Answer:
a. Momentum A = 280 Kgm/s.
b. Momentum B = 595 Kgm/s.
c. Total momentum = 875 Kgm/s.
Explanation:
Let the two people be A and B respectively.
Given the following data;
Mass A = 70kg
Mass B = 85kg
Velocity A = 4m/s
Velocity B = 7m/s
Momentum can be defined as the multiplication (product) of the mass possessed by an object and its velocity. Momentum is considered to be a vector quantity because it has both magnitude and direction.
Mathematically, momentum is given by the formula;
[tex] Momentum = mass * velocity [/tex]
a. To find the momentum of A;
[tex] Momentum \; A = 70 * 4 [/tex]
Momentum A = 280 Kgm/s.
b. To find the momentum of B;
[tex] Momentum \; B = 85 * 7 [/tex]
Momentum B = 595 Kgm/s.
c. To find the total momentum of the two persons;
[tex] Total \; momentum = Momentum \; A + Momentum \; B [/tex]
Substituting into the equation, we have;
[tex] Total momentum = 280 + 595 [/tex]
Total momentum = 875 Kgm/s.
what is a asteroid traveling rapidly called
Answer:
meteor
Explanation:
A asteroid stays still and a meteor goes fast
Answer:
meteor
Explanation:
or some people call it a shooting star
A hockey puck with mass 0.30 kg is sliding along the ice with initial speed of 12.68 m/s. A hockey player is heading toward the puck with his stick in hand. After the player strikes the puck, the puck reverses its direction and is traveling at double its speed before the strike. If the collision occurs in 0.05 s, what is the magnitude of the force the hockey player's stick applied to the puck
Answer:
F = 228.24 N
Explanation:
According Newton's 2nd Law, the impulse on one object is equal to the change in momentum of that object.I = F*Δt = Δp = pf - po (1)where pf = final momentum = m*vf
p₀ = initial momentum = m*v₀
Since after the strike, the puck reverses its direction and travels at double its speed before the strike, that means that vf = -2*v₀.Replacing in the right side of (1), we have:[tex]m*v_{f} - m*v_{o} = -2*v_{o} -m*v_{o} = -3*m*v_{o} = -3*0.3kg*12.68m/s = -11.41m/s (2)[/tex]
Replacing Δt = 0.05s, and solving for F in (1):[tex]F_{net} = \frac{-11.41m/s}{0.05s} = -228.24 N (3)[/tex]
which means that the force is applied in a direction opposite to the initial velocity of the puck.The magnitude of the force is just 228.24 N.ecosystem ecosystem science science
Answer:
c
Explanation:
Answer: C
Explanation:
If the kinetic energy of the 40kg box is 784 J, what is the velocity before it strikes the ground?
Answer:
Explanation:
[tex]KE=\frac{1}{2}mv^2[/tex]
[tex]784=\frac{1}{2}(40)v^2[/tex]
[tex]784=20v^2[/tex]
[tex]39.2=v^2[/tex]
[tex]v=6.26m/s[/tex]
Write any two uses of plane mirrors?
Answer:
Uses of plane mirrorsThey are used in periscopes, for signalling, in kaleidoscopes, to see round dangerous bends, in meters, as mirror tiles, in a sextant, in an overhead projector, an SLR camera, car wing mirrors, in microscopes and as reflecting number plates to mention only some!
Explanation:
Hope it is helpful....
