How much physical activity should an adult have each week?

7. A ball of mass m makes a head-on elastic collision with a second ball (at rest) and rebounds with a speed equal to 0.450 its

emmasim [6.3K]

Answer:

mass of the second ball is 0.379m

Explanation:

Given;

mass of first ball = m

let initial velocity of first ball = u₁

let final velocity of first ball = v₁ = 0.45u₁

let the mass of the second ball = m₂

initial velocity of the second ball, u₂ = 0

let the final velocity of the second ball = v₂

Apply the principle of conservation of linear momentum;

mu₁ + m₂u₂ = mv₁ + m₂v₂

mu₁ + 0 = 0.45u₁m + m₂v₂

mu₁ = 0.45u₁m + m₂v₂ -------- equation (i)

Velocity for elastic collision in one dimension;

u₁ + v₁ = u₂ + v₂

u₁ + 0.45u₁ = 0 + v₂

1.45u₁ = v₂ (final velocity of the second ball)

Substitute in v₂ into equation (i)

mu₁ = 0.45u₁m + m₂(1.45u₁)

mu₁ = 0.45u₁m + 1.45m₂u₁

mu₁ - 0.45u₁m = 1.45m₂u₁

0.55mu₁ = 1.45m₂u₁

divide both sides by u₁

0.55m = 1.45m₂

m₂ = 0.55m / 1.45

m₂ = 0.379m

Therefore, mass of the second ball is 0.379m (where m is mass of the first ball)

6 0

3 years ago

Inthisexperiment,aspringforcewasusedtokeep moving object travelingin a circularpath.The size of a spring force should be proport

poizon [28]

Answer:

By calculation, it can be shown that;

K = $\frac{F_{angular}}{2\times x\times c}$

Whereby for constant K, as ${F_{angular}}$ increases, x also increases.

Explanation:

The experiment set up consisted of the use of a spring force to maintain the object in circular path.

The energy in the spring is given by

$\frac{1}{2}\cdot k\cdot x^2$ .

Rotational kinetic energy = $\frac{1}{2}$ ·I·ω²

Inertia, I = $\frac{1}{2}$ ·m·r²

ω = $\frac{v}{r}$

Substituting gives

Rotational kinetic energy = $\frac{1}{2}$ ·

= $\frac{1}{4}$ ·m· v²

Equating both equations gives

K = $\frac{2\times m\times v^2}{4\times x^{2} }$ = $\frac{1\times m\times v^2}{2\times x^{2} }$

Within the proportionality limit, x ∝ r

therefore we can write x = c·r which gives

$\frac{ m\times v^2}{2\times x\times c\times r }$ = $\frac{v^{2} }{r} \times\frac{m}{2\times x\times c}$

Since $\frac{v^{2} }{r}$ = angular acceleration, α, then

m× $\frac{v^{2} }{r}$ = Angular force

Therefore K = $\frac{F_{angular}}{2\times x\times c}$

Therefore as Force, F increases, x also increases and the size of a spring force should be proportional to the amount of stretch in the spring.

3 0

3 years ago

The total charge a battery can supply is rated in mA⋅hmA⋅h, the product of the current (in mA) and the time (in h) that the batt

Nat2105 [25]

Answer:

118800 seconds

Explanation:

Given :

Voltage, V = 1.2 V

Resistance, R = 22 Ω

Applying Ohm's law, we get

Voltage, V = IR

Current $I=\frac{V}{R}$

$I=\frac{1.2}{22}$

I = 0.0545 A

Rate = 1800 mAh

Time taken, $t=\frac{1800 \times 10^{-3}}{0.0545}$

= 33 hr

= 118800 s

8 0

3 years ago

Describe the role of minerals in the formation of rocks

klio [65]

<span>Minerals make up rocks. The role of minerals in rock formation is largely dependent on how the rock is formed.</span>

3 0

3 years ago

If a transformer has 50 turns in the primary winding and 10 turns on the secondary winding, what is the reflected resistance in

Elza [17]

Answer:

The reflected resistance in the primary winding is 6250 Ω

Explanation:

Given;

number of turns in the primary winding, $N_P$ = 50 turns

number of turns in the secondary winding, $N_S$ = 10 turns

the secondary load resistance, $R_S$ = 250 Ω

Determine the turns ratio;

$K = \frac{N_P}{N_S} \\\\K = \frac{50}{10} \\\\K = 5$

Now, determine the reflected resistance in the primary winding;

$\frac{R_P}{R_S} = K^2\\\\R_P = R_SK^2\\\\R_P = 250(5)^2\\\\R_P = 6250 \ Ohms$

Therefore, the reflected resistance in the primary winding is 6250 Ω

6 0

3 years ago