For every _______ over 50 mph, your chances of being seriously injured, disfigured, or killed are doubled.

Answer:

3.9 m

Explanation:

The principle of work and energy

ΔE = W  Formula (1)

where:

ΔE: mechanical energy change (J)

W : work of the non-conservative forces (J)

ΔE = Ef - E₀  

Ef : final mechanical energy

E₀   : initial mechanical energy

Ef = K f+ Uf

E₀ = K₀ + U₀

K =(1/2 )mv² :  Kinetic energy (J)

U = mgh  :Potential energy (J)

m: mass (kg)

v : speed (m/s)

h: hight (m)

Known data

m = 1 kg  : mass of the block

v₀ =  mg(h).0 m/s Initial speed of the block

vf = 0 = Final speed of the block

θ =40°  :angle θ of the ramp with respect to the horizontal direction

μk=0  : coefficient of kinetic friction

g = 9.8 m/s² : acceleration due to gravity

 Problem development

W = 0 ,  Because the friction force (non-conservative force ) is zero

Principle of work and energy to the Block:

ΔE = W

Ef - E₀  = 0   Equation (1)

Ef = K f+ Uf =(1/2 )m(0)² + mg(h)=  mg(h)  (Joules)

E₀ = K₀ + U₀ = (1/2 )m(7)² + mg(0) = 24.5m (Joules)

In the equation (1) :

Ef =  E₀

mg(h) = 24.5m

We divided  by m both sides of the equation

g(h) = 24.5

h = 24.5 / g

h = 24.5 / 9.8

h = 2.5 m

We apply trigonometry at the ramp to calculate how far up the ramp (d) does the block slide before coming momentarily to rest :

sinθ = h/d

d = h / sinθ

d = 2.5 m / sin40°

d = 3.9 m

The distance the block slides up the ramp can be determined using the principle of energy conservation: initial kinetic energy is converted into gravitational potential energy. The distance along the ramp is the height (calculated using the energy conservation equation) divided by the sine of the ramp's angle of inclination.

This is a problem of energy conservation. Initially, the block has some kinetic energy due to its speed and no potential energy. When it starts climbing the ramp, kinetic energy is transformed into gravitational potential energy. The block will come to rest when all the kinetic energy has been converted into potential energy.

To determine how far up the ramp the block will slide, we use the equation of energy conservation: 1/2 * mass * speed^2 = mass * gravity * height. Given that we know the mass of the block (1.0 kg), its initial speed (7.0 m/s), and the acceleration due to gravity (approximately 9.8 m/s^2), we can solve this equation to get the height as: height=1/2 * mass * speed^2 / (mass * gravity), which simplifies to height= 1/2 * speed^2 / gravity.

Once we get the height, we need to determine the distance along the ramp, which is the height divided by the sine of the ramp’s angle of inclination (40°). Thus, the distance = height / sin(40°).

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Answer:

An increase in air temperature because of its compression.

Explanation:

The Gay-Lussac's Law states that a gas pressure is directly proportional to its temperature in an enclosed system to constant volume.  

[tex] P = kT [/tex]  

where P: is the gas pressure, T: is the gas temperature and k: is a constant.

Therefore, due to Gay-Lussac's Law, when the plunger is pushed down very rapidly, the pressure of the air increase, which leads to its temperature increase. That is why cotton flashes and burns.      

I hope it helps you!

Answer:

8.85 × 10 ⁻³ m

Explanation:

The electric field produced by infinite sheet of charge relates by

E = σ / 2ε₀

where  σ = surface charge density = 0.1 μC/m² = 0.1 × 10⁻⁶C/m², ε₀ = 8.85 × 10⁻¹² C⁻²/Nm²

also

V = E Δs where Δs is the distance apart in meters, V is the voltage of the

make E subject of the formula

V/Δs = E

equate the two equation

V / Δs = σ / 2ε₀

make Δs subject of the formula

V2ε₀ / σ = Δs

substitute the values into the expression

Δs = (50 × 2 × 8.85 × 10⁻¹²)  / (0.1 × 10⁻⁶) = 8.85 × 10 ⁻³ m

Answer:

Power when towing

Explanation:

The V8 engine has Direct Injection Gas (DIG) technology, which provides better wide-open throttle performance and improved fuel consumption and emissions performance (compared to a non-direct-injection system) by reducing engine knock, improving combustion stability and offering precise injection control.

The multi-control valve on the V8 engine helps to enhance power when towing by reducing the amount of torque the engine must expend to move the vehicle, allowing it to commit more to moving the load.

