You exert a 100-N force on a pulley system to lift 300-N. What is the mechanical advantage of this system? How many sections of rope support weight?

Answer:

F = 1,875 N

Explanation:

force=

[tex] \frac{change \: in \: momentum}{time \: taken} [/tex]

∆H = m∆V

where ∆H ----> change in momentum.

( final momentum - initial momentum )

and ∆V ----> change in velocity

( final velocity - initial velocity )

and m ----> is mass

then f =

[tex] \frac{1500 \times (25 - 15)}{8} [/tex]

= 1,875 N

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

Answer:

Quick Check:

12-17-20

Conservation of Energy

In the use of an energy resource like coal, do you get all

the energy back in one form(100%)? (circle one)

Yes

No

Explanation:

Not sure

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.

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

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:

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

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.

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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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: 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:

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:

4m/s^2

Explanation:

Answer:

The direction could change

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

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:

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