a block of mas \( m \) = 4.8 kg slides head on into a spring of spring constant \( k \) = 430 N/m. When the block stops, it has compressed the spring by 5.8 cm. The coefficient of kinetic friction between block and floor is 0.28. \( (g =9.8m/s^2) \)

Answer:

Because there was fewer fossils

Explanation:

Answer:

Actually the answer is "The stratum contains iridium.".

Explanation:

Answer:

the charge of the particle is -2.144 x 10⁻⁵ C.

Explanation:

The force acting on the particle is calculated as;

F = EQ = mg

[tex]Q = \frac{mg}{E}[/tex]

where;

Q is magnitude of the charge of the particle

[tex]Q = \frac{(1.4\times 10^{-3})(9.8)}{640} \\\\Q = 2.144 \ \times \ 10^{-5} \ C[/tex]

since the magnetic field is acting downward, the force must be acting upward in opposite direction.

Thus, the charge of the particle will be -2.144 x 10⁻⁵ C.

Answer:

a. 10.5 s b. 6.6 s

Explanation:

a. The driver's perception/reaction time before drinking.

To find the driver's perception time before drinking, we first find his deceleration from

v² = u² + 2as where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 0 m/s (since he stops), a = deceleration of driver and s = distance moved by driver = 385 ft = 385 × 0.3048 m = 117.35 m

So, a = v² - u²/2s

substituting the values of the variables into the equation, we have

a = v² - u²/2s

a = (0 m/s)² - (22.35 m/s)²/2(117.35 m)

a =  - 499.52 m²/s²/234.7 m

a = -2.13 m/s²

Using a = (v - u)/t where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 0 m/s (since he stops), a = deceleration of driver = -2.13 m/s² and t = reaction time

So, t = (v - u)/a

Substituting the values of the variables into the equation, we have

t = (0 m/s - 22.35 m/s)/-2.13 m/s²

t = - 22.35 m/s/-2.13 m/s²

t = 10.5 s

b. The driver's perception/reaction time after drinking.

To find the driver's perception time after drinking, we first find his deceleration from

v² = u² + 2as where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 30 mi/h = 30 × 1609 m/3600 s = 13.41 m/s, a = deceleration of driver and s = distance moved by driver = 385 ft = 385 × 0.3048 m = 117.35 m

So, a = v² - u²/2s

substituting the values of the variables into the equation, we have

a = v² - u²/2s

a = (13.41 m/s)² - (22.35 m/s)²/2(117.35 m)

a = 179.83 m²/s² - 499.52 m²/s²/234.7 m

a = -319.69 m²/s² ÷ 234.7 m

a = -1.36 m/s²

Using a = (v - u)/t where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 30 mi/h = 30 × 1609 m/3600 s = 13.41 m/s, a = deceleration of driver = -1.36 m/s² and t = reaction time

So, t = (v - u)/a

Substituting the values of the variables into the equation, we have

t = (13.41 m/s - 22.35 m/s)/-1.36 m/s²

t = - 8.94 m/s/-1.36 m/s²

t = 6.6 s

Answer: The correct statement is that the Last Quarter Moon (rises near midnight and sets near midday).

Explanation:

Phases of the moon also called the LUNAR PHASE can be defined as the different shades of illumination on the moon as seen from the earth. The moon is the natural satellite of the earth that illuminates upon reflection of light from the sun. This means it doesn't have power to shine on its own. When carefully observed, there are times it gets dark and beings to glow brighter over a period of time. This occurs because as the moon completes its four weeks lunar cycle round the earth, how much of its face we see illuminated by sunlight depends on the angle the Sun makes with the Moon.

There are 8 main types of the moon phases these includes:

--> New moon: This is when the moon is not visible to the earth because it's between the earth and the sun. It rises at sunrise and sets at sunset.

--> The waxing crescent: At this phase the moon gets brighter and illuminated from the sun that a crescent shape is seen.

--> First quarter: this occurs one week after the new moon. The moon rises at noon and sets at midnight.

--> The waxing gibbous: This occurs after the first quarter phase where more than half of the lit part of the moon is seen.