Answer:

(a) [tex]v^{2} _{x} = 36.508 m^2/s^2[/tex]

(b)  0.1235 Pa

(c) The volume of the ball is significant when compared with the volume of the closed room. Thus, the constant 'b' of the equation of Van der Waal is greater than zero.

Explanation:

In the question, we are given the following variables:

V = 453 m^3

[tex]m_{ball} = m = 42.0 g[/tex]

Kinetic energy (K.E) = 2.30 J

(a) The equation for kinetic energy is:

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

Since the average velocity components of the ball in the three dimensions are equal. Thus, we have:

[tex]v_{x} ^{2} = v^{2} _{y} =v^{2} _{z}[/tex]

In addition:

[tex]v^{2} =[/tex][tex]v_{x} ^{2} + v^{2} _{y} +v^{2} _{z}[/tex] = [tex]3v^{2} _{x}[/tex]

Therefore:

[tex]K.E = \frac{1}{2}m[3v^{2} _{x}][/tex] = [tex]\frac{3}{2}mv^{2} _{x}[/tex]

Thus:  

[tex]v^{2} _{x} = \frac{KE}{1.5*m} = \frac{2.3}{1.5*0.042} = 36.508 m^2/s^2[/tex]

It does not matter which direction we take as x-direction because the average velocity component in every direction is the same.

(b) The average pressure on the walls can be calculated using the equation below:

[tex]P =\frac{mv^{2} _{x} }{V}[/tex]

where V = 453 m^3

Thus: P = (0.042*36.508^2)/453 = 0.1235 Pa

(c) The volume of the ball is significant when compared with the volume of the closed room. Thus, the constant 'b' of the equation of van der Waal is greater than zero.

The average value of u² is 54.8 m²/s². It does not matter which direction is taken as x. The average pressure on the walls can be found using the force and area equations.

(a) The average value of u² can be found by dividing the average kinetic energy by the mass of the ball: u² = KE / m. In this case, u² = 2.30 J / 0.042 kg = 54.8 m²/s².

It does not matter which direction you take to be x. The average value of u² will be the same regardless of the direction.

(b) The average pressure on the walls can be found using the equation P = F / A, where P is the pressure, F is the force, and A is the area of the walls. The force exerted on the walls can be found using the equation F = ∆p / ∆t, where ∆p is the change in momentum and ∆t is the change in time. Since there are no pauses in the ball's motion, ∆p = m * ∆v, where m is the mass of the ball and ∆v is the change in velocity. The average pressure will be the force divided by the area of the walls.

(c) The main assumption in Pressure, Temperature, and RMS Speed that does not apply here is that there are a large number of gas molecules moving in random directions. In this problem, there is only one molecule, so the assumption of a large number of molecules does not hold.

Answer:

1. Equal.

Explanation:

Newton's third law of motion;

   According to this law every action have its reaction with same magnitude but in the opposite direction.

If we push a wall with some amount of force then wall pushes back with same magnitude but in the opposite direction.This law tell us about action and reaction forces which are in same magnitude but in the opposite direction.

Therefore the answer will be 1.

1. Equal.

When pushing a wall, the wall pushes back with equal force as stated by Newton's Third Law of Motion. This law means forces have equal magnitude, but opposite direction.

When you push against a wall, the wall pushes back against you with equal force. This is demonstrated by Newton's Third Law of Motion, which states that 'for every action, there is an equal and opposite reaction'. The force you exert on the wall and the force the wall exerts on you are action-reaction force pairs, meaning the force exerted on the wall is the same magnitude, but in the opposite direction to the force the wall exerts on you.

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Answer:

D)  θ₂= 36. 6º  

Explanation:

In this diffraction experiment it is described by the equation

               sin θ = m λ

The first dark strip occurs for m = 1 and since the angle is generally small we can approximate sine to the value of the angle

              θ₁ = λ/ a

This equation is valid for linear slits, in the case of a circular slit the problem must be solved in polar coordinates, so the equation changes slightly

             θ₂ = 1.22 λ / a

In the proposed exercise we start with a linear slit of width a, where tes1 = 30º and end with a circular slit of the same diameter

             θ₂ = 1.22 λ / a

Let's clear (Lam/a) of equalizing the two equations

             θ₁ = θ₂/ 1.22

             θ₂ = 1.22 θ₁

             θ₂ = 1.22 30

             θ₂= 36. 6º

When reviewing the correct results is D

Answer:

R=2.5 ohm

Explanation:

Given that   :

Voltage difference = 9 V

Current ,I= 3.6 A

As we know that from Ohm's law

V= I R

V =Voltage difference

I=Current

R=Resistance of the wire

Now by putting the values in the above equation we get

9 = 3.6 R

[tex]R=\dfrac{9}{3.6}[/tex]

R=2.5 ohm

Therefore the resistance of the wire will be 2.5 ohm.

