A 19.2 kg person climbs up a uniform ladder
with negligible mass. The upper end of the
ladder rests on a frictionless wall. The bottom
of the ladder rests on a floor with a rough
surface where the coefficient of static friction
is 0.1 . The angle between the horizontal and
the ladder is θ . The person wants to climb
up the ladder a distance of 0.49 m along the
ladder from the ladder’s foot.
What is the minimum angle θmin (between
the horizontal and the ladder) so that the
person can reach a distance of 0.49 m without
having the ladder slip? The acceleration of
gravity is 9.8 m/s^2
Answer in units of ◦

Answer:

π*R²*E

Explanation:

According to the definition of electric flux, it can be calculated integrating the product E*dA, across the surface.

As the electric field E is uniform and parallel to the hemisphere axis,  and no charge is enclosed within it, the net flux will be zero, so, in magnitude, the flux across the opening defining the hemisphere, must be equal to the one across the surface.

The flux across the open surface can be expressed as follows:

[tex]\int\ {E} \, dA = E*A = E*\pi *R^{2}[/tex]

As E is constant, and parallel to the surface vector dA at any point, can be taken out of the integral, which is just the area of the surface, π*R².

⇒Flux = E*π*R²

Final answer:

The electric flux through a hemisphere surface is calculated using Gauss's Law, where the electric field lines are parallel to the area element vector on the surface. The flux through a closed spherical surface is independent of its radius. Choosing a Gaussian surface with appropriate symmetry aids in evaluating the flux integral.

Explanation:

The magnitude of the electric flux through the hemisphere surface can be calculated using Gauss's Law. For a hemispherical surface of radius R in a uniform electric field,

Assuming the centripetal force is provided by the Coulomb attraction between the electron and the proton we have that,

[tex]F_c = k \frac{e^2}{R^2}[/tex]

Here,

k = Coulomb's Constant

R = Distance

e = Electron charge

And by the centripetal force,

[tex]F_c = m_e R\omega^2[/tex]

[tex]m_e[/tex] = Mass of electron

R = Radius

[tex]\omega[/tex] = Angular velocity

Equation both expression,

[tex]k \frac{e^2}{R^2} = m_e R\omega^2[/tex]

Replacing,

[tex]R^3 = \frac{(9*10^9)(1.6*10^{-19})^2}{(9.11*10^{-31})(10^{6})^2}[/tex]

[tex]R^3 = 2.52908*10^{-10} m[/tex]

[tex]R = 6.32394*10^{-4}m[/tex]

Therefore the radius of the electron orbit is [tex]6.32394*10^{-4}m[/tex]

We will use the trigonometric ratios to know the helicopter's deceleration speed. Later applying the concept of speed as a vector component of the value found we will find the vertical speed.

The 2% grade indicates that for every 100 meters traveled in the x direction, there is an ascent / descent of 2 meters. Therefore we will have the relationship

[tex]\rightarrow \frac{y}{x} = \frac{2}{100}[/tex]

Now you would have the value o[tex]v_y = 6.78m/s[/tex]f the angle tangent would be

[tex]tan \theta = \frac{y}{x}[/tex]

[tex]tan \theta = \frac{2}{100}{[/tex]

[tex]tan\theta = 0.02[/tex]

From this relationship we could conclude that the vertical speed would be

[tex]v_y = v*tan\theta[/tex]

[tex]v_y = 339*0.02[/tex]

[tex]v_y = 6.78m/s[/tex]

Therefore the component of the helicopter's velocity perpendicular to the sloping surface of the hill is 6.78m/s

The component of the helicopter's velocity perpendicular to the sloping surface of the hill is approximately 6.78 m/s.

In this scenario, we can analyze the helicopter's velocity vector into two components: the parallel component along the sloping surface of the hill and the perpendicular component. The component of the velocity perpendicular to the sloping surface is equal to the total velocity multiplied by the grade percentage as a decimal.

Given that the helicopter's speed is 339 m/s and the hill has a 2% grade, we can calculate the perpendicular component as follows:

Perpendicular component = Total velocity × Grade percentage
Perpendicular component = 339 m/s × 0.02
Perpendicular component ≈ 6.78 m/s

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Answer:

Time period at the moon will be 2.828 sec

Explanation:

Let initially the length of the pendulum is l and acceleration due to gravity is g

And time period is given T = 2 sec

So time period of the pendulum is equal to [tex]T=2\pi \sqrt{\frac{l}{g}}[/tex]-------------eqn 1

Now on the planet acceleration due to gravity [tex]g_{new}=\frac{g}{2}[/tex]

So time period of the pendulum at this planet [tex]T_{new}=2\pi \sqrt{\frac{l}{g_{new}}}[/tex]--------eqn 2

Now dividing eqn 1 by eqn 2

[tex]\frac{T}{T_{new}}=\sqrt{\frac{\frac{g}{2}}{g}}[/tex]

[tex]T_{new}=\sqrt{2}T=1.414\times 2=2.828sec[/tex]

So time period at the moon will be 2.828 sec

If the acceleration due to gravity decreases by a half, the period of a simple pendulum increases by √2, or approximately 1.41. Therefore, the period of the pendulum on the moon of Planet X, where gravity is half, would be 2.8 seconds.

The period of a simple pendulum in physics is influenced by its length and the acceleration due to gravity (g). It is calculated by the formula: T = 2π√(L/g), where T is the period, L is the length of the pendulum, and g is the gravity. To find the effect of a change in g on the period of the pendulum, we can look at the relationship between them in the formula. Because the formula for the period of a pendulum involves the square root of the gravity, if gravity is reduced by a half, the period will increase by a factor of √2, which is approximately 1.41.

