5. Did objects have to touch to interact? What causes this?

Answer:

50

Explanation:

Answer: The mean value = 9.85m/s².

Explanation:

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

The given measurements the acceleration of gravity (units of m/s²): 10.1, 9.87, 9.76, 9.91, 9.75, 9.88, 9.69, 9.83, and 9.90.

Number of measurements =9

Sum of measurements =  88.69

Mean = [tex]\dfrac{88.69}{9}=9.85444444\approx9.85[/tex]

Hence, the mean value = 9.85m/s².

Answer:

Explanation:

Satellite A and satellite B are approaching the earth from opposite directions , that means they are approaching each other . The velocity of satellite A and B are .648c and .795c respectively . Their velocities are comparable to velocity of light so they will follow relativistic laws .

Their relative velocity will be given by the following relation .

[tex]V_r=\frac{u+v}{1+\frac{uv}{c^2} } }[/tex]

where u and v are velocities of vehicles coming from opposite direction and c is velocity of light .

[tex]V_r=\frac{.795c+.648c}{1+\frac{.648c\times .795c}{c^2} } }[/tex]

[tex]V_r=\frac{1.443c}{1+.515 } }[/tex]

= .952c

Answer:

Speed of Agnes during her journey was 0.848c

Explanation:

Given that;

Age of Agnes t₀ = 20 years

distance d = 2 × distance of star from Earth = 2 × 16 light-years

= 32 light-years

so get her speed speed; we use the following expression

Yvt₀ = d

( v / √(1 - ([tex]\frac{v}{c}[/tex])²) )² = ( 32 light-years / 20 yrs )²

v² / (1 - ( v²/c²)) = ( 32 × c  / 20 )²

v² / (1 - ( v²/c²)) = 2.56 × c²  

v² / c²-v²/c² = 2.56 × c²  

v²c²  / c² - v²  = 2.56 × c²  

v²  / c² - v²  = 2.56

v²  = 2.56 (c² - v²)

v²  = 2.56 (c² - v²)

v²  = 2.56c² - 2.56v²

v² + 2.56v² = 2.56c²

3.56v² = 2.56c²

v² = (2.56/3.56)c²

v = √((2.56/3.56)c²)

so v = 0.848c

Therefore, Speed of Agnes during her journey was 0.848c

Answer: B>A=D>C

Explanation:

Kinetic Energy is the product of mass and square of the velocity

For Jogger A

[tex]K.E._a=\frac{1}{2}mv^2[/tex]

For Jogger B

[tex]K.E._b=\frac{1}{2}\times\frac{m}{2}\times(3v)^2=\frac{9}{4}mv^2\\K.E._b=2.25mv^2[/tex]

For Jogger C

[tex]K.E._c=\frac{1}{2}\times3m\times(\frac{v}{2})^2=\frac{3}{8}mv^2\\K.E._c=0.375mv^2[/tex]

For Jogger D

[tex]K.E._d=\frac{1}{2}\times4m\times(\frac{v}{2})^2=\frac{1}{2}mv^2[/tex]

Kinetic Energy of Joggers in increasing order

B>A=D>C

The total displacement of the dispatch rider is calculated as 43km.

Data;

To calculate the total displacement of the dispatch rider, we can simply add up the total distance covered by the rider.

This becomes;

[tex]10+11+22 = 43km[/tex]

The total displacement of the dispatch rider is calculated as 43km.

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Answer:

The speed (magnitude of the velocity) is 14.4 m/s

Explanation:

Vertical Launch Upwards

It occurs when an object is launched vertically up without taking into consideration any kind of friction with the air.

If vo is the initial speed and g is the acceleration of gravity, the speed vf at any time is calculated by:

[tex]v_f=v_o-g.t[/tex]

A tennis ball is launched vertically up with an initial speed of vo=34 m/s. At time t=2 s, its speed is:

[tex]v_f=34-9.8*2[/tex]

[tex]v_f=34-19.6[/tex]

[tex]v_f=14.4\ m/s[/tex]

The speed (magnitude of the velocity) is 14.4 m/s

The volume of snow on the lawn, in cubic m, is 44.63 or 44.63 m³ if the lawn outside your neighbor's house has an approximate area of 175 m².

