The water behind Grand Coulee Dam is 1000 m wide and 200 m deep. Find the hydrostatic force on the back of the dam. (Hint: the total force = average pressure × area)

Answer:

they meet from point o at distance 50.46 m and time taken is 11.6 seconds

Explanation:

given data

acceleration = 0.75 m/s²

speed B = 6 m/s

time B = 20 s

to find out

when and where the vehicles passed each other

solution

we consider here distance = x , when they meet after o point

and time = t for meet point z

we find first Bus B distance for 20 s ec

distance B = velocity × time

distance B = 6 × 20

distance B = 120 m

so

B take time to meet is calculate by distance formula

distance = velocity × time

120 - x = 6 × t

x = 120 - 6t   .................1

and

distance of A when they meet by distance formula

distance = ut + 1/2 × at²

here u is initial speed = 0 and t is time

x = 0 + 1/2 × 0.75 × t²

x = 0.375 × t²      .............2

so from equation 1 and 2

0.375 × t²  = 120 - 6t

t = 11.6

so time is 11.6 second

and

distance from point o from equation 2

x = 0.375 (11.6)²

x = 50.46

so distance from point o is 50.46 m

Automobile A and bus B pass each other approximately 11.6 seconds after automobile A started, at a position 50.5 meters from point O. This is determined by setting their displacement equations equal and solving the quadratic equation. The final position of the meeting point is found to be around 50.5 meters from point O.

To solve this problem, we need to determine the position and time at which automobile A and bus B pass each other.

Determine the position of automobile A:

Determine the position of bus B:

Set the positions equal to find when they pass each other:

Multiplying through by 8 to clear the decimal:

We have two potential solutions: t = (69.62 / 6) ≈ 11.6 seconds (since a negative time is not meaningful in this context)

Determine the meeting point:

Answer:

voltage across capacitor [tex]V_C=i\times X_C=0.022\times 0.04=9\times 10^{-4}V[/tex]

Voltage across resistor [tex]V_R=i\times R=0.022\times 2000=80V[/tex]

Explanation:

We have given resistance R = 2000 OHM

Capacitance [tex]C=1000\mu F=0.001F[/tex]

Voltage V = 45 volt

Frequency = 4000 Hz

Capacitive reactance [tex]X_C=\frac{1}{\omega C}=\frac{1}{2\times \pi \times 4000\times 0.001}=0.04ohm[/tex]

Impedance [tex]Z=\sqrt{R^2+X_C^2}=\sqrt{2000^2+0.04^2}=2000ohm[/tex]

Current [tex]i=\frac{V}{Z}=\frac{45}{2000}=0.022A[/tex]

Now voltage across capacitor [tex]V_C=i\times X_C=0.022\times 0.04=9\times 10^{-4}V[/tex]

Voltage across resistor [tex]V_R=i\times R=0.022\times 2000=80V[/tex]

The electrostatic force between an electron and a proton separated by 0.1mm, calculated using Coulomb's law, is approximately 2.3 × 10^-20N, and it is attractive due to the opposite charges.

The electrostatic force between charged particles can be calculated using Coulomb's law: F = k(Q1*Q2)/r², where F is the force, k is Coulomb's constant (~8.99 * 10^9 Nm^2/C^2), Q1 and Q2 are the charges, and r is the separation in meters. An electron and a proton have charges of -1.6*10^-19 C and +1.6*10^-19 C respectively. Plugging these values into Coulomb's law with a separation of 0.1 mm or 0.1*10^-3 meters, we find that the force is approximately 2.3 × 10^-20N, and it is attractive because the charges are opposite.

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"The correct answer is c. [tex]\(2.3 \times 10^{-18}\)[/tex] N, attractive.

To find the electrostatic force between an electron and a proton separated by a distance, we use Coulomb's law, which is given by:

[tex]\[ F = k \frac{|q_1 q_2|}{r^2} \][/tex]

The charge of an electron is approximately [tex]\(-1.602 \times 10^{-19}\) C[/tex]and the charge of a proton is approximately [tex]\(+1.602 \times 10^{-19}\) C[/tex]. The separation distance is given as 0.1 mm, which we need to convert to meters: [tex]\(0.1 \times 10^{-3}\)[/tex] m or [tex]\(1 \times 10^{-4}\)[/tex] m.

Plugging in the values, we get:

[tex]\[ F = (8.9875 \times 10^9) \frac{(1.602 \times 10^{-19})(1.602 \times 10^{-19})}{(1 \times 10^{-4})^2} \] \[ F = (8.9875 \times 10^9) \frac{2.5664 \times 10^{-38}}{1 \times 10^{-8}} \] \[ F = (8.9875 \times 10^9) (2.5664 \times 10^{-30}) \] \[ F = 2.3017 \times 10^{-19} \][/tex]

Since the force is attractive (opposite charges attract), the magnitude of the force is [tex]\(2.3 \times 10^{-19}\) N[/tex], but we need to express it in the format given in the options, which is [tex]\(2.3 \times 10^{-18}\) N[/tex].

