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ipn [44]
3 years ago
10

An open diving chamber rests on the ocean floor at a water depth of 60 meter. Find the air pressure (gage pressure relative to t

he local atmospheric pressure) in kPa inside of the diving chamber required to keep water from entering the chamber. (Assume SG=1.03)
Physics
1 answer:
Fofino [41]3 years ago
8 0

Answer:

Gauge Pressure required = 606.258 kPa

Explanation:

Water will not enter the chamber if the pressure of air in it equals that of the water which tries to enter it.

Thus at a depth of 60m we have pressure of water equals

P(z)=P_{0}+\rho _wgh

Now the gauge pressure is given by

P(z)-P_{0}=\rho _wgh

Applying values we get

P(z)-P_{0}=\rho _wgh\\\\P_{gauge}=1.03\times 1000\times 9.81\times 60Pa\\\\P_{gauge}=606258Pascals\\\\P_{gauge}=606.258kPa

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stiks02 [169]

The mass of the second car is 1434.21 kg

<u>Explanation:</u>

Using law of conservation of momentum,

          m_{1} u_{1}+m_{2} u_{2}=\left(m_{1}+m_{2}\right) v

Given:

m_{1} = 1090 kg

u_{1} = 11 m/s

u_{2} = 0

v = 4.75 m/s

We need to find m_{2}

When substituting the given values in the above equation, we get

(1090 \times 11)+\left(m_{2} \times 0\right)=\left(1090+m_{2}\right) 4.75

11990=5177.5+4.75 m_{2}

4.75 m_{2}=11990-5177.5

4.75 m_{2}=6812.5

m_{2}=\frac{6812.5}{4.75}=1434.21 \mathrm{kg}

6 0
2 years ago
A 200g air-track glider is attached to a spring. The glider is pushed in 10cm and released. A student with a stopwatch finds tha
cricket20 [7]

The spring constant will be k= 5.5N/m for a 200g air track glider attached to a spring.

<h3>What is spring constant?</h3>

The spring constant, k, is a measure of the stiffness of the spring. It is different for different springs and materials.

Calculation for What is the spring constant

First step is to calculate the time period

T = 12 second/10

T = 1.2 second

Now let calculate the spring constant using this formula

k=\dfrac{4\pi ^2m}{T^2}

Where,

m=0.2kg

T=1.2second

k represent spring constant=?

Let plug in the formula

k=\dfrac{4 \pi \times 0.2}{(1.2)^2}

k=\dfrac{39.48\times 0.2}{1.44}

k=\dfrac{7.90}{1.44}

k=5.48 N/m

k=5.5 N/m ( Approximately)

Therefore the spring constant will be 5.5 N/m

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4 0
1 year ago
a 70 kg man standing on ice throws a 3 kg body horizontally at 8 m/s. the friction coefficient between the ice and his feet is 0
GalinKa [24]

The distance at which the man slips is 0.3 m

Newton's Second Law, F = ma, is used to calculate the braking distance. By dividing the mass of the car by the gravitational acceleration, one may determine its weight. The weight of the car multiplied by the coefficient of friction equals the brake force.

Given-

mass of man= 70 kg

frictional coefficient μ=0.02

mass of body thrown= m2 = 3kg

let s be the stopping distance

we know that frictional force = F= μN

=μMg= 0.02 x 70 x 10

=14 N

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v1= m2v2 /m1= 0.3 m/s

we know,

v²- u² = -2as

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6 0
1 year ago
This picture shows human red blood cells. The human body contains about 5 million red blood cells per cubic milliliter. This mea
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✧・゚: *✧・゚:*    *:・゚✧*:・゚✧

                  Hello!

✧・゚: *✧・゚:*    *:・゚✧*:・゚✧

❖ The correct answer choice is B) is multicellular. When something is multicellular, it consists of two or more cells. When something is unicellular, it consists of only one cell and in this case we have 5 million red blood cells so that wouldn't make sense.

~ ʜᴏᴘᴇ ᴛʜɪꜱ ʜᴇʟᴘꜱ! :) ♡

~ ᴄʟᴏᴜᴛᴀɴꜱᴡᴇʀꜱ

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Which of the following is NOT an indicator of a physical change?
S_A_V [24]

Answer:

1.  C.  The change is easily reversible

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Explanation:

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