Thus, if the pendulum's period on Planet X was 2.0 s, then on its moon, where gravity is halved, the period will be approximately 2.0 s * 1.41 = 2.8 s.

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Answer:

   v = - 0.096 m/s

Explanation:

given,

mass of the robot, M = 100 Kg

mass of wrench, m = 0.8 Kg

speed of the wrench,v' = 12 m/s

recoil of the robot,v = ?

initial speed of the robot and the wrench is equal to zero

using conservation of motion

(M + m) V = M v + m v'

(M + m) x 0 = 100 x v + 0.8 x 12

100 v = -9.6

   v = - 0.096 m/s

negative sign shows that velocity of robot is in opposite direction of wrench.

hence, the recoil velocity of the robot is equal to 0.096 m/s

Answer:

Please see below as the answer is self-explanatory

Explanation:

Assuming that the electric field is pointing upward this will produce a downward force on the electron. Neglecting the effect of gravity, according to Newton's 2nd Law, the force on the electron due to the field, produces an acceleration, that can be found solving the following equation:

F = me*a = qe*E ⇒ a = qe*E / me

If the electric field is uniform, the acceleration that produces is constant, so, we can use the kinematic equation that relates displacement and acceleration with time:

x = v₀*t + 1/2*a*t² = v₀*t +1/2*(qe*E/me)*t²

Now, for a proton falling, if  the direction of the field  is reversed (pointing downward) it will accelerate the proton downward.

Using the  same reasoning as above, we get the value  of the  acceleration as follows:

F = mp*a = qp*E ⇒ a = qp*E / mp

The equation for displacement is just the following:

x = v₀*t + 1/2*a*t² = v₀*t +1/2*(qp*E/mp)*t²

We know that qe = qp = 1.6*10⁻¹⁹ coul, but mp = 1,836 me, so, for the same displacement, the time must be much less for the electron, that has an acceleration 1,836 times higher.

When both objects fall freely the same distance under  the sole influence of gravity, if the initial velocity is the same, the time must be the  same  also, as the fall time doesn't depend on the mass of the object.

Answer:

Explanation:

We know that the electric force can be expressed as: F=qE. According to Newton's Second Law of Motion, force can also be expressed as: F=ma. Therefore: a=F/m. We can substitute the electric force expression for "F" in this equation. We get: a=qE/m. We can see from this equation that acceleration is inversely proportional to mass and directly proportional to the electric field and charge. Since the electric field is being reversed and since the charges on the proton and electron differ only by the + or - sign respectively, the numerator of this fraction will remain constant in this scenario. The only variables that are effectively changing are the mass and the resultant acceleration. From the inverse relationship of acceleration and mass, we can say that the proton - having a significantly larger mass than the electron - should experience a smaller acceleration, and should thus take longer to fall distance "d." The electron, on the other hand, should experience a greater acceleration due to its significantly smaller mass, and should fall through distance "d" in a shorter amount of time.

Under the influence of gravity (on the surface of the Earth, for example), objects released from the same height should fall freely with the same acceleration at any given time, regardless of mass. It makes sense, however, that subatomic particles interacting with the electric field are hardly affected by gravity, given how weak gravitational forces are on the microscopic scale.

To solve this problem we will start from the constants given on the two celestial bodies, that is, the earth and the moon. From there we will seek the balance of forces, where the gravitational force is zero. That distance will remain incognito and will be the one we will seek to find, this is

Given that,

[tex]M_E =5.97*10^24 kg[/tex]

[tex]m_m = 7.36*10^22 kg[/tex]

[tex]r_E = 6370 km[/tex]

[tex]r_m = 1737 km[/tex]

The distance between Earth and Moon

[tex]x = 384403*10^3 m[/tex]

As the net gravitational force is zero we have that

[tex]\frac{GMm'}{d^2} = \frac{Gmm'}{(x -d)^2}[/tex]

[tex]\frac{M}{d^2} = \frac{m}{(x -d)^2}[/tex]

Replacing with our values

[tex]\frac{(5.97*10^{24} )}{d^2} = \frac{(7.36*10^{22})}{(x -d)^2}[/tex]

[tex]\frac{(x- d)^2}{d^2}= 0.0123[/tex]

[tex]x^2 + d^2 - 2xd = 0.0123 d^2[/tex]

Replacing x as the distance betwen moon and earth,

[tex](384403 x 10^3)^2+d^2-2(384403 x 10^3)d=0.0123 d^2[/tex]

Solving the polynomial we have that

[tex]d =4.3387*10^8 \text{ or } 3.4506*10^{8}[/tex]

So the minimum value is [tex]3.4506*10^{8}m[/tex]

Answer:

1.897825 seconds

Explanation:

L = Length of pendulum = 3.58 m

g = Acceleration due to gravity = 9.81 m/s²

The amplitude of a pendulum is given by

[tex]T=2\pi \sqrt{\dfrac{L}{g}}\\\Rightarrow T=2\pi\sqrt{\dfrac{3.58}{9.81}}\\\Rightarrow T=3.79565\ s[/tex]

It can be seen that the time period does not depend on amplitude

Time period includes the time to go from extreme left to right and back to extreme left.

So, time taken to go from extreme right to left is [tex]\dfrac{T}{2}=\dfrac{3.79565}{2}=1.897825\ s[/tex]

The period of a simple pendulum is determined by its length and the acceleration due to gravity. It is independent of other factors such as mass and amplitude.

