Answer:
a) Acceleration of runner is 1.33 m/s²
b) Acceleration of motorcycle is 2.85 m/s²
c) The motorcycle moves 84.21-2.94 = 81.06 m farther than the runner.
Explanation:
t = Time taken
u = Initial velocity = 0
v = Final velocity
s = Displacement
a = Acceleration
Equation of motion
[tex]v=u+at\\\Rightarrow a=\frac{v-u}{t}\\\Rightarrow a=\frac{2.8-0}{2.1}\\\Rightarrow a=1.33\ m/s^2[/tex]
Acceleration of runner is 1.33 m/s²
[tex]v=u+at\\\Rightarrow a=\frac{v-u}{t}\\\Rightarrow a=\frac{43-37}{2.1}\\\Rightarrow a=2.85\ m/s^2[/tex]
Acceleration of motorcycle is 2.85 m/s²
[tex]v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{2.8^2-0^2}{2\times 1.33}\\\Rightarrow s=2.94\ m[/tex]
The runner moves 2.94 m
[tex]v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{43^2-37^2}{2\times 2.85}\\\Rightarrow s=84.21\ m[/tex]
The motorcycle moves 84.21 m
The motorcycle moves 84.21-2.94 = 81.06 m farther than the runner.
A rocket is fired at a speed of 75.0 m/s from ground level, at an angle of 60.0° above the horizontal. The rocket is fired toward an 11-0m-high wall, which is located 27.0 m away. The rocket attains its launch speed in a negligibly short period of time, after which its engines shut down and the rocket coasts. By how much does the rocket clear the top of the wall?
Answer:
ΔH = 33.17m
Explanation:
By knowing the amount of time it takes the rocket to travel the horizontal 27m, we will be able to calculate the height when x=27m. So:
[tex]X = V_{o}*cos(60)*t[/tex] where [tex]X=27m; V_{o}=75m/s[/tex]
Solving for t:
t=0.72s
Now we calculate the height of the rocket:
[tex]Y_{f}=V_{o}*sin(60)*t-\frac{g*t^{2}}{2} = 44.17m[/tex]
If the wall was 11m-high, the difference is:
ΔH = 33.17m
Using the equation of projectile motion, the rocket cleared the top of the wall by 38.305m.
The equation of projectile motion is given by:
y= xtanθ +(1/2)gx²/(u²cos²θ )
Given information:
Initial velocity, u=75m/s
The angle of projection, θ = 60°
The horizontal distance, x=27m
Vertical distance will be calculated as:
y= xtanθ +(1/2)gx²/(u²cos²θ )
y=27 tan 60° + 0.5×9.8×27²/(75²cos²60°)
y=49.305m
So, the rocket cleared the top of the wall by
49.305-11 =38.305m
Therefore, using the equation of projectile motion the rocket cleared the top of the wall by 38.305m.
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A rock is dropped from the top of a tall building. How high is the building if it takes 3.25seconds for the rock to strike the ground? Ignore the effects of air resistance. A) 169m B)15.9m C)51.8m D) 32.2m
Answer:
C)51.8m
Explanation:
Kinematics equation:
[tex]y=v_{oy}*t+1/2*g*t^2[/tex]
in this case:
[tex]v_{oy}=0[/tex]
[tex]y=1/2*9.81*3.25^2=51.8m[/tex]
Light is incident normally on two narrow parallel slits a distance of 1.00 mm apart. A screen is placed a distance of 1.2 m from the slits. The distance on the screen between the central maximum and the centre of the n=4 bright spot is measured to be 3.1 mm. a Determine the wavelength of light. b This experiment is repeated in water (of refractive index 1.33). Suggest how the distance of 3.1 mm would change, if at all.
Answer:
Explanation:
distance between slits d = 1 x 10⁻³ m
Screen distance D = 1.2 m
Wave length of light = λ
Distance of n th bright fringe fro centre
= n λ D / d where n is order of bright fringe . Here n = 4
Given
3.1 x 10⁻³ = (4 x λ x 1.2) / 1 x 10⁻³
λ = 3.1 x 10⁻⁶ / 4.8
= .6458 x 10⁻⁶
6458 x 10⁻¹⁰m
λ= 6458 A.
The distance will reduce 1.33 times
New distance = 3.1 /1.33
= 2.33 mm.
A piano wire of length 2.5 m vibrates so that one-half wavelength is contained on the string. If the frequency of vibration is 35 Hz, the amplitude of vibration is 3.0 mm, and the density is 20 g/m, how much energy is transmitted per second down the wire?
Answer:
The energy transmitted per second down the wire is 0.761 watt.
Explanation:
Given that,
Length = 2.5 m
Amplitude = 3.0 mm
Density = 20 g/m
Frequency = 35 Hz
We need to calculate the wavelength
Using formula of wavelength
[tex]L = \dfrac{\lambda}{2}[/tex]
[tex]\lambda=2L[/tex]
Put the value into the formula
[tex]\lambda=2\times2.5[/tex]
[tex]\lambda=5\ m[/tex]
We need to calculate the speed
Using formula of speed
[tex]v = f\lambda[/tex]
Put the value into the formula
[tex]v =35\times5[/tex]
[tex]v =175\ m/s[/tex]
We need to calculate the energy is transmitted per second down the wire
Using formula of the energy is transmitted per second
[tex]P=\dfrac{1}{2}\mu A^2\omega^2\times v[/tex]
[tex]P=\dfrac{1}{2}\mu\times A^2\times(2\pi f)^2\times v[/tex]
Put the value into the formula
[tex]P=\dfrac{1}{2}\times20\times10^{-3}\times(3.0\times10^{-3})^2\times4\times\pi^2\times(35)^2\times175[/tex]
[tex]P=0.761\ watt[/tex]
Hence, The energy transmitted per second down the wire is 0.761 watt.
