The image to the left shows Earth’s major plates. A geologist is studying a plate boundary indicated by the arrow. Fill in the blank to complete the statement about the plate boundary. This boundary is between the African Plate and the Plate.

Answer:

4L

Explanation:

Data provided in the question according to the question is as follows

Length = L

Gravity = G

For friend

Length = ?

Growth = 4G

Moreover,

[tex]T_1 = T_2[/tex]

Based on the above information ,

Now we have to apply the simple pendulum formula which is shown below:

[tex]T = 2\pi \frac{L}{G}[/tex]

Now equates these equations in both sides

[tex]2\pi \frac{L}{G} = 2\pi \frac{L}{4G}[/tex]

So after solving this, the length of the pendulum is 4L

Answer:

the length of a pendulum on your friend s planet should be 4 times than that on earth

Explanation:

We know that time period of simple pendulum is given by

[tex]T= 2\pi\sqrt{\frac{L}{g} }[/tex]

L= length of pendulum

g= acceleration due to gravity

therefore, Let T_1 and T_2 be the time period of the earth and other planet respectively.

[tex]\frac{T_1}{T_2} =\sqrt(\frac{L_1}{L_2}\times\frac{g_2}{g_1})[/tex]

ATQ

T_1=T_2=T,   g_2=4g_1

Putting the values in above equation and solving we get

[tex]\frac{L_1}{L_2} =\frac{1}{4}[/tex]

Answer:

The speed of the resistive force is 42.426 m/s

Explanation:

Given;

mass of skydiver, m = 75 kg

terminal velocity, [tex]V_T = 60 \ m/s[/tex]

The resistive force on the skydiver is known as drag force.

Drag force is directly proportional to square of terminal velocity.

[tex]F_D = kV_T^2[/tex]

Where;

k is a constant

[tex]k = \frac{F_D_1}{V_{T1}^2} = \frac{F_D_2}{V_{T2}^2}[/tex]

When the new drag force is half of the original drag force;

[tex]F_D_2 = \frac{F_D_1}{2} \\\\\frac{F_D_1}{V_{T1}^2} = \frac{F_D_2}{V_{T2}^2} \\\\\frac{F_D_1}{V_{T1}^2} = \frac{F_D_1}{2V_{T2}^2} \\\\\frac{1}{V_{T1}^2} = \frac{1}{2V_{T2}^2}\\\\2V_{T2}^2 = V_{T1}^2\\\\V_{T2}^2= \frac{V_{T1}^2}{2} \\\\V_{T2}= \sqrt{\frac{V_{T1}^2}{2} } \\\\V_{T2}= \frac{V_{T1}}{\sqrt{2} } \\\\V_{T2}= 0.7071(V_{T1})\\\\V_{T2}= 0.7071(60 \ m/s)\\\\V_{T2}= 42.426 \ m/s[/tex]

Therefore, the speed of the resistive force is 42.426 m/s

At terminal speed, the speed of the resistive force will be:

"42.426 m/s".

According to the question,

Skydriver's mass, m = 75 kg

Terminal velocity, [tex]V_T[/tex] = 60 m/s

Constant = k

We know the relation,

→ [tex]F_D[/tex] = k[tex]V_T^2[/tex]

here, k = [tex]\frac{F_D_1}{V_T_1^2} = \frac{F_D_2}{V_T_2^2}[/tex]

Now,

  [tex]F_D_2[/tex] = [tex]\frac{F_D_1}{2}[/tex]

  [tex]\frac{F_D_1}{V_T_1^2}= \frac{F_D_2}{V_T_2^2}[/tex]

   [tex]\frac{1}{V_T_1^2} = \frac{1}{2V_T_2^2}[/tex]

By applying cross-multiplication,

  [tex]V_T_2^2 = \sqrt{\frac{V_T_1^2}{2} }[/tex]

By substituting the above values,

  [tex]V_T_2[/tex] = 0.7071 ([tex]V_T_1[/tex])

        = 0.7071 × 60

        = 42.426 m/s

Thus the above response is correct.

