Elements that typically give up electrons

Part 1: What is your Reaction Time?

nlexa [21]

Answer:

1. A ruler is a common instrument used for measuring the length of small objects

2. The acceleration due to gravity at the surface of Earth is about 9.8 metres per second per second.

8 0

3 years ago

Two metal disks, one with radius R1 = 2.45 cm and mass M1 = 0.900 kg and the other with radius R2 = 5.00 cm and mass M2 = 1.60 k

larisa86 [58]

Answer:

part (a) $a_1\ =\ 2.9\ kg$

Part (b) $a_2\ =\ 6.25\ kg$

Explanation:

Given,

Mass of the larger disk = $M_2\ =\ 1.60\ kg$
Mass of the smaller disk = $M_1\ =\ 0.900\ kg$
Radius of the larger disk = $R_2\ =\ 5.00\ cm\ =\ 0.05\ m$
Radius of the smaller disk = $R_1\ =\ 2.45\ cm\ =\ 0.0245\ m$
Mass of the block = M = 1.60 kg

Both the disks are welded together, therefore total moment of inertia of the both disks are the summation of the individual moment of inertia of the disks.

$\therefore I\ =\ I_1\ +\ I_2\\\Rightarrow I\ =\ \dfrac{1}{2}M_1R_1^2\ +\ \dfrac{1}{2}M_2R_2^2\\\Rightarrow I\ =\ \dfrac{1}{2} (0.9\times 0.0245^2\ +\ 1.60\times 0.05^2)\\\Rightarrow I\ =\ 2.27\times 10^{-3}\ kgm^2$

part (a)

Given that a block of mass m which is hanging with the smaller disk,

Let 'T' be 'a' be the tension in the string and acceleration of the block.

From the free body diagram of the smaller block,

$mg\ -\ T\ =\ ma\\\Rightarrow T\ =\ mg\ -\ ma\,\,\,\,eqn (1)$

From the pulley,

$\sum \tau\ =\ I\alpha\\\Rightarow T\times R_1\ =\ I\alpha\ =\ \dfrac{Ia}{R_1}\\\Rightarrow T\ =\ \dfrac{I\alpha}{R_1^2}\,\,\,eqn(2)$

From the equation (1) and (2),

$mg\ -\ ma\ =\ \dfrac{Ia}{R_1^2}\\\Rightarrow a\ =\ \dfrac{mg}{\dfrac{I}{R_1^2}\ +\ m}\\\Rightarrow a\ =\ \dfrac{1.60\times 9.81}{\dfrac{2.27\times 10^{-3}{0.0245^2}}\ +\ 1.60}\\\Rightarow a\ =\ 2.91\ m/s^2$

part (b)

Above expression for the acceleration of the block is only depended on the radius of the pulley.

Radius of the larger pulley = $R_2\ =\ 0.05\ m$

Let $a_2$ be the acceleration of the block while connecting to the larger pulley. $\therefore a\ =\ \dfrac{mg}{\dfrac{I}{R_2^2}\ +\ m}\\\Rightarrow a\ =\ \dfrac{1.60\times 9.81}{\dfrac{2.27\times 10^{-3}{0.05^2}\ +\ 1.60}}\\\Rightarow a\ =\ 6.25\ m/s^2$

4 0

3 years ago

Radium-226 emits alpha (α), beta (β) and gamma (γ) radiation.

Furkat [3]

Answer:

Explanation:

alpha

Alpha Radiation (α): A large, unstable nucleus decays to produce a smaller, more stable nucleus and an alpha particle (identical to a helium nucleus, ⁴₂He or ⁴₂α).

It has a very high ionizing energy and low penetrating power. It can be stopped by paper skin

Beta Radiation (β): A neutron in an unstable nucleus decays, forming a proton and emitting a beta (β) particle (identical to an electron, ⁰₋₁e or ⁰₋₁b) and resulting in a more stable nucleus.

It has high ionizing energy and penetrating power. It can be stopped by aluminium sheet

Gamma Radiation (γ): An unstable nucleus releases energy in the form of a high energy photon (no mass)to become more stable; this often accompanies other forms of radioactivity.

It has very high penetrating power and very low ionizing energy. It can be stopped by lead block.

3 0

4 years ago

a parallel plate capacitor has square plates that have edge length equal to 100cm and are separated by 1 mm. It is connected to

Flura [38]

Answer:

the energy is stored in the capacitor is 0.32 μJ

Explanation:

Given;

distance of separation, d = 1 mm = 0.001 m

edge length of the square = 100 cm

potential difference across the plates, V = 12 v

let the side of the square = L

This edge length is also the diagonal of the square which makes a right angle with the side of the square.

Applying Pythagoras theorem;

L² + L² = 100²

2L² = 100²

L² = 100²/2

Note area of a square is L²

A = L² = 100²/2 = 5000 cm²

A (m²) = 5000 cm² x 1m²/(100 cm)²

A = 5000 cm² x 1m²/10000 cm²

A = 0.5 m²

Energy stored in a parallel plate capacitor, E= ¹/₂CV²

C = ε₀A/d

where;

ε₀ is permittivity of free space = 8.85 x 10⁻¹² F/m

d is the distance of separation = 0.001 m

A is the area of the plate

C = ε₀A/d = (8.85 x 10⁻¹²)x0.5 / 0.001

C = 4425 x 10⁻¹² F

E = ¹/₂CV² = ¹/₂ x 4425 x 10⁻¹² x ( 12)²

E = 318600 x 10⁻¹² = 0.32 μJ

Therefore, the energy is stored in the capacitor is 0.32 μJ

8 0

3 years ago

Read 2 more answers

N/a<br>n/a<br>nnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnn

disa [49]

Answer:

hiii

Can I get the Brainiest?

7 0

3 years ago