A common parameter that can be used to predict turbulence in fluid flow is called the Reynolds number. The Reynolds number for fluid flow in a pipe is a dimensionless quantity defined as Re = rhovd μ where rho is the density of the fluid, v is its speed, d is the inner diameter of the pipe, and μ is the viscosity of the fluid. Viscosity is a measure of the internal resistance of a liquid to flow and has units of Pa · s. The criteria for the type of flow are as follows. • If Re < 2,300, the flow is laminar. • If 2,300 < Re < 4,000, the flow is in a transition region between laminar and turbulent. • If Re > 4,000, the flow is turbulent. (a) Let's model blood of density 1.06 103 kg/m3 and viscosity 3.00 10-3 Pa · s as a pure liquid, that is, ignore the fact that it contains red blood cells. Suppose it is flowing in a large artery of radius 1.25 cm with a speed of 0.0650 m/s. Show that the flow is laminar. (State the Reynolds number of the flow, which will be less than 2,300, indicating laminar flow.) (b) Imagine that the artery ends in a single capillary so that the radius of the artery reduces to a much smaller value. What is the radius of the capillary that would cause the flow to become turbulent? (Use the minimum Reynolds number for which flow is fully turbulent.)
Answer:
a) Re= 574.17
b) r= 8.71 cm
Explanation:
In order to solve this problem, we will need to use the formula for the Reynolds number:
Re=ρ*v*d/μ
All of the required data is already given in the problem, but before we use the above-mentioned formula, we need to convert the data to SI units, as follows:
The density already has SI units ( 1.06 *10³ kg/m³)The radius is 1.25 cm, which converts to 0.0125 m. Thus, the inner diameter is 0.025 mThe viscosity already has SI units (3.00 * 10⁻³ Pa · s)The speed already has SI units (0.0650 m/s)a) Now we proceed to calculate Reynolds number:
Re=1.06 *10³ kg/m³ * 0.0650 m/s * 0.025 m / (3.00 * 10⁻³ Pa · s)
Re=574.17
Re<2,300 ; thus the flow is laminar.
b) To answer this question we use the same equation, and give the Reynolds number a value of 4,000 in order to find out d₂:
4,000= 1.06 *10³ kg/m³ * 0.0650 m/s * d₂ / (3.00 * 10⁻³ Pa · s)
We solve for d₂:
d₂=0.174 m
Thus the radius of the capillary that would cause the flow to become turbulent is 0.174 m / 2= 0.0871 m or 8.71 cm, given that neither the speed nor the viscosity change.
However, in your question you wrote that the artery ends in a capillary so that the radius reduces its value. But the lower the radius, the lower the Reynolds number. And as such, it would not be possible for the flow to turn from laminar to turbulent, if the other factors (such as speed, or density) do not change.
At the instant the traffic light turns green, a car that has been waiting at an intersection starts ahead (from rest) with a constant acceleration of 2.9 m/s². At the same instant, a truck traveling with a constant speed of 36.3 mph passes the car. How much time in seconds does it take the car to catch up with the truck. Keep 2 decimal places.
Answer:
It take the car to catch up with the truck 111.91 s.
Explanation:
To solve this problem we have to use the formula for uniformly accelerated motion (for the car) and the formula for uniform rectilinear movement (for the truck).
We apply the corresponding formula for each vehicle, so we will have two equations. As the question is how much time, time is the unknown variable that we will call t from now on.
Equation for the car is:
[tex]x_{c}=\frac{1}{2}*a*t^{2}[/tex]
Equation for the truck is
[tex]x_{t} =v*t[/tex]
We know that t will be the same for the two vehicle on the instant the car catch up the truck.
On the time t the distance x traveled for both cars are the same, so we can equate the two formulas ans isolate t.
[tex]v*t=\frac{1}{2} *a*t^{2} \\t=(2*v)/a\\t=111,91s[/tex]
Note: all unit of measurement must be the same, for speed, we need to convert 36,3mph to m/s.
36,3mph=162.27m/s we use 162.27m/s in the formulas.
It will take the car approximately 11.17 seconds to catch up with the truck that passes it traveling at a constant speed when the car starts from rest and accelerates at 2.9 m/s².
Explanation:The question asks how long it takes for a car that starts from rest with a constant acceleration to catch up with a truck traveling at a constant speed. The car accelerates from rest at 2.9 m/s², while the truck travels at a constant speed of 36.3 mph, which is approximately 16.2 m/s (1 mph equals approximately 0.44704 m/s).
To solve this problem, we need to consider that the car and truck will have traveled the same distance when the car catches up to the truck. The equations for distance for the car (with acceleration) and the truck (traveling at constant speed) are:
Distancecar = ½ * acceleration * time²Distancetruck = speed * timeWe set the distances equal to each other:
½ * 2.9 m/s² * time² = 16.2 m/s * time
This gives us a quadratic equation:
½ * 2.9 * time² - 16.2 * time = 0
Factoring out the common term 'time', we get:
time(½ * 2.9 * time - 16.2) = 0
Ignoring the solution where time equals zero (since we want to know how long it takes after they have started moving), we get:
½ * 2.9 * time = 16.2
time = (16.2 / (0.5 * 2.9)) seconds
Solving this, time ≈ 11.17 seconds
Therefore, it will take the car approximately 11.17 seconds to catch up with the truck.
