When passing in a medium for a distance of 1.5 cm intensity

of the light decreased by 3 times. What will the distance x equal to when the intensity of the light decreases by 9 times?

Answer:

eˣ-8.

Explanation:

(eˣ-8x+7)'=eˣ-8.

The initial velocity of the arrow and the height of the building will be 29.05 m/s² and 781 m respectively.

The change of distance with respect to time is defined as speed. Speed is a scalar quantity. It is a time-based component. Its unit is m/sec.

The given data in the problem is;

u is the initial velocity of fall = ?  m/sec

h is the distance of fall = 200 m

g is the acceleration of free fall = 9.81 m/sec²

H is the height of the building

t is the time period = 10 second

According to Newton's second equation of motion,

[tex]\rm h= ut+\frac{1}{2} gt^2 \\\\\ h-\frac{1}{2} gt^2 =ut \\\\ 200 - 0.5 \times(9.81) \times 10^2 = 10 u \\\\ u = - 29.05 \ m/sec[/tex]

- ve shows the direction is downward.The magnitude of the initial velocity is found as;

u = 29.05 m/sec

The height of the building

[tex]\rm H= ut+\frac{1}{2} gt^2 \\\\\ H = 29.05 \times 10 + 0.5 \times 9.81 \times 10^2 \\\\ H = 781 \ m[/tex]

Hence the initial velocity of the arrow and the height of the building will be 29.05 m/s² and 781 m respectively.

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Answer:

8.4335 x 10^-²¹, 1.78204 x 10²⁶

Explanation:

p = mv

p = (5.05 x 10⁶)(1.67 x 10^-²⁷)

p = 8.4335 x 10^-²¹

p = mv

p = (5.98 x 10²⁴)(2.98 x 10)

p = 1.78204 x 10²⁶

The magnitude of the average force exerted on the ball by the wall is 225.469 N.

The change in momentum is equal to the product of impact force applied while colliding and time for that impact.

Impulse F. t = m (Vf -Vi)

where, Vf is the final velocity and Vi is the initial velocity.

Given is a 3.42 kg steel ball strikes a massive wall at 14.3 m/s at an angle of α = 61.5° with respect to the plane of the wall. It bounces off with the same speed and angle. The ball is in contact with the wall for 0.207 s

Substitute the values into the expression, we get

Impulse = 2 x mvcosθ

Impulse= 2 x 3.42 x 14.3 x cos 61.5°

Impulse = 46.672 kg.m/s

The impact force can be written as

F.t = I

Put the given values, we have

F = 46.672 /  0.207

F = 225.469 N

Thus, the magnitude of force exerted by the wall is 225.469 N

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The magnitude of the reaction force of the air pushing the balloon would be equal and opposite.

A push or a pull that an object experiences as a result of interacting with another item is known as a  reaction force.

Newton's third law is officially expressed as follows: There is an equal and opposite reaction to every action. The implication of the statement is that there are always two forces acting on the two interacting objects. The force acting on the first object is equal in size to the force acting on the second. The force acting on the first object is acting in the opposite direction to the force acting on the second object. Force pairs—equal and opposing action-reaction force pairs—always exist in pairs.

Knowing that everything has an equal and opposite reaction according to Newton's second law. The preasured air that a balloon had to push out into the free air acts as the reaction force.

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Answer:

D. s=3x+4y

Explanation:

The line is at the 3rd column in the 4th row.

Answer:

a force of acts on a body of mass 50 kg for 10 sec when the first of acting on the body the body covers 80 and in the next 10 second

Answer:

a = 0.4 ms⁻²

Explanation:

• First calculate the resultant force on the body:

If we consider upwards to be the positive direction and downwards to be the negative direction, then the resultant force will be:

F = +100 + (-80)

  = +20 N

The resultant force is positive, so it will act upwards.

• Next calculate the acceleration:
F = ma

a = F/m

a = 20/ 50

a = 0.4 ms⁻²

As the force is acting upwards, the acceleration will also be upwards.

This is an exercise of thermal scales.

T = 40 °C

T = °F ¿?