The period of a simple pendulum is determined by its length and the acceleration due to gravity. The period is independent of other factors such as mass and amplitude. For a small amplitude of less than 15 degrees, the period of a simple pendulum can be calculated using the formula:

T = 2π√(L/g)

where T is the period, L is the length of the pendulum, and g is the acceleration due to gravity.

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Final answer:

The percent uncertainty of a blood pressure measurement of 120 mm Hg with an uncertainty of ±2 mm Hg is 1.67%. Using this percentage, the uncertainty in a blood pressure measurement of 71 mm Hg would be ±1.2 mm Hg, after rounding to one decimal place.

Explanation:

To calculate the uncertainty in a blood pressure measurement of 71 mm Hg, given the percent uncertainty is the same as that for a measurement of 120 mm Hg with an uncertainty of ± 2 mm Hg, we first find the percent uncertainty of the original measurement.

The percent uncertainty is calculated by dividing the uncertainty by the actual measurement and then multiplying by 100 to get a percentage:

Percent Uncertainty = (Uncertainty / Measurement) × 100

For the measurement of 120 mm Hg, the percent uncertainty is:

(2 mm Hg / 120 mm Hg) × 100 = 1.67%

Assuming the same percent uncertainty for a blood pressure measurement of 71 mm Hg, you would use the same percentage to find the uncertainty:

Uncertainty = Percent Uncertainty × Measurement / 100

Uncertainty in 71 mm Hg = 1.67% × 71 mm Hg / 100 = 1.187 mm Hg

Since we typically round to the nearest whole unit or significant digit the uncertainty would be ± 1.2 mm Hg (rounded to one decimal place).

Answer:

[tex]8\mu C[/tex]

Explanation:

t = Time taken = [tex]2\mu s[/tex]

i = Current = 3 A

q(0) = Initial charge = [tex]2\mu C[/tex]

Charge is given by

[tex]q=\int_0^t idt+q(0)\\\Rightarrow q=\int_0^{2\mu s} 3dt+2\mu C\\\Rightarrow q=3(2\mu s-0)+2\mu C\\\Rightarrow q=6\mu C+2\mu C\\\Rightarrow q=8\mu C[/tex]

The magnitude of the net electric charge of the capacitor is [tex]8\mu C[/tex]

The total power emitted by a radio transmitter can be calculated by first determining the intensity from the electric field and then multiplying by the surface area of the sphere covered by the isotropic radiation.

The total power emitted by a radio transmitter can be calculated by using the relation between the electric field and the intensity of the radiation. The intensity (I) can be obtained by squaring the electric field strength (E) and dividing it by two times the impedance of free space (Z0). The formula is I = E²/2Z0. After finding the intensity, we can multiply it by the surface area of the sphere which is 4πr² (as it's isotropically radiated, it forms a sphere). The formula becomes Power (P) = I * 4πr². Substituting the given values, we calculate the total power.

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Answer:

as soon as they touch the sheet, it returns to its initial position due to the decrease in the load on the rubber ball.

Explanation:

When you rub a foam rod that is an insulator with felt it acquires some loads, when separated these two the barrel is loaded

When approaching the loaded rubber rod, suppose positively, to the aluminum ball that is a metal, the positive charge attracts the free electors of the aluminum until a balance of the electric force and the tension of the wire in which it is hung is established The aluminum foil therefore the aluminum foil approaches the rubber ball, due to the electrostatic attraction, the closer the aluminum wave deflects.

When the ball touches the sheet, the free charges of the metal pass to the insulator neutralizing the existing charges, whereby the electric field and the electric force decrease. Consequently the aluminum sheet descending and moving away from the rubber ball.

 In summary, as soon as they touch the sheet, it returns to its initial position due to the decrease in the load on the rubber ball.

Due do the friction charged develop in the ball, the foil and ball attract.

When the ball and rod contact the charge neutralized and the attraction force decreased, which separate the ball and rod.

When the insulating material are rubbed against each other, the electric charge generate between them.

The effect of friction charge between rod and ball leads to,

Hence due to the friction charged develop in the ball, the foil and ball attract.

When the ball and rod contact the charge neutralized and the attraction force decreased, which separate the ball and rod.

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Answer:

The element X is chlorine.

The element M is Ytterium.

Ytterium (III) chloride is the correct name for [tex]YCl_3[/tex]

Explanation:

Molecules of [tex]X_2, N = 8.92\times 10^{20} [/tex]

Moles of [tex]X_2, n= ?[/tex]

[tex]N=n\times N_A[/tex]

[tex]n=\frac{N}{N_A}=\frac{8.92\times 10^{20}}{6.022\times 10^{23} mol^{-1}}[/tex]

n = 0.001481 mol

Mass of [tex]X_2, m=0.105 g[/tex]

Molar mass of [tex]X_2=M[/tex]

[tex]m=M\times n[/tex]

[tex]M=\frac{m}{n}=\frac{0.105 g}{0.001481 mol}[/tex]

M = 70.87 g/mol

Atomic mass of X = [tex]\frac{70.87 g/mol}{2}=35.435 g/mol[/tex]

X is a chlorine atom.

The compound [tex]MX_3[/tex] consists of 54.47% X by mass.

M' = Molar mass of compound [tex]MX_3[/tex]

Percentage of X in compound = 54.47%

[tex]54.47\%=\frac{3\times 35.435 g/mol}{M'}\times 100[/tex]

M' = 195.16 g/mol

Atomic mass of M = a

[tex]195.16 g/mol=a+3\times 35.435 g/mol[/tex]

[tex]a = 195.16 g/mol-3\times 35.435 g/mol[/tex]

a = 88.86 g./mol

The element M is Ytterium.

