When an object is thrown vertically upward from the surface of the Earth: What is the instantaneous velocity in the point of maximum height?What is the acceleration in the point of maximum height?

Answer:

Explanation:

We can use conservation of momentum to find the initial velocities.

Taking the unit vector [tex]\hat{i}[/tex] pointing north and [tex]\hat{j}[/tex] pointing east, the final velocity will be

[tex] \vec{V}_f = 16.0 \frac{m}{s} \ ( \ cos(24.0 \°) \ , \ sin (24.0 \°) \ )[/tex]

[tex] \vec{V}_f = ( \ 14.617 \frac{m}{s} \ , \ 6.508 \frac{m}{s} \ )[/tex]

The final linear momentum will be:

[tex]\vec{P}_f = (m_{car}+ m_{truck}) * V_f[/tex]

[tex]\vec{P}_f = (950 \ kg \ + 1900 \ kg \ ) *  ( \ 14.617 \frac{m}{s} \ , \ 6.508 \frac{m}{s} \ )[/tex]

[tex]\vec{P}_f = (2.850 \ kg \ ) *  ( \ 14.617 \frac{m}{s} \ , \ 6.508 \frac{m}{s} \ )[/tex]

[tex]\vec{P}_f = ( \ 41,658.45 \frac{ kg \ m}{s} \ , \ 18,547.8 \frac{kg \ m}{s} \ )[/tex]

As there are not external forces, the total linear momentum must be constant.

So:

[tex]\vec{P}_0= \vec{P}_f [/tex]

As initially the car is travelling east, and the truck is travelling north, the initial linear momentum must be

[tex]\vec{P}_0= ( m_{truck} * v_{truck}, m_{car}* v_{car} )[/tex] 

so:

 [tex]\vec{P}_0= \vec{P}_f[/tex] 

[tex]( m_{truck} * v_{truck}, m_{car}* v_{car} ) = ( \ 41,658.45 \frac{ kg \ m}{s} \ , \ 18,547.8 \frac{kg \ m}{s} \ )[/tex]  

so

[tex]\left \{ {{m_{truck} \ v_{truck} = 41,658.45 \frac{ kg \ m}{s}  } \atop {m_{car} \ v_{car}=18,547.8 \frac{kg \ m}{s} }} \right.[/tex]

So, for the truck

[tex]m_{truck} \ v_{truck} = 41,658.45 \frac{ kg \ m}{s} [/tex]

[tex]1900 \ kg \ v_{truck} = 41,658.45 \frac{ kg \ m}{s} [/tex]

[tex] v_{truck} = \frac{41,658.45 \frac{ kg \ m}{s}}{1900 \ kg} [/tex]

[tex]v_{truck} = \frac{41,658.45 \frac{ kg \ m}{s}}{1900 \ kg} [/tex]

[tex]v_{truck} = 21.93 \frac{m}{s}[/tex]

And, for the car

[tex]950 \ kg \ v_{car}=18,547.8 \frac{kg \ m}{s}[/tex]

[tex]v_{car}=\frac{18,547.8 \frac{kg \ m}{s}}{950 \ kg}[/tex]

[tex]v_{car}=19.524 \frac{m}{s}[/tex]

Final answer:

To change 350 g of ice at -20 degrees Celsius to water at 20 Celsius, you need to calculate the heat required for two processes: warming the ice and melting the ice. The heat needed to warm the ice can be calculated using the equation: Q = mcΔT, and the specific heat capacity of ice is 2.09 J/g°C. The heat needed to melt the ice can be calculated using the equation: Q = mL, where L is the heat of fusion, which is 334 J/g for ice.

Explanation:

To calculate the heat necessary to change 350 g of ice at -20 degrees Celsius to water at 20 degrees Celsius, we need to consider the energy required for two processes: warming the ice from -20°C to 0°C and then melting the ice at 0°C to form water at 0°C.

The heat needed to warm the ice can be calculated using the equation: Q = mcΔT, where Q is the heat, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. The specific heat capacity of ice is 2.09 J/g°C.

The heat needed to melt the ice can be calculated using the equation: Q = mL, where L is the heat of fusion, which is 334 J/g for ice.

Let's calculate the heat needed for each process and add them together:

Finally, add both values to find the total heat required.

Answer:

Part I:

(a): [tex]7.889\times 10^2\ Nm.[/tex]

(b): [tex]0\ Nm.[/tex]

(c): [tex]6.52\times 10^2\ Nm.[/tex]

Part II:

(a): [tex]2.664\times 10^{-9}\ C.[/tex]

(b): Charge on the sphere is positive and only distributed on the surface of the sphere and not inside the sphere.

Explanation:

Assuming,

[tex]\hat i,\ \hat j,\ \hat k[/tex] are the unit vectors along positive x, y and z axes respectively.

Given, the electric field is of intensity 3.22 kN/C and is along x-axis.

