Answer:
vr = 142.90 m/s
the magnitude of its relative velocity is 142.90 m/s
Explanation:
Given;
A plane has an airspeed of 142 m/s (eastward)
vi = 142 m/s
16.0 m/s wind is blowing southward at the same time as the plane is flying
vb = 16.0m/s
Writing the relative velocity vector, we have;
Taking north and south as positive and negative y axis respectively, east and west as positive and negative x axis respectively.
v = 142i - 16j
The magnitude of the velocity is;
vr = √(vi^2 + vb^2)
vr = √(142^2 + 16^2)
vr = √(20420)
vr = 142.8985654231 m/s
vr = 142.90 m/s
the magnitude of its relative velocity is 142.90 m/s
What direct current will produce the same amount of thermal energy, in a particular resistor, as an alternating current that has a maximum value of 2.59 A?
Answer:
The direct current that will produce the same amount of thermal energy is 1.83 A
Explanation:
Given;
maximum current, I₀ = 2.59 A
The average power dissipated in a resistor connected in an AC source is given as;
[tex]P_{avg} = I_{rms} ^2R[/tex]
Where;
[tex]I_{rms} = \frac{I_o}{\sqrt{2} }[/tex]
[tex]P_{avg} = (\frac{I_o}{\sqrt{2} } )^2R\\\\P_{avg} = \frac{I_o^2R}{2} ----equation(1)[/tex]
The average power dissipated in a resistor connected in a DC source is given as;
[tex]P_{avg} = I_d^2R --------equation(2)[/tex]
where;
[tex]I_d[/tex] is direct current
Solve equation (1) and (2) together;
[tex]I_d^2R = \frac{I_o^2R}{2} \\\\I_d^2 = \frac{I_o^2}{2} \\\\I_d=\sqrt{\frac{I_o^2}{2} } \\\\I_d = \frac{I_o}{\sqrt{2}} \\\\I_d = \frac{2.59}{\sqrt{2} } \\\\I_d = 1.83 \ A[/tex]
Therefore, the direct current that will produce the same amount of thermal energy is 1.83 A
Two lenses of focal length 4.5cm and 1.5cm are placed at a certain distance apart, calculate the distance between the lenses if they form an achromatic combination
3.0cm
Explanation:
For lenses in an achromatic combination, the following condition holds, assuming the two lenses are of the same materials;
d = [tex]\frac{f_1 + f_2}{2}[/tex] ---------(i)
Where;
d= distance between lenses
f₁ = focal length of the first lens
f₂ = focal length of the second lens
From the question;
f₁ = 4.5cm
f₂ = 1.5cm
Substitute these values into equation (i) as follows;
d = [tex]\frac{4.5+1.5}{2}[/tex]
d = [tex]\frac{6.0}{2}[/tex]
d = 3.0cm
Therefore, the distance between the two lenses is 3.0cm
When a hydrometer (see Fig. 2) having a stem diameter of 0.30 in. is placed in water, the stem protrudes 3.15 in. above the water surface. If the water is replaced with a liquid having a specific gravity of 1.10, how much of the stem would protrude above the liquid surface
Answer:
5.79 in
Explanation:
We are given that
Diameter,d=0.30 in
Radius,r=[tex]\frac{d}{2}=\frac{0.30}{2}=0.15 in[/tex]
Weight of hydrometer,W=0.042 lb
Specific gravity(SG)=1.10
Height of stem from the water surface=3.15 in
Density of water=[tex]62.4lb/ft^3[/tex]
In water
Volume of water displaced [tex]V=\frac{mass}{density}=\frac{0.042}{62.4}=6.73\times 10^{-4} ft^3[/tex]
Volume of another liquid displaced=[tex]V'=\frac{V}{SG}=\frac{6.73\times 10^{-4}}{1.19}=5.66\times 10^{-4}ft^3[/tex]
Change in volume=V-V'
[tex]V-V'=\pi r^2 l[/tex]
Substitute the values
[tex]6.73\times 10^{-4}-5.66\times 10^{-4}=3.14\times (\frac{0.15}{12})^2l[/tex]
By using
1 ft=12 in
[tex]\pi=3.14[/tex]
[tex]l=\frac{6.73\times 10^{-4}-5.66\times 10^{-4}}{3.14\times (\frac{0.15}{12})^2}[/tex]
l=2.64 in
Total height=h+l=3.15+2.64= 5.79 in
Hence, the height of the stem protrude above the liquid surface=5.79 in
An electron has an initial velocity of (17.1 + 12.7) km/s, and a constant acceleration of (1.60 × 1012 m/s2) in the positive x direction in a region in which uniform electric and magnetic fields are present. If = (529 µT) find the electric field .
