You sit on ice and shove a heavy box with your feet with a given force. What will you and the box share? *
A) Same acceleration
B) equal and opposite acceleration
C) the equal and opposite force
D) same force
explain please

Answer: Mean weight = 2.81 N and  Measurement of uncertainty = 0.82 N

Explanation:

Mean =  [tex]\dfrac{\text{Sum of observations}}{\text{number of observations}}[/tex]

Mean weight is [tex]($\mu)=\frac{2.8+2.6+2.9+3.1+2.4+2.9+3.2+2.5+2.7+3.0}{10}[/tex]

[tex]=2.81$[/tex]

[tex]$\sum_{i=1}^{10}\left(x_{i}-\mu\right)^{2}=0.61[/tex]

[tex]$\sigma=\sqrt{\frac{1}{N-1} \sum_{i=1}^{10}\left(x_{i}-\mu\right)^{2}}[/tex]

[tex]=\sqrt{\frac{1}{10-1} 0.61}=0.082$[/tex]

Measurement of uncertainty will be  [tex]$\sigma=0.082$[/tex]  

hence,  Weight [tex]$W=2.81 \pm 0.082 N$[/tex]

When we exhale, 90% waste material is Carbon Dioxide ( CO2 ) , so, it gets exhaled out in the form of CO2 rich air and it gets removed from the body, therefore our internal body becomes more pure and helps in making our internal temperature constant at a suitable level.

Answer:

The center of gravity of the tile is at the center of the tile:

Examining 1/2 of the tile, each mass  point will have an equal mass point

on the other side of the tile (considering a line thru the center of the tile)

with coordinates opposite and equal of the point being considered.

Each mass point will have an identical mass point on the opposite side of the tile, with coordinates opposite and equal to the point being analyzed, while looking at only half of the tile.

A hypothetical location where the gravitational force is said to appear to act is the human body's centre of gravity (COG). That is the location where the entire body's bulk appears to be concentrated[2]. The COG need not be located inside the actual boundaries of an object or a person because it is a hypothetical point. Imagine the thing balanced on one finger to get an idea of its COG (there are objective measurements).

The COG is positioned anatomically so that it is about anterior to the second sacral vertebra. The specific placement of the COG, however, continually shifts with each new posture of the torso and limbs since humans do not remain stable in the anatomical position. The dimensions of the body

The center of the gravity is at half of the length of tile.

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Answer:

B. -20 m/s

Explanation:

Given the following data:

Initial velocity = 30 m/s

Acceleration = 5 m/s²

Deceleration = -5 m/s² (deceleration is the negative of acceleration)

Time = 10 seconds

To find the final velocity, we would use the first equation of motion;

[tex] V = U + at[/tex]

Where;

V is the final velocity.

U is the initial velocity.

a is the acceleration.

t is the time measured in seconds.

Substituting into the formula, we have;

[tex] V = 30 + (-5)*10[/tex]

[tex] V = 30 - 50[/tex]

Final velocity, V = -20 m/s

Answer:

In the above diagram, we move by horizontally with a velocity to the right. But due to gravity the hand falls a distance as we move .

Therefore the time it takes to fall by a distance is given by =.

If = as →∞, then from the laws of motion

=12()22

Therefore for the whole journey from left to right our muscle has to give a beat up at an amount equal to the gravity force. So, the work supplied by the muscle for one trip would be

∫=∫012()22

A man is going to rub the chalk off a blackboard, he is going to choose a way to rub off the chalk in two ways

Starting from the upper left corner of the board and moving horizontally to the right and moving slightly down and then again moving to the right and then when it reaches the left corner and so on

The second way is given in the following: He starts at the top and rubs down and then up again and so on.

He is going to choose the way that is less tiring in his arms. i.e., less work done by his muscles!

Somehow from vague intuitive notion I am inclined to agree that it would be less tiring to rub off the chalk if he moves the duster beginning at the top and rubbing horizontally and gradually decreasing the height when one reach the corners till everything is off.

I tried to calculate the muscle work by considering the following ideas(please correct me if I'm wrong):

The nerves inside our muscle has to fire its signal continuously throughout the entire interval of the process of rubbing and its has to oppose the gravitational force keeping the hand up the air.

When the hand is at the highest point on the blackboard, the muscle has to work against gravity =, as the muscle continues to work against gravity throughout the interval from left to right (the horizontal path of the first case) I couldn't find the total work withstood by our muscle.

