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2 years ago

7

a 50kg person is standing still in ice skates throws their 2.0kg helmet to the right at 25m/s while on a friction less surface.

what is the final speed of the person

1 answer:

Roman55 [17]2 years ago

8 0

Answer:

Vp = 1 [m/s]

Explanation:

In order to solve this problem, we must use the principle of conservation of the amount of movement. In the same way, analyze the before and after of the actions.

<u>The moment before</u>

The 50kg person is still hold (no movement) with the 2kg helmet

<u>The moment after</u>

The helmet moves at 25[m/s] in one direction, the person moves in the opposite direction, due to the launch of the helmet.

In this way we can apply the principle of conservation of movement, expressing the before and after. To the left we have the before and to the right of the equal sign we have the after.

Σm*V1 = Σm*V

where:

m = total mass = (2 + 50) = 52[kg]

V1 = velocity before the lunch = 0

(50 + 2)*V1 = (25*2) - (Vp*50)

0 = 50 - 50*Vp

50 = 50*Vp

Vp = 1 [m/s]

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A cylinder is sliced in half along its diagonal. Determine the location of the center of mass and the inertia properties relativ

Black_prince [1.1K]

Answer:

See the explanation below

Explanation:

To better understand this problem, a cylinder sketch is attached before and after the cut, we see that after the cut, the shape of this resembles that of a right triangle.

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3 0

2 years ago

Two electric force vectors act on a particle. Their x-components are 13.5 N and −7.40 N and their y-components are −12.0 N and −

guapka [62]

Answer:

Explanation:

Given two vectors as follows

E₁ = 13.5 i -12 j

E₂ = -7.4 i - 4.7 j

Resultant E = E₁ + E₂

= 13.5 i -12 j -7.4 i - 4.7 j

E = 6.1 i - 16.7 j

a ) X component of resultant = 6.1 N

b ) y component of resultant = -16.7 N

Magnitude of resultant = √ ( 6.1² + 16.7² )

= 17.75 N

d ) If θ be the required angle

tanθ = 16.7 / 6.1 = 2.73

θ = 70° .

counterclockwise = 360 - 70 = 290°

6 0

2 years ago

You are called as an expert witness to analyze the following auto accident: Car B, of mass 2000 kg, was stopped at a red light w

OLga [1]

Answer:

a). va=17.23 $\frac{m}{s}$ or 38.54 mph

b). v=38.54 mph and limit is 35 mph

c). Completely inelastic

d). Eka=192.967 kJ

Ekt=76.071 kJ

Explanation:

$m_{a}=1300kg\\m_{b}=2000kg\\x_{f}=7.25m\\u_{k}=0.65$

The motion is an inelastic collision so

$m_{a}*v_{a}+m_{b}*v_{b}=(m_{a}+m_{b})*v_{f}$

The force of the motion is contrarest by the force of friction so

$F-F_{uk} =0\\F=F_{uk}\\F_{uk}=u_{k}*m*g\\F=m*a\\a=\frac{F}{m}\\ a=\frac{F_{uk}}{m}\\a=\frac{u_{k}*m*g}{m}\\a=u_{k}*g\\a=0.65*9.8\frac{m}{s^{2}} \\a=6.39\frac{m}{s^{2}}$

Now with the acceleration can find the time and the velocity final that make the distance 7.25m being united

$x_{f}=x_{o}+v_{o}*t+2*a*t^{2}\\x_{o}=0\\v_{o}=0\\x_{f}=2*a*t^{2}\\t^{2}=\frac{x_{f}}{2*a}\\t=\sqrt{\frac{7.25m}{6.37\frac{m}{s^{2} } } } \\t=1.06s$

So the velocity final can be find using this time

$v_{f}=v_{o}+a*t\\v_{o}=0\\v_{f}=6.37\frac{m}{s^{2} } *1.06s\\v_{f}=6.79 \frac{m}{s}$

a).

Replacing in the first equation the final velocity can find the initial velocity

$m_{a}*v_{a}+m_{b}*v_{b}=(m_{a}+m_{b})*v_{f}$

$v_{b}=0$

$v_{a}= \frac{(m_{a}+m_{b)*v_{f}}}{m_{a}}\\v_{a}= \frac{(1300+2000)*6.37}{1300}\\v_{a}=17.23 \frac{m}{s}$

b).

$35mph*\frac{1m}{0.000621371mi} *\frac{1h}{3600s}=15.646\frac{m}{s}$

Velocity limit in m/s is 15.646 m/s and the initial velocity is 17.23 m/s

so is exceeding the speed limit in about 1.58 m/s

or in miles per hour

3.5 mph

c).

The collision is complete inelastic because any mass can be returned to the original mass, so even they are no the same mass however in the moment they move the distance 7.25m as a same mass the motion is considered completely inelastic

d).

$Ek=\frac{1}{2}*m*(v)^{2}\\ Eka=\frac{1}{2}*1300kg*(17.23\frac{m}{s})^{2}\\Eka=192.967 kJ\\Ekt=\frac{1}{2}*m*(v)^{2}\\Ekt=\frac{1}{2}*3300kg*(6.79\frac{m}{s})^{2}\\Ekt=76.071 kJ$

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2 years ago

A circular pipe of 25-mm outside diameter is placed in an airstream at 25C and 1-atm pressure. The air moves in cross flow over

kifflom [539]

Answer:

f_D = =3.24 N/m

Explanation:

data given

properties of air $\nu\ of air =19.31*10^{-6} m2/s$

$\rho = 1.048 kg/m3$

k = 0.0288 W/m.K

WE KNOW THAT

Reynold's number is given as

$Re =\frac{VD}{\nu}$

$= \frac{ 15*0.025}{19.31*10^{-6}}$

= 1.941 *10^4

drag coffecient is given as

$C_D = \frac{f_D}{A_f\frac{\rho v^2}{2}}$

solving for f_D

$f_D = C_D A_f*\frac{\rho v^2}{2}$

$=C_D D*\frac{\rho v^2}{2}$

Drag coffecient for smooth circular cylinder is 1.1

therefore Drag force is

$f_D = 1.1*0.025 *\frac{1.048*15^2}{2}$

f_D = =3.24 N/m

4 0

2 years ago

If R is the total resistance of three resistors, connected in parallel, with resistances R1, R2, R3, then 1 R = 1 R1 + 1 R2 + 1

ANTONII [103]

Answer:

$\Delta R_{max} = 0.46 ohm$

Explanation:

Resistance is given of different values

now we have

$R_1 = 64 ohm$

$R_2 = 20 ohm$

$R_3 = 8 ohm$

possible error in all resistance is 0.5 %

now we know that

$\Delta R_1 = 0.005\times 64 = 0.32 ohm$

$\Delta R_2 = 0.005 \times 20 = 0.1 ohm$

$\Delta R_2 = 0.005 \times 8 = 0.04 ohm$

Now the maximum possible error when all resistance is connected in series

$\Delta R_{max} = \Delta R_1 + \Delta R_2 + \Delta R_3$

$\Delta R_{max} = 0.32 + 0.1 + 0.04$

$\Delta R_{max} = 0.46 ohm$

7 0

2 years ago