Ask question

Ask question

All categories

2 years ago

6

In a skate park, you are trying to determine how to get the most speed at the bottom of the ramp. If the ramp is 4.5 m high, and

you start at the top of the ramp with an initial velocity of 1.5m/s, how fast will you be rolling at the bottom of the ramp, given there is no friction. Use conservation of energy to solve.

1 answer:

Oksanka [162]2 years ago

7 0

Answer:

If you have an initial velocity of 1.5m/s and the ramp height is 4.5, and an extreme fart is initiated at the bottom of the ramp exerting an didditional upward force, and the smell of the fart causes you to begin flailing, and everyone running from you causes an additional wind force, then you would be rolling at a speed of 3 m/s given there is no friction. :)

You might be interested in

Imagine a small child whose legs are half as long as her parent’s legs. If her parent can walk at maximum speed V, at what maxim

AnnZ [28]

Answer:

$\boxed{v=\frac {V}{\sqrt {2}}}$

Explanation:

We know that speed is given by dividing distance by time or multiplying length and frequency. The speed of the father will be given by Lf where L is the length of the father’s leg ad f is the frequency.

We know that frequency of simple pendulum follows that $f=\frac {1}{2\pi} \sqrt {\frac {g}{l}}$

Now, the speed of the father will be $V=Lf= L\times (\frac {1}{2\pi} \sqrt {\frac {g}{l}})$ while for the child the speed will be $v=\frac {L}{2}\times (\frac {1}{2\pi} \sqrt {\frac {g}{0.5l}})$

The ratio of the father’s speed to the child’s speed will be

$\frac {V}{v}=\frac {\frac {L}{2}\times (\frac {1}{2\pi} \sqrt {\frac {g}{0.5l}})}{ L\times (\frac {1}{2\pi} \sqrt {\frac {g}{l}})}\\\frac {V}{v}=\frac {\sqrt {2}}{2}\\\boxed{v=\frac {V}{\sqrt {2}}}$

8 0

3 years ago

WILL MARK BRAINLIEST!

boyakko [2]

It’s mass is 100 kg, because if you divide both numbers, you get 100

6 0

3 years ago

A parachutist of mass 56.0 kg jumps out of a balloon at a height of 1400 m and lands on the ground with a speed of 5.10 m/s. How

Vlad [161]

Answer:

769,048.28Joules

Explanation:

A parachutist of mass 56.0 kg jumps out of a balloon at a height of 1400 m and lands on the ground with a speed of 5.10 m/s. How much energy was lost to air friction during this bump

The energy lost due to friction is expressed using the formula;

Energy lost = Potential Energy + Kinetic Energy

Energy lost = mgh + 1/2mv²

m is the mass

g is the acceleration due to gravity

h is the height

v is the speed

Substitute the given values into the formula;

Energy lost = 56(9.8)(1400) + 1/2(56)(5.10)²

Energy lost = 768,320 + 728.28

Energy lost = 769,048.28Joules

<em>Hence the amount of energy that was lost to air friction during this jump is 769,048.28Joules</em>

6 0

3 years ago

How much voltage would be necessary to generate 10 amps of current in a circuit that has 5 ohms of

Lana71 [14]

$U = R * I \rightarrow U = 10 * 5 = 50 V$

8 0

3 years ago

Read 2 more answers

You are at the controls of a particle accelerator, sending a beam of 2.10×107 m/s protons (mass m) at a gas target of an unknown

Kipish [7]

Answer:

a

The mass is $m_2 =21.75*10^{-27} \ kg$

b

The velocity is $v_2 = 3.0*10^{6} m/s$

Explanation:

From the question we are told that

The speed of the protons is $u_1 = 2.10*10^{7} m/s$

The mass of the protons is $m$

The speed of the rebounding protons are $v_1 = -1.80 * 10^{7} \ m/s$

The negative sign shows that it is moving in the opposite direction

Now according to the law of energy conservation mass of one nucleus of the unknown element. is mathematically represented as

$m_2 = [\frac{u_1 -v_1}{u_1 + v_1} ] m_1$

Where $m_1$ is the mass of a single proton

So substituting values

$m_2 = \frac{2.10 *10^{7} - (-1.80 *10^{7})} {(2.10 *10^7) + (-1.80 *10^{7})} m_1$

$m_2 =13 m_1$

The mass of on proton is $m_1 = 1.673 * 10^{-27} \ kg$

So $m_2 =13 ( 1.673 * 10^{-27} )$

$m_2 =21.75*10^{-27} \ kg$

Now according to the law of linear momentum conservation the speed of the unknown nucleus immediately after such a collision is mathematically evaluated as

$m_1 u_1 + m_2u_2 = m_1 v_1 + m_2v_2$

Now $u_2$ because before collision the the nucleus was at rest

So

$m_1 u_1 = m_1 v_1 + m_2v_2$

=> $v_2 = \frac{m_1(u_1 -v_1)}{m_2}$

Recall that $m_2 =13 m_1$

So

$v_2 = \frac{m_1(u_1 -v_1)}{13m_1}$

=> $v_2 = \frac{(u_1 -v_1)}{13}$

substituting values

$v_2 = \frac{( 2.10*10^{7} -(-1.80 *10^{7}))}{13}$

$v_2 = 3.0*10^{6} m/s$

7 0

3 years ago