Therefore,

[tex]\vec E = 3.22\ kN/C\ \hat i=3.22\times 10^3\ N/C\ \hat i.[/tex]

The magnetic flux through a surface is defined as

[tex]\phi = \vec E\cdot \vec A[/tex]

where,

[tex]\vec A[/tex] is the area vector of the surface which is directed along the normal to the plane of the surface and its magnitude is equal to the area of the surface.

[tex]A = \text{length of the plane }\times \text{width of the plane}\\=0.700\times 0.350\\=0.245\ m^2.[/tex]

(a):

When the plane is parallel to yz plane, then its normal is along x axis, therefore,

[tex]\vec A = A\ \hat i.[/tex]

Electric flux,

[tex]\phi = \vec E\cdot \vec A \\=(3.22\times 10^3)\hat i\cdot (0.245)\hat i = 3.22\times10^3\times 0.245\times (\hat i\cdot \hat i) = 7.889\times 10^2\ Nm.[/tex]

(b):

When the plane is parallel to xy plane, then its normal is along z axis, therefore,

[tex]\vec A = A\ \hat k.[/tex]

Electric flux,

[tex]\phi = \vec E\cdot \vec A \\=(3.22\times 10^3)\hat i\cdot (0.245)\hat k = 3.22\times10^3\times 0.245\times (\hat i\cdot \hat k) = 0\ Nm.[/tex]

(c):

When the normal to the plane makes an angle [tex]34.2^\circ[/tex] with the x-axis.

Electric flux through the plane is given by,

[tex]\phi = \vec E\cdot \vec A\\=EA\cos(34.2^\circ) \\=(3.22\times 10^3)\times (0.245)\times \cos(34.2^\circ)\\=6.52\times 10^2\ Nm.[/tex]

(a):

Given:

The electric field is directed radially outward from the surface of the sphere at its every point and the normal to the surface of the sphere is also radially outwards at its every point therefore both the electric field and the area vector of the sphere is along the same direction.

The flux through the surface is given by

[tex]\phi = \vec E \cdot \vec A \\= EA\cos0^\circ\\=EA\\=E\ 4\pi R^2\\=570\times 4\pi \times 0.205^2\\=3.01\times 10^2\ Nm.[/tex]

According to Gauss law of electrostatics,

[tex]\phi = \dfrac q{\epsilon_o}[/tex]

where,

Therefore, the net charge on the sphere is given by

[tex]q=\epsilon_o\times \phi\\=8.85\times 10^{-12}\times 3.01\times 10^2\\=2.664\times 10^{-9}\ C.[/tex]

(b):

Since, the sphere is charged all of the charge resides on its surface. The electric field through the sphere is radially outwards which means the charge on the sphere should be positive.

There is no charge distributed inside the sphere.

Answer:

Distance, d = 6.75 meters

Explanation:

Given that,

Time taken by the eyes to shut during a hard sneeze, t = 0.32 s

Speed of the car, v = 76 km/h = 21.11 m/s

We need to find the distance covered by the car during this time. The product of speed and the time taken by the car is called the distance covered by the it. Mathematically,

[tex]d=v\times t[/tex]

[tex]d=21.11\times 0.32[/tex]

d = 6.75 meters

So, the car will cover 6.75 meters during that time. Hence, this is the required solution.

Answer:

q=heat flux=216.83W/m^2

Ts=36.45C

Explanation:

What you should do is raise the heat transfer equation from the inside of the granite wall to the air.

You should keep in mind that in the numerator you place the temperature differences, while the denominator places the sum of the thermal resistances by conduction and convection.

When you have the heat value, all you have to do is use the convection equation q = h (ts-t) and solve for the surface temperature.

I attached procedure

Answer:

(a) 62.57 L

(b) 801.94 kPa

Explanation:

Given:

[tex]n[/tex] = number of moles of gas = 10.3 mol

[tex]P_1[/tex] = initial pressure of the gas = [tex]519\ kPa = 5.19\times 10^5\ Pa[/tex]

[tex]T_1[/tex] = initial temperature of the gas = [tex]361.38^\circ C = (361.38+273)\K = 598.38\ K[/tex]

[tex]T_2[/tex] = final temperature of the gas = [tex]651.6^\circ C = (651.6+273)\K = 924.6\ K[/tex]

[tex]V[/tex] = volume of the tank

R = universal gas constant = [tex]8.314 J/mol K[/tex]

Part (a):

Using Ideal gas equation, we have

[tex]PV=nRT\\\Rightarrow V = \dfrac{nRT}{P}\\\Rightarrow V = \dfrac{10.3\times 8.314\times 598.38}{5.19\times 10^{5}}\\\Rightarrow V = 6.2567\times10^{-2}\ m^3\\\Rightarrow V = 62.567\ L[/tex]

Hence, the volume of the container is 62.567 L.

Part (b):

As the volume of the container remains constant.