Answer:108.71 mL

Explanation:

Given

Volume of sample V=150 mL

concentration of sucrose solution 35 % w/w i.e. In  100 gm of sample 35 gm is sucrose

specific gravity =1.115

Density of solution [tex]\rho _s=1.115\times density\ of\ water[/tex]

Thus

[tex]\rho _s=1.15\times 1 gm/mL=1.115\ gm/mL[/tex]

mass of sample [tex]M=1.115\times 150=167.25\ gm[/tex]

mass of sucrose [tex]m_s=0.35\times 167.25=58.53\ gm[/tex]

mass of Water [tex]m_w=108.71 gm[/tex]

Volume of water [tex]=108.71\times 1=108.781 mL[/tex]

Final answer:

The magnitude of the acceleration of the object at time t = 2.00 s is 6.40 m/s².

Explanation:

The magnitude of acceleration can be found by taking the second derivative of the position function with respect to time.

Given the position function r = [2.0 m + (5.00 m/s)t] i^ + [3.0m−(2.00 m/s²)t²] j^, we can differentiate it twice to get the acceleration function a(t) = 5.00 m/s i^ - 4.00 m/s² j^.

Plugging in t = 2.00 s gives us a(2.00 s) = 5.00 m/s i^ - 4.00 m/s² j^, and calculating the magnitude of this vector gives us 6.40 m/s² as the magnitude of acceleration at t = 2.00 s.

The magnitude of the acceleration of the object at [tex]\( t = 2.00 \, \text{s} \) is \( 4.00 \, \text{m/s}^2 \).[/tex]

To find the magnitude of the acceleration of the object at time [tex]\( t = 2.00 \, \text{s} \)[/tex], we first need to determine the acceleration vector from the given position vector. The position vector[tex]\( \mathbf{r}(t) \)[/tex]is given by:

[tex]\[\mathbf{r}(t) = [2.0 \, \text{m} + (5.00 \, \text{m/s})t] \hat{i} + [3.0 \, \text{m} - (2.00 \, \text{m/s}^2)t^2] \hat{j}\][/tex]

Find the Velocity Vector

The velocity vector [tex]\( \mathbf{v}(t) \)[/tex] is the time derivative of the position vector [tex]\( \mathbf{r}(t) \)[/tex]:

[tex]\[\mathbf{v}(t) = \frac{d\mathbf{r}(t)}{dt}\][/tex]

Differentiate each component with respect to time  t :

For the [tex]\( \hat{i} \)[/tex] component:

[tex]\[v_x(t) = \frac{d}{dt}[2.0 + (5.00 \, \text{m/s})t] = 5.00 \, \text{m/s}\][/tex]

For the[tex]\( \hat{j} \)[/tex] component:

[tex]\[v_y(t) = \frac{d}{dt}[3.0 - (2.00 \, \text{m/s}^2)t^2] = -2.00 \times 2t = -4.00 \, t \, \text{m/s}\][/tex]

So the velocity vector is:

[tex]\[\mathbf{v}(t) = (5.00 \, \text{m/s}) \hat{i} + (-4.00 \, t) \hat{j}\][/tex]

Find the Acceleration Vector

The acceleration vector[tex]\( \mathbf{a}(t) \)[/tex] is the time derivative of the velocity vector [tex]\( \mathbf{v}(t) \)[/tex]:

[tex]\[\mathbf{a}(t) = \frac{d\mathbf{v}(t)}{dt}\][/tex]

Differentiate each component with respect to time t:

For the [tex]\( \hat{i} \)[/tex]component:

[tex]\[a_x(t) = \frac{d}{dt}[5.00] = 0 \, \text{m/s}^2\][/tex]

For the [tex]\( \hat{j} \)[/tex] component:

[tex]\[a_y(t) = \frac{d}{dt}[-4.00 \, t] = -4.00 \, \text{m/s}^2\][/tex]

So the acceleration vector is:

[tex]\[\mathbf{a}(t) = 0 \hat{i} - 4.00 \hat{j}\][/tex]

Calculate the Magnitude of the Acceleration

The magnitude of the acceleration [tex]\( |\mathbf{a}(t)| \)[/tex] is:

[tex]\[|\mathbf{a}(t)| = \sqrt{a_x^2 + a_y^2}\][/tex]

Substitute[tex]\( a_x = 0 \) and \( a_y = -4.00 \)[/tex]:

[tex]\[|\mathbf{a}(t)| = \sqrt{0^2 + (-4.00)^2} = \sqrt{16.00} = 4.00 \, \text{m/s}^2\][/tex]

Answer:

68.829 N

Explanation:

The given parameters are:

Weight of ladder, [tex]W_l[/tex] = 120 N

Weight of object, [tex]W_o[/tex] = 98 N

Angle, [tex]\theta[/tex] = 53°

And we also know that, while

Length of ladder = L

Distance the object is placed = L/3

If we apply translational equilibrium horizontally, then [tex]T-N=0[/tex], so [tex]T = N[/tex]

If we apply rotational equilibrium about the distance the object is placed, then

[tex]W_o\frac{L}{3}cos(\theta) + W_l\frac{L}{2}cos(\theta) - NLsin(\theta) = 0\\2NLsin(\theta) = 2W_o\frac{L}{3}cos(\theta)+W_lLcos(\theta)\\2Tsin(\theta) = \frac{2}{3}W_ocos(\theta)+W_lLcos(\theta)\\T = \frac{1}{3}W_ocot(\theta) + \frac{1}{2}W_lcot(\theta)\\T = \frac{1}{3} * 98 * cot(53) + \frac{1}{2} * 120 * cot(53) = 69.829[/tex]

Answer:

The force that must be provided to accelerate the object = 302.196 N

Explanation:

Force: Force can be defined as the product of mass and acceleration. The S.I unit of force is Newton (N).

Ft - W = ma............ Equation 1

Ft = W + ma .......... Equation 2

Where Ft = force provided, W = weight of the object, m= mass of the object, a = acceleration of the object.

Given: mass = 64 lb, g = 32 ft/s² a = 2 ft/s²

Conversion: (i)  from 64 pounds to Kg = 64/2.2046

                      = 29.03 kg

  (ii)  from 2 ft/s² to m/s² = 0.3048 × 2 = 0.6098 m/s²

 (iii) from 32 ft/s² to m/s² = 0.3048×32 = 9.8 m/s².

and W = mg = 29.03 × 9.8 = 284.494 N,

Substituting these values into equation 2,

Ft = 284.494 + 29.03×0.6098

Ft = 284.494 + 17.7

Ft = 302.196 N

Therefore, the force that must be provided to accelerate the object = 302.196 N

To accelerate a 64-lb object upward at a rate of 2 ft/s², a force of approximately 3.974 lbf is required.

To find the force required to accelerate a 64-lb object upward at a rate of 2 ft/s², we need to use Newton's second law of motion. The formula for force is F = m * a, where F is the force, m is the mass, and a is the acceleration.

First, convert the mass of the object from pounds to slugs. Since 1 slug is equal to the mass that accelerates at 1 ft/s² when acted upon by 1 lb of force, 1 slug is equal to 32.2 lbs. Therefore, the mass of the object is 64 lbs divided by 32.2 lbs/slugs, which is approximately 1.987 slugs.

Now, substitute the mass (m = 1.987 slugs) and the acceleration (a = 2 ft/s²) into the formula F = m * a. The force required to accelerate the object upward is approximately 3.974 lb-ft/s² or 3.974 lbf.

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Answer:

cold cathode fluorescent lamp

Explanation:

A cold cathode fluorescent lamp (CCFL) is a lighting system that uses two phenomena: electron discharge and fluorescence.

CCFLs are mainly used as light sources for backlights, since they are compact and durable than ordinary fluorescent lamps. They exhibit a wide range of brightness and color (color temperature and chromaticity) that can be achieved by varying the pressure and type of the material injected into the glass tube. The thickness and type of the phosphor used to coat the inner wall of the tube also plays a role in altering the color and brightness.

Final answer:

Early LCD backlights used CCFL technology for its efficiency and longevity. LCDs rely on a backlight to illuminate the liquid crystals in pixels. Each pixel can control red, blue, or green filters to create the full color spectrum on the screen.

Explanation:

Early LCD backlights use CCFL (Cold Cathode Fluorescent Lamp) technology, popular for its low power use, even brightness, and long life. LCD, which stands for liquid crystal display, relies on the light modulating properties of liquid crystals. These crystals themselves do not emit light. In flat screen LCD televisions, the backlight illuminates the screen through millions of tiny units called pixels. Each pixel has three cells with red, blue, or green filters, each controlled independently to vary the picture contrast and color representation by adjusting the voltage applied to the liquid crystal.

Answer:

The velocity of the boy is 0.11 m/s to the right.

The velocity of the girl is 0.13 m/s to the left.

Explanation:

Hi there!

The momentum of the system is conserved. That means that the initial momentum is equal to the final momentum (after throwing the ice ball for the boy and after catching the ball for the girl).

