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4 years ago

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An object is dropped from a bridge. A second object is thrown downward 1.48 s later. They both reach the water 48.1 m below at t

he same instant. What was the initial speed of the second object

1 answer:

pashok25 [27]4 years ago

8 0

To solve this problem we will apply the linear motion kinematic equations. With the data provided we will calculate the time of the first object to fall. Later we will get the time difference between the two. This difference will allow us to find the free fall distance. Through the distance we will find the initial velocity, that is,

$x = v_0 t +\frac{1}{2}at^2$

$48.1 = 0*t + \frac{1}{2} (9.8)t^2$

$t = 3.13s$

The second object is thrown downward at one second later and it meets the first object at the water is

$t' = 3.13 -1.48$

$t' = 1.65s$

The distance of the object will travel due to free fall acceleration is

$x = v_0 t+\frac{1}{2} at^2$

$x = 0*(1.65) +\frac{1}{2}(9.8)(1.65)^2$

$x = 13.34m$

The distance of the object will travel due to its initial velocity is

$v_0 = \frac{d_0}{t}$

$d_0 = v_0 t$

$48.1-13.34 = v_0 (1.65)$

$v_0 = 21.06m/s$

Therefore the initial speed of the second object is 21.06m/s

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A 90 kg ice skater moving at 12.0 m/s on the ice encounters a region of roughed up ice with a coefficient of kinetic friction of

balandron [24]

Answer:

The skater covers a distance of <u>15 m</u> before stopping.

Explanation:

Let the distance traveled before stopping be 'd' m.

Given:

Mass of the skater (m) = 90 kg

Initial velocity of the skater (u) = 12.0 m/s

Final velocity of the skater (v) = 0 m/s (Stops finally)

Coefficient of kinetic friction (μ) = 0.490

Acceleration due to gravity (g) = 9.8 m/s²

Now, we know that, from work-energy theorem, the work done by the net force on a body is equal to the change in its kinetic energy.

Here, the net force acting on the skater is only frictional force which acts in the direction opposite to motion.

Frictional force is given as:

$f=\mu N$

Where, 'N' is the normal force acting on the skater. As there is no vertical motion, $N=mg$

∴ $f=\mu mg=0.490\times 90\times 9.8=432.18\ N$

Now, work done by friction is a negative work as friction and displacement are in opposite direction and is given as:

$W=-fd=-432.18d$

Now, change in kinetic energy is given as:

$\Delta K=\frac{1}{2}m(v^2-u^2)\\\\\Delta K=\frac{1}{2}\times 90(0-12^2)\\\\\Delta K=45\times (-144)=-6480\ J$

Therefore, from work-energy theorem,

$W=\Delta K\\\\-432.18d=6480\\\\d=\frac{6480}{432.18}\\\\d=14.99\approx 15\ m$

Hence, the skater covers a distance of 15 m before stopping.

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3 years ago

If a trapeze artist rotates once each second while sailing through the air, and contracts to reduce her rotational inertia to on

finlep [7]

Answer:

There are finally 4 rotations per second.

Explanation:

If a trapeze artist rotates once each second while sailing through the air, and contracts to reduce her rotational inertia to one fourth of what it was. We need to find the final angular velocity. It is a case of conservation of angular momentum such that :

$I_1\omega_1=I_2\omega_2$

Let $I_1=I$ , $I_2=\dfrac{I}{4}$ and $\omega_1=1$

So,

$\omega_2=\dfrac{I_1\omega_1}{I_2}$

$\omega_2=\dfrac{I\times 1}{(I/4)}$

$\omega_2=4\ rev/sec$

So, there are finally 4 rotations per second. Hence, this is the required solution.

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3 years ago

A rocket takes off from Earth's surface, accelerating straight up at 47.2 m/s2. Calculate the normal force (in N) acting on an a

lions [1.4K]

Answer:

Approximately $4.61\times 10^{3}\; {\rm N}$ upwards (assuming that $g = 9.81\; {\rm m\cdot s^{-2}}$ .)

Explanation:

External forces on this astronaut:

Weight (gravitational attraction) from the earth (downwards,) and
Normal force from the floor (upwards.)

Let $(\text{normal force})$ denote the magnitude of the normal force on this astronaut from the floor. Since the direction of the normal force is opposite to the direction of the gravitational attraction, the magnitude of the net force on this astronaut would be:

$\begin{aligned}(\text{net force}) &= (\text{normal force}) - (\text{weight})\end{aligned}$ .

Let $m$ denote the mass of this astronaut. The magnitude of the gravitational attraction on this astronaut would be $(\text{weight}) = m\, g$ .

Let $a$ denote the acceleration of this astronaut. The magnitude of the net force on this astronaut would be $(\text{net force}) = m\, a$ .

Rearrange $\begin{aligned}(\text{net force}) &= (\text{normal force}) - (\text{weight})\end{aligned}$ to obtain an expression for the magnitude of the normal force on this astronaut:

$\begin{aligned}(\text{normal force}) &= (\text{net force}) + (\text{weight}) \\ &= m\, a + m\, g \\ &= m\, (a + g) \\ &= 80.9\; {\rm kg} \times (47.2\; {\rm m\cdot s^{-2}} + 9.81\; {\rm m\cdot s^{-2}}) \\ &\approx 4.61 \times 10^{3}\; {\rm N}\end{aligned}$ .

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2 seconds is the period (time it takes to complete one cycle)

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