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Mashcka [7]
3 years ago
12

Glider one and glider two collided. The data table above shows the momentum of each before and after the collision. Perform an a

nalysis of the data to determine whether or not momentum was conserved in the collision and answer the following question.
WAS MOMENTUM CONSERVED IN THIS COLLISION? Yes or No Explain your ANSWER in TWO SENTENCES

Physics
1 answer:
k0ka [10]3 years ago
7 0

I think there was momentum conserved

Explanation: I took the test

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Section 2.2
Feliz [49]

Answer:

Speed changes at the rate of 24 m/s for each second over time.

Explanation:

We are told the object's acceleration is equal to 24 m/s²

Now we know that acceleration can also be defined as the rate of change of speed with time. Also speed has a unit known as m/s.

Thus, we can rephrase the acceleration in this question to mean;

Speed changes at the rate of 24 m/s for every second with time.

6 0
3 years ago
What happens to the motion of a baseball after a pitcher throws it and the batter hits it.
kolbaska11 [484]
THE MOTION IS GOOD BECAUSE IT MAKES THE BATTER IT THE BALL AND THE BATTER LINES UP THE BALL AND BAT AND HE HITS IT
7 0
4 years ago
A light bulb uses 0.5 amperes from a source of 120 volts. How much power is used by the bulb? 
Sladkaya [172]

Power = voltage(V) * current(I)

          = 120 * 0.5

Power = 60 watts

8 0
3 years ago
What was the average speed in km/h of a car that travels 490.0 km in 4.2 h?
Lana71 [14]

Answer:

116.67 km/h

Explanation:

avarge speed = total distance / total time

Answer: 490 / 4.2 = 116.67 km/h

7 0
4 years ago
Each plate of a parallel‑plate capacitor is a square of side 3.63 cm, and the plates are separated by 0.473 mm. The capacitor is
NARA [144]

Answer:

E= 55.53 x 10³ V/m

Explanation:

Given that

a=  3.63 cm

Area ,A= a²

distance ,d= 0.473 mm

Stored energy ,U = 8.49 nJ

Value of capacitor given as

C=\dfrac{\varepsilon _oA}{d}

By putting the values

C=\dfrac{8.85\times 10^{-12}\times 3.63^2\times 10^{-4}}{0.473\times 10^{-3}}

C=2.46 x 10⁻¹¹ F

U=\dfrac{1}{2}CV^2

V=Voltage difference

V=\sqrt{\dfrac{2U}{C}}

V=\sqrt{\dfrac{2\times 8.49\times 10^{-9}}{2.46\times 10^{-11}}}

V=26.27 V

V= E d

E=Electric filed

26.27 = E x 0.473 x 10⁻³

E= 55.53 x 10³ V/m

7 0
4 years ago
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