A hot-air balloon has just lifted off and is rising at the constant rate of 2.0m/s. Suddenly one of the passengers realizes she has left her camera on the ground. A friend picks it up and tosses it straight upward with an initial speed of 12m/s. If the passenger is 2.5m above her friend when the camera is tossed, how high is she when the camera reaches her?
To find out how high the passenger is when the camera reaches her, we use kinematic equations, taking into account the initial speed of the camera, the constant rise speed of the passenger, and gravity's acceleration. The solution requires equating the displacements of both camera and passenger to solve for time and therefore the height.
Explanation:A hot-air balloon is rising at a constant rate of 2.0m/s when a passenger's camera is tossed straight upward with an initial speed of 12m/s from a position 2.5m below her. To determine how high the passenger is when the camera reaches her, we can apply kinematic equations of motion, incorporating the constant acceleration due to gravity (approximately 9.81m/s² downwards).
For the camera: Its initial upward velocity is 12 m/s, and it is subject to gravity's acceleration. For the passenger: Rising at a constant 2.0 m/s, not accelerating since the rate is constant. Since the initial distance between them is 2.5 m, we need to calculate when the camera, starting from a lower point but moving faster, reaches the vertically moving passenger.
Using the formula s = ut + 0.5at² for both camera and passenger, where s is the displacement, u is initial velocity, a is acceleration, and t is time, we can set the equations equal to solve for t, then determine the height by applying it to the passenger's motion equation.
Due to the mathematical complexity and potential for variability in solving these equations, the exact numerical solution isn't presented here. However, the approach involves determining the time it takes for the camera to reach the same height as the passenger and using that to find her height at that moment.
Which of the following is always true about the magnitude of a displacement? A. It is less than the distance traveled. B. It is equal to the distance traveled. C. It is less than or equal to the distance traveled. D. It is greater than the distance traveled. E. It is greater than or equal to the distance traveled.
Final answer:
The magnitude of displacement is always less than or equal to the distance traveled. Displacement is the straight line distance between the starting and ending points, while distance is the total path length traveled. So the correct option is C
Explanation:
The magnitude of displacement is always less than or equal to the distance traveled. This is because displacement is the straight-line measurement from the initial point to the final point, regardless of the path taken, and it's a vector quantity with both magnitude and direction. On the other hand, distance is a scalar quantity that represents the total path length traveled without regard to direction. Therefore, if a path is straight and in one direction, the distance and the magnitude of displacement are equal. However, if the path involves changes in direction, the distance will be greater than the magnitude of displacement.
The correct answer to the question 'Which of the following is always true about the magnitude of a displacement?' is C. It is less than or equal to the distance traveled.
A rigid adiabatic container is divided into two parts containing n1 and n2 mole of ideal gases respectively, by a movable and thermally conducting wall. Their pressure and volume are P1, V1 for part 1 and P2, V2 for part 2 respectively. Find the final pressure P and temperature T after the two gas reaches equilibrium. Assume the constant volume specific heats of the two gas are the same.
Answer:
Explanation:
Given
Pressure, Temperature, Volume of gases is
[tex]P_1, V_1, T_1 & P_2, V_2, T_2 [/tex]
Let P & T be the final Pressure and Temperature
as it is rigid adiabatic container therefore Q=0 as heat loss by one gas is equal to heat gain by another gas
[tex]-Q=W+U_1----1[/tex]
[tex]Q=-W+U_2-----2[/tex]
where Q=heat loss or gain (- heat loss,+heat gain)
W=work done by gas
[tex]U_1 & U_2[/tex] change in internal Energy of gas
Thus from 1 & 2 we can say that
[tex]U_1+U_2=0[/tex]
[tex]n_1c_v(T-T_1)+n_2c_v(T-T_2)=0[/tex]
[tex]T(n_1+n_2)=n_1T_1+n_2T_2[/tex]
[tex]T=\frac{n_1+T_1+n_2T_2}{n_1+n_2}[/tex]
where [tex]n_1=\frac{P_1V_1}{RT_1}[/tex]
[tex]n_2=\frac{P_2V_2}{RT_2}[/tex]
[tex]T=\frac{\frac{P_1V_1}{RT_1}\times T_1+\frac{P_2V_2}{RT_2}\times T_2}{\frac{P_1V_1}{RT_1}+\frac{P_2V_2}{RT_2}}[/tex]
[tex]T=\frac{P_1V_1+P_2V_2}{\frac{P_1V_1}{T_1}+\frac{P_2V_2}{T_2}}[/tex]
and [tex]P=\frac{P_1V_1+P_2V_2}{V_1+V_2}[/tex]
Does percent error give indication of accuracy or precision? Discuss.
Answer:
Accuracy
Explanation:
Percent error is the ratio of the difference of the measured and actual value to the actual value multiplied by 100.
It gives the percent deviation of the value obtained from the actual value.
Accuracy is the measure of how close the readings are to the actual value or set standard and can be improved by increase the no. of readings in an experiment.
