Shooting a rock with a slingshot converts
energy to__________,
energy____________,

Answer:

The answer to this question is plasma

Answer:

Plasma

Explanation:

Answer:

1.67 m

Explanation:

The potential energy change of the small ball ΔU equals the work done by the buoyant force, W

ΔU = -W

Now ΔU = mgΔh where m = mass of small ball = ρV where ρ = density of small ball and V = volume of small ball. Δh = h - h' where h = final depth of small ball and h' = initial height of small ball = 5 m. Δh = h - 5

ΔU = mgΔh

ΔU = ρVgΔh

Now, W = ρ'VgΔh'   where ρ = density of water and V = volume of water displaced = volume of small ball. Δh' = h - h' where h = final depth of small ball and h' = initial depth of small ball at water surface = 0 m. Δh' = h - h' = h - 0 = h

So, ΔU = -W

ρVgΔh = -ρ'VgΔh'

ρVg(h - 5) = -ρ'Vgh

ρ(h - 5) = -ρ'h

Since the density of the small ball equals 1/2 the density of water,

ρ = ρ'/2

ρ(h - 5) = -ρ'h

(ρ'/2)(h - 5) = -ρ'h

ρ'(h - 5)/2 = -ρ'h

(h - 5)/2 = -h

h - 5 = -2h

h + 2h = 5

3h = 5

h = 5/3

h = 1.67 m

So, the maximum depth the ball reaches is 1.67 m.

Answer:

Explanation:

Intensity of sound = sound energy emitted by source / 4 π d² , where d is distance of the source .

A )

Intensity of sound at 1 m distance = 60 /4 π d²

d = 1 m

Intensity of sound at 1 m distance = 60 /(4 π 1²)

= 4.78 W m⁻² s⁻¹

B )

Intensity of sound at 1.5 m distance = 60 /4 π d²

d = 1.5  m

Intensity of sound at 1 m distance = 60 /(4 π 1.5²)

= 2.12 W m⁻² s⁻¹

C )

Intensity of sound due to 4 speakers at 1.5 m distance = 4 x 60 /4 π d²

d = 1.5  m

= 4 x 60 /(4 π 1.5²)

= 8.48 W m⁻² s⁻¹

D )

Intensity of sound due to .06 W speaker must be 10⁻¹² W s ⁻² . Let the distance be d .

.06 /4 π d² = 10⁻¹²

d² = .06 /4 π 10⁻¹²

d = 6.9 x 10⁴ m .

Answer:

Centripetal Acceleration 18.75 m/s^2, Rotational Kinetic Energy 843.75 J

Explanation:

a Linear acceleration (we cant find tangential acceleration with the givens so we will find centripetal)

a= ω^2*r

ω= 300rev/min

convert into rev/s

300/60= 5rev/s

a= 18.75m/s^2

b) use Krot= 1/2 Iω^2

plug in gives

1/2(30*2.25)(25)= 843.75 J

Answer:

the unit for ΔL/L is "unitless".

Explanation:

Given;

ΔL/L

by physics convection, the above parameters can be defined as;

delta L (ΔL) is change in length, with SI unit as meters (m),

L is the original length of the material, with SI unit as meters (m)

The ratio of the change in length to the original length has no unit since both units cancel out during the division.

[tex]\frac{\Delta L }{L } = \frac{(m)}{(m)} = \ unitless[/tex]

This ratio (ΔL/L), is also called tensile strain and it has no unit.

Therefore, the unit for ΔL/L is "unitless".

The image is missing and so i have attached it.

