Answer:
the radius of curvature of the track for this instant is 266 m
Explanation:
Given that;
The Initial Velocity u = 100 km/h = 100 × [tex]\frac{5}{18}[/tex] = 27.77 m/s
velocity of the train at t=12 s is;
[tex]V_{t=12}[/tex] = 50 km/h = 50 × [tex]\frac{5}{18}[/tex] = 13.89 m/s
now, we calculate the deceleration of the train
[tex]V_{t=12}[/tex] = u + at
13.89 = 27.77 + [tex]a_{t}[/tex]12
[tex]a_{t}[/tex] = (13.89 - 27.77) / 12
[tex]a_{t}[/tex] = -13.88 / 12
[tex]a_{t}[/tex] = - 1.1566 m/s²
Now, the velocity of the train at 6 seconds is;
[tex]V_{t=6}[/tex] = u + at
[tex]V_{t=6}[/tex] = 27.77 + ( - 1.1566 m/s²)6
[tex]V_{t=6}[/tex] = 27.77 - 6.9396
[tex]V_{t=6}[/tex] = 20.83 m/s
The acceleration at t=6 s is;
a = √[ ([tex]a_{t}[/tex] )² + ([tex]a_{n}[/tex])²]
a = √[ ([tex]a_{t}[/tex] )² + ([tex]a_{n}[/tex])²]
we substitute
2m/s² = √[ (- 1.15 )² + ([tex]a_{n}[/tex])²]
4 = (- 1.1566 )² + ([tex]a_{n}[/tex])²
4 = 1.3377 + ([tex]a_{n}[/tex])²
([tex]a_{n}[/tex])² = 4 - 1.3377
([tex]a_{n}[/tex])² = 2.6623
[tex]a_{n}[/tex] = √2.6623
[tex]a_{n}[/tex] = 1.6316 m/s²
Now the radius of curve is;
a = V² / p
[tex]p_{t=6}[/tex] = ( [tex]V_{t=6}[/tex] )² / [tex]a_{n}[/tex]
[tex]p_{t=6}[/tex] = ( 20.83 m/s )² / 1.6316 m/s²
[tex]p_{t=6}[/tex] = 433.8889 / 1.6316
[tex]p_{t=6}[/tex] = 265.9 m ≈ 266 m
Therefore; the radius of curvature of the track for this instant is 266 m
A hanging wire made of an alloy of titanium with diameter 0.05 cm is initially 2.7 m long. When a 15 kg mass is hung from it, the wire stretches an amount 1.68 cm. A mole of titanium has a mass of 48 grams, and its density is 4.54 g/cm3. Based on these experimental measurements, what is Young's modulus for this alloy of titanium
Answer:
Explanation:
Young's modulus of elasticity Y = stress / strain
stress = force / cross sectional area
= weight of 15 kg / π r²
= 15 x 9.8 / 3.14 x ( .025 x 10⁻² )²
stress = 74.9 x 10⁷ N / m²
strain = Δ L / L , Δ L is change in length and L is original length
Putting the values
strain = .0168 / 2.7 =.006222
Young's modulus of elasticity Y = 74.9 x 10⁷ / .006222
= 120.88 x 10⁹ N / m² .
A spacecraft is flying away from the moon toward earth.
What will be true of the moon’s gravitational pull on the spacecraft?
It will decrease.
It will increase.
It will repel the spacecraft.
It will remain the same.
Answer:
it will decrease
Explanation:
According to the law of universal gravitation, the gravitational force exerted by the moon on the spacecraft is equal to the product of their masses and inversely proportional to the square of the distance that separates them. Therefore, as the spacecraft moves away, its distance increases and the force of attraction exerted by the moon decreases.
Answer:
A. It will increase
Explanation:
I took the quiz on K12 and this was the correct option.
Hope I helped
Meandering valleylike features on the Moon's surface are called
Answer:
Meandering valley like features on the Moon's surface are called rilles
Explanation:
NOUN
rilles (plural noun)
a fissure or narrow channel on the moon's surface.
