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liberstina [14]
3 years ago
5

There are usually _________ collisions in a motor vehicle crash.

Physics
1 answer:
mel-nik [20]3 years ago
5 0

Answer:

3

Explanation:

During a crash 3 types of collisions can occur.

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A 10 kg mass car initially at rest on a horizontal track is pushed by a horizontal force of 10 N magnitude. If we neglect the fr
vlada-n [284]

Answer:

50 m

Explanation:

F = ma

10 N = (10 kg) a

a = 1 m/s²

Given:

v₀ = 0 m/s

a = 1 m/s²

t = 10 s

Find: Δx

Δx = v₀ t + ½ at²

Δx = (0 m/s) (10 s) + ½ (1 m/s²) (10 s)²

Δx = 50 m

5 0
2 years ago
What effect does increasing mass have on the amount of friction generated
LekaFEV [45]
The greater the mass, the greater the fiction generated
4 0
3 years ago
A driver travels 135 km, east in 1.5 h, stops for 45 minutes for lunch, and then resumes driving for the next 2.0 h through a di
Oksi-84 [34.3K]
<h2>82.353 km/hr</h2>

Explanation:

       The driver travels 135 km towards East in 1.5 hr. He stops for 45 min. He again travels 215 km towards East in 2.0 hr.

       The total displacement of the driver in the given time is ths sum of individual displacements, because all the displacements are in the same directon.

       Total displacement = 135\text{ }km\text{ }East\text{ }+\text{ }0\text{ }km\text{ }+\text{ }215\text{ }km\text{ }East\text{ }=\text{ }350\text{ }km\text{ }East

       Total time travelled = 1.5\text{ }hr\text{ }+\text{ }45\text{ }min\text{ }+\text{ }2.0\text{ }hr\text{ }=\text{ }90\text{ }min\text{ }+\text{ }45\text{ }min\text{ }+\text{ }120\text{ }min\text{ }=\text{ }255\text{ }min\text{ }=\text{ }4\text{ }hr\text{ }15\text{ }min\text{ }=\text{ }4.25\text{ }hr

      \text{Average velocity = }\dfrac{\text{Total displacement}}{\text{Total time taken}}=\dfrac{350\text{ }km}{4.25\text{ }hr}=82.353\text{ }\frac{km}{hr}

∴ Driver's average velocity = 82.353\text{ }\frac{km}{hr}

3 0
3 years ago
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mestny [16]

Answer: D

Explanation:

5 0
3 years ago
Read 2 more answers
A 2kg hockey puck is sliding across the ice skating rink at 2 m/s. A player hits the puck so it's velocity increases to 10 m/s.
konstantin123 [22]

The work done on the puck is 96 J

Explanation:

According to the work-energy theorem, the work done on the hockey puck is equal to the change in kinetic energy of the puck.

Mathematically:

W=K_f -K_i= \frac{1}{2}mv^2-\frac{1}{2}mu^2

where

K_f = \frac{1}{2}mv^2 is the final kinetic energy of the puck, with

m = 2 kg being the mass of the puck

v = 10 m/s is the final speed

K_i = \frac{1}{2}mu^2 is the initial kinetic energy of the puck, with

u = 2 m/s being the initial speed of the puck

Substituting numbers into the equation, we find the work done by the player on the puck:

W=\frac{1}{2}(2)(10)^2 - \frac{1}{2}(2)(2)^2=96 J

Learn more about work and kinetic energy:

brainly.com/question/6763771  

brainly.com/question/6443626  

brainly.com/question/6536722

#LearnwithBrainly

6 0
3 years ago
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