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Answer:

Explanation:

We know that efficiency is the ratio of output power by input power. i.e. Efficiency describes the quality of machine or system how good it is.

Solution,

Energy input of system = 120 J

Efficiency = 80% = [tex] \frac{80}{100} = 0.8[/tex]

Now,

According to definition,

Efficiency = [tex] \frac{output}{input} [/tex]

Cross multiplication:

[tex]output \: = \: 0.8 \times 120[/tex]

Calculate the product

[tex]output \: = 96 \: joules[/tex]

Hope this helps...

Good luck on your assignment...

Answer:

The kinetic energy of a spinning disk will be reduced to a tenth of its initial kinetic energy if its moment of inertia is made five times larger, but its angular speed is made five times smaller.

Explanation:

Let us first consider the initial characteristics of the angular motion of the disk

moment of inertia = [tex]I[/tex]

angular speed = ω

For the second case, we consider the characteristics to now be

moment of inertia = [tex]5I[/tex]  (five times larger)

angular speed = ω/5  (five times smaller)

Recall that the kinetic energy of a spinning body is given as

[tex]KE = \frac{1}{2}Iw^{2}[/tex]

therefore,

for the first case, the K.E. is given as

[tex]KE = \frac{1}{2}Iw^{2}[/tex]

and for the second case, the K.E. is given as

[tex]KE = \frac{1}{2}(5I)(\frac{w}{5} )^{2} = \frac{5}{50}Iw^{2}[/tex]

[tex]KE = \frac{1}{10}Iw^{2}[/tex]

this is one-tenth the kinetic energy before its spinning characteristics were changed.

This implies that the kinetic energy of the spinning disk will be reduced to a tenth of its initial kinetic energy if its moment of inertia is made five times larger, but its angular speed is made five times smaller.

A spinning disk's kinetic energy will change to one-tenth if its moment of inertia was five times larger but its angular speed was five times smaller.

Relation between Kinetic energy and Moment of Inertia:

Now, let's consider moment of inertia =  I  and angular speed = ω

It is asked that what would be change in Kinetic energy if

moment of inertia =   (five times larger)

angular speed = ω/5  (five times smaller)

The kinetic energy of a spinning body is given as:

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

On substituting the values, we will get:

[tex]K.E.= \frac{1}{2} (5I) (\frac{w}{5} )^2 \\\\K.E. =\frac{1}{10} I. w^2[/tex]

Kinetic energy will be one-tenth to the kinetic energy before its spinning characteristics were changed.

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Answer:

The value  of  the function is  [tex]q__{t }} = 1.878 *10^{35}[/tex]

Explanation:

From the question we are told that

     The temperature is  [tex]T = 1000 \ K[/tex]

      The volume  is  [tex]V = 1 m^3[/tex]

Generally the  transnational partition function is mathematically represented as

        [tex]q__{t }} = [\frac{2 * \pi * m * k * T }{ N_a * h} ]^{\frac{3}{2} } * V[/tex]

Where  m is the molar mass of oxygen with a constant value of  [tex]m = 32 *10^{-3} \ kg/mol[/tex]

 k  is the Boltzmann constant with a value  of  [tex]k = 1.38 *10^{-23 } \ J/K[/tex]

[tex]N_a[/tex] is the Avogadro Number with a constant value of  [tex]N_a = 6.022 *10^{23} \ atoms[/tex]

h is  the Planck's  constant  with value   [tex]h = 6.626 *10^{-34 } \ J\cdot s[/tex]

Substituting values

       [tex]q__{t }} = [\frac{2 * 3.142 * 32*10^{-3} * 1.38 *10^{-23} * 1000 }{ 6.022 *10^{23} * [6.626 *10^{-34}] ^2 }]^{\frac{3}{2} } * 1[/tex]

       [tex]q__{t }} = 1.878 *10^{35}[/tex]

Answer:

The effective (rms) current when the circuit is in resonance is 6 A

Explanation:

Given;

resistance of the resistor, R = 20 ohms

capacitance of the capacitor, C = 0.75 microfarads

inductance of the inductor, L =  0.12 H

effective rms voltage, [tex]V_{rms}[/tex] = 120

At resonance, the impedance Z = R, Since the capacitive reactance (Xc) is equal to inductive reactance (XL).