Precision is the measure of the closeness of the obtained values to one another.
Thus accuracy of the reading can be sensed by the percent error.
Calculate the speed of a proton that has moved in a uniform electric field of 180.0 N/C from rest over a distance of 12.5-cm. Assume it began at rest.
Answer:
Given:
Electric field = 180 N/C
[tex]Force\ on\ proton = 1.6\times10^{-19} C[/tex]
[tex]Force\ on\ proton = 180\times1.6\times10^{-19} =288\times10^{-19} N[/tex]
[tex]Mass\ of\ proton = 1.673\times10^{-27} kg[/tex]
[tex]Acceleration of proton = \frac{force}{mass}[/tex]
[tex]Acceleration\ of\ proton = \frac{288\times10^{-19}}{1.673*10^{-27}} =172\times108 m/s^{2}[/tex]
Let the speed of proton be "x"
x = [tex]\sqrt{Acceleration}[/tex]
[tex]x = \sqrt{(2\times172\times108\times0.125)}=65602.2 m/s[/tex]
Answer:
the velocity of the proton is 65574.38 m/s
Explanation:
given,
uniform electric field = 180 N/C
Distance = 12.5 cm = 0.125 m
charge of proton = 1.6 × 10⁻¹⁹ C
force = E × q
=180 × 1.6 × 10⁻¹⁹
F= 2.88 × 10⁻¹⁷ N
mass of proton = 1.673 × 10⁻²⁷ kg
acceleration =[tex]\dfrac{force}{mass}[/tex]
=[tex]\dfrac{2.88 \times 10^{-17}}{1.673\times 10^{-27}}[/tex]
=1.72 × 10¹⁰ m/s²
velocity = [tex]\sqrt{2\times 0.125 \times 1.72 \times 10^{10}}[/tex]
=65574.38 m/s
hence , the velocity of the proton is 65574.38 m/s
A perfectly flexible cable has length L, and initially it is at rest with a length Xo of it hanging over the table edge. Neglecting friction, compute the length hanging over the table edge after an elapsed time t, assuming cable sections remain straight during the subsequent motion.
Answer:
[tex]X=X_o+\dfrac{1}{2}gt^2[/tex]
Explanation:
Given that
Length = L
At initial over hanging length = Xo
Lets take the length =X after time t
The velocity of length will become V
Now by energy conservation
[tex]\dfrac{1}{2}mV^2=mg(X-X_o)[/tex]
So
[tex]V=\sqrt{2g(X-X_o)}[/tex]
We know that
[tex]\dfrac{dX}{dt}=V[/tex]
[tex]\dfrac{dX}{dt}=\sqrt{2g(X-X_o)}[/tex]
[tex]\sqrt{2g}\ dt=(X-X_o)^{-\frac{1}{2}}dX[/tex]
At t= 0 ,X=Xo
So we can say that
[tex]X=X_o+\dfrac{1}{2}gt^2[/tex]
So the length of cable after time t
[tex]X=X_o+\dfrac{1}{2}gt^2[/tex]
What is the minimum magnitude of an electric field that balances the weight of a plastic sphere of mass 14.2 g that has been charged to 7.7 nC. Give your answer to the nearest 0.1 MN/C (mega Newton per Coulomb)
Answer:
18.1 MN/C
Explanation:
The gravitational force of the plastic sphere is in equilibrium with the electric force.
Mass of the plastic sphere = m = 14.2 g = 0.0142 kg
Force of gravity = F = mg = (0.0142)(9.81) = 0.139 N
This force is balanced by the electric force due to the charge 7.7 nC
Charge = q = 7.7 x 10⁻⁹ C
Electric field = E = F / q = (0.139) /(7.7 x 10⁻⁹) = 18.1 MN/C
A 1100 kg car is traveling around a flat 82.3 m radius curve. The coefficient of static friction between the car tires and the road is .521. What is the maximum speed in m/s at which the car can take the curve?
Answer:
The maximum speed of car will be 20.5m/sec
Explanation:
We have given mass of car = 1100 kg
Radius of curve = 82.3 m
Static friction [tex]\mu _s=0.521[/tex]
We have to find the maximum speed of car
We know that at maximum speed centripetal force will be equal to frictional force [tex]m\frac{v^2}{r}=\mu _srg[/tex]
[tex]v=\sqrt{\mu _srg}=\sqrt{0.521\times 82.3\times 9.8}=20.5m/sec[/tex]
So the maximum speed of car will be 20.5m/sec
Answer:20.51 m/s
Explanation:
Given
Mass of car(m)=1100 kg
radius of curve =82.3 m
coefficient of static friction([tex]\mu [/tex])=0.521
here centripetal force is provided by Friction Force
[tex]F_c(centripetal\ force)=\frac{mv^2}{r}[/tex]
Friction Force[tex]=\mu N[/tex]
where N=Normal reaction
[tex]\frac{mv^2}{r}=\mu N[/tex]
[tex]\frac{1100\times v^2}{82.3}=0.521\times 1100\times 9.81[/tex]
[tex]v^2=0.521\times 9.81\times 82.3[/tex]
[tex]v=\sqrt{420.63}=20.51 m/s [/tex]
When a honeybee flies through the air, it develops a charge of +20 pC . Part A How many electrons did it lose in the process of acquiring this charge? Express your answer as a number of electrons.
