A brick is resting on a smooth wooden board that is at a 30° angle. What is one way to overcome the static friction that is holding the brick in place?

Answer:

0.2812

Explanation:

Given that

mass of skater 1, m1 = 45 kg

mass of skater 2, m2 = 60 kg

speed of skater 1, v1 = 0.375 m/s

To attempt this question, we would be using the Law of conservation of momentum That says the momentum is constant, before and after the movement.

Thus, momentum p = mv

Law of conservation of momentum infers that,

m1v1 = m2v2

Now we proceed to substitute our values into the formula.

45 * 0.375 = 60 * v2

v2 = 16.875 / 60

v2 = 0.2812 m/s

Therefore the speed of the second skater has to be 0.2812 m/s

Answer;

a 2.00-kg body will vibrate at 0.707Hz

Answer:-7.9

Explanation:

Answer:

The same amount of current flows through the battery and light bulb

Explanation:

Because for a single loop, the current is the same at every point in the loop. Thus, the amount of current that flows through the lightbulb is the same as the amount that flows through the battery

Answer:

The same amount of current flows through the battery and light bulb

Explanation:

Answer:

20.96 m/s^2 (or 21)

Explanation:

Using the formula (final velocity - initial velocity)/time = acceleration, we can plug in values and manipulate the problem to give us the answer.

At first, we know a car is going 8 m/s, that is its initial velocity.

Then, we know the acceleration, which is 1.8 m/s/s

We also know the time, 7.2 second.

Plugging all of these values in shows us that we need to solve for final velocity. We can do so by manipulating the formula.

(final velocity - initial velocity) = time * acceleration

final velocity = time*acceleration + initial velocity

After plugging the found values in, we get 20.96 m/s/s, or 21 m/s

Answer:

The acceleration in the block is 2.1 m/s²

Explanation:

Given that,

Mass = 50 kg

Angle = 54°

Force = 40 N

Coefficient of friction = 0.33

We need to calculate the acceleration in the block

Using balance equation

[tex]F_{net}=F_{f}-F\cos\theta[/tex]

[tex]ma=\mu mg\sin\theta-F\cos\theta[/tex]

[tex]a=\dfrac{\mu mg\sin\theta-F\cos\theta}{m}[/tex]

Put the value into the formula

[tex]a=\dfrac{0.33\times50\times9.8\sin54-40\cos54}{50}[/tex]

[tex]a=2.1\ m/s^2[/tex]

Hence, The acceleration in the block is 2.1 m/s²

Answer:

In a tuning fork, two basic qualities of sound are considered, they are

1) The pitch of the waveform: This pitch depends on the frequency of the wave generated by hitting the tuning fork.

2) The loudness of the waveform: This loudness depends on the intensity of the wave generated by hitting the tuning fork.

Hitting the tuning fork harder will make it vibrate faster, increasing the number of vibrations per second. The number of vibration per second is proportional to the frequency, so hitting the tuning fork harder increase the frequency. From the explanation on the frequency above, we can say that by increasing the frequency the pitch of the tuning fork also increases.

Also, hitting the tuning fork harder also increases the intensity of the wave generated, since the fork now vibrates faster. This increases the loudness of the tuning fork.

Answer:

HELLO your question has some missing parts below are the missing parts

note: The specific heat ratio and gas constant for air are given as k=1.4 and R=0.287 kJ/kg-K respectively.

--Given Values--

Inlet Temperature: T1 (K) = 325

Inlet pressure: P1 (kPa) = 560

Inlet Velocity: V1 (m/s) = 97

Throat Area: A (cm^2) = 5.3

Pressure upstream of (before) shock: Px (kPa) = 207.2

Mach number at exit: M = 0.1

Answer: A)  match number at inlet  = 0.2683

              B)  stagnation temperature at inlet =  329.68 k

              C)  stagnation pressure = 588.73 kPa

              D) ) Throat temperature = 274.73 k

Explanation:

Determining states at several locations in the system

A) match number at inlet

= V1 / C1 = 97/ 261.427 = 0.2683

C1 = sound velocity at inlet = [tex]\sqrt{K*R*T}[/tex] = [tex]\sqrt{1.4 *0.287*10^3}[/tex]  = 361.427 m/s

v1 = inlet velocity = 97

B) stagnation temperature at inlet

     = T1 + [tex]\frac{V1 ^2}{2Cp}[/tex]  = 325 + [tex]\frac{97^2}{2 * 1.005*10^{-3} }[/tex]

stagnation temperature = 329.68 k

C) stagnation pressure

= [tex]p1 ( 1 + 0.2Ma^2 )^{3.5}[/tex]

Ma = match number at inlet = 0.2683

p1 = inlet pressure = 560

hence stagnation pressure = 588.73 kPa

D) Throat temperature

= [tex]\frac{Th}{T} = \frac{2}{k+1}[/tex]

Th = throat temperature

T = stagnation temp at inlet = 329.68 k

k = 1.4

make Th subject of the relation

Th = 329.68 * (2 / 2.4 ) = 274.73 k

Answer:

no, the truck is not overloaded

Explanation:

The computation is shown below;

Let us assume the mass of the loan in the second truck be M

So, the equation is as follows

{(Mass + M) × second truck × 1000 ÷ 3,600} = {(Mass + M + mass + first truck) × Pair moves away  × 1,000 ÷ 3,600}

{(5500 + M) × 64 × 1,000 ÷ 3,600 = {(5,500 + M + 5,500 + 3,900) × 44 × 1,000 ÷ 3,600}

(5500 + M) × 64 = (14,900 + M) × 44

352,000 + 64 M = 655,600 + 44 M

After solving this

M = 15,180 kg

Therefore the second truck is not overloaded

Answer:

It does not violate the first law because the total energy taken is what is used 100J = 25J + 75J

But violates 2nd lawbecause the engine has a higher energy after doing work than the initial for e.g A cold object in contact with a hot one never gets colder, transferring heat to the hot object and making it hotter confirming the second law

Answer:

At 3 meter distance, the per-second count is 222.22 and at a 10 meter distance, the per-second count is 20.

Explanation:

The number of particles (N)  counts are inversely proportional to the distance between the source and the detector.  

By using the below formula we can find the number of counts.

[tex]N2 = \frac{(D1)^2}{(D2)^2} \times N1 \\N1 = 2000 \\D 1 = 1 \ meter \\D2 = 3 \\[/tex]

The number of count per second, when the distance is 3 meters.

[tex]= \frac{1}{3^2} \times 2000 \\= 222.22[/tex]

Number of count per second when the distance is 10 meters.

[tex]= \frac{1}{10^2} \times 2000 \\= 20[/tex]

Answer:

B.

Explanation:

The water collects in the ocean; it is then evaporated by the sun. After evaporation the water turns into water vapor, it then condenses to form clouds.

The ocean water prior to the part of the water cycle should be option B.

The ocean water should be collected back in the ocean prior to the part of the water cycle.

Because this should be done when it is evaporated by the sun.  When the evaporation is done so the water should be transformed into water vapor.

Find out more information about the  Water here:brainly.com/question/4381433?referrer=searchResults

Answer:

F = 4212 N

Explanation:

Given that,

Mass of a car, m = 1300 kg

Speed of car on the road is 9 m/s

Radius of curve, r = 25 m

We need to find the magnitude of the unbalanced force that steers the car out of its natural straight-  line path. The force is called centripetal force. It can be given by :

[tex]F=\dfrac{mv^2}{r}\\\\F=\dfrac{1300\times 9^2}{25}\\\\F=4212\ N[/tex]

So, the force has a magnitude of 4212 N

Answer:

a) 5.5×10^17 Hz

b) visible light

Explanation:

Since the wavelength of the electromagnetic radiation must be about the size of the about itself, this implies that;

λ= 5.5 × 10^-10 m

Since;

c= λ f and c= 3×10^8 ms-1

f= c/λ

f= 3×10^8/5.5 × 10^-10

f= 5.5×10^17 Hz

The electromagnetic wave is visible light

Answer:

2/5 m/s

Explanation:

There are two vectors  v and w . Let θ be angle b/w the two vector.

