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kkurt [141]
3 years ago
10

Starting from rest, a 2.3x10-4 kg flea springs straight upward. While the flea is pushing off from the ground, the ground exerts

an average upward force of 0.49 N on it. This force does 3.6x10-4 J of work on the flea. (a) What is the flea's speed when it leaves the ground
Physics
1 answer:
Harman [31]3 years ago
4 0

Answer:

3.13 m/s

Explanation:

From the question,

Since the flea spring started from rest,

Ek = W................... Equation 1

Where Ek = Kinetic Energy of the flea spring, W = work done on the flea spring.

But,

Ek = 1/2mv²............ Equation 2

Where m = mass of the flea spring, v = flea's speed when it leaves the ground.

substitute equation 2 into equation 1

1/2mv² = W.................... Equation 3

make v the subject of the equation

v = √(2W/m)................. Equation 4

Given: W = 3.6×10⁻⁴ J, m = 2.3×10⁻⁴ kg

Substitute into equation 4

v = √[2×3.6×10⁻⁴ )/2.3×10⁻⁴]

v = 7.2/2.3

v = 3.13 m/s

Hence the flea's speed when it leaves the ground  = 3.13 m/s

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dalvyx [7]

Answer:

(a) W_c=127.008 J

(b) W_g=148.176 J

(c) K.E. = 21.168 J

(d) v=3.4293m.s^{-1}

Explanation:

Given:

  • mass of a block, M = 3.6 kg
  • initial velocity of the block, u=0 m.s^{-1}
  • constant downward acceleration, a_d= \frac{g}{7}

\Rightarrow That a constant upward acceleration of \frac{6g}{7} is applied in the presence of gravity.

∴a=- \frac{6g}{7}

  • height through which the block falls, d = 4.2 m

(a)

Force by the cord on the block,

F_c= M\times a

F_c=3.6\times (-6)\times\frac{9.8}{7}

F_c=-30.24 N

∴Work by the cord on the block,

W_c= F_c\times d

W_c= -30.24\times 4.2

We take -ve sign because the direction of force and the displacement are opposite to each other.

W_c=-127.008 J

(b)

Force on the block due to gravity:

F_g= M.g

∵the gravity is naturally a constant and we cannot change it

F_g=3.6\times 9.8

F_g=35.28 N

∴Work by the gravity on the block,

W_g=F_g\times d

W_g=35.28\times 4.2

W_g=148.176 J

(c)

Kinetic energy of the block will be equal to the net work done i.e. sum of the two works.

mathematically:

K.E.= W_g+W_c

K.E.=148.176-127.008

K.E. = 21.168 J

(d)

From the equation of motion:

v^2=u^2+2a_d\times d

putting the respective values:

v=\sqrt{0^2+2\times \frac{9.8}{7}\times 4.2 }

v=3.4293m.s^{-1} is the speed when the block has fallen 4.2 meters.

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Cars A and B are racing each other along the same straight road in the following manner: Car A has a head start and is a distanc
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Answer:\frac{D_A}{v_B-v_A}

Explanation:

Given

car A had a head start of D_A

and it starts at x=0 and t=0

Car B has to travel a distance of D_A and d_a

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distance travel by car A is

d_a=v_A\times t

For car B with  speed v_B

d_B=D_A+d_a

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t=\frac{D_A}{v_B-v_A}

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ANSWER.no cause they have to be going a the same speed and I doubt that are gonna go at the same speed.
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faltersainse [42]
The correct answer is magnetic field, electric field, and charges.


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A point charge Q of mass M is located initially at the center of the ring. When it is displaced slightly, the point charge accel
dolphi86 [110]

Answer:

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Explanation:

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