Answer:
x=4.06m
Explanation:
A body that moves with constant acceleration means that it moves in "a uniformly accelerated movement", which means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.
When performing a mathematical demonstration, it is found that the equations that define this movement are as follows.
Vf=Vo+a.t (1)\\\\
{Vf^{2}-Vo^2}/{2.a} =X(2)\\\\
X=Xo+ VoT+0.5at^{2} (3)\\
Where
Vf = final speed
Vo = Initial speed
T = time
A = acceleration
X = displacement
In conclusion to solve any problem related to a body that moves with constant acceleration we use the 3 above equations and use algebra to solve
for this problem
Vf=7.6m/s
t=1.07
Vo=0
we can use the ecuation number one to find the acceleration
a=(Vf-Vo)/t
a=(7.6-0)/1.07=7.1m/s^2
then we can use the ecuation number 2 to find the distance
{Vf^{2}-Vo^2}/{2.a} =X
(7.6^2-0^2)/(2x7.1)=4.06m
To find the velocity and acceleration vectors for uniform circular motion and to recognize that this acceleration is the centripetal acceleration. Suppose that a particle's position is given by the following expression: r(t)=R[cos(ωt)i^+sin(ωt)j^] =Rcos(ωt)i^+Rsin(ωt)j^.The particle's motion can be described by ____________.(A) an ellipse starting at time t=0 on the positive x axis(B) an ellipse starting at time t=0 on the positive y axis(C) a circle starting at time t=0 on the positive x axis(D) a circle starting at time t=0 on the positive y axis
Answer:
(C) a circle starting at time t=0 on the positive x axis
Explanation:
particle's position is
r(t)=R[cos(ωt)i^+sin(ωt)j^] =Rcos(ωt)i^+Rsin(ωt)j^
this is a parametric equation of a circle, because the axis at x and y are the same = R.
for t=0:
r=Ri^
so: circle starting at time t=0 on the positive x axis
On the other hand:
[tex]v=\frac{dx}{dt}= Rw[-sin(wt)i+cos(wt)j]\\a=\frac{dv}{dt}= Rw^{2}[-cos(wt)i-sin(wt)j][/tex]
The value of the magnitude of the acceleration is:
[tex]a=Rw^{2}(cos^{2}(wt)+sin^{2}(wt))=Rw^{2}[/tex]
we can recognise that this represent the centripetal acceleration.
You irradiate a crystalline sample that has 2.9344 Å between atoms with electrons to do an electron diffraction experiment in reflection mode. You observe a first order (m = 1) diffraction peak at θ = 12.062°. What is the wavelength of the electrons?
Answer:
[tex]\lambda=1.23[/tex]Å
Explanation:
Bragg's Law refers to the simple equation:
[tex]n\lambda = 2d sin(\theta)[/tex]
In this case:
n=1
θ = 12.062°
d=2.9344 Å
[tex]\lambda = 2*2.9344sin(12.062)=1.23[/tex]Å
Two 2.3 cm -diameter disks face each other, 2.9 mm apart. They are charged to ±16nC . A) What is the electric field strength between the disks?
Express your answer to two significant figures and include the appropriate units.
B) A proton is shot from the negative disk toward the positive disk. What launch speed must the proton have to just barely reach the positive disk?
Express your answer to two significant figures and include the appropriate units.
Final answer:
To find the electric field strength between the disks, we can use Coulomb's law and the formula for electric field. The launch speed of the proton can be found using the conservation of energy.
Explanation:
To find the electric field strength between the disks, we can use the formula:
Electric Field = Force / Charge
The force between the disks is given by Coulomb's law:
Force = (k * q1 * q2) / r^2
Where k is the electrostatic constant (9 * 10^9 Nm^2/C^2), q1 and q2 are the charges on the disks, and r is the distance between them.
Substituting the given values, we have:
Force = (9 * 10^9 Nm^2/C^2) * (16 * 10^-9 C)^2 / (2.3 * 10^-2 m)^2
Calculating this expression, we get the force between the disks. Then, we can divide this force by the charge on either disk to find the electric field strength.
For part B, we can use the conservation of energy to find the launch speed of the proton. The potential energy difference between the disks can be calculated as:
Potential Energy = charge * voltage
Given the charge of the proton and the distance between the disks, we can find the voltage. Since the proton starts from rest, all of its initial potential energy will be converted into kinetic energy:
Kinetic Energy = (1/2) * mass * velocity^2
Where the mass of the proton is known.
Solving for velocity, we can find the launch speed of the proton.
A crate of eggs is located in the middle of the flat bed of a pickup truck as the truck negotiates a curve in the flat road. The curve may be regarded as an arc of a circle of radius 36.1 m. If the coefficient of static friction between crate and truck is 0.570, how fast can the truck be moving without the crate sliding?
Answer:
[tex] v_{max}=14.2\frac{m}{s} [/tex]
Explanation:
Hi!
