. Determine the pressure due to water at the
bottom of a tank which is 25cm deep and is
four-fifths full of water,
(Density of water = 1000kgm',g=10ms”).​

Answer:

yellow

Explanation:

yellow

make me brain

Answer:

friction as friend

friction as foe

hope it helps

Answer:

Wear and tear of soles of our shoes is due to friction.

When a tyre deflates, it is difficult to move the vehicle because of increased friction between the tyre and the road surface.

Machines get heated up and produce noise because of friction.

Answer:the answer is conduction

Answer:

The relation between velocity and time is a simple one during uniformly accelerated, straight-line motion. The longer the acceleration, the greater the change in velocity. Change in velocity is directly proportional to time when acceleration is constant.

~Hoped this helped~

~Brainiliest?~

Answer:

We describe motion in terms of velocity and acceleration. Velocity: The rate of change of displacement of an object (displacement over elapsed time) is velocity. Velocity is a vector since it has both magnitude (called speed) and direction. ... Acceleration: The rate of change of velocity is acceleration.

Explanation:

Displacement is a vector which points from the initial position of an object to its final position. ... Instantaneous velocity, on the other hand, describes the motion of a body at one particular moment in time. Acceleration is a vector which shows the direction and magnitude of changes in velocity.

Displacement is the vector difference between the ending and starting positions of an object. Velocity is the rate at which displacement changes with time. ... The average velocity over some interval is the total displacement during that interval, divided by the time.

Hope this helps      :)

Answer:

The trucks must have equal masses that is they must have same load.

Explanation:

Momentum refers to the product of the mass of the object and velocity.

As momentum has both magnitude and direction, it is a vector quantity.

In the given problem, it is given that after the collision, the trucks move at more than half the original speed of the moving truck. Therefore, the trucks must have equal masses that is they must have the same load.

Inner Planets:

- Rocky

- Small

- Only 3 moons total

- Very close to each other (for space)

- Underweight

Outer Planets:

- Gassy oop-

- GIANT

- Over 200 moons total

- VERY far from each other

- Obese

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Good luck foe.

Lb.

Answer:

classical

concert

exposition

orchesta

devolopment

recapitulation

sympathy

sonata

sinfonia

soloist

Explanation:

Answer:

  w = 16.73 rad / s,  v = - 1.04 m / s

Explanation:

This is an exercise in oscillatory motion of a system of a spring and a mass, the angular velocity is

         w = [tex]\sqrt{ \frac{k}{m} }[/tex]

To calculate the spring constant we use the data that Δx = 3.5 cm = 0.035 m with a mass of m = 0.38 kg, let's write Hooke's law

          F = -kx

in this case the applied force is the weight of the body

          F = W

we substitute

          mg =  k x

          k = m g / x

         k = 0.38 9.8 / 0.035

         k = 106.4 N / m

let's find the angular velocity

         w = Ra 106.4 / 0.38

         w = 16.73 rad / s

the movement of this system is described by the expression

          ax = A cos (wt + Ф)

the quantity A is called the amplitude of the movement in this case A = 7.5 cm = 0.075 m

to encode the constant we use that the system is released from its position of maximum displacement (x = A)

we substitute

          A = A cos (0 + Ф)

         cos Ф = 1

         Ф = cos-1 1

         Ф = 0º

by substituting in the equation of motion

            x = 0.075  cos (16.73 t)

Let's find the time it takes to reach the desired position x = 4.2 cm = 0.042 m

           cos 16.73 t = 0.042 / 0.075 = 0.56

            16.73 t = cos -1 0.56

remember that angles must be expressed in radians

            t = 0.976 / 16.73

            t = 0.0583 s

body velocity is

           v = [tex]\frac{dx}{dt}[/tex]

            v = -0.075 16.73 sin 16.73t

let's calculate for the time found

           v = - 1.25475 sin (16.73  0.0583)

           v = - 1.04 m / s

Answer:

papa's donuteria

Explanation:

:rawr:

Answer:

2m/s^2

Explanation:

Force is equal to;

mass × acceleration

therefore;

acceleration equals

force divided by mass

acceleration = 2m/s ^2

Answer:

f = 60N

m = 30kg

f= m×a

60N= 30 × a

a = 60/30

a = 2

Answer:

A. Gravitational force

C. Their charges

Explanation:

As we'll discuss in this lesson, he found that the force between charged particles was dependent on only two factors: the distance between the particles and the amount of electric charge that they carried.

Hope it help you

have a good day :)

Answer:

300 pascals

Explanation:

P=F/A

P=15N/0.05

P=300 pascals

Answer:

motion example of kinetic energy charged particle such as electrons and protons create electromagnetic field when they move in this field transport the type of energy vehicle electromagnetic radiation or light

Answer:

the number of donated valence electrons

Explanation:

  A metallic bond is formed when atoms come together in a metallic crystal lattice.

The metal atoms are usually ionized such that in a metallic crystal lattice consists of metal ions and a sea of mobile electrons. Attraction between these positively charged ions and the sea of electrons constitutes the metallic bond.

Hence, the more the number of donated valence electrons, the stronger the metallic bond.

Answer:

Assume that [tex]g =9.81\; \rm m\cdot s^{-1}[/tex], and that the air resistance on the stone is negligible.

a.

Height of the stone: [tex]389.5\; \rm m[/tex] (above the ground.)

Velocity of the stone: [tex]\left(-110.5\; \rm m \cdot s^{-1}\right)[/tex] (the stone is travelling downwards.)

b.

Height of the stone: [tex]629.5\; \rm m[/tex] (above the ground.)