The number of electrons lost by the by the honeybee in acquiring the charge of +20 pC is;
n = 1.25 × 10^(8) electrons
We are given;
Charge of honeybee; Q = 20 pC = 20 × 10^(-12) C
Now, formula for number of electrons is;
n = Q/e
Where;
e is charge on electron = 1.6 × 10^(-19) C
Thus;
n = (20 × 10^(-12))/(1.6 × 10^(-19))
n = 1.25 × 10^(8) electrons
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By evaluating their dot product, find the values of the scalar s for which the two vectors ~b = ˆx + syˆ and ~c = ˆx−syˆ are orthogonal. (Remember that two vectors are orthogonal if and only if their dot product is zero.) Explain your answers with a sketch.
Answer:
[tex]s=\pm 1[/tex]
Explanation:
The dot product of two vectors [tex]\vec{a}=a_1\vec{x}+b_1\vec{y}[/tex] and [tex]\vec{b}=a_2\vec{x}+b_2\vec{y}[/tex] is given by
[tex]\vec{a}\cdot\vec{b}=a_1\cdot a_2+b_1\cdot b_2[/tex]
The dot product of two orthogonal vector is always zero thus if [tex]\vec{a}=\vec{x}+s\vec{y}[/tex] and [tex]\vec{b}=\vec{x}-s\vec{y}[/tex], their dot product would be
[tex]\vec{a}\cdot\vec{b}=1\times1+s\times-(s)=1-s^2=0[/tex]
[tex]\implies 1-s^2=0[/tex]
[tex]\implies s^2=1[/tex]
[tex]\implies s=\pm 1[/tex]
Charge g is distributed in a spherically symmetric ball of radius a. (a) Evaluate the average volume charge density p. (b) Now assume p(r) is directly proportional to r. (i) Derive the formula for p(r) in terms of r, Q, and a. At what value of r does p(r)= ? [Ans: 340] (ii) Find q(r), and graph it.
Answer:
Explanation:
The volume of a sphere is:
V = 4/3 * π * a^3
The volume charge density would then be:
p = Q/V
p = 3*Q/(4 * π * a^3)
If the charge density depends on the radius:
p = f(r) = k * r
I integrate the charge density in spherical coordinates. The charge density integrated in the whole volume is equal to total charge.
[tex]Q = \int\limits^{2*\pi}_0\int\limits^\pi_0 \int\limits^r_0 {k * r} \, dr * r*d\theta* r*d\phi[/tex]
[tex]Q = k *\int\limits^{2*\pi}_0\int\limits^\pi_0 \int\limits^r_0 {r^3} \, dr * d\theta* d\phi[/tex]
[tex]Q = k *\int\limits^{2*\pi}_0\int\limits^\pi_0 {\frac{r^4}{4}} \, d\theta* d\phi[/tex]
[tex]Q = k *\int\limits^{2*\pi}_0 {\frac{\pi r^4}{4}} \, d\phi[/tex]
[tex]Q = \frac{\pi^2 r^4}{2}}[/tex]
Since p = k*r
Q = p*π^2*r^3 / 2
Then:
p(r) = 2*Q / (π^2*r^3)
Two 2.0 kg masses are 1.1 m apart (center to center) on a frictionless table. Each has + 9.6 μC of charge. PART A
What is the magnitude of the electric force on one of the masses?
Express your answer to two significant figures and include the appropriate units.
PART B
What is the initial acceleration of the mass if it is released and allowed to move?
Express your answer to two significant figures and include the appropriate units.
Answer:
A) Force = 0.69 N
B) Acceleration = 0.34 m/s^2
Explanation:
The electric force is given by the expression:
[tex]F_e= K *\frac{q_1*q_2}{r^2}[/tex]
K is the Coulomb constant equal to 9*10^9 N*m^2/C^2, q1 and 12 is the charge of the particles, and r is the distance:
[tex]F_e = 9*10^9 Nm^2/C^2 * \frac{(9.6*10^{-6}C)^2}{(1.1m)^2} = 0.69 N[/tex]
Part B.
For the acceleration, you need Newton's second Law:
F = m*a
Then,
[tex]a = \frac{F}{m} = \frac{0.69 N}{2 kg} = 0.34 m/s^2[/tex]
The magnitude of the electric force on one of the masses is 102.71 N. The initial acceleration of the mass is 51.36 m/s^2.
Explanation:PART A:
To find the magnitude of the electric force on one of the masses, we can use Coulomb's Law.
The formula for the magnitude of the electric force is:
F = k * (|q1| * |q2|) / r^2
where F is the force, k is the Coulomb's constant (9 x 10^9 Nm^2/C^2), |q1| and |q2| are the magnitudes of the charges (9.6 μC), and r is the distance between the charges (1.1 m).
Plugging in the values:
F = (9 x 10^9 Nm^2/C^2) * (9.6 μC * 9.6 μC) / (1.1 m)^2
F = 102.71 N
The magnitude of the electric force on one of the masses is 102.71 N.
PART B:
To find the initial acceleration of the mass when it is released and allowed to move, we can use Newton's second law.
The formula for the acceleration is:
a = F / m
where a is the acceleration, F is the force (102.71 N), and m is the mass (2.0 kg).
Plugging in the values:
a = 102.71 N / 2.0 kg
a = 51.36 m/s^2
A 5.0 A current is set up in a circuit for 6.0 min by a rechargeable battery with a 6.0 V emf. By how much is the chemical energy of the battery reduced?