If the crate is not sliding, its trajectory is the arc with 36.1 m radius. Then the crate has a centripetal acceleration:
[tex]a_c= \frac{v^2}{r} \\r = radius\\v = tangential \; velocity[/tex]
The centripetal force acting on the crate is the static friction force between crate and truck. The maximum value of this force is:
[tex]F_{max} = \mu N\\\mu = 0.570=static\;friction \;coefficient\\N =normal\; force\\[/tex]
The normal force has a magnitude equal to the weight of the crate:
[tex]N=mg[/tex]
Then the condition for not sliding is:
[tex]F_{centripetal} = M\frac{v^2}{r}<\mu N=\mu Mg\\ v^2<r \mu g = 36.1\;m*0.570*9.8\frac{m}{s^2}= 201.65 \frac{m^2}{s^2}\\ v<14.2\frac{m}{s}[/tex]
You know that a point charge is located somewhere along the x-axis. When you measure the electric field at 2.00 m the result is 3.000 N/C and points left while the electric field at 5.00 m is 0.750 N/C also pointing left. (a) Where is the charge located?
(b) What is the size of the charge, including sign?
The electric field measurements indicate a point charge located to the right of both measurement positions on the x-axis, being more than 5.00 m away. Using the electric field values (3.000 N/C at 2.00 m and 0.750 N/C at 5.00 m), and applying the equation E = kQ/r^2, we can solve for both the point charge's magnitude and location.
The student's question revolves around the electric field produced by a point charge and involves finding both the location of the point charge and its magnitude. To solve this, we need to apply Coulomb's Law and the principle that the intensity of the electric field (E) from a point charge can be described by the equation E = kQ/r^2, where k is Coulomb's constant (8.988 imes 10^9 N m^2/C^2), Q is the charge, and r is the distance from the charge to the point where the electric field is measured.
Given the electric field measurements at two different points both pointing to the left, we can deduce that the point charge is located to the right of both points on the x-axis. The point charge must thus be more than 5.00 m to the right of the origin.
By using the measured electric field values and the distances to set up two equations, we can determine the charge's value:
E1 = kQ/r1^2 (at 2.00 m)
E2 = kQ/r2^2 (at 5.00 m)
Where E1 = 3.000 N/C, E2 = 0.750 N/C, r1 = 2.00 m, and r2 = 5.00 m. By solving these equations simultaneously, we can find the value of Q and the exact location along the x-axis.
Tidal friction is slowing the rotation of the Earth. As a result, the orbit of the Moon is increasing in radius at a rate of approximately 4.0 cm/yr. Assuming this to be a constant rate, how many years will pass before the radius of the Moon's orbit increases by 3.84x 10^7 m (10%)?
Answer:
967500000 years
Explanation:
The Speed at which the radius of the orbit of the Moon is increasing is 4 cm/yr
Converting to m
1 m = 100 cm
[tex]1\ cm=\frac{1}{100}\ m[/tex]
[tex]4\ cm\y=\frac{4}{100}=0.04\ m/yr[/tex]
The distance by which the radius increases is 3.84×10⁷ m
Time = Distance / Speed
[tex]\text{Time}=\frac{3.87\times 10^7}{0.04}\\\Rightarrow \text{Time}=967500000\ yr[/tex]
967500000 years will pass before the radius of the orbit increases by 10%.
Final answer:
The radius of the Moon's orbit increases by approximately 4 cm/year. To calculate the number of years it will take for the radius to increase by 3.84 × 10^6 m, we can use the formula Time = Change in Distance / Rate of Increase. The answer is approximately 9.6 × 10^7 years.
Explanation:
The radius of the Moon's orbit is increasing at a rate of approximately 4 cm/year. To find out how many years will pass before the radius of the Moon's orbit increases by 3.84 × 10^6 m, we can use the formula:
Time = Change in Distance / Rate of Increase
Substituting the given values, we get:
Time = (3.84 × 10^6 m) / (4 cm/year)
Now, we need to convert the units so that they are consistent. 3.84 × 10^6 m is equivalent to 3.84 × 10^8 cm. Substituting this value into the equation, we get:
Time = (3.84 × 10^8 cm) / (4 cm/year)
Canceling out the units, we find that:
Time = 9.6 × 10^7 years.
Two students are sitting 1.50 m apart. One student has a mass of 70.0 kg and the other has a mass of 52.0 kg. What is the gravitational force between them?
Answer:
The gravitational force between them is [tex]1.079\times10^{-7}\ N[/tex].
Explanation:
Given that,
Distance = 1.50 m
Mass of one student = 70.0 kg
Mass of other student = 52.0 kg
We need to calculate the gravitational force
Using formula of gravitational force
[tex]F=\dfrac{Gm_{1}m_{2}}{r^2}[/tex]
Where, m₁ = mass of one student
m₂ = mass of other studen
r = distance between them
Put the value into the formula
[tex]F=\dfrac{6.67\times10^{-11}\times70.0\times52.0}{1.50^2}[/tex]
[tex]F=1.079\times10^{-7}\ N[/tex]
Hence, The gravitational force between them is [tex]1.079\times10^{-7}\ N[/tex].
To calculate the gravitational force between the two students, we use Newton's law of universal gravitation, substituting the given values for mass and distance into the formula. The result suggests that the gravitational force would be incredibly small, aligning with our daily experiences.
