a box has a mass of 4 kg and surface area 4 metre square calculate the pressure exerted by the box on the floor​

Answer:

440 m

Explanation:

S=(u+v) t / 2

S = (11+33) × 20/2

S= 44× 20/2

S=440 m

a.

The object's speed at 2.20 m below balcony level is 8.74 m/s

Let the balcony level be 0 m and the height above the balcony level be positive and height below the balcony level negative.

Using the principle of conservation of energy, the total energy at a vertical height of 1.89 m above the balcony level equals the total mechanical energy when the object is 2.20 m below the balcony level and

So, E = E'

U + K + f = U' + K' + f'

where U = initial potential energy at 1.89 m = mgh, K = initial kinetic energy at 1.89 m = 0 J(since it is released from rest), f = energy loss at 1.89 m = 0 J, U' = final potential energy at 2.20 m below balcony level = mgh', K = final kinetic energy at 2.20 m = 1/2mv², f' = energy loss at 1.89 m = 10%U = 0.10mgh(since 10% of the initial energy is lost).

So,

U + K + f = U' + K' + f'

mgh + 0 + 0 = mgh' + 1/2mv² + 0.10mgh

mgh = mgh' + 1/2mv² + 0.10mgh

Dividing through by m, we have

gh = gh' + 1/2v² + 0.10gh

So, gh -  0.10gh = gh' + 1/2v²

0.90gh = gh' + 1/2v²

1/2v² = 0.90gh - gh'

1/2v² = g(0.90h - h')

v² = 2g(0.90h - h')

Taking square-root of both sides, we have

v = √[2g(0.90h - h')]

where v = velocity of object at 2.20 m below balcony level, h = height above the balcony level = 1.89 m, h' = height below the balcony level = -2.20 m and  g = acceleration due to gravity = 9.8 m/s²

Substituting the values of the variables into the equation, we have

v = √[2g(0.90h - h')]

v = √[2 × 9.8 m/s²{0.90 × 1.89 m - (-2.20 m)}]

v = √[2 × 9.8 m/s²(1.701 m + 2.20 m)]

v = √[2 × 9.8 m/s²(3.901 m)]

v = √[76.4596 m²/s²]

v = 8.74 m/s

So, the object's speed at 2.20 m below balcony level is 8.74 m/s

b.

Yes it does matter when we apply 10% loss before V calculations

We need to apply the 10 % loss before V calculations because this would give us a proper value for V since the energy is lost before V is obtained.

So, yes it does matter when we apply 10% loss before V calculations

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Answer:

If the observer is stationary but the source moves toward the observer at a speed vs, the observer still intercepts more waves per second and the frequency goes up. This time it is the wavelength of the wave received by the observer that is effectively shifted by the motion, rather than the speed.

Answer:b

Explanation:

-29.4 m/s

The velocity of the ball dropped from 80 m if it reaches the ground within 3 seconds is 26.6 m/s. If it is in midway within this time, then the velocity will be 29.4 m/s.

Velocity of a moving object is the measure of its distance travelled per unit time. Velocity is a vector quantity having both magnitude and direction. Acceleration is the rate of change in velocity.

Given that, height of the building = 80 m

the ball is moving downwards by acceleration due to gravity g = 9.8 m/s².

Then after 3 seconds, the velocity of the ball is calculated as follows:

velocity = acceleration × time

v = 9.8 m/s²  × 3 s = -29.4 m/s

If the ball reaches the ground within the time of 3 s, then, the velocity is:

v = 80 m/3s = 26.6 m/s.

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#SPJ6

Answer:

6e-12

Explanation:

divide the time value by 1e+12

Answer:

The First 8 Presidents

For this exercise, we're going to use a silly story made of silly sentences. The letters that represent the last names of these presidents are W, A, J, M, M, A, J, V. One silly sentence to help you remember this sequence is: Wilma and John made merry and just vanished

working memory.

sensory memory.

short-term memory.

long-term memory.

Answer:

how many feet?