Answer:
two uses are:
they are using for looking glassthey are used to make periscope13. Now we will examine the electric field of a dipole. The magnitude and direction of the electric field depends on the distance and the direction. We will investigate in detail just two directions. With charges available in the simulation how do you create a dipole with dipole moment 1 x 10-9 Cm with a direction for the dipole moment pointing to the right
This question is incomplete, the complete question is;
Now we will examine the electric field of a dipole. The magnitude and direction of the electric field depends on the distance and the direction. We will investigate in detail just two directions. With charges available in the simulation (all the charges are either positive or negative 1 nC increments).
how do you create a dipole with dipole moment 1 x 10-9 Cm with a direction for the dipole moment pointing to the right. Make a table below that shows the amounts of charge and the distance between the charges. There are many correct answers
Answer:
Given the data in question;
Dipole moment P = 1 × 10⁻⁹ C.m
now dipole pointing to the right;
P→
[tex]_{-\theta }[/tex] (-) ---------------->(+) [tex]_{+\theta }[/tex]
d
so let distance between the dipoles be d
∴ P = d[tex]\Theta[/tex]
Let [tex]\Theta_{1}[/tex] = 1 nC
so
P = d[tex]\Theta[/tex]
1 × 10⁻⁹ = 1 × 10⁻⁹ × d
d = (1 × 10⁻⁹) / (1 × 10⁻⁹)
d = 1 m
Also Let [tex]\Theta_{2}[/tex] = 2 nC
so
P = d[tex]\Theta[/tex]
1 × 10⁻⁹ = 2 × 10⁻⁹ × d
d = (1 × 10⁻⁹) / (2 × 10⁻⁹)
d = 0.5 m
Also Let [tex]\Theta_{3}[/tex] = 3 nC
so
P = d[tex]\Theta[/tex]
1 × 10⁻⁹ = 3 × 10⁻⁹ × d
d = (1 × 10⁻⁹) / (3 × 10⁻⁹)
d = 0.33 m
such that;
charge distance
1 nC 1.00 m
2 nC 0.50 m
3 nc 0.33 m
4 nC 0.25 m
5 nC 0.20 m
Which mode of kinetic energy contributes to temperature?
Answer:
Kinetic energy Temperature
Explanation:
Microbes such as bacteria have small positive charges when in solution. Public health agencies are exploring a new way to measure the presence of small numbers of microbes in drinking water by using electric forces to concentrate the microbes. Water is sent between the two oppositely charged electrodes of a parallel-plate capacitor. Any microbes in the water will collect on one of the electrodes.
Required:
a. On which electrode will the microbes collect?
b. How could the microbes be easily removed from the electrodes for analysis?
Answer:
The answer is below
Explanation:
a) Coulomb's law of electric force for charges at rest states that Like charges repel each other while unlike charges attract one another.
Therefore since the microbes has small positive charges, the microbe would be repelled by the positively charged electrodes and attracted by the negative charged electrodes.
Hence, the microbes would collect on the negatively charged electrodes.
b) The microbes can easily removed from the negative electrode for analysis by discharging the electrode from the source. Thereby making the electrode to be incapable of attracting the microbe.
Which one of the following statements is true concerning the magnitude of the electric field at a point in space? It is a measure of the electric force on any charged object. It is a measure of the ratio of the charge on an object to its mass. It is a measure of the electric force per unit mass on a test charge. It is a measure of the electric force per unit charge on a test charge. It is a measure of the total charge on the object.
Answer:
It is a measure of the electric force per unit charge on a test charge.
Explanation:
The magnitude of the electric field is defined as the force per charge on the test charge.
Since we define electric field as the force per charge, it will have the units of force divided by the unit of charge. This implies that the SI unit of electric field is given as Newton/Coulomb (N/C).
Because from the definition of the electric field intensity, it can be defined as force per unit charge. The correct answer is option D
ELECTRIC FIELDElectric field is the region of space where electric force can be felt. It can also be expressed as electric field intensity E. Mathematically, it can be expressed as;
E = F/q or E = V/d
From the question, the statements that is true concerning the magnitude of the electric field at a point in space is " It is a measure of the electric force per unit charge on a test charge "
Because from the definition of the electric field intensity, is can be defined as force per unit charge.
Therefore, option D is the right answer.
Learn more about Electric Field here : https://brainly.com/question/14372859
A spring with a spring constant of 22 N/m is stretched from equilibrium to 2.9 m. How much work is done in the process?
