What aspect of solar radiation causes Cleveland to have more solar radiation than Miami in June and July

Answer:

Explanation:

Kinetic energy ( KE ) = 1/2 m v²

= m²v² / 2 m = p² / 2m where p is momentum

KE = p² / 2m

p² = 2m KE

KE is constant

p is proportional to mass

So car having higher mass will have higher momentum .

p₁ = √ ( 2 m x KE )

p₂ = √ ( 6 m x KE )

p₂ - p₁ =  √ ( 6 m x KE ) - √ ( 2 m x KE )

= √KE m ( √6 - √2 )

Kinetic energy ( K.E )

[tex]= \frac{1}{2} m v^2\\\\= \frac{m^2 v^2}{2 m} \\\\= \frac{p^2}{2m}[/tex]

where p is momentum

[tex]K.E =\frac{p^2}{2m}\\\\p^2 = 2m. KE[/tex]

KE is constant

p is proportional to mass

So car having higher mass will have higher momentum .  

[tex]p_1 =\sqrt{(2m*K.E)}\\\\p_2 = \sqrt{(6m*K.E)} \\\\p_2 - p_1 = \sqrt{(6m*K.E)} -\sqrt{(2M*K.E} \\\\p_2 - p_1 = \sqrt{K.E m(\sqrt{6}-\sqrt{2}) }[/tex]

The difference is shown above.

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Answer: Family

Explanation:

Answer:

2.5 m/s^2

Explanation:

Given the following data;

Force = 80N

Mass = 32kg

To find acceleration;

Force is given by the multiplication of mass and acceleration.

Mathematically, Force is;

[tex] F = ma[/tex]

Where;

F represents force.

m represents the mass of an object.

a represents acceleration.

Making acceleration (a) the subject, we have;

[tex]Acceleration (a) = \frac{F}{m}[/tex]

Substituting into the equation;

[tex]Acceleration (a) = \frac{80}{32}[/tex]

Acceleration = 2.5m/s²

Answer:

radiant

Explanation:

Answer:

kinetic

Explanation:

this is because the kinetic theory is when energy is used in motion.

Answer:

the correct answer is d

Explanation:

The laws of mechanics are related

          F = m a

the acceleration of the body is given by the kinematics

         a = [tex]\frac{dv}{dt}[/tex]

         v = [tex]\frac{dx}{dt}[/tex]

substituting

         a = \frac{d2x}{dt^2}

         F = m [tex]\frac{d^2x}{dt^2}[/tex]

Therefore, in order to obtain the force (interaction of a body), continuous curves are needed and derivable from the position and the speed, for which all change in the trajectory of a body must be smooth where smooth is understood to have until the second derived.

Consequently the correct answer is d

Answer:

The time that passes until the police catch the speeder is 82.6204 seconds.

Explanation:

A body performs a uniformly accelerated rectilinear motion or uniformly varied rectilinear motion when its path is a straight line and its acceleration is constant. This implies that the speed increases or decreases its modulus in a uniform way.

The position is calculated by the expression:

x = x0 + v0*t + 1/2*a*t²

where:

First, let’s look at the police car’s equations of motion. In this case:

So: x = 50 m/s*t + 1/2*2 m/s²*t²

Now for the speeder’s car’s equations of motion you know:

So: x = 3,000 m + 55 m/s*t + 1/2*1 m/s²*t²

When the police catch the speeder they are both in the same position. So:

50 m/s*t + 1/2*2 m/s²*t²= 3,000 m + 55 m/s*t + 1/2*1 m/s²*t²

Solving:

0= 3,000 m + 55 m/s*t + 1/2*1 m/s²*t² - 50 m/s*t - 1/2*2 m/s²*t²

0= 3,000  + 55 *t + 1/2*t² - 50*t - 1*t²

0= 3,000  + 55 *t - 50*t - 1*t² + 1/2*t²

0= 3,000  + 5*t - 1/2*t²

Applying the quadratic formula:

[tex]x1,x2=\frac{-5+-\sqrt{5^{2}-4*(-\frac{1}{2})*3000 } }{2*(-\frac{1}{2} )}[/tex]

x1= -72.6209

and x2= 82.6209

Since you are calculating the value of a time and it cannot be negative, then the time that passes until the police catch the speeder is 82.6204 seconds.

