A box of mass 20 kg has a base of area 4 m2
. Calculate the pressure exerted by it on the ground. (Use 1
kg-f = 10 N)

Answer:

D is the answer

Explanation:

D is the most highest one so

the answer is D

Answer:

10 kg rock at 15 degrees celcius

Answer:

Explanation:

A decrease in the density of something is rarefaction. ... Most of the time, rarefaction refers to air or other gases becoming less dense. When rarefaction occurs, the particles in a gas become more spread out. You may come across this word in the context of sound waves.

The definition of the center of mass allows to find the result for the position of the mass center of more than the H₂O molecule is;

         [tex]x_{cm} = 0 \ and \ y_{cm} = 6.9 10^{-3 } nm[/tex]  

the concept of center of mass of a system is the point where external forces are applied, it is given by the expression

             [tex]\frac{x}{y} =\frac{1}{M_{total}} \sum m_i r_i[/tex]  

Where M is the total mass of the systemr_i and m_i sums the position and masses of the element i of the system

In the attachment we have a diagram of the system where the axis and coordinates of the molecules are shown, in this case it is indicated that the origin is in the oxygen atom, so its distance is zero.

           [tex]r_{cm} = \frac{1}{2 m + M} \ (2 m r )[/tex] )

They indicate the mass of the hydrogen atom m = 1.008 amu, the bond length r = 0.103 nm and there is an angle 106º between the two hydrogens, therefore the angle from the vertical is:

           θ = 106/2 = 53º

Let's find the position of the center of mass for each axis.

x-axis

           [tex]x_{cm} = \frac{1}{2m+ M} \ ( m x_1 - m x_2)[/tex]  

y-axis

          [tex]y_{cm} = \frac{1}{2m + M} \ ( m y_1 + m_2)[/tex]

Let's use trigonometry to find the components of the bond length.

         cos θ = [tex]\frac{y}{L}[/tex]  

         sin θ = [tex]\frac{x}{L}[/tex]  

         y = L cos θ

         x = L sin θ  

We substitute.

          [tex]x_{cm} = \frac{1}{2m+M} \ (mL (sin 53 + sin (-53)) \\y_{cm} = \frac{1}{2m + M} \ ( mL cos 53 + mL sin 53)[/tex]

we use.  

          sin θ = - sin -θ

          cos θ = cos -θ

Let's calculate.

        [tex]x_{cm} = 0 \\y_{cm} = \frac{1}{2 \ 1.008 + 15.9999} \ ( 2 \ 1.008 \ 0.103 cos 53)[/tex]

We see that the center of mass is on the x axis and at a distance from the y-axis of 6.9 10-3 nm

In conclusion using the definition of the center of mass we can find the result for the position of the center of mass of the H₂O molecule is;

          [tex]x_{cm}=0[/tex]  and [tex]y_{cm}[/tex]cm = 6.9 10⁻³ nm

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Answer:

Explanation:

Let v be the velocity of object ,

Given that mass of object , m=100 g=  

1000

100

​

kg=0.1 kg

& kinetic energy of object =20 J

K.E  =  

2

1

​

 mv  

2

20=  

2

1

​

×0.1v  

2

⟹v  

2

=400

⟹v=20 m/s

Answer:

The nuclei of source have greater stability than those of source B.

Explanation:

In everyday use and in kinematics, the speed of an object is the magnitude of the rate of change of its position with time or the magnitude of the change of its position per unit of time; it is thus a scalar quantity.

Answer:

the graph plotted based on your data, you did not mention what are the variables on the x-axis and y- axis . and the graphs can be three small graphs of different material of one quality i.e. thickness or length

hope it helps

Explanation:

(a) The normal force on the sleigh is 596.36 N.

(b) The magnitude and direction of acceleration of the sleigh is 3.2 m/s² upwards.

The given parameters;

The normal force on the sleigh is calculated as follows;

[tex]F_n = mg \times cos(\theta)\\\\F_n = 63 \times 9.8 \times cos(15)\\\\F_n = 596.36 \ N[/tex]

The magnitude and direction of acceleration of the sleigh is calculated as follows;

[tex]\Sigma F= ma\\\\F - mgsin(\theta) - F_f = ma\\\\F - mgsin(\theta) - \mu F_n = ma\\\\510\ - \ 63 \times 9.8 \times sin15 \ -\ 0.25\times 596.36 = 63a\\\\201 .11 = 63a\\\\a = \frac{201.11}{63} \\\\a = 3.2 \ m/s^2 \ upwards[/tex]

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Answer:

Energy = Power × Time

[tex]{ \tt{E = ( \frac{60}{1000} ) \times ( \frac{5}{60}) }} \\ \\ { \tt{E = 0.005 \: kwh {}^{ } }}[/tex]

Answer:

Yes

Explanation:

It's still the same orange with the same amount of stuff.