Answer:

Explanation:

The first thing we are asked to find is the Force experienced by the box. That is found in the formula:

F - f = ma where F is the force exerted by the box, f is the friction opposing the box, m is the mass, and a is the acceleration (NOT the same as the pull of gravity). But F can be rewritten in terms of the angle of inclination also:

[tex]wsin\theta-f=ma[/tex] where w is the weight of the box. We will use this version of the formula because it will help us answer the second question, which is to solve for a. Filling in:

First we need the weight of the box. Having the mass, we find the weight:

w = mg so

w = 25.14(10) so

w = 251.4 N (I am not paying any attention at all to the sig fig's here, since I noticed no one on this site does!) Now we have the weight. Filling that in:

251.4sin(30) - f = ma Before we go on to fill in for f, let's answer the first question. F = 251.4sin(30) so

F = 125.7   And in order to answer what a is equal to, we find f:

f = μ[tex]F_n[/tex] where Fn is the weight of the object.

f = .25(251.4) so

f = 62.85. Filling everything in now altogether to solve for a, the only missing value:

125.7 - 62.85 = 25.14a and

62.85 = 25.14a so

a = 2.5 m/s/s

Now we have to move on to another set of equations to answer the last part. The last part involves the y-dimension. In this dimension, what we know is that

a = -10 m/s/s

v₀ = 0 (it starts from rest)

Δx = -13.30 m (negative because the box falls this fr below the point fro which it started). Putting all that together in the equation for displacement:

Δx = v₀t + [tex]\frac{1}{2}at^2[/tex] and we are solving for time:

[tex]-13.30=0t+\frac{1}{2}(-10)t^2[/tex] and

[tex]t=\sqrt{\frac{2(-13.30)}{-10} }[/tex] so

t = 1.6 seconds to reach the bottom of the slope from 13.30 m high.

Answer:

[tex]5,2[/tex]

Explanation:

From the question we are told that:

Speed of light [tex]C=3.0×10^8 m/s.[/tex]

Generally the equation for Average Speed is mathematically given by

[tex]V_{avg}=\frac{d}{t}[/tex]

Where

d=Distance between the Earth and the sun

[tex]d=1.5*10^11m[/tex]

Therefore

[tex]t=\frac{d}{V_{avg}}[/tex]

[tex]t=\frac{1.5*10^11m}{3.0×10^8 m/s.}[/tex]

[tex]t=5*10^2s[/tex]

Since m and n is given in the form of

[tex]m*10^n[/tex]

Therefore

[tex]m=5 & n=2[/tex]

[tex]5,2[/tex]

Answer:

C

Explanation:

The normal is the line which divides the angle between the incident ray (which is the ray of an object which strikes the mirror) and the reflected ray(the ray which is thrown back as the object hits the mirror surface) into two equal parts. The normal is always perpendicular to the surface. In the description agram Given , the Noa which is the line C, divides the reflected ray (line D) and the incident ray (line A) into two equal parts. The plane surface is line B and the other incident ray (line C) is perpendicular to B

Answer:

[tex]E=6.68\times 10^{-26}\ J[/tex]

Explanation:

Given that,

The frequency of FM ratio station, f = 100.8 MHz = 100.8 × 10⁶ Hz

We need to find the energy of the wave. We know that,

Energy, E = hf

Put all the values,

[tex]E=6.63\times 10^{-34}\times 100.8\times 10^6\\\\=6.68\times 10^{-26}\ J[/tex]

So, the energy of the wave is equal to [tex]6.68\times 10^{-26}\ J[/tex].

Answer:

A. the motion of electrically charged particles

Answer: A 10,000-Hz sound is 10 times more intense as compared to a 1000-Hz sound to be perceived as equal to 60 phons of loudness.

Explanation:

The formula used is as follows.

[tex]\beta = 10 dB log (\frac{I}{I_{o}})\\60 = 10 dB log (\frac{I}{I_{o}})[/tex]

[tex]I_{o} = 10^{-12}[/tex] normal threshold

The difference is sound level is as follows.

60 - 60 = 0

Hence,

[tex]0 = 10 dB [log (\frac{I_{f}}{I_{o}}) - log (\frac{I_{i}}{I_{o}})]\\log (\frac{1000}{I_{o}}) = log (\frac{10000 x}{I_{o}})\\log (10^{15}) = log (10^{16}x)\\15 = 16 + log x\\log x = 1\\x = 10[/tex]

This means that 10,000 Hz sound is 10 times more intense.

Thus, we can conclude that a 10,000-Hz sound is 10 times more intense as compared to a 1000-Hz sound to be perceived as equal to 60 phons of loudness.

