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mariarad [96]
3 years ago
15

During a trampoline routine, a gymnast is tumbling in the air at 20 rad/s in a tuck position. He then extends into a layout posi

tion and doubles his radius of gyration just before landing on the trampoline bed. How fast is his angular velocity at this instant, just before landing on the trampoline bed? Group of answer choices 5 rad/s 10 rad/s 5 m/s 4 rad/s 80 rad/s
Physics
1 answer:
Yuki888 [10]3 years ago
5 0

Answer:

\omega = 5 rad/s

Explanation:

As we know that the gymnast has no external torque on it

so here we can say that angular momentum of the system will be conserved

so here we have

L_1 = L_f

I_1\omega_1 = I_2\omega_2

m_1r_1^2\omega_1 = m_1r_2^2\omega

r_1^2\omega_1 = r_2^2\omega

here we know that

r_2 = 2r_1

r_1^2\omega_1 = (2r_1)^2\omega

20 = 4\omega

\omega = 5 rad/s

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If two cars A and B are moving with velocity 60 km/hr and 80 km/hr
Vitek1552 [10]

Answer:

VAB = 20km/hr

Explanation:

<u>Given the following data;</u>

Velocity of car A, VA = 60km/hr

Velocity of car B, VB = 80km/hr

To find the relative velocity of B w.r.t A, VAB;

Since the two cars are moving in the same direction, we have;

VAB = VB - VA

Substituting into the equation, we have;

VAB = 80 - 60

<em>VAB = 20km/hr</em>

Therefore, the relative velocity of car B with respect to car A is 20 kilometers per hour.

3 0
3 years ago
Two children are throwing a ball back-and-forth straight across the back seat of a car. The ball is being thrown 7 mph relative
Evgesh-ka [11]

Answer:

the ball will fly in AX direction, making angle of 8.84° from the motion of the car

Explanation:

Given the data in the question and as illustrated in the diagram below;

Now, Lets assume line AB represent the movement of the car,

AC is the movement of the ball been thrown back and forth in the back seat

Ax is the motion of the ball it flies off the window

so from the diagram, We can see triangle ABC

where AB is 45 mph and AC = 7 mph

and angle ∠CAB = 90°

using SOH CAH TOA

TOA; tanθ = Opposite / Adjacent

tanθ = Opposite / Adjacent

tan( ∠ ABC ) = AC /  AB

we substitute

tan( ∠ ABC ) = 7 /  45

tan( ∠ ABC ) = 0.15555

( ∠ ABC ) = tan⁻¹ 0.15555

( ∠ ABC ) = 8.84°

Therefor, angle ( ∠ ABC )  is 8.84°

Meaning angle ( ∠ XAA' ) is also 8.84°

Therefore, the ball will fly in AX direction, making angle of 8.84° from the motion of the car

4 0
3 years ago
An object with kinetic energy k explodes into two pieces, each of which moves with twice the speed of the original object.
zlopas [31]
<span>Assuming that the momenta of the two pieces are equal: when they have equal velocities, then the masses of the two pieces are also equal. Since there is no force from outside of the system, the center of mass moves on with the same velocity as before the equation. So the two pieces must fly at the side side of the mass center, i.e., they must always be at 90° to the side of the mass center. Otherwise it would not be the mass center, respectively the pieces would not have equal velocities. This is only possible, when the angle of their velocity with the initial direction is 60°. Because, cos (60°) = 1/2 = v/(2v).</span>
6 0
3 years ago
Describe what a hydrogen bond is. 6th grade answer
r-ruslan [8.4K]

Answer:

<em>Hydrogen bond is the attractive force between the hydrogen attached electronegative atom </em>

Explanation:

8 0
3 years ago
Given a particle that has the velocity v(t) = 3 cos(mt) = 3 cos (0.5t) meters, a. Find the acceleration at 3 seconds. b. Find th
dalvyx [7]

Answer:

Explanation:

a )  V = 3 cos(0.5t)

differentiating with respect to t

dv /dt = -3 x .5 sin0.5t

= -1.5 sin0.5t.

acceleration = - 1.5 sin 0.5t

when t = 3 s

acceleration = - 1.5 sin 1.5

= - 1.496 ms⁻²

v = 3 cos.5t

b )  dx/dt = 3 cos 0.5 t

dx = 3 cos 0.5 t dt

integrating on both sides

x = 3 sin .5t / .5

x = 6 sin0.5t

At t = 2 s

x = 6 sin 1

x = 5.05 m

4 0
3 years ago
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