Again using ideal gas equation,

[tex]PV=nRT\\\because V,\ n, R\ are\ constant\\\therefore \dfrac{P_1}{P_2}=\dfrac{T_1}{T_2}\\\Rightarrow P_2 = \dfrac{T_2}{T_1}P_1\\\Rightarrow P_2 = \dfrac{924.6}{598.38}\times 519\\\Rightarrow P_2 = 801.94\ kPa[/tex]

Hence, the final pressure of the gas is now 801.94 kPa.

Answer with Explanation:

We know from newton's second law that acceleration produced by a force 'F' in a body of mass 'm' is given by

[tex]a=\frac{Force}{Mass}=\farc{F}{m}[/tex]

In the given case the acceleration equals

[tex]a=\frac{F_{o}cos^{2}(\omega t)}{m}[/tex]

Now by definition of acceleration we have

[tex]a=\frac{dv}{dt}\\\\\int dv=\int adt\\\\v=\int adt\\\\v=\int \frac{F_{o}cos^{2}(\omega t)}{m}\cdot dt\\\\v=\frac{F_{o}}{m}\int cos^{2}(\omega t)dt\\\\v=\frac{F_{o}}{\omega m}\cdot (\frac{\omega t}{2}+\frac{sin(2\omega t)}{4}+c)[/tex]

Similarly by definition of position we have

[tex]v=\frac{dx}{dt}\\\\\int dx=\int vdt\\\\x=\int vdt[/tex]

Upon further solving we get

[tex]x=\int [\frac{F_{o}}{\omega m}\cdot (\frac{\omega t}{2}+\frac{sin(2\omega t)}{4}+c)]dt\\\\x=\frac{F_{o}}{\omega m}\cdot ((\int \frac{\omega t}{2}+\frac{sin(2\omega t)}{4}+c)dt)\\\\x(t)=\frac{F_{o}}{\omega m}\cdot (\frac{\omega t^{2}}{4}-\frac{cos(2\omega  t)}{8\omega }+ct+d)[/tex]

The plots can be obtained depending upon the values of the constants.

Answer:

a) 12.528 m/s

b) 15.344 m/s

Explanation:

Given:

Mass of the child, m = 34.9 kg

Height of the water slide, h = 16.0 m

Now,

a) By the conservation of energy,

loss in potential energy = gain in kinetic energy

mgh = [tex]\frac{\textup{1}}{\textup{2}}\textup{m}\times\textup{v}^2[/tex]

where,

g is the acceleration due to the gravity

v is the velocity of the child

thus,

at halfway down, h = [tex]\frac{\textup{16}}{\textup{2}}[/tex]= 8 m

therefore,

34.9 × 9.81 × 8 = [tex]\frac{\textup{1}}{\textup{2}}\textup{34.9}\times\textup{v}^2[/tex]

or

v = 12.528 m/s

b)

at three-fourth way down

height = [tex]\frac{\textup{3}}{\textup{4}}\times16[/tex] = 12 m

thus,

loss in potential energy = gain in kinetic energy

34.9 × 9.81 × 12 = [tex]\frac{\textup{1}}{\textup{2}}\textup{34.9}\times\textup{v}^2[/tex]

or

v = 15.344 m/s

Answer:

(a) 12.52 m/s

(b) 15.34 m/s

Explanation:

mass, m = 34.9 kg

h = 16 m

(a) Initial velocity, u = 0

height = h / 2 = 16 / 2 = - 8 m (downward)

let the speed of child is v.

acceleration, a = - 9.8 m/s^2 (downward)

Use third equation of motion

[tex]v^{2}=u^{2}+2as[/tex]

[tex]v^{2}=0^{2}+2\times 9.8\times 8[/tex]

v = 12.52 m/s

(b) Initial velocity, u = 0

height = 3 h / 4 = 12 m = - 12 m (downward)

let the speed of child is v.

acceleration, a = - 9.8 m/s^2 (downward)

Use third equation of motion

[tex]v^{2}=u^{2}+2as[/tex]

[tex]v^{2}=0^{2}+2\times 9.8\times 12[/tex]

v = 15.34 m/s

Answer:

a) The speed of the ball at the bottom of the first plane is 2.59 m/s

b) It takes the ball 6.56 s to roll down the first plane.

Explanation:

The equations for the position and velocity of the ball are as follows:

x = x0 + v0 · t + 1/2 · a · t²

v = v0 + a · t

Where:

x = position at time t

x0 = initial position

v0 = initial speed

t = time

a = acceleration

v = velocity at time t

b) First, let´s calculate the time it takes the ball to reach the bottom of the plane using the equation for the position:

x = x0 + v0 · t + 1/2 · a · t²

Placing the center of the frame of reference at the point where the ball starts rolling, x = 0. Since the ball starts from rest, v0 = 0. Then:

x = 1/2 · a ·t²

Let´s find the time when the ball reaches a position of 8.50 m

8.50 m = 1/2 · 0.395 m/s² · t²

t² = 2 · 8.50 m / 0.395 m/s²

t = 6.56 s

a) Now, using the equation of the velocity, we can calculate the velocity of the ball at the bottom of the plane (t = 6.56 s):

v = a · t

v = 0.395 m/s² · 6.56 s = 2.59 m/s

Friction:

When an object slips on a surface, an opposing force acts between the tangent planes which acts in the opposite direction of motion. This opposing force is called Friction. Or in other words, Friction is the opposing force that opposes the motion between two surfaces.