Answer:
The chemical energy of the battery was reduced in 10800J
Explanation:
The first thing to take into account is that the stored energy in a battery is in Watts per second or Joules ([tex]W\cdot s=J[/tex]). It means that the battery provides a power for a certain time.
The idea is to know how much [tex]W\cdot s[/tex] has been consumed by the circuit.
The first step is to know the power that is consumed by the circuit. It is [tex]P=V\cdot I[/tex]. The problem says that the circuit consumes a current of 5.0A with a voltage of 6.0V. It means that the power consumed is:
[tex]P=V\cdot I=(6.0V)\cdot (5.0A)=30W[/tex]
The previous value (30W) is the power that the circuit consumes.
Now, you must find the total amount of power that is consumed by the circuit in 6.0 minutes. You just have to multiply the power that the circuit consumed by the time it worked, it means, 6.0 minutes.
[tex]energy=P\cdot t=(30W)\cdot (6.0min)=180W\cdot min[/tex]
You must convert the minutes unit to seconds. Remember that 1 minute has 60 seconds.
[tex]energy=P\cdot t=(30W)\cdot (6.0min) \cdot \frac{60s}{1min}=10800W\cdot s=10800J[/tex]
Thus, the chemical energy of the battery was reduced in 10800J
A particle of charge Q is fixed at the origin of an xy coordinate system. At t = 0 a particle (m = 0.923 g, q = 4.52 µC is located on the x axis at x = 22.6 cm, moving with a speed of 45.7 m/s in the positive y direction. For what value of Q will the moving particle execute circular motion? (Neglect the gravitational force on the particle.)
Answer:
[tex]Q = -1.43\times 10^[-5} coulomb[/tex]
Explanation:
Given data:
particle mass = 0.923 g
particle charge is 4.52 micro C
speed of particle 45.7 m/s
In this particular case, coulomb attraction will cause centrifugal force and taken as +ve and Q is taken as -ve
[tex]-\frac{Qq}{4\pi \epsilon r^2} = \frac{mv^2}{r}[/tex]
solving for Q WE GET
[tex]Q = -\frac{mv^2}{r} \times r^2 \frac{4\pi \epsilon}{q}[/tex]
[tex]Q = -mv^2\times r \frac{4\pi \epsilon}{q}[/tex]
[tex]Q = - \frac{0.923\times 10^{-3} \times 45.7^2\times (22.6\times 10^{-2})} {4.52\times 10^{-6} \times 9\times 10^9}[/tex]
where[tex] \frac{1}{4\pi \epsilon} = 9\times 10^9[/tex]
[tex]Q = -1.43\times 10^[-5} coulomb[/tex]
Final answer:
To find the charge Q for circular motion, equate centripetal force m * v^2 / r with Coulomb's force k * |Q * q| / r^2, and solve for Q. Use m = 0.923 g, v = 45.7 m/s, q = 4.52 µC, and convert units accordingly.
Explanation:
To determine the value of charge Q that will allow the moving particle to execute circular motion, we use the concept of centripetal force. Centripetal force is the net force required to keep an object moving in a circle at a constant speed and is directed towards the center of the circle. For a charged particle moving in a circular path due to an electric force, the centripetal force is provided by the electric force between the charges.
The centripetal force (Fc) is equal to the mass (m) of the particle times the square of its speed (v) divided by the radius (r) of the circle:
Fc = m * v2 / r.
The electric force (Fe) acting on the particle is given by Coulomb's law:
Fe = k * |Q * q| / r2,
where k is Coulomb's constant (8.99 x 109 Nm2/C2), Q is the charge at the origin, q is the charge of the moving particle, and r is the separation between the charges.
Setting the centripetal force equal to the electric force yields:
m * v2 / r = k * |Q * q| / r2,
Solving for Q gives us:
Q = m * v2 / (k * q).
Plugging in the values:
Q = (0.923 g * 45.7 m/s2) / (8.99 x 109 Nm2/C2 * 4.52 µC)
Remembering to convert grams to kilograms and microcoulombs to coulombs, the final calculation will yield the required charge Q for circular motion.
Q = 1.03 mC
A muon has a kinetic energy equal to 4 times its rest energy of 105 MeV. (a) What is its velocity, in units of c?
(b) What is its momentum in energy units (i.e., units of MeV/c)?
Answer:
v = 0.9798*c
Explanation:
E0 = 105 MeV
The mass of a muon is
m = 1.78 * 10^-30 kg
The kinetic energy is:
[tex]Ek = \frac{E0}{\sqrt{1 - \frac{v^2}{c^2}}}-E0[/tex]
The kinetic energy is 4 times the rest energy.
[tex]4*E0 = \frac{E0}{\sqrt{1 - \frac{v^2}{c^2}}}-E0[/tex]
[tex]4 = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}-1[/tex]
[tex]5 = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}[/tex]
[tex]\sqrt{1 - \frac{v^2}{c^2}} = \frac{1}{5}[/tex]
[tex]1 - \frac{v^2}{c^2} = \frac{1}{25}[/tex]
v^2 / c^2 = 1 - 1/25
v^2 / c^2 = 24/25
v^2 = 24/25 * c^2
v = 0.9798*c
A 100 meter rope is 20 kg and is stretched with a tension of 20 newtons. If one end of the rope is vibrated with small amplitude at 10Hz, what would the velocity of waves traveling down it be? What would the velocity be if it rained and the rope soaked up 5 kg of water?