A basketball with a mass of 20 kg is accelerated with a force of 10 N. If resisting forces are ignored, what is the acceleration of the basketball?
Do You think History is the most important subject that deserves first place? Do you see a way that learning history could assist you in your future career?
yes
Explanation:
history is an important class and it helps to you understand what went on in the past so that we can learn from our mistakes and help us grow
A student stretches a spring, attaches a 1.20 kg mass to it, and releases the mass from rest on a frictionless surface. The resulting oscillation has a period of 0.750 s and an amplitude of 15.0 cm. Determine the oscillation frequency, the spring constant, and the speed of the mass when it is halfway to the equilibrium position.
Answer:
the speed of the mass when it is halfway to the equilibrium position is 1.26 m/s
Explanation:
Given that;
mass of the object m = 1.20 kg
period of oscillation = 0.750 s
Amplitude ( A/x) = 15.0 cm = 0.15 m
now;
a) Determine the oscillation frequency;
oscillation frequency f = 1/T
we substitute
f = 1 / 0.750 s
f = 1.33 Hz
Therefore, the oscillation frequency is 1.33 Hz
b) Determine the spring constant;
we solve for spring constant from the following expression;
T = 2π√(m/k)
k = 4π²m / T²
so we substitute
k = (4π² × 1.20) / (0.750)²
k = 47.3741 / 0.5625
k = 84.22 N/m
Therefore, the spring constant is 84.22 N/m
c) determine the speed of the mass when it is halfway to the equilibrium position
So, at equilibrium, the energy is equal to K.E
such that;
1/2mv² = 1/2kx²
mv² = kx²
v² = kx² / m
v = √( kx²/m)
we substitute
v = √( 84.22×(0.15 m)²/ 1.2 )
v = √( 1.89495 / 1.2 )
v = √ 1.579125
v = 1.26 m/s
Therefore, the speed of the mass when it is halfway to the equilibrium position is 1.26 m/s
Please answer the question
Answer:
D
Explanation:
He walked a shorter distance, she walked a longer distance but got that wing thingies
A car initially traveling 7 m/s speeds up uniformly at a rate of 3 m/s2 until it reaches a velocity of 22 m/s. How much time did it take the car to reach this final velocity?
Answer:
t = 5 s
Explanation:
Data:
Initial Velocity (Vo) = 7 m/sAcceleration (a) = 3 m/s²Final Velocity (Vf) = 22 m/sTime (t) = ?Use formula:
[tex]\boxed{t=\frac{Vf - Vo}{a}}[/tex]Replace:
[tex]\boxed{t=\frac{22\frac{m}{s} -7\frac{m}{s}}{3\frac{m}{s^{2}}}}[/tex]Solve the subtraction of the numerator:
[tex]\boxed{t=\frac{15\frac{m}{s}}{3\frac{m}{s^{2}}}}[/tex]It divides:
[tex]\boxed{t=5\ s}[/tex]How much time did it take the car to reach this final velocity?
It took a time of 5 seconds.
to what temperature it will a 30 KG of glass raise if it absorbs 4275 joules of heat in its specific heat is 0.5 J/KG degree celsius. The initial temperature of the glass is 35°C
Answer:
230° C
Explanation:
A substance's specific heat tells you how much heat much either be added or removed from 1 g of that substance in order to cause a 1∘C
Two forces P and Q act on an object of mass 7.00 kg with Q being the larger of the two forces. When both forces are directed to the left, the magnitude of the acceleration of the object is 1.40 m/s2. However, when the force P is directed to the left and the force Q is directed to the right, the object has an acceleration of 0.700 m/s2 to the right. Find the magnitudes of the two forces P and Q .
Answer:
Explanation:
Q is larger than P . When two forces act in the same direction , Resultant force
can be calculated by adding them up . When two forces act in the opposite direction , Resultant force can be calculated by subtracting them .