The main component of friction are:

Normal Reaction (R):

Suppose a block is placed on a table in the above picture, which is in resting state, then two forces are acting on it at that time.

The first is due to its weight mg which is working from its center of gravity towards the vertical bottom.

The second one is superimposed vertically upwards by the table on the block, called the reaction force (P). This force passes through the center of gravity of the block.

Due to P = mg, the box is in equilibrium position on the table.

Coefficient of friction ( μ ):

The ratio of the force of friction and the reaction force is called the coefficient of friction.

Coefficient of friction, µ = force of friction / reaction force

μ = F / R

The coefficient of friction is volume less and dimensionless.

Its value is between 0 to 1.

Advantage and disadvantage from friction force:

How to reduce friction:

Answer:

Option a

Explanation:

Galileo's experiment of inclined plane provides the measurement of acceleration with accuracy with the help of simple equipment.

The ultimate objective of the experiment was to prove that when other external forces are absent like the one due to air resistance, all the objects fall at a constant accelerated rate towards the Earth.

Thus he measured the influence of gravity on the free falling object accurately and observed that the distance covered by these objects when released from rest and moving at a steady rate, then the distance traveled by the object is proportional to the square of the time taken.

From his findings:

[tex]D = \frac{gH}{2L}t^{2}[/tex]

where

D = distance

L = Inclined plane length

H = Height of the inclined plane

Answer:

−5.0  

Explanation:

Answer:

The squirrel's initial velocity is 5 m/s.

Explanation:

It is given that,

Distance covered by the squirrel, d = 5 m

Time taken, t = 0.5 s

Final speed of the squirrel, v = 15 m/s

To find,

The squirrel's initial velocity.

Solution,

Let a is the acceleration of the squirrel. Using the first equation of motion to find it :

[tex]a=\dfrac{v-u}{t}[/tex]

[tex]a=\dfrac{15-u}{0.5}[/tex]..............(1)

Let u is the initial speed of the squirrel. Using again third equation of kinematics to find it :

[tex]v^2-u^2=2ad[/tex]

Use equation (1) to put value of a

[tex](15)^2-u^2=2\times \dfrac{15-u}{0.5}\times 5[/tex]

[tex]225-u^2=20(15-u)[/tex]

On solving the above quadratic equation, we get the value of u as :

u = 5 m/s

Therefore, the squirrel's initial velocity before getting started is 5 m/s.

The rate of heat loss by radiation is equal to -207.5kW

Why?

To calculate the heat loss rate (or heat transfer rate) by radiation, from the given situation, we can use the following formula:

[tex]HeatLossRate=E*S*A*((T_{cold})^{4} -(T_{hot})^{4} )[/tex]

Where,

E, is the emissivity of the body.

A, is the area of the body.

T, are the temperatures.

S, is the Stefan-Boltzmann constant, which is equal to:

[tex]5.67x10^{-8}\frac{W}{m^{2}*K^{-4} }[/tex]

Now, before substitute the given information, we must remember that the given formula works with absolute temperatures (Kelvin), so,  we need to convert the given values of temperature from Celsius degrees to Kelvin.

We know that:

[tex]K=Celsius+273.15[/tex]

So, converting we have:

[tex]T_{1}=1110\°C+273.15=1383.15K\\\\T_{2}=36.2\°C+273.15=309.35K[/tex]

Therefore, substituting the given information and calculating, we have:

[tex]HeatLossRate=E*S*A*((T_{cold})^{4} -(T_{hot})^{4} )[/tex]

[tex]HeatLossRate=1*5.67x10^{-8}\frac{W}{m^{2}*K^{-4} }*1m^{2} *((309.35K)^{4} -(1383.15})^{4} )\\\\HeatLossRate=5.67x10^{-8}\frac{W}{K^{-4} }*(95697.42K^{4} -3.66x10^{12}K^{4})\\ \\HeatLossRate=5.67x10^{-8}\frac{W}{K^{-4} }*(-3.66x10^{12} K^{4})=-207522W=-207.5kW[/tex]

Hence, we have that the rate of heat loss is equal to -207.5kW.