Ytterium (III) chloride is the correct name for [tex]YCl_3[/tex]

The element X is Chlorine and the element M is Ytterium.

Ytterium Chloride is the correct name for [tex]MX_3[/tex] that is [tex]YCl_3[/tex]

Number of molecules of [tex]X_2={8.92*10^{20}[/tex]

Number of moles of X:

[tex]n=\frac{8.92*10^{20}}{6.023*10^{23}}[/tex]  

n = 0.001481 mol = 0.105g

Molar mass of [tex]X_2[/tex] molecule

[tex]M= 0.105/0.001481\\\\M=70.87g/mol[/tex]

Therefore molar mass of X atom = molar mass of [tex]X_2[/tex]/2

m = 35.435g/mol

X is a chlorine atom.

Given that the compound  consists of 54.47% X by mass.

Let M' = Molar mass of compound [tex]MX_3[/tex]

Percentage of X in compound = 54.47%

54.47 = {amount of X}/{mola mass of compound}*100

[tex]54.47 = \frac{3*35.435}{M_{'} }*100\\\\M^{'}= 195.16g/mol[/tex]

let Atomic mass of M = A

A = 195.16 - 3 × 35.435

A = 88.86 g/mol

The element M is Ytterium.

Therefore the compound [tex]MX_3[/tex] is Ytterium Chloride [tex]YCl_3[/tex]

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The force constant of the spring is 1.932 N/m. The maximum mass of the glider can be calculated using Newton's second law. The magnitude of the maximum acceleration of the glider can be calculated using the equation for acceleration in SHM.

In simple harmonic motion (SHM), the force exerted by an ideal spring is directly proportional to the displacement from its equilibrium position. Let's calculate the force constant of the spring by using Hooke's Law. We know that the force exerted by the spring is given by F = kx, where F is the force, k is the force constant, and x is the displacement. At the maximum displacement, the force exerted is F = k * 2.05 m. At the minimum displacement, the force exerted is F = k * 1.06 m. Equating these two forces, we get k * 2.05 = k * 1.06. Solving this equation, we find the force constant of the spring to be k = 2.05 / 1.06 = 1.932 N/m.

The maximum mass of the glider can be found by calculating the maximum force that can be exerted by the spring. The maximum force is given by F = k * x, where F is the force, k is the force constant, and x is the maximum displacement. Plugging in the values, we get F = 1.932 N/m * 2.05 m = 3.961 N. Using Newton's second law, F = m * a, where F is the force, m is the mass, and a is the acceleration, we can calculate the maximum mass. Rearranging the equation gives m = F / a. Since the acceleration in SHM is given by a = 4π^2 * x / T^2, where a is the acceleration, x is the amplitude, and T is the period, we can plug in the given values to find the maximum mass.

The magnitude of the maximum acceleration of the glider can be found by using the equation a = 4π^2 * x / T^2, where a is the acceleration, x is the amplitude, and T is the period. Plugging in the given values, we can calculate the magnitude of the maximum acceleration.

Answer:

The number of excess electrons are on the negative surface is [tex]4.80\times10^{10}\ electrons[/tex]

Explanation:

Given that,

Distance =1.5 cm

Side = 22 cm

Electric field = 18000 N/C

We need to calculate the capacitance in the metal plates

Using formula of capacitance

[tex]C=\dfrac{\epsilon_{0}A}{d}[/tex]

Put the value into the formula

[tex]C=\dfrac{8.85\times10^{-12}\times(22\times10^{-2})^2}{1.5\times10^{-2}}[/tex]

[tex]C=0.285\times10^{-10}\ F[/tex]

We need to calculate the potential

Using formula of potential

[tex]V=Ed[/tex]

Put the value into the formula

[tex]V=18000\times1.5\times10^{-2}\ V[/tex]

[tex]V=270\ V[/tex]

We need to calculate the charge

Using formula of charge

[tex]Q=CV[/tex]

Put the value into the formula

[tex]Q=0.285\times10^{-10}\times270[/tex]

[tex]Q=76.95\times10^{-10}\ C[/tex]

Here, the charge on both the positive and negative  plates

[tex]Q=+76.95\times10^{-10}\ C[/tex]

[tex]Q=-76.95\times10^{-10}\ C[/tex]

We need to calculate the number of excess electrons are on the negative surface

Using formula of number of electrons

[tex]n=\dfrac{q}{e}[/tex]

Put the value into the formula

[tex]n=\dfrac{76.95\times10^{-10}}{1.6\times10^{-19}}[/tex]

[tex]n=4.80\times10^{10}\ electrons[/tex]

Hence, The number of excess electrons are on the negative surface is [tex]4.80\times10^{10}\ electrons[/tex]

Answer

given,

position of particle

x(t)= A t + B t²

A = -3.5 m/s

B = 3.9 m/s²

t = 3 s

a)  x(t)= -3.5 t + 3.9 t²

   velocity of the particle is equal to the differentiation of position w.r.t. time.

[tex]\dfrac{dx}{dt}=\dfrac{d}{dt}(-3.5t + 3.9t^2)[/tex]

[tex]v= -3.5 + 7.8 t [/tex]------(1)

velocity of the particle at t = 3 s

  v = -3.5 + 7.8 x 3

 v = 19.9 m/s

b) velocity of the particle at origin

  time at which particle is at origin

  x(t)= -3.5 t + 3.9 t²

   0 = t (-3.5  + 3.9 t )

   t = 0, [tex]t=\dfrac{3.5}{3.9}[/tex]

   t = 0 , 0.897 s

speed of the particle at t = 0.897 s

from equation (1)

 v = -3.9 + 7.8 t

 v = -3.9 + 7.8 x 0.897

  v = 3.1 m/s

To solve the problem we should know about velocity.

Velocity is the rate of change of its position with respect to time.

[tex]V = \dfrac{dy}{dt}[/tex]

Given to us

x(t)= At+Bt²

[tex]V(t) = \dfrac{dy}{dt} = \dfrac{d(At+Bt^2)}{dt} = A+2Bt[/tex]

A.)  the velocity, in meters per second. of the particle at the time t1= 3.0 s,

[tex]V(t) = A +2Bt[/tex]

Substituting the values,

[tex]V(t_1=3) = (-3.5) +2(3.9)(3.0)\\\\V(t_1=3) = 19.9\ m/s[/tex]

B.)  the velocity, in meters per second: of the particle when it is at the origin (x=0) at t ≥ 0

[tex]x(t)= At+Bt^2\\\\0 = At+Bt^2\\\\[/tex]

Taking t as common,

[tex]0 = t(A+Bt)\\\\[/tex]

[tex]0 = (A+Bt)\\\\[/tex]

Substituting the values and solving or t,

[tex]0 = A+ Bt\\0 = -3.5 + (3.9)t\\3.5=3.9t\\t= \dfrac{3.5}{3.9}\\\\t = 0.8974\ s[/tex]

Substituting the value in the formula of velocity,

[tex]V(t) = A +2Bt[/tex]

Substituting the values,

[tex]V(t_1=0.8974) = (-3.5) +2(3.9)(0.8974)\\\\V(t_1=3) = 3.5 \ m/s[/tex]

Hence, the velocity of the particle is 3.5 m/s.

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Answer:

[tex]f=7 \ Hz[/tex]

Explanation:

Simple Harmonic Motion

The spring-mass system is a typical case of a simple harmonic motion, since the distance traveled by the mass describes an oscillatory behaviour. The natural angular frequency of a spring-mass system is computed by

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

And the frequency is

[tex]{\displaystyle f=\frac {w}{2\pi}[/tex]

Thus

[tex]{\displaystyle f =\frac {1}{2\pi}{\sqrt {\frac {k}{m}}}}[/tex]

The total mass of the car and the driver is

[tex]m=1700+66=1766\ kg[/tex]

They both weigh

[tex]W=m.g=1766\ kg*9.8\ m/s^2[/tex]

[tex]W=17306.8\ N[/tex]

We need to know the constant of the spring. It can be found by using the formula of the Hook's law:

[tex]F=k.x[/tex]

We know the spring stretches 5 mm (0.005 m) when holding the total weight of the car and the driver. Solving for k

[tex]\displaystyle k=\frac{F}{x}[/tex]

[tex]\displaystyle k=\frac{17306.8}{0.005}[/tex]

[tex]k=3,461,360\ N/m[/tex]

Thus, the frequency of oscillations is

[tex]{\displaystyle f =\frac {1}{2\pi}{\sqrt {\frac {3,461,360}{1,766}}}}[/tex]

[tex]\boxed{f=7 \ Hz}[/tex]

The frequency of oscillation of a car going over a bump can be calculated using principles of simple harmonic motion. By calculating the spring constant from the car's weight and the amount the springs are compressed, we can find the angular frequency and then the normal frequency of the oscillation.

The question is about understanding the oscillation concepts of a car once it goes over a bump. It can be treated as a simple harmonic oscillator. To calculate the frequency, we need to first find the spring constant (k) based on Hooke's Law which states that the force F on the spring exerted by the car's weight equals k multiplied by the distance the spring is compressed (x), or F = kx.

In this case, the force F is the weight of the car plus the driver: (1700kg + 66kg) * 9.81m/s² gravitational acceleration. The spring compression x is 5.0mm, or 0.005m. Solving F = kx for k gives us the spring constant.

The angular frequency w of oscillation is equal to √(k/m), where m is the total mass: 1700 kg (car mass) + 66 kg (driver mass), and k is the spring constant we found earlier. The frequency (f) can then be found by dividing w by 2π.

This frequency measures the number of oscillations the car makes per second after going over a bump.

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Answer:

1777.92 m/s

Explanation:

R = Radius of asteroid = 545 km

M = Mass of planet

g = Acceleration due to gravity = 2.9 m/s²

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

Acceleration due to gravity is given by

[tex]g=\dfrac{GM}{R^2}\\\Rightarrow M=\dfrac{gR^2}{G}[/tex]

The expression of escape velocity is given by

[tex]v=\sqrt{\dfrac{2GM}{R}}\\\Rightarrow v=\sqrt{\dfrac{2G}{R}\dfrac{gR^2}{G}}\\\Rightarrow v=\sqrt{2gR}\\\Rightarrow v=\sqrt{2\times 2.9\times 545000}\\\Rightarrow v=1777.92\ m/s[/tex]

The escape speed is 1777.92 m/s

Answer:

1.78 km/s

Explanation:

radius, R = 545 km = 545000 m

acceleration due to gravity, g = 2.9 m/s²

The formula for the escape velocity is given by

[tex]v=\sqrt{2gR}[/tex]

[tex]v=\sqrt{2\times 545000\times 2.9}[/tex]

v = 1777.92 m/s

v = 1.78 km/s

Thus, the escape velocity on the surface of asteroid is 1.78 km/s.

Answer:

The speed of electron is [tex]v=4.52\times 10^6\ m/s[/tex] and the speed of proton is 2468.02 m/s.

Explanation:

Given that,

Electric field, E = 560 N/C

To find,

The speed of each particle  (electrons and proton) 46.0 ns after being released.

Solution,

For electron,

The electric force is given by :

[tex]F=qE[/tex]

[tex]F=1.6\times 10^{-19}\times 560=8.96\times 10^{-17}\ N[/tex]

Let v is the speed of electron. It can be calculated using first equation of motion as :

[tex]v=u+at[/tex]

u = 0 (at rest)

[tex]v=\dfrac{F}{m}t[/tex]

[tex]v=\dfrac{8.96\times 10^{-17}}{9.1\times 10^{-31}}\times 46\times 10^{-9}[/tex]

[tex]v=4.52\times 10^6\ m/s[/tex]

For proton,

The electric force is given by :

[tex]F=qE[/tex]

[tex]F=1.6\times 10^{-19}\times 560=8.96\times 10^{-17}\ N[/tex]

Let v is the speed of electron. It can be calculated using first equation of motion as :

[tex]v=u+at[/tex]

u = 0 (at rest)

[tex]v=\dfrac{F}{m}t[/tex]

[tex]v=\dfrac{8.96\times 10^{-17}}{1.67\times 10^{-27}}\times 46\times 10^{-9}[/tex]

[tex]v=2468.02\ m/s[/tex]

So, the speed of electron is [tex]v=4.52\times 10^6\ m/s[/tex] and the speed of proton is 2468.02 m/s. Therefore, this is the required solution.

Answer:

Option-(D):Thermal energy unchanged, temperature decreased.

Explanation:

Thermal energy and Temperature,T:

Thermal energy of the system is remained unchanged while the temperature,T of the objects present inside the system is more associated with the objects fundamental properties which each matter shows or exhibits in nature.

While the thermal energy is the flow or exchange of thermal or heat energy between the different objects in interaction to each other.

Answer:

[tex]T=366.23\ N[/tex]

Explanation:

Given:

Now follow the schematic to understand the symmetry and solution via Lami's theorem.

The weight of the monkey will be balanced equally by the tension in both the vines:

Using Lami's Theorem:

[tex]\frac{w}{sin\ 70^{\circ}} =\frac{T}{sin\ 145^{\circ}}[/tex]

[tex]\frac{600}{sin\ 70^{\circ}} =\frac{T}{sin\ 145^{\circ} }[/tex]

[tex]T=366.23\ N[/tex]

Answer

given,

mass of full back, M = 100 Kg

velocity of full back  in east = 3.5 m/s

mass of the defensive back,m = 80 Kg

velocity of defensive back, due east = 6 m/s

a) initial momentum of each player

  for full back

   P₁ = M v  =  100 x 3.5 = 350 kg.m/s

 for defensive back

   P₂= m v = 80 x 6 = 480 kg.m/s

b) total momentum before collision

     P = P₁ + P₂

taking west direction as positive

     P = 350  + (- 480 )

     P = -130 kg.m/s

c) speed of the them when they stick together.

    using conservation of momentum

   initial momentum = final momentum

    -130 = (M + m ) V

    180 V = -130

    V = -0.722 m/s

velocity after the collision is equal to 0.722 m/s in direction of east.

a) P₁=350 kg.m/s and P₂= 480 kg.m/s

b) Total momentum before collision is -130 kg.m/s.

c) 0.722 m/s in direction of east

Given,

Mass of full back, M = 100 Kg

Velocity of full back  in east = 3.5 m/s

Mass of the defensive back, m = 80 Kg

Velocity of defensive back, due east = 6 m/s

Momentum

It is the product of the mass and velocity of an object. It is a vector quantity, possessing a magnitude and a direction.

To find:

a) initial momentum of each player

For full back  

P₁ = M*v  =  100 * 3.5 = 350 kg.m/s

For defensive back

P₂= m*v = 80*6 = 480 kg.m/s

b) total momentum before collision

P = P₁ + P₂

taking west direction as positive

P = 350  + (- 480 )

P = -130 kg.m/s

c) speed of the them when they stick together.

Using conservation of momentum

initial momentum = final momentum

-130 = (M + m ) V

180 V = -130

V = -0.722 m/s

Velocity after the collision is equal to 0.722 m/s in direction of east.

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Answer:

v = 3.06 x 10⁻³ m/s

Explanation:

given,

mass of man,M = 90 kg

mass of stone , m= 69 g = 0.069 Kg

speed of the stone, u = 4 m/s

speed of the man, v = ?

using conservation of momentum

initial velocity of the man and stone is equal to zero

(M + m)V = M v + m u

(M + m) x 0 = 90 x v + 0.069 x 4

90 v = 0.276

 v = 0.00306 m/s

v = 3.06 x 10⁻³ m/s

speed of the man on the frictionless  surface is equal to 3.06 x 10⁻³ m/s

The man's velocity after shoving the stone is -0.003067 m/s, indicating movement in the direction opposite to the stone, as per the conservation of momentum.

This problem is a classic example of the conservation of momentum, which is a fundamental concept in physics. According to the law of conservation of momentum, if no external forces are acting on a system, the total momentum of the system remains constant. In this scenario, the man and the stone together constitute an isolated system because the surface is frictionless and there are no external forces acting on the system.

The total momentum before the man shoves the stone is zero since both are initially at rest. When the man shoves the stone, the stone acquires momentum in one direction, and the man must acquire momentum in the opposite direction to conserve the total momentum of the system. The magnitude of the momentum gained by the stone and the man will be equal because their initial total momentum was zero.

The man's velocity (v) can be calculated using the formula:

Momentum of stone = mass of stone (m_stone) * velocity of stone (v_stone)

Momentum of man = mass of man (m_man) * velocity of man (v)


Since total momentum is conserved, Momentum of stone = -Momentum of man.

We can then solve for the man's velocity:

(m_stone * v_stone) = -(m_man * v)


To find v, we rearrange the equation:

v = - (m_stone * v_stone) / m_man

Plugging in the given values:

v = - (0.069 kg * 4.0 m/s) / 90 kg

v = - (0.276 kg·m/s) / 90 kg
v = -0.003067 m/s

The negative sign indicates that the man's velocity is in the opposite direction of the stone's velocity.

Answer:

v₂ = -2.35m/s

This is an Inelastic collision

Explanation:

Law of conservation of momentum

This states that for a collision occurring between two object in an isolated system, the total momentum of the two objects before the collision is equal to the total momentum of the two objects after the collision. That is, the momentum lost by object 1 is equal to the momentum gained by object 2.

Given from the question

m₁=40 g

initial velocity of m₁: v₁=2 m/s

m₂=47 g

initial velocity of m₂: v₂ = -5 m/s

final velocity of m₁: v'₁ = -1.11 m/s.

final velocity of  m₂ = ?

second experiment final velocity of m₁ : v₁' = -3.8942 m/s

second experiment final velocity of m₂ : v₂' = 0.0163 m/s

Considering the first experiment

Apply the knowledge of conservation of momentum

[tex] m₁v₁ + m₂v₂ =m₁v₁ + m₂v₂[/tex]

[tex](0.040)(2) + (0.047)(-5) = (0.040)(-1.11) + (0.040)v₂[/tex]

[tex]v₂ = -2.35m/s[/tex]

Considering the second experiment

initial kinetic energy [tex]KEₓ  = (1/2)m₁v₁² + (1/2)m₂v₂²[/tex]

     [tex]KEₓ = (1/2)(0.040)(2)² + (1/2)(0.047)(-5)²[/tex]

     [tex]KEₓ = 0.6675 J[/tex]

Final kinetic energy[tex] KEₙ = (1/2)m₁v₁² + (1/2)m₂v₂² [/tex]

[tex]KEₙ = (1/2) (0.040) (-3.8942)² + (1/2) (0.047) (0.0163)² [/tex]

[tex]KEₙ = 0.3033 J  [/tex]

Loss in kinetic energy  [tex]ΔKE = KEₓ - KEₙ [/tex]

ΔKE = 0.6675 - 0.3033

ΔKE = 0.3642 J

This collision is a perfectly inelastic collision because the maximum  kinetic energy is lost this means that the kinetic energy before the collision is not the same as the kinetic energy after the collision.Though momentum is conserved kinetic energy is not conserved.

The voltage across the capacitor is 20.79 V.

Capacitance is a measure of how much charge can be stored in a capacitor per unit potential difference. The voltage across the capacitor can be calculated using the formula V = Q/C, where V is the voltage, Q is the charge, and C is the capacitance. In this case, the charge is 1.85×10-7 C and the capacitance is 8900 pF. Converting the capacitance to farads, we get C = 8900 × 10-12 F. Plugging in these values, we have V = 1.85×10-7 C / (8900 × 10-12 F), which simplifies to V = 20.79 V.

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Complete Question:

During a car collision, the knee, thighbone, and hip can sustain a force no greater than 4000 N/ Forces that exceed this amount could cause dislocations or fractures. Assume that in a collision a knee stops when it hits the car's dashboard. Also assume that the mass of the body parts stopped by the knee is about 20% of the total body mass, which is 62 kg.

a) What is the minimum stopping time interval in needed to avoid injury to the knee if the person is initially traveling at 15 m/s (34 mi/h)? b) what is the minimum stopping distance?

Answer:

a) t= 46.5 msec. b) 0.35 m

Explanation:

Applying Newton´s 2nd Law to the mass supported by the knee (20% of the total mass), we can get the maximum acceleration allowable in order to avoid an injury, as follows:

a = F/m = 4000 N / 0.2*62 kg = 322.6 m/s²  

Applying the definition of acceleration, and taking into account that the knee finally come to an stop, we have:

a = vf - v₀ / Δt = -15 m/s / Δt

Solving for Δt :

Δt = -15 m/s / -322.6 m/s² = 0.0465 sec = 46.5 msec.

b) Assuming the acceleration remains constant during this time interval, we can find the distance needed to come to an ⇒stop, applying any of the kinematic equations, as this one:

vf² - v₀² = 2*a*Δx

⇒Δx = (vf²-v₀²) / 2*a

⇒Δx = -(15 m/s)² / 2*(-322.6 m/s²) = 0.35 m

This question involves the concepts of the equations of motion and Newton's Second Law of Motion.

a) The minimum stopping time interval needed to avoid knee injury is "0.05 s".

b) The minimum stopping distance is "0.37 m".

a)

First, we will use Newton's Second Law of Motion to find out the acceleration:

[tex]F=ma\\\\a=\frac{F}{m}[/tex]

where,

a = acceleration = ?

m = mass supported by knee = 20% of total mass = (0.2)(65 kg) = 13 kg

Assuming the average mass of a person to be 65 kg.