Explanation:

Answer:

D

Explanation:

Answer:

192

Explanation:

Correct option is B 8 cm.

Let radius of gyration for the axis not passing through center of mass be r and that for the axis passing through the center of mass be k and the distance between the two parallel axes be a.

Parallel axes theorem gives:

Thus, option B is the correct answer.

Answer:

she will move in the same direction at the same speed forever.

Explanation:

If there are no outside forces like gravity the net force will never change, she will just keep flying for forever and ever! poor lady

A Torque is a twisting force, or turning moment, it is a vector quantity with both magnitude and direction e.g Turning the handle of a cork-screw clockwise and then counterclockwise will advance the screw first inward and then outward By convention, counterclockwise torques are positive and clockwise torques are negative.

The direction is perpendicular to both the radius from the axis and to the force. It is conventional to choose it in the right hand rule direction along the axis of rotation.

Counterclockwise is the positive rotation direction and clockwise is the negative direction.

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Answer:

Remember, NORTH ^, EAST >, SOUTH v, WEST <

Explanation:

It doesn't have to be a super complex answer. All you have to do is look to the left (west) of the North American plate. What are the 2 plates that you see? The Pacific and the Juan de Fuca, yeah? To the South, there is the Cocos amongst a few others.

I am not going to share the answer for sure as I haven't completed the test yet but that's how I'm solving it. You should write the answer in your own words anyways. Hope this helps! Have a good day :)

Answer:

The Juan de Fuca Plate and the Pacific Plate both border the west side of the North American Plate.

Explanation:

Edmentum

Answer:

c) cornea being wavy or not spherical

Answer:

a) Impuise of force =F∗?(t) = area of F-T graph area= impulse =triangle + rectangle + triangle = 0.5*4*2 + 4*1 + 0.5*4*2 = 12 N-s (b) impulse = change in momentum \(= mExplanation:

[tex]r = 1.29×10^8\:\text{m}[/tex]

Explanation:

According to Newton's law of universal gravitation, the gravitational force between Uranus and Miranda is

[tex]F_G = G\dfrac{M_UM_M}{r^2}[/tex]

where [tex]M_U[/tex] is the mass of planet Uranus, [tex]M_M[/tex] is the mass of its satellite Miranda, r is the distance between their centers and G is the universal gravitational constant. Moving the variable r to the left side, we get

[tex]r^2 = G\dfrac{M_UM_M}{F_G}[/tex]

Taking the square root of the equation above, we get

[tex]r = \sqrt{G\dfrac{M_UM_M}{F_G}}[/tex]

Plugging in the values, we get

[tex]r = \sqrt{(6.67×10^{-11}\:\text{N-m}^2{\text{/kg}}^2)\dfrac{(8.68×10^{25}\:\text{kg})(6.59×10^{19}\:\text{kg})}{2.28×10^{19}\:\text{N}}}[/tex]

[tex]\:\:\:\:\:=1.29×10^8\:\text{m}[/tex]

Hi there!

With a Force/Displacement curve, we must take the integral (area underneath the curve) to calculate the work done.

We know that:

W = ΔKE

Calculate the work by finding the area underneath the force curve from

x = 0 to 3 m:

We can use a trapezoid:

A = 1/2(3 + 2)(3) = 7.5 J

This is the amount of work done, and since the object starts from rest:

7.5J = KEf - KEi (0 J)

7.5J = KEf = 7.5 J

Answer:

All organisms begin their lives as single cells.Overtime,these organisms grow and take on the characteristics of their species...All organisms grow,and different parts of organisms may grow at different rates.Organisims made out of only one cell

may change little during their lives, but they do grow

Explanation:

brainlest me please

Answer:

Explanation:

v = at

t = v/a

t = 20 m/s / 9.8 m/s²

t = 2.0408163...

t = 2.0 s

Answer:

Explanation:

The deflection of a charged particle by a magnetic field is proportional to its electric charge and to its velocity.   The deflection is also inversely proportional to its mass.   So given a proton and an electron going at the same velocity in a magnetic field and having equal (but opposite) electric charge the electron will deflect much more since the ratio of the masses is 1836.

