Question 7

Name two energy transformations that occur as Adeline pedals her bicycle up a steep hill and then coasts down the other side.

Juliette [100K]

Potential energy is first transformed into kinetic energy as she pedals, then gravitational as she coasts down the hill.

6 0

3 years ago

Read 2 more answers

A merry-go-round of radius R, shown in the figure, is rotating at constant angular speed. The friction in its bearings is so sma

mel-nik [20]

The angular speed of the merry-go-round reduced more as the sandbag is

placed further from the axis than increasing the mass of the sandbag.

The rank from largest to smallest angular speed is presented as follows;

[m = 10 kg, r = 0.25·R]

${}$ ⇩

[m = 20 kg, r = 0.25·R]

${}$ ⇩

[m = 10 kg, r = 0.5·R]

${}$ ⇩

[m = 10 kg, r = 0.5·R] = [m = 40 kg, r = 0.25·R]

${}$ ⇩

[m = 10 kg, r = 1.0·R]

Reasons:

The given combination in the question as obtained from a similar question online are;

1: m = 20 kg, r = 0.25·R

2: m = 10 kg, r = 1.0·R

3: m = 10 kg, r = 0.25·R

4: m = 15 kg, r = 0.75·R

5: m = 10 kg, r = 0.5·R

6: m = 40 kg, r = 0.25·R

According to the principle of conservation of angular momentum, we have;

$I_i \cdot \omega _i = I_f \cdot \omega _f$

The moment of inertia of the merry-go-round, $I_m$ = 0.5·M·R²

Moment of inertia of the sandbag = m·r²

Therefore;

0.5·M·R²· $\omega _i$ = (0.5·M·R² + m·r²)· $\omega _f$

Given that 0.5·M·R²· $\omega _i$ is constant, as the value of m·r² increases, the value of $\omega _f$ decreases.

The values of m·r² for each combination are;

Combination 1: m = 20 kg, r = 0.25·R; m·r² = 1.25·R²

Combination 2: m = 10 kg, r = 1.0·R; m·r² = 10·R²

Combination 3: m = 10 kg, r = 0.25·R; m·r² = 0.625·R²

Combination 4: m = 15 kg, r = 0.75·R; m·r² = 8.4375·R²

Combination 5: m = 10 kg, r = 0.5·R; m·r² = 2.5·R²

Combination 6: m = 40 kg, r = 0.25·R; m·r² = 2.5·R²

Therefore, the rank from largest to smallest angular speed is as follows;

Combination 3 > Combination 1 > Combination 5 = Combination 6 >

Combination 2

Which gives;

[m = 10 kg, r = 0.25·R] > [m = 20 kg, r = 0.25·R] > [m = 10 kg, r = 0.5·R] > [m =

10 kg, r = 0.5·R] = [m = 40 kg, r = 0.25·R] > [m = 10 kg, r = 1.0·R].

Learn more here:

brainly.com/question/15188750

6 0

2 years ago

A mass of 1000 kg drops from a height of 10 m on a platform of negligible mass. It is desired to design a spring and dashpot on

Juliette [100K]

Answer:

k = $5\times 10^{4}\ N/m$

b = $0.707\times 10^{3}$

t = $7.1\times 10^{- 5}\ s$

Solution:

As per the question:

Mass of the block, m = 1000 kg

Height, h = 10 m

Equilibrium position, x = 0.2 m

Now,

The velocity when the mass falls from a height of 10 m is given by the third eqn of motion:

$v^{2} = u^{2} + 2gh$

where

u = initial velocity = 0

g = 10 $m/s^{2}$

Thus

$v = \sqrt{2\times 10\times 10} = 10\sqrt{2}\ m/s$

Force on the mass is given by:

F = mg = $1000\times 10 = 10000 N = 10\ kN$

Also, we know that the spring force is given by:

F = - kx

Thus

$k = \frac{F}{x} = \frac{10000}{0.2} = 5\times 10^{4}\ N/m$

Now, to find the damping constant b, we know that:

F = - bv

$b = \frac{F}{v} = \frac{10000}{10\sqrt{2}} = 0.707\times 10^{3}$

Now,

Time required for the platform to get settled to 1 mm or 0.001 m is given by:

$t = \frac{0.001}{v} = \frac{0.001}{10\sqrt{2}} = 7.1\times 10^{- 5}\ s$

4 0

3 years ago

A cylindrical resistor of length l is made from a metal of mass m. It has a resistance R.

Vikentia [17]

C, because it has a parellell cicruit

7 0

3 years ago

Shtirlitz [24]

Answer:

3272.4 Hz

Explanation:

L = 31 mH

XL = 637 ohm

XL = 2 π f L

f = XL / (2 π L)

f = 637 / ( 2 x 3.14 x 31 x 10^-3)

f = 3272.4 Hz

5 0

3 years ago