Answer:

The amount of charge at the center is -132.7 nC

Solution:

As per the question:

Electric flux from one face of cube, [tex]\phi_{E} = - 2.5 kN.m^{2}/C[/tex]

Edge, a = 19cm

Since, a cube has six faces, thus total flux from all the 6 faces = [tex]6\times (-2.5) = - 15 kN.m^{2}/C[/tex]

Also, from Gauss' law:

[tex]\phi_{E,net} = \frac{1}{\epsilon_{o}}Q_{enc}[/tex]

[tex]Q_{enc} = 8.85\times 10^{- 12}\times - 15\times 10^{3}[/tex]

[tex]Q_{enc} = - 132.7 nC[/tex]

Answer:

3.825*10^-6 N

Explanation:

As particle 1 and particle 3 has opposite types of charge, particle 3 will be attracted to particle 1. And as particle 2 and 3 has the same sign, they will repel each other. Due to the position of the particles, both the Force that 1 exerts on 3 and the force that 2 exerts on 3, will have the same direction. Now, you need the magnitude. You can use the following expression:

[tex]F_e = K\frac{q_1*q_3}{r^2}[/tex]

K is the Coulomb constant equal to 9*10^9 N*m^2/C^2, q1 and q2 is the charge of the particles, and r is the distance.

Force 1 on 3 is equal to:

[tex]F_e = K\frac{q_1*q_3}{r^2} = 9*10^9 \frac{Nm^2}{C^2} * \frac{2.3*10^{-9}C * 7.5*10^{-9}C}{(-0.6m - (-0.3m))^2} = 1.725 * 10^{-6} N[/tex]

Force 2 on 3:

[tex]F_e = K\frac{q_2*q_3}{r^2} = 9*10^9 \frac{Nm^2}{C^2} * \frac{2.8*10^{-9}C * 7.5*10^{-9}C}{(0m - (-0.3m))^2} = 2.1 * 10^{-6} N[/tex]

The magnitude of the resultant force is the addition of both forces:

[tex]F_e = 1.725*10^{-6} N + 2.1*10^{-6} N= 3.825 * 10^{-6} N[/tex]

The magnitude of the net force exerted by the two charges on the third charge is 1.38 N.

The magnitude of the net force exerted by q1 and q2 on q3 can be found using Coulomb's Law. Coulomb's Law states that the force between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

The formula for Coulomb's Law is:

F = k(q1 * q2) / r^2

Where F is the force, k is Coulomb's constant (k = 9 x 10^9 N * m^2 / C^2), q1 and q2 are the charges, and r is the distance between them.

In this case, q1 = -2.30 nC, q2 = 2.80 nC, q3 = 7.50 nC, y1 = -0.600 m, y3 = -0.300 m, and y2 = 0 m.

Using the formula and substituting the values, we can calculate the force:

F = (9 x 10^9 N * m^2 / C^2) * (q1 * q3) / ((y3 - y1)^2)

F = (9 x 10^9 N * m^2 / C^2) * (-2.30 nC * 7.50 nC) / ((-0.300 m + 0.600 m)^2)

F = -1.38 N

Therefore, the magnitude of the net force exerted by the two charges on the third charge is 1.38 N.

Answer:

45°

Explanation:

λ = 600 nm = 600 x 10^-9 m

d = 1.7 x 10^-6 m

For second order, N = 2

Use the equation

d Sinθ = N λ

1.7 x 10^-6 x Sinθ = 2 x 600 x 10^-9

Sinθ = 0.706

θ = 45°

Thus, the angle of diffraction is 45°.

The diffraction angle of the second-order line is approximately 44.22°.

The diffraction angle of the second order line can be found using the formula:

dsinθ = mλ

Where d is the ruling separation of the diffraction grating, θ is the diffraction angle, m is the order of the line, and λ is the wavelength of light. Plugging in the values given in the question:

1.7 × 10-6 m × sinθ = 2 × 600 × 10-9 m

Simplifying the equation:

sinθ = 2 × 600 × 10-9 m / (1.7 × 10-6 m)

Calculating the value of sinθ, we find:

sinθ ≈ 0.7059

Finally, taking the inverse sine of 0.7059, we find:

θ ≈ 44.22°

The second-order line occurs at a diffraction angle of approximately 44.22°.

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Final answer:

In a tug-of-war where forces are exerted in opposite directions (10 N south and 17 N north), the net force is calculated by subtracting the smaller force from the larger, resulting in a 7 N force towards the north. The correct answer is D.

Explanation:

The net force in a tug of war scenario is the vector sum of all the forces acting on the object. Since force is a vector quantity, it has both magnitude and direction. In this case, the force exerted by one team (17 N north) and the force exerted by the other team (10 N south) are in opposite directions along the same line, so we subtract the smaller force from the larger one to find the net force.

|17 N| - |10 N| = 17 N - 10 N = 7 N north.