The effective (rms) current, = [tex]V_{rms}[/tex] / R

                                              = 120 / 20

                                              = 6 A

Therefore, the effective (rms) current when the circuit is in resonance is 6 A

Answer:

see that the entire spectrum of the visible is between the integers from 3 to 5 so only three wavelengths are reflected with constructive interference

Explanation:

This is an interference problem in thin films, the refractive index of water is 1.33 and the refractive index of oil is 1.5

Let's analyze the light beam path emitted by the diver.

* when the beam passes from the water to the oil with the highest refractive index, it has a phase change of 180º

* also the wavelength of light in a material medium changes

      λ_n =  λ / n

where  λ_n is the wavelength in the material and  λ the wavelength in the vacuum air and n the refractive index.

If we include these aspects, the constructive interference equation is

       2t = (m + ½)  λ_n

       2nt = (m + ½)  λ

let's apply this equation to our case

            λ = 2nt / (m + ½)

The incidence of replacement of the oil with respect to water is

        n = n_oil / n_water = 1.5 / 1.33

        n = 1,128

let's calculate

        λ = 2 1,128 t / (m + ½)

        λ = 2,256 t / (m + ½)

In your statement you do not include the value of the oil layer that is the thin film, suppose a value to finish the calculation

          t = 0.001 mm = 1 10⁻⁶ m

the formula remains

        λ = 2,256 10⁻⁶ / (m + ½)

Let's find what values ​​of m we have to cut light in the visible range (400 to 700) 10⁻⁹ m

     m + ½ = 2,256 10⁻⁶ / λ

     m = 2,256 10⁻⁶ / λ - ½

light purple lan = 400 10⁻⁹m

     m = 2,256 10-6 / 400 10⁻⁹ - ½

     m = 5.64 - 0.5

     m = 5.14

     m = 5

red light  λ = 700 10⁻⁹m

      m = 2,256 1-6 / 700 10⁻⁹ - ½

      m = 3.22 - 0.5

      m = 2.72

      m = 3

we see that the entire spectrum of the visible is between the integers from 3 to 5 so only three wavelengths are reflected with constructive interference

Answer:

1.26 secs.

Explanation:

The following data were obtained from the question:

Force (F) = 20 N

Extention (e) = 0.2 m

Mass (m) = 4 Kg

Period (T) =.?

Next, we shall determine the spring constant, K for spring.

The spring constant, K can be obtained as follow:

Force (F) = 20 N

Extention (e) = 0.2 m

Spring constant (K) =..?

F = Ke

20 = K x 0.2

Divide both side by 0.2

K = 20/0.2

K = 100 N/m

Finally, we shall determine the period of oscillation of the 4 kg object suspended on the spring. This can be achieved as follow:

Mass (m) = 4 Kg

Spring constant (K) = 100 N/m

Period (T) =..?

T = 2π√(m/K)

T = 2π√(4/100)

T = 2π x √(0.04)

T = 2π x 0.2

T = 1.26 secs.

Therefore, the period of oscillation of the 4 kg object suspended on the spring is 1.26 secs.

Answer:

Explanation:

The voltage across the inductor VL = IXL

I is the total current flowing in the circuit and XL is the inductive reactance.

First we need to get the current flowing in the circuit.

From the expression above;

I = VL/XL

Since XL = 2πfL

I = VL/ 2πfL

Given VL = 2.8V, frequancy f = 100Hz and inductance L = 0.049-H

I = 2.8/2π*100*0.049

I = 2.8/30.79

I = 0.091A

Also;

Vrms = VL + VR

VR is the voltage across the resistor.

VR = Vrms - VL

VR = 7.9 - 2.8

VR = 5.1V

Then we can calculate the resistance of the resistor

According to ohms law VR = IR

Since the inductance and resistance ar connected in series, the same current will flow through them.