Answer:
Explanation:
Since B is perpendicular, it does no work on the electron but instead deflects it in a circular path.
q = 1.6 x 10-19 C
v = (17.1j + 12.7k) km/s = square root(17.1² + 12.7²) = 2.13 x 10⁴ m/s
the force acting on electron is
F= qvBsinΦ
F= (1.6 x 10⁻¹⁹C)(2.13.x 10⁴ m/s)(526 x 10⁻⁶ T)(sin90º)
F = 1.793x 10⁻¹⁸ N
The net force acting on electron is
F = e ( E+ ( vXB)
= ( - 1.6 × 10⁻¹⁹) ( E + ( 17.1 × 10³j + 12.7 × 10³ k)X( 529 × 10⁻⁶ ) (i)
= ( -1.6 × 10⁻¹⁹ ) ( E- 6.7k + 9.0j)
a= F/m
1.60 × 10¹² i = ( -1.6 × 10⁻¹⁹ ) ( E- 6.9 k + 7.56 j)/9.11 × 10⁻³¹
9.11 i = - ( E- 6.7 k + 9.0 j)
E = -9.11i + 6.7k - 9.0j
what are the rays that come of the sun called? A. Ultraviolent rays B. Gamma rays C. soundwaves D. sonic rays
Answer:
The answer is ultraviolet rays...
Explanation:
...because the ozone layer protects us from the UV rays of the sun.
1. A ski-plane with a total mass of 1200 kg lands towards the west on a frozen lake at 30.0
m/s. The coefficient of kinetic friction between the skis and the ice is 0.200. How far does
the plane slide before coming to a stop?
Answer:
d = 229.5 m
Explanation:
It is given that,
Total mass of a ski-plane is 1200 kg
It lands towards the west on a frozen lake at 30.0 m/s.
The coefficient of kinetic friction between the skis and the ice is 0.200.
We need to find the distance covered by the plane before coming to rest. In this case,
[tex]\mu mg=ma\\\\a=\mu g\\\\a=0.2\times 9.8\\\\a=1.96\ m/s^2[/tex]
It is decelerating, a = -1.96 m/s²
Now using the third equation of motion to find the distance covered by the plane such that :
[tex]v^2-u^2=2ad\\\\d=\dfrac{-u^2}{2a}\\\\d=\dfrac{-(30)^2}{2\times -1.96}\\\\d=229.59\ m[/tex]
So, the plane slide a distance of 229.5 m.
Which of the following gives the magnitude of the average velocity (over the entire run) of an athlete running on a circular track with a circumference of 0.5 km, if that athlete runs a total length of 1.0 km in a time interval of 4 minutes?
a. O m/s
b. 2 m/s
c. 4.2 m/s
d. 16.8 m/s
Answer:
c. 4.2 m/s
Explanation:
The definition of the average velocity, measured in meters per second, is given by the following expression:
[tex]\bar v = \frac{x_{f}-x_{o}}{t_{f}-t_{o}}[/tex]
Where:
[tex]x_{o}[/tex], [tex]x_{f}[/tex] - Initial and final positions, measured in meters.
[tex]t_{o}[/tex], [tex]t_{f}[/tex] - Initial and final instants, measured in seconds.
Positions and instants must be written in meters and seconds, respectively:
[tex]x_{o} = 0\,m[/tex], [tex]x_{f} = 1000\,m[/tex].
[tex]t_{o} = 0\,s[/tex], [tex]t_{f} = 240\,s[/tex].
Finally, the average velocity of the athlete that runs a total length of 1.0 kilometer in a time interval of 4 minutes is:
[tex]\bar v = \frac{1000\,m-0\,m}{240\,s-0\,s}[/tex]
[tex]\bar v = 4.167\,\frac{m}{s}[/tex]
Hence, the best option is C.
Use Kepler's third law to determine how many days it takes a spacecraft to travel in an elliptical orbit from a point 6 590 km from the Earth's center to the Moon, 385 000 km from the Earth's center.
Answer:
1.363×10^15 seconds
Explanation:
The spaceship travels an elliptical orbit from a point of 6590km from the earth center to the moon and 38500km from the earth center.
To calculate the time taken from Kepler's third Law :
T^2 = ( 4π^2/GMe ) r^3
Where Me is the mass of the earth
r is the average distance travel
G is the universal gravitational constant. = 6.67×10-11 m3 kg-1 s-2
π = 3.14
Me = mass of earth = 5.972×10^24kg
r =( r minimum + r maximum)/2 ......1
rmin = 6590km
rmax = 385000km
From equation 1
r = (6590+385000)/2
r = 391590/2
r = 195795km
From T^2 = ( 4π^2/GMe ) r^3
T^2 = (4 × 3.14^2/ 6.67×10-11 × 5.972×10^24) × 195795^3
= ( 4×9.8596/ 3.983×10^14 ) × 7.5059×10^15
= 39.4384/ 3.983×10^14 ) × 7.5059×10^15
= (9.901×10^14) × 7.5059×10^15
T^2 = 7.4321× 10^30
T =√7.4321× 10^30
T = 2.726×10^15 seconds
The time for one way trip from Earth to the moon is :
∆T = T/2
= 2.726×10^15 /2
= 1.363×10^15 secs
2. A 2.0-kg block slides down an incline surface from point A to point B. Points A and B are 2.0 m apart. If the coefficient of kinetic friction is 0.26 and the block is starting at rest from point A. What is the work done by friction force
Answer:a
Explanation:
A 40.0 kg ballet dancer stands on her toes during a performance with 25.0 cm2 in contact with the floor. What is the pressure exerted by the floor over the area of contact if the dancer is stationary
Explanation:
40×10ms^-2
400N.