Answer:

(a) A = [tex]3.90 \AA[/tex]

(b) [tex]A = 4.50 \AA[/tex]

(c) [tex]A = 5.51 \AA[/tex]

(d) [tex]A = 9.02 \AA[/tex]

Solution:

As per the question:

Radius of atom, r = 1.95 [tex]\AA = 1.95\times 10^{- 10} m[/tex]

Now,

(a) For a simple cubic lattice, lattice constant A:

A = 2r

A = [tex]2\times 1.95 = 3.90 \AA[/tex]

(b) For body centered cubic lattice:

[tex]A = \frac{4}{\sqrt{3}}r[/tex]

[tex]A = \frac{4}{\sqrt{3}}\times 1.95 = 4.50 \AA[/tex]

(c) For face centered cubic lattice:

[tex]A = 2{\sqrt{2}}r[/tex]

[tex]A = 2{\sqrt{2}}\times 1.95 = 5.51 \AA[/tex]

(d) For diamond lattice:

[tex]A = 2\times \frac{4}{\sqrt{3}}r[/tex]

[tex]A = 2\times \frac{4}{\sqrt{3}}\times 1.95 = 9.02 \AA[/tex]

The average speed for the entire trip is 83.6 km/h. The average velocity for the entire trip is zero.

The average speed for the entire trip can be calculated by finding the total distance traveled divided by the total time taken. In this case, the student traveled 39.0 km to school and then returned the same distance back home, resulting in a total distance of 78.0 km. The total time taken for the entire trip is the sum of the time taken to drive to school, the time taken to return home, and the time spent finding the report, which is 23.0 min + 23.0 min + 10.0 min = 56.0 min. To convert the total time to hours, divide by 60: 56.0 min / 60 = 0.933 hours. Therefore, the average speed for the entire trip is 78.0 km / 0.933 hours = 83.6 km/h.

The average velocity for the entire trip can be determined by considering both the magnitude and direction of the displacement. In this case, since the student drove in the same direction to school and back home, the displacement is zero, meaning there was no change in position. Thus, the average velocity for the entire trip is also zero.

Explanation:

when a person jumps on the trampoline he stores his potential energy in the form of elastic energy of trampoline. Potential Energy converts into Elastic energy when a person is at the bottom point during stretching of the trampoline. When he again going upwards, Elastic energy is converted to the potential energy of the person. This is the reason why a person goes higher each time. This process goes on until trampoline reaches its elastic limit and finally breaks or get permanently deformed.

So there is a limit up to which a person can be reached.

Final answer:

A person goes higher on a trampoline by landing and pushing off with their feet, making better use of leg strength to transform kinetic to potential energy. There is a maximum height due to energy losses like air resistance and imperfect energy transfers. The sweet spot in tennis rackets illustrates efficient energy transfer with reduced arm jarring.

Explanation:

A person on a trampoline goes higher with each bounce because they can harness the elastic potential energy stored in the trampoline material. When landing on the back or feet, the height reached can differ.

Typically, a person may reach greater heights when landing and launching off their feet because they can make use of the stronger leg muscles to add upward force, thus converting more kinetic energy to gravitational potential energy upon ascent. There is, indeed, a maximum height obtainable, as energy losses due to factors like air resistance and less-than-perfect energy transfer prevent infinite increases in height.

Addressing the professional application, the 'sweet spot' on a tennis racket is an example of an optimal point where energy transfer from the racket to the ball is the most efficient, and vibrating force transmitted to the player's arm is minimal, hence no jarring of the arm.

The physics of motion and energy conservation provides us the information that increasing the initial speed (kinetic energy) of the object would result in a higher ascent, yet this is bounded by practical limitations such as the strength of the jumper and the efficiency of the trampoline material.

For example, to double the impact speed of a falling object, one would have to quadruple the height from which it falls, due to the relationship between gravitational potential energy and kinetic energy.

Answer:

a) The average velocity is v = (60 km/h ; 0)

b) The average speed is 60 km/h

Explanation:

The velocity is a vector that has a magnitude and direction. The average speed is the distance traveled over time without taking into account the direction of the motion.

a)The average velocity is calculated as the displacement over time:

v = Δx/Δt

where

v = velocity

Δx = final position - initial position = traveled distance relative to the center of the reference system.

Δt = final time - initial time (initial time is usually = 0)

We know that the displacement is 84 km but we do not know the time. It can be calculated from the two parts of the trip.