Explanation:The subject of this question is Physics, specifically gravitational force. From Newton's law of universal gravitation, we know that the gravitational force between two masses is given by the equation F = G(M₁M₂)/r², where F is the gravitational force, G is the gravitational constant, M₁ and M₂ are the two masses, and r is the distance between them.
Given that one student has a mass of 70 kg (M₁), the other a mass of 52 kg (M₂) and the distance between them is 1.5 m (r), we can substitute these values into the formula. Using a gravitational constant (G) of approximately 6.67 × 10-¹¹ Nm²/kg², the gravitational force (F) becomes:
F = (6.67 × 10-¹¹ Nm²/kg²)(70 kg)(52 kg)/(1.50 m)²
Note, though, that the gravitational force between two people sitting 1.50 m apart would be incredibly small due to the immense smallness of the gravitational constant. This is inline with our daily experiences where we don't feel any noticeable gravitational pull from an ordinary object.
Learn more about Gravitational Force here:https://brainly.com/question/32609171
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How would you use the parallelogram method of vector addition when more than two forces are added?
Answer:
In order to use the parallelogram method, we have to select first two vectors. We move the vectors until their initial points coincide. Then we draw lines to form a complete parallelogram, as is shown in the figure annexed. The diagonal from the initial point to the opposite vertex of the parallelogram is the resultant.
We use the last vector resultant with a third vector, and we use again the parallelogram method to add them. We use the new resultant and we add it with a 4th vector. We repeat this task until all the vectors are used.
Three vectors →a, →b, and →c each have a magnitude of 50 m and lie in an xy plane. Their directions relative to the positive direction of the x axis are 30°, 195°, and 315°, respectively. What are (a) the magnitude and (b) the angle of the vector →a+→b+→c and (c) the magnitude and (d) the angle of →a−→b+→c? What are the (e) magnitude and (f) angle of a fourth vector →d such that (→a+→b)−(→c+→d)=0 ?
Answer:
(a): 37.94 m.
(b): [tex]323.16^\circ.[/tex]
(c): 126.957 m.
(d): [tex]0.93^\circ.[/tex]
(e): 49.92 m.
(f): [tex]130.08^\circ.[/tex]
Explanation:
Given:
Magnitude of [tex]\vec a[/tex] = 50 m.Direction of [tex]\vec a = 30^\circ.[/tex]Magnitude of [tex]\vec b[/tex] = 50 m.Direction of [tex]\vec b = 195^\circ.[/tex]Magnitude of [tex]\vec c[/tex] = 50 m.Direction of [tex]\vec c = 315^\circ.[/tex]Any vector [tex]\vec A[/tex], making an angle [tex]\theta[/tex] with respect to the positive x-axis, can be written in terms of its x and y components as follows:
[tex]\vec A = A\cos\theta\ \hat i+A\sin\theta \ \hat j.[/tex]
where, [tex]\hat i,\ \hat j[/tex] are the unit vectors along the x and y axes respectively.
Therefore, the given vectors can be written as
[tex]\vec a = 50\cos30^\circ \ \hat i+50\sin 30^\circ\ \hat j = 43.30\ \hat i +25\ \hat j\\\vec b = 50\cos195^\circ \ \hat i+50\sin 195^\circ\ \hat j = -48.29\ \hat i +-12.41\ \hat j\\\vec c = 50\cos 315^\circ \ \hat i+50\sin 315^\circ\ \hat j = 35.35\ \hat i +-35.35\ \hat j\\[/tex]
(a):
[tex]\vec a +\vec b + \vec c= (43.30\ \hat i +25\ \hat j)+(-48.29\ \hat i +-12.41\ \hat j)+(35.35\ \hat i +-35.35\ \hat j)\\=(43.30-48.29+35.35)\hat i+(25-12.41-35.35)\hat j\\=30.36\hat i-22.75\hat j.\\\\\text{Magnitude }=\sqrt{30.36^2+(-22.75)^2}=37.94\ m.[/tex]
(b):
Direction [tex]\theta[/tex] can be found as follows:
[tex]\tan\theta = \dfrac{\text{x component of }(\vec a + \vec b +\vec c)}{\text{y component of }(\vec a + \vec b +\vec c)}=\dfrac{-22.75}{30.36}=-0.749\\\Rightarrow \theta = \tan^{-1}(-0.749)=-36.84^\circ.[/tex]
The negative sign indicates that the sum of the vectors is [tex]36.84^\circ.[/tex] below the positive x axis.