25/100m^2
0.25m^2
1.P=F/A
=400N/0.25m^2
=100Nm^-2
=100Pa
A coil has resistance of 20 W and inductance of 0.35 H. Compute its reactance and its impedance to an alternating current of 25 cycles/s.
Answer:
Reactance of the coil is 55 WImpedance of the coil is 59 WExplanation:
Given;
Resistance of the coil, R = 20 W
Inductance of the coil, L = 0.35 H
Frequency of the alternating current, F = 25 cycle/s
Reactance of the coil is calculated as;
[tex]X_L=[/tex] 2πFL
Substitute in the given values and calculate the reactance [tex](X_L)[/tex]
[tex]X_L =[/tex] 2π(25)(0.35)
[tex]X_L[/tex] = 55 W
Impedance of the coil is calculated as;
[tex]Z = \sqrt{R^2 + X_L^2} \\\\Z = \sqrt{20^2 + 55^2} \\\\Z = 59 \ W[/tex]
Therefore, the reactance of the coil is 55 W and Impedance of the coil is 59 W
A crate of mass 9.2 kg is pulled up a rough incline with an initial speed of 1.58 m/s. The pulling force is 110 N parallel to the incline, which makes an angle of 20.2° with the horizontal. The coefficient of kinetic friction is 0.400, and the crate is pulled 5.10 m.A) How much work is done by the gravitational force on thecrate?
B) Determine the increase in internal energy of the crate-inclinesystem owing to friction.
C) How much work is done by the 100N force on the crate?
D) What is the change in kinetic energy of the crate?
E) What is the speed of the crate after being pulled 5.00m?
Given that,
Mass = 9.2 kg
Force = 110 N
Angle = 20.2°
Distance = 5.10 m
Speed = 1.58 m/s
(A). We need to calculate the work done by the gravitational force
Using formula of work done
[tex]W_{g}=mgd\sin\theta[/tex]
Where, w = work
m = mass
g = acceleration due to gravity
d = distance
Put the value into the formula
[tex]W_{g}=9.2\times(-9.8)\times5.10\sin20.2[/tex]
[tex]W_{g}=-158.8\ J[/tex]
(B). We need to calculate the increase in internal energy of the crate-incline system owing to friction
Using formula of potential energy
[tex]\Delta U=-W[/tex]
Put the value into the formula
[tex]\Delta U=-(-158.8)\ J[/tex]
[tex]\Delta U=158.8\ J[/tex]
(C). We need to calculate the work done by 100 N force on the crate
Using formula of work done
[tex]W=F\times d[/tex]
Put the value into the formula
[tex]W=100\times5.10[/tex]
[tex]W=510\ J[/tex]
We need to calculate the work done by frictional force
Using formula of work done
[tex]W=-f\times d[/tex]
[tex]W=-\mu mg\cos\theta\times d[/tex]
Put the value into the formula
[tex]W=-0.4\times9.2\times9.8\cos20.2\times5.10[/tex]
[tex]W=-172.5\ J[/tex]
We need to calculate the change in kinetic energy of the crate
Using formula for change in kinetic energy
[tex]\Delta k=W_{g}+W_{f}+W_{F}[/tex]
Put the value into the formula
[tex]\Delta k=-158.8-172.5+510[/tex]
[tex]\Delta k=178.7\ J[/tex]
(E). We need to calculate the speed of the crate after being pulled 5.00m
Using formula of change in kinetic energy
[tex]\Delta k=\dfrac{1}{2}m(v_{2}^2-v_{1}^{2})[/tex]
[tex]v_{2}^2=\dfrac{2\times\Delta k}{m}+v_{1}^2[/tex]
Put the value into the formula
[tex]v_{2}^2=\dfrac{2\times178.7}{9.2}+1.58[/tex]
[tex]v_{2}=\sqrt{\dfrac{2\times178.7}{9.2}+1.58}[/tex]
[tex]v_{2}=6.35\ m/s[/tex]
Hence, (A). The work done by the gravitational force is -158.8 J.
(B). The increase in internal energy of the crate-incline system owing to friction is 158.8 J.
(C). The work done by 100 N force on the crate is 510 J.
(D). The change in kinetic energy of the crate is 178.7 J.
(E). The speed of the crate after being pulled 5.00m is 6.35 m/s
The electric field must be zero inside a conductor in electrostatic equilibrium, but not inside an insulator. It turns out that we can still apply Gauss's law to a Gaussian surface that is entirely within an insulator by replacing the right-hand side of Gauss's law, Qin/eo, with Qin/e, where ε is the permittivity of the material. (Technically, Eo is called the vacuum permittivity.) Suppose that a 70 nC point charge is surrounded by a thin, 32-cm-diameter spherical rubber shell and that the electric field strength inside the rubber shell is 2500 N/C.
What is the permittivity of rubber?