O A. 186 )
OB. 47 J
O C. 933
OD. 121 )
what is the answer ?
Answer:
using W=1/2kW2
k=22N/m w=2.9
w=1/2×22×2.9×2.9
w=92.51Joules
Approximately 93J answer is C
A transverse standing wave is set up on a string that is held fixed at both ends. The amplitude of the standing wave at an antinode is 2.20 mm and the speed of propagation of transverse waves on the string is 260 m/s. The string extends along the x-axis, with one of the fixed ends at x= 0, so that there is a node at x =0. The smallest value of x where there is an antinode is x= 0.150m.
Required:
a. What is the maximum transverse speed of a point on the string at an antinode?
b. What is the maximum transverse speed of a point on the string at x = 0.075 m?
Answer:
a) the maximum transverse speed of a point on the string at an antinode is 5.9899 m/s
b) the maximum transverse speed of a point on the string at x = 0.075 m is 4.2338 m/s
Explanation:
Given the data in the question;
as the equation of standing wave on a string is fixed at both ends
y = 2AsinKx cosωt
but k = 2π/λ and ω = 2πf
λ = 4 × 0.150 = 0.6 m
and f = v/λ = 260 / 0.6 = 433.33 Hz
ω = 2πf = 2π × 433.33 = 2722.69
given that A = 2.20 mm = 2.2×10⁻³
so [tex]V_{max1}[/tex] = A × ω
[tex]V_{max1}[/tex] = 2.2×10⁻³ × 2722.69 m/s
[tex]V_{max1}[/tex] = 5.9899 m/s
therefore, the maximum transverse speed of a point on the string at an antinode is 5.9899 m/s
b)
A' = 2AsinKx
= 2.20sin( 2π/0.6 ( 0.075) rad )
= 2.20 sin( 0.7853 rad ) mm
= 2.20 × 0.706825 mm
A' = 1.555 mm = 1.555×10⁻³
so
[tex]V_{max2}[/tex] = A' × ω
[tex]V_{max2}[/tex] = 1.555×10⁻³ × 2722.69
[tex]V_{max2}[/tex] = 4.2338 m/s
Therefore, the maximum transverse speed of a point on the string at x = 0.075 m is 4.2338 m/s
How will the plants that grow from a plant fertilized seeds compare to the plants grown from its daughter tubers?
Answer:
A
Explanation:
the answer would be basicly A
The x-component and y-component of two vectors A & B are Ax = 9, Ay = 12,Bx =
15 & By = 20. Find:
/A+B/
Answer:
40Explanation:
Given the following
Ax = 9,
Ay = 12,
Bx = 15
By = 20
Get A and B
A = √9²+12²
A= √81+144
A = √225
A = 15
Get B;
B = √15²+20²
B = √225+400
B = √625
B = 25
get /A+B/
A+B = 15+25
/A+B/ = /40/
Hence the value of /A+B/ is 40
You've been hired to design the hardware for an ink jet printer. You know that these printers use a deflecting electrode to cause charged ink drops to form letters on a page. The basic mechanism is that uniform ink drops of about 30 microns radius are charged to varying amounts after being sprayed out towards the page at a speed of about 20 m/s. Along the way to the page, they pass into a region between two deflecting plates that are 1.6 cm long. The deflecting plates are 1.0 mm apart and charged to 1500 volts. You measure the distance from the edge of the plates to the paper and find that it is one-half inch. Assuming an uncharged droplet forms the bottom of the letter, how much charge is needed on the droplet to form the top of a letter 3 mm high (11 pt. type)
Answer:
the required charged is 7.06 × 10⁻¹³ C
Explanation:
Given that;
Radius = 30 microns = 30 × 10⁻⁶
Speed v = 20 m/s
length x = 1.6 cm = 0.016 m
spacing d = 1.0 mm = 0.001 m
Voltage V = 1500 V
from the question, the electric field between the plates is uniform and equal to Voltage divided by the distance between the plates.
Electric field E = V/d
E = 1500 V / 0.001 m
E = 1.5 × 10⁶ V/m
Mass of ink drop m = pv
m = 10³ kg/m³ × [tex]\frac{4}{3}[/tex]πr³
m = 1000 kg/m³ × [tex]\frac{4}{3}[/tex]π × (30 × 10⁻⁶)³
m = 1.131 × 10⁻¹⁰ Kg
Time taken to travel t = x / sped
t = 0.016 m / 20 m/s
t = 0.0008 s
From the kinematic equation
to form the top of a letter 3 mm ( 0.003 m )high
y = [tex]\frac{1}{2}[/tex]at²
2y = at²
a = 2y/t²
we substitute
a = (2 × 0.003 m) / (0.0008 s)²
a = 9375 m/s²
Now Force F = Eq = ma
so
q = ma / E
we substitute
q = ( 1.131 × 10⁻¹⁰ Kg × 9375 m/s² ) / ( 1.5 × 10⁶ V/m )
q = 7.06 × 10⁻¹³ C
Therefore, the required charged is 7.06 × 10⁻¹³ C
what part of the electromagnetic spectrum can our skin detect?
Visible and infrared light.
I WILL MARK BRAINLIST!!!!
a person drops their phone. The phone's mass is 0.115 kilograms and the bridge is 15 meters tall. The instant the they dropped the phone, what was its mechanical energy?
1.725
16.9
200
What is the main cause of ocean currents? Question 2 options:
The prevailing winds
The Coriolis effect
Waves
The sun and the moon
A disk with radius R and uniform positive charge density s lies horizontally on a tabletop. A small plastic sphere with mass M and positive charge Q hovers motionless above the center of the disk, suspended by the Coulomb repulsion due to the charged disk.
Required:
a. What is the magnitude of the net upward force on the sphere as a function of the height z above the disk?
b. At what height h does the sphere hover?
Answer:
a. F = Qs/2ε₀[1 - z/√(z² + R²)] b. h = (1 - 2mgε₀/Qs)R/√[1 - (1 - 2mgε₀/Qs)²]
Explanation:
a. What is the magnitude of the net upward force on the sphere as a function of the height z above the disk?
The electric field due to a charged disk with surface charge density s and radius R at a distance z above the center of the disk is given by
E = s/2ε₀[1 - z/√(z² + R²)]
So, the net force on the small plastic sphere of mass M and charge Q is
F = QE
F = Qs/2ε₀[1 - z/√(z² + R²)]
b. At what height h does the sphere hover?
The sphere hovers at height z = h when the electric force equals the weight of the sphere.
So, F = mg
Qs/2ε₀[1 - z/√(z² + R²)] = mg
when z = h, we have
Qs/2ε₀[1 - h/√(h² + R²)] = mg
[1 - h/√(h² + R²)] = 2mgε₀/Qs
h/√(h² + R²) = 1 - 2mgε₀/Qs
squaring both sides, we have
[h/√(h² + R²)]² = (1 - 2mgε₀/Qs)²
h²/(h² + R²) = (1 - 2mgε₀/Qs)²
cross-multiplying, we have
h² = (1 - 2mgε₀/Qs)²(h² + R²)
expanding the bracket, we have
h² = (1 - 2mgε₀/Qs)²h² + (1 - 2mgε₀/Qs)²R²
collecting like terms, we have
h² - (1 - 2mgε₀/Qs)²h² = (1 - 2mgε₀/Qs)²R²
Factorizing, we have
[1 - (1 - 2mgε₀/Qs)²]h² = (1 - 2mgε₀/Qs)²R²
So, h² = (1 - 2mgε₀/Qs)²R²/[1 - (1 - 2mgε₀/Qs)²]
taking square-root of both sides, we have
√h² = √[(1 - 2mgε₀/Qs)²R²/[1 - (1 - 2mgε₀/Qs)²]]
h = (1 - 2mgε₀/Qs)R/√[1 - (1 - 2mgε₀/Qs)²]
A scientist reports a measurement of the temperature of the surface of a newly discovered planet as negative 20 Kelvin. What conclusion can you draw from this report
Answer:
The temperature of this newly discovered planet violates the third law of thermodynamic, there is a mistake in this value.
