Answer:
Explanation:
(a)Solving for the acceleration of the bullet
acceleration = (vf^2 – vi^2) / 2d
acceleration = ((280 m/s)^2 – (420 m/s)^2) / (2 * 0.12 m)
acceleration = (78400 - 176400) / 0.24 m
acceleration = -98000 / 0.24
acceleration = -408333 m/s^2
(a)Solving for contact time with board
t^2 = 2d/a
t^2 = 2 * 0.12 m / 408333 m/s^2
t^2 = 0.24 m / 408333 m/s^2
t^2 = 5.8775558 x 10^-7
t = 0.0007666 s or 767 microseconds
The Earth’s orbit around the Sun is slightly elliptical. At Earth's closest approach to the Sun (perihelion) the orbital radius is 1.471×10^11m, and at its farthest distance (aphelion) the orbital radius is 1.521×10^11m.
a. Find the difference in gravitational potential energy between when the Earth is at its aphelion and perihelion radii.
b. If the orbital speed of the Earth is 29,290 m/s at aphelion, what is its orbital speed at perihelion?
Answer:
1.25
Explanation:
g A small object of mass 2.5 g and charge 18 uC is suspended motionless above the ground when immersed in a uniform electric field perpendicular to the ground. What is the magnitude and direction of the electric field
Answer: [tex]1361.11\ N/C,\text{upward}[/tex]
Explanation:
Given
Mass of particle is [tex]m=2.5\ gm[/tex]
Charge of particle is [tex]q=18\ \mu C[/tex]
Electrostatic force must balance the weight of the particle
[tex]\lim_{n \to \infty} a_n \Rightarrow mg=qE\\\\\Rightarrow E=\dfrac{2.5\times 9.8\times 10^{-3}}{18\times 10^{-6}}\\\\\Rightarrow E=1361.1\ N/C[/tex]
Direction of the electric field is in upward direction such that it opposes the gravity force.
What is science?Give two examples of living beings?
Answer:
the study of the past
Explanation:
dogs and cats
A surveyor measures the distance across a straight river by the following method: Starting directly across from a tree on the opposite bank, he walks x = 106 m along the riverbank to establish a baseline. Then he sights across to the tree. The angle from his baseline to the tree is = 32.8°. How wide is the river?
Answer:
x = 68.3 m
Explanation:
tan 32.8 = x / 106
Assume that the car on the left makes a quick turn to the left. According to inertia, your body will resist a change and still want to go in the original direction. In which direction with the passenger slide?
Answer:
to the right
Explanation:
if the car turns to the lift, the body forces energy to the left side, so according to the first law of Newton, the body will move to the right side to resist the sudden motion.
In the diagram, the crest of the wave is show by:
A
B
C
D
Answer:
D.
Explanation:
The crest of a wave refers to the highest point of a wave. This is illustrated by D.
A strontium vapor laser beam is reflected from the surface of a CD onto a wall. The brightest spot is the reflected beam at an angle equal to the angle of incidence. However, fringes are also observed. If the wall is 1.2 m from the CD, and the second bright fringe is 0.803 m from the central maximum, what is the spacing (in m) of grooves on the CD
Answer:
[tex]d=1.29*10^{-6}m[/tex]
Explanation:
From the question we are told that:
Distance of wall from CD [tex]D=1.4[/tex]
Second bright fringe [tex]y_2= 0.803 m[/tex]
Let
Strontium vapor laser has a wavelength \lambda= 431 nm=>431 *10^{-9}m
Generally the equation for Interference is mathematically given by
[tex]y=frac{n*\lambda*D}{d}[/tex]
Where
[tex]d=\frac{n*\lambda*D}{y}[/tex]
[tex]d=\frac{2*431 *10^{-9}m*1.4}{0.803}[/tex]
[tex]d=1.29*10^{-6}m[/tex]
Find out other examples of bodies showing more than one type of motion Tabulate your findings.
Answer:
down below
Explanation:
Image 1- wheels of train showing both translatory motion as well as rotatory motion.
Image 2- rotation of ball shows both rotatory motion as well as translatory motion.
Image 3- the earth rotates about its axis, same time it revolves around the sun thus showing both rotatory motion and curvilinear motion in a fixed time. (perodic motion)
Image 4- while cutting wood, the
carpenter's saw has both
translatory motion and oscillatory
motion, as it moves down while
oscillating.