Answer:

3.75s

Explanation:

We can use the equations for constant acceleration motion. Let's call x, the total length of the path, then x/2 will be half of path. After falling from rest and reaching the half of its total path, the velocity of the body will be:

[tex]v_f^2 =v_0^2 + 2a(x/2)[/tex]

vf is the final velocity, v0 is the initial velocity, 0m/s because the body starts from rest. a is the acceleration, gravity = 9.81m/s^2 in this case. Now, clearing vf we get:

[tex]v_f=\sqrt{(0m/s)^2 + 2g(x/2)}\\v_f = \sqrt{g*x}

In the second half:

[tex]x/2 = \frac{1}{2}gtx^{2}  + v_ot[/tex]

[tex]x/2 = \frac{1}{2}g*(t)^2 + \sqrt{g*x}*(t)[/tex]

[tex](\frac{1}{2}\frac{(x-gt^2)}{\sqrt{g}t})^2 = x\\\\\frac{1}{4gt^2}(x^2 - 2xgt^2 + g^2t^4) = x\\\\\frac{1}{4gt^2}x^2 - (\frac{2gt^2}{4gt^2}+1)x + \frac{g^2t^4}{4gt^2} = 0\\ \frac{1}{4gt^2}x^2 - \frac{3}{2}x + \frac{gt^2}{4} = 0\\[/tex]

[tex]0.0211 x^2 - 1.5x + 2.97 = 0\\[/tex]

Solving for x, you get that x is equal to 69.2 m or 2.03m. The total time of the fall would be:

[tex]x = \frac{1}{2}gt^2\\t=\sqrt{(2x/g)}[/tex]

Trying both possible values of x:

[tex]t_1 = 3.75 s\\t_2 = 0.64 s[/tex]

t2  is lower than 1.1s, therefore is not a real solution.

Therefore, the path traveled will be 69.2m and the total time 3.75s

Answer:

[tex]r=6.72\times 10^{-13}\ m[/tex]

Explanation:

Let r is the radius of the n = 1 orbit of the gold. According to Bohr's model, the radius of orbit is given by :

[tex]r=\dfrac{n^2h^2\epsilon_o}{Z\pi me^2}[/tex]

Where

n = number of orbit

h = Planck's constant

Z = atomic number (for gold, Z = 79)

m = mass of electron

e = charge on electron

[tex]r=\dfrac{(6.63\times 10^{-34})^2 \times 8.85\times 10^{-12}}{79\pi \times 9.1\times 10^{-31}\times (1.6\times 10^{-19})^2}[/tex]

[tex]r=6.72\times 10^{-13}\ m[/tex]

So, the radius of the n = 1 orbit of gold is [tex]6.72\times 10^{-13}\ m[/tex]. Hence, this is the required solution.

Answer:

[tex]t=6.4534 s[/tex]

Explanation:

This is an exercise where you need to use the concepts of free fall objects

Our knowable variables are initial high, initial velocity and the acceleration due to gravity:

[tex]y_{0}=75m[/tex]

[tex]v_{oy} =20m/s[/tex]

[tex]g=9.8 m/s^{2}[/tex]

At the end of the motion, the rock hits the ground making the final high y=0m

[tex]y=y_{o}+v_{oy}*t-\frac{1}{2}gt^{2}[/tex]

If we evaluate the equation:

[tex]0=75m+(20m/s)t-\frac{1}{2}(9.8m/s^{2})t^{2}[/tex]

This is a classic form of Quadratic Formula, we can solve it using:

[tex]t=\frac{-b ± \sqrt{b^{2}-4ac } }{2a}[/tex]

[tex]a=-4.9\\b=20\\c=75[/tex]

[tex]t=\frac{-(20) + \sqrt{(20)^{2}-4(-4.9)(75) } }{2(-4.9)}=-2.37s[/tex]

[tex]t=\frac{-(20) - \sqrt{(20)^{2}-4(-4.9)(75) } }{2(-4.9)}=6.4534s[/tex]

Since the time can not be negative, the reasonable answer is

[tex]t=6.4534s[/tex]

The rock will hit the ground approximately 2.37 seconds after it is thrown.

Given:

- Initial height, [tex]\( y_0 = 75 \)[/tex] m

- Initial velocity, [tex]\( v_0 = 20 \)[/tex] m/s (since the rock is thrown upwards, this velocity will be negative in the equation as it is in the opposite direction to the positive y-axis)

- Acceleration due to gravity, [tex]\( g = 9.8 \) m/s\( ^2 \)[/tex]

The motion equation becomes:

[tex]\[ 0 = 75 - 20t - \frac{1}{2}(9.8)t^2 \][/tex]

Multiplying through by 2 to clear the fraction, we get:

[tex]\[ 0 = 150 - 40t - 9.8t^2 \][/tex]

Rearranging the terms to align with the standard quadratic form [tex]\( at^2 + bt + c = 0 \)[/tex], we have:

[tex]\[ 9.8t^2 + 40t - 150 = 0 \][/tex]

Now we can apply the Quadratic Formula to solve for t:

[tex]\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]

Where ( a = 9.8 ), ( b = 40 ), and ( c = -150 ). Plugging in these values, we get:

[tex]\[ t = \frac{-40 \pm \sqrt{40^2 - 4(9.8)(-150)}}{2(9.8)} \][/tex]

Solving this, we get two values for t, but only the positive value is physically meaningful since time cannot be negative.

