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3 years ago

Compare the momentum of a 6,300-kg elephant walking 0.11 m/s and a 50-kg dolphin swimming 10.4 m/s. your answer

1 answer:

5 0

<span>First sum applied the Newton's second law motion: F = ma Force = mass* acceleration This motion define force as the product of mass times Acceleration (vs.Velocity). Since acceleration is the change in velocity divided by time, force=(mass*velocity)/time such that, (mass*velocity)/time=momentum/time Therefore we get mass*velocity=momentum Momentum=mass*velocity Elephant mass=6300 kg; velocity=0.11 m/s Momentum=6300*0.11 P=693 kg (m/s) Dolphin mass=50 kg; velocity=10.4 m/s Momentum=50*10.4 P=520 kg (m/s) The elephant has more momentum(P) because it is large.</span>

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They make work easier

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4 years ago

point) A circular swimming pool has a diameter of 12 m. The circular side of the pool is 3 m high, and the depth of the water is

Sergio [31]

Answer:

(a) 86.65 J

(b) 149.65 J

Solution:

As per the question:

Diameter of the pool, d = 12 m

⇒ Radius of the pool, r = 6 m

Height of the pool, H = 3 m

Depth of the pool, D = 2.5 m

Density of water, $\rho_{w} = 1000\ kg//m^{3}$

Acceleration due to gravity, g = $9.8\ m/s^{2}$

Now,

(a) Work done in pumping all the water:

Average height of the pool, h = $\frac{H + D}{2}$

h = $\frac{3 + 2.5}{2} = 2.75\ m$

Volume of water in the pool, V = $\pi r^{2}h = \pi \times 6^{2}\times 2.75 = 311.02\ m^{3}$

Mass of water, $m_{w} = \frac{\rho_{w}}{V}$

$m_{w} = \frac{1000}{311.02} = 3.215\ kg$

Work done is given by the potential energy of the water as:

$W = m_{w}gh = 3.215\times 9.8\times 2.75 = 86.65\ J$

(b) Work done to pump all the water through an outlet of 2 m:

Now,

Height, h = 2.75 + 2 = 4.75

Work done, $W = m_{w}gh = 3.215\times 9.8\times 4.75 = 149.65\ J$

7 0

3 years ago

Could I get help on this question please . My parents won’t help me /:

vovikov84 [41]

Answer:

Tarzan will be moving at 7.4 m/s.

Explanation:

From the question given above, the following data were obtained:

Height (h) of cliff = 2.8 m

Initial velocity (u) = 0 m/s

Final velocity (v) =?

NOTE: Acceleration due to gravity (g) = 9.8 m/s²

Finally, we shall determine how fast (i.e final velocity) Tarzan will be moving at the bottom. This can be obtained as follow:

v² = u² + 2gh

v² = 0² + (2 × 9.8 × 2.8)

v² = 0 + 54.88

v² = 54.88

Take the square root of both side

v = √54.88

v = 7.4 m/s

Therefore, Tarzan will be moving at 7.4 m/s at the bottom.

3 0

3 years ago

Is james west still alive?

mafiozo [28]

Greetings

and james west is ALIVE

8 0

3 years ago

Read 2 more answers

In an RLC series circuit that includes a source of alternating current operating at fixed frequency and voltage, the resistance

maw [93]

Answer:

Capacitive Reactance is 4 times of resistance

Solution:

As per the question:

R = $X_{L} = j\omega L = 2\pi fL$

where

R = resistance

$X_{L} = Inductive Reactance$

f = fixed frequency

Now,

For a parallel plate capacitor, capacitance, C:

$C = \frac{\epsilon_{o}A}{x}$

where

x = separation between the parallel plates

Thus

C ∝ $\frac{1}{x}$

Now, if the distance reduces to one-third:

Capacitance becomes 3 times of the initial capacitace, i.e., x' = 3x, then C' = 3C and hence Current, I becomes 3I.

Also,

$Z = \sqrt{R^{2} + (X_{L} - X_{C})^{2}}$

Also,

Z ∝ I

Therefore,

$\frac{Z}{I} = \frac{Z'}{I'}$

$\frac{\sqrt{R^{2} + (R - X_{C})^{2}}}{3I} = \frac{\sqrt{R^{2} + (R - \frac{X_{C}}{3})^{2}}}{I}$

${R^{2} + (R - X_{C})^{2}} = 9({R^{2} + (R - \frac{X_{C}}{3})^{2}})$

${R^{2} + R^{2} + X_{C}^{2} - 2RX_{C} = 9({R^{2} + R^{2} + \frac{X_{C}^{2}}{9} - 2RX_{C})$

Solving the above eqn:

$X_{C} = 4R$

6 0

3 years ago