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2 years ago

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A firecracker is thrown downward from a height of 2.75m above the ground, with a speed of 3.15m/s. Ignore air resistance, determ

ine the height above the ground that the firecracker would be moving at 5.23m/s.

1 answer:

3241004551 [841]2 years ago

4 0

Here in this question as we can see there is no air friction so we can use the principle of energy conservation

$PE_i + KE_i = PE_f + KE_f$

$mgh_1 + \frac{1}{2}mv_i^2 = mgh_2 + \frac{1}{2}mv_f^2$

now here we know that

$h_1 = 2.75 m$

$v_i = 0$

$v_f = 5.23 m/s$

now plug in all values in above equation

$mg*2.75 + 0 = mgh + \frac{1}{2}m(5.23)^2$

divide whole equation by mass "m"

$9.8*2.75 = 9.8*h + \frac{1}{2}*27.35$

$9.8*h = 13.27$

$h = 1.35 m$

so height of the ball from ground will be 1.35 m

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A cyclist moving towards right with an acceleration of 4m/s² at t = 0 he has travelled 5 m moving towards the right at 15 m/s wh

ira [324]

Answer:

$x=2t^2+15t+5$

Explanation:

$x=\frac{1}{2}at^2+v_0t+x_0$

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2 years ago

Q 32.35: The isotope 235U decays by alpha emission with a half-life of 7.0 x 108 y. It also decays (rarely) by spontaneous fissi

Julli [10]

Answer:

rate of fission =5.89*10^3 1\Year

Explanation:

we know that

rate of fission is given as $rate of fission = \frac{0.69}{(T_{1/2})_{fission}} *\frac{ Mass* Avogardo\ number}{Molar\ mass}$

$(T_{1/2})_{fission} = 3*10^{17} y$

mass = 1.0 g

avogardo number = 6.02*10^23

molar mass of isotopes 235U =235

Putting all value to get rate of emission

rate of fission $= \frac{0.69}{3*10^{17}} *\frac{1.0*6.02*10^{23}}{235}$

rate of fission =5.89*10^3 1\Year

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2 years ago

Emmy kicks a soccer ball up at an angle of 45° over a level field. She watches the ball's trajectory and notices that it lands,

Elenna [48]

Let $u$ be the initial velocity of the soccer ball at an angle of inclination of $\theta_0$ with the positive x-axis.

Given that:

$\theta_0=45^{\circ}$

The horizontal distance covered by the projectile=20 m

Time of flight, $t_f=2$ seconds

Acceleration due to gravity, $g= 10 m/s^2$ downward.

As "north" and "up" as the positive x ‑ and y ‑directions, respectively.

So, $g= -10 m/s^2$

As the acceleration due to gravity is in the vertical direction, so the horizontal component of the initial velocity remains unchanged.

The x-component of the initial velocity, $u_x=u\cos\theta_0$ .

The horizontal distance covered by the projectile $= u_x\times t_f$

$\Rightarrow u_x\times t_f=20$

$\Rightarrow u_x\times 2=20$

$\Rightarrow u_x=10$ m/s

So, the horizontal component of the velocity is 10 m/s which is constant and the graph has been shown in the figure (i).

Now, $u\cos(45^{\circ})=10$ [as $u_x=u\cos\theta_0$ ]

$\Rightarrow u=10\sqrt{2}$ m/s.

The vertical component of the initial velocity,

$u_y= u\sin\theta_0$

$\Rightarrow u_y=10\sqrt{2}\sin(45^{\circ})$

$\Rightarrow u_y=10$ m/s

Let v be the vertical component of the velocity at any time instant t.

From the equation of motion,

$v=u+at$

where u: initial velocity, v: final velocity, a: constant acceleration, and t: time taken to change the velocity from u to v.

In this case, we have $u=u_y, a= -10 m/s^2$ .

So at any time instant, t.

$v=u_y+(-10)t$

$\Rightarrow v=10-10t$

The vertical component of the velocity, v, is the function of time and related as $v=10-10t$ .

This is a linear equation.

At 2 second, the vertical component of the velocity

v=10-10x2=-10 m/s.

The graph has been shown in figure (ii).

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2 years ago

If a force always acts perpendicular to an object's direction of motion, that force cannot change the object's kinetic energy.

laiz [17]

This is very good conceptual question and can clear your doubts regarding work-energy theorem.
Whenever force is perpendicular to the direction of the motion, work done by that force is zero.
According to work-energy theorem,
Work done by all the force = change in kinetic energy.

here, work done = 0.
Therefore,
0=change in kinetic energy
This means kinetic energy remains constant.
Hope this helps

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2 years ago

High‑speed ultracentrifuges are useful devices to sediment materials quickly or to separate materials. An ultracentrifuge spins

pantera1 [17]

Answer:

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Explanation:

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ω = 50000 rev/min * 1rev /( 2π rad ) * 1min / 60s = 132.63 rad/s

Now we can calculate centripetal force as:

$Fc = m * \frac{V^2}{R} = m * \frac{(\omega*R)^2}{R}=m*R*\omega ^2$

Replacing the values we get the answer:

Fc = 7.14N

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2 years ago