Velocity of the stone: [tex]\left(-86.5\; \rm m \cdot s^{-1}\right)[/tex] (the stone is travelling downwards.)

Explanation:

If air resistance on the stone is negligible, the stone would be accelerating downwards at a constant [tex]a = -g = -9.81\; \rm m \cdot s^{-2}[/tex].

Let [tex]h_0[/tex] denote the initial height of the stone (height of the stone at [tex]t = 0[/tex].)

Similarly, let [tex]v_0[/tex] denote the initial velocity of the stone.

Before the stone reaches the ground, the height [tex]h[/tex] (in meters) of the stone at time [tex]t[/tex] (in seconds) would be:

[tex]\displaystyle h(t) = -\frac{1}{2}\, g \cdot t^{2} + v_0 \cdot t + h_0[/tex].

Similarly, before the stone reaches the ground, the velocity [tex]v[/tex] (in meters-per-second) of the stone at time [tex]t[/tex] (in seconds) would be:

[tex]v(t) = -g\cdot t + v_0[/tex].

In section a., [tex]h_0 = 1000\; \rm m[/tex] while [tex]v_0 = -12\; \rm m\cdot s^{-1}[/tex] (the stone is initially travelling downwards.) Evaluate both [tex]h(t)[/tex] and [tex]v(t)[/tex] for [tex]t = 10\; \rm m \cdot s^{-1}[/tex]:

[tex]\begin{aligned} h(t) &= -\frac{1}{2}\, g \cdot t^{2} + v_0 \cdot t + h_0 \\ &= -\frac{1}{2}\ \times 9.81\; \rm m\cdot s^{-2}\times (10\; \rm s)^{2} \\&\quad\quad + \left(-12\; \rm m \cdot s^{-1}\right) \times 10\; \rm s + 1000\; \rm m \\[0.5em] &= 389.5\; \rm m \end{aligned}[/tex].

Indeed, the value of [tex]h(t)[/tex] at [tex]t = 10\; \rm m \cdot s^{-1}[/tex] is greater than zero. The stone hasn't yet hit the ground, and both the representation for the height of the stone and that for the velocity of the stone are valid.

[tex]\begin{aligned} v(t) &= -g\cdot t + v_0 \\ &= -9.81\; \rm m\cdot s^{-2}\times 10\; \rm s - 12\; \rm m\cdot s^{-1} \\ &= -110.5\; \rm m \cdot s^{-1} \end{aligned}[/tex].

The value of [tex]v(t)[/tex] at [tex]t = 10\; \rm m \cdot s^{-1}[/tex] is negative, meaning that the stone would be travelling downwards at that time.

In section b., [tex]h_0 = 1000\; \rm m[/tex] while [tex]v_0 = 12\; \rm m\cdot s^{-1}[/tex] (the stone is initially travelling upwards.) Evaluate both [tex]h(t)[/tex] and [tex]v(t)[/tex] for [tex]t = 10\; \rm m \cdot s^{-1}[/tex]:

[tex]\begin{aligned} h(t) &= -\frac{1}{2}\, g \cdot t^{2} + v_0 \cdot t + h_0 \\ &= -\frac{1}{2}\ \times 9.81\; \rm m\cdot s^{-2}\times (10\; \rm s)^{2} \\&\quad\quad + 12\; \rm m \cdot s^{-1} \times 10\; \rm s + 1000\; \rm m \\[0.5em] &= 629.5\; \rm m \end{aligned}[/tex].

Verify that the value of [tex]h(t)[/tex] at [tex]t = 10\; \rm m \cdot s^{-1}[/tex] is indeed greater than zero.

[tex]\begin{aligned} v(t) &= -g\cdot t + v_0 \\ &= -9.81\; \rm m\cdot s^{-2}\times 10\; \rm s + 12\; \rm m\cdot s^{-1} \\ &= -86.5\; \rm m \cdot s^{-1} \end{aligned}[/tex].

Similarly, the value of [tex]v(t)[/tex] at [tex]t = 10\; \rm m \cdot s^{-1}[/tex] is negative because the stone would be travelling downwards at that time.

Answer:

Follows are the soplution to this question:

Explanation:

In the given scenario it would be like a fluid in a simple harmonic in contact with the earth. Whenever a cheetah hops quicker, oscillatory amplitude rises, while the duration stays the same since it does not depend on frequency, which mostly means that time will be the same if you're in contact with the substrate.

Answer:

nuclear energy

Explanation:

[tex]\lambda=\frac{v}{f}=\frac{400}{200}=2[m][/tex]

Answer:

Explanation:

20 watts

Answer:

sound waves

Explanation:

this is because sound waves are longitudinal waves, and longitudinal waves are waves that travel parallel to the direction of the wave motion

thus it cannot be light or electromagnetic waves but only sound waves

hope this helps, please mark it

Answer:

the answer is all of the above

Explanation:

brainliest please

Answer: The answer is A

Explanation:

poste en français s’il vous plaît

Answer:

[tex]m=1.01\times 10^6\ kg[/tex]

Explanation:

Given that,

Pressure, P = 1 atm = 101325 Pa

Area of the square surface, A = 10² = 100 m²

We need to find the mass of vertical column of air. We know that, pressure is equal to the force acting per unit area. So,

[tex]P=\dfrac{mg}{A}\\\\m=\dfrac{PA}{g}\\\\m=\dfrac{101325\times 10^2}{10}\\\\m=1.01\times 10^6\ kg[/tex]

So, the required mass of the vertical column of air is [tex]1.01\times 10^6\ kg[/tex].

Answer:

1.55

Explanation:

that it is the standard refractive index.