Let´s calculate the momentum of the system boy-ball:

The equation of the momentum is the following:

p = m · v

Where:

p = momentum.

m = mass of the object.

v = velocity.

Initially, the ball and the boy are at rest, so that the momentum of the system boy-ball is zero:

initial momentum = mib · vib + mb · vb

Where:

mib = mass of the ice ball.

vib = velocity of the ice ball.

mb =mass of the boy.

vb = velocity of the boy.

The initial momentum is zero because both velocities are zero.

After throwing the ball, the momentum of the system will be:

final momentum = mib · vib + mb · vb

final momentum = 0.75 kg · (- 6.2 m/s) + 43 kg · vb (Let´s consider the left as negative).

Since final momentum = initial momentum

0 kg · m/s =- 4.65 kg · m/s + 43 kg · vb

Solving for vb:

4.65 kg · m/s/43 kg = vb

vb = 0.11 m/s

The velocity of the boy after releasing the ball is 0.11 m/s. Since there is no friction, this velocity is constant. When the girl catches the ball, the velocity of the boy will be 0.11 m/s (to the right).

Now let´s do the same for the system girl-ball.

initial momentum = mg · vg + mib · vib

initial momentum = 35 kg · 0 m/s + 0.75 kg · (-6.2 m/s)

initial momentum = -4.65 kg · m/s

The final momentum is the momentum of the girl plus the ball:

final momentum = (mg + mb) · vgb

final momentum = (35 kg + 0.75 kg) · vgb

Since initial momentum = final momentum

-4.65 kg · m/s = (35 kg + 0.75 kg) · vgb

-4.65 kg · m/s / (35 kg + 0.75 kg) = vgb

vgb = -0.13 m/s

The velocity of the girl after catching the ball is -0.13 m/s (0.13 m/s to the left).

The velocity of the boy after catching the ice is 0.108m/s right and the velocity of the girl after catching the ice is 0.132m/s left.

Data;

The conservation of linear momentum expects that the momentum of the boy or girl must be equal to the momentum of the ball.

for the boy

[tex]m_1v_1 = m_2v_2\\43*v = 0.75*6.2\\v = 0.108m/s[/tex]

The velocity of the boy after catching the ice is 0.108m/s right

for the girl

[tex]m_1v_1 = m_2 v_2\\0.75*6.2 = 35 * v\\v = 0.132 m/s[/tex]

The velocity of the girl after catching the ice is 0.132m/s left.

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a) The time of flight is 3.78 s

b) The initial speed is 17.0 m/s

c) The speed at impact is 46.4 m/s at [tex]70.3^{\circ}[/tex] below the horizontal

Explanation:

The picture of the previous problem (and some data) is missing: find it in attachment.

a)

The motion of the ball is a projectile motion, which consists of two independent motions:

- A uniform motion (constant velocity) along the horizontal direction  

- A uniformly accelerated motion, with constant acceleration (acceleration of gravity) in the downward direction  

We start by considering the vertical motion, to find the time of flight of the ball. We do it by using the following suvat equation: for the y-displacement:

[tex]y=u_y t+\frac{1}{2}at^2[/tex]

where we have:

y = -45.0 m is the vertical displacement of the ball (the height of the building)

[tex]u_y=u sin \theta[/tex] is the initial vertical velocity, with u being the initial velocity (unknown) and [tex]\theta=-27.0^{\circ}[/tex] the angle of projection

t is the time of the fall

[tex]a=g=-9.8 m/s^2[/tex] is the acceleration of gravity

Along the x-direction, the equation of motion is instead

[tex]x=(u cos \theta)t[/tex]

where [tex]ucos \theta[/tex] is the horizontal component of the velocity. Rewriting this equation as

[tex]t=\frac{x}{ucos \theta}[/tex]

And substituting into the previous equation, we get

[tex]y=xtan \theta + \frac{1}{2}gt^2[/tex]

And using the fact that the horizontal range is

x = 59.0 m

And solving for t, we find the time of flight:

[tex]t=\sqrt{\frac{y-x tan \theta}{g}}=\sqrt{\frac{-45-(59.0)(tan(-27^{\circ}))}{-9.8}}=3.78 s[/tex]

b)

We can now find the initial speed, u, by using the equation of motion along the x-direction

[tex]x=u cos \theta t[/tex]

where we know that:

x = 59.0 m is the horizontal range

[tex]\theta=-23^{\circ}[/tex] is the angle of projection

[tex]t=3.78 s[/tex] is the time of flight

Solving for u, we find the initial speed:

[tex]u=\frac{x}{cos \theta t}=\frac{59.0}{(cos (-23^{\circ}))(3.78)}=17.0 m/s[/tex]

c)

First of all, we notice that the horizontal component of the velocity remains constant during the motion, and it is equal to

[tex]v_x = u cos \theta = (17.0)(cos (-23^{\circ})=15.6 m/s[/tex]

The vertical velocity instead changes according to the equation

[tex]v_y = u sin \theta + gt[/tex]

Substituting all the values and t = 3.78 s, the time of flight, we find the vertical velocity at the time of impact:

[tex]v_y = (17.0)(sin (-23^{\circ}))+(-9.8)(3.78)=-43.7 m/s[/tex]

Where the negative sign means it is downward.

Therefore, the speed at impact is

[tex]v=\sqrt{v_x^2+v_y^2}=\sqrt{(15.6)^2+(-43.7)^2}=46.4 m/s[/tex]

while the direction is given by

[tex]\theta = tan^{-1}(\frac{v_y}{v_x})=tan^{-1}(\frac{-43.7}{15.6})=-70.3^{\circ}[/tex]

So, [tex]70.3^{\circ}[/tex] below the horizontal.

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Answer:

[tex]3.00\times10^{6}\,tons [/tex]

Explanation:

We can resolve this problem using a proportion because the ratio tons burned and energy produced is constant, using x for tons burned of coal to produce [tex] 9.00\times10^{16}J [/tex]:

[tex]\frac{x}{9.00\times10^{16}J}=\frac{1.00\,kg}{3.00\times10^{7}J} [/tex]

solving the proportion:

[tex]x*3.00\times10^{7}J=1.00\,kg*9.00\times10^{16}J [/tex]

[tex] \frac{x*\cancel{3.00\times10^{7}J}}{\cancel{3.00\times10^{7}J}}=\frac{1.00\,kg*9.00\times10^{16}J}{3.00\times10^{7}J}[/tex]

[tex]x=\frac{9.00\times10^{16}kg*\cancel{J}}{3.00\times10^{7}\cancel{J}}=3.00\times10^{9}\,kg [/tex]

Now we have the answer but in kilograms we should convert this knowing that 1 ton = 1000 kg:

[tex]3.00\times10^{9}\,kg=3.00\times10^{9}\,\cancel{kg}*\frac{1\,ton}{1000\,\cancel{kg}} [/tex]

[tex]3.00\times10^{9}\,kg=3.00\times10^{6}\,tons [/tex]

Answer:

3.3 x 106 tons

Explanation:

Since you know that E = mc2, and both equations possess the constant speed of light, set the two equations as equivalencies. For instance, E1/m1 = E2/m2. Solve for m2.

Took the test and got it correct

Answer:

Depending upon the size of the irregularities on the surface and the wavelength of the surface a surface can be smooth for the waves of smaller wavelengths than the size of the irregularities.

Explanation:

Depending upon the size of the irregularities on the surface and the wavelength of the surface a surface can be smooth for some of the waves but not for the others.

Contrarily, when the surface has the irregularities less than the size of the wavelength then it reflects most of the light in a uniform manner.

Explanation:

Antoine lavoisier is most famous for his role in  discovering of oxygen. He recognized and name two important element oxygen(1778) and hydrogen(1779).

In a chemical reaction, Lavoisier observed the mass is retained. A chemical reaction's total mass of the products will always be the same as the total mass of the reactant materials used in the reaction. His results led to one of the basic laws of chemical behavior: the law of material preservation, which states that matter is preserved in a chemical reaction.

Final answer:

Antoine Lavoisier's major contributions to atomic theory include the discovery of the law of conservation of mass and transforming chemistry into a quantitative science by disproving the phlogiston theory and identifying the role of oxygen in combustion. These discoveries paved the way for the future development of the atomic theory.

Explanation:

Contributions of Antoine Lavoisier to Atomic Theory

Antoine Lavoisier, often regarded as the 'father of modern chemistry', made several groundbreaking contributions to atomic theory. He is particularly famous for discovering the law of conservation of mass. This principle, established through careful experimentation, states that mass in an isolated system is conserved over time, regardless of the type of chemical reactions happening within. This fundamental law formed a cornerstone for John Dalton to later develop the modern atomic theory.

Additionally, Lavoisier conducted experiments that disproved the phlogiston theory and instead suggested that oxygen was a key element in combustion. His innovative approach to chemical reactions included the use of precise weighing measures and conducting reactions in sealed environments to accurately track the mass of gases involved. Lavoisier's meticulous quantitative methods transformed chemistry from a qualitative to a quantitative science.

