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Romashka [77]
3 years ago
10

The total momentum of the two cars before a collision is 100 kg m/s. what is the total momentum of the two cars after they colli

de?
Physics
2 answers:
OlgaM077 [116]3 years ago
4 0
The answer is c. 100 kg m/s
Dahasolnce [82]3 years ago
3 0

total momentum of the two cars after they collide= 100 kg m/s

Explanation:

We use law of conservation of momentum which says that , for an isolated system , total momentum before collision is equal to the total momentum after the collision.

Total momentum of two cars before collision= 100 kg m/s

so the total momentum of the two cars after the collision= 100 kg m/s

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How would you find the resistance of a parallel circuit with n identical resistors?
VMariaS [17]

Answer:

write the resistance of one resistor

Explanation:

This is because when resistors are connected in parallel current flows in different directions.

3 0
3 years ago
Read 2 more answers
Conversion fraction 1$=4q, how many are in 20$
Tom [10]

Answer:

\boxed{\sf 20 \$ = 80q}

Given:

1$ = 4q

To Find:

How many quarters are in 20$

Explanation:

To find out how many quarters are in 20$ we need to multiple 4 × 20.

\sf 1\$  = 4q

\sf  20\$  = 4 \times 20q

\sf = 80q

8 0
3 years ago
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When a golf club hits a 0.0459 kg ball at rest, it exerts a 2380 N force for 0.00100 s. What is the speed of the ball afterwards
aksik [14]

Answer:

51.85m/s

Explanation:

Given parameters:

Mass of ball  = 0.0459kg

Force  = 2380N

Time taken  = 0.001s

Unknown:

Speed of the ball afterwards  = ?

Solution:

To solve this problem, we use Newton's second law of motion:

   F = m x \frac{v - u}{t}  

F is the force

m is the mass

v is the final velocity

u is the initial velocity

t is the time taken

        2380  = 0.0459 x \frac{v- 0}{0.001}  

        0.0459v  = 2.38

                   v = 51.85m/s

8 0
3 years ago
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How much energy (in kilojoules) is required to convert 200 mL of diethyl ether at its boiling point from liquid to vapor if its
LiRa [457]

Answer:

55.96kJ

Explanation:

Energy = mass of diethyl ether × enthalpy of vaporization of diethyl ether

Volume (v) = 200mL, density (d) = 0.7138g/mL

Mass = d × v = 0.7138 × 200 = 142.76g

Enthalpy of vaporization of diethyl ether = 29kJ/mol

MW of diethyl ether (C2H5)2O = 74g/mol

Enthalpy in kJ/g = 29kJ/mol ÷ 74g/mol = 0.392kJ/g

Energy = 142.76g × 0.392kJ/g = 55.96kJ

4 0
4 years ago
A comet is in an elliptical orbit around the sun. its closest approach to the sun is a distance of 4.5 1010 m (inside the orbit
barxatty [35]

r1 = 5*10^10 m , r2 = 6*10^12 m

v1 = 9*10^4 m/s

From conservation of energy

K1 +U1 = K2 +U2

0.5mv1^2 - GMm/r1 = 0.5mv2^2 - GMm/r2

0.5v1^2 - GM/r1 = 0.5v2^2 - GM/r2

M is mass of sun = 1.98*10^30 kg

G = 6.67*10^-11 N.m^2/kg^2

0.5*(9*10^4)^2 - (6.67*10^-11*1.98*10^30/(5*10^10)) = 0.5v2^2 - (6.67*10^-11*1.98*10^30/(6*10^12))

v2 = 5.35*10^4 m/s

4 0
4 years ago
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