Answer:

Explanation:

Let the velocity after the 4 gram pencil strikes is v .

kinetic energy of the combination = 1/2 m v²

= .5 x ( 4 + 221 ) x 10⁻³ x v² = work done by friction

friction force acting on the combination = 225 x10⁻³x  .325 x 9.8 = .7166 N

work done by friction

= .7166 x .119 = .085 J

.5 x 225 x 10⁻³ v² = .085

v² = .085 / .1125 = .7555

v = .8692 m = 86.92 cm /s

Velocity of combination after collision = 86.92 cm /s

Let velocity of pencil before collision be V

Applying law of conservation of momentum at the time of collision ,

4 x V = 225 x 86.92

V = 4889.25 cm / s

= 48.9 m /s .

Answer:

3 bending

Explanation:

Answer:

1200 Sm^2mol^-1

Explanation:

Given data :

conductivity of water ( kwater ) = 76 mS m^-1 = 0.076 Sm^-1

conductivity of kcl (aq)( Kkcl ) = 1.1639 Sm^-1

Kkcl = 1.1639 - 0.076 = 1.0879  Sm^-1

Resistance = 33.21 Ω

where conductivity can be expressed as = [tex]\frac{Cell constant}{Resistance }[/tex]

hence cell constant = conductivity * Resistance

                                 = 1.0879 * 33.21 = 36.13m^-1

conductivity of  CH3COOH ( kCH3COOH ) =  36.13 / 300

                                                                       = 0.120 Sm^-1

Determine the molar conductivity of acetic acid

= ( kCH3COOH * 1000 ) / C

C = 0.1 mol dm

=  (0.120 * 1000) / 0.1  =  1200 Sm^2mol^-1

Answer: marine organisms

Explanation:

i just took the test

Answer:

36.1 nC

Explanation:

The electrostatic force F = kqq'/r² since q = q', F = kq²/r² where q = charge on tape, r = distance between tapes = 1.00 cm = 1 × 10⁻² m and k = 9 × 10⁹ Nm²/C².

Given that F = weight of 12.0 mg piece of tape, F = mg where m = mass of tape = 12.0 mg = 12 × 10⁻³ kg and g = acceleration due to gravity = 9.8 m/s²

So, kq²/r² = mg

q²/r² = mg/k

q² = mgr²/k

taking square-root of both sides,

q = √(mg/k)r

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

q = √(mg/k)r

q = √(12 × 10⁻³ kg × 9.8 m/s²/ 9 × 10⁹ Nm²/C²)1 × 10⁻² m

q = √(117.6 × 10⁻³ kgm/s²/9 × 10⁹ Nm²/C²)1 × 10⁻² m

q = √(13.07 × 10⁻¹² C²/m²)1 × 10⁻² m

q = 3.61 × 10⁻⁶ C/m × 1 × 10⁻² m

q =  3.61 × 10⁻⁸ C

q = 36.1 × 10⁻⁹ C

q = 36.1 nC

Answer:

1.Lactose

I Hope its help for you

Have a good day

Answer:

[tex]F_N>F_N'[/tex]

Explanation:

From the question we are told that

Radius of curvature[tex]r=100m[/tex]

Mass [tex]M=300kg[/tex]

initial Speed of Motorcycle  [tex]V_1=30m/s[/tex]

Final  Speed of Motorcycle  [tex]V_2=33m/s[/tex]

Generally the equation Force at initial speed is  mathematically given as

[tex]F_N=mg-\frac{mv^2}{R}[/tex]

[tex]F_N=300*9.8-\frac{(300*30)^2}{100}[/tex]

[tex]F_N=240N[/tex]

Generally the equation Force at Final speed is  mathematically given as

[tex]F_N'=mg-\frac{mv'^2}{R}[/tex]

[tex]F_N'=300*9.8-\frac{(300*33)^2}{100}[/tex]

[tex]F_N'=-327N[/tex]

Therefore

[tex]F_N>F_N'[/tex]

Following are the calculation to the given question:

Solution:

Using formula:

[tex]\to mg - F^{'}_N=\frac{mv^{2}}{R} \\[/tex]

Calculating the Initial value:    

[tex]\to F_N = mg - \frac{mv^2}{R}\\\\[/tex]

          [tex]= 300 \times (9.8 - \frac{(30)^2}{100}) \\\\= 300 \times (9.8 - \frac{900}{100}) \\\\= 300 \times (9.8 - 9) \\\\= 300 \times (0.8) \\\\ = 240\ N \\\\[/tex]

Calculating the Final value:

[tex]\to F^{'}_{N}= 300 (9.8 -\frac{33^2}{100})\\\\[/tex]

          [tex]= 300 (9.8 -\frac{1089}{100})\\\\= 300 (9.8 - 10.89)\\\\= 300 (- 1.09)\\\\=-327[/tex]

Therefore, the answer is "the new normal force is less than [tex]F_{N}[/tex]" or [tex]\bold{F^{'}_{N}< F_{N}}[/tex].