Answer:
The permittivity of rubber is [tex]\epsilon = 8.703 *10^{-11}[/tex]
Explanation:
From the question we are told that
The magnitude of the point charge is [tex]q_1 = 70 \ nC = 70 *10^{-9} \ C[/tex]
The diameter of the rubber shell is [tex]d = 32 \ cm = 0.32 \ m[/tex]
The Electric field inside the rubber shell is [tex]E = 2500 \ N/ C[/tex]
The radius of the rubber is mathematically evaluated as
[tex]r = \frac{d}{2} = \frac{0.32}{2} = 0.16 \ m[/tex]
Generally the electric field for a point is in an insulator(rubber) is mathematically represented as
[tex]E = \frac{Q}{ \epsilon } * \frac{1}{4 * \pi r^2}[/tex]
Where [tex]\epsilon[/tex] is the permittivity of rubber
=> [tex]E * \epsilon * 4 * \pi * r^2 = Q[/tex]
=> [tex]\epsilon = \frac{Q}{E * 4 * \pi * r^2}[/tex]
substituting values
[tex]\epsilon = \frac{70 *10^{-9}}{2500 * 4 * 3.142 * (0.16)^2}[/tex]
[tex]\epsilon = 8.703 *10^{-11}[/tex]
Which statement describes one feature of a mineral's definite chemical composition?
It always occurs in pure form.
It always contains certain elements.
It cannot form from living or once-living materials.
It cannot contain atoms from more than one element.
N
Answer:
It always contains certain elements
Explanation:
Minerals can be defined as natural inorganic substances which possess an orderly internal structural arrangement as well as a particular, well known chemical composition, crystal structures and physical properties. Minerals include; quartz, dolomite, basalt, etc. Minerals may occur in isolation or in rock formations.
Minerals contain specific, well known chemical elements in certain ratios that can only vary within narrow limits. This is what we mean by a mineral's definite chemical composition. The structure of these minerals are all well known as well as their atom to atom connectivity.
The statement describes one feature of a mineral's definite chemical composition - It always contains certain elements.
A mineral is a naturally occurring chemical compound, usually of a crystalline form.
A mineral has one specific chemical composition.chemical composition that varies within a specific limited range and the atoms that make up the mineral must occur in specific ratiosthe proportions of the different elements and groups of elements in the mineral.Thus, The statement describes one feature of a mineral's definite chemical composition - It always contains certain elements.
Learn more:
https://brainly.com/question/690965
Q 6.30: What is the underlying physical reason for the difference between the static and kinetic coefficients of friction of ordinary surfaces
Answer:
the coefficient of static friction is larger than kinetic coefficients of friction
Explanation:
The coefficient of static friction is usually larger than the kinetic coefficients of friction because an object at rest has increasingly settled agreements with the surface it's resting on at the molecular level, so it takes more force to break these agreement.
Until this force is been overcome, kinetic coefficient of friction is not going to surface.
Note: coefficient of static friction is the friction between two bodies when the bodies aren't moving. Meanwhile, kinetic coefficient is the ratio of frictional force of a moving body to the normal reaction.
[tex]F_{s}[/tex] ≤μ[tex]_{s}[/tex]N(coefficient of static friction)
where [tex]F_{s}[/tex] is the static friction, μ[tex]_{s}[/tex] is the coefficient of static friction and N is the normal reaction
μ = [tex]\frac{F}{N}[/tex](kinetic coefficient of friction)
attached is diagram illustrating the explanation
A parallel-plate capacitor with circular plates of radius R is being discharged. The displacement current through a central circular area, parallel to the plates and with radius R/2, is 9.2 A. What is the discharging current?
Answer:
The discharging current is [tex]I_d = 36.8 \ A[/tex]
Explanation:
From the question we are told that
The radius of each circular plates is R
The displacement current is [tex]I = 9.2 \ A[/tex]
The radius of the central circular area is [tex]\frac{R}{2}[/tex]
The discharging current is mathematically represented as
[tex]I_d = \frac{A}{k} * I[/tex]
where A is the area of each plate which is mathematically represented as
[tex]A = \pi R ^2[/tex]
and k is central circular area which is mathematically represented as
[tex]k = \pi [\frac{R}{2} ]^2[/tex]
So
[tex]I_d = \frac{\pi R^2 }{\pi * [ \frac{R}{2}]^2 } * I[/tex]
[tex]I_d = \frac{\pi R^2 }{\pi * \frac{R^2}{4} } * I[/tex]
[tex]I_d = 4 * I[/tex]
substituting values
[tex]I_d = 4 * 9.2[/tex]
[tex]I_d = 36.8 \ A[/tex]
How can global warming lead to changes to the Earth’s surface? a. Global warming can lead to an increased number of earthquakes, which change the Earth’s surface. b. Global warming can lead to glaciers melting, causing flooding to areas and the decrease of glacial land masses. c. Global warming leads to a decrease in water levels of coastal wetlands. d. Global warming cannot lead to changes to the Earth’s surface.
Answer:
Option: b. Global warming can lead to glaciers melting, causing flooding to areas and the decrease of glacial land masses.
Explanation:
Global warming is the reason for the changes in environment and climate on earth. Melting of glaciers leads to an increase in water level and a decrease in landmass. One of the most climactic consequences is the decrease in Arctic sea ice. Melting polar ice along with ice sheets and glaciers across Greenland, North America, Europe, Asia, and South America suspected to increase sea levels slowly. There is an increase in the glacial retreat due to global warming, which leaves rock piles that covered with ice.
