6. A swimming pool has the dimensions (1.w.h)
28.5 mx 10.0 mx 2.30 m. When it is full of water,
what is the force on the bottom?

Answer:

One idea is how, on an astronomical level, the planets can be observed to spread apart continuously.

Explanation:

Throughout various studies, many have observed that the planets continuously grow distant from each other. In fact, it is observed through our technological advances that the very galaxies are slowly drifting apart. Thus, indicating there must have been some big bang to set these galaxies and planets in motion.

Answer:

D

Explanation:

"Simple machines cannot change the amount of work done, but they can reduce the effort force that is required to do the work!"

source: https://www.smore.com/u609q-simple-machines

The resultant displacement of the Polly protractor is 25 km.

The displacement of an object is the change in the position of the object.

The displacement of the Polly protractor is calculated as follows;

d² = 15² + 20²

d = √(15² + 20²)

d = 25 km

Thus, the resultant displacement of the Polly protractor is 25 km.

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The complete question is below:

Find the resultant displacement of the Polly protractor.

The acceleration of the boat is 2.5 m/s².

The weight of a small sailboat is 200 kg.

The accelerating force of the wind on the boat is 707 N northwest and 500 N acting west.

Taking into consideration the wind's accelerating force. It can be visualized as the combination of two elements:

707 /√2 = 500 N acting north, and

707 /√ 2 = 500 N acting east.

It is acting both to the North and to the East because it is supposed to be acting in the Northeast wards. The drag of the keel's eastward component will then cancel out its westward counterpart.

Essentially, 500 N north will then be the result of two forces acting on the boat.

Then, we divide the force by the specified mass in order to obtain the boat's acceleration.

500/200 = 2.5 m/s² northward.

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The supplied statement states that B has the dimensions L/T².

Measurements like length, width, or height are referred to as dimensions. When you discuss an object or location's dimensions, you're referring to its size and proportions.

Dimensions include length, breadth, and height. One dimension is that of a line, two dimensions are those of a square, and three dimensions are those of a cube (3D).

Just rearranging the equation will get the dimensions.

x = Bt² ----> B= x/t²

Now you are told that x is displacement (L) and t is time (T) so sub these in.

B= L/T² Therefore the dimensions are L/T².

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Answer:

God bless you are a hacker the following reas of

Explanation:

God bless you forever to get

The mass of liquid in the container is 94 grams and the density of the liquid is 0.78 grams/cm³ or 780 Kg/m³.

Given in the question

The volume of the liquid = Volume of the container

                                  =  120 cm³

Total Mass = 145 grams

Mass of the empty Beaker = 51 grams

To find

Mass of the liquid

The density of the liquid

Now,

Total Mass = Mass of LIquid + Mass of the empty Beaker

Mass of LIquid = Total Mass -  Mass of the empty Beaker  

Put in the value, we get

Mass of liquid = 145 - 51 grams

Mass of liquid = 94 grams

Now, Density is the ratio of the mass of the substance and the volume, the substance occupies in the space. It is not a constant value and is variable with the variation in the temperature of the substance. Its S.I. unit is Kg/m³. B ut its other commonly used unit is gram/cm³.

The density of liquid = Mass of the liquid/ Volume of the liquid

Put in the value, we get

The density of the liquid = 94/120  grams/cm³

The density of the liquid = 0.78 grams/cm³ = 780 Kg/m³

Therefore, the mass of liquid in the container is 94 grams and the density of the liquid is 0.78 grams/cm³ or 780 Kg/m³.

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Your question was incomplete. Please find the full content below.

A Measuring Cylinder liquid contains a volume of 120cm³ the liquid was then poured into an empty beaker of mass 51g. The total mass was then found to be 145g . Calculate the mass of the liquid and the density of the liquid.

The distance travelled in that month is 870.92 miles

We'll begin by calculating the circumference of the tyre. This can be obatined as follow:

C = 2πr

C = 2 × 3.14 × 13

C = 81.64 in

63360 inches = 1 mile

Therefore,

81.64 inches = (81.64 inches × 1 mile) / 63360 inches

81.64 inches = 0.00129 mile

Distance = number × circumference

Distance = 0.00129 × 675135

Distance = 870.92 miles

Thus, the distance travelled in that month is 870.92 miles

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red depicts alkali metals, green depicts alkaline earth metals, blue depicts transition metals, and purple depicts inner transition metals.

alkali metal, any of the six chemical elements that make up Group 1 (Ia) of the periodic table—namely, lithium (Li), sodium (Na), potassium (K), rubidium (Rb), cesium (Cs), and francium (Fr).