In part 1:

v = 45 km/h = 42 km / t

t = 0.93 h

In part 2:

v = 90 km/h = 42 km / t

t = 42 km / 90 km/h

t = 0.47 h

The time of travel is 0.47 h + 0.93 h = 1.4 h

The average velocity will be:

v = 84 km / 1.4 h = 60 km/h

Expressed as a vector in a 2-dimension plane:

v = (60 km/h; 0)

b) The average speed is calculated as the distance traveled over time. Note that in this case, the distance is equal to the displacement since the direction of the motion is always in one direction. But if the direction of the second part of the trip would have been the opposite to the direction of the first part, the displacement would have been 0 (final position - initial position = 0, because final position = initial position), then, the average velocity would have been 0. In change, the average speed would have been the distance traveled (84 km, 42 km in one direction and 42 km in the other) over time.

Then:

average speed = 84 km / 1.4 h = 60 km/h

c) see attached figure.    

Answer:

B)  x-component of the total force exerted on the third charge by the other two (Fn₃x)

Fn₃x =8,56*10⁻⁵ N (+x)

B)  y-component of the total force exerted on the third charge by the other two (Fn₃y)

Fn₃y = 5.7*10⁻⁵ N  (-y)

Explanation:

Theory of electrical forces

Because the particle q₃ is close to two other electrically charged particles, it will experience two electrical forces and the solution of the problem is of a vector nature.

Equivalences:

1nC= 10⁻⁹C

1cm = 10⁻²m

Data

q₁=4.96 nC = +4.96*10⁻⁹C

q₂=-1.99 nC = -1.99*10⁻⁹C

[tex]d_{1} =\sqrt{ 4.01^{2}+2.98^{2}  } = 4.996 cm = 4.996*10^{-2} m=49.96*10^{-3} m[/tex]

d₂= 2.98 cm

Graphic attached

The directions of the individual forces exerted by q₁ and q₂ on q₃ are shown in the attached figure.

The force (F₁₃) of q₁ on q₃ is repulsive because the charges have equal signs.

 the force F₂₃ of q₂ and q₃ is attractive because the charges have opposite signs.

Calculation of the forces exerted on the charge q₁ and q₂ on q₃

To calculate the magnitudes of the forces exerted by the charges q₁, q₂, on q₃ we apply Coulomb's law:

F₁₃=k*q₁*q₃/d₁² =9*10⁹*4.96*10⁻⁹*5.99*10⁻⁹/(49.96*10⁻³)²=10.7**10⁻⁵ N

F₂₃=k*q₂*q₃/d₂² =9*10⁹*1.99*10⁻⁹*5.99*10⁻⁹/(2.98*10⁻²)²=12.08*10⁻⁵ N

F₁₃x=F₁₃cosβ=10.7*10⁻⁵* (4.01/4.996)=8,56*10⁻⁵ N  (+x)

F₁₃y=F₁₃sinβ= 10.7*10⁻⁵* (2.98/4.996)=6,38*10⁻⁵ N  (+x)

F₂₃x= 0

F₂₃y= F₂₃=12.08*10⁻⁵ N (-y)

A) x-component of the total force exerted on the third charge by the other two (Fn₃x)

Fn₃x= F₁₃x + F₂₃x= 8,56*10⁻⁵ N + 0

Fn₃x =8,56*10⁻⁵ N (+x)

B)  y-component of the total force exerted on the third charge by the other two (Fn₃y)

Fn₃y = F₁₃y + F₂₃y= 6,38*10⁻⁵ N - 12.08*10⁻⁵ N

Fn₃y = 5.7*10⁻⁵ N  (-y)

The charges, 4.96 nC, -1.99 nC, and 5.99 nC, forming a triangle gives

the following approximate values of the force at the 5.99 nC charge.

The given charges are;

Q₁ = 4.96 nC, at point (0, 0)

Q₂ = -1.99 nC, at point (4.01, 0)

Q₃ = 5.99 nC  at point (4.01, 2.98)

According to Coulomb's Law, we have;

[tex]F_{13} = \mathbf{\dfrac{9 \times 10^9 \times \left(4.96 \times 10^{-9} \right)\times \left(5.99\times 10^{-9} \right)}{4.01^2 + 2.98^2} } }\approx 1.07 \times 10^{-4}[/tex]

F₁₃ ≈ 1.07 × 10⁻⁴ N

The components of the force are;

[tex]cos\left(arctan \left(\dfrac{2.96}{4.01} \right) \right) \times 1.07 \times 10^{-4} \, \mathbf{\hat i} + sin\left(arctan \left(\dfrac{2.96}{4.01} \right) \right) \times 1.07 \times 10^{-4} \, \mathbf{\hat j}[/tex]

Which gives;

[tex]\vec{F_{13}} \approx \mathbf{8.6 \times 10^{-5} \, \mathbf{\hat i} + 6.39 \times 10^{-5} \, \mathbf{\hat j}}[/tex]

Therefore;

[tex]F_{23}= \dfrac{9 \times 10^9 \times \left((-1.99)\times 10^{-9} \right)\times \left(5.99\times 10^{-9} \right)}{2.98^2} } \approx\mathbf{ 1.208 \times 10^{-4}}[/tex]