Answer:

the correct answer is D

Explanation:

In this exercise, the vectors are in the same west-east direction, so we can assume that the positive direction is east and perform the algebraic sum.

       R = δ + ε

where

       δ = 2.0 m

       ε = 7.0 m

the positive sign indicates that it is heading east

        R = 2.0 + 7.0

        R = 9.0 m

the direction is east

the correct answer is D

Answer:

the process of finding the age of fossils is called carbon dating because they use samples of carbon and compare it's characteristics to the samples of carbon from each set time

Answer:

The downwards acceleration is 3.53 m/s2.

Explanation:

Let the true weight is m g.

The reading of the balance, R = 0.64 mg

Let the acceleration is a.

As the apparent weight is less than the true weight so the elevator goes down wards with some acceleration.

Use Newton's second law

m g - R = m a

m g - 0.64 m g = m a

0.36 g = a

a = 3.53 m/s2

Answer:

The acceleration of the object is -69.78 m/s²

Explanation:

Given;

postion of the particle:

[tex]x = 32t - 38\frac{t^3}{3} \\\\[/tex]

The velocity of the particle is calculated as the change in the position of the  particle with time;

[tex]v = \frac{dx}{dt} = 32 - 38t^2\\\\when \ the \ particle \ is \ at \ rest, \ v = 0\\\\32-38t^2 = 0\\\\38t^2 = 32\\\\t^2 = \frac{32}{38} \\\\t = \sqrt{\frac{32}{38} } \\\\t = 0.918 \ s[/tex]

Acceleration is the change in velocity with time;

[tex]a = \frac{dv}{dt} = -76t\\\\recall , \ t = 0.918 \ s\\\\a = -76(0.918)\\\\a = -69.78 \ m/s^2[/tex]

Answer:

Coefficient of friction =  Magnitude depends on the interacting materials

Friction = A force that acts parallel to the surface.

kinetic friction = A force that is constant regardless of the applied force

Normal = A force that acts perpendicular to the surface

Static friction = A force that increases as applied force increases up to some maximum value

Explanation:

Let's define all the forces, and then let's solve the problem.

Normal force:

When an object rests on some place (like a book on the table) the force that causes the book to not fall through the table is called the normal force, which is usually equal to the weight of the object and acts perpendicular to the surface where the object is resting.

Friction force.

When an object moves (or tries to move) parallel to a surface, such that the object is in contact with that surface, there appears a force that opposes to the movement (the force is parallel to the surface, and in the opposite direction to the movement).

And this force can be written as:

F = -N*μ

Where μ is the coefficient of friction and N is the normal force.

If the object is not moving yet (but there is applied a force that would move the object) the coefficient is called the coefficient of static friction which increases in a given range, until it can't keep increasing and the object starts to move, while if the object is moving, the coefficient is called the coefficient of kinetic friction and it is constant, where usually the first one is larger than the second, and these coefficients depend on both materials, the surface one and the object one.

Then we have two friction forces, one called the kinetic friction and the other called the static friction.

Then:

Coefficient of friction =  Magnitude depends on the interacting materials

Friction = A force that acts parallel to the surface.

kinetic friction = A force that is constant regardless of the applied force

Normal = A force that acts perpendicular to the surface

Static friction = A force that increases as applied force increases up to some maximum value

Answer:

The dipole always tries to align in the direction of electric field so that the net torque is zero.

Explanation:

When a dipole placed in an electric field at any arbitrary angle, it experiences a torque.

Let the dipole moment is p, electric field is E and the angle is A.  

The torque is given by

Torque = p E sin A

where, A is the angle between the electric field and dipole moment.

So, the dipole always tries to align in the direction of electric field so that the net torque is zero.

Answer:

(a) The acceleration is 2.35 m/s^2.

(b) The time is 5.53 s.

Explanation:

coefficient of friction = 0.24

(a) The acceleration of the car is

[tex]a =\mu g\\\\a = 0.24 \times9.8\\\\a = 2.35 m/s^2[/tex]

(b) initial velocity, u =0, final velocity, v = 13 m/s

Let the time is t.

Use first equation of motion

v = u + a t

13 = 0 + 2.35 x t

t = 5.53 seconds

Answer:

sorry dear

Explanation:

sorry dear but ur question is hard to understand can u try to edit it so i can tell u the answer?

