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weeeeeb [17]
3 years ago
10

Which of the following would have more momentum than fiona, a 313kg hippo, running at 8m/s?

Physics
1 answer:
kap26 [50]3 years ago
5 0
Let's calculate the momentum of Fiona, given by the product between its mass and its speed:
p=mv=(313 kg)(8 m/s)=2504 kg m/s

Now let's compare it with the momentum of the other animals:
a) the mass of the sea turtle is missing, so we cannot calculate its momentum.
b) the momentum of the dolphin is 
p=mv=(150 kg)(18 m/s)=2700 kg m/s
c) the momentum of the horse is
p=mv=(380 kg)(4 m/s)=1520 kg m/s
d) the momentum of the lion is
p=mv=(190 kg)(11 m/s)=2090 m/s

And we can see that the correct answer is b), because the momentum of the dolphin is greater than the momentum of Fiona.
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True or false? the potential energy of a membrane potential comes solely from the difference in electrical charge across the mem
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The distance from Earth to the North Star is about 430 light-years. Which of the following statements describes why scientists u
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12)A black body is heated from 27°C to 127° C. The ratio of their energies of radiations emitted will be
Nat2105 [25]

Answer:

81:256.

Explanation:

Let T denote the absolute temperature of this object.

Calculate the value of T before and after heating:

T(\text{before}) = 27 + 273 = 300\; \rm K.

T(\text{after}) = 127 + 273 = 400\; \rm K.

By the Stefan-Boltzmann Law, the energy that this object emits (over all frequencies) would be proportional to T^4.

Ratio between the absolute temperature of this object before and after heating:

\displaystyle \frac{T(\text{before})}{T(\text{after})} = \frac{3}{4}.

Therefore, by the Stefan-Boltzmann Law, the ratio between the energy that this object emits before and after heating would be:

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4 0
3 years ago
An airplane starts from A and flies to B at a constant speed. After reaching B it returns to A at the same speed. There was no w
Dafna1 [17]

Answer:

When there is wind it takes longer

Explanation:

With no wind, the round trip time is

t_1=\frac{d}{v}+\frac{d}{v}=\frac{2d}{v}

When we have a constant wind speed w

t_2=\frac{d}{v-w} +\frac{d}{v+w} =\frac{2vd}{v^{2} -w^{2}}

comparing the reciprocal times;

\frac{1}{t_2}=\frac{v^{2}-w^{2} }{2vd}=\frac{v}{2d}-\frac{w^{2}}{2vd}   \leq \frac{v}{2d}=\frac{1}{t_1}

This means that t1 is smaller than t2, ergo, it takes longer with wind

4 0
3 years ago
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