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3 years ago

URGENT DUE IN HALF AN HOUR:

1 answer:

7 0

If all the sides are equal, then it is an equilateral triangle. This means after walking 100 meters, you had to turn 1/3 of 180 degrees (because that's what the interior angles of the triangle add up to).

So, 180 * 1/3 = 60 degrees.

We aren't done yet.

60 degrees is not the angle which you have turned from your path.

You turned 120 degrees to describe an angle of 60 degrees on the interior side.

The answer is B.

You might be interested in

Assoli18 [71]

Answer: 40.4M/s

Solution: 46.6/1.15 = 40.4347826 then round it to a single decimal point, since 3 is lower than 5 it will be rounded to 40.4

8 0

3 years ago

During your summer internship for an aerospace company, you are asked to design a small research rocket. The rocket is to be lau

Thepotemich [5.8K]

Answer:

$T=6.75s$

Explanation:

We must separate the motion into two parts, the first when the rocket's engines is on and the second when the rocket's engines is off. So, we need to know the height ( $h_1$ ) that the rocket reaches while its engine is on and we need to know the distance ( $h_2$ ) that it travels while its engine is off.

For solving this we use the kinematic equations:

In the first part we have:

$h_1=v_0T+\frac{1}{2}aT^2\\h_1=0*T+\frac{1}{2}(16\frac{m}{s^2})T^2\\h_1=8\frac{m}{s^2}T^2\\$

and the final speed is:

$v_f=v_0+aT\\v_f=0+16\frac{m}{s^2}T\\v_f=16\frac{m}{s^2}T$

In the second part, the final speed of the first part it will be the initial speed, and the final speed is zero, since gravity slows it down the rocket.

So, we have:

$v_f^2=v_0^2+2gh_2\\2gh_2=v_f^2-v_0^2\\h_2=\frac{v_f^2-v_0^2}{2g}\\h_2=\frac{0^2-(16\frac{m}{s^2}T)^2}{2(-9.8\frac{m}{s^2})}\\h_2=\frac{-256\frac{m^2}{s^4}T^2}{-19.6\frac{m}{s^2}}\\h_2=13.06\frac{m}{s^2}T^2$

The sum of these heights will give us the total height, which is known:

$h=h_1+h_2\\960m=8\frac{m}{s^2}T^2+13.06\frac{m}{s^2}T^2\\960m=21.06\frac{m}{s^2}T^2\\T^2=\frac{960m}{21.06\frac{m}{s^2}}\\T^2=45.58s^2\\T=\sqrt{45.58s^2}\\T=6.75s$

This is the time that its needed in order for the rocket to reach the required altitude.

5 0

3 years ago

A uniform rod of mass M and length L can pivot freely at one end. Initially, the rod is oriented vertically above the pivot, in

Leya [2.2K]

Answer:

The speed of its center of mass = $\sqrt{\frac{3}{2}gL}$

Explanation:

Consider the potential energy at the level of center of mass of rod below the pivot=0

Mass of uniform rod=M

Length of rod=L

The rotational inertia about the end of a uniform rod= $\frac{1}{3}ML^2$

Kinetic energy at the level of center of mass of rod below the pivot= $\frac{1}{2}I\omega^2$

Kinetic energy at the level of center of mass of rod above the pivot=0

Potential energy at the level of center of mass of rod above the pivot=mgh

We have to find the center of mass ( in terms of g and L).

According to conservation of law of energy

Initial P.E+Initial K.E=Final P.E+Final K.E

$mgh+0=0+\frac{1}{2} I\omega^2$

Where $K.E=\frac{1}{2} I\omega^2$

I=Moment of inertia

$\omega$ =Angular velocity

Substitute the values then we get

$MgL=\frac{1}{2}\times \frac{1}{3}ML^2\omega^2$

$\omega^2=\frac{6g}{L}$

Now, we know that $\omega=\frac{v}{r}$ , $r=\frac{L}{2}$

Substitute the values then we get

$\frac{v^2}{(\frac{L}{2})^2}=\frac{6g}{L}$

$\frac{v^24}{L^2}=\frac{6g}{L}$

$v^2=\frac{6g\times L^2}{4L}$

$v^2=\frac{3gL}{2}$

$v=\sqrt{\frac{3}{2}gL}$

Hence, the speed of its center of mass = $\sqrt{\frac{3}{2}gL}$

4 0

4 years ago

What is its speed after 3. 83 as if it accelerates uniformly at −3. 04 m/s 2 ? answer in units of m/s.

weeeeeb [17]

The velocity equation is $v_{final} =v_{initial} +at\\$

Known facts:

t = 3.83s
a= -3.04
intial velocity = 0

Plug into equation known quantities:

$v_{final} = (-3.04) * 3.83 = -11.6432m/s$

Thus the final velocity is -11.6432m/s

Hope that helps!

6 0

3 years ago

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