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Lena [83]
3 years ago
10

a pennyfarthing is a style of a bicycle with a very large front wheel and a small real wheel, the cyclist who sit high above and

behind the front wheel pedals this wheel directly the distance the pedals are turned input distance in one rotation is about 0.64 m. if the mechanical advantage of the pennyfarthing is 0.16 how far does the large wheel turn in one rotation.
Physics
1 answer:
s344n2d4d5 [400]3 years ago
4 0

Answer:

In one rotation, the large wheel turns 4m.

Explanation:

The given values are:

Input distance,

= 0.64 m

Mechanical advantage,

= 0.16

As we know,

⇒ Out. \ Distance = \frac{Inp. \ distance}{Mechanical \ advantage}

On putting the values, we get

⇒                         =\frac{0.64}{0.16}

⇒                         =4 \ m

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When a boat is placed in liquid, two forces act on the boat. Gravity pulls the boat down with a force equal to the weight of the
wel

Answer:

the fraction of submerged volume is equal to the ratio of the densities of the body between the density of the fluid.

Explanation:

This is a fluid mechanics problem, where as the boat is in equilibrium with the pushing force we can write Newton's second law

                  B- W = 0

                  B = W

           

the thrust force is equal to the weight of the liquid that is dislodged

                  B = ρ g V

we substitute

             ρ g V = m g

             V = m /ρ_fluid          1

we can write the mass of the pot as a function of its density

             ρ_body = m / V_body

            m = ρ_body  V_body

             V_fluid / V_body = ρ_body / ρ _fluid         2

Equations 1 and 2 are similar, although 2 is easier to analyze, the fraction of submerged volume is equal to the ratio of the densities of the body between the density of the fluid.

The effect appears the pot as if it had a lower apparent weight

3 0
3 years ago
A concrete block is hung from an ideal spring that has a force constant of 100 N/m . The spring stretches 0.129 m .A- What is th
madam [21]

Answer:

1.31498 kg

0.72050 s

0.72050 s

Explanation:

m = Mass of block

g = Acceleration due to gravity = 9.81 m/s²

k = Spring constant = 100 N/M

x = Displacement = 0.129 m

The force balance is

mg=kx\\\Rightarrow m=\dfrac{kx}{g}\\\Rightarrow m=\dfrac{100\times 0.129}{9.81}\\\Rightarrow m=1.31498\ kg

The mass of the block is 1.31498 kg

Time period is given by

T=2\pi\sqrt{\dfrac{m}{k}}\\\Rightarrow T=2\pi\sqrt{\dfrac{1.31498}{100}}\\\Rightarrow T=0.72050\ s

The period of oscillations is 0.72050 s

The time period does not depend on the acceleration due to gravity. It varies with the mass and the spring constant.

Hence, the time period would be the same

8 0
3 years ago
Ciara is swinging a 0.015 kg ball tied to a string around her head in a flat, horizontal circle. The radius of the circle is 0.7
Sophie [7]

Answer:

B) 1.2 N, toward the center of the circle

Explanation:

The circumference of the circle is:

C = 2πr

C = 2π (0.70 m)

C = 4.40 m

So the velocity of the ball is:

v = C/t

v = 4.40 m / 0.60 s

v = 7.33 m/s

Sum of the forces in the radial direction:

∑F = ma

T = m v² / r

T = (0.015 kg) (7.33 m/s)² / (0.70 m)

T = 1.2 N

The tension force is 1.2 N towards the center of the circle.

4 0
3 years ago
Read 2 more answers
A group of students prepare for a robotic competition and build a robot that can launch large spheres of mass M in the horizonta
jeyben [28]

Answer:

V_0

Explanation:

Given that, the range covered by the sphere, M, when released by the robot from the height, H, with the horizontal speed V_0 is D as shown in the figure.

The initial velocity in the vertical direction is 0.

Let g be the acceleration due to gravity, which always acts vertically downwards, so, it will not change the horizontal direction of the speed, i.e. V_0 will remain constant throughout the projectile motion.

So, if the time of flight is t, then

D=V_0t\; \cdots (i)

Now, from the equation of motion

s=ut+\frac 1 2 at^2\;\cdots (ii)

Where s is the displacement is the direction of force, u is the initial velocity, a is the constant acceleration and t is time.

Here, s= -H, u=0, and a=-g (negative sign is for taking the sigh convention positive in +y direction as shown in the figure.)

So, from equation (ii),

-H=0\times t +\frac 1 2 (-g)t^2

\Rightarrow H=\frac 1 2 gt^2

\Rightarrow t=\sqrt {\frac {2H}{g}}\;\cdots (iii)

Similarly, for the launched height 2H, the new time of flight, t', is

t'=\sqrt {\frac {4H}{g}}

From equation (iii), we have

\Rightarrow t'=\sqrt 2 t\;\cdots (iv)

Now, the spheres may be launched at speed V_0 or 2V_0.

Let, the distance covered in the x-direction be D_1 for V_0 and D_2 for 2V_0, we have

D_1=V_0t'

D_1=V_0\times \sqrt 2 t [from equation (iv)]

\Rightarrow D_1=\sqrt 2 D [from equation (i)]

\Rightarrow D_1=1.41 D (approximately)

This is in the 3 points range as given in the figure.

Similarly, D_2=2V_0t'

D_2=2V_0\times \sqrt 2 t [from equation (iv)]

\Rightarrow D_2=2\sqrt 2 D [from equation (i)]

\Rightarrow D_2=2.82 D (approximately)

This is out of range, so there is no point for 2V_0.

Hence, students must choose the speed V_0 to launch the sphere to get the maximum number of points.

7 0
3 years ago
What is less dense water or oil steel or water helium or air or oil or water
Leni [432]

Answer:

air because their is nothing contained within the air other than all the solutions that you have listed

Explanation:

4 0
3 years ago
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