[tex]cos\theta =\frac{\overleftarrow{v}\cdot \overleftarrow{w}}{\left | v \right |\left | w \right |}\\=\frac{6-4}{\sqrt(2^2+1^2)\sqrt(3^2+4^2)} =\frac{2}{5\sqrt(5)}[/tex]

velocity of the ball in direction of the the wind

[tex]\left | vcos\theta \right |\\\left | v \right |cos\theta\\\sqrt(2^2+1^2)\frac{2}{5\sqrt(5)} = \frac{2}{5}[/tex]

The size of the velocity of the ball in the direction of the wind is 2/5 ms.

Since there are two vectors v and w

Also, here we assume θ be angle b/w the two vector.

So

Cos θ = 6-4 / √(2^2 + 1^2) √(3^2 + 4^2)

= 2/5√5

Now the velocity of the ball should be

= √(2^2 + 1^2) 2 ÷ 5√(5)

= 2 /5

hence, The size of the velocity of the ball in the direction of the wind is 2/5 ms.

Learn more about velocity here: https://brainly.com/question/1303810

Explanation:

Mary had 21 plants when she went on vacation.

When she got back, she only had 14 left alive.

We need to find the percent decrease in the number of plants.

Decrease in plants = 21 - 14 = 7

Percent decrease is given by :

[tex]\%=\dfrac{7}{21}\times 100\\\\\%=33.33\%[/tex]

So, there is 33% pf decrease in the number of plants.

Answer:

Option 2 (observing, measuring, testing, and explaining their ideas) is the correct choice.

Explanation:

The other three choices have no relation to a particular task. So the option given here is just the right one.

Answer:

 P = 2923.89 W  

Explanation:

Power is

     P = F v

for which we must calculate the force, let's use Newton's second law, let's set a coordinate system with a flat parallel axis and the other axis (y) perpendicular to the plane

X Axis  

         F - Wₓ = 0

         F = Wₓ

Y Axis

         N -  [tex]W_{y}[/tex] = 0

let's use trigonometry for the components of the weight

         sin 6 = Wₓ / W

         cos 6 = W_{y} / W

         Wₓ = W sin 6

         W_{y} = W cos 6

          F = mg cos 6

          F = 75 9.8 cos 6

          F = 730.97 N

let's calculate the power

        P = F v

        P = 730.97 4.0

        P = 2923.89 W

Answer

The capacitor should be connected in parallel as parallel connection gives the arithmetic sum of capacitance which will give a corresponding sum of energy while capacitors in series gives the sum of the reciprocal if the individual capacitance

Answer:

The force of the radiation on the surface is 3.33 x 10⁻¹⁰ N

Explanation:

Given;

intensity of light, I = 1kw/m² = 1000 W/m²

area of the surface, A = 1 cm² = 1 x 10⁻⁴ m²

Since the light is completely absorbed, the force of the radiation is given by;

F = P/c

where;

c is the speed of light = 3 x 10⁸ m/s

But P = IA

F = IA /c

F = (1000 X 1 X 10⁻⁴) / 3 x 10⁸

F = 3.33 x 10⁻¹⁰ N

Therefore, the force of the radiation on the surface is 3.33 x 10⁻¹⁰ N

The force of radiation will be "3.33 × 10⁻¹⁰ N"

According to the question,

Intensity of force, I = 1 kW/m² or,

                               = 1000 W/m²

Area of surface, A = 1 cm² or,

                              = 1 × 10⁻⁴ m²  

Speed of light, c = 3 × 10³ m/s

As we know the relation,

→ F = [tex]\frac{P}{c}[/tex]

or,

  P = IA

or,

  F = [tex]\frac{IA}{c}[/tex]

By substituting the values, we get

     = [tex]\frac{1000\times 1\times 10^{-4}}{3\times 10^3}[/tex]

     = 3.33 × 10⁻¹⁰ N

Thus the response above is correct.

