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OLga [1]
3 years ago
10

As part of a safety investigation, two 1900 kg cars traveling at 20 m/s are crashed into different barriers. Part A Find the ave

rage force exerted on the car that hits a line of water barrels and takes 1.3 s to stop.
Physics
1 answer:
DedPeter [7]3 years ago
8 0

Answer:

-29.2\times 10^{3} N

Explanation:

We are given that

Mass of cars= m=1900 kg

Initial speed of car=u=20 m/s

Final speed of car=v=0

Time=\Delta t=1.3 s

We have to find the average force exerted on the car.

Average force=\frac{change\;in\;momentum}{\Delta t}

F_{avg}=\frac{mv-mu}{1.3}

F_{avg}=\frac{1900(0)-1900(20)}{1.3}

F_{avg}=\frac{-38000}{1.3}=-29.2\times 10^{3} N

Hence, the average force exerted on the car that hits a line of water barrels=-29.2\times 10^{3} N

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3 years ago
Choose the letter of the best description of the nature of light:
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Why do asteroids and comets differ in composition? asteroids and comets formed at different times. asteroids formed inside the f
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3 years ago
Find the change in thermal energy of a 25kg severed clown doll head that heats up from 25°C to 35°C, and has the specific heat o
timama [110]

Answer:

Q = 425 kJ

Explanation:

Given that,

Mass, m = 25 kg

The clown doll head that heats up from 25°C to 35°C

The specific heat is 1700 J/kg°C

We need to find the internal energy of it. The heat required to raise the temperature is given by the formula as follows :

Q=mc\Delta T\\\\Q=25\times 1700\times (35-25)\\\\Q=425000\ J\\\\Q=425\ kJ

So, 425 kJ of thermal energy is severed.

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3 years ago
Dos cargas puntuales (q1 y q2 ) se atraen inicialmente entre sí con una fuerza de 550 N , si la separación entre ellas es de 90
antiseptic1488 [7]

Answer:

q2 = 9.02*10^{-4}C

Explanation:

To find the value of the other charge you use the Coulomb's law:

F=k\frac{q_1q_2}{r^2}

k: Coulomb's constant = 8.98*10^{9}Nm^2/C^2

q1: charge 1 = 55*10^{-6}C

r: distance between charges = 90cm = 0.9m

F: electric force = 550N

By doing q2 the subject of the formula and replacing you obtain:

q_2=\frac{r^2F}{kq_1}=\frac{(0.9m)^2(550N)}{(8.98*10^9Nm^2/C^2)(55*10^{-6}C)}=9.02*10^{-4}C

hence, the value of the other charge q2 is 9.02*10^{-4}C

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3 years ago
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