Use Newton’s First Law of Motion and the concept of inertia to explain what happens to a person in a head-on car accident who is not wearing a seat belt. How does wearing a seat belt help?
Use Newton’s First Law of Motion and the concept of - 1

Answers

Answer 1
Answer:

According to Newtons first law of motion, a passenger tends to continue in the state of uniform motion even after the accident which causes the passenger to move forward.

According to the Newtons first law of motion, an object will continue in a state of rest or uniform motion unless it is acted upon by an external force. This is also known as the law of inertia.

While in a car undergoing uniform motion and suddenly the car is involved in an accident which causes the vehicle stop abruptly, a person suddenly moves forward because of the tendency to continue in the former state of uniform motion according to Newtons first law.

A seat belt helps to prevent the person from suddenly moving forward in a way that may lead to injury as a result of the accident.

Learn more: brainly.com/question/974124

Answer 2
Answer:

Answer:

Newtons first law of motion explains what happens in a car crash because it basically states that the passenger will continue to travel at the same velocity until an unbalanced force acts on he or she. The force that will act upon he or she would be the window, so you should always wear a seat belt!

Explanation:


Related Questions

If you see a spacecraft moving past you at 90% of the speed of light, and if you are able to watch a clock on the spacecraft, then according to special relativity you will see the clock
As the mechanical advantage increases what happens to the length of the string you have to pull to raise the block?
Charged objects have a _________ charge. A. net B. moving C. positive D. negative
A car with a mass of 1800 kilograms is traveling at a speed of 29 m/s on a horizontal flat road. Calculate the radius of its path. Which formula must be used to solve this?
What is an object's change in position, or displacement, and the change in time over which that displacement occurred used to calculate?a. average velocity b. inertia c. average force d. force of gravity

A thin rod rotates at a constant angular speed. Consider the tangential speed of each point on the rod for the case when the axis of rotation is perpendicular to the rod at its center and at one end. Explain for each case whether there are any points on the rod that have the same tangential speeds.

Answers

Answer:

    v = R w    

With this expression we see that for each point at different radius the tangential velocity is different

Explanation:

They indicate that the angular velocity is constant, that is

            w = dθ / dt

Where θ is the radius swept angle and t the time taken.

The tangential velocity is linear or

           v = dx / dt

Where x is the distance traveled in time (t)

 

In the definition of radians

          θ = s / R

Where s is the arc traveled and R the radius vector from the pivot point, if the angle is small the arc (s) and the length (x) are almost equal

         θ = x / R

We substitute in the speed equation

         v = d (θ R) / dt

The radius is a constant for each point

         v = R dθ / dt

         v = R w

With this expression we see that for each point at different radius the tangential velocity is different

The _______ the object, the _______ effect a force will have on its motion.A.
softer; less
B.
harder; less
C.
lighter; less
D.
heavier; less

Answers

The correct choice is D. But it's only correct if all the experiments are done on the same planet.

In a machine shop, a hydraulic lift is used to raise heavy equipment for repairs. The system has a small piston with a cross-sectional area of 0.075 m2 and a large piston with a cross-sectional area of 0.237 m2 . An engine weighing 3400 N rests on the large piston. What force must be applied to the small piston in order to lift the engine? Answer in units of N.

Answers

Answer:

F_s=1075.9493\ N

Explanation:

Given:

  • area of piston on the smaller side of hydraulic lift, a_s=0.075\ m^2
  • area of piston on the larger side of hydraulic lift, a_l=0.237\ m^2
  • Weight of the engine on the larger side, W_l=3400\ N

Now, using Pascal's law which state that the pressure change in at any point in a confined continuum of an incompressible fluid is transmitted throughout the fluid at its each point.

P_s=P_l

(F_s)/(a_s)=(W_l)/(a_l)

(F_s)/(0.075) =(3400)/(0.237)

F_s=1075.9493\ N is the required effort force.

Answer:

   F = 1076 N

Explanation:

given,

small piston area, a = 0.075 m²

large piston area, A = 0.237 m²

weight on the large piston, W = 3400 N

force applied on the second piston, F = ?

using pascal law for the force calculation

(F)/(W)=(a)/(A)

(F)/(3400)=(0.075)/(0.237)

   F = 0.3165 x 3400

   F = 1076 N

The force applied to the small piston in order to lift the engine is equal to 1076 N.

A standing wave has a frequency of 471 Hz and a wavelength of 1.9. What is the speed of thewave? (Round to the 2nd number after the decimal)
1
I REALLY NEED HELP !

Answers

Answer:

c = 894.90 m/s

Explanation:

Given data:

Frequency of wave = 471 Hz

Wavelength of wave = 1.9 m

Speed of wave = ?

Solution:

Formula:

Speed of wave = frequency × wavelength

c = f×λ

c = 471 Hz × 1.9 m

  Hz = s⁻¹

c = 471s⁻¹ × 1.9 m

c = 894.90 m/s

The speed of wave is 894.90 m/s.

Compare the sounds produce by wave A and wave B. The human ear would hearA) wave A as a louder sound than wave B.

B) wave A as a higher pitched sound than wave B.

C) wave A and B with the same pitch, but wave A as louder.

D) wave B as a louder and higher pitched sound than wave A.

Answers

The right answer for the question that is being asked and shown above is that: "wave B as a louder and higher pitched sound than wave A." Compare the sounds produce by wave A and wave B. The human ear would hear wave B as a louder and higher pitched sound than wave A.

Answer:

It's A

Explanation:

The human ear would hear wave A as a louder sound than wave B. Wave A has a greater amplitude than wave B. That is perceived as a louder sound. Wave B has a higher frequency; that would be perceived as a higher pitch.

Suppose that now you want to make a scale model of the solar system using the same ball bearing to represent the sun. How far from it would you place a sphere representing the earth? (Center to center distance please.) The distance from the center of the sun to the center of the earth is 1.496×10111.496×1011 m and the radius of the sun is 6.96×1086.96×108 m.

Answers

In this exercise we have to use the knowledge in distance, in this way we will find that the proportional distance found is:

d = 0.645 m

So from the information given in the text we find that:

  • The distance from the center of the sun to the center of the earth is 1.496*10^(11) \ m
  • The radius of the sun is 6.96*10^(8)m
  • We need to assume a radius for the ball bearing, so suppose that the radius is 3 mm  

First, we need to find in what way or manner often the radius of the brightest star exist considerable respect to the range of the ball significance, that exist given apiece following equating:

(r_a)/(r_b)= \frac{6.96*10^8}{3*10{-3}}  =2.32*10^(11)

Now, we can calculate the distance from the center of the sun to the center of the sphere representing the earth:  

d_(s) = (d_(e))/(r_(s)/r_(b)) = (1.496 \cdot 10^(11) m)/(2.32\cdot 10^(11)) = 0.645 m

See more about distance at brainly.com/question/989117

Answer:

d = 0.645 m(assuming a radius of the ball bearing of 3 mm)

Explanation:

The given information is:

  • The distance from the center of the sun to the center of the earth is 1.496x10¹¹m = d_(e)
  • The radius of the sun is 6.96x10⁸m = r_(s)

We need to assume a radius for the ball bearing, so suppose that the radius is 3 mm = r_(b).  

First, we need to find how many times the radius of the sun is bigger respect to the radius of the ball bearing, which is given by the following equation:

(r_(s))/(r_(b)) = (6.96\cdot 10^(8)m)/(3\cdot 10^(-3)m) = 2.32\cdot 10^(11)

Now, we can calculate the distance from the center of the sun to the center of the sphere representing the earth, d_(s):  

[tex] d_{s} = \frac{d_{e}}{r_{s}/r_{b}} = \frac{1.496 \cdot 10^{11} m}{2.32\cdot 10^{11}} = 0.645 m

I hope it helps you!