I’m assuming you meant latitude:

The angular distance north or south of the earth's equator, measured in degrees along a meridian, as on a map or globe.

a region based on its distance to the equator uses latitude measurements

basically: up and down , north to south from a certain point

Explanation:

voltage is 50v and capacitance is 30c what is the charge

q=1500c

Answer:

    v₂ = 63.62 m / s

Explanation:

For this exercise in fluid mechanics we will use Bernoulli's equation

         P₁ + ρ g v₁² +  ρ g y₁ = P₂ +  ρ g v₂² +  ρ g y₂

where the subscript 1 refers to the inside of the wing and the subscript 2 to the top of the wing.

We will assume that the distance between the two parts is small, so y₁ = y₂

        P₁-P₂ =  ρ g (v₂² - v₁²)

pressure is defined by

        P = F / A

we substitute

        ΔF / A =  ρ g (v₂² - v₁²)

         v₂² = [tex]\frac{\Delta F}{A \ \rho \ g} + v_1^2[/tex]

suppose that the area of ​​the wing is A = 1 m²

we substitute

         v₂² = [tex]\frac{1000}{1 \ 1.29 \ 9.8} + 63^2[/tex]

         v₂² = 79.10 + 3969

         v₂ = √4048.1

         v₂ = 63.62 m / s

Answer:

We need, mass, gravity and height.

Explanation:

When a body falls freely from a height, its initial velocity is zero, but due to the height it has some potential energy at the top and the kinetic energy is zero.

As it falls, the potential energy is gradually converted in to the kinetic energy so that the total energy of the falling body is conserved.

At the time as the body strikes the ground, the entire potential energy is converted into the kinetic energy.

Potential energy is given by

U = m g h

where, m is the mass, g is the gravity and h is the height,

So, to get the kinetic energy we require mass, gravity and height of the body.

Answer:

an unconscious needs rooted in childhood are the foundation of psychoanalytic theory.

Explanation:

Answer:

The output electric power is 1338876 W.

Explanation:

Volume, V = 690 cubic meter

height, h = 220 mm = 0.22 m

efficiency = 90 %

time , t = 1 s

Let the mass is m.

m = volume x density  

m = 690 x 1000 = 690000 kg

The input power is

P = m g h = 690000 x 9.8 x 0.22 = 1497640 W

The electric power out put is

[tex]P' = 90 % of 1487640\\\\\\P' = 1338876 W[/tex]

Answer:

160 meters

Relative of speed Vr = 10 -2 = 8 m/s    (horizontal speed)

20 sec * 8 m/s = 160 m      since ball travels 20 sec

Answer:

160 meters

Explanation:

Answer:

False

Explanation:

Though fiber active cable is based on the concept of internal reflection but it is achieved by refractive index which transmit data through fast traveling pulses of light. It has a layer of glass and insulating casing called “cladding,”and this is is wrapped around the central fiber thereby causing light to continuously bounce back from the walls of the Cable.

Answer:

D. Centripetal Force :)

(a) 2.001N

(b) 2.001N

A sketch of the scenario has been attached to this response.

Since only the force vector F₁ is required, the only force shown in the sketch is F₁.

As shown in the sketch;

The x-component of the force vector F₁ = [tex]F_{x}[/tex]

The y-component of the force vector F₁ = [tex]F_{y}[/tex]

The magnitude of F₁ as given in the question = 2.83N

The angle that the force makes with respect to the x-axis = 45.0°

Using the trigonometric ratio, we see that;

(a) cos 45.0° = [tex]\frac{F_x}{F_1}[/tex]

=>  [tex]F_{x}[/tex] =  F₁ cos 45.0°

=> [tex]F_{x}[/tex] =  2.83 cos 45.0°

=> [tex]F_{x}[/tex] =  2.83 x 0.7071

=> [tex]F_{x}[/tex] =  2.001N

(b) Also;

sin 45.0° = [tex]\frac{F_y}{F_1}[/tex]

=>   [tex]F_{y}[/tex] =  F₁ sin 45.0°

=>  [tex]F_{y}[/tex] =  2.83 sin 45.0°

=>  [tex]F_{y}[/tex] =  2.83 x 0.7071

=>  [tex]F_{y}[/tex] =  2.001N

Therefore, the x-component and y-component of the force vector F₁ is 2.001N

The x and y component of vector F1 is mathematically given as

F_x =  2.001N

F_y=  2.001N

Question Parameters:

Generally, the equation for the x-component  is mathematically given as

x=Fsin\theta

Therefore

F_x =  F₁ cos 45.0°

F_x =  2.83 x 0.7071

F_x =  2.001N

For y component

x=Fcos\theta

F_y =  F₁ sin 45.0

F_y =  2.83 x 0.7071

F_y=  2.001N

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