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riadik2000 [5.3K]

2 years ago

10

An elevator is moving upward at 1.20 m/s when it experiences an acceleration of 0.31 downward over a distance of 0.75m. What wil

l its final velocity be?

2 answers:

iren2701 [21]2 years ago

5 0

<span>Info.: Vi = 1.20 m/s, a = 0.31 m/s^2, Δx = 0.75 m, Vf = ?

Vf = sqrt (Vi^2 + 2aΔx) = sqrt ((1.20 m/s)^2 + 2(0.31 m/s^2 * 0.75 m))
= 0.99 m/s</span>

Gemiola [76]2 years ago

5 0

Answer:

The final velocity is 0.99 m/s.

Explanation:

Given that,

Speed = 1.20 m/s

Acceleration = -0.31 m/s²

Distance = 0.75 m

We need to calculate the final velocity

Using equation of motion

$v^2=u^2+2as$

Where, v = final velocity

u = initial velocity

a = acceleration

s = distance

Put the value in the equation

$v^2=1.20^2-2\times0.31\times0.75$

$v=\sqrt{0.975}$

$v=0.99\ m/s$

Hence, The final velocity is 0.99 m/s.

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Answer:

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Explanation:

d₁ = Smaller diameter paintball = 5 cm

d₂ = Larger diameter paintball = 9 cm

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Volume of smaller diameter paintball

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Similarly

$V_2=\frac{4}{24}\pi d_2^3$

Dividing the above two equations, we get

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2 years ago

A small rock is launched straight upward from the surface of a planet with no atmosphere. The initial speed of the rock is twice

Scorpion4ik [409]

If gravitational effects from other objects are negligible, the speed of the rock at a very great distance from the planet will approach a value of $\sqrt{3} v_{e}$

<u>Explanation:</u>

To express velocity which is too far from the planet and escape velocity by using the energy conservation, we get

Rock’s initial velocity , $v_{i}=2 v_{e}$ . Here the radius is R, so find the escape velocity as follows,

$\frac{1}{2} m v_{e}^{2}-\frac{G M m}{R}=0$

$\frac{1}{2} m v_{e}^{2}=\frac{G M m}{R}$

$v_{e}^{2}=\frac{2 G M}{R}$

$v_{e}=\sqrt{\frac{2 G M}{R}}$

Where, M = Planet’s mass and G = constant.

From given conditions,

Surface potential energy can be expressed as, $U_{i}=-\frac{G M m}{R}$

R tend to infinity when far away from the planet, so $v_{f}=0$

Then, kinetic energy at initial would be,

$k_{i}=\frac{1}{2} m v_{i}^{2}=\frac{1}{2} m\left(2 v_{e}\right)^{2}$

Similarly, kinetic energy at final would be,

$k_{f}=\frac{1}{2} m v_{f}^{2}$

Here, $v_{f}=\text { final velocity }$

Now, adding potential and kinetic energies of initial and final and equating as below, find the final velocity as

$U_{i}+k_{i}=k_{f}+v_{f}$

$\frac{1}{2} m\left(2 v_{e}\right)^{2}-\frac{G M m}{R}=\frac{1}{2} m v_{f}^{2}+0$

$\frac{1}{2} m\left(2 v_{e}\right)^{2}-\frac{G M m}{R}=\frac{1}{2} m v_{f}^{2}$

'm' and $\frac{1}{2}$ as common on both sides, so gets cancelled, we get as

$4\left(v_{e}\right)^{2}-\frac{2 G M}{R}=v_{f}^{2}$

We know, $v_{e}=\sqrt{\frac{2 G M}{R}}$ , it can be wriiten as $\left(v_{e}\right)^{2}=\frac{2 G M}{R}$ , we get

$4\left(v_{e}\right)^{2}-\left(v_{e}\right)^{2}=v_{f}^{2}$

$v_{f}^{2}=3\left(v_{e}\right)^{2}$

Taking squares out, we get,

$v_{f}=\sqrt{3} v_{e}$

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2 years ago

An electric drill transfers 200 J of energy into a useful kinetic energy store. It also transfers 44 J of energy by sound and 48

kramer

Answer:

Er = 108 [J]

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To solve this problem we must understand that the total energy is 200 [J]. Of this energy 44 [J] are lost in sound and 48 [J] are lost in heat. In such a way that these energy values must be subtracted from the total of the kinetic energy.

200 - 44 - 48 = Er

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2 years ago

From the edge of a roof you throw a snowball downward that strikes the ground with 100J of kinetic energy. then you throw a seco

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Answer:

The second snowball hits the ground with a kinetic energy of 100 Joules

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From above two equation, we get :

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Answer:

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2 years ago