Which gives;

[tex]\vec{F_{23}} \approx \mathbf{-1.208 \times 10^{-4} \, \mathbf{\hat j}}[/tex]

The components of the force at Q₃ is therefore;

The magnitude of the total force is therefore;

|F| ≈ √((8.6×10⁻⁵)² + (-5.69 × 10⁻⁵)²) ≈ 1.03 × 10⁻⁴

The direction of the total force is found as follows;

[tex]The \ direction, \ \theta \approx \mathbf{arctan \left(\dfrac{-5.69}{8.6} \right)} \approx -33.5 ^{\circ}\alpha[/tex]

Learn more about Coulomb's Law on electric force here

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Final answer:

The volume of the air bubble just before it reaches the surface is 3.32 cm^3.

Explanation:

To find the volume of the air bubble just before it reaches the surface, we can use the ideal gas law equation: PV = nRT. Since the pressure and amount of gas remain constant, we can rewrite the equation as V/T = k, where V is the volume, T is the temperature, and k is a constant.

Using the given temperatures at the bottom and top of the lake (5.9°C and 16.0°C) and the initial volume of the bubble (1.22 cm^3), we can set up the following equation:

(1.22 cm^3) / (5.9°C) = V / (16.0°C).

Solving for V, the volume of the bubble just before it reaches the surface, we get:

V = (1.22 cm^3)(16.0°C) / (5.9°C) = 3.32 cm^3.

Final answer:

To calculate the change in output power handling capability of a power electronics package, you can use the formula Power = Output power / Efficiency. By comparing the original power dissipation with the new power dissipation, you can determine the change in output power. Efficiency is important because it affects the amount of input power required and the output power capability of a system.

Explanation:

To find out the change in output power handling capability, we need to calculate the new power dissipation and compare it with the original value. The formula for power is given as:

Power = Output power / Efficiency

Using the formula, we can calculate the original power dissipation:

Original Power = 400W / 0.89 = 449.43W

Next, we can calculate the new power dissipation:

New Power = 400W / 0.94 = 425.53W

The change in output power handling capability is the difference between the original power and the new power:

Change in Output Power = 449.43W - 425.53W = 23.9W

The increase in efficiency from 89% to 94% leads to a decrease in the power dissipation by 23.9W. Efficiency is important because it determines the amount of input power required to achieve a certain output power. Higher efficiency means less power is wasted as heat, resulting in a higher output power capability for a given input power.

To find the position, velocity, and acceleration of the snowboarder at t = 12 seconds, evaluate the given displacement equation and its first and second derivatives at t = 12 seconds, then round the final answers to one decimal place.

To determine the position, velocity, and acceleration of the snowboarder at t = 12 seconds, we need to evaluate the given displacement function x(t) = 0.5t³ + t² + 2t and its derivatives at that specific time value.

Position at t = 12 seconds

By substituting t = 12 into the displacement function, we get:
x(12) = 0.5(12)³ + (12)² + 2(12), which can be calculated to give the position x.

Velocity at t = 12 seconds

The velocity v(t) is the first derivative of the displacement function, v(t) = 1.5t² + 2t + 2. Evaluate v(12) to find the velocity at t = 12 seconds.

Acceleration at t = 12 seconds

Acceleration a(t) is the second derivative of the displacement function, a(t) = 3t + 2. Evaluate a(12) to find the acceleration at t = 12 seconds.

Using these steps, you can calculate the exact values and round them to one decimal place as asked in the question.

Answer:

The power is 0.73 kW.

(B) option is correct.

Explanation:

Given that,

Charge = 9000 C

Potential difference = 220 V

Time = 45 min

We need to calculate the energy required to transfer charge Q through V

Using formula of energy

[tex]E =QV[/tex]

Put the value into the formula

[tex]E=9000\times220[/tex]

[tex]E=1980000\ J[/tex]

We need to calculate the power

Using formula of power

[tex]P=\dfrac{1980000}{45\times60}[/tex]

[tex]P=733.33\ W[/tex]

[tex]P=0.73\ kW[/tex]

Hence, The power is 0.73 kW.

Answer:

Image distance is -52.5 cm

Image is virtual and forms on the same side of the lens and upright image is formed.

Explanation:

u = Object distance

v = Image distance

f = Focal length = 35

m = Magnification = 2.5

[tex]m=-\frac{v}{u}\\\Rightarrow 2.5=-\frac{v}{u}\\\Rightarrow v=-2.5 u[/tex]

Lens equation

[tex]\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{35}=\frac{1}{u}+\frac{1}{-2.5u}\\\Rightarrow \frac{1}{35}=\frac{3}{5u}\\\Rightarrow u=21\ cm[/tex]

[tex]v=-2.5\times 21=-52.5\ cm[/tex]

Image distance is -52.5 cm

Image is virtual and forms on the same side of the lens and upright image is formed.

