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2 years ago

This problem, a squid at rest suddenly sees a predator coming toward it and needs to escape. Assume the following:______.

Physics

1 answer:

makkiz [27]2 years ago

8 0

Answer:

6.79 m/s

Explanation:

By applying the principle of conservation of momentum.

The total momentum = MV - mv = 0 (since the squid is beginning at rest)

the mass of the squid (M) in absence of water in its cavity = (6.5 - 1.75) kg

= 4.75 kg

speed of the squid (V) = 2.5 m/s

mass of the water expelled (m) = 1.75 kg

speed of the water (v) = ???

∴

4.75 × 2.5 = 1.75 × v

$v = \dfrac{4.75 \times 2.5}{1.75 }$

v = 6.79 m/s

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A 1.0 kg football is given an initial velocity at ground level of 20.0 m/s [37 above horizontal]. It gets blocked just after re

stepan [7]

Refer to the diagram shown below.

Neglect air resistance.
The horizontal component of the launch velocity is
(20 m/s)*cos(37°) = 15.973 m/s
The vertical component of the launch velocity is
(20 m/s)*sin(37°) = 12.036 m/s

The acceleration due to gravity is g =9.8 m/s².
The time, t s, for the ball to reach a height of 3 m is given by
(12.036 m/s)*(t s) - (1/2)*(9.8 m/s²)*(t s)² = (3 m)
12.036t - 4.9t² - 3 = 0
2.4543t - t² - 0.6122 = 0
t² - 2.4563t + 0.6122 = 0
Solve with the quadratic formula.
t = (1/2)[2.4563 +/- √(6.0334 - 2.4488)]
t = 2.1748 or 0.2815 s
The ball reaches a height of 3 m twice.
The first time it reaches 3 m height is 0.2815 s.

Part a.
The vertical velocity when t = 0.2815 s is
Vy = 12.036 - 9.8*0.2815
= 9.2773 m/s
The horizontal component of velocity is Vx = 15.973 m/s
The resultant velocity is
√(9.2773² + 15.973² ) = 18.47 m/s
Answer:
The velocity at a height of 3.0 m is 18.5 m/s (nearest tenth)

Part b.
The horizontal distance traveled is
d = (15.973 m/s)*(0.2815 s) = 4.4964 m
Answer:
The horizontal distance traveled is 4.5 m (nearest tenth)

6 0

2 years ago

Kinetic energy problemset

Sonja [21]

Answer:

KE = 4 mv2 m = 2xKE valami. V m.

Explanation:

8 0

3 years ago

A car starts from rest and undergoes an acceleration of 4.0 m/s/s for a time of 5.0 s. What is the final velocity of the car?

Kruka [31]

Answer:

20m/s

Explanation:

acceleration=final velocity-initial velocity/time

4.0m/s²=v m/s-0m/s/5.0sec

5.0sec×4.0m/s²=v m/s-0m/s×5.0m/s/5.0m/s

20m/s=v

7 0

2 years ago

The speed of an aircraft is measure in the unit_

slega [8]

Answer:

knot...

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8 0

2 years ago

A 51-kg packing crate is pulled with constant speed across a rough floor with a rope that is at an angle of 37.0Â° above the hor

fiasKO [112]

work is done by the pulling force which is same as the tension force in the rope. the net work done is zero for the crate since crate moves at constant velocity. but there is work done by the tension force which is equal in magnitude to the work done by the frictional force.

T = tension force in the rope = 115 N

d = displacement of the crate = 7.0 m

θ = angle between the direction of tension force and displacement = 37 deg

work done on the crate is given as

W = F d Cosθ

inserting the values given above

W = (115) (7.0) Cos37

W = 643 J

3 0

3 years ago