Answer:

A) E = 8740 N/C

B) E = -6536 N/C

Explanation:

The formula for electric field is;

E = kq/r²

Where;

q is charge

k is a constant with value 8.99 x 10^(9) N•m²/C²

A) Now, to find the net electric field at point A, the formula would now be;

E = (kq1/(r1)²) - (kq2/(r2)²)

Where;

r1 is distance from charge q1 to point A

r2 is distance from charge q2 to point A.

q1 = -6.25 nC = -6.25 × 10^(-9) C

q2 = -12.5 nC = -12 5 × 10^(-9) C

From the attached image, r1 = 25 cm - 10 cm = 15 cm = 0.15 m

r2 = 10 cm = 0.1 m

Thus;

E = (8.99 x 10^(9)) × ((-6.25 × 10^(-9))/0.15^(2)) - ((-12.5 × 10^(-9))/0.1^(2))

E = 8740 N/C

B) similarly, electric field at point B;

E = (kq1/(r1)²) + (kq2/(r2)²)

Where;

r1 is distance from charge q1 to point B

r2 is distance from charge q2 to point B.

q1 = -6.25 nC = -6.25 × 10^(-9) C

q2 = -12.5 nC = -12 5 × 10^(-9) C

From the attached image, r1 = 10 cm = 0.1 m

r2 = 25cm + 10 cm = 35 cm = 0.35 m

Thus;

E = (8.99 x 10^(9)) × ((-6.25 × 10^(-9))/0.1^(2)) + ((-12.5 × 10^(-9))/0.35^(2))

E = -6536 N/C

Answer:

Henry Ford

Explanation:

he built the first ford

Answer:

[tex]B=2.91\ \mu T[/tex]

Explanation:

Given that,

The current in the loop, I = 2 A

The radius of the loop, r = 0.4 m

We need to find the magnetic field at a distance 0.09 m along the axis and above the center of the loop. The formula for the magnetic field at some distance is given as follows :

[tex]B=\dfrac{\mu_o}{4\pi }\dfrac{2\pi r^2 I}{(r^2+d^2)^{3/2}}[/tex]

Put all the values,

[tex]B=10^{-7}\times \dfrac{2\pi \times 0.4^2 \times 2}{(0.4^2+0.09^2)^{3/2}}\\\\=2.91\times 10^{-6}\ T\\\\or\\\\B=2.91\ \mu T[/tex]

So, the required magnetic field is equal to [tex]2.91\ \mu T[/tex].

Answer:

[tex]M.E=41J[/tex]

Explanation:

From the question we are told that:

Amplitude [tex]a=4.20cm[/tex]

Spring Constant [tex]K=262N/m[/tex]

Mass [tex]m=560g[/tex]

Generally the equation for mechanical energy is mathematically given by

[tex]M.E=\frac{1}{2}km^2[/tex]

[tex]M.E=0.5*262*0.56^2[/tex]

[tex]M.E=41J[/tex]

Answer:

multiply mp and c^2

Explanation:

e=mc^2

Answer:

243 N

Explanation:

The formula for electromagnetic force is F= Kq1q2/r^2

where r is the distance between the charges, if the distance between the charges is reduced by 1/3 then F will increase by 9 [(1/3r)^2 becomes 1/9r which is 9F] so 27*9 is 243N

Answer:

Look at explanation

Explanation:

a) Kinetic energy= ΔW. W=Fd, and since in both scenarios the same force and same distance is travelled. K1=K2. I am assuming that the objects are at non zero height so by P=mgh, P1>P2

b. Again I am assuming that the objects are at non zero height so by P=mgh, P1>P2.  A heavier mass, a constant force means a smaller acceleration. So a1<a2. We can then use x-x0=v0t+1/2at² and since v0=0, x-x0(d)=1/2at². Solve for t²=2d/a. Since t is the same for both but a1<a2, d1<d2. And since Kinetic Energy=ΔW, W=Fd and F is constant while d1<d2, K1<K2.

According to the question,

(a)

Word done towards both the block will be similar.

So,

→ [tex]P1 = P2[/tex]

→ [tex]K1= K2[/tex]

(b)

We know,

→ [tex]a = \frac{F}{M}[/tex]

or,

→ [tex]V = a\times t[/tex]

Now,

→ [tex]K = \frac{1}{2} MV^2[/tex]

       [tex]= 0.5\times M\times V^2[/tex]

       [tex]=0.5\times M\times (\frac{F^2}{M^2} )\times t^2[/tex]

       [tex]= 0.5\times F^2\times \frac{t^2}{M}[/tex]

The force and t will be same. So K of the smaller mass will be greater than the larger mass.

hence,

→ [tex]K1<K2[/tex]

Thus the above responses are correct.        

