What's an impure potassium carbonate

Two metal disks are welded together and are mounted on a frictionless axis through their common centers. One disk has a radius R

Airida [17]

The Inertia is 22. 488 kg. m² and the speed just before it hits the ground is 6. 4 m/s

<h3>How to determine the inertia</h3>

Using the formula:

I = 1/2 M₁R₁² + 1/2 M₂R₂²

Where I = Inertia

I = 1/2 * 0.810* (2. 60)² + 1/2 * 1. 58 * (5)²

I = 1/2 * 5. 476 + 1/2 * 39. 5

I = 2. 738 + 19. 75

I = 22. 488 kg. m²

To determine the block's speed, use the formula

v = $\sqrt{2gh}$

v = $\sqrt{2* 10 * 2. 10}$

v = $\sqrt{42}$

v = 6. 4 m/s

Therefore, the Inertia is 22. 488 kg. m² and the speed just before it hits the ground is 6. 4 m/s

Learn more about law of inertia here:

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7 0

2 years ago

An object attached to an ideal spring oscillates with an angular frequency of 2.81 rad/s. the object has a maximum displacement

NikAS [45]

Ω = 2.81
A = 0.232
k = 29.8

x = A cos(ωt + Ф)

at t = 0:
x = A = A cos(ωt + Ф) = A cos(Ф)
Ф = 0

at t = 1.42, with Ф = 0:
x = A cos(ωt)

U = 1/2 k x² = 1/2 k [A cos(ωt)]²

4 0

2 years ago

A mechanic needs to replace the motor for a merry-go-round. The merry-go-round should accelerate from rest to 1.5 rad/s in 7.0 s

marta [7]

Answer

given,

initial speed of merry-go-round = 0 rad/s

final speed of merry-go-round = 1.5 rad/s

time = 7 s

Radius of the disk = 6 m

Mass of the merry-go-round = 25000 Kg

Moment of inertia of the disk

$I = \dfrac{1}{2}MR^2$

$I = \dfrac{1}{2}\times 25000\times 6^2$

I = 450000 kg.m²

angular acceleration

$\alpha = \dfrac{\omega_f-\omega_0}{t}$

$\alpha = \dfrac{1.5-0}{7}$

$\alpha =0.214\ rad/s^2$

we know,

$\tau= I \alpha$

$\tau= 450000\times 0.214$

$\tau=96300\ N.m$

8 0

3 years ago

Three equal charge 1.8*10^-8 each are located at the corner of an equilateral triangle ABC side 10cm.calculate the electric pote

Arlecino [84]

Answer:

If all these three charges are positive with a magnitude of $1.8 \times 10^{-8}\; \rm C$ each, the electric potential at the midpoint of segment $\rm AB$ would be approximately $8.3 \times 10^{3}\; \rm V$ .

Explanation:

Convert the unit of the length of each side of this triangle to meters: $10\; \rm cm = 0.10\; \rm m$ .

Distance between the midpoint of $\rm AB$ and each of the three charges:

$d({\rm A}) = 0.050\; \rm m$ .
$d({\rm B}) = 0.050\; \rm m$ .
$d({\rm C}) = \sqrt{3} \times (0.050\; \rm m)$ .

Let $k$ denote Coulomb's constant ( $k \approx 8.99 \times 10^{9}\; \rm N \cdot m^{2} \cdot C^{-2}$ .)

Electric potential due to the charge at $\rm A$ : $\displaystyle \frac{k\, q}{d({\rm A})}$ .

Electric potential due to the charge at $\rm B$ : $\displaystyle \frac{k\, q}{d({\rm B})}$ .

Electric potential due to the charge at $\rm A$ : $\displaystyle \frac{k\, q}{d({\rm C})}$ .

While forces are vectors, electric potentials are scalars. When more than one electric fields are superposed over one another, the resultant electric potential at some point would be the scalar sum of the electric potential at that position due to each of these fields.

Hence, the electric field at the midpoint of $\rm AB$ due to all these three charges would be:

$\begin{aligned}& \frac{k\, q}{d({\rm A})} + \frac{k\, q}{d({\rm B})} + \frac{k\, q}{d({\rm C})} \\ &= k\, \left(\frac{q}{d({\rm A})} + \frac{q}{d({\rm B})} + \frac{q}{d({\rm C})}\right) \\ &\approx 8.99 \times 10^{9}\; \rm N \cdot m^{2} \cdot C^{-2} \\ & \quad \quad \times \left(\frac{1.8 \times 10^{-8} \; \rm C}{0.050\; \rm m} + \frac{1.8 \times 10^{-8} \; \rm C}{0.050\; \rm m} + \frac{1.8 \times 10^{-8} \; \rm C}{\sqrt{3} \times (0.050\; \rm m)}\right) \\ &\approx 8.3 \times 10^{3}\; \rm V\end{aligned}$ .

4 0

3 years ago

A force of 350n Acts on an object of mass 17.5 kg what acceleration does it produce

UNO [17]

Answer:

20 m/s/s

Explanation:

F=ma, 350=17.5 * a, a=20 m/s/s

3 0

2 years ago