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Tcecarenko [31]
3 years ago
12

(a) If a flea can jump straight up to a height of 0.440 m, what is its initial speed as it leaves the ground? (b) How long is it

in the air?
Physics
1 answer:
pickupchik [31]3 years ago
6 0

Answer

given,

height of the jump = 0.44 m

acceleration due to gravity, g = 9.8 m/s²

velocity at the height point = 0 m/s

initial speed = ?

Using equation of motion for speed calculation

v² = u² + 2 g h

0 = u² - 2 x 9.8 x 0.44

u = √8.624

u = 2.94 m/s

time taken to reach the highest point

v = u + g t

0 = 2.94 - 9.8 x t

t = 0.3 s

total time of flight will be equal to double of the time taken to reach the maximum height.

Total time = 2 x 0.3 = 0.6 s

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erica [24]

Explanation:

Given that,

Mass of a freight car, m_1=30,000-kg

Speed of a freight car, u_1=0.85\ m/s

Mass of a scrap metal, m_2=110,000\ kg

(a) Let us assume that the final velocity of the loaded freight car is V. The momentum of the system will remain conserved as follows :

30000\times 0.85=(30000+110000)V\\\\V=\dfrac{30000\times 0.85}{30000+110000}\\\\=0.182\ m/s

So, the final velocity of the loaded freight car is 0.182 m/s.

(b) Lost on kinetic energy = final kinetic energy - initial kinetic energy

\Delta K=\dfrac{1}{2}[(m_1+m_2)V^2-m_1u_1^2)]\\\\=\dfrac{1}{2}\times [(30,000+110,000 )0.182^2-30000(0.85)^2]\\\\=-8518.82\ J

Lost in kinetic energy is 8518.82. Negative sign shows loss.

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2 years ago
5. A 5.0 kg object accelerates uniformly from rest for 5.0 s and reaches a final velocity of 20.0 m/s. At 3.0 s, what is the obj
irina [24]

Answer:

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Explanation:

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3 years ago
A railroad car of mass 2.50*10^4 kg is moving with a speed of 4.00 m/s. It collides and couples with three other coupled railroa
NemiM [27]

Answer:

(a) 2.5 m/s

(b) 37.5 KJ

Explanation:

(a)

From the law of conservation of momentum, Initial momentum=Final momentum

mV_1+3mV_2=(m+3m)V_f=4mV_f

V_1+3V_2=4V_f and making V_f the subject then

V_f=\frac {V_1+3V_2}{4} and since V_1 is initial velocity of car, value given as 4 m/s, V_2 is the initial velocity of the three cars stuck together, value given as 2 m/s and v_f is the final velocity which is unknown. By substitution

V_f=\frac {4+(3\times2)}{4}=2.5 m/s

(b)

Initial kinetic energy is given by

\frac {mV_1^{2}}{2}+\frac {3mV_2^{2}}{2}=\frac {m(V_1^{2}+3V_2^{2}}{2}=\frac {2.5\times 10^{4}(4^{2}+3(2^{2}))}{2}=350\times10^{3} J= 350 KJ

Final kinetic energy is given by

\frac {4mV_f^{2}}{2}=\frac {4\times 2.5\times 10^{4}\times 2.5^{2}}{2}=312.5\times 10^{3} J=312.5 KJ

The energy lost is given by subtracting the final kinetic energy from the initial kinetic energy hence

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3 years ago
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Are the two atoms shown below in the image from the same element?
kvasek [131]

The answer is yes

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7 0
3 years ago
A finch rides on the back of a Galapagos tortoise, which walks at the stately pace of 0.060 m/s. After 1.1 minutes, the finch ti
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Answer:

Average Speed = 6.37 m/s

Explanation:

The average speed is simply given by the following formula:

Average Speed = Total Distance Traveled/Total Time Spent

here,

Total Time Spent = 1.1 min + 1.5 min = (2.6 min)(60 s/min) = 156 s

Now, for total distance, we have to calculate the distance traveled on tortoise and distance traveled while flying, separately. Therefore,

Distance Traveled on Tortoise = (Time spent on Tortoise)(Speed of Tortoise)

Distance Traveled on Tortoise = (1.1 min)(60 s/min)(0.06 m/s) = 3.96 m

Similarly,

Flying Distance = (Flying Time)(Flying Speed) = (1.5 min)(60 s/min)(11 m/s)

Flying Distance =  990 m

Since, total distance is the sum of both distances, therefore,

Total Distance = 3.96 m + 990 m = 993.96 m

Now, using the values in equation of average speed, we get:

Average Speed = 993.96 m/156 s

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