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Answer:

When a teen driver drives with a lot of his peers as passengers they may lead to distraction which may later end up in accident as the driver was distracted

Overconfidence, lack of focus, and phone while driving are the factors  contribute to accidents when a teen driver controls other teens as passengers,

When a teen driver drives with a lot of his peers as passengers they may direct to distraction which may later end up in casualty as the driver was distracted.

Several studies have indicated that passengers substantially increase the chance of crashes for young, novice drivers. This improved risk may result from distractions that young passengers complete for drivers.Teens driving with a teen or young adult passengers existence of teen or young adult passengers raises the crash risk of unsupervised teen drivers. This risk grows with each additional teen or a young adult passenger.

Crash risk is two- to six times more significant for those who utilize a cellphone while driving resembled for drivers who are not distracted. Using a phone delays reaction time increases lane deviations, and forces drivers to look away from the road for extended times.

Overconfidence, lack of focus, and phone while driving are the factors  contribute to accidents when a teen driver controls other teens as passengers,

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Answer:

The number of revolutions is 44.6.

Explanation:

We can find the revolutions of the wheel with the following equation:

[tex]\theta = \omega_{0}t + \frac{1}{2}\alpha t^{2}[/tex]

Where:

[tex]\omega_{0}[/tex]: is the initial angular velocity = 13 rad/s              

t: is the time = 8 s

α: is the angular acceleration

We can find the angular acceleration with the initial and final angular velocities:

[tex] \omega_{f} = \omega_{0} + \alpha t [/tex]

Where:

[tex] \omega_{f} [/tex]: is the final angular velocity = 57 rad/s

[tex] \alpha = \frac{\omega_{f} - \omega_{0}}{t} = \frac{57 rad/s - 13 rad/s}{8 s} = 5.5 rad/s^{2} [/tex]

Hence, the number of revolutions is:

[tex] \theta = \omega_{0}t + \frac{1}{2}\alpha t^{2} = 13 rad/s*8 s + \frac{1}{2}*5.5 rad/s^{2}*(8 s)^{2} = 280 rad*\frac{1 rev}{2\pi rad} = 44.6 rev [/tex]

Therefore, the number of revolutions is 44.6.

I hope it helps you!

Answer:

Initially let n cannonballs with a total mass of m be to the left of the center of mass at L /2 and the mass of the car at L/2

x1 =  [-m / (m + M)] * L / 2   is the original position of the CM

x2 = (m (x + L/2) + M x) / (m + M) * L/2 final position of CM with all cannon balls to the right

[-m x - m L / 2 + m x - M x] / (M + m) * L/2

= - ( m L / 2 + M x) / (m + M) * L/2 = Xcm

Check the math, but maximum distance occurs when the cannonballs of mass m move from -L/2 to L/2 and the car of mass M moves from zero to -x

Answer:

Explanation:

Friction can be defined as a force that resists the relative motion of two objects when there surface comes in contact. Thus, it prevents two surface from easily sliding over or slipping across one another. Also, friction usually reduces the efficiency and mechanical advantage of machines but can be reduced through lubrication.

Generally, there are four (4) main types of friction and these includes;

I. Static friction.

II. Rolling friction.

III. Sliding friction.

IV. Fluid friction.

Answer:

Reverse Osmosis

Explanation:

Reverse osmosis is a type of filtration that involves passing a solvent through a semipermeable membrane in the opposite direction that natural osmosis does. Separation is always enforced through the use of pressure in this process. Ions, fine dust particles, molecules, and larger particles are typically removed from solvents using this method. The technique is particularly popular in the treatment and purification of water.

Answer:

filtration is used to separate

Answer:

AM broadcasts occur on North American airwaves in the medium wave frequency range of 525 to 1705 kHz (known as the “standard broadcast band”). The band was expanded in the 1990s by adding nine channels from 1605 to 1705 kHz.

Let A be the position of the block at the top of the first incline; B its position at the bottom of the first incline; C its position at the bottom of the second incline; and D its position at the top of the second incline. I'll denote the energy of the block at a given point by E (point).

