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attashe74 [19]
3 years ago
11

The graph shows two runners participating in a race.

Physics
2 answers:
luda_lava [24]3 years ago
8 0

Answer: The answer is A

Explanation: Daniela has a 5 meter head start but leonard caught up to her.

qaws [65]3 years ago
3 0

The answer is the first one, Daniela has a 5 meter head start but leonard caught up to her.

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What is the logical relationship between the following two categorical propositions? Some food is not edible. Some food is inedi
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The answer is A. Contrapositive
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when a tuning fork vibrates over an open pipe and the air in the pipe starts to vibrate, the vibrations in the tube are caused b
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Sound waves....................................
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A person makes a cup of coffee by first placing a 200 W electric immersion heater in 0.32 kg
alexdok [17]

Answer:

Answer is D. 8.04 x 10^4 J

Explanation:

1. D

2. A

3. D

4. B

5. C

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All correct i promise you that

8 0
3 years ago
A 2 kg rubber ball is thrown at a wall horizontally at 3 m/s, and bounces back the way it came at an equal speed. A 2 kg clay ba
Lyrx [107]

Answer:

THE RUBBER BALL

Explanation:

From the question we are told that

      The mass of the rubber ball is m_r   =  2 \ kg

      The  initial  speed of the rubber ball is  u =  3 \ m/s

      The final speed at which it bounces bank v  - 3 \ m/s

      The mass of the clay ball  is  m_c =  2  \ kg

       The  initial  speed of the clay  ball is u = 3 \ m/s

       The final speed of the clay ball is  v = 0 \  m/s

Generally Impulse is mathematically represented as

       I  =  \Delta p

where \Delta  p is the change in the linear momentum so  

       I  =  m(v-u)

For the rubber  is  

        I_r  =  2(-3 -3)

       I_r  = -12\ kg \cdot  m/s

=>     |I_r|  = 12\ kg \cdot  m/s

For the clay ball

       I_c  =  2(0-3)

        I_c =  -6 \ kg\cdot \ m/s

=>    | I_c| =  6 \ kg\cdot \ m/s

So from the above calculation the ball with the a higher magnitude of impulse is the rubber ball

       

8 0
3 years ago
A boat is drifting with a speed of 5.0\,\dfrac{\text m}{\text s}5.0 s m ​ 5, point, 0, space, start fraction, m, divided by, s,
zepelin [54]

Answer:

d=42m(-\hat{i})

Explanation:

For this problem, we need to apply the formulas of constant accelerated motion.

To obtain the boat displacement we need to calculate the displacement because of the river flow and the displacement done because of the boat motor.

for the river:

d_r=v*t\\d_r=5m/s*6s\\d_r=30m(\hat{i})

for the boat:

x=\frac{1}{2}*a*t^2\\\\x=\frac{1}{2}*4.0m/s*(6s)^2\\\\\\x=72m(-\hat{i})

So the final displacement is given by:

d=dr+x\\d=30m-72m\\d=42m(-\hat{i})

8 0
3 years ago
Read 2 more answers
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