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2 years ago

5

If a 3-kg rabbit's leg muscles act as imperfectly elastic springs, how much energy will they hold if the rabbit lands from a hei

ght of 0.5 m and its legs are compressed by 0.2 m?

2 answers:

JulsSmile [24]2 years ago

5 0

Answer:

U = 14.7 J

Explanation:

Loss in gravitational potential energy is stored in the form of potential energy of rabbit legs

So here we will have by energy conservation

gravitational potential energy = energy stored in the legs

so here we will have

gravitational potential energy given as

$U = mgh$

m = 3 kg

g = 9.81 m/s/s

h = 0.5 m

now we have

$U = 3(9.81)(0.5)$

$U = 14.7 J$

DiKsa [7]2 years ago

3 0

Answer;

- 15 J

Explanation;

-Potential energy is defined as mechanical energy, stored energy, or energy caused by its position.

-For the gravitational force the formula is P.E. = mgh, where m is the mass in kilograms, g is the acceleration due to gravity (9.8 m /s² at the surface of the earth) and h is the height in meters.

Potential energy of the rabbit at the peak of its height is

PE = (3)(10)(0.5) = 15 J

(around 14.7 but because energy is lost, it is less than that)

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The Bernoulli equation is valid for steady, inviscid, incompressible flows with a constant acceleration of gravity. Consider flo

irina1246 [14]

Answer:

$p+\frac{1}{2}ρV^{2}+ρg_{0}z-\frac{1}{2}ρcz^{2}=constant$

Explanation:

first write the newtons second law:

F $_{s}$ =δma $_{s}$

Applying bernoulli,s equation as follows:

∑ $δp+\frac{1}{2} ρδV^{2} +δγz=0\\$

Where, $δp$ is the pressure change across the streamline and $V$ is the fluid particle velocity

substitute $ρg$ for {tex]γ[/tex] and $g_{0}-cz$ for $g$

$dp+d(\frac{1}{2}V^{2}+ρ(g_{0}-cz)dz=0$

integrating the above equation using limits 1 and 2.

$\int\limits^2_1 \, dp +\int\limits^2_1 {(\frac{1}{2}ρV^{2} )} \, +ρ \int\limits^2_1 {(g_{0}-cz )} \,dz=0\\p_{1}^{2}+\frac{1}{2}ρ(V^{2})_{1}^{2}+ρg_{0}z_{1}^{2}-ρc(\frac{z^{2}}{2})_{1}^{2}=0\\p_{2}-p_{1}+\frac{1}{2}ρ(V^{2}_{2}-V^{2}_{1})+ρg_{0}(z_{2}-z_{1})-\frac{1}{2}ρc(z^{2}_{2}-z^{2}_{1})=0\\p+\frac{1}{2}ρV^{2}+ρg_{0}z-\frac{1}{2}ρcz^{2}=constant$

there the bernoulli equation for this flow is $p+\frac{1}{2}ρV^{2}+ρg_{0}z-\frac{1}{2}ρcz^{2}=constant$

note: $ρ$ =density(ρ) in some parts and change(δ) in other parts of this equation. it just doesn't show up as that in formular

4 0

2 years ago

A hungry 169169 kg lion running northward at 77.377.3 km/hr attacks and holds onto a 31.731.7 kg Thomson's gazelle running eastw

navik [9.2K]

Answer: 75,242.9 m/s

Explanation:

from the question we are given the following parameters

mass of Lion (ML) = 169,169 kg

velocity of lion (VL) = 777,377.7 m/s

mass of Gazelle (Mg) = 31,731.7 kg

velocity of Gazelle (Vg) = 63,863.8 kg

mass of Lion and Gazelle (M) = 200,900.7 kg

velocity of Lion and Gazelle (V) = ?

The first figure below shows the motion of the Lion and Gazelle with their direction.

The second diagram shows the motion of the Lion and Gazelle with their directions rearranged to form a right angle triangle.

from the triangle formed we can get the velocity of the Lion and Gazelle immediately after collision using their momentum and Phytaghoras theorem

momentum = mass x velocity

momentum of the Lion = 169,169 x 77,377.3 = 13,089,840,463.7 kgm/s

momentum of the Gazelle = 31,731.7 x 63,863.8 = 2,026,506,942.46 kgm/s

momentum of the Lion and Gazelle = 200,900.7 x V

now applying Phytaghoras theorem we have

13,089,840,463.7 + 2,026,506,942.46 = 200,900.7 x V

15,116,347,406.16 = 200,900.7 x V

V = 75,242.9 m/s

7 0

2 years ago

Read 2 more answers

A Porsche challenges a Honda to a 200-m race.Because the Porsche's acceleration of 3.5 m/s2 is larger than the Honda's 3.0 m/s2,

Blizzard [7]

Answer:

Honda won by 0.14 s

Explanation:

We are given that

Distance =S=200 m

Initial velocity of Honda=u=0m/s

Initial velocity of Porsche=u'=0m/s

Acceleration of Honda= $3.0m/s^2$

Acceleration of Porsche's= $3.5m/s^2$

Time taken by Honda to start=1 s

$s=ut+\frac{1}{2}at^2$

Substitute the values

$200=0(t)+\frac{1}{2}(3)t^2$

$200=\frac{3}{2}t^2$

$t^2=\frac{200\times 2}{3}=\frac{400}{3}$

$t=\sqrt{\frac{400}{3}}=11.55s$

Time taken by Honda=11.55 s

Now, time taken by Porsche

$200=\frac{1}{2}(3.5)t^2$

$t^2=\frac{200\times 2}{3.5}$

$t=\sqrt{\frac{400}{3.5}}=10.69 s$

Total time taken by Porsche=10.69+1=11.69 s

Because it start 1 s late

Time taken by Honda is less than Porsche .Therefore, Honda won and

Time =11.69-11.55=0.14 s

Honda won by 0.14 s

3 0

2 years ago

A plastic rod is rubbed against a wool shirt, thereby acquiring a charge of −4.9 µc. how many electrons are transferred from the

creativ13 [48]

To find the number of electrons transferred, we should divide the total charge acquired by the rod $Q=-4.9 \mu C=-4.9 \cdot 10^{-6}C$ by the charge of a single electron ( $e=-1.6 \cdot 10^{-19}C$ ), and we find:
$N= \frac{Q}{e}= \frac{-4.9 \cdot 10^{-6}C}{-1.6 \cdot 10^{-19}C} =3.1 \cdot 10^{13}$

5 0

2 years ago

A car is traveling at 120 km/h (75 mph). When applied the braking system can stop the car with a deceleration rate of 9.0 m/s2.

Bumek [7]

Answer:

the number of additional car lengths approximately it takes the sleepy driver to stop compared to the alert driver is 15

Explanation:

Given that;

speed of car V = 120 km/h = 33.3333 m/s

Reaction time of an alert driver = 0.8 sec

Reaction time of an alert driver = 3 sec

extra time taken by sleepy driver over an alert driver = 3 - 0.8 = 2.2 sec

now, extra distance that car will travel in case of sleepy driver will be'

S_d = V × 2.2 sec

S_d = 33.3333 m/s × 2.2 sec

S_d = 73.3333 m

hence, number of car of additional car length n will be;

n = S_n / car length

n = 73.3333 m / 5m

n = 14.666 ≈ 15

Therefore, the number of additional car lengths approximately it takes the sleepy driver to stop compared to the alert driver is 15

8 0

2 years ago