--> Full moon: This occurs when the moon and the sun are opposite each other. That is way it is said to rise at sunset and sets at sun rise.

--> The waning gibbous: this occurs when more than half of the lit part of the moon gradually becomes darker

--> Third quarter ( Last Quarter): The moon rises at midnight and sets at noon. This occurs a week after the full moon.

--> The waning crescent: This occurs after the last quarter phase where a very thin fading crescent shaped moon is seen, just before the Moon is invisible again at the start of the cycle, the new moon.

Answer:

6.69 m/s

4.483 m

1.42s

Explanation:

Given that:

Initial Velocity, u = 0

Final velocity, v =?

Acceleration, a = 35m/s²

1.) using the relation :

v² = u² + 2as

v² = 0 + 2(35) * 64*10^-2m

v² = 70 * 0.64

v = sqrt(44.8)

v = 6.693

v = 6.69 m/s

B.) height from the ground, h0 = 2.2

How high ball went , h:

Using :

v² = u² + 2as

Upward motion, g = - ve

0 = 6.69² + 2(-9.8)*(h - 2.2)

0= 6.69² - 19.6(h - 2.2)

44.7561 + 43.12 - 19.6h = 0

19.6h = 44.7561 - 43.12

h = 87.8761 / 19.6

h = 4.483 m

C.)

vt - 0.5gt² = h - h0

6.69t - 0.5(9.8)t²

6.69t - 4.9t² = 1.83 - 2.2

-4.9t² + 6.69t + 0.37 = 0

Using the quadratic equation solver :

Taking the positive root:

1.4185 = 1.42s

Answer:

The torsion constant for the wire is [tex]2.856\times 10^{-4}\,N\cdot m[/tex].

Explanation:

The angular frequency of the torsional pendulum ([tex]\omega[/tex]), measured in radians per second, is defined by the following expression:

[tex]\omega = \sqrt{\frac{\kappa}{I} }[/tex] (1)

Where:

[tex]\kappa[/tex] - Torsional constant, measured in newton-meters.

[tex]I[/tex] - Moment of inertia, measured in kilogram-square meters.

The angular frequency and the moment of inertia are represented by the following formulas:

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

[tex]I = \frac{m\cdot L^{2}}{12}[/tex] (3)

Where:

[tex]T[/tex] - Period, measured in seconds.

[tex]m[/tex] - Mass of the stick, measured in kilograms.

[tex]L[/tex] - Length of the stick, measured in meters.

By (2) and (3), (1) is now expanded:

[tex]\frac{2\pi}{T} = \sqrt{\frac{12\cdot \kappa}{m\cdot L^{2}} }[/tex]

[tex]\frac{2\pi}{T} = \frac{2}{L}\cdot \sqrt{\frac{3\cdot \kappa}{m} }[/tex]

[tex]\frac{\pi\cdot L}{T} = \sqrt{\frac{3\cdot \kappa}{m} }[/tex]

[tex]\frac{\pi^{2}\cdot L^{2}}{T^{2}} = \frac{3\cdot \kappa}{m}[/tex]

[tex]\kappa = \frac{\pi^{2}\cdot m\cdot L^{2}}{3\cdot T^{2}}[/tex]

If we know that [tex]m = 5\,kg[/tex], [tex]L = 1\,m[/tex] and [tex]T = 240\,s[/tex], then the torsion constant for the wire is:

[tex]\kappa = \frac{\pi^{2}\cdot (5\,kg)\cdot (1\,m)^{2}}{3\cdot (240\,s)^{2}}[/tex]

[tex]\kappa = 2.856\times 10^{-4}\,N\cdot m[/tex]

The torsion constant for the wire is [tex]2.856\times 10^{-4}\,N\cdot m[/tex].

Non-Malleable and Ductile: Non-metals are very brittle, and cannot be rolled into wires or pounded into sheets. Conduction: They are poor conductors of heat and electricity. Luster: These have no metallic luster and do not reflect light.