Explanation:
The third law of the Thermodynamic says:
At zero kelvin all molecular movement stops, which means that the entropy will be zero at this temperature.
So we can say there is no thermodynamic system that has temperature values less than 0 K.
The conclusion of the report will be.
The temperature of this new planet violates the third law of thermodynamic, there is a mistake in this value.
I hope it helps you!
How much power does it take to lift 30.0 kg 100 m high in 5.00 s?
._.Answer:.
ijji._. ji
Explanation:.
The electron gun in an old TV picture tube accelerates electrons between two parallel plates 1.3 cm apart with a 21 kV potential difference between them. The electrons enter through a small hole in the negative plate, accelerate, then exit through a small hole in the positive plate. Assume that the holes are small enough not to affect the electric field or potential.
Required:
a. What is the electric field strength between the plates?
b. With what speed does an electron exit the electron gun if its entry speed is close to zero?
Answer:
a. Electric field strength is 1.6*10^6 V/m = 1.6 Megavolt/meter
b. speed of an electron is 5.2*10^14m/s
Explanation:
The electric field strength is given by:
E = ΔV/d
E = electric field strength,
ΔV = potential difference,
d = plate spacin
ΔV = 21×10^3V, d = 1.3×10^-2m
a) E = V / d = (21*10^3)/(1.3×10^-2) = 1.6*10^6 V/m = 1.6 Megavolt/meter
b) Exit energy = V . e
Where 'e' is the charge of electron (1.6 * 10^(-19)
Exit energy = (21*10^3)*(1.6*10^-19) = 3.36*10−15
From the above, speed v = √2*Exit Energy/mass
Where 'm' is the mass of electron (9.1 * 10^(-31)
v = √2*energy/mass
v = √2*(3.36*10^-15)/(9.1*10^-31) = 5.2*10^14m/s
A girl standing on a bridge throws a stone vertically downward with an initial velocity of 15.0 m/s into the river below. If the stone hits the water 2.00 seconds later, what is the height of the bridge above the water
Answer:
the height of the bridge above the water is 49.6 m.
Explanation:
Given;
initial velocity of the stone, u = 15 m/s
time of motion of the stone, t = 2 s
The height of the bridge above the water is calculated from the following kinematic equation as follows;
h = ut + ¹/₂gt²
h = (15 x 2) + ¹/₂(9.8)(2²)
h = 30 + 19.6
h = 49.6 m
Therefore, the height of the bridge above the water is 49.6 m.
In trial 1 of an experiment, a cart moves with a speed of vo on a frictionless, horizontal track and collides with another cart that is initially at rest. In trial 2, the setup is identical except the carts stick together during the collision. How does the speed of the two-cart system's center of mass change, if at all, during the collision in each trial
Answer:
1) elastic shock, the velocity of the center of mass does not change
2) inelastic shock, he velocity of the mass center change
Explanation:
The position of the center of mass of your system is defined by
[tex]x_{cm}[/tex] = [tex]\frac{1}{M} \sum x_i m_i[/tex]
in this case we have two bodies
x_{cm} = [tex]\frac{1}{M}[/tex] (x₁m₁ + x₂ m₂)
the velocity of the center of mass is
x_{cm} = dx_{cm} / dt = [tex]\frac{1}{M} ( m_1 \frac{dx_1}{dt} \ + m_2 \frac{dx_2}{dt} )[/tex]
x_{cm} = [tex]\frac{1}{M} ( m_1 v_1 + m_2 v_2 )[/tex]
where M is the total mass of the system.