Answer:
B: Global warming can lead to glaciers melting, causing flooding to areas and the decrease of glacial land masses.
Explanation:
Global warming is primarily caused by the increase in greenhouse gases, such as carbon dioxide, in the Earth's atmosphere. This leads to a rise in global temperatures, which has various impacts on the Earth's surface. One significant effect is the melting of glaciers and ice caps in polar regions and mountainous areas.
As temperatures increase, glaciers and ice sheets start to melt at a faster rate. This melting results in the release of massive amounts of water into rivers, lakes, and oceans. Consequently, there can be an increase in the frequency and intensity of flooding events in regions downstream from these melting glaciers.
Moreover, the melting of glaciers and ice caps contributes to a rise in sea levels. As the melted ice enters the oceans, it adds to the overall volume of water, leading to a gradual increase in sea levels worldwide. This rise in sea levels poses a threat to coastal areas, as they become more vulnerable to coastal erosion, storm surges, and saltwater intrusion into freshwater sources.
Additionally, the loss of glacial land masses due to melting can have long-term effects on ecosystems. Glaciers act as freshwater reservoirs, releasing water gradually throughout the year. With their decline, the availability of freshwater for agriculture, drinking water, and other human needs can be significantly affected.
Therefore, global warming can indeed lead to changes in the Earth's surface, particularly through the melting of glaciers and subsequent impacts on sea levels, flooding, and glacial land masses.
E23 verified.
your washer has a power of 350 watts and your dryer has a power of 1800 watts how much energy do you use to clean a load of clothes in 1 hour of washing and 1 hour of drying?
A. 1.29 x 10^3 J
B. 2.58 x 10^3 J
C. 1.55 x 10^7 J
D. 7.74 x 10^6 J
Answer:
7.74 x 10⁶ Joules
Explanation:
recall that "Watts" is the SI unit used for "energy per unit time"
Hence "Watts" may also be expressed as Joules / Second (or J/s)
We are given that the washer is rated at 350W (i.e. 350 Joules / s) and the dryer is rated at 1800W (i.e. 1800 Joules / s).
We are also given that the appliances are each run for 1 hour
1 hour = 60 min = (60 x 60) seconds = 3600 seconds
Hence the total energy used,
= Energy used by Washer in 1 hour + Energy used by dryer in 1 hour
= (350 J/s x 3600 s) + (1800 J/s x 3600 s)
= 3600 ( 350 + 1800)
= 3600 (2150)
= 7,740,000 Joules
= 7.74 x 10⁶ Joules
Given small samples of three liquids, you are asked to determine their refractive indexes. However, you do not have enough of each liquid to measure the angle of refraction for light retracting from air into the liquid. Instead, for each liquid, you take a rectangular block of glass (n= 1.52) and Place a drop of the liquid on the top surface f the block. you shine a laser beam with wavelength 638 nm in vacuum at one Side of the block and measure the largest angle of incidence for which there is total internal reflection at the interface between the glass and the liquid. Your results are given in the table.
Liquid A B C
θ 52.0 44.3 36.3
Required:
a. What is the refractive index of liquid A at this wavelength?
b. What is the refractive index of liquid B at this wavelength?
c. What is the refractive index of liquid C at this wavelength?
Answer:
A — 1.198B — 1.062C — 0.900Explanation:
The index of refraction of the liquid can be computed from ...
[tex]n_i\sin{(\theta_t)}=n_t[/tex]
where ni is the index of refraction of the glass block (1.52) and θt is the angle at which there is total internal refraction. nt is the index of refraction of the liquid.
For the given incidence angles, the computed indices of refraction are ...
A: n = 1.52sin(52.0°) = 1.198
B: n = 1.52sin(44.3°) = 1.062
C: n = 1.52sin(36.3°) = 0.900
In this problem we will consider the collision of two cars initially moving at right angles. We assume that after the collision the cars stick together and travel off as a single unit. The collision is therefore completely inelastic. Two cars of masses m1 and m2 collide at an intersection. Before the collision, car 1 was traveling eastward at a speed of v1, and car 2 was traveling northward at a speed of v2. After the collision, the two cars stick together and travel off in the direction.
Required:
a. Write the momentum conservation equation for the east-west components.
b. Write the momentum conservation equation for the north-south components.
c. Find the tangent of the angle.
