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A scientist has invented a robot to work on the seabed. According to his calculation, the armour of the robot can withstand a maximum pressure of 10⁵ Pa exerted by the sea water. If the density of the sea water is 1025 kg/m3, what is the maximum depth of the seabed that this robot can work? [Given g = 9.81 m/s2 and rho water = 1000 kg/m3] ​

Explanation:

We use the Theorem of conservation of mechanical energy for finding the velocity when it strikes the ground:

Ei = Ef

Ki + Ui = Kf + Uf

Ui = Kf

m g h = 1/2 m v^2

v = sqrt(2gh)

So the momentum will be:

p = mv = m * sqrt(2gh)

Answer:

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Answer:A.the Oort cloud

Explanation:

Answer:

try answer A..the Ootor cloud

Answer:

A closed organ pipe is λ/4  (node-antinode) long.

λ = speed / frequency = 331.5 / 259.6 = 1.28 m

λ/4 = .319 m     length of closed pipe

An open pipe has a fundamental wavelength of A-N-A or λ/2

The second harmonic would be A-N-A-N-A   or   λ = 1.28 m for the second harmonic   331.5 / 259.6 = 1.28  (the fundamental would be 331.5 / .628

Answer:

10 seconds.

Explanation:

We can use a kinematic equation where we know the final velocity, initial velocity, acceleration, and need to determine the time t:

[tex]\displaystyle v_f = v_i + at[/tex]

The initial velocit is 30 m/s, the final velocity is 0 m/s (as we stopped), and the acceleration is -3 m/s².

Substitute and solve for t:

[tex]\displaystyle \begin{aligned} (0\text{ m/s}) & = (30 \text{ m/s}) + (-3 \text{ m/s$^2$}) t \\ \\ t & = \frac{-30\text{ m/s}}{-3 \text{ m/s$^2$}} \\ \\ & = 10 \text{ s} \end{aligned}[/tex]

Hence, it will take the car 10 seconds to come to a stop.

The magnitude of the potential difference between the two points is 840 V

To solve the problem above, we need to use the formula of Potential difference as related to distance and electric field.

Potential difference: This can be defined as the work required to move a unit charge from a point to another in an electric field.

⇒ Formula:

⇒ Where:

From the question,

⇒ Given:

⇒ Substitute these values into equation 1

Hence, The magnitude of the potential difference between the two points is 840 V

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The kinetic energy of the roller coaster at the bottom of the hill is 12250000 J

To answer this question, we must understand that the energy of a system is always conserved as explained by the law of conservation of energy.

The kinetic energy of the roller coaster will be equivalent to the potential energy of the roller coaster.

The kinetic energy of the roller coaster can be obtained as follow:

Kinetic energy = Potential energy

KE = mgh

KE = 2500 × 9.8 × 500

KE = 12250000 J

Therefore, the kinetic energy of the roller coaster is 12250000 J

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Based on the information given, it can be noted that the bat was either shorter or longer than what he expected.

From the information given, it was stated that Raghu wants a cricket bat to be made by a carpenter and he tells the carpenter that the length of the bat should be 7 hand spans.

Since he got disappointed when he collected the bat, the reason for this will be because the bat was either shorter or longer than what he requested.

Learn more about length on:

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The magnitudes of the forces that the ropes must exert on the knot connecting are :

Given data :

Mass ( M ) = 12 kg

∅₂ = 63°

∅₃ = 45°

a) Force exerted by the first rope = weight of rope

∴  F₁ = mg

     = 12 * 9.81 ≈  118 kg

b) Force exerted by the second rope

applying equilibrium condition of force in the vertical direction

F₂ sin∅₂ + F₃ sin∅₃ - mg = 0  ---- ( 1 )

where: F₃ = ( F₂ cos∅₂ / cos∅₃ ) --- ( 2 )  applying equilibrium condition of force in the horizontal direction

Back to equation ( 1 )

F₂ =  [ ( mg / cos∅₂ ) / tan∅₂ + tan∅₃ ]

   = [ ( 118 / cos 63° ) / ( tan 63° + tan 45° ) ]

   = 89.21 N

C ) Force exerted by the third rope

Applying equation ( 2 )

F₃ = ( F₂ cos∅₂ / cos∅₃ )

    = ( 89.21 * cos 63 / cos 45 )

    = 57.28 N

Hence we can conclude that The magnitudes of the forces that the ropes must exert on the knot connecting are :

F₁ = 118 N, F₂ = 89.21 N, F₃ = 57.28 N

Learn more about  static equilibrium : https://brainly.com/question/2952156

The mass of block 1 will be 1.99 kg.The tension force is applied along the whole length of the wire, pulling energy equally on both ends.

The tension force is described as the force transferred through a rope, string, or wire as it is pulled by opposing forces.