Therefore to answer this question we have to find the velocity of the body after the collision.
Let's use momentum conservation, where the system is formed by the two bodies, so that the forces have been internal during the collision.
Let's solve each case separately.
2) inelastic shock
initial instant. Before the crash
p₀ = m₁ v₀ + 0
final instant. After the collision with the cars together
p_f = (m₁ + m₂) v
p₀ = p_f
m₁ v₀ = (m₁ + m₂) v
v = [tex]\frac{m_1}{m_1+m_2}[/tex] v₀
let's find the velocity of the center of mass
M = m₁ + m₂
initial.
[tex]v_{cm o}[/tex] = [tex]\frac{1}{m_1 +m_2}[/tex] (m₁ vo)
final
[tex]v_{cm f}[/tex] = [tex]\frac{1}{M} ( \frac{m_1}{m_1 + m_2} v_o )[/tex] ( v) = v
v_{cm f} = [tex]\frac{m_1}{M^2} v_o[/tex]
Let's find the ratio of the velocities of the center of mass
vcmf / vcmo = [tex]\frac{1}{M} = \frac{1}{m_1 +m_2}[/tex]
therefore the velocity of the mass center change
1) elastic shock
initial instant.
p₀ = m₁ v₀
final moment
p_f = m₁ v_{1f} + m₂ v_{2f}
p₀ = p_f
m₁ v₀ = m₁ v_{1f} + m₂ v_{2f}
m₁ (v₀ - v_{2f}) = m₂ v_{2f}
in this case the kinetic energy is conserved
K₀ = K_f
½ m₁ v₀² = ½ m₁ v_{1f}² + ½ m₂ v_{2f}²
m₁ (v₀² - v_{1f}²) = m₂ v_{2f}²
m₁ (v₀ + v_{1f}) (v₀ - v_{1f}) = m₂ v_{2f}
we write our system of equations
m₁ (v₀ - v_{1f}) = m₂ v_{2f} (1)
m₁ (v₀ - v_{1f}) (v₀ + v_{1f}) = m₂ v_{2f}²
we solve the system
v₀ + v_{1f} = v_{2f}
we substitute and look for the final speeds
v_{1f} = [tex]\frac{m_1 -m_2}{m1 +m2 } v_o[/tex]
v_{2f} = [tex]\frac{2 m_1}{m-1+m_2} vo[/tex]
now let's find the velocity of the center of mass
initial
[tex]v_{cm o}[/tex] = [tex]\frac{1}{M}[/tex] m₁ v₀
final
[tex]v_{cm f}[/tex] = [tex]\frac{1}{M}[/tex] (m₁ v_{1f} + m₂ v_{2f} )
v_{cm f} = [tex]\frac{1}{M}[/tex] [ [tex]m_1 \frac{m_2}{M}[/tex] + [tex]m_2 \frac{2 m_1}{M}[/tex] ] v₀
v_{cm f} = [tex]\frac{1}{M^2}[/tex] ( m₁² - m₁m₂ +2 m₁m₂) v₂
v_{cm f} = [tex]\frac{1}{M^2}[/tex] (m₁² + m₁ m₂) v₀
let's look for the relationship
v_{cm f} / v_{cm o} = [tex]\frac{1}{M}[/tex] M
v_{cm f} / v_{cm o} = 1
therefore the velocity of the center of mass does not change
we see in either case the velocity of the center of mass does not change.
Where the air conditioner disconnecting means is not within sight from the equipment, the provision for locking or adding a lock to the disconnecting means shall be on the switch or circuit breaker and remain in place _____ the lock installed.
Answer:
With or without.
Explanation:
According to the National Electrical Code (NEC), here the air conditioner disconnecting means is not within sight from the equipment, the provision for locking or adding a lock to the disconnecting means shall be on the switch or circuit breaker and remain in place with or without the lock installed. Thus, this is in accordance with section 110.25 of the National Electrical Code (NEC).