R = VR/I

R = 5.1/0.091

R = 56.04 ohms

Hence the resistance of the resistor in this circuit is 56.04 ohms

The complete question is missing, so i have attached the complete question.

Answer:

A) FBD is attached.

B) The condition that must be satisfied is for ω_min = √(g/r)

C) The tension in the string would be zero. This is because at the smallest frequency, the only radially inward force at that point is the weight(force of gravity).

Explanation:

A) I've attached the image of the free body diagram.

B) The formula for the net force is given as;

F_net = mv²/r

We know that angular velocity;ω = v/r

Thus;

F_net = mω²r

Now, the minimum downward force is the weight and so;

mg = m(ω_min)²r

m will cancel out to give;

g = (ω_min)²r

(ω_min)² = g/r

ω_min = √(g/r)

The condition that must be satisfied is for ω_min = √(g/r)

C) The tension in the string would be zero. This is because at the smallest frequency, the only radially inward force at that point is the weight(force of gravity).

Answer:

a) P = 10.27 kW

b) Pmax = 10.65 kW

c) E = 5.47 MJ

Explanation:

Mass of the loaded car, m = 950 kg

Angle of inclination of the shaft, θ = 28°

Acceleration due to gravity, g = 9.8 m/s²

The speed of the car, v = 2.35 m/s

Change in time, t = 14.0 s

a) The power that must be provided by the winch motor when the car is moving at constant speed.

P = Fv

The force exerted by the motor, F = mg sinθ

P = mgv sinθ

P = 950 * 9.8 *2.35* sin28°

P = 10,271.3 W

P = 10.27 kW

b) Maximum power that the motor must provide:

[tex]P = mv\frac{dv}{dt} + mgvsin \theta\\dv/dt = \frac{2.35 - 0}{14} \\dv/dt = 0.168 m/s^2\\P = (950*2.35*0.168) + (950*9.8*2.35* sin28)\\P = 374.74 + 10271.3\\P = 10646.04 W\\10.65 kW[/tex]

c) Total energy transferred:

Length of the track, d = 1250 m

[tex]E = 0.5 mv^2 + mgd sin \theta\\E = (0.5 * 950 * 2.35^2) + (950 * 9.8 * 1250 * sin 28)\\E = 2623.19 + 5463475.31\\E = 5466098.50 J\\E = 5.47 MJ[/tex]

Answer:

F₁ = 100 N

Explanation:

The pressure must be equally transmitted from the piston to the narrow arm. Therefore,

P₁ = P₂

F₁/A₁ = F₂/A₂

F₁/F₂ = A₁/A₂

where,

F₁ = Force Required to be applied to piston = ?

F₂ = Force to push cork at narrow arm = 16 N

A₁ = Area of wider arm = πd₁²/4

A₂ = Area of narrow arm = πd₂²/4

Therefore,

F₁/16 N = (πd₁²/4)/(πd₂²/4)

F₁ = (16 N)(d₁²/d₂²)

but, it is given that the diameter of wider arm is 2.5 times the diameter of the narrow arm.

d₁ = 2.5 d₂

Therefore,

F₁ = (16 N)[(2.5 d₂)²/d₂²]

F₁ = (16 N)(6.25)

F₁ = 100 N

Answer:

[tex]F=1.8\times 10^{-3}\ N[/tex]

Explanation:

We have,

Length of wires is 11 m

Separation between wires is 0.033 m

Current in both the wires is 5.2 A

It is required to find the magnitude of force between two wires. The force between wires is given by :

[tex]F=\dfrac{\mu_o I_1I_2l}{2\pi r}\\\\F=\dfrac{4\pi \times 10^{-7}\times 5.2\times 5.2\times 11}{2\pi \times 0.033}\\\\F=1.8\times 10^{-3}\ N[/tex]

So, the magnitude of force between wires is [tex]1.8\times 10^{-3}\ N[/tex]

