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tekilochka [14]

3 years ago

9

A skater is spinning at 15.0 rad/s with a rotational inertia of 8.0 kg m? If she changes her rotational inertia to 5.0 kg mº, wh

at is her new angular velocity (in rad/s)?

1 answer:

mixas84 [53]3 years ago

3 0

Answer:Idrk man

Explanation:

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an object weighs 98 n on earth. How much does it weigh on planet x where the acceleration due to gravity in 6 m/s^2

Degger [83]

60 N because 98N=mg (here g= 9.8 on earth) thus mass can be calculated which is 98/9.8 = 10kg Now,new weight with g = 6m/s^2 =m×g' (here g' is new acceleration of the new planet) = 10×6=60N

7 0

3 years ago

Read 2 more answers

In a lab experiment, a student is trying to apply the conservation of momentum. Two identical balls, each with a mass of 1.0 kg,

Studentka2010 [4]

Answer:

Second Trial satisfy principle of conservation of momentum

Explanation:

Given mass of ball A and ball B $=\ 1.0\ Kg.$

Let mass of ball $A$ and $B\ is\ m$

Final velocity of ball $A\ is\ v_1$

Final velocity of ball $B\ is\ v_2$

initial velocity of ball $A\ is\ u_1$

Initial velocity of ball $B\ is\ u_2$

Momentum after collision $=mv_1+mv_2$

Momentum before collision $= mu_1+mu_2$

Conservation of momentum in a closed system states that, moment before collision should be equal to moment after collision.

Now, $mu_1+mu_2=mv_1+mv_2$

Plugging each trial in this equation we get,

First Trial

$mu_1+mu_2=mv_1+mv_2\\1(1)+1(-2)=1(-2)+1(-1)\\1-2=-2-1\\-1=-3$

momentum before collision $\neq$ moment after collision

Second Trial

$mu_1+mu_2=mv_1+mv_2\\1(.5)+1(-1.5)=1(-.5)+1(-.5)\\.5-1.5=-.5-.5\\-1=-1$

moment before collision $=$ moment after collision

Third Trial

$mu_1+mu_2=mv_1+mv_2\\1(2)+1(1)=1(1)+1(-2)\\2+1=1-2\\3=-1$

momentum before collision $\neq$ moment after collision

Fourth Trial

$mu_1+mu_2=mv_1+mv_2\\1(.5)+1(-1)=1(1.5)+1(-1.5)\\.5-1=1.5-1.5\\-.5=0$

momentum before collision $\neq$ moment after collision

We can see only Trial- 2 shows the conservation of momentum in a closed system.

4 0

3 years ago

Read 2 more answers

How do you find the velocity after a collision

Evgen [1.6K]

Answer:

In a collision, the velocity change is always computed by subtracting the initial velocity value from the final velocity value. If an object is moving in one direction before a collision and rebounds or somehow changes direction, then its velocity after the collision has the opposite direction as before.

Explanation:

7 0

4 years ago

A diver springs upward from a board that is 2.70 m above the water. At the instant she contacts the water her speed is 10.9 m/s

TiliK225 [7]

Answer:

vo=5.87m/s

Explanation:

Hello! In this problem we have a uniformly varied rectilinear movement.

Taking into account the data:

α =69.2

vf = 10m / s

h=2.7m

g=9.8m/s2

We know we want to know the speed on the y axis.

We calculate vfy

vfy = 10m / s * (sen69.2) = 9.35m / s

We can use the following equation.

$vf^{2} =vo^{2}+2*g*h\\$

We clear the vo (initial speed)

$vo=\sqrt{vf^{2}-2*g*h }$

$v0=\sqrt{(9.35m/s)^{2}-2*9.8m/s^{2} *2.7m}$

vo=5.87m/s

7 0

3 years ago

A5 pound object is moving horizontally with a speed of 15m/s. How much force is needed to keep the object

Sindrei [870]

15ms cause look at the ans

6 0

3 years ago