Answer:

[tex][F]=[MLT^{-2}][/tex]

Explanation:

Newton’s second law states that the acceleration a of an object is proportional to the force F acting on it is inversely proportional to its mass m. The mathematical expression for the second law of motion is given by :

F = m × a

F is the applied force

m is the mass of the object

a is the acceleration due to gravity

We need to find the dimensions of force. The dimension of force m and a are as follows :

[tex][m]=[M][/tex]

[tex][a]=[LT^{-2}][/tex]

So, the dimension of force F is, [tex][F]=[MLT^{-2}][/tex]. Hence, this is the required solution.

Final answer:

In physics, the dimensions of force are derived from Newton's second law of motion, which can be represented as F=ma. Force has dimensions of mass (M), length (L), and time (T), leading to the dimensional formula [F] = [M][L][T]^-2. The SI unit for force is the Newton (N).

Explanation:

According to Newton's second law of motion, the acceleration a of an object is directly proportional to the net external force Fnet acting upon it and inversely proportional to its mass m. This law is mathematically represented by the equation Fnet = m × a, where Fnet is the net force, m is the mass, and a is the acceleration of an object.

The weight w of an object is another type of force, which is defined as the gravitational force acting on an object with mass m. The object experiences an acceleration due to gravity g, and this is represented by the equation w = m × g. In the International System of Units (SI), force is measured in Newtons (N), and one Newton is defined as the force required to accelerate a one-kilogram mass at a rate of one meter per second squared (1 N = 1 kg × m/s2).

Therefore, the dimensions of force in terms of the base physical quantities are mass (M), length (L), and time (T), and the dimensional formula for force is [F] = [M][L][T]-2.

Answer:

3.22 s

Explanation:

The ball would describe a projectile motion, where the horizontal velocity will remain constant, as there are no forces that act on the x-axis, and the vertical velocity will vary because of gravity in the following way:

1. First, the ball will go up, but the vertical velocity will decrease until it has a value of 0.

2. After the vertical velocity has reached the value of 0, the ball will start to fall, with the vertical velocity increasing because of gravity.

You need to know that the time that the ball's verical velocity takes to reach 0 is exactly the same that it takes to go from 0 to its original vertical velocity:

[tex]a = \frac{v_f - v_o}{t}\\t = \frac{v_f - v_o}{a}[/tex]

And not only the time will be the same, but also the distance traveled. Therefore, we can conclude that the time that the ball remain in the air is simply two times the time it takes for the ball to desacelerate:

[tex]t_{desaceleration} = \frac{v_f - v_o}{a} = \frac{0m/s - 15.8m/s}{-9.81 m/s^2} =1.61 s\\t_{air} = 2* t_{desaceleration} = 2*1.61s = 3.22 s[/tex]

Answer:

[tex]X - Xo = 54m[/tex]

k = 1/18

Explanation:

Data:

a = -k[tex]t^{2}[/tex][tex]\frac{m}{s^{2} }[/tex]

to = 0s    Vo = 12m/s

t = 6s the particle chage it's moviment, so v = 0 m/s

We know that acceleration is the derivative of velocity related to time:

[tex]a = \frac{dV}{dT}[/tex]

rearranging...

[tex]a*dT = dV[/tex]

Then, we must integrate both sides:

[tex]\int\limits^f_i {dV} \, dV =-k \int\limits^f_i {t^{2} } \, dT[/tex]

[tex]V - Vo = -k\frac{t^{3} }{3}[/tex]

V = 0 because the exercise says that the car change it's direction:

[tex]0 - 12 = -k\frac{6^{3} }{3}[/tex]

k = 1/6

In order to find X - Xo we must integer v*dT = dX

[tex]V - Vo = -k\frac{t^{3} }{3}[/tex]

so...

[tex](Vo -k\frac{t^{3} }{3})dT = dX[/tex]

[tex]\int\limits^f_i {dX} \, dX = \int\limits^f_i {Vo -k\frac{t^{3} }{3} } \, dT[/tex]

integrating...

