Given that,
Mass of roller coaster car = 700 kg
Radius = 12 m
Height from the ground h₂= 5 m
Normal force = 580 N
We need to calculate the speed of roller coaster at top
Using balance equation
[tex]N+mg=\dfrac{mv^2}{r}[/tex]
Put the value into the formula
[tex]580+700\times9.8=\dfrac{700v^2}{12}[/tex]
[tex]v^2=\dfrac{12(580+700\times9.8)}{700}[/tex]
[tex]v=\sqrt{\dfrac{12(580+700\times9.8)}{700}}[/tex]
[tex]v=11.29\ m/s[/tex]
We need to calculate the value of maximum height
Using conservation of energy
[tex]mgh_{1}=\dfrac{1}{2}mv^2+0.15(mgh_{1})+mgh_{2}[/tex]
[tex]mgh_{1}-0.15(mgh_{1})-mgh_{2}=\dfrac{1}{2}mv^2[/tex]
Put the value into the formula
[tex]9.8(h_{1}-0.15h_{1}-5)=\dfrac{1}{2}\times(11.29)^2[/tex]
[tex]0.85h_{1}-5=\dfrac{(11.29)^2}{9.8}[/tex]
[tex]h_{1}=\dfrac{13.00+5}{0.85}[/tex]
[tex]h_{1}=21.18\ m[/tex]
Hence, The max height of release h₁ for the roller coaster car is 21.18 m
In a device called the ballistic pendulum, a compressed spring is used to launch a steel ball horizontally into a soft target hanging from a string. The ball embeds in the target and the two swing together from the string. Describe the energy transfers and/or transformations that take place during the use of the ballistic pendulum and at what points they occur
Answer:
When the spring in the ballistic pendulum is compressed, energy is stored up in the spring as potential energy. When the steel ball is launched by the spring, the stored up potential energy of the compressed spring is transformed and transferred into the kinetic energy of the steel ball as it flies off to hit its target. On hitting the soft target, some of the kinetic energy of the steel ball is transferred to the soft target (since they stick together), and they both start to swing together. During the process of swinging, the system's energy is transformed between kinetic and potential energy. At the maximum displacement of the ball from its point of rest, all the energy is converted to potential energy of the system. At the lowest point of travel (at the rest point), all the energy of the system is transformed into kinetic energy. In between these two points, energy the energy of the system is a combination of both kinetic and potential energy.
In the end, all the energy will be transformed and lost as heat to the surrounding; due to the air resistance around; bringing the system to a halt.
Prof. Marcia Grail, supervillain and superscientist, wishes to spy on her foes (they'll all pay!) from orbit. She wishes to be able to resolve detail on order the size of the letters on a license plate (approx. 5cm), in a wavelength of approximately 550 nm, from an orbit approximately 4.2 x 10^7 m above the surface of the Earth. If her resolution is limited only by diffraction, then what is the smallest aperture telescope that she can get away with
Answer:
563.64 m
Explanation:
Given that as per the question
x = 5 cm = 0.05 m
D = 4.2 × 107 m
d = smallest aperture size
As per the situation the solution of the smallest aperture telescope that she can get away with is below :-
We will use Rayleigh's diffraction limit which is
[tex]d\frac{x}{D} = 1.22\lambda[/tex]
The equation will be
[tex]d\frac{0.05}{4.2\times 10^7} = 1.22[550\times 10^{-9}][/tex]
d = 563.64 m
So, the answer is d = 563.64 m
A glass flask whose volume is 1000 cm^3 at a temperature of 1.00°C is completely filled with mercury at the same temperature. When the flask and mercury are warmed together to a temperature of 52.0°C , a volume of 8.50 cm^3 of mercury overflows the flask.Required:If the coefficient of volume expansion of mercury is βHg = 1.80×10^−4 /K , compute βglass, the coefficient of volume expansion of the glass. Express your answer in inverse kelvins.
Answer:
the coefficient of volume expansion of the glass is [tex]\mathbf{ ( \beta_{glass} )= 1.333 *10^{-5} / K}[/tex]
Explanation:
Given that:
Initial volume of the glass flask = 1000 cm³ = 10⁻³ m³
temperature of the glass flask and mercury= 1.00° C
After heat is applied ; the final temperature = 52.00° C
Temperature change ΔT = 52.00° C - 1.00° C = 51.00° C
Volume of the mercury overflow = 8.50 cm^3 = 8.50 × 10⁻⁶ m³
the coefficient of volume expansion of mercury is 1.80 × 10⁻⁴ / K
The increase in the volume of the mercury = 10⁻³ m³ × 51.00 × 1.80 × 10⁻⁴
The increase in the volume of the mercury = [tex]9.18*10^{-6} \ m^3[/tex]
Increase in volume of the glass = 10⁻³ × 51.00 × [tex]\beta _{glass}[/tex]
Now; the mercury overflow = Increase in volume of the mercury - increase in the volume of the flask
the mercury overflow = [tex](9.18*10^{-6} - 51.00* \beta_{glass}*10^{-3})\ m^3[/tex]
[tex]8.50*10^{-6} = (9.18*10^{-6} -51.00* \beta_{glass}* 10^{-3} )\ m^3[/tex]
[tex]8.50*10^{-6} - 9.18*10^{-6} = ( -51.00* \beta_{glass}* 10^{-3} )\ m^3[/tex]
[tex]-6.8*10^{-7} = ( -51.00* \beta_{glass}* 10^{-3} )\ m^3[/tex]
[tex]6.8*10^{-7} = ( 51.00* \beta_{glass}* 10^{-3} )\ m^3[/tex]
[tex]\dfrac{6.8*10^{-7}}{51.00 * 10^{-3}}= ( \beta_{glass} )[/tex]
[tex]\mathbf{ ( \beta_{glass} )= 1.333 *10^{-5} / K}[/tex]
Thus; the coefficient of volume expansion of the glass is [tex]\mathbf{ ( \beta_{glass} )= 1.333 *10^{-5} / K}[/tex]
When mapping the equipotentials on the plates with different electrode configurations you may find that some have significant areas with uniform distribution of the equipotential lines. If the distance between such lines is 0.5 cm, what is the electric field there (in units SI)
Answer:
E = V/5 x10⁻³
Explanation:
if the potential difference is V
then electric field E is given by
E = V/d
d = 0.5cm = 5 x 10⁻³m
E = V/5 x10⁻³
Two small pith balls, each of mass m = 14.2 g, are suspended from the ceiling of the physics lab by 0.5 m long fine strings and are not moving. If the angle which each string makes with the vertical is θθ = 29.1°, and the charges on the two balls are equal, what is the magnitude of that charge
Answer:
1.424 μC
Explanation:
I'm assuming here, that the charged ball is suspended by the string. If the string also is deflected by the angle α, then the forces acting on it would be: mg (acting downwards),
tension T (acting along the string - to the pivot point), and
F (electric force – acting along the line connecting the charges).
We then have something like this
x: T•sin α = F,
y: T•cosα = mg.
Dividing the first one by the second one we have
T•sin α/ T•cosα = F/mg, ultimately,
tan α = F/mg.
Since we already know that
q1=q2=q, and
r=2•L•sinα,
k=9•10^9 N•m²/C²
Remember,
F =k•q1•q2/r², if we substitute for r, we have
F = k•q²/(2•L•sinα)².
tan α = F/mg =
= k•q²/(2•L•sinα)² •mg.
q = (2•L•sinα) • √(m•g•tanα/k)=
=(2•0.5•0.486) • √(0.0142•9.8•0.557/9•10^9) =
q = 0.486 • √(8.61•10^-12)
q = 0.486 • 2.93•10^-6
q = 1.424•10^-6 C
q = 1.424 μC.
When a charge q is placed at a certain point in an electric field, it experiences a force toward the west of magnitude F. If instead a change 2q were placed at that same point what force would it experience?
Answer:
If instead a charge 2q were placed at that same point the force will be 2F.
Explanation:
The electric force is equal to:
[tex] F = q*E [/tex] (1)
Where:
F: is the electric force
q: is the charge
E: is the electric field
We can see that in equation (1) the electric force (F) is proportional to the charge q, thus, if now the charge it's the double (2q) then the force will be the double too:
Initially:
[tex] F_{1} = q_{1}*E [/tex]
Now,
[tex](2q_{1}})*E = 2(q_{1}*E) = 2F_{1}[/tex]
Therefore, if instead a charge 2q were placed at that same point the force will be 2F.
I hope it helps you!
The block on this incline weighs 100 kg and is connected by a cable and pulley to a weight of 10 kg. If the coefficient of friction between the block and incline is o.3, the block will:
Answer:
a. 94.54 N
b. 0.356 m/s^2
Explanation:
Given:-
- The mass of the inclined block, M = 100 kg
- The mass of the vertically hanging block, m = 10 kg
- The angle of inclination, θ = 20°
- The coefficient of friction of inclined surface, u = 0.3
Find:-
a) The magnitude of tension in the cable
b) The acceleration of the system
Solution:-
- We will first draw a free body diagram for both the blocks. The vertically hanging block of mass m = 10 kg tends to move "upward" when the system is released.
