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dalvyx [7]
3 years ago
15

Name THREE things that impact a collision.

Physics
1 answer:
Lapatulllka [165]3 years ago
4 0

Answer:

The three types of impact that occur are those involving the vehicle, the body of the vehicle occupant, and the organs within the body of the occupant.

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How much energy can be stored in a spring with k = 470 n/m if the maximum possible stretch is 17.0 cm ?
MrRissso [65]
The strain energy stored in a linear spring is
SE = (1/2)*k*x²
where
k =  the spring constant
x =  the extension (or compression) of the spring

Given:
k = 470 N/m
x = 17.0 cm = 0.17 m

Therefore
SE = 0.5*(470 N/m)*(0.17 m)² = 6.7915 J

Answer:  6.8 J (nearest tenth)

6 0
4 years ago
PLS HELP ME<br> How does one stage in a stars life lead to another
Oxana [17]
Just as during formation, when the material contracts, the temperature and pressure increase. This newly generated heat temporarily counteracts the force of gravity, and the outer layers of the star are now pushed outward ( in not sure tho ) I hope this helps
3 0
3 years ago
Read 2 more answers
We can model a pine tree in the forest as having a compact canopy at the top of a relatively bare trunk. Wind blowing on the top
solmaris [256]

Answer:

114.075 N

798.525 Nm

Explanation:

C = Drag coefficient = 0.5

ρ = Density of air = 1.2 kg/m³

A = Surface area = 9 m²

v = Velocity of wind = 6.5 m/s

r = Height of the tree = 7 m

Drag equation

F=\frac{1}{2}\rho CAv^2\\\Rightarrow F=\frac{1}{2}\times 1.2\times 0.5\times 9\times 6.5^2\\\Rightarrow F=114.075\ N

Magnitude of the drag force of the wind on the canopy is 114.075 N

Toque is given by the product of force and radius

\tau=F\times r\\\Rightarrow \tau=114.075\times 7\\\Rightarrow \tau=798.525\ Nm

Torque exerted on the tree, measured about the point where the trunk meets the ground is 798.525 Nm

5 0
3 years ago
A 3kg ball moving at 8 m/s strikes a 2kg ball at rest,if the collision is elastic,what is the speed of the lighter ball if the h
zloy xaker [14]

Answer:

Speed of lighter ball is 4 m/s.

Explanation:

Applying the principle of conservation of linear momentum,

momentum before collision = momentum after collision.

m_{1}u_{1} + m_{2} u_{2} = m_{1}v_{1} - m_{2}v_{2}

m_{1} = 3 kg, u_{1} = 8 m/s, m_{2} = 2 kg, u_{2} = 0 m/s ( since it is at rest), v_{1} = 2 m/s, v_{2} = ?

(3 x 8) + (2 x 0) = (8 x 2) - (2 x v_{2})

24 + 0 = 16 - 2v_{2}

2v_{2} = 16 - 24

2v_{2} = -8

v_{2} = \frac{-8}{2}

   = -4 m/s

This implies that the light ball moves at the speed of 4 m/s in the opposite direction of the heavier ball after collision.

7 0
3 years ago
10 points and brainlyest if possible and right.
Delicious77 [7]
THE answer the table is holding the book up would be a normal force 
6 0
4 years ago
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