[tex]X - Xo = Vot -k\frac{t^{4} }{12}[/tex]

[tex]X - Xo = 12*6 -\frac{1}{6}* \frac{6^{4} }{12}[/tex]

X - Xo = 54m

Answer:

rock A: [tex]y=90-1/2*g*t^2[/tex]

rock B: [tex]y=30*t-1/2*g*t^2[/tex]

g=9.81m/s^2

Explanation:

Kinematics equation:

[tex]y=y_{o}+v_{oy}*t+1/2*a*t^2[/tex]

in our case the acceleration is the gravity and it has a negative direction.

a=-g

rock A, yo=90m, Voy=0m/s:

[tex]y=90-1/2*g*t^2[/tex]

rock B, yo=0m, Voy=30m/s:

[tex]y=30*t-1/2*g*t^2[/tex]

Answer:

The mass of the steam is 91.2 g

Mass of the steam=91 grams

Explanation:

Given:

When steam is mixed with the ice then the heat loss by the steam will be gained by the ice so there will no overall heat gain or loss during the mixing

So According to question

Let M be the mass of the steam mixed with ice then we have

[tex]M\times2256\times10^3+M\times4186\times(100-33)=0.499\times222\times10^3+0.499\times4186\times(33-0)\\M\times2.58\times10^6=2.35\times10^5\\M=91.2\ \rm g[/tex]

Answer:

The no. of electrons is [tex]7.59\times 10^{21}[/tex]

Solution:

According to the question:

The rate at which the charge is delivered is given by:

[tex]\frac{dQ}{dt} = - 0.75[/tex]

Now,

[tex]\int_{0}^{Q}dQ = - 0.75\int_{0}^{27 min} dt[/tex]

[tex]Q = -0.75t|_{0}^{27 min}[/tex]

[tex]Q= -0.75\times 27\times 60 = - 1215 C[/tex]

No. of electrons, n can be calculated from the following relation:

Q = ne

where

e = electronic charge =[tex]1.6\times 10^{- 19} C[/tex]

Thus

[tex]n = \frac{Q}{e}[/tex]

[tex]n= \frac{1215}{1.6\times 10^{- 19}}[/tex]

[tex]n = 7.59\times 10^{21}[/tex]

Answer:

81.42 km/h

Explanation:

t = Time taken for the car to slow down

u = Initial velocity = 100 km/h

v = Final velocity

s = Displacement = 20 m = 0.02 km

a = Acceleration = -6.5 m/s² = -0.0065×60×60×60×60 = -84240 km/h²

Equation of motion

[tex]v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times -84240\times 0.02+100^2}\\\Rightarrow v=81.42\ km/h[/tex]

Speed of the truck at the end of this distance is 81.42 km/h

Using the equation of motion v^2 = u^2 + 2as, and converting the initial speed from km/hr to m/s, the final velocity is approximately 15.22 m/s when deceleration and distance are considered. Converted back to km/hr, this is approximately 54.8 km/hr.

The final speed of the truck can be calculated using one of the equations of motion, specifically v^2 = u^2 + 2as, where 'v' is final velocity, 'u' is initial velocity, 'a' is acceleration (which in this case is -6.50 m/s^2 due to deceleration), and 's' is distance. Convert the initial speed from km/hr to m/s: 100 km/hr = 27.78 m/s.

Then, we can plug the values into the formula: v^2 = (27.78 m/s)^2 - 2(6.50 m/s^2)(20.0 m). Solving the equation gives a final velocity of approximately 15.22 m/s. This speed in km/h is then obtained by converting m/s back, 15.22 m/s * (3600 / 1000) = 54.8 km/h. So, the speed of the truck at the end of the distance is approximately 54.8 km/h.

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Answer:

B) 33 m C) 27 m

Explanation:

considering that the two pucks are sliding toward each other we can understand that they are on a collision course.

Since the total distance between them is 26 m, the common sense dictates that the distance traveled by each puck must be less than 26 m regardless of the speed of the two pucks.

so the options B) 33 m and C) 27 m are definitely wrong since they are greater than 26 m.

We can also easily find the distance traveled by each pucks also.

let [tex]v_{A}[/tex] and [tex]v_{B}[/tex] be the velocity of the pucks A and B respectively

[tex]v_{A}t = 26- v_{B}t\\\\2.30t = 26 - 3.90t\\\\6.2t = 26\\\\t = \frac{26}{6.2} \\\\x_{A} =v_{A}t\\\\x_{A} =2.3 \times \frac{26}{6.2}\\\\x_{A} =9.65\\\\x_{B} =v_{B}t\\\\x_{B} =3.9 \times \frac{26}{6.2}\\\\x_{B} =16.35\\[/tex]

Answer:

a) The car doesn't hit the barrier because after he sees the lights of a barrier he only travels for 27.95m, enough to miss the barrier. b) The maximum speed at which the car could be moving and not hit the barrier 40 meters ahead is 21m/s or 75.6km/hr.

Explanation:

a)

In order to solve this problem we must first do a drawing of the situation for us to visualize it better. (See picture attached).

As you may see on the drawing, the car still travels some distance when the driver notices the lights of the barrier. This distance is calculated for a constant velocity V:

x=Vt

the car has a velcity of 60km/h which is equivalent to:

[tex]\frac {60km}{hr} * \frac{1000m}{1km}*\frac{1hr}{3600s}=16.667m/s[/tex]

so on the first 0.80s the car travels a distance of:

x=(16.667m/s)(0.8s)=13.333m

Next, the car breaks, so we can say it moves with a constant acceleration of -9.5m/s, so the distance is found by using the following formula:

[tex]x=\frac {V_{f}^{2}-V_{0}^{2}}{2a}[/tex]

We know the final velocity will happen when the car stops, so the final velocity is zero, leaving us with:

[tex]x=\frac {-V_{0}^{2}}{2a}[/tex]

so we can substitute the provided values to find the distance traveled by the car during this time.

