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3 years ago

A clam of mass 0.12 kg dropped by a seagull takes 3.0 s to hit the ground. [Neglect friction.]

1 answer:

6 0

-- The acceleration of gravity is 9.8 m/s².
So if there's no air resistance, the speed of a falling object
always increases by 9.8 m/s for every second it falls.

   Speed = (original speed) + (gravity x falling time)

-- If it has no vertical speed when it started, then at the end
of 3 seconds, its speed is

   = (0)    + (9.8 m/s² x 3 sec)

   Velocity = 29.4 m/s downward .

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Which is harder, air tight or water tight seals?

Vladimir [108]

Should be an air tight seal

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3 years ago

75 kg + 1352 g = ______________g?

JulsSmile [24]

Answer is 76,352 just look it up

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2 years ago

Read 2 more answers

A 950-kg car strikes a huge spring at a speed of 22m/s (fig. 11-54), compressing the spring 5.0m. (a) what is the spring stiffne

alukav5142 [94]

(a) The spring stiffness constant of the spring is 18,392 N/m.

(b) The time the car was in contact with the spring before it bounces off in the opposite direction is 0.23 s.

<h3>Kinetic energy of the car</h3>

The kinetic energy of the car is calculated as follows;

K.E = ¹/₂mv²

K.E = ¹/₂ x 950 x 22²

K.E = 229,900 J

<h3>Stiffness constant of the spring</h3>

The stiffness constant of the spring is calculated as follows;

K.E = U = ¹/₂kx²

k = 2U/x²

k = (2 x 229,900)/(5)²

k = 18,392 N/m

<h3>Force exerted on the spring</h3>

F = kx

F = 18,392 x 5

F = 91,960 N

<h3>Time of impact</h3>

F = mv/t

t = mv/F

t = (950 x 22)/(91960)

t = 0.23 s

Learn more about spring constant here: brainly.com/question/1968517

#SPJ4

3 0

1 year ago

Ballistic data obtained on a firing range show that aerodynamic drag reduces the speed of a .44 magnum revolver bullet from 250

m_a_m_a [10]

Answer:

0.363999909622

Explanation:

F = Force

m = Mass = 15.6 g

C = Drag coefficient

ρ = Density of air = 1.21 kg/m³

A = Surface area = $\dfrac{\pi}{4}d^2$

v = Terminal velocity = $v=210\ m/s$

s = Displacement = 150 m

$a=\dfrac{v^2-u^2}{2s}$

Force is given by

F = ma

$F=\dfrac{1}{2}\rho CAv^2\\\Rightarrow ma=\dfrac{1}{2}\rho CAv^2\\\Rightarrow m\dfrac{v^2-u^2}{2s}=\dfrac{1}{2}\rho CAv^2\\\Rightarrow C=2\times m\dfrac{v^2-u^2}{2s}\times\dfrac{1}{\rho Av^2}\\\Rightarrow C=2\times15.6\times 10^{-3}\dfrac{210^2-250^2}{2\times 150}\times\dfrac{1}{1.21\times\dfrac{\pi}{4}\times (11.2\times 10^{-3})^2(210)^2}\\\Rightarrow C=-0.363999909622$

The drag coefficient is 0.363999909622 (ignoring negative sign)

4 0

3 years ago

Roxy is driving down East Capitol Street. She drives 1380 meters in 45 seconds. Assuming she does not speedup or slow down, what

Natasha_Volkova [10]

Answer:

30.66667

Explanation:

should be the answer if the calculator is correct

7 0

3 years ago