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larisa86 [58]
2 years ago
13

A skater moves with 15 m/s velocity in a circle of radius circle of radius 30 m. The ice exerts a center force of 450 N. What is

the mass of the skater?
Physics
1 answer:
wariber [46]2 years ago
6 0
He Hsb hejs us su she ejjj
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Which situation is the best analogy for the doppler effect?
Rudiy27
The best scenario to describe the doppler effect would be listening to the siren of a passing ambulance or fire truck

then it is coming towards you, the pitch is higher, it gets higher as it approaches and peaks as it gets right in front of you. then it drop at once when it passes you and continues to drop till it fades away. this is a classic descrption of the doppler effect
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Think about it: suppose a meteorite collided head-on with mars and becomes buried under mars's surface. what would be the elasti
Ket [755]

Answer:

Perfectly inelastic collision

Explanation:

There are two types of collision.

1. Elastic collision : When the momentum of the system and the kinetic energy of the system is conserved, the collision is said to be elastic. For example, the collision of two atoms or molecules are considered to be elastic collision.

2. Inelastic collision: When the momentum  the system is conserved but the kinetic energy is not conserved, the collision is said to be inelastic. For example, collision of a ball with the mud.

For a perfectly elastic collision, the two bodies stick together after collision.

Here, the meteorite collide with the Mars and buried inside it, the collision is said to be perfectly inelastic. here the kinetic energy of a body lost completely during the collision.

4 0
3 years ago
PLEASE HELP!!! GIVING BRAINLIEST!! ill also answer questions that you have posted if you answer this correctly!!!! (40pts)
melisa1 [442]
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2 years ago
What type of motion occurs when an object spins around an axis without altering its linear position?
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8 0
3 years ago
Read 2 more answers
Three point charges lie in a straight line along the y-axis. a charge of q1 = -9.10 µc is at y = 6.30 m, and a charge of q2 = -7
inysia [295]

Answer:

 Electric field E = kQ/r^2  

Distance between charges = 6.30 - (-4.40) = 10.70m  

Say the neutral point, P, is a distance d from q1. This means it is a distance (10.70 - d) from q2.  

Field from q1 at P = k(-9.50x^10^-6) / d^2  

Field from q2 at P = k(-8.40x^10^-6) / (10.70-d)^2  

These fields are in opposite directions and are equal magnitudes if the resultant field = 0  

k(-9.50x^10^-6) / d^2 = k(-8.40x^10^-6) / (10.70-d)^2  

9.50 / d^2 =8.40 / (10.70-d)^2  

d^2 / (10.70-d)^2 = 9.50/8.40 = 1.131  

d/(10.70-d) = sqrt(1.1331) = 1.063  

d = 1.063 ((10.70-d)  

= 10.63 - 1.063d  

2.063d = 10.63  

d = 5.15m  

The y coordinate where field is zero is 6.30 - 5.15 = 1.15m

Explanation:

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