A Geiger counter registers a count rate of 8,000 counts per minute from a sample of a radioisotope. The count rate 24 minutes later is 1,000 counts per minute. What is the half-life of the radioisotope?
11.54 minutes
Explanation:
The decay rate equation is given by
[tex]N = N_0e^{-\frac{t}{\lambda}}[/tex]
where [tex]\lambda[/tex] is the half-life. We can rewrite this as
[tex]\dfrac{N}{N_0} = e^{-\frac{t}{\lambda}}[/tex]
Taking the natural logarithm of both sides, we get
[tex]\ln \left(\dfrac{N}{N_0}\right) = -\left(\dfrac{t}{\lambda}\right)[/tex]
Solving for [tex]\lambda[/tex],
[tex]\lambda = -\dfrac{t}{\ln \left(\frac{N}{N_0}\right)}[/tex]
[tex]\:\:\:\:= -\dfrac{(24\:\text{minutes})}{\ln \left(\frac{1000\:\text{counts/min}}{8000\:\text{counts/min}}\right)}[/tex]
[tex]\:\:\:\:=11.54\:\text{minutes}[/tex]
Consider a 200-ft-high, 1200-ft-wide dam filled to capacity. Determine (a) the hydrostatic force on the dam and (b) the force per unit area of the dam near the top and near the bottom. Note: we will see that the resultant hydrostatic force will be
Answer:
a) [tex]F_g=1.5*10^9Ibf[/tex]
b) [tex]F_t=12490Ibf/ft^2[/tex]
[tex]F_b=0[/tex]
Explanation:
From the question we are told that:
Height [tex]h=200ft[/tex]
Width [tex]w=1200ft[/tex]
a)
Generally the equation for Dam's Hydro static force is mathematically given by
[tex]F_g=\rho*g*\frac{h}{2}(w*h)[/tex]
Where
[tex]\rho=Density\ of\ water[/tex]
[tex]\rho=62.4Ibm/ft^3[/tex]
Therefore
[tex]F_g=62.4*32.2*\frac{200}{2}(1200*200)[/tex]
[tex]F_g=1.5*10^9Ibf[/tex]
b)
Generally the equation for Dam's Force per unit area is mathematically given by
[tex]F=\rho*g*h[/tex]
For Top
[tex]F_t=\rho*g*h[/tex]
[tex]F_t=62.4*32.2*200[/tex]
[tex]F_t=12490Ibf/ft^2[/tex]
For bottom
[tex]Here \\H=0 zero[/tex]
Therefore
[tex]F_b=0[/tex]
The hydrostatic force on the dam is [tex]2.995 \times 10^9 \ lbF[/tex].
The force per unit area near the top is 86.74 psi.
The force per unit area near the bottom is zero.
Hydrostatic force
The hydrostatic force on the dam is the force exerted on the dam by the column of the water.
[tex]F = PA\\\\F = (\rho gh) \times (wh)\\\\F = (62.4 \times 32.17 \times 200) \times (1200 \times 200)\\\\F = 9.636 \times 10^{10} \ lb-ft/s^2\\\\1 \ lbF = 32.17\ lb-ft/s^2\\\\F = 2.995 \times 10^9 \ lbF[/tex]
Force per unit area near the topThe force per unit area is the pressure exerted near the top of the dam.
[tex]P = \rho gh\\\\P = 0.052 \times \rho h[/tex]
where;
P is pressure in PSI
ρ is density of water in lb/gal
h is the vertical height in ft
[tex]P = 0.052 \times 8.34 \times 200\\\\P = 86.74 \ Psi[/tex]
The pressure near the bottom is zero, become the vertical height is zero.
Learn more about hydrostatic pressure here: https://brainly.com/question/11681616
a volcano that may erupt again at some time in the distant future is
Two streams merge to form a river. One stream has a width of 8.3 m, depth of 3.2 m, and current speed of 2.2 m/s. The other stream is 6.8 m wide and 3.2 m deep, and flows at 2.4 m/s. If the river has width 10.4 m and speed 2.8 m/s, what is its depth?
Answer:
The depth of the resulting stream is 3.8 meters.