Answer:

a.     E = 1.02*10^-27 J

b.     E = 6.39*10^-9eV

Explanation:

a. In order to calculate the energy of the radio photon, you use the following formula:

[tex]E=hf[/tex]             (1)

h: Planck's constant = 6.626*10^-34 Js

f: frequency of the photon = 1545kHz = 1.545*10^6 Hz

Then, by replacing you obtain the energy of the photon:

[tex]E=(6.626*10^{-34}Js)(1.545*10^6s^{-1})=1.02*10^{-27}J[/tex]

b. In electron volts, the energy of the photon is:

[tex]E=1.02*10^{-27}J*\frac{6.242*10^{18}eV}{1J}=6.39*10^{-9}eV[/tex]

Answer:

Mass of the rope = 2.8 kg

Explanation:

The speed of waves travelling through a rope with linear density (μ) and under tension T is given as v = √(T/μ)

The speed of waves in the rope is also calculated as

v = (d/t)

d = L = length of the rope = 15 m

t = time taken for the wave to move through the rope = 0.545 s

Speed = v = (15/0.545) = 27.523 m/s

Speed = v = √(T/μ)

T = tension in the rope = 140 N

μ = linear density = ?

27.523 = √(140/μ)

27.523² = (140/μ)

(140/μ) = 757.512

μ = (140/757.512) = 0.1848155556 = 0.1848 kg/m

Linear density = μ = (m/L)

m = mass of the rope = ?

L = length of the rope = 15 m

0.1848 = (m/15)

m = 0.1848 × 15 = 2.77 kg = 2.8 kg to 1 d.p.

Hope this Helps!!!

Answer:

t = 0.22 s

Explanation:

The rate of curvature can be defined as the ratio of the radius of curvature to the time taken to bend the wire to that radius of curvature. Therefore,

v = r/t

where,

v = rate of curvature

r = radius of wire

t = time taken

Here, in our case:

v = 9 units/s

r = 2 units

t = ?

Therefore,

9 units/s = 2 units/t

t = (2 units)/(9 units/s)

t = 0.22 s

Therefore, it takes 0.22 second to bend a straight wire into a circle of radius 2 units

Answer:

Explanation:

From the question, the kinectic energy of the train will be equal to the energy stored in the spring.

Kinetic energy = 1/2 mv² and energy stored in a spring E = 1/2 ke².

Equating both we will have;

1/2 mv² = 1/2ke²

mv² = ke²

m is the mass of the train

v is the velocity of then train

k is the spring constant

e is the extension caused by the spring.

Given m = 30000kg, v = 4 m/s, e = 4 - 2.4 = 1.6m

Substituting this values into the formula will give;

30000*4² =  k*1.6²

[tex]k = \frac{30,000*16}{1.6^2}\\ \\k = \frac{480,000}{2.56}\\ \\k = 187,500Nm^{-1}[/tex]

The value of the spring constant is 187,500N/m

Answer:

r  = 3.8 × 10 ⁻¹⁴ m

Explanation:

given data

alpha nucleus = 5.1 MeV

Charge of the alpha particle q₁= 2 × 1.6 × 10⁻¹⁹ C = 3.2 × 10⁻¹⁹ C

Charge of the gold nucleus q₂= 79 × 1.6 × 10⁻¹⁹ = 1.264 × 10⁻¹⁷ C

Kinetic energy of  the alpha particle = 5.97 × 10⁶ × 1.602 × 10⁻¹⁹ J ( 1 eV) =  9.564 × 10⁻¹³

k electrostatic force constant = 9 × 10⁹ N.m²/c²

solution

we know that when its kinetic energy is equal to the potential energy than  alpha particle will deflect \

so

Kinetic energy = potential energy =   k q₁q₂ ÷ r   ..................1

here  r is close distance the alpha particle

so r will be put here value

r = (  9 × 10⁹  × 3.2 × 10⁻¹⁹  × 1.264 × 10⁻¹⁷ ) ÷ ( 9.564 × 10⁻¹³  )

r  = 3.8 × 10 ⁻¹⁴ m

Answer:

a) E = 2.7x10⁶ N/C

b) F = 54 N

Explanation:

a) The electric field can be calculated as follows:

[tex] E = \frac{Kq}{d^{2}} [/tex]

Where:

K: is the Coulomb's constant = 9x10⁹ N*m²/C²

q: is the charge

d: is the distance

Now, we need to find the electric field due to charge 1:

[tex] E_{1} = \frac{9 \cdot 10^{9} N*m^{2}/C^{2}*50 \cdot 10^{-6} C}{(0.5 m)^{2}} = 1.8 \cdot 10^{6} N/C [/tex]

The electric field due to charge 2 is:

[tex]E_{2} = \frac{9 \cdot 10^{9} N*m^{2}/C^{2}*(-25) \cdot 10^{-6} C}{(0.5 m)^{2}} = -9.0 \cdot 10^{5} N/C[/tex]

The electric field at a point midway between them is given by the sum of E₁ and E₂ (they are in the same direction, that is to say, to the right side):

[tex]E_{T} = E_{1} + E_{2} = 1.8 \cdot 10^{6} N/C + 9.0 \cdot 10^{5} N/C = 2.7 \cdot 10^{6} N/C to the right side[/tex]                                                                                                

Hence, the electric field at a point midway between them is 2.7x10⁶ N/C to the right side.  

b) The force on a charge q₃ situated there is given by:

[tex]E_{T} = \frac{F_{T}}{q_{3}} \rightarrow F_{T} = E_{T}*q_{3}[/tex]

[tex] F = 2.7 \cdot 10^{6} N/C*20 \cdot 10^{-6} C = 54 N [/tex]

Therefore, the force on a charge q₃ situated there is 54 N.  

I hope it helps you!

(a) The electric field at a point midway between [tex]q_1[/tex] and [tex]q_2[/tex] is obtained to be [tex]2.7\times 10^6 \,N/C[/tex].

(b) The electrostatic force on the third charge [tex]q_3[/tex] situated between [tex]q_1[/tex] and [tex]q_2[/tex] is obtained as 54 N.

The answer can be explained as follows.

Given that the two charges are;

(a) At the midpoint; [tex]r = 0.5\,m[/tex].

We know that the electric field due to charge [tex]q_1[/tex].

Where, [tex]k=9\times 10^9\,Nm^2/C[/tex]

The electric field due to charge [tex]q_2[/tex] is given by;

Therefore, the net electric field in the midpoint is given by;

The direction is towards the right side.

(b) Now, there is another charge [tex]q_3=20\times 10^{-6}[/tex] in the midpoint.

So the force on the charge is ;

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Answer:  16 meters per second

Explanation:

 The derivative of the position is the velocity.

 The derivative of the velocity is the acceleration.

                  x(t) = 2t² + t³ + 1

         x'(t) = v(t) = 4t + 3t²

x''(t) = v'(t) = a(t) = 4 + 6t

                  a(2) = 4 + 6(2)

                         = 4 + 12

                         = 16

Answer:

f = 357.29Hz

Explanation:

In order to calculate the fundamental frequency in the closed tube, you use the following formula:

[tex]f_n=\frac{nv}{4L}[/tex]       (1)

n: order of the mode = 1

v: speed of sound = 343m/s

L: length of the tube = 24cm = 0.24m

You replace the values of the parameters in the equation (1):

[tex]f_1=\frac{(1)(343m/s)}{4(0.24m)}=357.29Hz[/tex]

The fundamental frequency of in the tube is 357.29Hz

Answer:

I believe it's A.)

Explanation:

Although light comes into our atmosphere through refraction, it reaches our eyes only through reflection from objects. So when light rays reflect off an object and enter the eyes through the cornea you can then see that object.