Group 2A of the periodic table are the alkaline earth metals: beryllium (Be), magnesium (Mg), calcium (Ca), strontium (Sr), barium (Ba), and radium (Ra). They are harder and less reactive than the alkali metals of Group 1A.

transition metal, any of various chemical elements that have valence electrons—i.e., electrons that can participate in the formation of chemical bonds—in two shells instead of only one.

The period 6 inner transition metals (lanthanides) and period 7 are actinides.

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Answer:

13 m/s

Explanation:

Given average acceleration a and initial velocity  v0 we can compute the final velocity  using the equation

a = (v - v0)/t where t is the time taken to reach final velocity from initial velocity

Here a = 2 m/s² and v0 is the initial velocity when she passes her younger brother v0 = 5m/s.

It takes t = 4 s to pass her older brother at which time her speed is v, say

Then

a = (v - v0)/t

Plugging in known values,

2 = (v - 5)/4

Switch sides:
(v- 5)/4 = 2

Multiply both sides by 4:
v - 5  = 8

v =5 + 8 = 13 m/s - the speed at which she passes her older brother

Answer:

200m west

Explanation:

Displacement = v.t where v is the velocity of the object ad t = time

At 50m/hr for 4 hours

Displacement = 50 x 4 = 200m west of origin

The total displacement of the object was 7.5 meters.

There's a relation between the common movement analysis graphs that we can always use. These relations are the following:

The derivative (slope) of a displacement vs. time graph gives you the function of speed vs. time. Also, when you take the speed vs. time graph and find its derivative, you will get the acceleration function of the observed body.

Now, doing to opposite of derivation (integration), we are able to calculate the displacement of an object when we integrate the function of velocity vs. time. This is not a complicated process when we have a linear function, because integration is just calculating the area under the curve or line of a graph.

In this case, the graph forms a right triangle with the x axis, therefore, we are able to calculate displacement by finding the area under the line of the graph. It could also be made with integration, we are going to do it with both processes as they are both used, and in some cases you have to strictly use one of the 2 in order to get the most precise answers.

When we observe the line, we can crearly tell that the area under the graph constitutes a right triangle. Hence, the area under the graph is given by the following formula:

[tex]A=\frac{bh}{2}[/tex]

Substitute the values and calculate.

(Check attached image 1 to see where the values of b and h are taken from).

[tex]A=\frac{(5)(3)}{2}\\ \\A=7.5[/tex]

Hence, the total displacement of the object was 7.5 meters.

In this method, we'll need to find the equation of the line and then generate and solve a definite integral.

Take 2 points from the line.

(0, 0) and (5, 3).

Find the slope.

Formula: [tex]m=\frac{y_{2} -y_{1} }{x_{2}- x_{1} }[/tex]

[tex]m=\frac{3-0}{5-0}= \\ \\m= \frac{3}{5}[/tex]

Find the equation of the line.

Formula: [tex]y-y_{1} =m(x-x_{1} )[/tex]

[tex]y-(0) =\frac{3}{5} (x-(0) )\\ \\y=\frac{3}{5} x\\ \\[/tex]

Integrate within the interval of interest (from 0 to 5).

[tex]\int\limits^5_0 {(\frac{3}{5} x)} \, dx \\ \\=\frac{3}{5} x^{2} *\frac{1}{2} \\ \\=\frac{3}{10} x^{2} \\\\ (\frac{3}{10} (5)^{2})-(\frac{3}{10} (0)^{2})\\ \\(\frac{3}{10} *25)-(0)\\ \\\frac{75}{10} \\ \\7.5[/tex]

Once again, the total displacement of the object was 7.5 meters.

The Moment of inertia of the cylinder is equal to 8.82 Kgm².

The propensity of a force to produce or modify the rotational motion of a body is known as torque, also known as moment, or moment of force. It is a force that causes an object to twist or turn. Force and distance are multiplied to calculate torque. It has both a direction and a magnitude because it is a vector quantity. Newton-meters, or N*m, is the torque unit in the SI system.