Explanation:
Under the assumption that streams are formed by incompressible fluids, so that volume flow can observed conservation:
[tex]\dot V_{1} + \dot V_{2} = \dot V_{3}[/tex] (1)
All volume flows are measured in cubic meters per second.
Dimensionally speaking, we can determine the depth of the resulting stream ([tex]h_{3}[/tex]), in meters, by expanding (1) in this manner:
[tex]w_{1}\cdot h_{1}\cdot v_{1} + w_{2}\cdot h_{2}\cdot v_{2} = w_{3}\cdot h_{3}\cdot v_{3}[/tex]
[tex]h_{3} = \frac{w_{1}\cdot h_{1}\cdot v_{1}+w_{2}\cdot h_{2}\cdot v_{2}}{w_{3}\cdot v_{3}}[/tex] (2)
[tex]v_{1}, v_{2}[/tex] - Speed of the merging streams, in meters per second.
[tex]h_{1}, h_{2}[/tex] - Depth of the merging streams, in meters.
[tex]w_{1}, w_{2}[/tex] - Width of the merging streams, in meters.
[tex]w_{3}[/tex] - Width of the resulting stream, in meters.
[tex]v_{3}[/tex] - Speed of the resulting stream, in meters per second.
If we know that [tex]w_{1} = 8.3\,m[/tex], [tex]h_{1} = 3.2\,m[/tex], [tex]v_{1} = 2.2\,\frac{m}{s}[/tex], [tex]w_{2} = 6.8\,m[/tex], [tex]h_{2} = 3.2\,m[/tex], [tex]v_{2} = 2.4\,\frac{m}{s}[/tex], [tex]w_{3} = 10.4\,m[/tex] and [tex]v_{3} = 2.8\,\frac{m}{s}[/tex], then the depth of the resulting stream is:
[tex]h_{3} = \frac{(8.3\,m)\cdot (3.2\,m)\cdot \left(2.2\,\frac{m}{s} \right) + (6.8\,m)\cdot (3.2\,m)\cdot \left(2.4\,\frac{m}{s} \right)}{(10.4\,m)\cdot \left(2.8\,\frac{m}{s} \right)}[/tex]
[tex]h_{3} = 3.8\,m[/tex]
The depth of the resulting stream is 3.8 meters.
which one is more powerful hydrogen bomb or atom bomb and why?
Hydrogen bomb is more powerful than atom bomb
Hydrogen has a calorie value of 150000KJ .It is very much than nuclear bomb or atom bombScientists also told that Hydrogen bomb is more powerful.But both bombs are destructive.SI units are used for the scientific works,why?
Answer:
SI is used in most places around the world, so our use of it allows scientists from disparate regions to use a single standard in communicating scientific data without vocabulary confusion
If a proton and electron both move through the same displacement in an electric field, is the change in potential energy associated with the proton equal in magnitude and opposite in sign to the change in potential energy associated with the electron?
a. The magnitude of the change is smaller for the proton.
b. The magnitude of the change is larger for the proton.
c. The signs Of the two changes in potential energy are opposite.
d. They are equal in magnitude.
e. The signs of the two changes in potential energy are the same.
Answer: They are equal in magnitude.
- The signs of the two changes in potential energy are opposite
Explanation:
When the proton and electron both move through the same displacement in an electric field, the change in potential energy that is associated with the proton is equal in magnitude.
Also, it should be noted that the signs of the two changes in potential energy are opposite.
One charge is fixed q1 = 5 µC at the origin in a coordinate system, a second charge q2 = -3.2 µC the other is at a distance of x = 90 m from the origin.
What is the potential energy of this pair of charges?