Given that,

Mass of block 1=? kg

The coefficient of the kinetic friction,μ=0.200

Now consider the weight component in the uphill direction.The weight is found as;

[tex]\rm W=m_1gsin \theta[/tex]

The force is balanced in the vertical direction as;

[tex]\rm T=F_f-W[/tex]

When the force of friction is;

[tex]\rm F_F=\mu_k N[/tex]

[tex]\rm F_f=(m_1 gcos \theta)[/tex]

Substitute the value in the vertical balanced equation;

[tex]\rm T=m_1gsin(27.5)^0-\mu_kmgcos27.5^0[/tex]

[tex]\rm T-m_2g=0\\\\T=m_2g[/tex]

[tex]\rm (10.1) g=m_1g(0.699-0.2 \times(-0.714) ) \\\\ (10.1) g=m_1g (0.699+0.1428) \\\\\ (10.1) g= m_1 \times 0.8418 \\\\ m_1 =11.99 \ kg[/tex]

Hence the mass of block 1 will be 1.99 kg.

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Answer:

What is the atomic number of an atom that has 5 neutrons and 4 electrons? A neutral atom will have the same number of electrons is does protons. Since it has 4 protons, it must have an atomic number of 4. (That makes it beryllium.)

Explanation:

mark me brainliest please and thank you Ma'aM/Sir

Answer:

4 protons, 5 neutrons, and 4 electrons are there in an atom of beryllium.

The final position of the object after 2 s is 11 m.

Motion: This can be defined as the change in position of a body.

⇒ Formula:

⇒ Where:

From the question,

⇒ Given:

⇒ Substitute these values into equation 1

Hence, The final position of the object after 2 s is 11 m

Learn more about motion here: https://brainly.com/question/15531840

Answer:

from someone watching the bulldozer from the cab of the truck or from elsewhere on the flatbed.

Explanation:

Answer:

Life-threatening human pathogens carried by sewage include cholera, typhoid and dysentery. Other diseases resulting from sewage contamination of water include schistosomiasis, hepatitis A, intestinal nematode infections, and numerous others.

Explanation:

︴︴

____

Answer:

iron combines with oxygen to produce rust, which is the compound named iron oxide.

Explanation:

Answer:

1110111010010110

Explanation:

I am not able to upload the working out using the sum of weights method sorry

Answer:What is the half life of the element Lokium?

The half-live of the element Lokium is 4.

Explanation:

sorry if im wrong, have good day

Answer:

The half-life of the element Lokium is 4

What is half-life?

Whether or not a given isotope is radioactive is a characteristic of that particular isotope. Some isotopes are stable indefinitely, while others are radioactive and decay through a characteristic form of emission. As time passes, less and less of the radioactive isotope will be present, and the level of radioactivity decreases. An interesting and useful aspect of radioactive decay is the half-life. The half-life of a radioactive isotope is the amount of time it takes for one half of the radioactive isotope to decay. The half-life of a specific radioactive isotope is constant; it is unaffected by conditions and is independent of the initial amount of that isotope.

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Answer:

Its angular velocity is  

π

30

radians per second (about 0.105 radians/s.

Explanation:

Explanation:

Think of it like this:

There are  

2

π

radians in one complete rotation, and that takes the second hand 60 seconds to complete.

So, the rate of rotation (the angular velocity) is

2

π

/60 = pi/30 radians per second# which is about 0.105 radians per second.

To solve the problem we will first calculate the time taken by the second arm to travel 1 rad.

We know that the arm of the second travels 2π radians in 60 seconds. therefore, the time is taken to travel 1 rad by the second arm,

[tex]1\rm\ rad = \dfrac{60}{2\pi}[/tex]

Time taken by the second arm to travel 3.57 rad

[tex]=3.57\times \dfrac{60}{2\pi}\\\\= 34.09\rm\ sec[/tex]

Hence, the time taken by the second arm to travel 3.57 rad is 34.09 sec.

Learn more about Radians:

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Explanation:

A. The centripetal force experienced by an electron as it goes around a hydrogen nucleus is given by

[tex]F_c = m_e\dfrac{v^2}{r}[/tex]

where [tex]m_e = \text{electron\:mass} = 9.11×10^{-31}\:\text{kg}[/tex]

[tex]r = 5.17×10^{-11}\:\text{m}[/tex] = orbital radius

[tex]v = 2.16×10^6\;\text{m/s}[/tex] = orbital velocity

so the centripetal force is

[tex]F_c = (9.11×10^{-31}\:\text{kg})\dfrac{(2.16×10^6\;\text{m/s})^2}{5.17×10^{-11}\:\text{m}}[/tex]

[tex]\;\;\;=8.22×10^{-8}\:\text{N}[/tex]

B. The electron's centripetal acceleration is given by

[tex]a_c = \dfrac{v^2}{r}[/tex]

Using the values from (A), we get

[tex]a_c = \dfrac{(2.16×10^6\;\text{m/s})^2}{5.17×10^{-11}\:\text{m}}[/tex]

[tex]\;\;\;=9.02×10^{22}\:\text{m/s}^2[/tex]

Answer: they have one path to flow

Explanation: share the same current

Answer:

t is time in s For example, a car accelerates in 5 s from 25 m/s to 3 5m/s. Its velocity changes by 35 - 25 = 10 m/s. Therefore its acceleration is 10 ÷ 5 = 2 m/s2

Explanation:

Answer:

[tex]16m/s^{2}[/tex]

Explanation:

Please ask if you have more questions!