 Calculating the discriminant:

[tex]\[ \sqrt{40^2 - 4(9.8)(-150)} = \sqrt{1600 + 5880} = \sqrt{7480} \][/tex]

Now, we find the two possible values for t:

[tex]\[ t = \frac{-40 \pm \sqrt{7480}}{2(9.8)} \] \[ t = \frac{-40 \pm 86.52}{19.6} \][/tex]

The two solutions for t are:

[tex]\[ t_1 = \frac{-40 + 86.52}{19.6} \approx 2.37 \text{ s} \] \[ t_2 = \frac{-40 - 86.52}{19.6} \approx -6.43 \text{ s} \][/tex]

Since time cannot be negative, we discard [tex]\( t_2 \)[/tex] and take [tex]\( t_1 \)[/tex] as the time when the rock hits the ground.

The time taken is 2.37 s.

Answer:

The average velocity of the car is 32 m/s.

Explanation:

Given that,

Initial position = 50 m

Final position = 210 m

Time = 5 sec

(a). We need to calculate the average velocity of the car

Using formula of average velocity

[tex]v_{avg}=\dfrac{v_{f}-v_{i}}{t}[/tex]

Put the value into the formula

[tex]v_{avg}=\dfrac{210-50}{5}[/tex]

[tex]v_{avg}=32\ m/s[/tex]

(b). We need to draw the position vs. time graph

We need to calculate the slope of the line of best fit.

Using formula of slop

[tex]slope =\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}[/tex]

Put the value into the formula

[tex]slope =\dfrac{210-50}{5-0}[/tex]

[tex]slope =32[/tex]

Hence, The average velocity of the car is 32 m/s.

Answer: a) close together

Explanation: The electric field lines also  represent the intensity of the field, in this sense  for strong electric fields it is usually draw the lines close to each other. In constrast when they are far apart the electric field is weak.  

Final answer:

The electric field lines are close together when the electric field strength is large, indicating a stronger field due to the direct proportionality between field strength and the density of the field lines. The correct answer is a) close together.

Explanation:

When the electric field strength is large, electric field lines are close together. This is because the strength of the electric field is directly proportional to the number of lines per unit area perpendicular to the field lines. The more electric field lines there are in a given area, the stronger the electric field at that point.

Hence, near a positive charge or around a negative charge, where the field lines begin or terminate, the field lines are densely packed when the charge is large, indicating a strong electric field. Field lines provide a visual representation of the field's strength and direction, and they will never intersect because the electric field at a point must have a unique direction.

Answer:

The answer is  a=b=c=3.833 cm

Explanation:

Lets call the variables a=side a b=side b c=side c

We have that the formula of the equilateral triangle for its height is:

1)h=(1/2)*root(3)*a

2) If we resolve the equation we have

 2.1)2h=root(3)*a

 2.2)(2h/root(3))=a

3) After the replacement of each value we have that

a=2*3.32/1.73205

a=3.833 cm

And we know that the equilateral triangle has the same value for each side so a=b=c=3.833 cm

Answer:

Explanation:

The standard equation of the sinusoidal wave in one dimension is given by

[tex]y = A Sin\left ( \frac{2\pi }{\lambda }\left ( vt-x \right )+\phi  \right )[/tex]

Here, A be the amplitude of the wave

λ be the wavelength of the wave

v be the velocity of the wave

Φ be the phase angle

x be the position of the wave

t be the time

this wave is travelling along positive direction of X axis

The frequency of wave is f which relates with velocity and wavelength as given below

v = f x λ

The relation between the time period and the frequency is

f = 1 / T.

Answer:

- 273.77 rad/s^2

Explanation:

fo = 3800 rev/min = 3800 / 60 rps = 63.33 rps

f = 0

ωo = 2 π fo = 2 x 3.14 x 63.33 = 397.71 rad/s

ω = 2 π f = 0

θ = 46 revolutions = 46 x 2π radian = 288.88 radian

Let α be the angular acceleration of the centrifuge

Use third equation of motion for rotational motion

[tex]\omega^{2}=\omega _{0}^{2}+2\alpha \theta[/tex]

[tex]0^{2}=397.71^{2}+2 \times \alpha \times 288.88[/tex]

α = - 273.77 rad/s^2

Answer:

4.66 x 10^7 N/C

Explanation:

radius of drop, r = 35 micro meter = 35 x 10^-6 m

number of electrons, n = 26

density of water, d = 1000 kg / m^3

The gravitational force is equal to the product of mass of drop and the acceleration due to gravity.

