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VashaNatasha [74]

3 years ago

6

What will be the final velocity of a 5.0 g bullet starting from rest, if a net force of 45 N is applied over a time of 0.02s?

1 answer:

Wewaii [24]3 years ago

3 0

We know there’s a change in momentum due to a force applied over a time interval. Ft= m[v(final)-v(initial)]. Now simply plug in know values: (45)(0.02)=.005[v(final)-0]. Remember converting grams to kilograms. Solve for v final

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We have a toy gun with a spring constant of 50 N/m. The spring is compressed by 0.2 m. If you neglect friction and the mass of t

Arisa [49]

Answer:

$31.6\:\mathrm{m/s}$

Explanation:

The elastic potential energy of a spring is given by $Us=\frac{1}{2}kx^2$ , where $k$ is the spring constant of the spring and $x$ is displacement from point of equilibrium.

When released, this potential energy will be converted into kinetic energy. Kinetic energy is given by $KE=\frac{1}{2}mv^2$ , where $m$ is the mass of the object and $v$ is the object's velocity.

Thus, we have:

$Us=KE,\\\frac{1}{2}kx^2=\frac{1}{2}mv^2$

Substituting given values, we get:

$\frac{1}{2}\cdot 50\cdot 0.2^2=\frac{1}{2}\cdot 0.002\cdot v^2,\\v^2=\frac{50\cdot 0.2^2}{0.002},\\v^2=1000,\\v\approx \boxed{31.6\:\mathrm{m/s}}$

4 0

3 years ago

A spring gun is made by compressing a spring in a tube and then latching the spring at

inn [45]

Answer:

a) $v=13.2171\,m.s^{-1}$

b) $H=8.9605\,m$

Explanation:

Given:

mass of bullet, $m=4.97\times 10^{-3}\,kg$

compression of the spring, $\Delta x=0.0476\,m$

force required for the given compression, $F=9.12 \,N$

(a)

We know

$F=m.a$

where:

a= acceleration

$9.12=4.97\times 10^{-3}\times a$

$a\approx 1835\,m.s^{-2}\\$

we have:

initial velocity, $u=0\,m.s^{-1}$

Using the eq. of motion:

$v^2=u^2+2a.\Delta x$

where:

v= final velocity after the separation of spring with the bullet.

$v^2= 0^2+2\times 1835\times 0.0476$

$v=13.2171\,m.s^{-1}$

(b)

Now, in vertical direction we take the above velocity as the initial velocity "u"

so,

$u=13.2171\,m.s^{-1}$

∵At maximum height the final velocity will be zero

$v=0\,m.s^{-1}$

Using the equation of motion:

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

where:

h= height

g= acceleration due to gravity

$0^2=13.2171^2-2\times 9.8\times h$

$h=8.9129\,m$

is the height from the release position of the spring.

So, the height from the latched position be:

$H=h+\Delta x$

$H=8.9129+0.0476$

$H=8.9605\,m$

4 0

3 years ago

Visible light represents a very small portion of the electromagnetic spectrum. Radiation to the left in the image, such as micro

amid [387]

X rays have a higher frequency so A

4 0

3 years ago

Read 2 more answers

When the object is at half its amplitude from equilibrium, is the magnitude of its acceleration at half its maximum value?

Nezavi [6.7K]

We have F = kx or ma = kx where m and k are constants. Therefore, if x is halved, a must be halved too.

5 0

3 years ago

A 90 kg body is taken to a planet where the acceleration due to

Serggg [28]

Answer:

2250N

Explanation:

W= mg,

where W= weight

m= mass

g= acceleration due to gravity

Given that the body is 90kg, m= 90kg.

Acceleration due to gravity of planet

= 2.5(10)

= 25 m/s²

Weight of body on planet

= 90(25)

= 2250N

*Mass is the amount of matter an object has and is constant (same on earth and the planet).

6 0

3 years ago

Read 2 more answers