Use your periodic table to answer the following question.

A 33 kg gun is standing on a frictionless surface. The gun fires a 57.7g bullet with a muzzle velocity of 325m/s. The positive d

Stella [2.4K]

Kinetic Energy = (1/2) mv^2.

m = 57.7 g = 57.7/1000 = 0.00577 kg.
v = 325 m/s.

E = 0.5 * 0.00577 * 325^ 2. Use your calculator.

E = 304.728125 J.

That's the kinetic energy.

7 0

3 years ago

A 480 g falcon reaches a speed of 75 m/s in a vertical dive. if we assume that the falcon speeds up under the influence of gravi

poizon [28]

Answer:

286.7 m

Explanation:

So we are assuming the PE of the falcon is converted to KE

KE = PE

1/2 (.480)(75)^2 = .480 (9.81)(h ) solve for h = 286.7 m

8 0

2 years ago

At an outdoor market, a bunch of bananas is set into oscillatory motion with an amplitude of 25 cm on a spring with a spring con

KiRa [710]

Answer:

0.11m/s

Explanation:

To solve the exercise it is necessary to apply the concepts related to the conservation of both: kinetic and spring energy(Elastic potential energy), in this way

Kinetic Energy = Elastic potential energy

$KE = SE$

$\frac{1}{2}mv^2 = \frac{1}{2}kX^2$

Where,

m=mass

v=velocity

k=spring constant

x=amount of compression

Re-arrange the equation to find the velocity we have,

$v^2 = \frac{kX^2}{m}$

$v^2 = \frac{10(0.25)^2}{50}$

$v = \sqrt{0.0125}$

$v = 0.11m/s$

Therefore the maximum speed of the bananas is 0.11m/s

8 0

3 years ago

The severity of a fall depends on your speed when you strike the ground. All factors but the acceleration from gravity being the

Diano4ka-milaya [45]

Answer:

<em>The object could fall from six times the original height and still be safe</em>

Explanation:

<u>Free Falling</u>

When an object is released from rest in free air (no friction), the motion is completely dependant on the acceleration of gravity g.

If we drop an object of mass m near the Earth surface from a height h, it has initial mechanical energy of

$U=m.g.h$

When the object strikes the ground, all the mechanical energy (only potential energy) becomes into kinetic energy

$\displaystyle K=\frac{1}{2}m.v^2$

Where v is the speed just before hitting the ground

If we know the speed v is safe for the integrity of the object, then we can know the height it was dropped from

$\displaystyle m.g.h=\frac{1}{2}m.v^2$

Solving for h

$\displaystyle h=\frac{m.v^2}{2mg}=\frac{v^2}{2g}$

If the drop had occurred in the Moon, then

$\displaystyle h_M=\frac{v_M^2}{2g_M}$

Where hM, vM and gM are the corresponding parameters on the Moon. We know v is the safe hitting speed and the gravitational acceleration on the Moon is g_M=1/6 g

$\displaystyle h_M=\frac{v^2}{2\frac{1}{6}g}$

$\displaystyle h_M=6\frac{v^2}{2g}=6h$

This means the object could fall from six times the original height and still be safe

6 0

3 years ago

An object executing simple harmonic motion has a maximum speed of 4.3 m/s and a maximum acceleration of 0.65 m/s2 . find (a) the

Alexxx [7]

(1) The position around equilibrium of an object in simple harmonic motion is described by
$x(t) = A \cos (\omega t)$
where
A is the amplitude of the motion
$\omega$ is the angular frequency.

The velocity is the derivative of the position:
$v(t)=-\omega A \sin(\omega t) = -v_0 \sin (\omega t)$
where
$v_0 = \omega A$ is the maximum velocity of the object.

The acceleration is the derivative of the velocity:
$a(t)=- \omega^2 A \cos (\omega t) = -a_0 \cos (\omega t)$
where
$a_0=\omega^2 A$ is the maximum acceleration of the object.

We know from the problem both maximum velocity and maximum acceleration:
$v_0 = \omega A = 4.3 m/s$
$a_0 = \omega^2 A = 0.65 m/s^2$
From the first equation, we get
$A= \frac{4.3 }{\omega}$ (1)
and if we substitute this into the second equation, we find the angular fequency
$\omega=0.15 rad/s$
while the amplitude is (using (1)):
$A=28.7 m$

(b) We found in the previous step that the angular frequency of the motion is
$\omega=0.15 rad/s$
But the angular frequency is related to the period by
$T= \frac{2 \pi}{\omega}$
and so, the period is
$T= \frac{2 \pi}{\omega}= \frac{2 \pi}{0.15 rad/s}=41.9 s$

5 0

3 years ago