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Answer:

concave mirror

Explanation:

Why This is Correct Because, concave mirrors only form a virtual image  when a object is larger than the other. when the object is produced between the focus, object and the image object, it becomes a  virtual image.

Answer:

The answer is concave mirror

Answer:

0.97 id the correct answer

Explanation:

Answer:

the speed after 3 seconds is 10 m/s

Explanation:

The computation of the speed is shown below:

As we know that

V = U  + at

Here,

U = 34 m/s

a =  - 8 m/s²

t = 3 Sec

V = velocity after 3 sec

V  = 34 + (-8)3

 = 34 - 24

 V = 10 m/s

Hence, the speed after 3 seconds is 10 m/s

Answer:

P = 360 Watts

Explanation:

Given that,

The weight of a student, [tex]F=5.4\times 10^2\ N[/tex]

It takes 15 seconds to run up a hill.

The top of the hill is 10 meters vertically above her starting point.

We need to find the power develop during her run. We know that te power developed is given by :

[tex]P=\dfrac{W}{t}\\\\P=\dfrac{mgh}{t}\\\\P=\dfrac{5.4\times 10^2\times 10}{15}\\\\P=360\ W[/tex]

So, the power develop during her run is 360 W.

Answer:

The answer is "[tex]1.271 \ \frac{KJ}{kg}\\[/tex]"

Explanation:

[tex]\Delta e_{mech} =\frac{P_2-P_1}{P} + \frac{v_{2}^2-v_{1}^2}{2}+g(z_2-z_1)\\\\\Delta e_{mech} =\frac{ 1380 -138 \times 1000 }{1000} + 0+g(3-0)\\\\P = \frac{1}{v}= \frac{1}{0.001} = 1000 \frac{kg}{m} \\\\ \Delta e_{mech} =1242 +9.81(3)= 1271.43 \frac{J}{kg} \\\\\text{work per unit pass}= 1.271 \ \frac{KJ}{kg}\\[/tex]

Answer: The Boat will rise

Explanation: Because high amplitude means high in heights.

Answer:

189 m/s

Explanation:

The pilot will experience weightlessness when the centrifugal force, F equals his weight, W.

So, F = W

mv²/r = mg

v² = gr

v = √gr where  v = velocity, g = acceleration due to gravity = 9.8 m/s² and r = radius of loop = 3.63 × 10³ m

So, v = √gr

v = √(9.8 m/s² × 3.63 × 10³ m)

v = √(35.574 × 10³ m²/s²)

v = √(3.5574 × 10⁴ m²/s²)

v = 1.89 × 10² m/s

v = 189 m/s

Answer: a positive test charge

Explanation:

An electric field is a vector measure. Its direction is determined as the direction that a positive test charge would experience a force when it is placed in the field.

A positive charge has a higher potential and a negative charge is lower potential.

Thus, an electric field is represented by the force on a positive test charge.

Hence, the correct option is "a positive test charge".

Answer:

D. a positive test charge.

Explanation:

Edge 2021

Answer:

t = 0.39 s

Explanation:

       [tex]105.1 mi/hr * (\frac{1hr}{3600s})*\frac{1609m}{1mi} = 47.0 m/s (1)[/tex]

       [tex]60.5 ft * \frac{0.3058m}{1ft} = 18.5 m (2)[/tex]

       [tex]t = \frac{18.5m}{47m/s} = 0.39 s (3)[/tex]

Answer:

Distance, S = 440 meters.

Explanation:

Given the following data;

Initial velocity, u = 17m/s

Time, t = 20 seconds

Final velocity, v = 27m/s

To find the distance;

First of all, we would determine the acceleration of the truck.

Acceleration = (v-u)/t

Substituting the given values into the equation, we have;

Acceleration = (27 - 17)/20

Acceleration = 10/20

Acceleration = 0.5m/s²

Now, we would use the second equation of motion to find the distance traveled.

S = ut + ½at²

S = 17*20 + ½*0.5*20²

S = 340 + 0.25*400

S = 340 + 100

S = 440m

The equations of motion can be used to obtain the distance covered as 440 m.

We have to use of the equations that are used for uniformly accelerated motion in solving the problem. The chosen equation must be;

v^2 = u^2 + 2as and v = u + at

v = final velocity

u = initial velocity

a = acceleration

s = distance

To obtain the acceleration;

27 = 17 + 20(a)

27 - 17 = 20a

a = 0.5 ms-2

Now, to obtain the distance;

v^2 = u^2 + 2as

v^2 - u^2/as = s

s = (27)^2 - (17)^2/2(0.5)

s = 440 m

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Answer:

I think it should be C, which is one