Therefore, direction of this vector sum counterclockwise with respect to positive x-axis = [tex]360^\circ-36.84^\circ=323.16^\circ.[/tex]
(c):
[tex]\vec a -\vec b + \vec c= (43.30\ \hat i +25\ \hat j)-(-48.29\ \hat i +-12.41\ \hat j)+(35.35\ \hat i +-35.35\ \hat j)\\=(43.30+48.29+35.35)\hat i+(25+12.41-35.35)\hat j\\=126.94\hat i+2.06\hat j.\\\\\text{Magnitude }=\sqrt{126.94^2+2.06^2}=126.957\ m.[/tex]
(d):
Direction [tex]\theta[/tex] can be found as follows:
[tex]\tan\theta = \dfrac{\text{x component of }(\vec a - \vec b +\vec c)}{\text{y component of }(\vec a - \vec b +\vec c)}=\dfrac{2.06}{126.94}=0.01623\\\Rightarrow \theta = \tan^{-1}(0.01623)=0.93^\circ.[/tex]
(e):
[tex](\vec a + \vec b)-(\vec c + \vec d)=0\\(\vec a + \vec b)=(\vec c + \vec d)\\\vec d = \vec a + \vec b -\vec c.[/tex]
[tex]\vec d = \vec a +\vec b - \vec c= (43.30\ \hat i +25\ \hat j)+(-48.29\ \hat i +-12.41\ \hat j)-(35.35\ \hat i +-35.35\ \hat j)\\=(43.30-48.29-35.35)\hat i+(25-12.41+35.35)\hat j\\=-40.34\hat i+47.94\hat j.\\\\\text{Magnitude }=\sqrt{(-40.34)^2+47.94^2}=62.65\ m.[/tex]
(f):
Direction [tex]\theta[/tex] can be found as follows:
[tex]\tan\theta = \dfrac{\text{x component of }\vec d}{\text{y component of }\vec d}=\dfrac{47.94}{-40.34}=-1.188\\\Rightarrow \theta = \tan^{-1}(-1.188)=-49.92^\circ.[/tex]
The x component of this vector is negative and y component is positive therefore the vector lie in second quadrant, which means, the direction of this vector, counterclockwise with respect to positive x axis = [tex]180^\cir.
c-49.92^\circ=130.08^\circ.[/tex]
We attach two blocks of masses m1 = 7 kg and m2 = 7 kg to either end of a spring of spring constant k = 1 N/m and set them into oscillation. Calculate the angular frequency ω of the oscillation.
Answer:
The angular frequency [tex]\omega[/tex] of the oscillation is [tex]0.58s^{-1}[/tex]
Explanation:
For this particular situation, the angular frequency of the system is given by
[tex]\omega=\sqrt{\frac{m_1+m_2}{m_1m_2}k}=\sqrt{\frac{7 kg+5 kg }{7kg *5 kg}1\frac{N}{m}}=\sqrt{\frac{3}{35s^2}}\approx 0.58s^{-1}[/tex]
An arrow is shot straight up in the air with an initial speed of 250 ft/s. If on striking the ground, it embeds itself 4.00 in into the ground, find the magnitude of the acceleration (assumed constant) required to stop the arrow in units of ft/sec^2
Answer:
93750 ft/s²
Explanation:
t = Time taken
u = Initial velocity = 250 ft/s (It is assumed that it is speed of the arrow just when it enter the ground)
v = Final velocity = 0
s = Displacement = 4 in = [tex]\frac{4}{12}=\frac{1}{3}\ feet[/tex]
a = Acceleration
Equation of motion
[tex]v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-250^2}{2\times \frac{1}{3}}\\\Rightarrow a=-93750\ ft/s^2[/tex]
The magnitude of acceleration is 93750 ft/s²
A 5.0-V battery is places in series with two 1.25-Ω resistors. Determine the current through each resistor.
Answer:
Current through each resistor is 2 A.
Explanation:
Given that,
Voltage of a battery, V = 5 volts
Resistance 1, R = 1.25 ohms
Resistance 2,R' = 1.25 ohms
Both resistors are connected in series. The equivalent resistance is given by :
R" = R + R'
R" = 1.25 + 1.25
R" = 2.5 ohms
The current flowing throughout all resistors is same in series combination of resistors. Current can be calculated using Ohm's law as :
[tex]I=\dfrac{V}{R"}[/tex]
[tex]I=\dfrac{5\ V}{2.5}[/tex]
I = 2 A
So, the current through each resistor is 2 A. Hence, this is the required solution.
At a construction site a pipe wrench struck the ground with a speed of 23 m/s. (a) From what height was it inadvertently dropped? (b)How long was it falling?
Answer:26.96 m,2.34 s
Explanation:
Given
Wrench hit the ground with a speed of 23 m/s
Applying equation of motion
[tex]v^2-u^2=2as[/tex]
Here u=0 because it is dropped from a height of S m
[tex]23^2-0=2\times 9.81\times s[/tex]
[tex]s=\frac{529}{2\times 9.81}=26.96 m[/tex]
Time required by wrench to hit the ground
v=u+at
[tex]23=9.81\times t[/tex]
[tex]t=\frac{23}{9.81}=2.34 s[/tex]
A 7.80-nc charge is located 1.81 m from a 4.30-nc point charge. (a) Find the magnitude of the electrostatic force that one charge exerts on the other. N (b) Is the force attractive or repulsive? O attractive O repulsive
Explanation:
Given that,
Charge 1, [tex]q_1=7.8\ nC=7.8\times 10^{-9}\ C[/tex]
Charge 2, [tex]q_2=4.3\ nC=4.3\times 10^{-9}\ C[/tex]
Distance between charges, r = 1.81 m
(a) The electrostatic force that one charge exerts on the another is given by :
[tex]F=k\dfrac{q_1q_2}{r^2}[/tex]
[tex]F=9\times 10^9\times \dfrac{7.8\times 10^{-9}\times 4.3\times 10^{-9}}{(1.81)^2}[/tex]
[tex]F=9.21\times 10^{-8}\ N[/tex]
(b) As both charges are positively charged. So, the force of attraction between them is repulsive.
Over a time interval of 1.99 years, the velocity of a planet orbiting a distant star reverses direction, changing from +20.7 km/s to -22.0 km/s. Find (a) the total change in the planet's velocity (in m/s) and (b) its average acceleration (in m/s2) during this interval. Include the correct algebraic sign with your answers to convey the directions of the velocity and the acceleration.
Answer:
(a) - 42700 m/s
(b) - 6.8 x 10^-4 m/s^2
Explanation:
initial velocity of star, u = 20.7 km/s
Final velocity of star, v = - 22 km/s
time, t = 1.99 years
Convert velocities into m/s and time into second
So, u = 20700 m / s
v = - 22000 m/s
t = 1.99 x 365.25 x 24 x 3600 = 62799624 second
(a) Change in planet's velocity = final velocity - initial velocity
= - 22000 - 20700 = - 42700 m/s
(b) Accelerate is defined as the rate of change of velocity.
Acceleration = change in velocity / time
= ( - 42700 ) / (62799624) = - 6.8 x 10^-4 m/s^2
Air enters a heat exchanger at a rate of 5000 cubic feet per minute at a temperature of 55 °F and pressure of 14.7 psia. The air is heated by hot water flowing in the same exchanger at a rate of 11,200 pounds per hour with a decrease in temperature of 10 °F. At what temperature does the air leave the heat exchanger?
Answer:
75 °F
Explanation:
Air has a specific heat at constant pressure of:
Cpa = 0.24 BTU/(lbm*F)
The specific heat of water is:
Cpw = 1 BTU/(lbm*F)
The first law of thermodynamics:
Q = L + ΔU
The heat exchanger is running at a steady state, so ΔU = 0. Also does not perform or consume any work L = 0.
Then:
Q = 0.
We split the heat into the heat transferred by the air and the heat trnasferred by the water:
Qa + Qw = 0
Qa = -Qw
The heat exchanged by the air is
Qa = Ga * Cpa * (tfin - ti)
And the heat exchanged by the water is:
Qw = Gw * Cpw * Δt
Replacing:
Ga * Cpa * (tfin - ti) = -Gw * Cpw * Δt
tfin - ti = (-Gw * Cpw * Δt) / (Ga * Cpa)
tfin = (-Gw * Cpw * Δt) / (Ga * Cpa) + ti
The G terms are mass flows, however we have volume flow of air.
With the gas state equation we calculate the mass:
p * V = m * R * T
m = (p * V) / (R * T)
55 °F = 515 °R
The gas constant for air is R = 53.35 (ft*lb)/(lbm* °R)
14.7 psi = 2117 lb/ft^2
m = (2117 * 5000) / (53.35 * 515) = 385 lbm
The mass flow is that much amount per minute
The mass flow of water is
11200 lbm/h = 186.7 lbm/min
Then:
tfin = (-186.7 * 1 * (-10)) / (385 * 0.24) + 55 = 75 °F
Consider an electron that is 10-10 m from an alpha particle (9 = 3.2 x 10-19 C). (Enter the magnitudes.) (a) What is the electric field in N/C) due to the alpha particle at the location of the electron? N/C (b) What is the electric field (in N/C) due to the electron at the location of the alpha particle? N/C (c) What is the electric force in N) on the alpha particle? On the electron? electric force on alpha particle electric force on electron
Answer:
a)[tex]E=2.88*10^{11}N/C[/tex]
b)[tex]E=1.44*10^{11}N/C[/tex]
c)[tex]F=4.61*10^{-8}N[/tex]
Explanation:
We use the definition of a electric field produced by a point charge:
[tex]E=k*q/r^2[/tex]
a)Electric Field due to the alpha particle:
[tex]E=k*q_{alpha}/r^2=9*10^9*3.2*10^{-19}/(10^{-10})^2=2.88*10^{11}N/C[/tex]
b)Electric Field due to electron:
[tex]E=k*q_{electron}/r^2=9*10^9*1.6*10^{-19}/(10^{-10})^2=1.44*10^{11}N/C[/tex]
c)Electric Force on the alpha particle, on the electron:
The alpha particle and electron feel the same force but with opposite direction:
[tex]F=k*q_{electron}*q_{alpha}/r^2=9*10^9*1.6*10^{-19}*3.2*10^{-19}/(10^{-10})^2=4.61*10^{-8}N[/tex]
A ball is thrown with velocity of 10 m/s upwards. If the ball is caught 1 m above its initial position, what is the speed of the ball when it is caught?
Answer:
v = 8.96 m/s
Explanation:
Initial speed of the ball, u = 10 m/s
It caught 1 meter above its initial position.
Acceleration due to gravity, [tex]g=-9.8\ m/s^2[/tex]
We need to find the final speed of the ball when it is caught. Let is equal to v. To find the value of v, use third equation of motion as :
[tex]v^2-u^2=2as[/tex]
[tex]v^2=2as+u^2[/tex]
[tex]v^2=2(-9.8)\times 1+(10)^2[/tex]
v = 8.96 m/s
So, the speed of the ball when it is caught is 8.96 m/s. Hence, this is the required solution.
Answer:
8.96 m/s, upward direction
Explanation:
Given that, the initial velocity of the ball is,
[tex]u=10m/s[/tex]
And the acceleration in the downward direction is positive but in this situation the acceleration will be negative so,
[tex]a=9.8\frac{m}{s^{2} }[/tex]
And according to question vertical displacement is,
[tex]s=1m[/tex]
Now suppose v be the final velocity of the ball.
Applying third equation of motion,
[tex]v^{2}=u^{2}+2as[/tex]
Here, u is the initial velocity, a is the acceleration, s is the displacement.
Substitute all the variables.
[tex]v=\sqrt{10^{2}+2(-9.8)\times 1 } \\v=\sqrt{80.4}\\ v=8.96\frac{m}{s}[/tex]
Therefore, the speed of ball when it is caught is 8.96 m/s in the upward direction.
The masses of the earth and moon are 5.98 x 1024 and 7.35 x 1022 kg, respectively. Identical amounts of charge are placed on each body, such that the net force (gravitational plus electrical) on each is zero. What is the magnitude of the charge placed on each body?
Answer:
The magnitude of charge on each is [tex]5.707\times 10^{13} C[/tex]
Solution:
As per the question:
Mass of Earth, [tex]M_{E} = 5.98\times 10^{24} kg[/tex]
Mass of Moon, [tex]M_{M} = 7.35\times 10^{22} kg[/tex]
Now,
The gravitational force of attraction between the earth and the moon, if 'd' be the separation distance between them is:
[tex]F_{G} = \frac{GM_{E}M_{M}}{d^{2}}[/tex] (1)
Now,
If an identical charge 'Q' be placed on each, then the Electro static repulsive force is given by:
[tex]F_{E} = \frac{1}{4\pi\epsilon_{o}}\frac{Q^{2}}{d^{2}}[/tex] (2)
Now, when the net gravitational force is zero, the both the gravitational force and electro static force mut be equal:
Equating eqn (1) and (2):
[tex]\frac{GM_{E}M_{M}}{d^{2}} = \frac{1}{4\pi\epsilon_{o}}\frac{Q^{2}}{d^{2}}[/tex]
[tex](6.67\times 10^{- 11})\times (5.98\times 10^{24})\times (7.35\times 10^{22}) = (9\times 10^{9}){Q^{2}}[/tex]
[tex]\sqrt{\farc{(6.67\times 10^{- 11})\times (5.98\times 10^{24})\times (7.35\times 10^{22})}{9\times 10^{9}}} = Q[/tex]
Q = [tex]\pm 5.707\times 10^{13} C[/tex]
A 60.0 kg astronaut is freely floating in space and pushes on a freely floating 120.0 kg spacecraft with a force of 30.0 N for 1.50 s. (a) Compare the forces exerted on the astronaut and the spacecraft, and (b) compare the acceleration of the astronaut to the acceleration of the spacecraft
Answer:
Explanation:
mass of astronaut, m = 60 kg
mass of space craft, M = 120 kg
t = 1.5 s
Force on space craft = 30 N
(a) According to Newton's third law
Force on spacecraft by the astronaut = Force on astronaut by the space craft
Force on astronaut by the space craft = 30 N
(b) According to Newton's second law
Force = mass x acceleration
Let a be the acceleration of the astronaut
30 = 60 x a
a = 0.5 m/s^2
Let A be the acceleration of the spacecraft
30 = 120 x A
a = 0.25 m/s^2
During the execution of a play, a football player carries the ball for a distance of 33 m in the direction 58° north of east. To determine the number of meters gained on the play, find the northward component of the ball's displacement.
Answer:28 m
Explanation:
Given
Direction is [tex]58^{\circ}[/tex] North of east i.e. [tex]58 ^{\circ}[/tex] with x axis
Also ball moved by 33 m
therefore its east component is 33cos58=17.48 m
Northward component [tex]=33sin58=27.98 m\approx 28 m[/tex]
A piano string having a mass per unit length of 5.00 g/m is under a tension of 1350 N. Determine the speed of transverse waves in this string.
Answer:
The speed of transverse waves in this string is 519.61 m/s.
Explanation:
Given that,
Mass per unit length = 5.00 g/m
Tension = 1350 N
We need to calculate the speed of transverse waves in this string
Using formula of speed of the transverse waves
[tex]v=\sqrt{\dfrac{T}{\mu}}[/tex]
Where, [tex]\mu[/tex] = mass per unit length
T = tension
Put the value into the formula
[tex]v = \sqrt{\dfrac{1350}{5.00\times10^{-3}}}[/tex]
[tex]v =519.61\ m/s[/tex]
Hence, The speed of transverse waves in this string is 519.61 m/s.
the wavelength of a certain light source is "0.535" where 1 micrometer = 1.0 x 10^-6m. express this wavelentg in nanometers.
Answer:
The answer is 535 nanometers.
Explanation:
[tex]1\ micrometer = 1\ \mu m = 1.0\times 10^{-6}\ m.[/tex]
and
[tex]1\ nanometer = 1\ nm = 1.0\times 10^{-9}\ m.[/tex],
so
[tex]1\ \mu m = 1.0 \times 10^{3}\ nm[/tex]
which means that
[tex]\lambda = 0.535\ \mu m = 535\ nm[/tex].
In fact we can say that the light is green, because its wavelength is in the range of 500 nm to 565 nm.
The question ask:
A block is hung by a string from the inside roof of a van.When
the van goes straight ahead at a speed of 28m/s, theblock hangs
vertically down. But when the van maintains this samespeed around
an unbanked curve (radius=150m), the block swingstoward the outside
of the curve. Then the string makes anangle theta with the
vertical. Find theta.
Answer:
[tex]\theta=28.07^{\circ}[/tex]
Explanation:
Speed of van, v = 28 m/s
Radius of unbanked curve, r = 150 m
Let [tex]\theta[/tex] is the angle with the vertical. In case of banking of road,
[tex]T\ cos\theta=mg[/tex].............(1)
And
[tex]T\ sin\theta=\dfrac{mv^2}{r}[/tex]..........(2)
From equation (1) and (2) :
[tex]tan\theta=\dfrac{v^2}{rg}[/tex]
[tex]tan\theta=\dfrac{(28)^2}{150\times 9.8}[/tex]
[tex]\theta=28.07^{\circ}[/tex]
So, the string makes an angle of 28.07 degrees with the vertical. Hence, this is the required solution.
A basketball referee tosses the ball straight up for the starting tip-off. At what velocity (in m/s) must a basket ball player leave the grownd to rist 1.24 m above the floor in an attempt to get the ball?
Answer:4.93 m/s
Explanation:
Given
height to reach is (h )1.24 m
here Let initial velocity is u
using equation of motion
[tex]v^2-u^2=2ah[/tex]
here Final Velocity v=0
a=acceleration due to gravity
[tex]0-u^2=2\left ( -g\right )h[/tex]
[tex]u=\sqrt{2gh}[/tex]
[tex]u=\sqrt{2\times 9.81\times 1.24}[/tex]
[tex]u=\sqrt{24.328}[/tex]
u=4.93 m/s
The pilot of an airplane carrying a package of mail to a remote outpost wishes to release the package at the right moment to hit the recovery location A. What angle θ with the horizontal should the pilot’s line of sight to the target make at the instant of release? The airplane is flying horizontally at an altitude of 86 m with a velocity of 283 km/h.
Answer:
The angle is [tex]\theta\approx 14.61[/tex] degrees.
Explanation:
Se the attached drawing if you need a visual aid for the explanation. Let [tex]\theta[/tex] be the angle of elevation of the plante which in itself is the same drop angle that the pilot measures. Let [tex]d[/tex] be the horizontal distance from the target and [tex]h[/tex] the height of the plane. We know that the package is dropped without any initial vertical speed, that means that it has a y-position equation of the form:
[tex]y(t)=-\frac{1}{2}gt^2+h[/tex]
If we set [tex]y(t)=0[/tex] we are setting the condition that the package is in the ground. We can then solve for t and get the flight time of the package.
[tex]0=-\frac{1}{2}gt^2+h\implies t_f=\sqrt{\frac{2h}{g}}[/tex].
If the flight time is -[tex]t_f[/tex] then the distance b can be found in meters by taking into account that the horizontal speed of the plane is [tex]v=283\, Km/h=78.61 \, m/s[/tex].
[tex]d=v\cdot t_f=78.61\cdot \sqrt{\frac{2h}{g}}[/tex]
The angle is thus
[tex]\theta=\arctan{\frac{h}{v\cdot t_f}}=\arctan{\frac{h}{v\cdot \sqrt{\frac{2\cdot h}{g}}}\approx 14.61 [/tex] degrees.
The Z0 boson, discovered in 1985, is themediator of
the weak nuclear force, and it typically decays veryquickly. Its
average rest energy is 91.19 GeV, but its shortlifetime shows up as
an intrinsic width of 2.5 GeV (rest energyuncertainty). What is the
lifetime of this particle?
Answer:
The life time of the particle is [tex]2.491\times 10^{- 25} s[/tex]
Solution:
As per the question:
Average rest energy of [tex]Z^{0}boson = 91.19 GeV[/tex]
Uncertainty in rest energy, [tex]\Delta E_{r} = 2.5 GeV = 2.5\times 10^{9}\times 1.6\times 140^{- 19} J = 4\times 10^{- 10} J[/tex]
Now,
From the Heisenberg's Uncertainty Principle, we can write:
[tex]\Delta E_{r}\times \Delta T \geq \frac{h}{2\pi}[/tex]
where
T = Life time of the particle
[tex]\Delta T \geq \frac{h}{2\pi\Delta E_{r}}[/tex]
[tex]\Delta T \geq \frac{6.262\times 10^{- 34}}{2\pi\times 4\times 10^{- 10}}[/tex]
[tex]\Delta T \simeq 2.491\times 10^{- 25} s[/tex]
The lifetime of the Z0 boson is approximately 8.95 x 10^-17 seconds.
