A car traveling at 45 m/s starts to decelerate steadily. It comes to a complete stop in 10 seconds. What is it’s acceleration

A ball with mass 0.11 kg is thrown upward with initial velocity 10 m/s from the roof of a building 20 m high. Neglect air resist

inessss [21]

Answer:

a) $H=25.1020\ m$

b) $t=3.2838\ s$

Explanation:

Given:

mass of the ball thrown up, $m=0.11\ kg$

initial velocity of the ball thrown up, $u=10\ m.s^{-1}$

height above the ground from where the ball is thrown up, $h=20\ m$

a)

Maximum height attained by the ball above the roof level can be given by the equation of motion.

As,

$v^2=u^2-2g.h'$

where:

$v=$ final velocity at the top height of the upward motion $=0\ m.s^{-1}$

$g=$ acceleration due to gravity

$h'=$ height of the ball above the roof

Now,

$0^2=10^2-2\times 9.8\times h'$

$h'=5.10\ m$

Therefore total height above the ground:

$H=h+h'$

$H=20+5.1020$

$H=25.1020\ m$

b)

Now we find the time taken in raching the height $h'$ :

$v=u-gt'$

$v=$ final velocity at the top of the motion $=0\ m.s^{-1}$

So,

$0=10-9.8\times t'$

$t'=1.0204\ s$

Now the time taken in coming down to the ground from the top height:

$H=u'.t_d+\frac{1}{2} g.t_d^2$

where:

$u'=$ is the initial velocity of the ball in course of coming down to ground from the top $=0\ m.s^{-1}$

Here the direction acceleration due to gravity is same as that of motion so we are taking them positively.

$25.1020=0+0.5\times 9.8\times t_d^2$

$t_d=2.2634\ s$

Therefore the total time taken in by the ball to hit the ground after it begins its motion:

$t=t'+t_d$

$t=1.0204+2.2634$

$t=3.2838\ s$

4 0

3 years ago

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].

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