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DENIUS
1 month ago
14

For a short time the position of a roller-coaster car along its path is defined by the equations r=25 m, θ=(0.3t) rad, and z=(−8

cosθ) m, where t is measured in seconds, Determine the magnitudes of the car's velocity and acceleration when t=4s .

Physics
1 answer:
ValentinkaMS [3.4K]1 month ago
8 0

Answer:

Velocity = v = 2.24 m/s

Acceleration = a = 0.20 m/s²

Explanation:

Refer to the attachment below.

Given

z=(−8 cosθ) and θ = 0.3t

z = -8Cos (0.3t)

V = dz/dt

a = v²/R.

Please see full solution below.

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The acceleration of segment D is m/s2. Rank segments A, B , C from least accelerations to greatest acceleration. Least
Ostrovityanka [3204]

Answer:

D, C, B, A

Explanation:

The derivative from a velocity-time graph provides the acceleration value.

Segment A

\frac{dy}{dx} = \frac{15m/s}{1s} = 15m/s^2

Segment B

\frac{dy}{dx} = \frac{5m/s}{1s} = 5m/s^2

Segment C

\frac{dy}{dx} = \frac{0m/s}{2s} = 0m/s^2

Segment D

\frac{dy}{dx} = \frac{-20m/s}{1s} = -20m/s^2

Sorted from the lowest to the highest acceleration:

D, C, B, A

8 0
1 month ago
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A window washer on a hanging platform cleans the outside of windows on a skyscraper. The washer hoists the platform up the side
kicyunya [3294]

Answer:

The acceleration of the platform is - 1.8 m/s²

Explanation:

The net force on a body causes that body to accelerate in the direction of the resultant force applied.

Setting up the force equilibrium for the configuration:

ma = 800 - mg

100a = 800 - 100×9.8

100a = - 180

100a = - 180

a = - 1.8 m/s²

This indicates that the body is falling downward.

6 0
2 months ago
A skateboarder with mass ms = 54 kg is standing at the top of a ramp which is hy = 3.3 m above the ground. The skateboarder then
serg [3582]

Response:

A) W_{ff} =-744.12J

B) F_f=-W_{ff}*sin\theta /hy = 112.75N

C) F_{f2}=207.58N

Clarification:

The question is not fully provided. The complete question was:

A skateboarder with a mass of ms = 54 kg is at the top of a ramp with a height of hy = 3.3 m. He then jumps on his skateboard and goes down the ramp. His speed at the base is vf = 6.2 m/s.  

Part (a) Formulate an expression for the work, Wf, done by the frictional force on the skateboarder in terms of the variables listed in the problem.

Part (b) The ramp is at an angle θ with the ground, where θ = 30°. Formulate an expression for the frictional force's magnitude, fr, between the skateboarder and the ramp.

Part (c) Upon reaching the bottom, the skateboarder continues with speed vf onto a grass-covered flat surface. The friction between the grass and the skateboarder brings him to a halt after 5.00 m. Determine the frictional force, Fgrass in newtons, between the skateboarder and the grass.

For part A), we assess the energy balance to determine the work done by the friction:

W_{ff}=\Delta E

W_{ff}=1/2*m*vf^2-m*g*hy

W_{ff}=-744.12J

For part B), we utilize the previously calculated work:

W_{ff}=-F_f*(hy/sin\theta)   Rearranging for friction force:

F_f=-W_{ff}*sin\theta /hy

F_f=112.75N

For part C), we first ascertain the acceleration through kinematic equations and subsequently find the frictional force using dynamic methods:

Vf^2=Vo^2+2*a*d

Rearranging for 'a':

a=-3.844m/s^2

Now, using dynamics:

|F_f|=|m*a|

|F_f|=207.58N

8 0
2 months ago
"As the Voyager spacecraft penetrated into the outer solar system, the illumination from the Sun declined. Relative to the situa
serg [3582]

Answer:

\frac{I_{2}}{I_{1}} = 0.04

Explanation:

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I_{1}\cdot r_{1}^{2} = I_{2}\cdot r_{2}^{2}

The brightness of sunlight on Jupiter as compared to Earth is:

\frac{I_{2}}{I_{1}} = \frac{r_{1}^{2}}{r_{2}^{2}}

\frac{I_{2}}{I_{1}} = \left(\frac{1\,AU}{5.2\,AU} \right) ^{2}

\frac{I_{2}}{I_{1}} = 0.04

3 0
1 month ago
A transverse wave is traveling from north to south. Which statement could be true for the motion of the wave particles in the me
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Transverse waves propagate in a direction that is at right angles to the movement of the particles (or the medium involved). Hence, the particles would be shifting from east to west, which is perpendicular to the north-south direction of the wave.
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