Answer:

please check the image

Explanation:

I hope this image helps you please follow the steps. Thank you

Answer:

1kg/```````2

Explanation:

solution

Answer:

i think its going to be 150 because its half of 300

Explanation:

this is my attachment answer hope it's helpful to you

Answer:

0

Explanation:

The bicyclist is not speeding up or slowing down so there is no acceleration.

Hi there!

Since we are dealing with such large distances, we can approximate.

Recall that the equation for speed is:

displacement / time = velocity

OR:

d/t = v

Begin by converting one hour to seconds:

1 hr  = 3600 sec

Now, we can solve for velocity:

(-2.0 × 10⁷) / 3600 = -5555.6 m/s ⇒ D: -5.55 × 10³ m/s

Answer: 100 pascals times 0.2 m^2= 20 N force is applied.

Explanation:

The kinetic energy of the toy before hitting the ground is 294.3 J

From the question given above, the following data were obtained:

Mass (m) = 3 Kg

Height (h) = 10 m

Acceleration due to gravity (g) = 9.81 m/s²

Before the toy will hit the ground, the potential and kinetic energy of the toy will be the same.

KE = PE = mgh

KE = 3 × 9.81 × 10

Therefore, the kinetic energy of the toy is 294.3 J

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Answer:

ETotal=38505J

Explanation:

Known values:

m=150kg

h=20m

v=11m/s

KE=mv^2/2

PE=mgh

ETotal=PE+KE

KE=(150(11^2))/2

KE=9075

PE=150(9.81)(20)

PE=29430

ETotal=9075+29430

ETotal=38505J

Answer:

11.1

Explanation:

2 hours 15 mins = 2 1/4 hours

2 1/4 hours = 25 miles

1 hour = 25 ÷ 2 1/4 = 25 ÷ 9/4 = 25 x 4/9 = 11.1 miles

Answer:

4.8 N*s

Explanation:

P = F*t

P = Impulse/ momentum

F = force (Newtons)

t = time (seconds)

P = 9.5N*0.5s

P = 4.75 N*s

Hi there!

[tex]\large\boxed{I = 4.8 N}[/tex]

Impulse = Force (N) × Time (s)

We can also express this as:

I = Δp = Ft

Plug in the given values:

I = 9.5(.50) = 4.75 ≈ 4.8 Ns

Answer:

Explanation:

BMI= weight/(height × height)          ; weight in kilogram and height in metter

     = 58kg / (1.61m  × 1.61m )

     = (58/ 2.5921) kg/[tex]m^{2}[/tex]

     = 22.375  kg/[tex]m^{2}[/tex]

     ≈ 22.4 kg/[tex]m^{2}[/tex]

Answer:

180 meters

Explanation:

Convert km to meters:

1km = 1000 m

1.5 km x 1000 = 1500 m

1500m / 4.8 m/s = 312.5 seconds

312.5 seconds / 60 seconds per minute =5.2 minutes = 5 minutes 12.5 seconds

Answer:

Friction or tension

Explanation:

Friction stops an object from moving in the presence of force

Answer:

Chemical modification of starch is based on the chemical reactivity of the constituent glucose monomers which are polyhydroxyl and can undergo several reactions. Starch can undergo reactions such as hydrolysis, esterification, etherification and oxidation.

Explanation:

Hope this helps!

The question is fishing for "balanced force".

But the description in the question is terrible.

Answer:

If and only if the object has 0 speed. Any other case is a no.

Explanation:

In increasing math complexity:

If the graph of a distance/time is a straight line with a non-zero slope it means that the velocity-time graph is an horizontal line where every point has the same second coordinate, equal to the value of the slope of the first line (ie, if the first line has equation [tex]y=2t[/tex] the second line will have equation [tex]y=2[/tex] - I'm using y to denote either the distance or the speed, read it as "the non-time variable")

If the graph of a speed/time is a straight line, the graph of the distance/time is the area of the region of space beneath the straight line in the first graph, ie a triangle. And as you increase the sides of that triangle keeping the angles fixed, you get a parabolic curve.

Bonus case: if [tex]v(t)= 0; d(t) = 0[/tex], they are both straight lines, they both pass through the origin, and they are the exact same line.

With calculus (or: math becomes fun after you add the alphabet to it)

You are asking if the answer of the IVP

[tex]\left \{ {{x'=x} \atop {x(0)=0}} \right.[/tex]

is a linear function. The answer is: besides the obvious [tex]x(t)'=x(t)=0 \ \forall t[/tex] there are no other solutions, linear or otherwise.