Force = mass x acceleration .
In the first case
Resultant force = mass x acceleration
P + Q = 7 x 1.4 = 9.8 N
In the second case
Q - P = 7 x 0.7 = 4.9
Adding up these two equations
2 Q = 14.7
Q = 7.35 N
P = 9.8 - 7.35 = 2.45 N .
What inspired Ronald McNair to do science
Answer:
While working as a staff physicist at hughes Research Laboratories McNair learned that the National Aeronautics and Space Administration (NASA) was looking for scientist to join the shuttle program;)
Calculate the electric field associated to an electric dipole for two charges separated 10-8 m with a dipole moment of 10-33 C m. Do not use unit of measure, just a whole number. Give the result in standard notation, not in scientific notation. Use for the Coulomb constant the value k
Answer:
18 N/C
Explanation:
Given that:
Electric field constant, k = 9*10^9 N/c
Distance, r = 10^-8 m
Dipole moment, p = 10^-33
Using the relation for electric field due to dipole :
E = [2KP / r³]
E = (2 * (9*10^9) * 10^-33) ÷ (10^-8)^3
E = (18 * 10^9 * 10^-33) ÷ 10^-24
E = [18 * 10^(9-33)] ÷ 10^-24
E = (18 * 10^-24) / 10^-24
E = 18 * 10^-24+24
E = 18 * 10^0
E = 18 N/C
The study of heat is ____?
Explanation:
thermodynamics is the study of heat.
Answer The study of heat and its relationship to useful work is called thermodynamics and involves macroscopic quantities such as pressure, temperature, and volume without regard for the molecular basis of these quantitie
Explanation:
Why can’t a real machine ever have 100% efficiency
Answer:
Almost all machines require energy to offset the effects of gravity, friction, and air/wind resistance. Thus, no machine can continually operate at 100 percent efficiency.
Astronomers study the electromagnetic radiation emitted by distant stars and planets to determine things like: how far away they are, their temperatures and speed, etc. Based on what you learned in this class, explain why the NASA Hubble Space Telescope is better for observing the electromagnetic radiation emitted from stars and planets at 560 km above sea level compared to the Keck telescope in Hawaii, which is 4 km above sea level
Answer:
This same Hawaii telescope, which would be 4 km across water level, can't provide an appropriate version of distanced planetary bodies. A further overview is provided below.
Explanation:
The surface area of that same earth's orbit seems to be approximately 480 km heavy. The atmosphere isn't translucent to the only certain wavelength range of the radioactivity. Not because all-stars, as well as gliders, emit specific wavelengths, but several of them generate ultraviolet as well as infrared. Those same radiations have either been mediated primarily as well as passes through the atmosphere. Due to the Blockage, they can't even be interpreted with such a similar quality unless the telescope would be positioned throughout the portion of the atmosphere.How much force does it take to give a 70 kg object an acceleration of 20 mls2
Answer:
heyy
Explanation:how r uuu
The nearest neighbor interaction force is of magnitude 481 nanoNewtons, e.g., the magnitude of the force of the leftmost electron on the proton, or the magnitude of the force of any of the three on its nearest neighbor electron. Calculate the size of the net force on the leftmost proton.
Answer:
F = 120.25 10⁻⁹ N
Explanation:
In this exercise, the force between the closest neighbors is indicated by f = 481 10⁻⁹ N, in general between the one-dimensional solid the distances remain the same, if the distance between the first neighbor is d, the distance between the second neighbors is 2d.