- The block experiences a tension force ( T ) in the upward direction due the attached cable. The tension in the cable is combated with the weight of the vertically hanging block.
- We will employ the use of Newton's second law of motion to express the dynamics of the vertically hanging block as follows:
[tex]T - m*g = m*a\\\\[/tex] ... Eq 1
Where,
a: The acceleration of the system
- Similarly, we will construct a free body diagram for the inclined block of mass M = 100 kg. The Tension ( T ) pulls onto the block; however, the weight of the block is greater and tends down the slope.
- As the block moves down the slope it experiences frictional force ( F ) that acts up the slope due to the contact force ( N ) between the block and the plane.
- We will employ the static equilibrium of the inclined block in the normal direction and we have:
[tex]N - M*g*cos ( Q )= 0\\\\N = M*g*cos ( Q )[/tex]
- The frictional force ( F ) is proportional to the contact force ( N ) as follows:
[tex]F = u*N\\\\F = u*M*g *cos ( Q )[/tex]
- Now we will apply the Newton's second law of motion parallel to the plane as follows:
[tex]M*g*sin(Q) - T - F = M*a\\\\M*g*sin(Q) - T -u*M*g*cos(Q) = M*a\\[/tex] .. Eq2
- Add the two equation, Eq 1 and Eq 2:
[tex]M*g*sin ( Q ) - u*M*g*cos ( Q ) - m*g = a* ( M + m )\\\\a = \frac{M*g*sin ( Q ) - u*M*g*cos ( Q ) - m*g}{M + m} \\\\a = \frac{100*9.81*sin ( 20 ) - 0.3*100*9.81*cos ( 20 ) - 10*9.81}{100 + 10}\\\\a = \frac{-39.12977}{110} = -0.35572 \frac{m}{s^2}[/tex]
- The inclined block moves up ( the acceleration is in the opposite direction than assumed ).
- Using equation 1, we determine the tension ( T ) in the cable as follows:
[tex]T = m* ( a + g )\\\\T = 10*( -0.35572 + 9.81 )\\\\T = 94.54 N[/tex]
A Nearsighted Eye. A certain very nearsighted person cannot focus on anything farther than 36.0 cm from the eye. Consider the simplified model of the eye. In a simplified model of the human eye, the aqueous and vitreous humors and the lens all have a refractive index of 1.40, and all the refraction occurs at the cornea, whose vertex is 2.60 cm from the retina.
Required:
a. If the radius of curvature of the cornea is 0.65 cm when the eye is focusing on an object 36.0 cm from the cornea vertex and the indexes of refraction are as described before, what is the distance from the cornea vertex to the retina?
b. What does this tell you about the shape of the nearsighted eye?
1. This distance is greater than for the normal eye.
2. This distance is shorter than for the normal eye.
Answer:
a) The distance from the cornea vertex to the retina is 2.37 cm
b) This distance is shorter than for the normal eye.
Explanation:
a) Let refractive index of air,
n(air) = x = 1
Let refractive index of lens,
n(lens) = y = 1.4
Object distance, s = 36 cm
Radius of curvature, R = 0.65 cm
The distance from the cornea vertex to the retina is the image distance because image is formed in the retina.
Image distance, s' = ?
(x/s) + (y/s') = (y-x)/R
(1/36) + (1.4/s') = (1.4 - 1)/0.65
1.4/s' = 0.62 - 0.028
1.4/s' = 0.592
s' = 1.4/0.592
s' = 2.37 cm
Distance from the cornea vertex to the retina is 2.37 cm
(b) For a normal eye, the distance between the cornea vertex and the retina is 2.60 cm. Since 2.37 < 2.60, this distance is shorter than for normal eye.
3.Cuanto Calor pierden 514 ml de agua si su temperatura desciende de 12°C a 11°C. Expresa el resultado en calorias.
