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3 years ago

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A 332 kg mako shark is moving in the positive direction at a constant velocity of 2.30 m/s along the bottom of a sea when it enc

ounters a lost 19.5 kg scuba tank. Thinking the tank is a meal, it has lunch. Assuming momentum is conserved in the collision, what is the velocity of the shark immediately after it swallows the tank?

1 answer:

Digiron [165]3 years ago

4 0

To solve this problem we will apply the concepts related to the conservation of momentum. By definition we know that the initial moment must be equivalent to the final moment of the two objects therefore

$p_1 = p_2$

$m_1u_1+m_2u_2 = m_1v_1+ m_2v_2$

Here,

$m_{1,2} =$ Mass of each object

$u_{1,2} =$ Initial velocity of each object

$v_{1,2}$ = Final velocity of each object

Since the initial velocity relative to the metal tank is at rest, that velocity will be zero. And considering that in the end, the speed of the two bodies is the same, the equation would become

$m_1u_1 = (m_1+m_2)v_f$

Rearranging to find the velocity,

$v_f = \frac{m_1u_1}{ (m_1+m_2)}$

Replacing we have that,

$v_f = \frac{(332)(2.3)}{ (332+19.5)}$

$v_f = 2.17 m/s$

Therefore the velocity of the shark immediately after it swallows the tank is $2.17m/s$

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If you lose control of your vehicle and collide with a fixed object, such as a tree, at 60 m.p.h., the force of impact is the sa

My name is Ann [436]

You can compare the velocity of the car, 60 mph, with the velocity that a mass would acquire when falls from certain height.

First, convert 60 mph to m/s:

60 miles/h * 1.60 km/mile * 1000 m/km * 1h/3600s = 26.67 m/s

Second, calculate from what height a body in free fall reachs 26.67 m/s velocity when hits the floor.

free fall => Vf^2 = 2g*H => H = Vf^2 / (2g)

H = (26.67m/s)^2 / (2*9.8 m/s) = 36.2 m

If you consider that the height between the floors of a building is approximately 3.6 m, you get 36.2 m / 3.6 m/floor = 10 floors.

Then, you conclude that the force of impact is the same as driving you vehicle off a 10 story building.

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3 years ago

A sculpture is suspended in equilibrium by two cables, one from a wall and the other

aleksandrvk [35]

Answer:

$T_1=6655.295917 \approx 6655.3N$

Explanation:

From the question we are told that

Angle of cable 2 $\theta=37.0\textdegree$

Weight of sculpture $W=5000 N$

Generally the Tension from cable 2 $T_2$ is mathematically given by

$T_2sin37\textdegree=5000N$

$T_2=5000N/sin37\textdegree$

$T_2=8308.2N$

Generally the Tension from Cable 1 $T_1$ is mathematically given by

$T_1=T_2 cos37\textdegree$

$T_1=8308.2* cos 37\textdegree$

$T_1=6655.295917 \approx 6655.3N$

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3 years ago

A physics teacher performing an outdoor demonstration suddenly falls from rest off a high cliff and simultaneously shouts "Help"

Digiron [165]

Answer: a) The cliff is 532.05m high

b) Her speed just before hitting the ground is 102.12 m/s

Explanation: To solve This, I'll use a sketch diagram, attached to this solution,

In 3seconds, the teacher heard the echo of her initial scream back. We can obtain the distance the teacher had fallen at the end of 3 seconds using the equations of motion,

Y1 = ut + 0.5g(t^2)

Since she's falling under the influence of gravity, her initial velocity, u = 0m/s, g = 9.8m/s2, t = 3s

Y1, distance she fell through in 3 seconds = 0.5×9.8(3^2) = 44.1m

Let the total height of the cliff be (44.1 + x); where is the remaining height of cliff that the teacher will fall through.

Using the equations of motion again, we can obtain distance travelled by the sound waves in 3s. sound waves travel with a constant speed of 340m/s, no acceleration,

Y2 = ut + 0.5g(t^2) where g = 0, u = 340m/s, t = 3seconds

Y2 = 340 × 3 = 1020m

But in 3 secs, the sound waves would have travelled through the total height of the cliff (44.1 + x) and back to the teacher's current height, x. That is, 1020 = 44.1 + x + x

x = 487.95m

So, total height of cliff = 44.1 + 487.95 = 532.05m

b) the speed of the teacher just before she hits the ground.

Using the equations of motion again,

(V^2) = (U^2) + 2gs

Where v is the final velocity to be calculated

U is the initial velocity = 0m/s

g is acceleration due to gravity = 9.8m/s2

S is the total height she fell through, that is, the height of the cliff = 532.05m

(V^2) = 0 + 2×9.8×532.05 = 10428.18

V = √(10428.18) = 102.12m/s

QED!

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