Find out more information about intensity here:

https://brainly.com/question/1444040

Answer:

77.96dB

Explanation:

Recall that decibels are a unit of measuring intensity of sound, and depend on the logarithm of the intensity

the intensity, measured in decibels is given by:

I(db)=10log(I/I0)

I is the intensity in MKS units; I0 is the threshold intensity for human hearing (10^-12 W/m^2)

Thus, if the two sounds together have a dB of 81, we know:

81=10log(I/I0)

using the data above, we can find the intensity of the two sounds to be

0.000125 W/m^2

therefore, one firecracker has an intensity half of that, or 0.0000625W/m^2

now use this value to find the dB of one firecracker:

I(dB0=10log(0.0000625/10^-12)=77.96dB

Answer:

Explanation:

Concave mirrors is otherwise known as converging mirrors: These are mirrors that are caved inwards (reflecting surface is on the outside curved part). It is called a converging mirror due to the fact that light converges to a point when it strikes and reflects from the surface of the mirror. This type of mirror is used to focus light; parallel rays that are directed towards it will be concentrated to a point.

For a concave mirror to reflect light with properties that are the same as a spotlight (directed light rays parallel to each other), one has to consider its property to gather light to a point after reflecting. Meaning that, we can achieve the spotlight by locatng the point where the rays will be parallel, this point is called the focal point.

Therefore, the light bulb should be placed at the focal point of the mirror.

Answer:

The difference in angle of refraction between the red and blue light is 0.2°

Explanation:

Here is the complete question

Sunlight strikes a piece of crown glass at an angle of incidence of 37.4°. Calculate the difference in the angle of refraction between a red (660 nm) and a blue (470 nm) ray within the glass. The index of refraction is n=1.520 for red and n=1.531 for blue light.

Solution

From Snell's law refractive index n = sini/sinr where i = angle of incidence and r = angle of refraction.

Now for the red light n₁ = 1.520, i = 37.4° and r₁ = angle of refraction of red light

So, n₁ = sini/sinr₁

n₁sinr₁ = sini

sinr₁ = sini/n₁

r₁ = sin⁻¹(sini/n₁) = sin⁻¹(sin37.4°/1.52) = sin⁻¹(0.6074/1.52) = sin⁻¹(0.3996) = 23.55°

Now for the blue light n₂ = 1.531, i = 37.4° and r₂ = angle of refraction of blue light

So, n₂ = sini/sinr₂

n₂sinr₂ = sini

sinr₂ = sini/n₂

r₂ = sin⁻¹(sini/n₂) = sin⁻¹(sin37.4°/1.531) = sin⁻¹(0.6074/1.531) = sin⁻¹(0.3967) = 23.37°

So the difference in angle of refraction between the red and blue light is r₁ - r₂ = 23.55° - 23.37° = 0.18° ≅ 0.2°

Answer:

  K = 1.6 10⁻¹⁵ J

Explanation:

In an electron microscope, electrons are used to form images, these electrons are accelerated in electric fields so that they have a kinetic energy that allows obtaining a good amplification with the microscope.

electrical potential energy is converted to kinetic energy

                U = K

                e V = ½ m v²

                v = √2eV /m

the wavelength of these electrons we obtain from the de Broglie equation

                λ = h / p

p = mv

               λ = h / mv

               λ = h / mra 2eV / m

               λ = h / ra 2eVm

where we can see that as the potential energy increases, it electrifies the shorter the wavelength of the electrons and consequently the greater the magnification of the microscope

in general these microscopes use from 10000X onwards therefore for this saponification

                K = e V

               K = 1.6 10⁻¹⁹ 10000

               K = 1.6 10⁻¹⁵ J

Answer:

Explanation:

dipole moment = qs = q x s

= charge x charge separation

charge = q

separation between charge = s

half separation l = s / 2

dipole has two charges + q and - q separated by distance s .

