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2 years ago

Over time Pangaea broke apart to form other continents.

Physics

1 answer:

Travka [436]2 years ago

6 0

Answer:

jwhgrewhuejqiwmkosjcdihwbfuqjiwdkmojcshidvwuf hiiii againnnn :)) good luck

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What is the kinetic energy in joules of a 0.05. kg bullet traveling 310 m/s

Wittaler [7]

The formula is=1/2(m x v^2)

so = 1/2*(0.05)*(310)^2

ans is =2402.5 joules

3 0

3 years ago

What is the kinetic energy of a 3-kilogram bali that is rolling at 2 meters per second

alex41 [277]

6 jules

hope this helps

3 0

2 years ago

Read 2 more answers

Which of the following Illustrates 2 resistors in a series circult? A, B, C, D.

Kaylis [27]

Answer:

It's either B or D, I'm not positive which it is

Explanation:

6 0

2 years ago

An unfortunate astronaut loses his grip during a spacewalk and finds himself floating away from the space station, carrying only

Gnesinka [82]

Answer:

15.75 m/s

Explanation:

v = Velocity of the combined mass of astronaut and tools = 1.8 m/s

$m_1$ = Mass of astronaut = 124 kg

$m_2$ = Mass of tools = 16 kg

$v_1$ = Velocity of astronaut = 0

$v_2$ = Velocity of tools

As linear momentum is conserved

$m_1v_1 + m_2v_2 =(m_1 + m_2)v\\\Rightarrow v_2=\frac{(m_1 + m_2)v-m_1v_1}{m_2}\\\Rightarrow v_2=\frac{(124+16)\times 1.8-124\times 0}{16}\\\Rightarrow v_2=15.75\ m/s$

The velocity of the tools is 15.75 m/s

3 0

2 years ago

Baseball player a bunts the ball by hitting it in such a way that it acquires an initial velocity of 2.4 m/s parallel to the gro

LUCKY_DIMON [66]

Let $\mathbf r_A$ denote the position vector of the ball hit by player A. Then this vector has components

$\begin{cases}r_{Ax}=\left(2.4\,\frac{\mathrm m}{\mathrm s}\right)t\\r_{Ay}=1.2\,\mathrm m-\frac12gt^2\end{cases}$

where $g=9.8\,\dfrac{\mathrm m}{\mathrm s^2}$ is the magnitude of the acceleration due to gravity. Use the vertical component $r_{Ay}$ to find the time at which ball A reaches the ground:

$1.2\,\mathrm m-\dfrac12\left(9.8\,\dfrac{\mathrm m}{\mathrm s^2}\right)t^2=0\implies t=0.49\,\mathrm s$

The horizontal position of the ball after 0.49 seconds is

$\left(2.4\,\dfrac{\mathrm m}{\mathrm s}\right)(0.49\,\mathrm s)=12\,\mathrm m$

So player B wants to apply a velocity such that the ball travels a distance of about 12 meters from where it is hit. The position vector $\mathbf r_B$ of the ball hit by player B has

$\begin{cases}r_{Bx}=v_0t\\r_{By}=1.6\,\mathrm m-\frac12gt^2\end{cases}$

Again, we solve for the time it takes the ball to reach the ground:

$1.6\,\mathrm m-\dfrac12\left(9.8\,\dfrac{\mathrm m}{\mathrm s^2}\right)t^2=0\implies t=0.57\,\mathrm s$

After this time, we expect a horizontal displacement of 12 meters, so that $v_0$ satisfies

$v_0(0.57\,\mathrm s)=12\,\mathrm m$

$\implies v_0=21\,\dfrac{\mathrm m}{\mathrm s}$

5 0

3 years ago