A spinning object's change in angular velocity per unit of time is expressed quantitatively as angular acceleration, also known as rotational acceleration. It has a vector form.

Given in the question

Torque = 75 Nm

Angular acceleration = 8.5 rad/s²

Torque = Angular acceleration × moment of inertia

Put in the value, we get

Moment of inertia = 75/ 8.5

Moment of inertia = 8.82 Kgm²

So the Moment of inertia of the cylinder is equal to 8.82 Kgm².

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There is no such things as an exact measurement there can never be a perfect instrument or measurer.

The term measurement has to do with the process of obtaining the value of a quantity. We know that measurements are done in order to give us an idea of the magnitude of a quantity. In every measurement, there is an instrument that is used for the measurement. For instance, if I want to measure length, I may have to use a meter rule to do it.

Now, there are two ways that error could be introduced into a measurement;

1) By the instrument that is used

2) By the error of the measurer

Hence, there is no such things as an exact measurement there can never be a perfect instrument or measurer.

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(a) Time taken by the stone to fall into the water will be t = 2.99 sec

(b) The speed with which the stone strikes the water will be v=32.91 m/s

(a)

As the stone is kicked horizontally so that the initial velocity of the stone will be [tex]v_{0} = 0[/tex]

For finding the time taken by the stone to fall into the water will be found using the following equation

[tex]y=v_{0}t + \frac{1}{2}at^{2}[/tex]

[tex]y= \frac{1}{2}at^{2}[/tex]

[tex]t=\sqrt{\frac{2y}{a} }[/tex]

[tex]t=\sqrt{\frac{2*(-44 m)}{(-9.8 m/m^{2} )} }[/tex]

[tex]t=2.99 s[/tex]

Time taken by the stone to fall into the water will be t = 2.99 sec

(b)

Since the ball was thrown horizontally, its velocity in the horizontal axis is constant and equal to its beginning velocity.

[tex]v_{0x} = v_{0} = 15 m/s[/tex]

Final velocity at vertical axis is

[tex]v_{y} = v_{0y} + at[/tex]

[tex]v_{y} = at[/tex]

therefore,

time = 2.99 sec

[tex]v_{y} = (-9.8 m/s^{2})(2.99 sec)\\ v_{y} = -29.3 m/s^{}[/tex]

The final velocity is,

[tex]v_{}=\sqrt{v_{0x} ^{2} + v_{y} ^{2} }[/tex]

[tex]v_{}=\sqrt{(15 m/s) ^{2} + (-29.3 m/s) ^{2} }[/tex]

v=32.91 m/s

The speed with which the stone strikes the water will be v=32.91 m/s

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Answer:

The mass of the person is 56.83 kg

Explanation:

W= 557 N

We know that acceleration due to gravity (g)= 9.8 m/s^2

Now,

W= mass * acceleration due to gravity

557=m * 9.8

m= 557/9.8

m= 56.83

Answer:

her mass is 55.7 Kg.

Explanation:

Weight = Mass × Acceleration

N =ma [a= g for gravitational force acting down word]

N=Mg

find M so make mass become the subject therefore,

Mass = force / gravitational acceleration [let g=10m/s^2]

m=557 / 10

m= 55.7 kg

The maximum height reached by the baseball is 31.89 m and the time taken to reach there is 1.3 seconds

Linear motions are straight line motions. It can be horizontal motion, or vertical motion which is motion under gravity.

Given that a baseball is hit almost straight up into the air with a speed of 25m/s estimate how high it goes.

At the maximum height, final velocity V = 0

Using third equation of motion, that is, V² = U² - 2gH, we can estimate how high it goes.

0 = 25² - 2 × 9.8 × H

625 = 19.6H

H = 625/19.6

H = 31.89 m

To know how long it is in the air, the factors that make this estimate will be

Speed = s / t

25 = 31.89 / t

t = 31.89 / 25

t = 1.2756

t = 1.3 s

Therefore, the maximum height reached by the baseball is 31.89 m and the time taken to reach there is 1.3 seconds

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Since the graph provided represents voltage and current data for resistor R, the resistance of R is 2.0 Ω

The resistance of a metallic material is its opposition to the flow of electric current.

Resistance is given by R = V/I where

To find the resistance of R, we take the gradient of the graph since reistance R = V/I.