Answer:
a) vfₓ = m₁ / (m₁ + m₂) v₁, b) tan θ = m₂ / m₁ v₂ / v₁, c)
Explanation:
Momentum is a vector quantity, so the consideration must be fulfilled in all axes
a) conservation of the moment east-west direction
the system is formed by the two cases, so that the forces during the sackcloth have been internal and therefore the mummer remains
before the crash
p₀ = m₁ v₁
after the crash
[tex]p_{f}[/tex]= (m1 + m2) vfₓ
p₀ = pf
m₁ v₁ = (m₁ + m₂) vfₓ
vfₓ = m₁ / (m₁ + m₂) v₁
b) conservation of the North-South axis moment
before the shock
p₀ = m₂ v₂
after the crash
p_{f} = ( m₁ +m₂) [tex]vf_{y}[/tex]
p₀ = p_{f}
me 2 v₂ = (m₁ + m₂) vfy
[tex]vf_{y}[/tex] = m₂ / (m₁ + m₂) v₂
c) the angle with which the car moves is
tan θ = Vfy / Vfₓ
tan θ = [m₂ / (m₁ + m₂) v] / [m₁ / (m₁ + m₂) v₁]
tan θ = m₂ / m₁ v₂ / v₁
The momentum conservation equation for the north-south components is [tex]m_1u_1 = v(m_1 + m_2)[/tex]
The momentum conservation equation for the north-south components is [tex]m_2u_2 = v(m_1 + m_2)[/tex]
The tangent of the angle is 1.
The given parameters;
angle between the initial velocity of the cars, θ = 90Apply the principle of conservation of linear momentum of inelastic collision as shown below;
[tex]m_1u_1 + m_2u_2 = v(m_1 + m_2)[/tex]
The momentum conservation equation for the east-west components is written as follows;
[tex]m_1(u_1cos \ 0) + m_2(u_2 cos 90)= v(m_1 + m_2)\\\\m_1u_1 = v(m_1 + m_2)[/tex]
The momentum conservation equation for the north-south components is written as follows;
[tex]m_1(u_1sin 0) + m_2(u_2sin90) = v(m_1 + m_2)\\\\m_2u_2 = v(m_1 + m_2)[/tex]
The tangent of the angle is calculated as follows;
[tex]tan \ \theta = \frac{p_y}{p_x} = \frac{v(m_1 + m_2)}{v(m_1 + m_2)} \\\\tan \ \theta = 1\\\\\theta = tan^{-1} (1) \\\\\theta = 45\ ^0[/tex]
Learn more here:https://brainly.com/question/24424291
Oceanographers often express the density of sea water in units of kilograms per cubic meter. If the density of sea water is 1.025 g/cm3 at 15ºC, what is its density in kilograms per cubic meter?
Answer: The density in kilograms per cubic meter is 1025
Explanation:
Density is defined as mass contained per unit volume.
Given : Density of sea water = [tex]1.025g/cm^3[/tex]
Conversion : [tex]1.025g/cm^3=?kg/m^3[/tex]
As 1 g = 0.001 kg
Thus 1.025 g =[tex]\frac{0.001}{1}\times 1.025=0.001025kg[/tex]
Also [tex]1cm^3=10^{-6}m^3[/tex]
Thus [tex]1.025g/cm^3=\frac{0.001025}{10^{-6}kg/m^3}=1025kg/m^3[/tex]
Thus density in kilograms per cubic meter is 1025
A typical arteriole has a diameter of 0.080 mm and carries blood at the rate of 9.6 x10-5 cm3/s. What is the speed of the blood in (cm/s) the arteriole
Answer:
v= 4.823 × 10⁻⁹ cm/s
Explanation:
given
flow rate = 9.6 x10-5 cm³/s, d = 0.080mm
r = d/2= 0.080/2= 0.0040 cm
speed = rate of blood flow × area
v = (9.6 x 10⁻⁵ cm³/s) × (πr²)
v = (9.6 x 10⁻⁵ cm³/s) × π(0.0040 × cm)²
v= 1.536 × 10⁻⁹π cm/s
v= 4.823 × 10⁻⁹ cm/s
(a) According to Hooke's Law, the force required to hold any spring stretched x meters beyond its natural length is f(x)=kx. Suppose a spring has a natural length of 20 cm. If a 25-N force is required to keep it stretched to a length of 30 cm, how much work is required to stretch it from 20 cm to 25 cm?
(b) Find the area of the region enclosed by one loop of the curve r=2sin(5θ).
Answer:
a) The work required to stretch the spring from 20 centimeters to 25 centimeters is 0.313 joules, b) The area of the region enclosed by one loop of the curve [tex]r(\theta) = 2\cdot \sin 5\theta[/tex] is [tex]4\pi[/tex].
Explanation:
a) The work, measured in joules, is a physical variable represented by the following integral:
[tex]W = \int\limits^{x_{f}}_{x_{o}} {F(x)} \, dx[/tex]
Where
[tex]x_{o}[/tex], [tex]x_{f}[/tex] - Initial and final position, respectively, measured in meters.
[tex]F(x)[/tex] - Force as a function of position, measured in newtons.