Group 15, the nitrogen family, contains two nonmetals: nitrogen and phosphorus. These non-metals usually gain or share three electrons when reacting with atoms of other elements. Group 16, the oxygen family, contains three nonmetals: oxygen, sulfur, and selenium.

Elements: Nitrogen; Oxygen; Phosphorus; Selenium...

Without friction, NO.

The speed at D depends only on the difference in height between A and D. Whatever happens between them doesn't matter.

The speed of the coaster car at point D will be affected if  the height of point C is changed.

Potencial Energy:

It is the enrgy in a body due to the position of differnt part of the object or system.

As we increase the the hight of the car the potetial enrgy increase, the gravitational acceleration on car will be more due to the high of the point C.

Therefore, the speed of the coaster car at point D will be affected if  the height of point C is changed.

To know more about  speed of the coaster car,

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The  charge on the inner surface of the shell is -Q

The  charge on the outer surface of the shell is Q

After this contact, the charge on the tack is  0

The charge on the inner surface of the shell now is  0

The charge on the outer surface of the shell now is Q

The charge on a shell depends on the situation and the conditions of the shell. If the shell is an electrically neutral object, such as a metallic spherical shell, it has no net charge, meaning that the total positive charge is equal to the total negative charge. However, if the shell has an excess or deficit of electrons, it will have a net charge, either positive or negative, depending on whether it has an excess of electrons or a deficit of electrons.

Read more on charges here:https://brainly.com/question/18102056

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Answer:

This idea helps students explain why more rain forms over West Ferris than East Ferris. ... Therefore, when students explain that water vapor condenses higher in the atmosphere, they are actually explaining that water vapor condenses high in the troposphere, which is relatively low in the atmosphere.

Explanation:

Plz mark me brainliest thank u> have a good day

Answer:

5766.7 K

Explanation:

We are given that

Radius of Sun , R=[tex]6.96\times 10^{8} m[/tex]

Distance between the Sun and the Earth, D=[tex]1.50\times 10^{11}m[/tex]

Irradiance arriving on the Earth is the value for AMO=[tex]1350W/m^2[/tex]

We have to find the temperature at the surface of the Sun.

We know that

Temperature ,T=[tex](\frac{K_{sc}D^2}{\sigma R^2})^{\frac{1}{4}}[/tex]

Where [tex]K_{sc}=1350 W/m^2[/tex]

[tex]\sigma=5.67\times 10^{-8}watt/m^2k^4[/tex]

Using the formula

[tex]T=(\frac{1350\times (1.5\times 10^{11})^2}{5.67\times 10^{-8}\times (6.96\times 10^{8})^2})^{\frac{1}{4}}[/tex]

T=5766.7 K

Hence, the temperature at the surface of the sun=5766.7 K

Answer:

Explanation:

Fluid A :

Δ V = Change in volume = (3000 - 2804) x 10⁻⁶ m³ = 196 x 10⁻⁶ m³

volume strain = Δ V / V  = 196 x 10⁻⁶ / 3000 x 10⁻⁶

= .06533

Δ P = increase in pressure = 6 x 10⁷ Pa

E = Δ P / volume strain = 6 x 10⁷ / .06533 = 91.84 x 10⁷ Pa = .92 GPa .

It is Acetone .

Fluid B :

Δ V = Change in volume = (3000 - 2862) x 10⁻⁶ m³ = 138 x 10⁻⁶ m³

volume strain = Δ V / V  = 138 x 10⁻⁶ / 3000 x 10⁻⁶

= .046

Δ P = increase in pressure = 6 x 10⁷ Pa

E = Δ P / volume strain = 6 x 10⁷ / .046 = 130.43  x 10⁷ Pa = 1.3  GPa .

It is Gasoline  .

Fluid C :

Δ V = Change in volume = (3000 - 2916) x 10⁻⁶ m³ = 84 x 10⁻⁶ m³

volume strain = Δ V / V  = 84 x 10⁻⁶ / 3000 x 10⁻⁶

= .028

Δ P = increase in pressure = 6 x 10⁷ Pa

E = Δ P / volume strain = 6 x 10⁷ / .028 = 214.28 x 10⁷ Pa = 2.14  GPa .