So, the resistance R = ΔV/ΔI where

R = ΔV/ΔI

= (V' - V)/(I' - I)

Taking points from the graph

So, substituting the values of the variables into the equation, we have

R  = (V' - V)/(I' - I)

= (9.0 V - 0 V)/(4.5 A - 0 A)  

= 9.0 V/4.5 A

= 2.0 Ω

So, the resistance of R is 2.0 Ω

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Answer:

tr

Explanation:

ewetwer

Answer:

Im pretty sure its B

Explanation:

correct me if wrong

Answer:

Explanation:

the rilation shhip betveen boats annd volummes are ver expennsive

The buoyancy of a boat and the density of the materials used in its construction have an impact on the connection between its volume and the amount of mass it can support before sinking.

Thus, The buoyancy principle asserts that the upward force a fluid (in this case, water) exerts on a submerged or floating item is equal to the weight of the fluid displaced by the object, which is why a boat floats in water.

A boat moves as much water as its own volume while it is submerged in water.

A boat will float if its total weight, including its passengers and cargo, is less than the weight of the water it moves.

Thus, The buoyancy of a boat and the density of the materials used in its construction have an impact on the connection between its volume and the amount of mass it can support before sinking.

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1 The ball travel 58.29 m high

2. The time the ball spend in the air is 8.67 s

The maximum height reached by the ball can be obtained as follow:

v² = u² – 2gh (since the ball is going against gravity)

0² = 33.8² – (2 × 9.8 × h)

0 = 1142.44 – 19.6h

Collect like terms

0 – 1142.44 = –19.6h

–1142.44 = –19.6h

Divide both side by –19.6

h = –1142.44 / –19.6

h = 58.29 m

We'll begin by calculating the time to reach the maximum height . This can be obtained as follow:

v = u – gt (since the ball is going against gravity)

0 = 33.8 – (9.8 × t₁)

0 = 33.8 – 9.8t₁

Collect like terms

0 – 33.8 = –9.8t₁

–33.8 = –9.8t

Divide both side by –9.8

t₁ = –33.8 / –9.8

t₁ = 3.45 s

Next, we shall determine the time take to fall to the ground

h = ½gt²

133.29 = ½ × 9.8 × t₂²

133.29 = 4.9 × t₂²

Divide both side by 4.9

t₂² = 133.29 / 4.9

Take the square root of both side

t₂ = √(133.29 / 4.9)

t₂ = 5.22 s

Finally, we shall determine the time spend by the ball in the air. This is shown below:

T = t₁ + t₂

T = 3.45 + 5.22

T = 8.67 s

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The correct answer is 17.12 meters.

S = VoT - (1/2)g[tex]T^2[/tex]

where

S = distance travelled by the body

Vo = initial velocity

T = travel time

g = acceleration due to gravity = 9.8 m/sec^2 (constant)

For the ball,

Sb = 20T - (1/2)(9.8)(  )

and for the stone,

Ss = 24(T - 2) + (1/2)(9.8)(T - 2[tex])^2[/tex]

and for them to pass each other from the release point,

Sb = Ss

and therefore

20T - (1/2)(9.8)[tex]T^2[/tex] = 24(T - 2) - (1/2)(9.8)(T - 2[tex])^2[/tex]

20T - 4.9[tex]T^2[/tex] = 24T - 48 - 4.9([tex]T^2[/tex] - 4T + 4)

20T - 4.9[tex]T^2[/tex] = 24T - 48 - 4.9[tex]T^2[/tex] + 19.6T - 19.6

Combining like terms,

24T + 19.6T - 20T = 48 +19.6

23.6T = 67.6

T = 2.86 sec.

In other words, the stone will pass the ball 2.86 seconds after the ball has been tossed (or 0.86 sec after the stone has been tossed).

Solving for Sb,

Sb = 20(2.86) - (1/2)(9.8)([tex]2.86^2[/tex])

Sb = 17.12 meters -the ball will be passed by the stone at this elevation above the release point. This is the first option in your given choices.

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(i) The angle of inclination of the string to the vertical is 65⁰.

(ii) The tension in the string is determined as 2.11 N.

The angular speed of the string is calculated as follows;

ω = 2πN/60

ω = 2π x 80/60

ω = 8.38 rad/s

v = ωr

v = 8.38 x 0.3 m

v = 2.514 m/s

tanθ = v²/rg

tan θ = (2.514²) / (0.3 x 9.8)

tan θ = 2.15

θ = arc tan(2.15)

θ = 65⁰

T = mv²/r

T = (0.1 x 2.514²) / (0.3)

T = 2.11 N

Thus, the tension in the string is determined as 2.11 N.