For most solids the attractive forces are electrical, therefore force is proportional to the electrical charges and the inverse of the distance squared,
F = [tex]k \frac{q_1 q_2}{r^2}[/tex]
if we call fo the force for the first neighbors
F₀ = k \frac{q_1 q_2}{d^2}
the force for the second neighbors r= 2d
F = k \frac{q_1 q_2}{(2d)^2}
F = F₀ / 4
let's calculate
F = 481 10⁻⁹ / 4
F = 120.25 10⁻⁹ N
Jojo and Roro begin side-by-side at one end of the playground. At the same moment, they begin to move toward the other end of the playground, Jojo at a constant velocity of 3.0 m/s, Roro at a constant velocity of 2.0 m/s. Sometime during her trip, Roro stops to rest for 2.0 s, but then starts again at her original constant speed. When Jojo reaches the end of the playground, she is 10 m ahead of Roro.
(a) For how much time did Roro move?
(b) How far did Roro move? (Set it up, good notation, equations in symbols first, etc.)
Answer:
Roro's total travel time = 6 seconds out of which he rested for 2 seconds
Distance covered by Roro = 8 meters
Explanation:
Given that :
Jojo:
Constant velocity, v = 3m/s
Travel time = h
Roro:
Constant velocity, v = 2m/s
Roro rest for 2 seconds
Travel time = h - 2
Recall:
Distance = speed * time
Distance covered by Jojo:
3 * h = 3h
At this distance ;
Roro's distance = 3h - 10
Using formula :
Roro's distance = 2 * (h - 2)
Hence,
2(h - 2) = 3h - 10
2h - 4 = 3h - 10
2h - 3h = - 10 + 4
-h = - 6
h = 6
Hence, Roro moved for :
h - 2 = 6 - 2 = 4seconds
Distance moved by Roro:
2(h - 2) = 2(6 - 2) = 2(4) = 8 meters
What are used to measure temperature.
Answer:
A thermometer is an instrument that measures temperature.
Explanation:
A 6.00 nC is 2.00 m from a 3.00 nC charge. Find the magnitude of the electric field at a point midway between
the two charges? Which way does the electric field point, towards the positive or the negative charge?
Answer:
E_total = 26.97 N/C
Electric field points towards the positive charge
Explanation:
We are given;
Charge 1; q1 = 6 nC = 6 × 10^(-9) C
Charge 2; q2 = 3 nC = 3 × 10^(-9) C
Distance between both charges; R_o = 2 m
Since we want to find electric field midway, the distance midway is r = 2/2 = 1 m
Using coulumbs law;
E = kq/r²
Where k is a constant with a value of 8.99 × 10^(9) N.m/C²
Thus;
E1 = kq1/r²
E1 = (8.99 × 10^(9) × 6 × 10^(-9))/1²
E1 = 53.94 N/C
Similarly;
E2 = kq2/r²
E2 = (8.99 × 10^(9) × 3 × 10^(-9))/1²
E2 = 26.97 N/C
Since both electric fields are positive, it means that they are both moving towards the midpoint of the distance between both charges.
This implies they will have opposite directions.
Thus, total electric field at the midway point is;
E_total = E1 - E2
E_total = 53.94 - 26.97
E_total = 26.97 N/C
Starting from rest, a wheel with constant angular acceleration turns through an angle of 25 rad in a time t. What will its angular velocity be after 3t?
Answer:
θ = 225 rad
Explanation:
given data
angle = 25 rad
to find out
angular velocity after 3t?
solution
let angular acceleration α in t
θ = ω × t + 0.5 × α × t² ........................1
here ω = 0 (initial velocity )
so put this value here
25 = 0 + 0.5 × α × t² ..........................2
α = 25 ÷ (0.5 t²)
α = 50 ÷ t² .........................3
now here we take in 3t
θ = ω × 3t + 0.5 × α × (3t)²
for ω = 0
θ = 0 + 0.5 × α × 9t²
now put value in eq 2
so
θ = (0.5) × (50 ÷ t²) × (3t)²
θ = 25 × 9
θ = 225 rad
Please answer the question
Answer:
Option B. 300 m/s².
Explanation:
From the question given above, the following data were obtained:
Mass (m) of student = 100 Kg
Mass (m) of ball = 1.5 Kg
Force (F) applied on the ball = 450 N
Acceleration (a) of ball =?