Answer:
The velocity waves before rain is 10 m/s
The velocity of wave after the rope soaked up 5 kg more is 8.944 m/s
Solution:
As per the question:
Length of the rope, l = 100 m
Mass of the rope, m = 20 kg
Force due to tension in the rope, [tex]T_{r} = 20 N[/tex]
Frequency of vibration in the rope, f = 10 Hz
Extra mass of the rope after being soaked in rain water, m' = 5 kg
Now,
In a rope, the wave velocity is given by:
[tex]v_{w} = \sqrt{\frac{T_{r}}{M_{d}}}[/tex] (1)
where
[tex]M_{d}[/tex] = mass density
Mass density before soaking, [tex]M_{d} = \frac{m}{l} = \frac{20}{100} = 0.20[/tex]
Mass density after being soaked, [tex]M_{d} = \frac{m + m'}{l} = \frac{25}{100} = 0.25[/tex]
Initially, the velocity is given by using eqn (1):
[tex]v_{w} = \sqrt{\frac{20}{0.20}} = 10 m/s[/tex]
The velocity after being soaked in rain:
[tex]v_{w} = \sqrt{\frac{20}{0.25}} = 8.944 m/s[/tex]
A heat pump has a coefficient of performance that is 60% of the Carnot heat pump coefficient of performance. The heat pump is used to heat a home to 24.0°C during the winter with the low temperature reservoir at the outdoor temperature. At which outdoor temperature would it be more efficient to add the energy directly to the interior of the home than use it to run the heat pump? -154°C -40.0°C -4.00°C -83.4°C -25.2°C
Answer:[tex]T_L=-154.2^{\circ}[/tex]
Explanation:
Given
COP= 60 % of carnot heat pump
[tex]COP=\frac{60}{100}\times \frac{T_H}{T_H-T_L}[/tex]
For heat added directly to be as efficient as via heat pump
[tex]Q_s=W[/tex]
[tex]COP=\frac{Q_s}{W}=\frac{60}{100}\times \frac{T_H}{T_H-T_L}[/tex]
[tex]1=\frac{60}{100}\times \frac{T_H}{T_H-T_L}[/tex]
[tex]1=\frac{60}{100}\times \frac{24+273}{24+273-T_L}[/tex]
[tex]T_L=118.8 K[/tex]
[tex]T_L=-154.2^{\circ}[/tex]
A fireworks shell is accelerated from rest to a velocity of 55.0 m/s over a distance of 0.210 m. (a) How long (in s) did the acceleration last? s
(b) Calculate the acceleration (in m/s2). (Enter the magnitude.) m/s^2
Answer:
a) The acceleration took 0.0076s
b) The aceleration was of 7202.4 m/s^2
Explanation:
We need to use the formulas for acceleration movement in straight line that are:
(1) [tex]a = \frac{V}{t}[/tex] and (2)[tex]x=x_{0} +V_{0}t + \frac{1}{2} at^2[/tex]
Where
a = acceleration
V = Velocity reached
Vo = Initial velocity
t = time
x = distance
xo = initial distance.
We have the following information:
a = We want to find V = 55.0 m/s
Vo = 0m/s because it starts from rest t = we want to find
x = 0.210 m xo= 0 m we beging in the point zero.
We have to variables in two equations, so we are going to replace in the second equation (2) the aceleration of the first one(1):
[tex]x=x_{0} +V_{0}t + \frac{1}{2} ( \frac{V}{t})t^2[/tex] We can cancel time because it is mutiplying and dividing the same factor so we have
[tex]x=x_{0} +V_{0}t + \frac{1}{2} Vt[/tex]
In this equation we just have one variable that we don't know that is time, so first we are going to replace the values and after that clear time.
[tex]0.210=0 +0*t + \frac{1}{2} 55t[/tex]
[tex]0.210=27.5t[/tex]
[tex]\frac{0.21}{27.5} = t\\[/tex]
t = 0.0076s
a) The acceleration took 0.0076s
Now we replace in the (1) equation the values of time and velocity
[tex]a = \frac{V}{t}[/tex]
[tex]a = \frac{55}{0.0076}[/tex]
a = 7202.4 m/s^2
b) The aceleration was of 7202.4 m/s^2
A runner of mass 51.8 kg starts from rest and accelerates with a constant acceleration of 1.31 m/s^2 until she reaches a velocity of 5.47 m/s. She then continues running at this constant velocity. How long does the runner take to travel 165 m?
Answer:32.24 s
Explanation:
Given
mass of runner (m)=51.8 kg
Constant acceleration(a)=[tex]1.31 m/s^2[/tex]
Final velocity (v)=5.47 m/s
Time taken taken to reach 5.47 m/s
v=u+at
[tex]5.47=0+1.31\times t[/tex]
[tex]t=\frac{5.47}{1.31}=4.17 s[/tex]
Distance traveled during this time is
[tex]s=ut+\frac{1}{2}at^2[/tex]
[tex]s=\frac{1}{2}\times 1.31\times 4.17^2=11.42 m[/tex]
So remaining distance left to travel with constant velocity=153.57 m
thus time [tex]=\frac{distance}{speed}[/tex]
[tex]t_2=\frac{153.57}{5.47}=28.07 s[/tex]
Total time=28.07+4.17=32.24 s
(a) With what speed must a ball be thrown vertically from ground level to rise to a maximum height of 52 m? (b) How long will it be in the air?