Answer:
5.4uC
Explanation:
A long string is moved up and down with simple harmonic motion with a frequency of 46 Hz. The string is 579 m long and has a total mass of 46.3 kg. The string is under a tension of 3423 and is fixed at both ends. Determine the velocity of the wave on the string. What length of the string, fixed at both ends, would create a third harmonic standing wave
Answer:
a) [tex]v=206.896m/s[/tex]
b) [tex]L=6.749m[/tex]
Explanation:
From the question we are told that:
Frequency [tex]F=46Hz[/tex]
Length [tex]l=579m[/tex]
Total Mass [tex]T=4.3kg[/tex]
Tension [tex]T=3423[/tex]
a)
Generally the equation for velocity is mathematically given by
[tex]v=\sqrt{\frac{T}{\rho}}[/tex]
Where
[tex]\pho=m*l\\\\\pho=46*579\\\\\pho=0.0799kg/m[/tex]
Therefore
[tex]v=\sqrt{\frac{3423}{0.0799}}[/tex]
[tex]v=206.896m/s[/tex]
b)
Generally the equation for length of string is mathematically given by
[tex]L=\frac{3\lambda}{2}[/tex]
Where
[tex]\lambda=\frac{v}{f}[/tex]
[tex]\lambda=\frac{206.89}{46}[/tex]
[tex]\lambda=4.498[/tex]
Therefore
[tex]L=\frac{3*4.498}{2}[/tex]
[tex]L=6.749m[/tex]
2. A parallel-plate capacitor has a capacitance of C. If the area of the plates is doubled and
the distance between the plates is doubled, what is the new capacitance?
A) C/4
B) C/2
C)C
D) 4C
(C)
Explanation:
The capacitance C of a parallel plate capacitor is given by
[tex]C = \epsilon_0 \dfrac{A}{d}[/tex]
Let C' be the new capacitance where the area and the plate separation distance are doubled. This gives us
[tex]C' = \epsilon_0\dfrac{A'}{d'} = \epsilon_0\left(\dfrac{2A}{2d}\right) = \epsilon_0 \dfrac{A}{d} = C[/tex]
True or false : conservation of energy gives a relationship between the speed of a falling object and the height from which it was dropped
Answer:
truee
Explanation:
what is science ? what qualities do we deal in deal in physic ?
science is all about the world around us
For example, we can take Water
In (A) Water has same mass and great volume
In (B) Water has same mass and lower volume
Will there be any change in its density then?
Answer:
yes there will be change in its density
Assume that a friend hands you a 10-newton box to hold for her. If you hold the box without moving it at a height of 10 meters above the ground, how much work do you do
Answer:
100 Joules
Explanation:
Applying,
W = mgh................... Equation 1
Where W = workdone to hold the box above the ground, mg = weight of the box, h = height of the box.
From the question,
Given: mg = 10 newtons, h = 10 meters.
Substitute these values into equation 1
W = 10×10
W = 100 Joules.
Hence the amount of workdone is 100 Joules
A large metal sphere has three times the diameter of a smaller sphere and carries three times the charge. Both spheres are isolated, so their surface charge densities are uniform. Compare (a) the potentials (relative to infinity) and (b) the electric field strengths at their surfaces.
Answer:
A. Equals to that of the smaller sphere
B. 3 times less than that of the smaller sphere
Explanation:
(a) Equals to that of the smaller sphere
The potential of an isolated metal sphere, with charge Q and radius R, is kQ=R, so a sphere with charge 3Q and radius 3R has the same potential
b) 3 times less than that of the smaller sphere
However, the electric field at the surface of the smaller sphere is ?=? 0 = kQ=R2 , so tripling Q and R reduces the surface field by a factor of 1/3
A proton traveling at 17.6° with respect to the direction of a magnetic field of strength 3.28 mT experiences a magnetic force of 9.14 × 10-17 N. Calculate (a) the proton's speed and (b) its kinetic energy in electron-volts.
Answer:
a) The proton's speed is 5.75x10⁵ m/s.
b) The kinetic energy of the proton is 1723 eV.
Explanation:
a) The proton's speed can be calculated with the Lorentz force equation:
[tex] F = qv \times B = qvBsin(\theta) [/tex] (1)
Where:
F: is the force = 9.14x10⁻¹⁷ N
q: is the charge of the particle (proton) = 1.602x10⁻¹⁹ C
v: is the proton's speed =?
B: is the magnetic field = 3.28 mT
θ: is the angle between the proton's speed and the magnetic field = 17.6°
By solving equation (1) for v we have:
[tex]v = \frac{F}{qBsin(\theta)} = \frac{9.14 \cdot 10^{-17} N}{1.602\cdot 10^{-19} C*3.28 \cdot 10^{-3} T*sin(17.6)} = 5.75 \cdot 10^{5} m/s[/tex]
Hence, the proton's speed is 5.75x10⁵ m/s.
b) Its kinetic energy (K) is given by:
[tex] K = \frac{1}{2}mv^{2} [/tex]
Where:
m: is the mass of the proton = 1.67x10⁻²⁷ kg
[tex] K = \frac{1}{2}mv^{2} = \frac{1}{2}1.67 \cdot 10^{-27} kg*(5.75 \cdot 10^{5} m/s)^{2} = 2.76 \cdot 10^{-16} J*\frac{1 eV}{1.602 \cdot 10^{-19} J} = 1723 eV [/tex]
Therefore, the kinetic energy of the proton is 1723 eV.
