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____ [38]
3 years ago
11

Which is moving faster a car traveling 150 km in 3 hours or one traveling 100 km in 2 hours?

Physics
1 answer:
ki77a [65]3 years ago
7 0
Neither they’re both moving the same . 150/3 = 50km
100/2 = 50km
therefor they’re traveling at the same rate.
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An engine moves a motorboat through water at a constant velocity of 22 meters/second. If the force exerted by the motor on the b
trapecia [35]

Answer: Option B: 1.3×10⁵ W

Explanation:

Power = \frac{Work \hspace{1mm} done}{Time}

P=\frac{W}{t}

Work Done, W= F.s

Where s is displacement in the direction of force and F is force.

\Rightarrow P = \frac{F.s}{t} =F \times \frac{s}{t}=F.v

where, v is the velocity.

It is given that, F = 5.75 × 10³N

v = 22 m/s

P = 5.75 × 10³N×22 m/s = 126.5 × 10³ W ≈1.3×10⁵W

Thus, the correct option is B

6 0
3 years ago
(will give brainliest to the correct person!)
Neporo4naja [7]

Answer:

Explanation:

When a force hits something, an equal amount of force is exerted back on it.

3 0
3 years ago
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An important aspect of fission reactions is that they produce _______________ which allow for chain reactions.
Alex777 [14]

An important aspect of fission reactions is that they produce free neutrons,  which causes chain reactions.

4 0
3 years ago
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At its closest point, Mercury is approximately 46 million kilometers from the sun. What is this distance in AU?
alina1380 [7]
1 astronomical unit 1 AU = 1.4960 * 10^11 meters
it is the average distance between earth and sun
mercury to sun distance is = 46,000,000 * 1000 meters
               = 4.6 * 10^9 meters = 4.6 * 10^9 / 1.4960 * 10^11  AU
               =  3.0.74 / 100  = 0.0374 AU

5 0
3 years ago
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A boat of mass 250 kg is coasting, with its engine in neutral, through the water at speed 1.00 m/s when it starts to rain with i
Vera_Pavlovna [14]

Answer:

Explanation:

1. We use the conservation of momentum for before the raining and after. And also we take into account that in 0.5h the accumulated water is

100kg/h*0.5h = 50kg

p_{b}=p_{a}\\M_{b}v_{b}=(M_{b}+m_{r})v_{a}\\v_{a}=\frac{M_{b}v_{b}}{M_{b}+m_{r}}=\frac{250kg*1\frac{m}{s}}{250kg+50kg}=0.83\frac{m}{s}

2. the momentum does not conserve because the drag force of water makes that the boat loses velocity

3. If we assume that the force of the boat before the raining is

F=ma=m\frac{v-v_{0}}{t-t_{0}}=250kg\frac{1m}{s^{2}}=250N

where we have assumed that the acceleration of the boat is 1m/s{2} just before the rain starts

And if we take the net force as

F_{net}=M_{b}a_{net}=F-F_{d}=250N-bv^{2}\\F_{net}=250N-0.5N\frac{s^{2}}{m^{2}}(1\frac{m}{s})^{2}=249.5N\\a_{net}=\frac{249.5N}{M_{b}}=\frac{249.5N}{250kg}=0.99\frac{m}{s^{2}}

where we take v=1m/s because we are taking into account tha velocity just after the rain stars.

I hope this is useful for you

regards

5 0
3 years ago
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