To convert degrees Celsius to degrees Fahrenheit, we have the formula:

[tex]\bf{^{\circ}F=\dfrac{9}{5} \ ^{\circ}C+32 \ \ \to \ \ \ Formula }[/tex]

We substitute our data into the formula.

[tex]\bf{^{\circ}F=\dfrac{9}{5} \times 40+32}[/tex]  

First we divide 9/5 multiplied by 40.

[tex]\bf{=75+32}[/tex]

Add 75 plus 32

[tex]\bf{=104 \ ^{\circ}F}[/tex]

Therefore, the temperature change of 40 °C on the °F scale is equal to 104.

[tex]\huge \red{\boxed{\green{\boxed{\boldsymbol{\purple{Piscis04}}}}}}[/tex]

HIV can be contracted from contact with bloodborne pathogens.

Other bloodborne diseases are HBV, malaria, syphilis and brucellosis

Bloodborne pathogens can be defined as those microorganisms or pathogenic organisms that cause disease and are present in human blood.

Blood borne pathogens can also be contacted through the following means

In conclusion; HIV can be contracted from contact with bloodborne pathogens.

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The speed at which the medicine leaves the needle is  2.462 m/s

When an incompressible, ideal fluid is flowing through a tube or pipe, the total energy remains constant.

p₁ /ρg + v₁²/2g +z₁ = p₂ /ρg + v₂²/2g +z₂

Where, p/ρg = pressure energy

            v²/2g = kinetic energy

                   z = potential energy

Given is during an injection, pressure in the barrel of syringe is 1.03  atm while pressure in the needle section is 1.00 atm. Assuming the syringe lays horizontally and mass density of the liquid medicine ρ =1000 kg/m³.

The fluid is initially at rest.

Using the Bernoulli's equation, we have

v₁² -  v₂²= 2 x (p₂ -p₁) / ρ

Substituting the values, we get

0 -  v₂² = 2 x (1.00 -1.03) x 1.01 x 10⁵ /1000

v₂² = 6.06

v₂ = 2.462 m/s

Thus, the speed at which the medicine leaves the needle is 2.462 m/s

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Answer:

Approximately [tex]9.39 \; {\rm m\cdot s^{-1}}[/tex] after the sphere has travelled a distance of [tex]5\; {\rm m}[/tex].

Approximately [tex]16.3\; {\rm m\cdot s^{-1}}[/tex] right before touching the ground (a distance of [tex]15\; {\rm m}[/tex].)

Assumption: [tex]g = 9.81\; {\rm N\cdot kg^{-1}}[/tex].

Explanation:

Weight of the sphere: [tex]m\, g = 9.81\; {\rm N \cdot kg^{-1}} \times 10\; {\rm kg} = 98.1\; {\rm N}[/tex], downwards.

Drag on the sphere: [tex]10.0\; {\rm N}[/tex] upwards.

Net force on the sphere: [tex]98.1\; {\rm N} - 10\; {\rm N} = 88.1\; {\rm N}[/tex] downwards.

Acceleration of the sphere: [tex]a = F_\text{net} / m = 88.1\; {\rm N} / (10\; {\rm kg}) = 8.81\; {\rm m\cdot s^{-2}}[/tex].

Apply the SUVAT equation [tex]v^{2} - u^{2} = 2\, a\, x[/tex], where [tex]v[/tex] is the final velocity, [tex]u[/tex] is the initial velocity ([tex]0[/tex] in this case, as the sphere was released from rest,) and [tex]x[/tex] is the distance (displacement) that the sphere has travelled so far.

Rearrange this equation to obtain an expression for [tex]v[/tex]:

[tex]\displaystyle v = \sqrt{2\, a\, x + u^{2}}[/tex].

For example, after the ball travelled a distance of [tex]5.00\; {\rm m}[/tex], [tex]x = 5.00 \; {\rm m}[/tex]:

[tex]\begin{aligned} v &= \sqrt{2\, a\, x + u^{2}} \\ &= \sqrt{2 \times 8.81\; {\rm m\cdot s^{-2}} \times 5.0\; {\rm m} + 0} \\ &\approx 9.39\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].

Similarly, [tex]x = 15.0\; {\rm m}[/tex] right before landing, such that:

[tex]\begin{aligned} v &= \sqrt{2\, a\, x + u^{2}} \\ &= \sqrt{2 \times 8.81\; {\rm m\cdot s^{-2}} \times 15.0\; {\rm m} + 0} \\ &\approx 16.3\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].

The final velocity of the projectile when it strikes the ground below is 198.51 m/s.

The time taken for the projectile to fall to the ground is calculated as follows;

h = vt + ¹/₂gt²

where;

265 = (185 x sin45)t + (0.5)(9.8)t²

265 = 130.8t + 4.9t²

4.9t² + 130.8t - 265 = 0

solve the quadratic equation using formula method,

t = 1.89 s

vyf = vyi + gt

where;

vyf = (185 x sin45) + (9.8 x 1.89)

vyf = 149.322 m/s

vxf = vxi

where;

vxf = 185 x cos(45)

vxf = 130.8 m/s

vf = √(vyf² + vxf²)

where;

vf = √(149.322²  +  130.8²)

vf = 198.51 m/s

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The stress on the wire is determined as  6.37 x 10⁶ N/m².

The stress on the wire is calculated as follows;

σ = F/A

where;

Let the diameter of the wire = 2 mm

Area = πr²

Area = π(1 x 10⁻³)²

Area = 3.142 x 10⁻⁶ m²

Stress = 20/(3.142 x 10⁻⁶)

Stress = 6.37 x 10⁶ N/m²

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Explanation:

an object's gravitational potential energy Eg is m×g×h where:

m=mass

g=9.8m/s²

h=height relative to the closest object below it (because it cannot potentially fall through it

so Eg = 15×9.8×5=735J

The amount of time it will take to have 125 grams of erbium-165 left

in the sample is; C: 31.2 hours

The formula for amount of substance after decay is;

N(t) = N₀(¹/₂)^(t/t_¹/₂)

We are given;

Half life; t_¹/₂ = 10.4 hours

Initial amount; N₀ = 1000 g

Amount left; N(t) = 125 g

Thus;

125 = 1000 * (¹/₂)^(t/10.4)

125/1000 = (¹/₂)^(t/10.4)

In 0.125 = (t/10.4) In 0.5

-2.08 = -0.693(t/10.4)

t/10.4 = -2.08/-0.693

t = 10.4 * 3

t = 31.2 hours

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The weight of an object on pluto will be 15.56. Mass multiplied by the gravitational acceleration gives weight.

Gravitation is a natural law by which all things with all matter are attracted towards one another. Gravity is responsible for large-scale structures present in the Universe.

By dividing the object's weight on Earth by 9.8 m/s², as illustrated below, one may calculate the object's mass.

m = 250 N / 9.8 m/s²

m = 25.51 kg

Multiply the acquired mass by Pluto's gravitational acceleration (g);

W = (25.51 kg) x (0.61 m/s²)  

W = 15.56 N

Thus, the item will only be 15.56 N in weight.

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Answer: B.

15.6 newtons

Explanation: edmentum

Answer:

a) 0.2m/s

b) 5m/s

Explanation:

a) acceleration=∆v/∆t

v=10

t=50

10/50=1/5

=0.2m/s

b) time= d/s

d=50m

s=10

50/10

=5m/s

Answer:

See below

Explanation:

Average speed =  (0+ 10)/2 = 5 m/s

  then to cover 50 m     will take    50 m / 5 m/s  = 10 seconds

      change in velocity/ change in time = acceleration = 10/10 = 1 m/s^2

Answer:

Using 3 periods to get an accurate reading:

3T = (6.40 S - 0.90 s) = 5.50 S So, T = 1.83 s

m = 0.250 kg

Using algebra:

k= [tex]\frac{4\pi ^{2}m }{T^{2} }[/tex]=[tex]\frac{4\pi ^{2}0.250 }{1.90^{2} }[/tex]=2.95kg/sec

Answer:

See below....the question is unclear....pick the answer you think they want

Explanation:

Car goes from 30 m/s    to   24 m/s   in 10 seconds

acceleration = change in velocity/change in time

                     =   -6 m/s   / 10 s   =  - 3/5   m/s^2  

The wording of the question is not very good.....perhaps they meant to say that after the 5 seconds the velocity is now 24 m/s ?

a1 *10   +   a2 *5   = -6 m/s  

            ( 6 m/s is the change in velocity from 30 to 24 m/s)    

    a1/a2 = 1.5    so   a2 = a1/1.5   (substitute this in)

10 a1   + 5 ( a1/1.5)  =-6

15 a1 + 5 a1  = -9

a1 =  - .45 m/s^2

The velocity of the car at the end of the initial ten-second interval, given that the car further slows down for the next five seconds, is 25.5 m/s

From the first statement, we have:

v₁ = u₁ + a₁t₁

v₁ = 30 + (a₁ × 10)

v₁ = 30 + 10a₁ ......(1)

From the second statement, we have:

v₂ = u₂ + a₂t₂

24 = u₂ + (a₂ × 5)

24 = u₂ + 5a₂   .......(2)

But,

u₂ = v₁ = 30 + 10a₁

Thus, we have:

24 = 30 + 10a₁ + 5a₂

But,

a₁/a₂ = 1.5

a₁ = 1.5a₂ ..... (3)

Thus, we have:

24 = 30 + 10a₁ + 5a₂

24 = 30 + 10(1.5a₂) + 5a₂

24 = 30 + 15a₂ + 5a₂

24 = 30 + 20a₂

Collect like terms

24 - 30 = 20a₂

-6 = 20a₂

Divide both sides by 20

a₂ = -6 / 20

= -0.3 m/s²

Substituting the value of a₂ into equation 3, we have

a₁ = 1.5a₂

= 1.5 × -0.3

= -0.45 m/s²

Substitute the value of a₁ into equation 1 to obtain the velocity, v₁ at the end of the initial to 10 s

v₁ = 30 + 10a₁

= 30 + (10 × -0.45)

= 30 - 4.5

= 25.5 m/s

Thus, velocity of the car at the end of the initial ten-second interval is 25.5 m/s

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The magnitude of the angular momentum of air will be 4.128 x 10^(-3) kg·m^2/s

The rotating equivalent of linear momentum in physics is called angular momentum. Because it is a conserved quantity—the total angular momentum of a closed system stays constant—it is significant in physics. Both the direction and the amplitude of angular momentum are preserved.

Given the density of air to be 1.29 kg/m3 and a wind speed of 73.0 mi/h

We have to find the magnitude of the angular momentum

Let,

ρ = Density of air = 1.29 kg/m^3

v = Speed of wind = 73.0 mi/h = 0.032 km/s

M = angular momentum of air

Let the volume of air be 1 m^3

Mass = Volume x ρ = 1 x 1.29 = 1.29 kg

Momentum = M = mass x velocity

Momentum = 1.29 x 0.0032

Momentum = 4.128 x 10^(-3) kg·m^2/s

Hence the magnitude of the angular momentum of air will be 4.128 x 10^(-3) kg·m^2/s

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(a) The kind of frictional force acts on the ring upon contact with the surface is kinetic friction before rolling and afterwards it is static friction.

(b) If the ring begins to roll smoothly to the right with speed v = 10 m/s after 5 seconds of contact (the After state), the coefficient of friction corresponding to the frictional force is 0.612

(c) The increase in the thermal energy of the ring/surface system from right before the ring contacts the surface, to when the ring starts to roll smoothly is 7500 J.

(d) The horizontal distance, from the end of the ramp, at which the ring lands is 7.78 m

When an object is moving on a rough surface, it experiences opposition. This opposing force is called the friction force.

Given is a uniform solid ring of mass M = 10 kg and radius R = 2 m is initially rotating in mid-air with angular speed 20 rad/s [the Before state in figure (I) below]. It is then placed in contact with a surface (assume that the surface immediately feels the full weight of the ring).

(a) The kind of frictional force acts on the ring upon contact with the surface is kinetic friction before rolling and afterwards it is static friction.