I hope it helps you!
A cylinder is given a push and then rolls up an inclined plane. If the origin is the starting point, sketch the position, velocity, and acceleration of the cylinder vs. time as it goes up and then down the plane.
A 100 kg man is one fourth of the way up a 4.0 m ladder that is resting against a smooth, frictionless wall. The ladder has mass 25 kg and makes an angle of 56 degrees with the ground. What is the magnitude of the force of the wall on the ladder at the point of contact, if this force acts perpendicular to the wall and points away from the wall
Answer:
[tex]N_f=248N[/tex]
Explanation:
From the question we are told that:
Mass [tex]m=100kg[/tex]
Ladder Length [tex]l=4.0m[/tex]
Mass of Ladder [tex]m_l=25kg[/tex]
Angle [tex]\theta=56 \textdegree[/tex]
Generally the equation for Co planar forces is mathematically given by
[tex]mgcos \theta *2+Mgcos\theta*1 -N_fsin \theta*4=0[/tex]
Therefore
[tex]25*9.81cos 56 *2+100*9.81cos56*1 -N_fsin 56*4=0[/tex]
[tex]N_f=248N[/tex]
Derive the dimension of coefficient of linear expansivity
Answer:
The SI unit of coefficient of linear expansion can be expressed as °C-1 or °K-1. ... The dimension of coefficient of linear expansion will be M0L0T0K−1.
A force of 200 N, acting at 60° to the horizontal, accelerates a block of mass 50 kg along a horizontal plane. Calculate the component of the 200N force that accelerates the block horizontally
Answer:
Explanation:
a) Fx = F cos (θ)
= (200) cos(60)
= 100 N
b) FR = ma
Fx + Ff = ma
100 + Ff = (50)(1,5)
Ff = 75 - 100
= -25 N
c) Fy = F sin θ
= (200) sin(60)
= 173,2 N
Andrea's near point is 20.0 cm and her far point is 2.0 m. Her contact lenses are designed so that she can see objects that are infinitely far away. What is the closest distance that she can see an object clearly when she wears her contacts?
Answer:
the closest distance that she can see an object clearly when she wears her contacts is 22.2 cm
Explanation:
Given the data in the question,
near point = 20 cm
far point = 2 m = 200 cm
Now, for an object that is infinitely far away, the image is at is its far point.
so using the following expression, we can determine the focal length
1/f = 1/i + 1/o
where f is the focal length, i is the image distance and o is the object distance.
here, far point i = 2 m = 200 cm and v is ∞
so we substitute
1/f = 1/(-200 cm) + 1/∞
f = -200 cm
Also, for object at its closest point, the image appear at near point,
so
1/f = 1/i + 1/o
we make o the subject of formula
o = ( i × f ) / ( i - f )
given that near point i = 20 cm
we substitute
o = ( -20 × -200 ) / ( -20 - (-200) )
o = 4000 / 180
o = 22.2 cm
Therefore, the closest distance that she can see an object clearly when she wears her contacts is 22.2 cm
The cart travels the track again and now experiences a constant tangential acceleration from point A to point C. The speeds of the cart are 11.0 ft/s at point A and 18.0 ft/s at point C. The cart takes 5.00 s to go from point A to point C, and the cart takes 1.30 s to go from point B to point C. What is the cart's speed at point B
Answer:
The speed at B is 16.18 ft/s .
Explanation:
Speed at A, u = 11 ft/s
Speed at C, v' = 18 ft/s
Time from A to C = 5 s
Time from B to C = 1.3 s
Let the speed of car at B is v.
Let the acceleration is a.
From A to B
Use first equation of motion
v = u + a t
18 = 11 + a x 5
a = 1.4 ft/s^2
Let the time from A to B is t' .
t' = 5 - 1.3 = 3.7 s
Use first equation of motion from A to B
v = 11 + 1.4 x 3.7 = 16.18 ft/s