Force = - 4000 N (reaction force)

Therefore,

[tex]a=\frac{-4000\ N}{13\ kg}\\\\a=-307.7\ m/s^2[/tex] (negative sign shows decelration)

Now, for the minimum time interval, we will use the first equation of motion:

[tex]v_f=v_i+at\\\\t=\frac{v_f-v_i}{a}[/tex]

where,

t = time interval = ?

vf = final speed = 0 m/s

vi = initial speed = 15 m/s

Therefore,

[tex]t=\frac{0\ m/s-15\ m/s}{-307.7\ m/s^2}[/tex]

t = 0.05 s = 50 ms

b)

Now, we will use the second equation of motion to find out the stopping distance:

[tex]s=v_it+\frac{1}{2}at^2\\s=(15\ m/s)(0.05\ s)+\frac{1}{2}(-307.7\ m/s^2)(0.05\ s)^2\\[/tex]

s = 0.75 m - 0.38 m

s = 0.37 m

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The attached picture shows the equations of motion.

Answer:

B. QC > 0; QH < 0

Explanation:

Given that there are two reservoir of energy.

Sign convention for heat and work :

1.If the heat is adding to the system then it is taken as positive and if heat is going out from the system then it is taken as negative.

2. If the work is done on the system then it is taken as negative and if the work is done by the system then it is taken as positive.

From hot reservoir heat is going out that is why it is taken as negative

[tex]Q_H<0[/tex]

From cold reservoir heat is coming inside the reservoir that is why it is taken as positive

[tex]Q_C>0[/tex]

That is why the answer will be

[tex]Q_H<0[/tex] ,[tex]Q_C>0[/tex]

In the context of energy transfers with hot and cold reservoirs, the sign convention is

Given that there are two reservoir of energy. Sign convention for heat and work :

1) If the heat is adding to the system then it is taken as positive and if heat is going out from the system then it is taken as negative.

2) If the work is done on the system then it is taken as negative and if the work is done by the system then it is taken as positive.

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To solve this problem we will apply the considerations of equilibrium on a body, we will decompose the tensions into their respective components and apply the sum of forces and solving equations for the given system.

According to the graph, the sum of the horizontal components would be

[tex]\sum T_x = 0[/tex]

[tex]TSin35\° -TSin35\° = 0[/tex]

[tex]0 = 0[/tex]

The sum of the vertical components is

[tex]\sum T_y = 0[/tex]

[tex]2TCos35\° - mg = 0[/tex]

[tex]2TCos35\° = 600N[/tex]

[tex]T = \frac{600N}{2Cos35\°}[/tex]

[tex]T = 366.2N[/tex]

Therefore the net force on the new vine is 366.2N

Answer:

The detect frequency is 1622.72 Hz.

Explanation:

Given that,

Frequency = 1500 Hz

Suppose you move with a speed of 27.0 m/s toward the fire engine. what frequency do you detect ?

We need to calculate the frequency

Using formula of frequency

[tex]f=(\dfrac{v+v_{0}}{v})f_{0}[/tex]

Where, v = speed of sound

v₀ = speed of source

f₀ = frequency of siren

Put the value into the formula

[tex]f=\dfrac{330+27.0}{330}\times1500[/tex]

[tex]f=1622.72\ Hz[/tex]

Hence, The detect frequency is 1622.72 Hz.

Answer:

[tex]W_{total}=4.5*10^{-8}J[/tex]

Explanation:

Remember that electric potential can be written as:

[tex]V=\frac{kQ}{r}[/tex],

Where V is the electric potential, k is Coulomb's constant, Q is a point charge, and r is the distance from the point charge. Also, we can write the electric potential as:

[tex]V=\frac{W}{q}[/tex],

where W is the work made to move a charge from infinitely far apart to a certain distance, and q the point charge were are moving.

From all this we can get an expression for the work:

[tex]W=\frac{kQq}{r}[/tex]

We are going to let

[tex]q_{1}=-1.00nC\\q_{2}=2.00nC\\q_{3}=3.00nC[/tex]

To take the first charge [tex]q_{1}[/tex] from infinitely far apart to one of the vertices of the triangle, since there is no electric field  and charges, we make no work.

Next, we will move [tex]q_{2}[/tex] . Now, [tex]q_{1}[/tex] is a vertice of the triangle, and we want [tex]q_{2}[/tex] to be 20.cm apart from [tex]q_{1 }[/tex] so the work we need to do is

[tex]W_{12}=\frac{kq_{1}q_{2}}{(0.20)}\\\\W_{12}=\frac{(9*10^{9})(-1*10^{-9})(2*10^{-9})}{0.20}\\\\W_{12}=-9*10^{-8}J[/tex]

Now, we move the last point charge. Here, we need to take in account the electric potential due to [tex]q_{1}[/tex] and [tex]q_{2}[/tex]. So

[tex]W=W_{13}+W_{23}\\\\\\W=\frac{kq_{1}q_{3}}{0.20}+\frac{kq_{2}q_{3}}{0.20}\\\\W=q_{3}k(\frac{q_{1}}{0.20}+\frac{q_{2}}{0.20})\\\\W=(3*10^{-9})(9*10^{9})(\frac{-1*10^{-9}}{0.20}+\frac{2*10^{-9}}{0.20})\\\\[/tex]

[tex]W=1.35*10^{-7}J[/tex]

Now, the only thing left to do is to find the total work, this can be easily done by adding [tex]W_{12}[/tex] and [tex]W[/tex]:

[tex]W_{total}=W_{12}+W\\W_{total}=1.35*10^{-7}-9*10^{-8}\\W_{total}=4.5*10^{-8}J[/tex]