A system with a mass of 10 kg, initially moving horizontally with a velocity of 20 m/s, experiences a constant horizontal force of 25 N opposing the direction of motion. As a result, the system comes to rest. Determine the amount of energy transfer by work, in kJ, for this process and the total distance, in m, that the system travels
Answer:
Explanation:
Kinetic energy of the mass of 10 kg
= 1/2 m v² , m is mass and v is velocity .
= .5 x 10 x 20²
= 2000 J
The opposing force stops it . so work done by opposing force will be equal to this energy and it will be negative .
So energy transfer will be - 2000 J .
= 2 kJ .
If distance travelled by mass is d , force 25 N will have a displacement of d . so work done by force of 25 N
= 25 x d
25 d = 2000
d = 80 m .
Hence system travels a distance of 80 m .
As the distance between the sun and earth decreases, the gravity force between them
a
Increases
b
decreases
c
stays the same
Answer:
a increases
Explanation:
as distance between two objects increases the gravitational force decreases so when distance decreases the gravitational force increases
What happens to the force attraction of the distance two objects is increased?
Answer:
Explanation:
The attraction weakens. Two objects that are farther apart are not drawn together as strongly as if they were close together.
The resolution of a lens can be estimated by treating the lens as a circular aperture. The resolution is the smallest distance between two point sources that produce distinct images. This is similar to the resolution of a single slit, related to the distance from the middle of the central bright band to the firstorder dark band; however, the aperture is circular instead of a rectangular slit which introduces a scale factor. Suppose the Hubble Space Telescope, 2.4 m in diameter, is in orbit 90.4 km above Earth and is turned to look at Earth. If you ignore the effect of the atmosphere, what is the resolution of this telescope for light of wavelength 557 nm?
Answer:
y = 2.56 10⁻² m
Explanation:
The resolution of this telescope is given by the Rayleigh criterion, for the phenomenal diffraction the first minimum for a linear slit is in
a sin θ = λ
in general the angles are very small, so we approximate
sin θ = θ
we substitute
θ = λ / a
in the case of circular slits we must use polar coordinates, which introduces a numerical factor, leaving the equation
θ = 1.22 [tex]\frac{\lambda }{D}[/tex]
where D is the diameter of the circular opening
In this case they indicate the lens diameter D = 2.4 m, the observation distance r = 90.4 km = 90.4 10³ m
how angles are measured in radians
θ = y / r
we substitute
y / r = 1.22\frac{\lambda }{D}
y = 1.22 \frac{\lambda r }{D}
let's calculate
y = [tex]1.22 \frac{ 557 \ 10^{-9} \ \ 90.4 \ 10^{3} }{2.4}[/tex]
y = 2.56 10⁻² m
this is the minimum distance that can differentiate two objects on Earth
Two masses are being pulled up a 30.0-degree incline by a force F parallel to the incline. The acceleration up the incline is 1.00 m/s2 and the velocity is down the incline. The force is applied to a 200-kg mass and a string connects the 200-kg mass to a 150-kg mass. The coefficient of kinetic friction is 0.200. The force F is
Answer:
Explanation:
The acceleration up the incline is 1.00 m/s²
Net force acting on two masses = total mass x acceleration
= 350 x 1 = 350 N
weight acting down the plane = m g sinФ
= 350 x 9.8 x sin30 = 1715 N
Friction force acting down the plane = mg cosФ x μ where μ is coefficient of friction
= 350 x 9.8 x cos30 x .2 = 594N
Net force acting on total mass
= F - 1715 - 594 = 350 , where F is required force
F = 2659 N .
If the mass of the book is 50 sliding with acceleration 1.2 m/s ^ 2 then the friction force is
364N
185N
173N
73N
Answer and I will give you brainiliest