Physics College

A car company is doing crash tests and has installed forcesensors in the chairs and seatbelts to measure the forces thatare exerted on crash test dummies during a crash. In oneparticular test the sensors measure a horizontal net force onthe dummy of 39,400 N when the car comes to a complete stopfrom an initial speed of 35 m/s in 1.4 m. Treating the dummyas a point particle and assuming it comes to rest in the samedistance, what is the mass of the test dummy used during thistest

Answers

Answer 1

Answer:

[tex]m=90.1kg[/tex]

Explanation:

From the question we are told that

Force F=39400

Initial speed [tex]V_1=35m/s[/tex]

Distance[tex]d=1.4m[/tex]

Generally the equation for acceleration is mathematically given by

[tex]a=\frac{V^2-U^2}{2d}[/tex]

[tex]a=\frac{0^2-35^2}{21.4}[/tex]

[tex]a=\frac{0^2-35^2}{21.4}[/tex]

[tex]a=-437.5m/sec^2[/tex]

Generally the equation for mass is mathematically given by

[tex]F=ma\\m=F/a[/tex]

[tex]m=\frac{-39400}{-437.5}[/tex]

[tex]m=90.1kg[/tex]

Related Questions

Calculate the x-component of the electric field at point P due to charge Q1. Write your answer in units of N/C.

Answers

This question is incomplete, the complete question is;

A positive charge of magnitude Q1 = 6.5 nC is located at the origin.

A negative charge Q2 = -3.5 nC is located on the positive x-axis at x = 16.5 cm from the origin. The point P is located y = 10.5 cm above charge Q2.

Calculate the x-component of the electric field at point P due to charge Q1. Write your answer in units of N/C.

Answer:

the x-component of the electric field at point P due to charge Q1 is 1291.33 N/C    

Explanation:

Given the data in the question;

Q1 = 6.5 nC, Q2 = -3.5 nC

from the image below, to get our angle ∅

tan∅ = opp/adj

tan∅ = 10.5 / 16.5

tan∅ = 0.636363

∅ = tan⁻¹( 0.636363 )

∅ = 32.47°

also, r1 = √( 16.5² + 10.5²)

r1 = √( 272.25 + 110.25 )

r1 = √382.5

r1 = 19.55 cm = 0.1955 m

Now, the x-component of the electric field at point P due to charge Q1 will be;

Ex = E2cos32.47°

= (kQ1/r1²)cos32.47°

we know that;  k is Coulomb's law constant ( 9 × 10⁹ N.m²/ C²

Q1 = 6.5 nC = 6.5 × 10⁻⁹ C

so we substitute

= ((9 × 10⁹ × 6.5 × 10⁻⁹) / (0.1955)²) cos32.47°

= (58.5 / 0.03822025) × 0.843672

= 1291.33 N/C

Therefore, the x-component of the electric field at point P due to charge Q1 is 1291.33 N/C

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