[tex]x=\frac{-(16.667m/s)^{2}}{2(-9.5m/s^{2})}[/tex]

which yields:

x=14.62m

so in total the car traveled:

[tex]x_{tot}=13.333m+14.62m = 27.95m[/tex]

Which is not enough for the car to hit the barrier.

b)

In order to solve this part of the problem, we must combine the two equations we got on the previous part to find a single equation that will represent the total displacement of the car:

[tex]x=vt-\frac{v^{2}}{2a}[/tex]

so we can now substitute the known values so we get:

[tex]40=0.8v-\frac{v^{2}}{2(-9.5)}[/tex]

which simplifies to:

[tex]40=0.8v+0.05263v^{2}[/tex]

this is a quadratic equation so it can be solved by using the quadratic formula, but first we must rewrite it in standard form, so we get:

[tex]0.05263v^{2}+0.8v-40=0[/tex]

now we can use the quadratic formula:

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

So we can substitute the given values:

[tex]v=\frac{-0.8\pm\sqrt{(0.8)^{2}-4(0.05263)(-40)}}{2(0.05263)}[/tex]

which returns two answers:

v=21m/s  and   v=-36.20m/s

We take the positive answer since is the one that represents a moving towards the right side of the drawing.

So the maximum speed at which the car could be moving and not hit the barrier 40 meters ahead is 21m/s

Answer:

Truck's speed = 5.21 m/s

Car's speed = 20.2 m/s

Explanation:

Given:

Mass of truck = M = 1650 kg

Speed of the truck initially = U = 15 m/s

Mass of the car = m = 779 kg

Initial speed of the car =u = 0

From the momentum conservation, Total initial momentum = Total final momentum.

M V+m U = M V +m v

⇒ (1650)(15) + 779×0 = (1650)V + 779 v

⇒ 24750 = 1650 V+779 v →(1)

Since the collision is elastic, relative velocity of approach = relative velocity of separation. 15 = v - V

⇒ v =V + 15; This is now substituted in the equation(1) above.

24750 = 1650 V + (799) (V+15)

⇒ 24750 = 1650 V + 799 V + 11985

⇒ 2449 V = 12765

⇒ Final velocity of the truck = [tex]\frac{12765}{2449}[/tex] = 5.21 m/s

Final velocity of the car = v = V+15 = 5.21 + 15 = 20.2 m/s

The student's average speed on her trip to university was 0.78 km/min or 46.67 km/h, while the magnitude of her average velocity was 0.57 km/min or 34.3 km/h.

In this scenario, we have to find the average speed and the magnitude of her average velocity. Average speed is calculated by dividing the total distance travelled by the total time taken. So, her average speed is 14.0 km / 18.0 min = 0.78 km/min or 46.67 km/h (note: convert minutes to hours for speed in km/h).

The next part of the question involves average velocity, which is the displacement (straight-line distance) divided by time. It can be different from the average speed when the path travelled is not a straight line. In this case, the magnitude of her average velocity is 10.3 km / 18.0 min = 0.57 km/min or 34.3 km/h.

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Answer:

0.21 m/s/s.

Explanation:

Whenever there is an action force acting on a body, there will be a reaction force.

Here the force with which the astronaut throws the tool is given as 16 N.

Force is measured in newtons and is equal to the rate of change of momentum.

Since the astronaut has a mass, she experience a reaction force. It is given by F = ma, according to Newton's 2nd law.

16 = 75 a

⇒ Acceleration = a = F/m = 16/75 = 0.21 m/s/s

Answer:

It will take 33 seconds to stop the car.

Explanation:

Using the first equation of kinematics we have

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

where

'v' is final speed of object

'u' is initial speed of object

'a' is acceleration of object

't' is time of acceleration of object

Now since it is given that [tex]a=-2km/s^{2}[/tex] since acceleration is negative  and [tex]u=66km/s[/tex]

We know that the object will stop when it's velocity reduces to zero hence in the equation above setting v = 0 we get

[tex]0=66-2\times t\\\\\therefore t=\frac{66}{2}=33seconds[/tex]

Final answer:

Using the kinematic equation for velocity and acceleration, it is calculated that the car, decelerating at a rate of 2 km/s² from an initial speed of 66 km/s, will take 33 seconds to come to a complete stop.