Explanation:The Z0 boson is a particle that mediates the weak nuclear force. Its average rest energy is 91.19 GeV and it has an intrinsic width of 2.5 GeV. The lifetime of a particle can be calculated using the Heisenberg uncertainty principle, which relates the energy uncertainty to the time uncertainty. The relationship is given by the equation ΔE × Δt ≥ ℏ/2, where ℏ is the reduced Planck constant. By rearranging the equation and substituting the values, we can calculate the lifetime of the Z0 boson to be approximately 8.95 x 10^-17 seconds.
Learn more about Z0 boson here:https://brainly.com/question/35880581
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A parallel-plate capacitor consists of two plates, each with an area of 27 cm^2 separated by 3.0 mm. The charge on the capacitor is 4.8 nC . A proton is released from rest next to the positive plate. How long does it take for the proton to reach the negative plate? Steps please with right answer.
Answer:
Explanation:
Capacity of a parallel plate capacitor C = ε₀ A/ d
ε₀ is permittivity whose value is 8.85 x 10⁻¹² , A is plate area and d is distance between plate.
C =( 8.85 X10⁻¹² X 27 X 10⁻⁴ ) / 3 X 10⁻³
= 79.65 X 10⁻¹³ F.
potential diff between plate = Charge / capacity
= 4.8 X 10⁻⁹ / 79.65 X 10⁻¹³
= 601 V
Electric field = V/d
= 601 / 3 x 10⁻³
= 2 x 10⁵ N/C
Force on proton
= charge x electric field
1.6 x 10⁻¹⁹ x 2 x 10⁵
= 3.2 x 10⁻¹⁴
Acceleration a = force / mass
= 3.2 x 10⁻¹⁴ / 1.67 x 10⁻²⁷
= 1.9 x 10¹³ m s⁻²
Distance travelled by proton = 3 x 10⁻³
3 x 10⁻³ = 1/2 a t²
t = [tex]\sqrt{\frac{3\times2\times10^{-3}}{1.9\times10^{13}} }[/tex]
t = 1.77 x 10⁻⁸ s
The first accurate measurements of the properties of high-pressure gases were made by E. H. Amagat in France between 1869 and 1893. Before developing the dead-weight gauge, he worked in a mineshaft and used a mercury manometer for measurements of pressure to more than 400 bar. Estimate the height of the manometer required.
Answer:
Height, h = 300.27 meters
Explanation:
Given that,
Pressure of the gas, [tex]P=400\ bar=4\times 10^7\ Pa[/tex]
We need to find the height of the manometer required. The pressure at a height is given by :
[tex]P=\rho gh[/tex]
Where
[tex]\rho[/tex] is the density of mercury, [tex]\rho=13593\ kg/m^3[/tex]
h is the height of the manometer required.
[tex]h=\dfrac{P}{\rho g}[/tex]
[tex]h=\dfrac{4\times 10^7}{13593\times 9.8}[/tex]
h = 300.27 meters
So, the height of the manometer required is 300.27 meters. Hence, this is the required solution.
A runner in a relay race runs 20 m north, turns around and runs south for 30 m, then turns north again and runs 40 m. The entire run took 30 seconds. What was the average speed of the runner? What was the average velocity of the runner?
Answer:
The average speed its 1 m/sThe average velocity its 1 m/s to the north.Explanation:
So, lets say the runner stars from the position [tex]x_0[/tex]. Lets make this point the origin of a coordinate system in which the vector i points north.
[tex]x_0 = (0,0)[/tex]
Now, in the first sections of the race, he runs 20 meters north, so, he finds himself at:
[tex]x_1 = x_0 + 20 m * i = (0,0) \ + (20 \ m,0)[/tex].
[tex]x_1 = (20 \ m,0)[/tex].
The, he runs 30 meters south
[tex]x_2 = x_1 - 30 \ m * i = (20 \ m,0)-(30 \ m,0)[/tex]
[tex]x_2 = (-10 \ m,0)[/tex]
Finally, he runs 40 meter north
[tex]x_3 = x_2 + 40 \ m * i = (-10 \ m,0)+(40 \ m,0)[/tex]
[tex]x_3 = (30 \ m,0)[/tex].
This is our displacement vector. Now, the average speed will be:
[tex]\frac{distance}{time}[/tex].
The distance its the length of the displacement vector,
[tex]d=\sqrt{x^2+y^2}[/tex]
[tex]d=\sqrt{(30 \ m)^2+0^2}[/tex]
[tex]d=30 \ m[/tex]
So, the average speed its:
[tex]\frac{30 \ m }{30 \ s} = 1\frac{m}{s}[/tex].
The average velocity, instead, its:
[tex]\vec{v} = \frac{displacement}{time}[/tex]
[tex]\vec{v} = \frac{(30 \ m ,\ 0)}{30 \ s}[/tex]
[tex]\vec{v} = (1 \ \frac{m}{s} ,\ 0)[/tex]
This is, 1 m/s north.