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nikdorinn [45]
3 years ago
10

The diagram shows Ned’s movement as he left his house and traveled to different places throughout the day. Which reference point

should be used to find out how far Ned traveled when he went from the pet store to the swimming pool?
Physics
2 answers:
aleksley [76]3 years ago
6 0

The pet store would be the reference point because it is where he started and it will not move. Hope this helped.

vodomira [7]3 years ago
6 0

Answer:

i think its w

Explanation:

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A disk shaped grindstone of mass 3.0 kg and radius 8 cm is spinning at 600 rpm. After the power is shut off the frictional torqu
seropon [69]

Answer:

\theta=50\ revolution

Explanation:

It is given that,

Mass of the grindstone, m = 3 kg

Radius of the grindstone, r = 8 cm = 0.08 m

Initial speed of the grindstone, \omega_i=600\ rpm=62.83\ rad/s

Finally it shuts off, \omega_f=0

Time taken, t = 10 s

Let \alpha is the angular acceleration of the grindstone. Using the formula of rotational kinematics as :

\alpha =\dfrac{\omega_f-\omega_i}{t}

\alpha =\dfrac{0-62.83}{10}

\alpha =-6.283\ rad/s^2

Let \theta is the number of revolutions of the grindstone after the power is shut off. Now using the third equation of rotational kinematics as :

\omega_f^2-\omega_i^2=2\alpha \theta

\theta=\dfrac{\omega_f^2-\omega_i^2}{2\alpha }

\theta=\dfrac{-62.83^2}{2\times -6.283}

\theta=314.15\ radian

\theta=49.99\ revolution

or

\theta=50\ revolution

So, the number of revolutions of the grindstone after the power is shut off is 50.

7 0
3 years ago
As a raindrop falls from a cloud to the surface of Earth,
Sunny_sXe [5.5K]
The correct answer would be C. it moves at a constant speed. The troposphere(the layer our weather is in) is not nearly high enough for gravity to be different at different altitudes.
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A 91-kg astronaut and a 1300-kg satellite are at rest relative to the space shuttle. The astronaut pushes on the satellite, givi
WARRIOR [948]

Answer:

18.2145 meters

Explanation:

Using the conservation of momentum, we have that:

m1v1 + m2v2 = m1'v1' + m2'v2'

m1 = m1' is the mass of the astronaut, m2=m2' is the mass of the satellite, v1 and v2 are the inicial speed of the astronaut and the satellite (v1 = v2 = 0), and v1' and v2' are the final speed of the astronaut and the satellite. Then we have that:

0 + 0 = 91*v1' + 1300*0.17

v1' = -1300*0.17/91 = -2.4286\ m/s

The negative sign of this speed just indicates the direction the astronaut goes, which is the opposite direction of the satellite.

If the astronaut takes 7.5 seconds to come into contact with the shuttle, their initial distance is:

distance = 2.4286 * 7.5 = 18.2145\ meters

8 0
4 years ago
If the mass of the earth and all objects on it were suddenly doubled, but the size remained the same, the acceleration due to gr
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A circular test track for cars has a circumference of 3.5 kmkm . A car travels around the track from the southernmost point to t
elixir [45]

Answer:

(A) Distance will be equal to 1.75 km

(B) Displacement will be equal to 1.114 km

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We have given circumference of the circular track = 3.5 km

Circumference is given by 2\pi r=3.5

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(a) It is given that car travels from southernmost point to the northernmost point.

For this car have to travel the distance equal to semi perimeter of the circular track

So distance will be equal to =\frac{3.5}{2}=1.75km

(b) If car go along the diameter of the circular track then it will also go from southernmost point to the northernmost point. and it will be equal to diameter of the track

So displacement will be equal to d = 2×0.557 = 1.114 m

8 0
3 years ago
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