Explanation:

To find out how long it will take the car to come to a complete stop, we need to use the kinematic equation that relates initial velocity, acceleration, and time without displacement:

Final velocity (v) = Initial velocity (u) + (Acceleration (a) × Time (t))

Here, the final velocity (v) is 0 km/s, since the car is coming to a stop, the initial velocity (u) is 66 km/s, and the acceleration (a) is -2 km/s² (negative because it is deceleration).

The equation becomes:

0 = 66 + (-2 × t)
We can solve for t as follows:

0 = 66 - 2t

2t = 66

t = 33 seconds

So it will take the car 33 seconds to come to a complete stop.

Answer:16.59 m

Explanation:

Given

initial horizontal speed of ball(u)=15 m/s

Height of building =6 m

Consider vertical motion first

[tex]h=u_yt+\frac{1}{2}gt^2[/tex]

here initial vertical velocity is zero

[tex]6=0+\frac{1}{2}\times 9.81\times t^2[/tex]

[tex]t=\sqrt{\frac{12}{9.81}}=1.106 s[/tex]

Thus time taken will also be 1.106 s in horizontal motion

[tex]R_x=u_xt+\frac{1}{2}at^2[/tex]

here a=0

[tex]R_x=15\times 1.106=16.59 m[/tex]

The total time that the ball is in the air can be calculated using the formula for free fall, which is 1.10 seconds. Given that time and the constant horizontal speed, we can then calculate the total horizontal distance traveled by the ball, which is 16.5 meters.

To calculate the horizontal distance the ball travels before striking the ground, we must first determine how long the ball is in the air. This time is only affected by the vertical motion, thus we can treat the ball as if it were dropped from the hill with no initial horizontal velocity. The formula used to calculate the time is based on free fall, where t = sqrt((2*h)/g). In this case, h = 6.0 m (the height of the hill) and g = 9.8 m/s² (the acceleration due to gravity). This gives us t = sqrt((2*6)/9.8) = 1.10 s.

Once we have the time of flight, we can calculate the horizontal distance traveled using the horizontal speed of the ball. As the horizontal motion occurs at a constant velocity we can use the formula d = v*t, where v = 15 m/s (the speed the ball is thrown) and t = 1.10 s. This gives us the horizontal distance d = 15*1.10 = 16.5 m.

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Answer:

E=[tex]k*\frac{q}{21}*u[/tex]

[tex]u=\frac{1}{\sqrt{21}} *(-1,-2,-4)m[/tex]

[tex]u_{y}=\frac{-2}{\sqrt{21}} m[/tex]

Explanation:

q: particle's charge

k: coulomb constant

E=E*u

r=r*u

r=distancia vectorial entre P y S

r=distancia escalar entre P y S

E: Electric field vector

E: magnitud of magnetic field vector

u: unit vector radial

then:

[tex]E=k*q/r^{2}[/tex]

r=r*u

r=P-S=(-1,-2,-4)m

[tex]r^{2}=(Magnitude(P-S))^2=(-1)^2+(-2)^2+(-4)^2=21[/tex]

[tex]r=\sqrt{21}[/tex]

E=[tex]k*\frac{q}{21}*u[/tex]

u=r/r=[tex]\frac{1}{\sqrt{21}} *(-1,-2,-4)m[/tex]

[tex]u_{y}=\frac{-2}{\sqrt{21}} m[/tex]

Answer:

the building is 34.408 m tall

Explanation:

given,

initial velocity of brick = 15.1 m/s

at an angle of = 25.7°

vertical component of the velocity

 Vy = 15.1 sin 25.7°            

       = 6.54 m/s                      

we know                                

[tex]s = u t + \dfrac{1}{2} a t^2[/tex]

[tex]s = 6.54\times 3.4 + \dfrac{1}{2}\times -9.8 \times 3.4^2[/tex]

s = -34.408 m

hence, the building is 34.408 m tall .

Answer:

The components of the moving frame is (8.07c, -2, 3, 9.493)

Solution:

As per the question:

Velocity of moving frame w.r.t original frame [tex]v_{m}[/tex] 0.85c

Point 'a' of an event in one reference frame corresponds to the (x, y, z, t) coordinates of the plane

a = (0, - 2, 3, 5)

Now, according the the question, the coordinates of moving frame, say (X, Y, Z, t'):

New coordinates are given by:

X = [tex]\frac{x - v_{m}t}{\sqrt{1 - \frac{v_{m}^{2}}{c^{2}}}}[/tex]

X = [tex]\frac{0 - 0.85c\times 5}{\sqrt{1 - \frac{(0.85c)^{2}}{c^{2}}}}[/tex]

X = [tex]8.07 c[/tex]

Now,

Y = y = - 2

Z = z = 3

Now,

[tex]t' = \frac{t - \frac{vx}{c}^{2}}{\sqrt{1 - (\frac{v}{c})^{2}}}[/tex]

[tex]t' = \frac{5 - 0}{\sqrt{1 - (\frac{0.85c}{c})^{2}}} = 9.493 s[/tex]