Answer:

lol

Explanation:

. Power

Patintero

, also known as harangang-taga or tubigan, (Intl. Translate: Escape from the hell or Block the runner) is a traditional Filipino children's game. Along with tumbang preso, it is one of the most popular outdoor games played by children in the Philippines.[1]

is an ultramarathon trail race held in Frozen Head State Park near Wartburg, Tennessee. If runners complete 60 miles (97 km) this is known as a "fun run." The full course is about 100 miles (160 km). The race is limited to a 60-hour period and takes place in late March or early April of each year.

Answer:

Explanation:

α = Δω/t = (0 - ωi)/0.50 = -2ωi rad/s²

ωf² = ωi² + 2αθ

θ = (ωf² - ωi²) / 2α

2π/4 = (0² - ωi²) / (2(-2ωi))

2π/4 = ωi / 4

ωi = 2π rad/s

α = -2(2π) = -4π rad/s²

a = rα = 0.23(-4π) = 0.92π m/s² ≈ -2.89 m/s²

as the time to stop approaches zero, acceleration goes toward infinity.

Answer:

in general context yes it is closest to 1.04

Explanation:

theres no right or wrong way to scientifically prove this though.

Overall in scale its closest to 1.04 hope that helped

Answer:

Explanation:

The work will equal the increase in potential energy.

PE = mgh

m = PE/gh = W/gh = 201/(9.81(3.85)) = 5.32 kg

Hi there!

We can use work and energy to solve this problem.

We know that:

Ei = Ef

Ei = Potential energy = mgh

Ef = Rotational kinetic + Translational kinetic = 1/2Iω² + 1/2mv²

The barrel is comprised of a hollow cylinder and disk-shaped bottom, so:

I (hollow cylinder) = mr²

I (disk) = 1/2mr²

Calculate the moment of inertias of each.

Since the mass on the base is one-fourth of its side:

x = mass of side

x + x/4 = 15

4x + x = 60

5x = 60

x = 12 kg

end mass = 3 kg

Solve for each moment of inertia:

Side: (12)(0.4²) = 1.92 Kgm²

Bottom: 1/2(3)(0.4²) = 0.24 Kgm²

Side + bottom = 2.16 Kgm²

We can now solve:

mgh = 1/2mv² + 1/2(2.16)v²/r²

(15)(9.8)(33) = 1/2(15)v² + 1/2(13.5)v²

4851 = 14.25v²

v = 18.45 m/s

The slope of the road can be given as the ratio of the change in vertical

distance per unit change in horizontal distance.

Reasons:

Width of the truck = 2.4 meters

Height of the truck = 4.0 meters

Height of the center of gravity = 2.2 meters

Required:

The allowable steepness of the slope the truck can be parked without tipping over.

Solution:

Let, C represent the Center of Gravity, CG

At the tipping point, the angle of elevation of the slope = θ

Where;

[tex]tan\left(\theta \right) = \dfrac{\overline{AM}}{\overline{CM}}[/tex]

The steepness of the slope is therefore;

[tex]\mathrm{The \ steepness \ of \ the \ slope}= \dfrac{\overline{AM}}{\overline{CM}} \times 100[/tex]

Where;

[tex]\overline{AM}[/tex] = Half the width of the truck = [tex]\dfrac{2.4 \, m}{2}[/tex] = 1.2 m

[tex]\overline{CM}[/tex] = The elevation of the center of gravity above the ground = 2.2 m

[tex]\mathrm{The \ steepness \ of \ the \ slope}= \dfrac{1.2}{2.2} \times 100 \approx 54.55\%[/tex]

[tex]tan\left(\theta \right) = \mathbf{\dfrac{2.2}{1.2}} = \dfrac{11}{6}[/tex]

[tex]Elevation \ of \ the \ road \ \theta = arctan\left( \dfrac{6}{11} \right) \approx 28.6 ^{\circ}[/tex]

The maximum steepness of the slope where the truck can be parked is 54.55 %.

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