Potential at distance x along x axis due to + q

[tex]v_1=\frac{1}{4\pi \epsilon } \times\frac{q}{x-l}[/tex]

Potential at distance x along x axis due to - q

[tex]v_2=\frac{1}{4\pi \epsilon } \times\frac{-q}{x+l}[/tex]

Total potential

v = v₁ + v₂

[tex]v=\frac{1}{4\pi \epsilon } \times( \frac{q}{x-l}-\frac{q}{x+l})[/tex]

[tex]v=\frac{1}{4\pi \epsilon } \times\frac{2ql}{x^2-l^2}[/tex]

[tex]v=\frac{1}{4\pi \epsilon } \times\frac{qs}{x^2-(\frac{s}{2}) ^2}[/tex]

Potential at distance y along y axis due to + q

[tex]v_1=\frac{1}{4\pi \epsilon } \times\frac{qs}{(y^2+\frac{s^2}{4})^\frac{1}{2} }[/tex]

Potential at distance y along y axis due to - q

[tex]v_1=\frac{1}{4\pi \epsilon } \times\frac{-qs}{(y^2+\frac{s^2}{4})^\frac{1}{2} }[/tex]

Total potential

v = v₁ + v₂

[tex]v= 0[/tex]

Answer:

300 W

Explanation:

power of each bulb P = 75 W

voltage in the circuit = 120 V

we know that electrical power P = IV    ....1

and V = IR

we can also say that I = V/R

substituting for I in  equation 1, we have

P = [tex]V^{2}/R[/tex]    ....2

The total total power in the circuit = 75 x 2 = 150 W

from equation 2, we have

150 = [tex]120^{2} /R[/tex]

R = [tex]120^{2}/150[/tex] = 96 Ω    this is the resistance of the whole circuit.

This resistance is due to the two light bulbs, for each light bulb since they are arranged in series

R = 96/2 = 48 Ω

From P =  [tex]V^{2}/R[/tex]  

for each light bulb, power is

P = [tex]120^{2} /48[/tex] = 300 W

Answer:   2

Explanation:

1/2=1/1 +1/x

x=-2

magnification= 2/1

magnification=2

Answer:

v = - 14.08 m / s

Explanation:

The definition of center of mass is

        [tex]x_{cm}[/tex] = 1 /M  ∑sun [tex]x_{i} m_{i}[/tex]

where M is the total mass of the system and [tex]x_{i}[/tex] and [tex]m_{i}[/tex] are the position and mass of each component.

The velocity of the center of mass can be found by deriving this expression with respect to time

         [tex]v_{cm}[/tex] = 1 / M ∑ m_{i} [tex]v_{i}[/tex] vi

let's find the total mass

          M = m₁ + m₂ + m₃ + m₄

          M = 1.45 + 2.81 +3.89 + 5.03

          m = 13.18 kg

let us substitute in the velocity of the center of mass [tex]v_{cm}[/tex] = 0

          0 = 13.18 (m₁ v₁ + m₂ v₂ + m₃v₃ + m₄v₄)

          v₄ = - (m₁ v₁ + m₂ v₂ + m₃v₃) / m₄

let's substitute the given values

v₄ = -[1.45 (6.09 +0.299 t) +2.81 (7.83 + 0.357t) +3.89 (8.09 + 0.405 t)] / 5.03

They ask us for the calculations for a time t = 2.83 s

          v₄ = - [8.8305 + 1.227 + 22.00 + 2.839 + 31.47 +4.4585] / 5.03

          v = - 14.08 m / s

The velocity of the particle 4 at time, t = 2.83 s, is -14.1 m/s.