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Knowing that Newton's first law is the law of inertia.

The law of inertia helps us when we have problems with bodies that are motionless or moving at constant speed.

The first law tells us that: "A body remains at rest or moves with constant speed if a resultant force equal to zero acts on it."

If the body is at rest or moves with constant speed without acceleration, then the sum of the forces in both the x, y or z axis is equal to zero.

Answer:

[tex]{ \tt{ {v}^{2} = {u}^{2} + 2as}} \\ { \tt{20 = 0 + (2 \times a \times 100)}} \\ { \tt{20 = 200a}} \\ { \tt{a = 0.1 \: {ms}^{ - 2} }}[/tex]

Answer:

Si un cuerpo de masa m se encuentra a una altura h sobre la superficie de la Tierra, la fuerza gravitacional que actúa sobre él es: De acuerdo con la segunda ley de Newton, esta fuerza gravitatoria imprimirá una aceleración en el cuerpo

Explanation:

According to the research, the correct option are protection, thermoregulation and immune surveillance. The three functions of the integumentary system are protection, thermoregulation, and immune surveillance.

It is made up of a deep layer of connective tissue that is directly related to the environment in which it has the peculiarity of the abundance of collagen fibers.

In this sense, its functions include protection against mechanical, thermal, and chemical-toxic injuries and against microorganisms, regulation of body temperature, and immune surveillance and response to external agents.

Therefore, we can conclude that according to the research, the correct option are protection, thermoregulation and immune surveillance. The three functions of the integumentary system are protection, thermoregulation, and immune surveillance.

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Answer:

The pelican was travelling at approximately [tex]9.5\; {\rm m\cdot s^{-1}}[/tex] when it dropped the fish.

The fish hit the water at approximately [tex]16\; {\rm m\cdot s^{-1}}[/tex].

(Assumption: air resistance on the fish is negligible; [tex]g = 9.81\; {\rm m\cdot s^{-2}}[/tex].)

Explanation:

During the descent:

Since acceleration in the vertical direction is constant, make use of the SUVAT equation [tex]x = (1/2)\, a\, t^{2} + v_{0}\, t[/tex] to find the duration [tex]t[/tex] of the descent:

Rearrange this equation to find the duration of the descent, [tex]t[/tex]:

[tex]\begin{aligned} t &= \sqrt{\frac{2\, x}{a}} \\ &= \sqrt{\frac{2 \times 7.8\; {\rm m}}{9.81 \; {\rm m\cdot s^{-2}}} \\ &\approx 1.261\; {\rm s}\end{aligned}[/tex].

The vertical speed of the fish right before impact would be:

[tex]\begin{aligned} v_{y} &= a\, t \\ &\approx 9.81\; {\rm m\cdot s^{-2}} \times 1.261\; {\rm s} \\ &\approx 12.37\; {\rm m\cdot s^{-1}} \end{aligned}[/tex].

In the horizontal direction, the speed of the fish was constant- same as the initial speed of the pelican. The fish travelled a horizontal distance of [tex]x = 12.0 \; {\rm m}[/tex] within the [tex]t \approx 1.26\; {\rm s}[/tex] of the descent. As a result, the horizontal velocity of the fish would be:

[tex]\begin{aligned} v_{x} &= \frac{x}{t}\\ &\approx \frac{12.0\; {\rm m}}{1.261\; {\rm s}} \\ &\approx 9.52\; {\rm m\cdot s^{-1}} \end{aligned}[/tex].

Hence, the initial speed of the pelican would be approximately [tex]9.5\; {\rm m\cdot s^{-1}}[/tex].

Thus, right before impact:

Apply the Pythagorean Theorem to find the overall velocity of the fish at that moment:

[tex]\begin{aligned} v &= \sqrt{{v_{x}}^{2} + {v_{y}}^{2}} \\ &\approx \sqrt{{(9.52\; {\rm m\cdot s^{-1}})}^{2} + {(12.37\; {\rm m\cdot s^{-1}})}^{2}} \\ &\approx 16\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].

In other words, the fish would hit the water at approximately [tex]16\; {\rm m\cdot s^{-1}}[/tex].