Hope this helps you out : )

Answer:

The power exerted by the student is 51.2 W

Explanation:

Given;

extension of the elastic band, x = 0.8 m

time taken to stretch this distance, t = 0.5 seconds

the spring constant, k = 40 N/m

Apply Hook's law;

F = kx

where;

F is the force applied to the elastic band

k is the spring constant

x is the extension of the elastic band

F = 40 x 0.8

F = 32 N

The power exerted by the student is calculated as;

P = Fv

where;

F is the applied force

v is velocity = d/t

P = F x (d/t)

P = 32 x (0.8 /0.5)

P = 32 x 1.6

P = 51.2 W

Therefore, the power exerted by the student is 51.2 W

Answer:

0.242

Explanation:

m = 13.3 kg

a =d= 2.42 m/s²

g = 10 m/s²

from the laws of friction F = ¶R

===> ¶ = F/R = ma/mg = a/g

¶ = a/g = 2.42/10 = 0.242

Jerome solves a problem using the law of conservation of momentum. What should Jerome always keep constant for each object after the objects collide and bounce apart?

a-velocity

b-mass

c-momentum

d-direction

Answer:

b. Mass

Explanation:

This question has to do with the principle of the law of conservation of momentum which states that the momentum of a system remains constant if no external force is acting on it.

As the question states, two objects collide with each other and eventually bounce apart, so their momentum may not be conserved but the mass of the objects is constant for each non-relativistic motion. Because of this, the mass of each object prior to the collision would be the same as the mass after the collision.

Therefore, the correct answer is B. Mass.

This question involves the concept of the law of conservation of momentum.

Jerome should always keep the "mass" of each object constant after the objects collide and bounce apart.

The law of conservation of momentum states that the momentum of a system of objects must remain constant before and after the collision has taken place.

Mathematically,

[tex]m_1u_1+m_2u_2=m_1v_1+m_v_2[/tex]

where,

m₁ = mass of the first object

m₂ = mass of the second object

u₁ = velocity of the first object before the collision

u₂ = velocity of the second object before the collision

v₁ = velocity of the first object after the collision

v₂ = velocity of the second object after the collision

Hence, it is clear from the formula that the only thing unchanged before and after the collision is the mass of each object.

Learn more about the law of conservation of momentum here:

brainly.com/question/1113396?referrer=searchResults

The attached picture illustrates the law of conservation of momentum.

Answer:

Gain heat and evaporate

Explanation:

As water molecules are exposed to sunlight, they begin to heat up. This means that the molecules begin to jiggle faster, and as such take up less space. Since they take up less space they are less dense, and therefore more bouyant. This means that they begin to rise into the air, and evaporate. Hope this helps!

Answer:

The motion of the lighter child would look faster than that of the heavier child, but both have the same period of oscillation.

Explanation:

Oscillation is a type of simple harmonic motion which involves the to and fro movement of an object. The oscillation takes place at a required time called the period of oscillation.

Since the swings are similar, the period of oscillation of the two children are the same and they would complete one oscillation in the same time. Though the oscillation of the lighter child seems faster than that of the heavy child, their masses does not affect the period of oscillation.

When a heavy object oscillates, its mass increases the drag or damping force, but not the period of oscillation. Thus, it oscillate slowly.

Answer:

a. the magnitude of the force experienced by the muon is 2.55 × 10⁻¹⁹N

b.  this force compare to the weight of the muon; the force is 1.38 × 10⁸ greater than muon

Explanation:

F= ma

v²=u² -2aS

(1.56 ✕ 10⁶)²=(2.40 ✕ 10⁶)²-2a(1220)

a=1.36×10⁹m/s²

recall

F=ma

F = 1.88 ✕ 10⁻²⁸ kg × 1.36×10⁹m/s²

F= 2.55 × 10⁻¹⁹N

the magnitude of the force experienced by the muon is 2.55 × 10⁻¹⁹N

b.  this force compare to the weight of the muon

F/mg= 2.55 × 10⁻¹⁹/ (1.88 ✕ 10⁻²⁸ × 9.8)

= 1.38 × 10⁸

Explanation:

Einsteins theory of relativity explains how space and time are linked for objects that are moving at a consistent speed in a straight line.