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3 years ago

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Two cars started at the same position and were driven through a city using different streets. The red car traveled north 2 miles

, then east 4 miles, and then south 6 miles. The green car traveled south 4 miles and then east 4 miles. Both cars ended at the same location. Which best compares the motion of the cars?

1 answer:

inysia [295]3 years ago

4 0

Answer:

The green car traveled a shorter distance, but the displacement of the cars was equal.

Explanation:

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A person of mass 70 kg stands at the center of a rotating merry-go-round platform of radius 2.9 m and moment of inertia 900 kg⋅m

Cloud [144]

Explanation:

It is given that,

Mass of person, m = 70 kg

Radius of merry go round, r = 2.9 m

The moment of inertia, $I_1=900\ kg.m^2$

Initial angular velocity of the platform, $\omega=0.95\ rad/s$

Part A,

Let $\omega_2$ is the angular velocity when the person reaches the edge. We need to find it. It can be calculated using the conservation of angular momentum as :

$I_1\omega_1=I_2\omega_2$

Here, $I_2=I_1+mr^2$

$I_1\omega_1=(I_1+mr^2)\omega_2$

$900\times 0.95=(900+70\times (2.9)^2)\omega_2$

Solving the above equation, we get the value as :

$\omega_2=0.574\ rad/s$

Part B,

The initial rotational kinetic energy is given by :

$k_i=\dfrac{1}{2}I_1\omega_1^2$

$k_i=\dfrac{1}{2}\times 900\times (0.95)^2$

$k_i=406.12\ rad/s$

The final rotational kinetic energy is given by :

$k_f=\dfrac{1}{2}(I_1+mr^2)\omega_1^2$

$k_f=\dfrac{1}{2}\times (900+70\times (2.9)^2)(0.574)^2$

$k_f=245.24\ rad/s$

Hence, this is the required solution.

5 0

3 years ago

the angular speed of an automobile engine is increased at a constant rate from 1300rev/min to 2000rev/min in 3s (a) what is its

GalinKa [24]

Answer:

please find attached pdf

Explanation:

8 0

3 years ago

Two blocks are connected by a light weight, flexible cord that passes over a frictionless pulley.Ifm1=2 kg and m2 = 3 kg, and bl

Vera_Pavlovna [14]

Answer:

t = 1.41 sec.

Explanation:

If we assume that the acceleration of the blocks is constant, we can apply any of the kinematic equations to get the time since the block 2 was released till it reached the floor.

First, we need to find the value of acceleration, which is the same for both blocks.

If we take as our system both blocks, and think about the pulley as redirecting the force simply (as tension in the strings behave like internal forces) , we can apply Newton's 2nd Law, as they were moving along the same axis, aiming at opposite directions, as follows:

F = m₂*g - m₁*g = (m₁+m₂)*a (we choose as positive the direction of the acceleration, will be the one defined by the larger mass, in this case m₂)

⇒ a = ( $\frac{(m₂-m₁)}({m₁+m₂} * g$ = g/5 m/s²

Once we got the value of a, we can use for instance this kinematic equation, and solve for t:

Δx = 1/2*a*t² ⇒ t² = (2* 1.96m *5)/g = 2 sec² ⇒ t = √2 = 1.41 sec.

6 0

3 years ago

If the ball shown in the figure lands in 0.5 s, about what height was it thrown from?

Ede4ka [16]

Answer: 1.22 m

Explanation:

The equation of motion in this situation is:

$y=y_{o}+V_{oy}t-\frac{g}{2}t^{2}$ (1)

Where:

$y=0$ is the final height of the ball

$y_{o}=h$ is the initial height of the ball

$V_{oy}=V_{o}sin(0\°)=0$ is the vertical component of the initial velocity (assuming the ball was thrown vertically and there is no horizontal velocity)

$t=0.5 s$ is the time at which the ball lands

$g=9.8 m/s^{2}$ is the acceleration due gravity

So, with these conditions the equation is rewritten as:

$h=\frac{g}{2}t^{2}$ (2)

$h=\frac{9.8 m/s^{2}}{2}(0.5 s)^{2}$ (3)

Finally:

$h=1.22 m$

4 0

3 years ago

Explain what happens to a light wave that hits the surface of a pool.

brilliants [131]

A light wave that hits the surface of a pool gets refracted and gives us an apparent image of the surface of the pool, following the concepts of refraction.

<u>Explanation:</u>

Let’s recall the concept of refraction when a light wave passes from medium of rarer to denser. There is a change in the speed of light while travelling from medium of rarer to denser.

There can be a change in the direction as well. This property is known as “Refraction” and the best example to see refraction is watching the surface of a clean pond, lake or pool.

When the light travels from a rarer medium (air) to a denser medium (water), it changes its angle of direction and gets refracted and hit to our eye lenses. With this, we see the surface of the pool at a changed angle and it seems to be a bit shallow than its original depth.

5 0

3 years ago