Explanation:
Maximum height reached by the ball, s = 52 m
Let u is the initial speed of the ball and v is the final speed of the ball, v = 0 because at maximum height the final speed goes to 0. We need to find u.
(a) The third equation of motion as :
[tex]v^2-u^2=2as[/tex]
Here, a = -g
[tex]0-u^2=-2gs[/tex]
[tex]u^2=2\times 9.8\times 52[/tex]
u = 31.92 m/s
(b) Let t is the time when the ball is in air. It is given by :
[tex]v=u+at[/tex]
[tex]u=gt[/tex]
[tex]t=\dfrac{31.92\ m/s}{9.8\ ms/^2}[/tex]
t = 3.25 seconds
Hence, this is the required solution.
A skiy diver, with parachute unopened, falls 625 m in 15.0s.
Then she opens her parachute and falls another 356 m in142 s. What
is her average velocity (both magnitude anddirection) for the
entire fall?
Answer:
average velocity = 6.25m/sec
Explanation:
given data:
for unopened
height = 625 m
time = 15 sec
for opened
height = 356 m
time = 142 sec
Unopened:
[tex]V1 = \frac{625\ m}{15\ sec} = 41.67m/sec[/tex]
Opened:
[tex]V2 = \frac{356\ m}{142\ sec} = 2.51m/sec[/tex]
we know that
Total Average Velocity[tex] = \frac{Total\ distance}{Total\ time}[/tex]
average velocity[tex] = \frac{981\ m}{157\ sec}[/tex]
average velocity = 6.25m/sec
downward direction.
(a) what wavelength photon would you need to ionize a hydrogen atom (ionization energy = 13.6 eV)? (b) Compute the temperature of the blackbody whose spectrum peaks at wavelength you found in (a).
Answer:
(a). The wavelength of photon is 914 A.
(b). The temperature of the black body whose spectrum peaks at wavelength is 31706.78 K.
Explanation:
Given that,
Ionization energy = 13.6 eV
(a). We need to calculate the wavelength
Using formula of wavelength
[tex]E=\dfrac{hc}{\lambda}[/tex]
[tex]\lambda=\dfrac{hc}{E}[/tex]
Where, h = Planck constant
c = speed of light
E = energy
Put the value into the formula
[tex]\lambda=\dfrac{6.63\times10^{-34}\times3\times10^{8}}{13.6\times1.6\times10^{-19}}[/tex]
[tex]\lambda=9.14\times10^{-8}\ m[/tex]
[tex]\lambda=914\ \AA[/tex]
The wavelength of photon is 914 A.
(b). We need to calculate the temperature of the black body whose spectrum peaks at wavelength
Using Wien's displacement law
[tex]\lambda_{max} T=2.898\times10^{-3}[/tex]
[tex]T=\dfrac{2.898\times10^{-3}}{\lambda}[/tex]
Put the value of wavelength
[tex]T=\dfrac{2.898\times10^{-3}}{914\times10^{-10}}[/tex]
[tex]T=31706.78\ K[/tex]
The temperature of the black body whose spectrum peaks at wavelength is 31706.78 K.
Hence, This is the required solution.
An object object travels 24 meters in the +x direction for 3.5 seconds, and then immediately travels some distance in the -x direction. After a total time of 9.3 seconds, the object's average velocity is 0 m/s. What distance did the object travel in the -x direction?
Answer:
The distance in -x axis is 39.77 m
Explanation:
Given that,
Distance in +x axis = 24 m
Time = 3.5 sec
Total time = 9.3 sec
Average velocity = 0
We need to calculate the velocity in +x axis
[tex]v=\dfrac{d}{t}[/tex]
Put the value into the formula
[tex]v=\dfrac{24}{3.5}[/tex]
We need to calculate the velocity in -x axis
[tex]v=\dfrac{x}{9.3-3.5}[/tex]
We need to calculate the distance
Using formula of average velocity
[tex]\dfrac{\dfrac{24}{3.5}-\dfrac{x}{9.3-3.5}}{2}=0[/tex]
[tex]x=\dfrac{24}{3.5}\times5.8[/tex]
[tex]x=39.77\ m[/tex]
Hence, The distance in -x axis is 39.77 m.
A car starts from rest and accelerates at 4 m/s^2 for 8 s, after which it continues at constant speed for 19 s. How far would it have traveled from its starting point by that time.
Answer:
The total distance traveled is 736 m
Solution:
According to the question:
Initial velocity, v = 0
(since, the car is starting from rest)
[tex]acceleration, a = 4 m/s^{2}[/tex]
Time taken, t = 8 s
Now, the distance covered by it in 8 s is given by the second eqn of motion:
[tex]d = vt + \farc{1}{2}at^{2}[/tex]
[tex]d = 0.t + \farc{1}{2}4\times 8^{2} = 128 m[/tex]
Now, to calculate the velocity, we use eqn 1 of motion:
v' = v + at
v' = 0 + 4(8) = 32 m/s
Now, the distance traveled by the car with uniform velocity of 32 m/s for t' = 19 s:
distance, d' = v't'
[tex]d' = 32\times 19 = 608 m[/tex]
Total distance traveled = d + d' = 128 + 608 = 736 m