Mass of drop, m = Volume of drop x density of water

[tex]m = \frac{4}{3}\times 3.14 \times r^{3}\times d[/tex]

[tex]m = \frac{4}{3}\times 3.14 \times \left ( 35\times10^{-6} \right )^{3}\times 1000[/tex]

m = 1.795 x 10^-10 kg

Gravitational force, Fg = m x g = 1.795 x 10^-10 x 9.8 = 1.759 x 10^-9 N

Charge,q = number of electrons x charge of one electron

q = 236 x 1.6 x 10^-19 = 3.776 x 10^-17 C

Electrostatic force, Fe = q E

Where, E is the strength of electric field.

here electrostatic force is balanced by the gravitational force

Fe = Fg

3.776 x 106-17 x E = 1.759 x 10^-9

E = 4.66 x 10^7 N/C

Answer:

[tex]r_{al}=2.598cm[/tex]

Explanation:

Density =mass/volume

[tex]D_{al}=2.70103kg/m^{3}[/tex]

[tex]D_{iron}=7.86103/m^{3}[/tex]

condition for balance:

[tex]M_{iron}=M_{al}[/tex]

M=D*Volum

then:

[tex]D_{iron}*4/3*pi*r_{ir}^{3}=D_{al}*4/3*pi*r_{al}^{3}[/tex]

[tex]r_{al}=r_{ir}*\sqrt[3]{\frac{D_{iron}}{D_{al}}}[/tex]

[tex]r_{al}=2.598cm[/tex]

Answer:

1.28 cm

Explanation:

Density (ρ) is an intensive property resulting from the quotient of mass (m) and volume (V).

The density of Al is ρAl = 2.70 × 10³ kg/m³

The density of Fe is ρFe = 7.86 × 10³ kg/m³

A Fe sphere of radius 1.82 cm has the following volume.

V = 4/3 × π × r³

V = 4/3 × π × (0.0182 cm)³

V = 2.53 × 10⁻⁵ m³

The mass corresponding to this sphere is:

2.53 × 10⁻⁵ m³ × 2.70 × 10³ kg/m³ = 0.0683 kg

The Al sphere has the same mass, so its volume is:

0.0683 kg × (1 m³/7.86 × 10³ kg) = 8.69 × 10⁻⁶ m³

The radius corresponding to an Al sphere of 8.69 × 10⁻⁶ m³ is:

V = 4/3 × π × r³

8.69 × 10⁻⁶ m³ = 4/3 × π × r³

r = 0.0128 m = 1.28 cm

Answer:

a) 10.2 eV

b) 122 nm

Explanation:

a) First we must obtain the energy for each of the states, which is given by the following formula:

[tex]E_{n}=\frac{-13.6 eV}{n^2}[/tex]

So, we have:

[tex]E_{1}=\frac{-13.6 eV}{1^2}=-13.6 eV\\E_{2}=\frac{-13.6 eV}{2^2}=-3.4 eV[/tex]

Now we find the energy that the electron loses when it falls from state 2 to state 1, this is the energy carried away by the emitted photon.

[tex]E_{2}-E_{1}=-3.4eV-(-13.6eV)=10.2eV[/tex]

b) Using the Planck – Einstein relation, we can calculate the wavelength of the photon:

[tex]E=h\nu[/tex]

Where E is the photon energy, h the Planck constant and  [tex]\nu[/tex] the frequency.

Recall that [tex]\nu=\frac{c}{\lambda}[/tex], Rewriting for [tex]\lambda[/tex]:

[tex]E=\frac{hc}{\lambda}\\\lambda=\frac{hc}{E}\\\lambda=\frac{(4.13*10^{-15}eV)(3*10^8\frac{m}{s})}{10.2eV}=1.22*10^{-7}m[/tex]

Recall that [tex]1 m=10^9nm[/tex], So:

[tex]1.22*10^{-7}m*\frac{10^9nm}{1m}=122nm[/tex]

We have that for the Question "What is the minimum charge required to pick up the tissue paper?" it can be said that minimum charge is

From the question we are told

Generally the equation for the Electro static force  is mathematically given as

[tex]\frac{kqq'}{r^2}=mg\\\\Therefore\\\\q=\frac{5*10^{-3}*9.8*(0.06^)^2}{9*10^{9}*14*10^{-6}}\\\\q=1.4*10^{-9}C\\\\q=1.4mc\\\\[/tex]

q=0.0014\muC

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Answer:

1. Atoms are the smallest particles of an element that retain the element's chemical properties.

2. Substances composed of only one type of atom are classified as elements.

Explanation:

Statement 3 is wrong because out of the 118 identified elements, the first 94 of them occur naturally on Earth and the remaining 24 are synthetic elements.

Final answer:

Atoms are the smallest units retaining an element's properties, and elements consist exclusively of one type of atom. The provided statement about natural occurrence of elements is incorrect, as 92 elements, not 48, occur naturally.

Explanation:

Regarding the student's question on the properties of atoms and elements:

Atoms are the smallest particles of an element that retain the element's chemical properties. For instance, a hydrogen atom possesses all properties of the element hydrogen.

Substances made up of only one type of atom are considered elements. Each element has atoms with a characteristic mass and identical chemical properties.

The statement about the occurrence of elements is partially incorrect. Although 118 elements have been identified, some sources suggest that 92 occur naturally, not just 48.