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Effectus [21]
2 years ago
7

Need help asap

Physics
1 answer:
pochemuha2 years ago
8 0

Answer:

Explanation:

Well which is it ? ρ = 1000 kg/m³ or ρ = 1025 kg/m³?

Obviously the sea is salt water so we can ignore  ρ = 1000 kg/m³

1025 kg/m³(d  m)(9.81 N/kg) = 1 x 10⁵ N/m² = Pa

d = 9.9450535...

d = 10 meters

That's if we only account for the pressure due to the water. On top of that pressure would be atmospheric pressure which is about 101000 Pa

so the robot would be a hair above its pressure limit before it even got in the water.

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​A piston–cylinder assembly contains 5.0 kg of air, initially at 2.0 bar, 30 oC. The air undergoes a process to a state where th
vlada-n [284]

Answer:

Explanation:

The process is isothermic,  as P V = constant .

work done = 2.303 n RT log P₁ / P₂

= 2.303 x 5 / 29 x 8.3 x 303  log 2 / 1 kJ

= 300.5k J

This energy in work done by the gas will come fro heat supplied as internal energy is constant due to constant temperature.

heat supplied  = 300.5k J

specific volume is volume per unit mass

v / m

pv = n RT

pv  = m / M  RT

v / m = RT / p M

specific volume = RT / p M

option B is correct.

5 0
3 years ago
In a lever, the effort arm is two times as long as the load arm. The resultant force will be
jasenka [17]
Answer is B. 

In a lever, the effort arm is 2 times as a long as the load arm. The resultant force will be twice the applied force.

Hope it helped you.

-
Charlie
5 0
3 years ago
Read 2 more answers
What is the wavelength of radar waves with the frequency of 6.0 x 10 ^10 Hz?
Serjik [45]

Yes yes multiply hurry up

8 0
3 years ago
You look down an old well, cannot see the bottom, and mutter to yourself "Oh well!". In order to estimate the depth of the well,
telo118 [61]

Answer:

The best estimate of the depth of the well is 2.3 sec.

Explanation:

Given that,

Record time,

t_{1}=2.19\ sec

t_{2}=2.30\ sec

t_{3}=2.26\ sec

t_{4}=2.29\ sec

t_{5}=2.27\ sec

We need to find the best estimate of the depth of the well

According to record time,

We can write of the record time

t_{1}=2.19\approx 2.2\ sec

t_{2}=2.30\approx 2.3\ sec

t_{3}=2.26\approx 2.3\ sec

t_{4}=2.29\approx 2.3\ sec

t_{5}=2.27\approx 2.3\ sec

Here, all time is nearest 2.3 sec.

So, we can say that the best estimate of the depth of the well is 2.3 sec.

Hence, The best estimate of the depth of the well is 2.3 sec.

6 0
3 years ago
The magnetic field in a cyclotron is 1.25 T, and the maximum orbital radius of the circulating protons is 0.40 m. (a) What is th
Darya [45]

Answer:

1.92 x 10⁻¹²J

Explanation:

The magnetic force from the magnetic field gives the circulating protons gives the particle the necessary centripetal acceleration to keep it orbiting round the circular path. And from Newton's second law of motion, the force(F) is equal to the product of the mass(m) of the proton and the centripetal acceleration(a). i.e

F = ma

Where;

a = \frac{v^2}{r}             [v = linear velocity, r = radius of circular path]

=> F = m\frac{v^2}{r}           ------------(i)

We also know that the magnitude of this magnetic force experienced by the moving charge (proton) in a magnetic field is given by;

F = q v B sin θ       ----------(ii)

Where;

q = charge of the particle

v = velocity of the particle

B = magnetic field

θ = the angle between the velocity and the magnetic field.

Combining equations (i) and (ii) gives

m\frac{v^2}{r} = q v B sin θ           [θ = 90° since the proton is orbiting at the maximum orbital radius]

=> m\frac{v^2}{r} = q v B sin 90°

=> m\frac{v^2}{r} = q v B

Divide both side by v;

=> m\frac{v}{r} = qB

Make v subject of the formula

v = \frac{qBr}{m}

From the question;

B = 1.25T

m = mass of proton = 1.67 x 10⁻²⁷kg

r = 0.40m

q = charge of a proton = 1.6 x 10⁻¹⁹C

Substitute these values into equation(iii) as follows;

v = \frac{(1.6*10^{-19})(1.25)(0.4)}{(1.67*10^{-27})}

v = 4.79 x 10⁷m/s

Now, the kinetic energy, K, is given by;

K = \frac{1}{2}mv²

m = mass of proton

v = velocity of the proton as calculated above

K = \frac{1}{2}(1.67*10^{-27} * (4.79 * 10^7)^2 )

K = 1.92 x 10⁻¹²J

The kinetic energy is 1.92 x 10⁻¹²J

8 0
3 years ago
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