Given that [tex]F = k\cdot x[/tex] and the fact that [tex]F = 25\,N[/tex] when [tex]x = 0.3\,m - 0.2\,m[/tex], the spring constant ([tex]k[/tex]), measured in newtons per meter, is:
[tex]k = \frac{F}{x}[/tex]
[tex]k = \frac{25\,N}{0.3\,m-0.2\,m}[/tex]
[tex]k = 250\,\frac{N}{m}[/tex]
Now, the work function is obtained:
[tex]W = \left(250\,\frac{N}{m} \right)\int\limits^{0.05\,m}_{0\,m} {x} \, dx[/tex]
[tex]W = \frac{1}{2}\cdot \left(250\,\frac{N}{m} \right)\cdot [(0.05\,m)^{2}-(0.00\,m)^{2}][/tex]
[tex]W = 0.313\,J[/tex]
The work required to stretch the spring from 20 centimeters to 25 centimeters is 0.313 joules.
b) Let be [tex]r(\theta) = 2\cdot \sin 5\theta[/tex]. The area of the region enclosed by one loop of the curve is given by the following integral:
[tex]A = \int\limits^{2\pi}_0 {[r(\theta)]^{2}} \, d\theta[/tex]
[tex]A = 4\int\limits^{2\pi}_{0} {\sin^{2}5\theta} \, d\theta[/tex]
By using trigonometrical identities, the integral is further simplified:
[tex]A = 4\int\limits^{2\pi}_{0} {\frac{1-\cos 10\theta}{2} } \, d\theta[/tex]
[tex]A = 2 \int\limits^{2\pi}_{0} {(1-\cos 10\theta)} \, d\theta[/tex]
[tex]A = 2\int\limits^{2\pi}_{0}\, d\theta - 2\int\limits^{2\pi}_{0} {\cos10\theta} \, d\theta[/tex]
[tex]A = 2\cdot (2\pi - 0) - \frac{1}{5}\cdot (\sin 20\pi-\sin 0)[/tex]
[tex]A = 4\pi[/tex]
The area of the region enclosed by one loop of the curve [tex]r(\theta) = 2\cdot \sin 5\theta[/tex] is [tex]4\pi[/tex].
Find the terminal velocity (in m/s) of a spherical bacterium (diameter 1.81 µm) falling in water. You will first need to note that the drag force is equal to the weight at terminal velocity. Take the density of the bacterium to be 1.10 ✕ 103 kg/m3. (Assume the viscosity of water is 1.002 ✕ 10−3 kg/(m · s).)
Answer:
The terminal velocity of a spherical bacterium falling in the water is 1.96x10⁻⁶ m/s.
Explanation:
The terminal velocity of the bacterium can be calculated using the following equation:
[tex] F = 6\pi*\eta*rv [/tex] (1)
Where:
F: is drag force equal to the weight
η: is the viscosity = 1.002x10⁻³ kg/(m*s)
r: is the radium of the bacterium = d/2 = 1.81 μm/2 = 0.905 μm
v: is the terminal velocity
Since that F = mg and by solving equation (1) for v we have:
[tex] v = \frac{mg}{6\pi*\eta*r} [/tex]
We can find the mass as follows:
[tex] \rho = \frac{m}{V} \rightarrow m = \rho*V [/tex]
Where:
ρ: is the density of the bacterium = 1.10x10³ kg/m³
V: is the volume of the spherical bacterium
[tex] m = \rho*V = \rho*\frac{4}{3}\pi*r^{3} = 1.10 \cdot 10^{3} kg/m^{3}*\frac{4}{3}\pi*(0.905 \cdot 10^{-6} m)^{3} = 3.42 \cdot 10^{-15} kg [/tex]
Now, the terminal velocity of the bacterium is:
[tex] v = \frac{mg}{6\pi*\eta*r} = \frac{3.42 \cdot 10^{-15} kg*9.81 m/s^{2}}{6\pi*1.002 \cdot 10^{-3} kg/(m*s)*0.905 \cdot 10^{-6} m} = 1.96 \cdot 10^{-6} m/s [/tex]
Therefore, the terminal velocity of a spherical bacterium falling in the water is 1.96x10⁻⁶ m/s.
I hope it helps you!
Legacy issues $570,000 of 8.5%, four-year bonds dated January 1, 2019, that pay interest semiannually on June 30 and December 31. They are issued at $508,050 when the market rate is 12%.
1. Determine the total bond interest expense to be recognized.
Total bond interest expense over life of bonds:
Amount repaid:
8 payments of $24,225 $193,800
Par value at maturity 570,000
Total repaid 763,800
Less amount borrowed 645 669
Total bond interest expense $118.131
2. Prepare a straight-line amortization table for the bonds' first two years.
Semiannual Period End Unamortized Discount Carrying Value
01/01/2019
06/30/2019
12/31/2019
06/30/2020
12/31/2020
3. Record the interest payment and amortization on June 30. Note:
Date General Journal Debit Credit
June 30
4. Record the interest payment and amortization on December 31.
Date General Journal Debit Credit
December 31
Answer:
1) Determine the total bond interest expense to be recognized.
Total bond interest expense over life of bonds:
Amount repaid:
8 payments of $24,225: $193,800
Par value at maturity: $570,000
Total repaid: $763800 (193,800 + 570,000)
Less amount borrowed: $508050
Total bond interest expense: $255750 (763800 - 508,050)
2)Prepare a straight-line amortization table for the bonds' first two years.
Semiannual Interest Period End; Unamortized Discount; Carrying Value
01/01/2019 61,950 508,050
06/30/2019 54,206 515,794
12/31/2019 46,462 523,538
06/30/2020 38,718 531,282
12/31/2020 30,974 539,026
3) Record the interest payment and amortization on June 30:
June 30 Bond interest expense, dr 31969
Discount on bonds payable, Cr (61950/8) 7743.75
Cash, Cr ( 570000*8.5%/2) 24225
4) Record the interest payment and amortization on December 31:
Dec 31 Bond interest expense, Dr 31969
Discount on bonds payable, Cr 7744
Cash, Cr 24225
PLS HELP ILL MARK U BRAINLIEST I DONT HAVE MUCH TIME!!