It is defined as a three-dimensional space enclosed by an object or thing.

It is given that:

The lawn outside your neighbor's house has an approximate area of 175 m². One night it snows so that the snow on the lawn has a uniform depth of 25.5 cm.

As we know, unit conversion can be defined as the conversion from one quantity unit to another quantity unit followed by the process of division, and multiplication by a conversion factor.

25.5 cm =  0.255 m

Volume = area×depth

Volume = 175×0.255

Volume = 44.63 cubic meters

Thus, the volume of snow on the lawn, in cubic m is 44.63 or 44.63 m³ if the lawn outside your neighbor's house has an approximate area of 175 m².

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#SPJ2

Answer:

The direction could change

Answer:

Explanation:

Most environments have many niches. If a niche is "empty" (no organisms are occupying it), new species are likely to evolve to occupy it. This happens by the process of natural selection. By natural selection, the nature of the species gradually changes to become adapted to the niche.

Answer:

33.5J

Explanation:

Given:

Mass of the canister= 3.1 kg

Initial velocity of canister= v(i)= 4.4i m/s

Final velocity of canister= v(f)= 6.4j m/s

Force magnitude( xy plane)= 5 N

The magnitude of vector V'= Vxi + Vyj + Vzk

|V|= √( Vx^2 + Vy^2 + Vz^2

From Kinectic energy and work theorem.

Net work = Kinectic energy of the canister

ΔK= W

(Kf - Ki)= W

Where Kf= final Kinectic energy

= 1/2 mv^2

If we input the given values we have,

= 1/2 × 3.1 ×√(4.4^2 + 0^2 + 0^2)^2 = 30J

Ki= initial Kinectic energy

= 1/2 mv^2

If we substitute the given values we have

=1/2 × 3.1 ×√(0^2 + 6.4^2 + 0^2)^2 = 63.5 J

Work done by canister = (final Kinectic energy - initial energy)

= 63.5- 30

=33.5J

Hence, work done on the canister 33.5J

Answer:

The material that Emily invented can be easily repaired by shining ultraviolet light on it.

Explanation:

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Answer:

The material that Emily invented can be easily repaired by shining ultraviolet light on it.

Explanation:

Answer:

So increasing the voltage increases the charge in direct proportion to the voltage. If the voltage exceeds the capacitors rated voltage, the capacitor may fail due to breakdown of the dielectric between the two plates that make up the capacitor.

Explanation:

A option.

Answer:

Explanation:

Time period of oscillation T = 3 s .

Frequency of oscillation = 1 / T = 1 / 3 = .333 per second .

The mass is  pulled down an additional 0.25 meters so amplitude of oscillation A = .25 m .

Full range of vertical motion = .25 x 2 = 0.5 m .

The period of the spring = 3 s .

Answer:

Nitrogen and Oxygen are the two major atmospheric gases.

Answer:

10.88kJ

Explanation:

Given data

mass= 10kg

heigth= 111m

Applying

PE= mgh

assume g= 9.81m/s^2

substitute

PE= 10*9.81*111

PE=10889.1 Joules

PE=10.881kJ

Hence the potential energy is 10.88kJ

Answer:

Theory

Explanation:

Conservation of energy is explained as a scientific law and not a theory because it does not explain why energy is conserved.

A law is a the statement of a scientific fact. It is a product of repeated experiment and observation through time. Most laws do not explain the reason for the logic behind their premise.

A theory on the other hand provides an explanation for an observed phenomenon. Most theories are no immutable. They are often changed when new finds are reported or made.

Laws are immutable and they stand still.

The gravitational force exerted by the moon on the satellite is such that

F = G M m / R ² = m a   →   a = G M / R ²

where a is the satellite's centripetal acceleration, given by

a = v ² / R

The satellite travels a distance of 2πR about the moon in complete revolution in time T, so that its tangential speed is such that

v = 2πR / T   →   a = 4π ² R / T ²

Substitute this into the first equation and solve for T :

4π ² R / T ² = G M / R ²

4π ² R ³ = G M T ²

T ² = 4π ² R ³ / (G M )

T = √(4π ² R ³ / (G M ))

T = 2πR √(R / (G M ))

The correct expression for the time T it takes the satellite to make one complete revolution around the moon is [tex] T = 2\pi R\sqrt{\frac{R}{GM}} [/tex].

We can find the period T (the time it takes the satellite to make one complete revolution around the moon) from the gravitational force:

[tex] F = \frac{GmM}{R^{2}} [/tex]    (1)

Where:

G: is the gravitational constant = 6.67x10⁻¹¹ Nm²/kg²

R: is the distance between the satellite and the center of the moon

m: is the satellite's mass

M: is the moon's mass

The gravitational force is also equal to the centripetal force:

[tex] F = ma_{c} [/tex]   (2)

The centripetal acceleration ([tex]a_{c}[/tex]) is equal to the tangential velocity (v):

[tex] a_{c} = \frac{v^{2}}{R} [/tex]   (3)

And from the tangential velocity we can find the period:

[tex] v = \omega R = \frac{2\pi R}{T} [/tex]   (4)

Where:

ω: is the angular speed = 2π/T

By entering equations (4) and (3) into (2), we have:

[tex] F = m\frac{v^{2}}{R} = m\frac{(\frac{2\pi R}{T})^{2}}{R} = \frac{mR(2\pi)^{2}}{T^{2}} [/tex]   (5)

By equating (5) and (1), we get:

[tex] \frac{mR(2\pi)^{2}}{T^{2}} = \frac{GmM}{R^{2}} [/tex]

[tex] T^{2} = \frac{R^{3}(2\pi)^{2})}{GM} [/tex]  

[tex] T = \sqrt{\frac{R^{3}(2\pi)^{2})}{GM}} [/tex]

[tex] T = 2\pi R\sqrt{\frac{R}{GM}} [/tex]

Therefore, the expression for the time T is [tex] T = 2\pi R\sqrt{\frac{R}{GM}} [/tex].

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Explanation:

In physics, a force is any interaction that, when unopposed, will change the motion of an object. A force can cause an object with mass to change its velocity, i.e., to accelerate. Force can also be described intuitively as a push or a pull. A force has both magnitude and direction, making it a vector quantity.

Formula

Newton's Second Law

F = m * a

F = force

m = mass of an object

a = acceleration

Answer:

1470J

Explanation:

Given parameters:

Height of stairs  = 2.5m

Time taken  = 4.25s

Mass of Bart  = 60kg

Unknown:

Work done by Bart  = ?

Solution:

Work done is a function of the force acting to move a body through a distance.

    Work done  = Weight  x height

  Work done  = mass x acceleration due to gravity x height

Work done  = 60 x 9.8 x 2.5  = 1470J

Answer:

4m/s^2

Explanation:

Answer:

 the statement is False       [tex]\frac{x_{ball} }{ x_{rec} }[/tex] = cos θ

Explanation:

Let's analyze this problem, the ball and the receiver leave the same point and we want to know if at the same moment they reach the same point, for this we must have both the ball and the receiver travel the same distance.

Let's start by finding the time it takes for the ball to reach the ground

        y = [tex]v_{oy}[/tex] t - ½ g t²

when it reaches the ground its height is y = 0

       0 = vo sin θ  - ½ g t²

       0 = t (vo sin θ - ½ g t)

The results are

       t = 0                             exit point

       t = 2 v₀ sin θ/g            arrival point

at this point the ball traveled

       [tex]x_{ball}[/tex]= v₀ₓ t

       x_{ball} = v₀ cos θ  2v₀ sin θ / g

       x_{ball}= 2 v₀² cos θ   sin θ/ g

Now let's find that distantica traveled the receiver in time

        [tex]x_{rec}[/tex] = v₀ t

        x_{rec} = v₀ (2 v₀ sin θ / g)

        x_{rec} = 2 v₀² sin θ / g

without dividing this into two distances

          [tex]\frac{x_{ball} }{ x_{rec} }[/tex] = cos θ

therefore the distances are not equal to the ball as long as behind the receiver

Therefore the statement is False

Answer:

1.19m

Explanation:

Given parameters:

Speed of the trumpet sound  = 351m/s

Frequency of the note = 294Hz

Unknown:

Wavelength of the sound  = ?