From Newton's 2nd law,
F = ma
Where
F => Force applied
m => mass of object
a => acceleration of object.
With the above formula, we can obtain the acceleration of the ball as follow:
Mass (m) of ball = 1.5 Kg
Force (F) applied on the ball = 450 N
Acceleration (a) of ball =?
F = ma
450 = 1.5 × a
Divide both side by 1.5
a = 450 / 1.5
a = 300 m/s²
Therefore, the acceleration of the ball is 300 m/s²
If a skaters mass increases how does that effect kinetic energy
Answer:
By paying close attention to the formula for average kinetic energy, we can see that by increasing the mass by a proportional amount will lead to an increase in the total average kinetic energy. There is a direct relationship being observed between the values.
A 7300 kg rocket blasts off vertically from the launch pad with a constant upward acceleration of 2.20 m/s2 and feels no appreciable air resistance. When it has reached a height of 575 m , its engines suddenly fail so that the only force acting on it is now gravity. Part A What is the maximum height this rocket will reach above the launch pad
Answer:
Explanation:
We shall first calculate the velocity at height h = 575 m .
acceleration a = 2.2 m /s²
v² = u² + 2 a s
u is initial velocity , v is final velocity , s is height achieved
v² = 0 + 2 x 2.2 x 575
v = 50.3 m /s
After 575 m , rocket moves under free fall so g will act on it downwards
If it travels further by height H
from the relation
v² = u² - 2 g H
v = 0 , u = 50.3 m /s
H = ?
0 = 50.3² - 2 x 9.8 H
H = 129.08 m
Total height attained by rocket
= 575 + 129.08
= 704.08 m .
A baseball player hits a baseball. The mass of the ball is 0.15 kg. The ball accelerates at a rate of 60 m/s 2 . What is the net force on the ball to the nearest newton?
Answer:
Please find attached pdf
Explanation:
hmu if u brave shawtys
Answer:
BET, & done ✌
Answer:
boop
Explanation:
A pendulum has a period of 6.98s. Calculate the length of the pendulum. Use
9.8m/s^2 for gravity. *
Answer:
Length, l = 0.126 meters.
Explanation:
Given the following data;
Period = 6.98s
Acceleration due to gravity, g = 9.8m/s²
To find the length, l;
[tex] Period, T = 2 \pi \sqrt {lg} [/tex]
Substituting into the equation, we have;
[tex] 6.98 = 2*3.142 \sqrt {l*9.8} [/tex]
[tex] 6.98 = 6.284 \sqrt {9.8l} [/tex]
[tex] \frac {6.98}{6.284} = \sqrt {9.8l} [/tex]
[tex] 1.1108 = \sqrt {9.8l} [/tex]
Taking the square of both sides
[tex] 1.1108^{2} = 9.8l [/tex]
[tex] 1.2339 = 9.8l [/tex]
[tex] l = \frac {1.2339}{9.8} [/tex]
Length, l = 0.126m.
A 0.500 cm diameter plastic sphere, used in a static electricity demonstration, has a uniformly distributed 44.0 pC charge on its surface. What is the potential near its surface
Answer:
Explanation:
Radius of sphere R = .250 x 10⁻² m
Potential on the surface V = k Q / R , where Q is charge on the surface , R is radius of the surface and k = 9 x 10⁹
Q = 44 x 10⁻¹² C
V = 9 x 10⁹ x 44 x 10⁻¹² / ( .25 x 10⁻²)
= 1584 x 10⁻¹ Volt .
= 158.4 Volt
Which subatomic particle is NOT found in the nucleus of an atom? *
protons
neutrons
electrons
Answer:
Electrons
Explanation:
Only Protons and Neutrons are found in the nucleus
noooooooooooooooooooooooo
Answer:
yes
Explanation:
Answer:yeeeeeeeeeeeeeeeeeeeeeees
Explanation:
