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Basile [38]
3 years ago
9

You Fire A Bullet Into A 3 Kg Wooden Block Attached To A Spring With Spring Constant 70 N/m. When The Bullet Strikes The Block,

The Spring Compresses. The Bullet Has A Mass Of 0.03 Kg And Strikes The Block With A Speed Of 200 M/s. What Is The Speed Of The Block And Bullet When They First Start Moving Together?
Physics
1 answer:
brilliants [131]3 years ago
7 0

here at the moment when bullet strikes the block it will have no external force on it

so here we can use momentum conservation

so we will have

P_i = P_f

m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}

0.03(200) + 3(0) = 0.03v + 3v

6 = 3.03v

v = \frac{6}{3.03}

v = 1.98 m/s

so they move together with speed 1.98 m/s

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Speed and Motion you went from the starting line to the finish line at different rates. If you repeated the activity while carry
Phantasy [73]

Answer:

It will cause kinetic energy to increase.

Explanation:

Given that Speed and Motion you went from the starting line to the finish line at different rates.

If you repeated the activity while carrying weights but keeping your times the same, the weight carried will add up to the mass of the body.

And since Kinetic energy K.E = 1/2mv^2

Increase in the mass of the body will definitely make the kinetic energy of the body to increase.

Since the time is the same, that means the speed V is the same.

Weight W = mg

m = W/g

The new kinetic energy will be:

K.E = 1/2(M + m)v^2

This means that there will be increase in kinetic energy.

3 0
3 years ago
1. Si comprimes un globo hasta reducir su volumen a un tercio de su valor original, ¿cuánto aumenta la presión en su interior?
harina [27]

Answer:

See step by step sexplanation

Explanation:

1.-Sabemos que la relación:

P₁ * V₁  = P₂ * V₂

Para una temperatura constante debe mantenerse entonces si el globo se comprime hasta llevarlo a 1/3 de su valor inicial, entonces necesariamente para cumplir con la relación mencionada, la presión aumenta tres veces su valor original

2.-La definición de presión es fuerza por unidad de superficie, entonces la fuerza es determinada por la altura de la columna de liquido en el recipiente y no por la cantidad total de liquido, de acuerdo a esto habrá más presión en la base del florero, ya que la columna de agua tiene más altura.

3.-No se puede estar de acuerdo con el criterio del plomero. En su solución no plantea el aumento de la altura del tanque,  para el logro del aumento de la presión que es realmente lo que hay que hacer

4 0
3 years ago
Describe how one plays Dr.Dogeball​
JulsSmile [24]
•To play Dr. Dodgeball you need to have 2 teams to verse each other.
•Next, select one person from each team to be the doctor (depending on the size of the teams you can have varying amounts of doctors)
•Continue to play dodgeball how you normally would
•When a player gets hit and is “out” they have to sit on the ground and wait for the doctor to “revive them” (this usually requires the doctor dragging,touching, or moving the player that is out to a “revival place” which is usually decided on by the advisor or person in charge.
•Finally, try to get all the doctors and players out from the other team. Get the doctors first, for they cannot revive themselves. Which means the other players are out after they get hit with a ball since the doctors are out. (Some games are played where if all doctors are out the game ends)
Hope this helped! Play on! And plz mark brainliest lol this was long to write :D
4 0
3 years ago
A projectile is launched diagonally into the air and has a hang time of 24.5 seconds. Approximately how much time is required fo
Rasek [7]

Answer:

t=12.25\ seconds

Explanation:

<u>Diagonal Launch </u>

It's referred to as a situation where an object is thrown in free air forming an angle with the horizontal. The object then describes a known path called a parabola, where there are x and y components of the speed, displacement, and acceleration.

The object will eventually reach its maximum height (apex) and then it will return to the height from which it was launched. The equation for the height at any time t is

x=v_ocos\theta t

\displaystyle y=y_o+v_osin\theta \ t-\frac{gt^2}{2}

Where vo is the magnitude of the initial velocity, \theta is the angle, t is the time and g is the acceleration of gravity

The maximum height the object can reach can be computed as

\displaystyle t=\frac{v_osin\theta}{g}

There are two times where the value of y is y_o when t=0 (at launching time) and when it goes back to the same level. We need to find that time t by making y=y_o

\displaystyle y_o=y_o+v_osin\theta\ t-\frac{gt^2}{2}

Removing y_o and dividing by t (t different of zero)

\displaystyle 0=v_osin\theta-\frac{gt}{2}

Then we find the total flight as

\displaystyle t=\frac{2v_osin\theta}{g}

We can easily note the total time (hang time) is twice the maximum (apex) time, so the required time is

\boxed{t=24.5/2=12.25\ seconds}

4 0
3 years ago
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Rzqust [24]
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.
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3 years ago
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