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3 years ago

Please help with these i dont know how to do them

Physics

1 answer:

Svetradugi [14.3K]3 years ago

3 0

20.0M/S. IS THE HORIZONTAL VELOCITY OF THE BALL JUST BEFORE IT REACHES THE GROUND

12.2 SECONDS IS THE APPROXIMATE TOTAL TIME REQUIRED FOR THE BALL

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In an attempt to impress its friends, an acrobatic beetle runs and jumps off the bottom step of a flight of stairs. The step is

Aleks04 [339]

Answer:

0.3677181864 m

Explanation:

u = Velocity = 1.5 m/s

$\theta$ = Angle = 20°

y = -20 cm

Velocity components

$u_x=ucos\theta\\\Rightarrow u_x=1.5cos20\\\Rightarrow u_x=1.40953\ m/s$

$u_y=usin\theta\\\Rightarrow u_y=1.5sin20\\\Rightarrow u_y=0.51303\ m/s$

Acceleration components

$a_x=0$

$a_y=-9.81\ m/s^2$

$y=u_yt+\dfrac{1}{2}a_yt^2\\\Rightarrow -0.2=0.51303\times t+\dfrac{1}{2}\times -9.81t^2\\\Rightarrow 4.905t^2-0.51303t-0.2=0$

$t=\frac{-\left(-0.51303\right)+\sqrt{\left(-0.51303\right)^2-4\cdot \:4.905\left(-0.2\right)}}{2\cdot \:4.905}, \frac{-\left(-0.51303\right)-\sqrt{\left(-0.51303\right)^2-4\cdot \:4.905\left(-0.2\right)}}{2\cdot \:4.905}\\\Rightarrow t=0.26088, -0.15629$

Time taken is 0.26088 seconds

$x=u_xt+\dfrac{1}{2}a_xt^2\\\Rightarrow x=1.40953\times 0.26088\\\Rightarrow x=0.3677181864\ m$

The distance the beetle travels on the ground is 0.3677181864 m

6 0

3 years ago

8.)The potential energy of a 40-kg cannonball is 14000 J. How high was the cannon ball to have this much potential energy?

Delvig [45]

Hey! I empathize with you, as I just started the energy unit in physics as well!

Alright, lets list what we know, and which equation(s) we need.

PE = Gravitational Potential Energy

$PEg = mgh$

m = mass

g = gravity (9.8 m/s^2)

h = height

Guess what?

We need to find the height!

Lets make this more organized:

Known:

m = 40 kg

PE = 14000

g = 9.8 m/s^2

Unknown:

h = ?

================================

Now, if we take a look at the equation PEg = mgh, you will see that we have everything besides the height! So lets solve for the height by substituting in for the variables we know:

14000 J = 40 kg * (9.8 m/s^2) * h

40 * 9.8 = 392

14000 J = 392 * h

14000/392 = h

35.71 m = h

There we go! If you simply list what you know and don't know, you will find the equation you need to solve the problem.

Have a great day and good luck!

7 0

3 years ago

Does static electricity attract dead skin cells

lesantik [10]

Answer:

no i dont think so

Explanation:

6 0

3 years ago

A small car with mass of 0.800 kg travels at a constant speed

Alexandra [31]

Answer:

The equation of equilibrium at the top of the vertical circle is:

\Sigma F = - N - m\cdot g = - m \cdot \frac{v^{2}}{R}

The speed experimented by the car is:

\frac{N}{m}+g=\frac{v^{2}}{R}

v = \sqrt{R\cdot (\frac{N}{m}+g) }

v = \sqrt{(5\,m)\cdot (\frac{6\,N}{0.8\,kg} +9.807\,\frac{kg}{m^{2}} )}

v\approx 9.302\,\frac{m}{s}

The equation of equilibrium at the bottom of the vertical circle is:

\Sigma F = N - m\cdot g = m \cdot \frac{v^{2}}{R}

The normal force on the car when it is at the bottom of the track is:

N=m\cdot (\frac{v^{2}}{R}+g )

N = (0.8\,kg)\cdot \left(\frac{(9.302\,\frac{m}{s} )^{2}}{5\,m}+ 9.807\,\frac{m}{s^{2}} \right)

N=21.690\,N

7 0

2 years ago

When you go on a hunting trip, you should leave a hunting plan with someone you trust. What information should the plan include?

bulgar [2K]

You should put when you will leave, where you will be, and what time you will get back.

4 0

3 years ago

Read 2 more answers