(b) angular frequency = velocity / radius

ωf = 10/2 = 5 rad/s and ωi = 20 rad/s

Angular acceleration, α = ωf - ωi /t

Put the values, we get

α = -15/5 = -3 rad/s²

Coefficient of friction, μ = a/g = rα/g

Plug the values, we get

μ = 0.612

Thus, the coefficient of friction corresponding to the frictional force is 0.612.

(c) The energy lost = heat generated

 energy lost = 1/2 Iω² + 1/2 Mv²

 energy lost = 1/2 MR²ω² + 1/2 Mv²

Plug the values, we get

 energy lost = 7500 J

Thus, the  increase in the thermal energy is 7500 J.

(d) The horizontal distance, from the end of the ramp, at which the ring lands

s= (v- 2gH) sin2θ /g

s = (10 - 2x (-9.81)x5 ) sin (2x π/6) / 9.81

s = 7.78m

Thus, the horizontal distance is  7.78 m

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Answer:

Explanation:

Comment

It's high. It takes quite a bit of heat to get water to change its heat content. Most substances don't reach much more the 0.5 J/(gram*oC)

If the displacement is tripled then the spring force also is tripled.

The amount of force required to stretch or compress the spring is known as the spring force. Its unit is Newton(N).

Froom the hooks law;

F = kx

Where,

F is the applied force

x is the displacement in case 1

Case 2;

x'=3x

F'=kx'

F'=3KX

F'=3x'

Where,

x' is the displacement when the force is tripled

F' is the force in the case2

Hence, if the displacement is tripled then the spring force also be tripled.

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Answer:

V = 6.53 × 10^14

h = 6.626 × 10^-34

E = hV

E = (6.626 × 10^-34) (6.53 × 10^14)

E = 4.33 × 10^-19

[tex]q = -21 * 10^{-6} C[/tex]

The acceleration is constant and equal to the gravitational acceleration g which is 9.8 m/s at sea level on the Earth.

As we know that charge and its sign that remains in equilibrium is under gravity must be such that it will balance the gravitational force by electric force,

mg =qE

[tex]1.45 * 10^{-3}(9.80)=q(600)\\\\q = 21 *10^{6} C\\\\[/tex]

and its sign must be negative so that it will have upward electric force

so it is

[tex]q = -21 * 10^{-6} C[/tex]

The charge of a particle of mass is [tex]-21 * 10^{-6} C[/tex]

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The atomic bomb explosion, depends on the energy released, W, the elapsed time, t, and the air density which altogether lead to the vaporisation of the contents within the mushroom cloud created.

An atomic bomb is type of nuclear weapon that releases a vast amount of energy upon explosion in the form of fission reaction.

During an explosion of atomic bomb, a cloud of mushroom fire is formed which vaporises anything found within it. The extent of destruction depends on:

Therefore, the atomic bomb explosion, depends on the energy released, W, the elapsed time, t, and the air density which altogether lead to the vaporisation of the contents within the mushroom cloud created.

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At pH = 2 the ratio of the concentration of the hydrogen to the hydroxide ion is 10¹⁰. The hydrogen ion concentration at pH, 2 is 10⁻² and the hydroxide ion is 10⁻¹².

The parameter of measuring the concentration of the hydroxide and the hydrogen ion in the solution in order to determine the basic and the acidic nature is known as pH.

The concentration of H⁺ ions is calculated as:

pH = -log [H⁺]

2 = -log [H⁺]

[H⁺] = 10⁻²

The concentration of OH⁻ ions is calculated as:

pH + pOH = 14

pOH = 14 - 2

pOH = 12

Solving further:

pOH = -log [OH⁻]

12 =-log [OH⁻]

[OH⁻] = 10⁻¹²

At a pH = 2, the ratio of hydrogen ions to hydroxide ions:

10⁻² ÷ 10⁻¹² = 10¹⁰

Therefore, at pH = 2 the ratio is 10¹⁰.

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According to the pavlovian perspective, people who suffer from PTSD are classically conditioned.

This is referred to as post traumatic stress disorder which is a mental disorder from a terrifying incident.

The pavlovian perspective however view this disorder as being a type of classical conditioned one.

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Answer:

a (2m)

Explanation:

according to wavelenght = speed / frequency,

the wavelength is 2 m