Through these contributions, Lavoisier laid down the basis for atomic theory as well as the concept of chemical elements, thereby fundamentally changing our understanding of chemical processes. He also impacted other areas such as explaining the role of oxygen in respiration and the composition of air.

Answer:

The object will continue to move in straight line at 50 mi/hr except it is acted upon by an external force

Explanation:

Newton First Law of motion: A body will continue in its present state of  rest or, if it is in motion will continue to move with uniform speed in a straight line unless it is acted upon by a force. Newton first law of motion is also called the law of inertia.

Inertia: This is the tendency of a body to remain in its states of rest or uniform motion.

From the question, The object moving in a straight line at 50 mi/hr will continue to move in a straight line at 50 mi/hr, except it is acted upon by external force that will change the speed of the object.

Final answer:

An object moving in a straight line at a constant speed will continue this state of motion until an external force is applied, according to Newton's first law.

Explanation:

According to Newton's first law of motion, an object moving at a constant speed in a straight line will continue to do so until a net external force acts upon it. In the case of an object moving at 50 mi/hr in a straight line, it will maintain this motion and speed as long as no forces such as friction, air resistance, or another applied force interfere with its movement. This principle indicates that without external forces, there is no change in velocity (speed and direction).

Answer:

The images output from your new color laser printer seem to be a little too blue. to fix this problem we need to calibrate the printer.

Explanation:

This can be done by opening the toolbox, clicking in the device setting folder their you get print quality page click on it. Under the print quality option click on the calibrate next to calibrate now. Then click OK unless when the 'your request has been sent to the device' appears on the screen. When the calibration ends again try printing. calibrating is useful for managing the proper alignment of the inkjet cartridge nozzle to the paper and each other, without proper calibration the print quality deteriorates.

Answer:20 cal/s

Explanation:

Given

Heat transfer rate is [tex]\dot{Q}=10 cal/s[/tex]

Also heat rate is given by

[tex]\dot{Q}=kA\frac{dT}{dx}[/tex]

where [tex]k=thermal\ conductivity[/tex]

[tex]A=area\ of\ cross-section[/tex]

[tex]dT=change\ in\ temperature[/tex]

[tex]dx=change\ in\ length[/tex]

[tex]10=k\frac{\pi }{4}d^2\frac{dT}{L}----1[/tex]

For [tex]d'=2d, Length L'=2L[/tex]

[tex]\dot{Q}=k\frac{\pi }{4}(2d)^2\frac{dT}{2L}---2[/tex]

dividing 1 and 2 we get

[tex]\frac{10}{\dot{Q}}=\frac{2d^2}{4d^2}[/tex]

[tex]\dot{Q}=20 cal/s[/tex]                            

The power rate of the steel is mathematically given as

P=20cal/s

Question Parameter(s):

The rod conducts heat from one end to the other at a rate of 10 cal/s

Generally, the equation for the Power   is mathematically given as

Power = (kA∆T)/l

Therefore

Power = [k × 4(πd^2)/4 × ∆T]/2l

Therefore

Power  = 2(kA∆T)/l

Hence, Initial Power

iP = (kA∆T)/l

iP= 10cal/s

Where

dT is the same

2[(kA∆T)/l] = 2 × 10

P=20cal/s

In conclusion, The power rate is

P=20cal/s

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Answer:

Minimum velocity is 25m/s

Explanation:

Given force F =1000N

Mass flow rate m/t= 40kg/s

Velocity=v

Recall Newton second law of motion F =MV/t

1000=40v

V=1000/40

25m/s

Answer:

(b) she is using fishing apparatus which restricts her manoeuvrability

Explanation:

The explanation is spelled out in Rule 3(d) of the inland and water ways  laws that state that: A vessel engage in fishing is one whose fishing apparatus restricts manoeuvrability. I f the fishing gear does not hamper her movements, then she is engage in fishing. The rules state categorically that a troller  with hook-spangled lines following in her wake is not restricted by her gear. Rule 3(d) might appear to be splitting hairs, but it has a point to make. Although a vessel might not be hampered as to be unable to keep clear of others, her fishing gear could still make following a script difficulty

Answer:

Option d

d. 16 times the ke

Explanation:

The kinetic energy is defined as:

[tex]Ke = \frac{1}{2}mv^{2}[/tex]  (1)

Where m and v are the mass and velocity of the object.

The kinetic energy represents how much energy have an object as a consequence of its movement.

For the case of a car moving 4 times as fast as another identical car.

By means of equation 1 it is gotten:

[tex]Ke = \frac{1}{2}m\cdot(4v)^{2}[/tex]

[tex]Ke = \frac{1}{2}m\cdot16v^{2}[/tex]

Hence, the faster car has 16 times the kinetic energy.

