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3 years ago

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1) An ice skater with a moment of inertia of 2.2 kg m^2 rotates at a frequency of 0.8 rotations per second. The ice skater tucks

his/her arms and legs in to achieve a final frequency of 1.5 rotations per second. Find the final moment of inertia of the ice skater.

1 answer:

trapecia [35]3 years ago

4 0

Answer:

Explanation:

2.3 kg·m/s²

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I need help ASAP plzzzz

Fiesta28 [93]

Answer:

a) 1.75s b) 17.2 m/s (down)

Explanation:

d1= 15m d2= 0m (because it hits ground)

a= -9.81 m/s^2 t=???

Equation

the triangle means change in so d2-d1

Δd= v1 * t + 1/2 * a * t^2

0m-15m= v1*t + 1/2 a t^2

-15 m= 0m/s*t (goes away) + 1/2* a *t^2

-15mx2= t^2

-15mx2/a= t^2

Square root (-30/-9.81m/s^2)

t=1.75 s

b) now v2!!

Im going to use v2= v1 + a*t

v2= 0m/s + -9.81 x 1.75s

v2 = -17.2 m/s or you can say 17.2 m/s down!!!

7 0

3 years ago

A shell is fired at an angle of 35° above the horizontal at a velocity of 40 m/s. (a) What is it's speed at the highest point of

Fudgin [204]

Answer:

Part a)

$v_f = v_x = 32.77 m/s$

Part b)

T = 4.68 s

Explanation:

Part a)

Shell is fired at speed of 40 m/s at angle of 35 degree

so here we have

$v_x = 40 cos35 = 32.77 m/s$

$v_y = 40 sin35 = 22.94 m/s$

since gravity act opposite to vertical speed of the shell so at the highest point of its trajectory the vertical component of the speed will become zero

so at the highest point the speed is given

$v_f = 32.77 m/s$

Part b)

After completing the motion we know that the displacement of the object will be zero in Y direction

so we have

$\Delta y = 0$

$0 = v_y t - \frac{1}{2}gt^2$

$T = \frac{2v_y}{g}$

$T = \frac{2(22.94)}{9.81} = 4.68 s$

7 0

3 years ago

Two electrons are initially at rest separated by a distance of 2nm. At time t=0, they start to move apart due to Coulombic repul

Gnom [1K]

Answer:

$t=2.5\times 10^{-14}\ s$

Explanation:

We know that charge on electron

$q=1.6\times 10^{-19}\ C$

r= 2 nm

We know that force between two charge given

$F=K\dfrac{Q_1Q_2}{r^2}$

Now by putting the value

$F=9\times10^9\dfrac{1.6\times 10^{-19}\times 1.6\times 10^{-19}}{(2\times 10^{-9})^2}$

$F=5.67\times 10^{-11}\ N$

We know that mass of electron

The mass of electron

$m=9.1\times 10^{-31}\ kg$

F= m a

a= Acceleration of electron

a= F/m

$a=\dfrac{5.67\times 10^{-11}}{9.1\times 10^{-31}}\ m/s^2$

$a=6.2\times 10^{19} m/s^2$

$S=ut+\dfrac{1}{2}at^2$

initial velocity given that zero ,u=0

$20\times 10^{-9}=\dfrac{1}{2}\times 6.2\times 10^{19} t^2$

$t=\sqrt {\dfrac{40\times 10^{-9}}{6.2\times 10^{19}}}$

$t=2.5\times 10^{-14}\ s$

3 0

3 years ago

2.Cars were previously manufactured to be as sturdy as possible, whereas today's cars

Marina CMI [18]

Answer:

Crumple zones are designed to absorb and redistribute the force of a collision. ... Also known as a crush zone, crumple zones are areas of a vehicle that are designed to deform and crumple in a collision. This absorbs some of the energy of the impact, preventing it from being transmitted to the occupants.

4 0

3 years ago

The motion of a car on a position-time graph is represented with a horizontal line. What does this indicate about the car's moti

Art [367]

Answer:

B

Explanation:

on USAtestprep

7 0

2 years ago