Explanation:

We know that,

→ F = (6 × 6) N

→ F = 36 N

Therefore, the net force is 36N.

Answer:

[tex]\huge\color{skyblue}\boxed{\colorbox{black}{Answer ☘}}[/tex]

[tex]total \: resistance \: ( R_{t} ) = (3 + 1)ohm \\ (since \: the \: resistors \: are \: connected \\ in \: series)[/tex]

[tex]current \: flowing \: through \: circuit(I) = 2A \\ \\ now... \: by \: ohms \: law \\ V = IR\\ V = (2)(4) \\ = > 8v[/tex]

hope helpful~

Answer:

Earth's magnetic field (and the surface magnetic field) is approximately a magnetic dipole, with the magnetic field S pole near the Earth's geographic north pole (see Magnetic North Pole) and the other magnetic field N pole near the Earth's geographic south pole (see Magnetic South Pole). This makes the compass usable for navigation. The cause of the field can be explained by dynamo theory. A magnetic field extends infinitely, though it weakens with distance from its source. The Earth's magnetic field, also called the geomagnetic field, which effectively extends several tens of thousands of kilometres into space, forms the Earth's magnetosphere. A paleomagnetic study of Australian red dacite and pillow basalt has estimated the magnetic field to be at least 3.5 billion years old

Answer:

The new speed of the car is 30 m/s.

Hope you could get an idea from here.

Doubt clarification - use comment section.

Hi there!

a.

We can use the initial conditions to solve for w₀.

It is given that:

[tex]\frac{d\theta}{dt} = w_0e^{-\sigma t}[/tex]

We are given that at t = 0, ω =  3.7 rad/sec. We can plug this into the equation:

[tex]\omega(0)= \omega_0e^{-\sigma (0)}\\\\3.7 = \omega_0 (1)\\\\\omega_0 = \boxed{3.7 rad/sec}[/tex]

Now, we can solve for sigma using the other given condition:

[tex]2 = 3.7e^{-\sigma (8.6)}\\\\.541 = e^{-\sigma (8.6)}\\\\ln(.541) = -\sigma (8.6)\\\\\sigma = \frac{ln(.541)}{-8.6} = \boxed{0.0714s^{-1}}[/tex]

b.

The angular acceleration is the DERIVATIVE of the angular velocity function, so:

[tex]\alpha(t) = \frac{d\omega}{dt} = -\sigma\omega_0e^{-\sigma t}\\\\\alpha(t) = -(0.0714)(3.7)e^{-(0.0714) (3)}\\\\\alpha(t) = \boxed{-0.213 rad\sec^2}[/tex]

c.

The angular displacement is the INTEGRAL of the angular velocity function.

[tex]\theta (t) = \int\limits^{t_2}_{t_1} {\omega(t)} \, dt\\\\\theta(t) = \int\limits^{2.5}_{0} {\omega_0e^{-\sigma t}dt\\\\[/tex]

[tex]\theta(t) = -\frac{\omega_0}{\sigma}e^{-\sigma t}\left \| {{t_2=2.5} \atop {t_1=0}} \right.[/tex]

[tex]\theta = -\frac{3.7}{0.0714}e^{-0.0714 t}\left \| {{t_2=2.5} \atop {t_1=0}} \right. \\\\\theta= -\frac{3.7}{0.0714}e^{-0.0714 (2.5)} + \frac{3.7}{0.0714}e^{-0.0714 (0)}[/tex]

[tex]\theta = 8.471 rad[/tex]

Convert this to rev:

[tex]8.471 rad * \frac{1 rev}{2\pi rad} = \boxed{1.348 rev}[/tex]

d.

We can begin by solving for the time necessary for the angular speed to reach 0 rad/sec.

[tex]0 = 3.7e^{-0.0714t}\\\\t = \infty[/tex]

Evaluate the improper integral:

[tex]\theta = \int\limits^{\infty}_{0} {\omega_0e^{-\sigma t}dt\\\\[/tex]

[tex]\lim_{a \to \infty} \theta = -\frac{\omega_0}{\sigma}e^{-\sigma t}\left \| {{t_2=a} \atop {t_1=0}} \right.[/tex]

[tex]\lim_{a \to \infty} \theta = -\frac{3.7}{0.0714}e^{-0.0714a} + \frac{3.7}{0.0714}e^{-0.0714(0)}\\\\ \lim_{a \to \infty} \theta = \frac{3.7}{0.0714}(1) = 51.82 rad[/tex]

Convert to rev:

[tex]51.82 rad * \frac{1rev}{2\pi rad} = \boxed{8.25 rev}[/tex]