At point A, the block has total energy

E (A) = (10.0 kg) (9.80 m/s²) (7.00 m) + 1/2 (10.0 kg) v₀²

E (A) = 686 J + 1/2 (10.0 kg) v₀²

At point B, the block's potential energy is converted into kinetic energy, so that its total energy is

E (B) = 1/2 (10.0 kg) v₁²

The block then slides over the horizontal surface with constant speed v₁ until it reaches point C and slides up a maximum height of 26.0 m to point D. Its total energy at D is purely potential energy,

E (D) = (10.0 kg) (9.80 m/s²) (26.0 m) = 2548 J

Throughout this whole process, energy is conserved, so

E (A) = E (B) = E (C) = E (D)

(a) Solve for v₀ :

686 J + 1/2 (10.0 kg) v₀² = 2548 J

==>   v₀ ≈ 19.3 m/s

(b) Solve for v₁ :

1/2 (10.0 kg) v₁² = 2548 J

==>   v₁ ≈ 22.6 m/s

Now if the horizontal surface is not frictionless, kinetic friction will contribute some negative work to slow down the block between points C and D. Check the net forces acting on the block over this region:

• net horizontal force:

∑ F = -f = ma

• net vertical force:

∑ F = n - mg = 0

where f is the magnitude of kinetic friction, a is the block's acceleration, n is the mag. of the normal force, and mg is the block's weight. Solve for a :

n = mg = (10.0 kg) (9.80 m/s²) = 98.0 N

f = µn = 0.500 (98.0 N) = 49.0 N

==>   - (49.0 N) = (10.0 kg) a

==>   a = - 4.90 m/s²

The block decelerates uniformly over a distance 2.00 m and slows down to a speed v₂ such that

v₂² - v₁² = 2 (-4.90 m/s²) (2.00 m)

==>   v₂² = 490 m²/s²

and thus the block has total/kinetic energy

E (C) = 1/2 (10.0 kg) v₂² = 2450 J

(c) The block then slides a height h up the frictionless incline to D, where its kinetic energy is again converted to potential energy. With no friction, E (C) = E (D), so

2450 J = (10.0 kg) (9.80 m/s²) h

==>   h = 25.0 m

(d) At half the maximum height, the block has speed v₃ such that

2450 J = (10.0 kg) (9.80 m/s²) (h/2) + 1/2 (10.0 kg) v₃²

==>   v₃ ≈ 15.7 m/s

The block loses speed and thus energy as it moves between B and C, but its energy is conserved elsewhere. If we ignore the inclines and pretend that the block is sliding over a long horizontal surface, then its velocity v at time t is given by

v = v₁ + at = 22.6 m/s - (4.90 m/s²) t

The block comes to a rest when v = 0 :

0 = 22.6 m/s - (4.90 m/s²) t

==>   t ≈ 4.61 s

It covers a distance x after time t of

x = v₁t + 1/2 at ²

so when it comes to a complete stop, it will have moved a distance of

x = (22.6 m/s) (4.61 s) + 1/2 (-4.90 m/s²) (4.61 s)² = 52.0 m

(e) The block crosses the rough region

(52.0 m) / (2.00 m) = 26 times

If both forces are measured in Newtons, then the net force is

F = (18, 7, -12) N + (16, -10, 16) N = (34, -3, 4) N

The toolbox undergoes a displacement (i.e. change in position) in the direction of the vector

d = (17, 14, -1) m - (15, 14, -8) m = (2, 0, -9) m

The total work done by the astronauts on the toolbox is then

F • d = (34, -3, 4) N • (2, 0, -9) m = (68 + 0 - 36) N•m = 32 J

The work done by the two astronauts is equal to 96 J.

work done?Work done is defined as the product of force applied and the distance moved by the force.

The forces applied = 18+16 N, 7+ -10 N, and -12 + 16N

Forces = 34 N, -3 N, and 4N

Distances = (17 - 15, 14 - 14, -1 - - 8) m

Distances = 2, 0, 7

Work done = 34 × 2 + -3 × 0 + 4 × 7

Work done = 96 J

Therefore, the work done by the two astronauts is equal to 96 J.