Answer:

130 cm

Explanation:

After each bounce, the mechanical energy decreases by 18%, which means the maximum height decreases by 18%.  Modeling this as exponential decay:

h = 290 (1 − 0.18)ⁿ

h = 290 (0.82)ⁿ

h = 290 (0.82)⁴

h = 130

The ball reaches a height of 130 cm.

Answer:

130 cm

Explanation:

Gradpoint

Answer:

The quantity of heat required to to heat the cup of water through a temperature change of 80°C = 200.83 kJ

Explanation:

Density = mass/Volume.

∴ mass = Density ×  volume

Where Density of water = 1.00 g/mL, volume of water = 600 mL

Mass = 1 × 600 = 600 g.

(Q) = cm(ΔT)...................... equation 1

Where Q = quantity of heat, c = specific heat capacity of water, m = mass of water, ΔT = T₁-T₂ = change in temperature.

Where c = 4.184 J/g°C, m = 600 g, ΔT = T₁-T₂ = 100 - 20 = 80°C

Substituting these values into equation 1,

Q = 4.184 × 600 × 80

Q = 200832 J = 200.83 kJ

Therefore the quantity of heat required to to heat the cup of water through a temperature change of 80°C = 200.83 kJ

Explanation:

1200 feet.

In the given scenario we would be operation under VFR.

VFR is visual flight rules are set of regulations under which the pilot operates aircraft in a weather which generally clear enough that facilitates pilot to see the flight where it is going.

Answer:

This is an incomplete question. The diagram and a part of the question is missing.

Explanation:

In the context, it is given that the particle tight to a string is whirled in a circular direction in a constant speed. And there is no gravitational field.

Now since the particle's speed is constant, hence the tangential acceleration is zero. The particle exhibits radial acceleration only which will act along the radius towards the center of the circle (towards the left).

Therefore the correct representation of the component of the acceleration is given below --  

A particle being whirled in a circle at a constant speed demonstrates uniform circular motion. The changing direction of velocity results in a centripetal acceleration, determined by an external centripetal force. Similar motion is observed in centrifuges and orbiting satellites.

A top view of a particle on a string whirled in a circle at a constant speed would show uniform circular motion. In this simplest form of curved motion, the particle moves along a fixed radius at a constant speed. Although the magnitude of its velocity doesn't change, the direction of the velocity is constantly changing, leading to what's referred to as centripetal acceleration. This is caused by a net external force - known as the centripetal force.

This type of motion is similar to other examples of circular motion, like a satellite orbiting Earth, a race car moving around a racetrack, or a ball being swung in a circle. Whenever there's no centripetal force to cause circular motion, inertia carries the object along a line tangent to the circle observed as outward motion in a centrifugal frame of reference. This inertial effect is seen in various physical phenomena.

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Answer:

[tex]T'=1.58K[/tex]

Explanation:

As the Universe expands, the photons of the cosmic microwave background increase its wavelength, making its temperature inversely proportional to the scale factor of the Universe. That is, as the average distances between galaxies increase, the temperature of the cosmic microwave background decreases by the same factor. Therefore, the temperature when the distances between galaxies are 1.7 times as large as they are today will be:

[tex]T'=\frac{T_{now}}{1.73}\\T'=\frac{2.725K}{1.73}\\T'=1.58K[/tex]

When the average distances between galaxies are 1.7 times their present distances, the temperature of the cosmic microwave background (CMB), as a result of Wien's Law, will be about 1.61 Kelvin, which is 1.7 times lower than its current close to 2.73 Kelvin figure.

The cosmic microwave background radiation (CMBR), an aftermath of the Big Bang event, exhibits a relationship between wavelength and temperature called Wien's Law. At present, the temperature of the CMBR is approximately 2.73 Kelvin (K), a finding made possible by scientists Penzias and Wilson who associated radiation intensity at different wavelengths to a temperature of roughly 3.5 K.

As the galaxies continue to recede from one another, their emitted radiation wavelengths become redshifted, effectively stretching them out. This movement is also reflected in the increase in the peak wavelength of the CMBR. If the average distances between galaxies were to magnify by 1.7 times, the peak wavelength of the CMBR would equally elongate by the same factor.

Applying Wien's Law, this increase will consequently result in a decrease in the CMB's temperature. Thus, when galaxies have expanded such that their average distances are 1.7 times larger than their current distances, the temperature of the cosmic microwave background will be 1.7 times lower than its current temperature of approximately 2.73 K. Hence, the new temperature would approximately be 1.61 K.

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