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Nadya [2.5K]
3 years ago
9

Do you think ill look better in a french braid or a bun for a figure skating show.

Physics
2 answers:
avanturin [10]3 years ago
8 0
Depends on what music you are skating too. I've done both but I prefer a french braid. 
Setler [38]3 years ago
6 0
Hmmm...both seem beautiful, but if you're good at keeping them in then either on if fine, but buns make you seem elegant. While french braids make you look...I don't know...natural? But in my opinion, buns are easy to do and make people look good.
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A bumblebee darts past at 3 m/s. The frequency of the hum made by its wings is 152 Hz. Assume the speed of sound to be 342 m/s.
Veronika [31]

It'll be 152 Hz at the exact instant the bumblebee
is right at the tip of your nose, on his way past you.

Before he gets there, while he's coming at you,
he sounds like a frequency higher than 152 Hz.

After he passes by, and is going away from you,
he sounds like a frequency lower than 152 Hz.

8 0
3 years ago
Read 2 more answers
What is the gravitational force between two identical 5000 kg asteroids whose centers of mass are separated by 100 m?
Alenkinab [10]

Answer: 1.67 x 10^-7N

Explanation:

5 0
2 years ago
A racehorse is running with a uniform speed of 69 km/hr along a straightaway. what is the time it takes for the horse to cover 4
Alex
Hello there,
400 meters= 0.4 km
Time= Distance / speed
        = 0.4 / 69
        = 0.0057971014492754 hr
        = 0.35 min

Hope this helps :))

~Top
 
8 0
3 years ago
Read 2 more answers
You have been assigned to investigate a traffic accident. The masses of car A and car B are 1300 kg and 1200 kg, respectively. C
jarptica [38.1K]

Answer:

The velocity of A before impact = 17.90 m/s

Explanation:

Coefficient of restitution = (speed of seperation)/(speed of approach)

= (v₁ - v₂)/(u₂ - u₁)

where v₁ = velocity of the car A after the impact = ?

v₂ = velocity of the car B after the impact = ?

u₂ = velocity of the car B before the impact = 0 m/s (it was initially at rest)

u₁ = velocity of car A before the impact = ?

First of, we can solve for v₂, the velocity of car B after the impact, from some of the information given in the question.

- Skid marks indicate car B slid 10 m after the impact

- The coefficient of kinetic friction the tires and road is 0.8.

According to the work energy theorem, the work done by frictional force in stopping the car B is equal to the change in kinetic energy of the car B. (All after collision)

W = ΔK.E

ΔK.E = (1/2)(1200)(v₂²) - 0 (final kinetic energy is 0 since the car comes to stop eventually)

ΔK.E = (600v₂²) J

W = F × d

where F = frictional force = μmg = 0.8×1300×9.8 = 10,192 N

d = distance the car skids over before stopping = 10 m

W = 10,192 × 10 = 101,920 J

W = ΔK.E

101,920 = 600v₂²

v₂² = (101920/600) = 169.867

v₂ = 13.03 m/s

But recall,

Coefficient of restitution = (v₁ - v₂)/(u₂ - u₁)

For the sake of convention, we take the direction of car A's initial velocity to be the positive direction.

u₁ = ?

u₂ = 0 m/s

v₁ = ?

v₂ = +13.03 m/s

Coefficient of restitution = 0.4

0.4 = (v₁ - 13.03)/(0 - u₁)

-0.4u₁ = v₁ - 13.03

v₁ = 13.03 - 0.4u₁

But this is a collision. In a collision, the linear momentum is usually conserved.

Momentum before collision = Momentum after collision

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

1300u₁ + (1200×0) = 1300v₁ + (1200×13.03)

1300u₁ + 0 = 1300v₁ + 15639.95

1300u₁ = 1300v₁ + 15639.95

But recall, from the coefficient of restitution relation,

v₁ = 13.03 - 0.4u₁

Substituting this into the momentum balance equation.

1300u₁ = 1300v₁ + 15639.95

1300u₁ = 1300(13.03 - 0.4u₁) + 15639.95

1300u₁ = 16943.28 - 520u₁ + 15639.95

1820u₁ = 32,583.23

u₁ = (32,583.23/1820)

u₁ = 17.90 m/s

Therefore, the velocity of A before impact = 17.90 m/s

Hope this Helps!!!

4 0
3 years ago
Which landform is the farthest north?
quester [9]
The North Pole would be your answer
7 0
2 years ago
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