It is Water   .

Answer:

y = -2.69 m

the negative sign indicates that it is descending and the distance is less than the difference in height between the two buildings, therefore the person would be saved in the jump.

Explanation:

his problem must be solved with the missile launch equations.

Let's start by looking for the jumper's initial velocity

          R = v₀² sin 2θ / g

for the long jump the angle used is tea = 45º, in the exercise they indicate that the best record is R = 7.9m

          v₀² = R g / sin 2te

          v₀ = [tex]\sqrt{ \frac{7.9 \ 9.8}{1 }[/tex]

          v₀ = 8.80 m / s

Now suppose you jump with this speed to get to the other building, let's use trigonometry for the components of the speed

          sin 45 = [tex]v_{oy}[/tex] /v₀

          cos 45 = v₀ₓ / v₀

         v_{oy} = v₀ sin 45

          v₀ₓ = v₀ cos 45

          v_{oy} = 8.8 sin 45 = 6.22 m / s

          v₀ₓ = 8.8 cos 45 = 6.22 m / s

now let's calculate the sato with these speeds

           x = [tex]v_{ox}[/tex] t

the minimum jump is x = 10 m

           t = x / v₀ₓ

           t = 10 / 6.22

           t = 1.61 s

let's find the vertical distance for this time

           y = v_{oy} t - ½ g t²

where zero is placed on the jump building

           y = 6.22 1.61 - ½ 9.8 1.61²

            y = -2.69 m

Let's analyze this result, the negative sign indicates that it is descending and the distance is less than the difference in height between the two buildings, therefore the person would be saved in the jump.

Answer:

2.16 inch

Explanation:

area under water = 66 km²

= 66 x ( 3280.84 x 12 )² inch²

= 1.023 x 10¹¹ sq inch

volume of rain = 9.57 x 10⁸  gallon = 9.57 x 10⁸ x 231 inch³

= 2.21 x 10¹¹ inch³

If depth of rainfall be t

volume of rain = surface area x depth

= 1.023 x 10¹¹ x t

So ,

1.023 x 10¹¹ x t  = 2.21 x 10¹¹

t = 2.16 inch

Answer:

C) 128 kg*m/s

Explanation:

When you double something you multiply it by 2 most of the time. 64*2=128 or you can add it 64+64=128. Hope this helps.

Answer:

 k = 1400.4 N / m

Explanation:

When the springs are oscillating a simple harmonic motion is created where the angular velocity is

          w² = k / m

          w = [tex]\sqrt{ \frac{k}{m} }[/tex]

where angular velocity, frequency and period are related

          w = 2π f = 2π / T

we substitute

          2π / T = \sqrt{ \frac{k}{m} }

         T² = 4π² [tex]\frac{m}{k}[/tex]

          k =  π²  [tex]\frac{m}{T^{2} }[/tex]

in this case the period is T = 1.14s, the combined mass of the children is

m = 92.2 kg and the constant of the two springs is

          k = 4π² 92.2 / 1.14²

          k = 2800.8 N / m

to find the constant of each spring let's use the equilibrium condition

          F₁ + F₂ - W = 0

           k x + k x = W

indicate that the compression of the two springs is the same, so we could replace these subtraction by another with an equivalent cosecant

           (k + k) x = W

            2k x = W

            k_eq = 2k

            k = k_eq / 2

            k = 2800.8 / 2

            k = 1400.4 N / m

Answer:

force F = 1.66 × [tex]10^{-13}[/tex] N

Explanation:

given data

proton and an electron = 865 nm

solution

we get here force that is express as

force F = k q1 q2 ÷ r²      ......................1

put here value and we get

force F = 9 × [tex]10^{9}[/tex] × [tex]\frac{1.6\times (10^{-19})^{2}}{865 \times (10^{-9})^{2}}[/tex]    

force F = 1.66 × [tex]10^{-13}[/tex] N

Answer: he can only make it over 5 buses

Explanation:

Given the data in the question;

we know that range is expressed as;

R = (V₀²sin2∅₀)/g

V₀ is the initial velocity( 25.4 m/s), ∅₀ is the angle of projection( 64.8°), g is acceleration due to gravity( 9.8 m/s²),

so we substitute

R = ((25.4)²sin2(64.8))/9.8

R = 50.7 m

now, them number of buses will be;

n = R / bus length

given that bus length is 10.0 m

we substitute

n = 50.7 m / 10.0

n = 5.07 ≈ 5

Therefore, he can only make it over 5 buses

Explanation:

False, it is oppisite the shorter the wavelength the more energy it carries.

