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2 years ago

Why are there more fat, shaggy bison in Time 3 than in Time 1?

Physics

1 answer:

tatiyna2 years ago

8 0

Answer:

I believe the answer is D.

Explanation:

It would have been helpful if there had been more information included with this problem, but it would make sense that the answer is D. If there is a winter between time 1 and 3, bison that are not well-suited to the weather, (skinny and smooth) will die off, and only the ones well-suited to winter weather will survive to produce offspring.

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Describe the energy transformations that occur in a waterfall...

Leno4ka [110]

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2 years ago

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The Doppler Effect means that all observers of a moving wave source detect the same wave frequency.

ELEN [110]

Index fossils (also known as guide
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2 years ago

A wagon with an initial velocity of 2 m/s and a mass of 60 kg, gets a push with 150 joules of

Aleks04 [339]

Answer:

v_f = 3 m/s

Explanation:

From work energy theorem;

W = K_f - K_i

Where;

K_f is final kinetic energy

K_i is initial kinetic energy

W is work done

K_f = ½mv_f²

K_i = ½mv_i²

Where v_f and v_i are final and initial velocities respectively

Thus;

W = ½mv_f² - ½mv_i²

We are given;

W = 150 J

m = 60 kg

v_i = 2 m/s

Thus;

150 = ½×60(v_f² - 2²)

150 = 30(v_f² - 4)

(v_f² - 4) = 150/30

(v_f² - 4) = 5

v_f² = 5 + 4

v_f² = 9

v_f = √9

v_f = 3 m/s

7 0

1 year ago

Consider a cloudless day on which the sun shines down across the United States. If 2073 kJ of energy reaches a square meter ( m

Mnenie [13.5K]

The total amount of energy per hour is $2.039\cdot 10^{16} kJ$

Explanation:

In this problem we are told that the amount of energy reaching a square meter in the United States per hour is

$E_1 = 2073 kJ$

The total surface area of the United States is

$A=9.834\cdot 10^6 km^2$

And converting into squared metres,

$A=9.834\cdot 10^6 \cdot 10^6 = 9.834\cdot 10^{12} m^2$

Therefore, the total energy reaching the entire United States per hour is given by:

$E=AE_1 = (9.834\cdot 10^{12})(2073)=2.039\cdot 10^{16} kJ$

Learn more about energy and power:

brainly.com/question/7956557

#LearnwithBrainly

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2 years ago

The moment of inertia of a uniform-density disk rotating about an axle through its center can be shown to be . This result is ob

Naddik [55]

(a) $0.2888 kg m^2$

The moment of inertia of a uniform-density disk is given by

$I=\frac{1}{2}MR^2$

where

M is the mass of the disk

R is its radius

In this problem,

M = 16 kg is the mass of the disk

R = 0.19 m is the radius

Substituting into the equation, we find

$I=\frac{1}{2}(16 kg)(0.19 m)^2=0.2888 kg m^2$

(b) 142.5 J

The rotational kinetic energy of the disk is given by

$K=\frac{1}{2}I\omega^2$

where

I is the moment of inertia

$\omega$ is the angular velocity

We know that the disk makes one complete rotation in T=0.2 s (so, this is the period). Therefore, its angular velocity is

$\omega=\frac{2\pi}{T}=\frac{2\pi}{0.2 s}=31.4 rad/s$

And so, the rotational kinetic energy is

$K=\frac{1}{2}(0.2888 kg m^2)(31.4 rad/s)^2=142.5 J$

(c) $9.07 kg m^2 /s$

The rotational angular momentum of the disk is given by

$L=I\omega$

where

I is the moment of inertia

$\omega$ is the angular velocity

Substituting the values found in the previous parts of the problem, we find

$L=(0.2888 kg m^2)(31.4 rad/s)=9.07 kg m^2 /s$

8 0

2 years ago