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Answer:

I DON'T UNDERSTAND

Explanation:

GUESS A MISUNDERSTANDING PLZ PUT A UNDERSTANDABLE QUESTION.

Answer:

A(3.56m)

Explanation:

We have a conservation of energy problem here as well. Potential energy is being converted into linear kinetic energy and rotational kinetic energy.

We are given ω= 4.27rad/s, so v = ωr, which is 6.832 m/s. Place your coordinate system at top of the hill so E initial is 0.

Ef= Ug+Klin+Krot= -mgh+1/2mv^2+1/2Iω^2

Since it is a solid uniform disk I= 1/2MR^2, so Krot will be 1/4Mv^2(r^2ω^2=  v^2).

Ef= -mgh+3/4mv^2

Since Ef=Ei=0

Mgh=3/4mv^2

gh=3/4v^2

h=0.75v^2/g

plug in givens to get h= 3.57m

Answer:

1 month

Explanation:

Answer:

a)  U = 506 J, b)  U = 37.11 J, c) U = 0

Explanation:

The gravitational power energy is given by the expression

         U = m g (y -y₀)

In general, a reference system is set that allows the expression to be simplified, in this case let's assume the reference system at the child's lowest point, therefore y₀ = 0

Let's use trigonometry to find the child's height

          h = y = L - L cos θ

we substitute

           U = m g L (1 - cos θ)

a) when the chain is horizontal θ = 90 and cos 90 = 0

           U = mg L

weight and mass are related

            W = mg

            m = W / g

           U = 230 2.20

           U = 506 J

b) θ = 33.0º

           cos 33 = 0.83867

           U = 230 (1 - 0.83867)

           U = 37.11 J

c) in this case θ = 0 cos 0 = 1

            U = 0

Answer:

The inductance is 17784.96 ohm and rms current is 4.77 mA.

Explanation:

Voltage, V = 120 V

frequency, f = 60 Hz

Inductance, L = 47.2 H

The rms  voltage is

[tex]V_{rms}=\frac{V_o}{\sqrt 2}\\\\V_{rms}=\frac{120}{\sqrt 2}\\\\V_{rms} = 84.87 V[/tex]

The reactance is given by

[tex]X_L = 2\pi f L\\\\X_L = 2\times 3.14\times 60\times 47.2 \\\\X_L = 17784.96 ohm[/tex]

The rms current is

[tex]I_{rms} =\frac{V_{rms}}{X_L}\\\\I_{rms}=\frac{84.87}{17784.96}\\\\I_{rms} = 4.77\times 10^{-3} A = 4.77 mA[/tex]

Answer:

Reaction force

Explanation:

Newton´s 3rd law says that every force exerted in nature has an equal and opposite force.

For example here, when Al exerts force on Ben, Ben exerts the same amount of force (30N) on Al.

Al exerts the action force and Ben exerts the reaction force.

Hi there!

a.

Use the formula d = st to solve:

d = 20 × 0.5 = 10m

150 - 10 = 140m away when brakes are applied

b.

Use the following kinematic equation to solve:

vf² = vi² + 2ad

Plug in known values:

0 = 20² + 2(150)(a)

Solve:

0 = 400 + 300a

-300a = 400

a = -4/3 (≈ -1.33) m/s² required

c.

Use the following kinematic equation to solve:

vf = vi + at

0 = 20 - 4/3t

Solve:

4/3t = 20

Multiply both sides by 3/4 for ease of solving:

t = 15 sec

Answer: -5×10-3

Explanation:

E=kq/r

Answer:

7.46×10⁻⁶ m

Explanation:

Applying,

F = kqq'/r²............ Equation 1

make r the subject of the equation

r = √(F/kqq').......... Equation 2

From the question,

Given: F = 8 N, q' = q= 4 C

Constant: k = 8.98×10⁹ Nm²/C²

Substitute these values into equation 2

r = √[8/(4×4×8.98×10⁹)]

r = √(55.7×10⁻¹²)

r = 7.46×10⁻⁶ m