens probe land in 2004, providing our first view of the varied and active surface
yawa3891 [41]

On Titan, the largest moon of of Saturn did the Cassini-Huygens probe land in 2004.

To find the answer, we have to know more about the Cassini-Huygens Mission.

<h3>What is Cassini-Huygens mission?</h3>
  • Before arriving at its final destination of Saturn in 2004 and beginning a series of flybys of Saturn's moons, the spacecraft contributed to studies of Jupiter for six months in 2000.
  • In the same year, it launched the Huygens probe to explore Titan's atmosphere and surface makeup on Saturn's moon.
  • During its second extended mission, Cassini sailed between the rings, entered the planet's atmosphere, and obtained the first measurements of a whole seasonal period for Saturn and its moons.

Thus, we can conclude that, on Titan, the largest moon of of Saturn did the Cassini-Huygens probe land in 2004.

Learn more about the Cassini-Huygens mission here:

brainly.com/question/27907891

#SPJ4

4 0
2 years ago
During typical urination, a man releases about 400 mL of urine in about 30 seconds through the urethra, which we can model as a
gregori [183]

Answer:

Explanation:

Given:

volume of urine discharged, V=400~mL=0.4~L=4\times 10^{-4}~m^3

time taken for the discharge, t=30~s

diameter of cylindrical urethra, d=4\times10^{-3}~m

length of cylindrical urethra, l=0.2~m

density of urine, \rho=1000~kg/m^3

a)

we have volume flow rate Q:

Q=A.v & Q=\frac{V}{t}

where:

A= cross-sectional area of urethra

v= velocity of flow

A.v=\frac{V}{t}

\frac{\pi d^2}{4}\times v=\frac{4\times 10^{-4}}{30}

v=\frac{4\times4\times 10^{-4}}{30\times \pi (4\times 10^{-3})^2}

v=1.06~m/s

b)

The pressure required when the fluid is released at the same height as the bladder and that the fluid is at rest in the bladder:

P=\rho.g.l

P=1000\times 9.8\times 0.2

P=1960~Pa

5 0
3 years ago
What is the speed of a wave that has a frequency of 6 Hz and a wavelength of 4 m?
notsponge [240]

Answer:

a ) 24 m/s

Explanation:

Given,

Frequency ( f ) = 6 Hz

Wavelength ( λ ) = 4 m

To find : Speed ( v ) = ?

Formula : -

v = f x λ

v

= 4 x 6

= 24 m/s

Therefore, the speed of a wave that has a frequency of 6 Hz and a wavelength of 4 m

is 24 m/s.

4 0
3 years ago
Calculate the acceleration due to gravity at Earth due to the Moon. (b) Calculate the acceleration due to gravity at Earth due t
HACTEHA [7]

Answer:

a)  g’= 3.44 10⁻⁵ m / s²

b)  g ‘’ = 5.934 10⁻³ m / s²

Explanation:

For this exercise let's use the law of universal gravitation

         F = G \frac{m M}{r^2}

where m is the mass of the body under study, M the mass of the body that creates the force and r the distance between the bodies

         F = m \ ( G \frac{M}{r^2} )

the attractive force is called weight W = m g,

Thus

         g = G \frac{M}{r^2}

is called the acceleration of gravity

a) the acceleration created by the moon

       g' = G \frac{M}{r^2}

the mass of the moon is M = 7.36 10²² kg

the distance from the moon to the Earth's surface is

       r = D -R_e

       r = 3.84 10⁸ -6.37 10⁶

       r = 3.7763 10⁸ m

we calculate

     g’= 6.67 \ 10^{-11} \ \frac{7.36 \ 10^{22}}{ (3.7763 \ 10^8)^2}

     g ’= 3.44 10⁻⁵ m / s²

b) the acceleration created by the sun

mass of the sun M = 1,9991 10³⁰ ka

the distance from the sun wears down the Earth's surface

        r = D -R_e

        r = 1.496 10¹¹ -6.37 10⁶

        r = 1.4959 10¹¹ m

let's calculate

       g ’’ = 6.67 \ 10^{-11} \frac{1.991 \ 10^{30}}{ (1.4959 \ 10^{11})^2 }

       g ‘’ = 5.934 10⁻³ m / s²

4 0
3 years ago
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