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3 years ago

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A skateboarder rolls off a ledge that is 1.12 m high, and lands 1.48 m from the base of the edge. How much time was he in the ai

r?

1 answer:

nignag [31]3 years ago

3 0

The rider's horizontal motion, and how much ground he covers before he hits it, have nothing to do with how long he takes to hit the ground. The problem is simply: "How long does it take an object to fall 1.12 m from rest ?"

This seems like a good time to use this formula:

Distance fallen from rest = (1/2) (acceleration) (time)²

The problem doesn't tell us what planet the skateboarder is exercising on. I'm going to assume it's on Earth, where the acceleration of gravity is 9.8 m/s². And now, here's the solution to the problem I just invented:

1.12 m = (1/2) (9.8 m/s²) (time)²

Time² = (1.12 m) / (9.8 m/s²)

Time² = 0.1143 sec²

Time = √(0.1143 sec² )

<em>Time = 0.34 second</em>

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The Tevatron acceleator at the Fermi National Accelerator Laboratory (Fermilab) outside Chicago boosts protons to 1 TeV (1000 Ge

Eva8 [605]

Answer:

a) $v = c \cdot 0.04 = 1.2\cdot 10^{7} m/s$

b) $v = c \cdot 0.71 = 2.1\cdot 10^{8} m/s$

c) $v = c \cdot 0.994 = 2.97\cdot 10^{8} m/s$

d) $v = c \cdot 0.999 = 2.997\cdot 10^{8} m/s$

e) $v = c \cdot 0.9999 = 2.999\cdot 10^{8} m/s$

Explanation:

At that energies, the speed of proton is in the relativistic theory field, so we need to use the relativistic kinetic energy equation.

$KE=mc^{2}(\gamma -1) = mc^{2}(\frac{1}{\sqrt{1-\beta^{2}}} -1)$ (1)

Here β = v/c, when v is the speed of the particle and c is the speed of light in vacuum.

Let's solve (1) for β.

$\beta = \sqrt{1-\frac{1}{\left (\frac{KE}{mc^{2}}+1 \right )^{2}}}$

We can write the mass of a proton in MeV/c².

$m_{p}=938.28 MeV/c^{2}$

Now we can calculate the speed in each stage.

a) Cockcroft-Walton (750 keV)

$\beta = \sqrt{1-\frac{1}{\left (\frac{0.75 MeV}{938.28 MeV}+1 \right )^{2}}}$

$\beta = 0.04$

$v = c \cdot 0.04 = 1.2\cdot 10^{7} m/s$

b) Linac (400 MeV)

$\beta = \sqrt{1-\frac{1}{\left (\frac{400 MeV}{938.28 MeV}+1 \right )^{2}}}$

$\beta = 0.71$

$v = c \cdot 0.71 = 2.1\cdot 10^{8} m/s$

c) Booster (8 GeV)

$\beta = \sqrt{1-\frac{1}{\left (\frac{8000 MeV}{938.28 MeV}+1 \right )^{2}}}$

$\beta = 0.994$

$v = c \cdot 0.994 = 2.97\cdot 10^{8} m/s$

d) Main ring or injector (150 Gev)

$\beta = \sqrt{1-\frac{1}{\left (\frac{150000 MeV}{938.28 MeV}+1 \right )^{2}}}$

$\beta = 0.999$

$v = c \cdot 0.999 = 2.997\cdot 10^{8} m/s$

e) Tevatron (1 TeV)

$\beta = \sqrt{1-\frac{1}{\left (\frac{1000000 MeV}{938.28 MeV}+1 \right )^{2}}}$

$\beta = 0.9999$

$v = c \cdot 0.9999 = 2.999\cdot 10^{8} m/s$

Have a nice day!

4 0

4 years ago

Rounded to the nearest whole number what is the atomic mass of platinum?

NISA [10]

195 is the rounded atomic atomic mass of platinum

6 0

3 years ago

Two vectors A and B are added to give a resultant R. The components of A are Ax = -8.0 units and Ay = 6.0 units and the componen

Georgia [21]

Answer:

<em>10.09 units</em>

Explanation:

For the A

Ax = -8.0 units

Ay = 6.0 units

The resultant vector = Ra = $\sqrt{A^2_{x} + A^2_{y} }$

Ra = $\sqrt{(-8)^2 + 6^2 }$ = 10 units

For B

Bx = 1.0 units

By = -1.0 units

The resultant vector = Rb = $\sqrt{B^2_{x} + B^2_{y} }$

Rb = $\sqrt{1^2 + (-1)^2 }$ = $\sqrt{2}$ units

Adding these two vectors A and B together, magnitude of vector R is

R = $\sqrt{R^2_{a} + R^2_{b} }$

R = $\sqrt{10^2 + (\sqrt{2} ) ^2}$ = <em>10.09 units</em>

5 0

4 years ago

NEED HELP ASAP PLEASE

Zarrin [17]

Answer:

D. 3.0 m/s

Explanation:

because I did this in my class

8 0

3 years ago

Change in speed over a given period of time is

madam [21]

Explanation:

Acceleration is the change in speed over a given time period

6 0

3 years ago

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