Do y’all know what A is?! I really need help!

The ice cap at the North Pole may be 1000 m thick, with a density of 920 kg/m3. Find the pressure at the bottom and the correspo

bija089 [108]

<span>Pice=920kg/m^3 deltaP=PgH=920kg/m^3 X 9.80665m/s^2 X 1000m = 9022118 Pa P=Po + deltaP=101.325 + 9022 = 9123kPa</span>

7 0

3 years ago

What do the mitochondria do??

Ksju [112]

Mitochondria breaks down sugar to release energy to the cell.

4 0

4 years ago

What elements make gas

kari74 [83]

<span>carbon, hydrogen, and oxygen onLY</span>

7 0

3 years ago

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Horseshoe crabs, or Limulus polyphemus, are very common along the east coast of the US. Fossils of horseshoe crabs have revealed

liubo4ka [24]

The answer to your question is C

5 0

2 years ago

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Block with mass m =7.6 kg is hung from a vertical spring. when the mass hangs in equilibrium, the spring stretches x = 0.29 m. w

PSYCHO15rus [73]

1) 256.9 N/m

The force applied to the spring is equal to the weight of the block hanging on the spring:

$F=mg=(7.6 kg)(9.8 m/s^2)=74.5 N$

And the spring constant can be found by using Hook's law, because we know that the displacement caused by this force is x = 0.29 m:

$F=kx\\k=\frac{F}{x}=\frac{74.5 N}{0.29 m}=256.9 N/m$

2) 1.08 Hz

The angular frequency of oscillation of the spring is given by the formula:

$\omega=\sqrt{\frac{k}{m}}=\sqrt{\frac{256.9 N/m}{7.6 kg}}=5.81 rad/s$

And the frequency of oscillation is given by:

$\omega=2\pi f\\f=\frac{2 \pi}{\omega}=\frac{2\pi}{5.81 rad/s}=1.08 Hz$

3) 2.19 m/s

The velocity at time t of the block is given by:

$v=v_0 cos (\omega t)$

where

$v_0 = 4.4 m/s$ is the initial velocity of the block

$\omega=5.81 rad/s$ is the angular frequency

t is the time

Substituting t=0.36 s, we find the speed of the block at that time:

$v(0.36 s)=(4.4 m/s) ( cos ((5.81 rad/s)(0.36 s)) = -2.18 m/s$

And the negative sign means that the direction of the velocity is upward (because the initial velocity was downward)

4) 25.6 m/s^2

The maximum acceleration is given by:

$a_0 = \omega^2 A$

where A is the amplitude of the oscillation.

We can find the amplitude by using the law of conservation of energy: in fact, the kinetic energy at the equilibrium point must be equal to the elastic potential energy at the point of maximum displacement:

$K=U\\\frac{1}{2}mv_0^2 = \frac{1}{2}kA^2\\A=\sqrt{\frac{mv_0^2}{k}}=\sqrt{\frac{(7.6 kg)(4.4 m/s)^2}{256.9 N/m}}=0.76 m$

So, the maximum acceleration is

$a_0 = \omega^2 A=(5.81 rad/s)^2 (0.76 m)=25.6 m/s^2$

5) 95.1 N

The magnitude of the net force acting on the block is given by the difference between the weight and the restoring force of the spring:

$F=mg-kx$

First, we need to find the position x at t=0.36 s, which is given by

$x(t)=A sin(\omega t)=(0.76 m)(sin ((5.81 rad/s)(0.36 s))=0.66 m$

And so, the net force is

$F=(7.6 kg)(9.8 m/s^2)-(256.9 N/m)(0.66 m)=-95.1 N$

And the negative sign means the direction of the force is upward.

8 0

3 years ago

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