A football player of mass 103 kg running with a velocity of 2.0 m/s [E] collides head-
on with a 110 kg player on the opposing team travelling with a velocity of 3.2 m/s
[W]. Immediately after the collision the two players move in the same direction.
Calculate the final velocity of the two players.
Answer:
The final velocity of the two players is 0.69 m/s in the direction of the opposing player.
Explanation:
Since the players are moving in opposite directions, from the principle of conservation of linear momentum;
[tex]m_{1} u_{1}[/tex] - [tex]m_{2}u_{2}[/tex] = [tex](m_{1} + m_{2} )[/tex] v
Where: [tex]m_{1}[/tex] is the mass of the first player, [tex]u_{1}[/tex] is the initial velocity of the first player, [tex]m_{2}[/tex] is the mass of the second player, [tex]u_{2}[/tex] is the initial velocity of the second player and v is the final common velocity of the two players after collision.
[tex]m_{1}[/tex] = 103 kg, [tex]u_{1}[/tex] = 2.0 m/s, [tex]m_{2}[/tex] = 110 kg, [tex]u_{2}[/tex] = 3.2 m/s. Thus;
103 × 2.0 - 110 × 3.2 = (103 + 110)v
206 - 352 = 213 v
-146 = 213 v
v = [tex]\frac{-146}{213}[/tex]
v = -0.69 m/s
The final velocity of the two players is 0.69 m/s in the direction of the opposing player.
What is the work done in stretching a spring by a distance of 0.5 m if the restoring force is 24N?
Answer:
3Nm
Explanation:
work = 0.5 x 12 x 0.5 = 3
The work done in stretching the spring by a distance of 0.5 m, with a restoring force of 24 N, is 6 joules.
To calculate the work done in stretching a spring, we can use the formula for work done by a spring:
Work = (1/2) * k *[tex]x^2[/tex]
where:
k = spring constant
x = distance the spring is stretched
Given that the restoring force (F) acting on the spring is 24 N, and the distance the spring is stretched (x) is 0.5 m, we can find the spring constant (k) using Hooke's law:
F = k * x
k = F / x
k = 24 N / 0.5 m
k = 48 N/m
Now, we can calculate the work:
Work = (1/2) * 48 N/m * [tex](0.5 m)^2[/tex]
Work = (1/2) * 48 N/m * [tex]0.25 m^2[/tex]
Work = 6 joules
Therefore, the work done in stretching the spring by a distance of 0.5 m, with a restoring force of 24 N, is 6 joules.
To know more about work done, here
brainly.com/question/2750803
#SPJ2
A circular coil of wire of 200 turns and diameter 2.0 cm carries a current of 4.0 A. It is placed in a magnetic field of 0.70 T with the plane of the coil making an angle of 30° with the magnetic field. What is the magnetic torque on the coil?
Answer:
0.087976 Nm
Explanation:
The magnetic torque (τ) on a current-carrying loop in a magnetic field is given by;
τ = NIAB sinθ --------- (i)
Where;
N = number of turns of the loop
I = current in the loop
A = area of each of the turns
B = magnetic field
θ = angle the loop makes with the magnetic field
From the question;
N = 200
I = 4.0A
B = 0.70T
θ = 30°
A = π d² / 4 [d = diameter of the coil = 2.0cm = 0.02m]
A = π x 0.02² / 4 = 0.0003142m² [taking π = 3.142]
Substitute these values into equation (i) as follows;
τ = 200 x 4.0 x 0.0003142 x 0.70 sin30°
τ = 200 x 4.0 x 0.0003142 x 0.70 x 0.5
τ = 200 x 4.0 x 0.0003142 x 0.70
τ = 0.087976 Nm
Therefore, the torque on the coil is 0.087976 Nm
what is mean by the terminal velocity
Terminal Velocity is the constant speed that a falling thing reaches when the resistence of a medium prevents the thing to reach any further speed.
Best of Luck!
Calculate the angular momentum of a solid uniform sphere with a radius of 0.150 m and a mass of 13.0 kg if it is rotating at 5.70 rad/s about an axis through its center.
Answer:
The angular momentum of the solid sphere is 0.667 kgm²/s
Explanation:
Given;
radius of the solid sphere, r = 0.15 m
mass of the sphere, m = 13 kg
angular speed of the sphere, ω = 5.70 rad/s
The angular momentum of the solid sphere is given;
L = Iω
Where;
I is the moment of inertia of the solid sphere
ω is the angular speed of the solid sphere
The moment of inertia of solid sphere is given by;
I = ²/₅mr²
I = ²/₅ x (13 x 0.15²)
I = 0.117 kg.m²
The angular momentum of the solid sphere is calculated as;
L = Iω
L = 0.117 x 5.7
L = 0.667 kgm²/s
Therefore, the angular momentum of the solid sphere is 0.667 kgm²/s