Answer:

Explanation:

I would use a hand truck dollies

Answer:

9.8m/s²

Explanation:

The average acceleration due to gravity on Earth for a 2000kg boulder is 9.8m/s².

Every object on earth is accelerated towards the center by a rate of change of velocity with time value of 9.8m/s².

The acceleration due to gravity on earth is a constant value from places to places.

For other planetary bodies, the value varies and it differs.

 But on earth every object is accelerated at 9.8m/s².

Answer:

This is my answer

Explanation:

First convert 150 kPa to Pa:

150 × 1,000 = 150,000.

Next substitute the values into the equation:

force normal to a surface area = pressure × area of that surface.

force = 150,000 × 180.

force = 27,000,000 N.

1.First convert 150kPato Pa:

2.150 x 1,000 + 150,000

3.next substitute the values into the equations:

4.force normal to a surface area =pressure x area of that surface.

5.force=150,000 x 180.

6.force = 27,000,000N.

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Answer:

To have the same kinetic energy the speed of the marble must be 9 times the speed of rock.

Explanation:

The general formula of kinetic energy is given as follows:

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

where,

K.E = Kinetic Energy

m = mass of the object

v = speed of the object

So, for the marble and rock to have same kinetic energy, we can write:

[tex]K.E_{marble} = K.E_{rock}\\\\\frac{1}{2}m_{marble}v_{marble}^{2} = \frac{1}{2}m_{rock}v_{rock}^{2}\\\\(0.03\ kg)v_{marble}^{2} = (2.43\ kg)v_{rock}^{2}\\\\taking\ square\ root\ on\ both\ sides:\\v_{marble} = \sqrt{\frac{2.43\ kg}{0.03\ kg}}v_{rock}\\\\v_{marble} = 9\ v_{rock}[/tex]

Hence, to have the same kinetic energy the speed of the marble must be 9 times the speed of rock.

Answer:

the magnitude of gravitational force is 6 x 10⁻⁸ N.

Explanation:

Given;

mass of the two people, m₁ and m₂ = 90 kg

distance between them, r = 3.0 m

The magnitude of gravitational force exerted by one person on another is calculated as;

[tex]F = \frac{Gm_1m_2}{r^2} \\\\[/tex]

where;

G is gravitational constant = 6.67 x 10⁻¹¹ Nm²/kg²

[tex]F = \frac{Gm_1m_2}{r^2} \\\\F = \frac{6.67\times 10^{-11} \times \ 90 \ \times \ 90}{3^2} \\\\F = 6\times 10^{-8} \ N[/tex]

Therefore, the magnitude of gravitational force is 6 x 10⁻⁸ N.

Answer:

See explanation below

Explanation:

Psychoactive drugs are drugs that affect the central nervous system. They alter cognitive function by changing mood and consciousness.

Examples;

Depressants: Inhibit the function of the central nervous system and neural activity.

Stimulants: Excite neural activity and temporarily elevate awareness.

Amphetamines: Increase dopamine activity and produce schizophrenic-like symptoms.

Hallucinogens: Distort perceptions and effects on thinking.

A drug is any substance that alters how the body functions.

A drug is any substance that alters how the body functions. There are different types of drugs that affect different parts of the body.

We shall now explain the following classifications of drugs;

Learn more about drugs: https://brainly.com/question/6022349

Answer:

Chemical energy

Explanation:

The energy in the torch is stored as chemical energy before the torch gets switch on.

The chemical energy energy in the battery of cell will power the cell and allows it to produce light.