Solution:

To solve this problem, we use the wave - velocity equation.

    V  = F x ∧  

V is the velocity of the body

F is the frequency

∧ is the wavelength

 So;

          351   = 294 x ∧  

       ∧   = [tex]\frac{351}{294}[/tex]   = 1.19m

Answer:

x = 1382.9 km

Explanation:

The speed of the wave is constant, so we can use the uniform motion relationships

p wave

          [tex]v_p[/tex] = x / t₁

          t₁ = x /v_p

S wave

          v_s = x / t₂

           t₂ = x / v_s

indicate that the time difference between the two waves is

          t₂ - t₁ = 3.25 min (60 s / 1 min)

          t₂ -t₁ = 195 s

let's substitute

         [tex]\frac{x}{v_s} - \frac{x}{v_p}[/tex] = 195

         x ([tex]\frac{1}{v_s} - \frac{1}{v_p}[/tex] = 195

let's calculate

         x [tex]( \frac{1}{4.2} - \frac{1}{10.3} )[/tex] = 195

         x (0.1410) = 195

         x = 195 /0.141

         x = 1382.9 km

Answer:

a)   F = 1.70 10⁻⁹N,   F = 1.47 10⁻⁸ N,

b) * the electronegative repulsion, from the repulsion by quantum effects

Explanation:

a) The atraicione force comes from the electric force given by Coulomb's law,

           F = [tex]k \frac{ q_1 q_2}{r^2}[/tex]

divalent atoms

In this case q = 2q₀ where qo is the charge of the electron -1,6 10⁻¹⁹ C and the separation is given

           F = k q² / r²

           F = [tex]2 \ 10^9 \ \frac{2 (1.6 \ 10^{-19} )^2}{ (0.52 10^{-9} )^2 }[/tex]

           F = 1.70 10⁻⁹N

monovalent atoms

in this case the load is q = q₀

           F = 2 \ 10^9 \  \frac{ (1.6 \  10^{-19} )^2}{ (0.125 10^{-9} )^2 }

           F = 1.47 10⁻⁸ N

b) repulsive forces come from various sources

* the electronegative repulsion of positive nuclei

* the electrostatic repulsion of the electrons when it comes to bringing the electron clouds closer together

* from the repulsion of electron clouds, by quantum effects

Answer:

20 C to F

Ans: 68F

-40 C to F

Ans:-40F

40C to F

Ans:104F

Answer:

weight = 900 N

acceleration  = 11.25 m/s²

Explanation:

given data

mass m1 = 75 kg

mass m2 = 80 kg

gravitational acceleration = 12 m/s²

solution

As we know weight of a mass that is

weight of mass = Mass × Acceleration due to gravity .................1

so Astro 1 weight is

weight = 75kg  × 12 m/s²

weight = 900 N

and

so, when Astro 2 needs this much weight the planet on which he is will have the acceleration

acceleration = Weight ÷ Mass of Astro 2        .....................2

acceleration  = 900 ÷ 80 m/s²

acceleration  = 11.25 m/s²

Answer:

a. P.E = 3430Joules.

b. Workdone = 3430Nm

Explanation:

Given the following data;

Mass = 70kg

Distance = 5m

We know that acceleration due to gravity is equal to 9.8m/s²

To find the potential energy;

Potential energy = mgh

P.E = 70*9.8*5

P.E = 3430J

b. To find the workdone;

Workdone = force * distance

But force = mass * acceleration

Force = 70*9.8

Force = 686 Newton.

Workdone = 686 * 5

Workdone = 3430Nm