The given parameters;

[tex]m_1 = 1.45 \ kg, \ \ v_1(t) = (6.09 \ m/s) + (0.299 \ m/s^2)\times t\\\\m_2 = 2.81 \ kg, \ \ v_2(t) = (7.83 \ m/s) + (0.357 \ m/s^2)\times t \\\\m_3 = 3.89 \ kg, \ \ v_3(t) = (8.09 \ m/s) + (0.405 \ m/s^2)\times t\\\\m_4 = 5.03 \ kg[/tex]

The velocity of the center mass of the particles is calculated as;

[tex]M_{cm}V_{cm} = m_1v_1 + m_2 v_2 + m_3v_3 + m_4v_4\\\\V_{cm} = \frac{m_1v_1 + m_2 v_2 + m_3v_3 + m_4v_4}{M_{cm}} \\\\0 = \frac{m_1v_1 + m_2 v_2 + m_3v_3 + m_4v_4}{M_{cm}}\\\\m_1v_1 + m_2 v_2 + m_3v_3 + m_4v_4 = 0\\\\m_4v_4 = -(m_1v_1 + m_2 v_2 + m_3v_3)\\\\v_4 = \frac{-(m_1v_1 + m_2 v_2 + m_3v_3)}{m_4}[/tex]

The velocity of particle 1 at time, t = 2.83 s;

[tex]v_1 = 6.09 \ + \ 0.299\times 2.83\\\\v_1 = 6.94 \ m/s[/tex]

The velocity of particle 2 at time, t = 2.83 s;

[tex]v_2 = 7.83\ + \ 0.357\times 2.83\\\\v_2 = 8.84 \ m/s[/tex]

The velocity of particle 3 at time, t = 2.83 s;

[tex]v_3 = 8.09\ + \ 0.405 \times 2.83\\\\v_3 = 9.24 \ m/s[/tex]

The velocity of the particle 4 at time, t = 2.83 s;

[tex]v_4 = \frac{-(m_1v_1 + m_2v_2 + m_3v_3)}{m_4} \\\\v_4 = \frac{-(1.45\times 6.94\ + \ 2.81\times 8.84\ + \ 3.89 \times 9.24)}{5.03} \\\\v_4 = -14 .1 \ m/s[/tex]

Thus, the velocity of the particle 4 at time, t = 2.83 s, is -14.1 m/s.

Learn more here:https://brainly.com/question/22698801

Answer:

5.2791264*10¹³

Explanation:

Convert the 9 years to seconds and then multiple it by 186000

The star is 4.62 x 10¹⁶ miles away from Earth.

The speed of light is 186,000 miles per second. It takes light from a particular star approximately 9 years to reach Earth. There are 365 days in 1 year, so it takes 9 x 365 = 3285 days for light from the star to reach Earth.

The distance between the star and Earth is 3285 x 186,000 = 608,810,000 miles. In scientific notation, this is 4.62 x 10¹⁶ miles.

Here is the calculation:

distance = speed * time

distance = 186,000 miles/second * 3285 days * 24 hours/day * 60 minutes/hour * 60 seconds/minute

distance = 608,810,000 miles

distance = 4.62 x 10¹⁶ miles

To know more about the Light, here

https://brainly.com/question/20296439

#SPJ2

Answer:

4.245s

Explanation:

Given that,

Hypothetical value of speed of light in a vacuum is 18 m/s

Speed of the car, 14 m/s

Time given is 6.76 s, and we're asked to find the observed time, T

The relationship between the two times can be given as

T = t / √[1 - (v²/c²)]

The missing variable were looking for is t, and we can find it if we rearrange the formula and make t the subject

t = T / √[1 - (v²/c²)]

And now, we substitute the values and insert into the equation

t = 6.76 * √[1 - (14²/18²)]

t = 6.76 * √[1 - (196/324)]

t = 6.76 * √(1 - 0.605)

t = 6.76 * √0.395

t = 6.76 * 0.628

t = 4.245 s

Therefore, the time the driver measures for the trip is 4.245s

Answer:

the large amplitude wave keeps moving to the right

the small amplitude wave continues to move to the left.

When checking the answers, the correct ones are 1, 2

Explanation:

The waves fulfill the principle of superposition, which states that the value of the function at a point is the algebraic sum of the waves at a given instant.

The two waves in this exercise travel in the opposite direction, so when they are close, the resulting wave is the sum of the two waves, having a complicated shape. But when the waves follow their movement, they give in the same way as the initial a,

the large amplitude wave keeps moving to the right

the small amplitude wave continues to move to the left.

When checking the answers, the correct ones are 1, 2