Which option is the best blackbody radiator?

An experiment is performed aboard the International Space Station to verify that linear momentum is conserved during collisions

Rom4ik [11]

Answer:

The speed of the combined honey drop after the collision is 9.05 m/s.

Explanation:

Given that,

The masses and velocities of the drops are

$m_{1}=31.5\ g$

$v_{1}=12.7\ m/s$

$m_{2}=49.9\ g$

$v_{1}=12.5\ m/s$

$m_{1}=78.1\ g$

$v_{1}=15.9\ m/s$

We need to calculate the total mass

Using formula of masses

$m=m_{1}+m_{2}+m_{3}$

Put the value into the formula

$m=31.5+49.9+78.1$

$m=159.5\ g$

We need to calculate the velocity in x- direction

Using conservation of momentum

$m_{1}v_{1}=(m_{1}+m_{2}+m_{3})v_{x}$

$v_{x}=\dfrac{m_{1}v_{1}}{m_{1}+m_{2}+m_{3}}$

Put the value into the formula

$v_{x}=\dfrac{31.5\times12.7}{159.5}$

$v_{x}=2.50\ m/s$

We need to calculate the velocity in y- direction

Using conservation of momentum

$m_{2}v_{2}=(m_{1}+m_{2}+m_{3})v_{y}$

$v_{y}=\dfrac{m_{2}v_{2}}{m_{1}+m_{2}+m_{3}}$

Put the value into the formula

$v_{y}=\dfrac{49.9\times12.5}{159.5}$

$v_{y}=3.91\ m/s$

We need to calculate the velocity in z- direction

Using conservation of momentum

$m_{3}v_{3}=(m_{1}+m_{2}+m_{3})v_{z}$

$v_{z}=\dfrac{m_{3}v_{3}}{m_{1}+m_{2}+m_{3}}$

Put the value into the formula

$v_{z}=\dfrac{78.1\times15.9}{159.5}$

$v_{z}=7.78\ m/s$

We need to calculate the combined honey drop after the collision

Using formula of velocity

$v=v_{x}+v_{y}+v_{z}$

Put the value into the formula

$v=2.50i+3.91j+7.78k$

The magnitude of velocity

$v=\sqrt{(2.50)^2+(3.9)^2+(7.78)^2}$

$v=9.05\ m/s$

Hence, The speed of the combined honey drop after the collision is 9.05 m/s.

3 0

3 years ago

You have been pulled over by a police officer. He begins performing the following tests: horizontal gaze nystagmus, walk and tur

bezimeni [28]

1st is A, don't know the rest :(

5 0

3 years ago

Read 2 more answers

An 80.0-kg man jumps from a height of 2.50 m onto a platform mounted on springs. As the springs compress, he pushes the platform

frosja888 [35]

Answer:

6.16 m/s

0.0105 m

Explanation:

Let the ground 0 for potential reference be at where the spring is compress 0.24 m. The the man would jump from a height h = 2.5 + 0.24 = 2.74 m from it. We can apply the law of energy conservation knowing that as the man jumps, his potential energy converts to kinetic energy, then finally to elastic energy:

$E_p = E_e$

$mgh = kx^2/2$

where m = 80 kg is the man mass, g = 9.81 m/s2 is the gravitational acceleration, h = 2.74 m is the potential distance he travels, k N/m is the spring constant and x = 0.24 is the distance it compresses

$80*9.81*2.74 = k0.24^2/2$