514 cal
51.4 Kcal
514J/cal
5.149 Cal
Answer:
514 cal
Explanation:
In order to calculate the lost heat by the amount of water you first take into account the following formula:
[tex]Q=mc(T_2-T_1)[/tex] (1)
Q: heat lost by the amount of water = ?
m: mass of the water
c: specific heat of water = 1cal/g°C
T2: final temperature of water = 11°C
T1: initial temperature = 12°C
The amount of water is calculated by using the information about the density of water (1g/ml):
[tex]m=\rho V=(1g/ml)(514ml)=514g[/tex]
Then, you replace the values of all parameters in the equation (1):
[tex]Q=(514g)(1cal/g\°C)(11\°C-12\°C)=-514cal[/tex]
The amount of water losses a heat of 514 cal
wo parallel conducting plates are separated by 10.0 cm, and one of them is taken to be at zero volts. (a) What is the magnitude of the electric field strength between them, if the potential 7.05 cm from the zero volt plate (and 2.95 cm from the other) is 393 V?
Answer:
-18896.49 V/m
Explanation:
Distance between the two plates = 10 cm = 10 x [tex]10^{-2}[/tex] m = 0.1 m
Also, one of the plates is taken as zero volt.
a. The potential strength between the zero volt plate, and 7.05 cm (0.0705 m) away is 393 V
b. The potential strength between the other plate, and 2.95 cm (0.0295 m) away is 393 V
Potential field strength = -dV/dx
where dV is voltage difference between these points,
dx is the difference in distance between these points
For the first case above,
potential field strength = -393/0.0705 = -5574.46 V/m
For the second case ,
potential field strength = -393/0.0295 = -13322.03 V/m
Magnitude of the field strength across the plates will be
-5574.46 + (-13322.03) = -5574.46 + 13322.03 = -18896.49 V/m
An air bubble underwater has the same pressure as that of the surrounding water. As the air bubble rises toward the surface (and its temperature remains constant), the volume of the air bubble.
a) increase or decrease, depending on the rate it rises.
b) decreases.
c) increases.
d) remains constant
Answer:
D Remains constant
Explanation:
You are playing ice hockey with friends on a frozen pond. The hockey puck (mass 0.2 kg) is sliding on the ice at 10 m/s when it starts to hail. A hailstone lands on the puck, causing it to slow to 8 m/s. What mass does the hailstone have
Answer:
The mass of the hailstone is [tex]m_2 = 0.05[/tex]
Explanation:
From the question we are told that
The mass of the hockey puck is [tex]m_1 = 0.2 \ kg[/tex]
The initial speed of the puck is [tex]u_1 = 10 \ m/s[/tex]
The final speed of the puck is [tex]v_f = 8 \ m/s[/tex]
According to the principles of linear momentum
[tex]m_1u_1 + m_2u_2 = (m_1 + m_2)v_f[/tex]
At initial the velocity of the hailstone was zero so
[tex]u_2 = 0 \ m/s[/tex]
So
[tex]m_1 u_1 = (m_1 + m_2 )v_f[/tex]
substituting values
[tex]0.2 * 10 = (0.2 + m_2 )* 8[/tex]
=> [tex]2 = (0.2 + m_2 )* 8[/tex]
=> [tex]0.25 = 0.2 + m_2[/tex]
=> [tex]m_2 = 0.05[/tex]
A total charge Q is distributed uniformly over a large flat insulating surface of area A . If the electric field magnitude is equal to 1000 NC/ at a point located a perpendicular distance of 0.1 m away from the center of the sheet, then the electric field at a point a perpendicular distance 0.2 m from the center of the sheet is:_______
a. 1000N/C
b. 500N/C
c. Impossible to say since we are not given Q and A
d. 250 N/C
Answer:
a. 1000N/C
Explanation:
Data mentioned in the question
Electrical field magnitude = 1000 NC
Perpendicular distance = 0.1 m
Perpendicular distance = 0.2 m
Based on the above information, the electric field is
As we know that
[tex]E = \frac{\sigma}{2\times E_o}[/tex]
where,
[tex]\sigma[/tex] = surface charge density
E = distance from nearby point to sheet i.e be independent
The distance at 0.1 and 0.2, the electric field would remain the same
So,
Based on the above explanation, the first option is correct
A hawk is flying horizontally at 18.0 m/s in a straight line, 230 m above the ground. A mouse it has been carrying struggles free from its grasp. The hawk continues on its path at the same speed for 2.00 s before attempting to retrieve its prey. To accomplish the retrieval, it dives in a straight line at constant speed and recaptures the mouse 3.00 m above the ground. (a) Assuming no air resistance, find the diving speed (magnitude of the total velocity vector) of the hawk. (b) What angle did the hawk make with the horizontal during its descent? (c) For how long did the mouse "enjoy" free fall?
Answer:
a) vd = 47.88 m/s
b) θ = 80.9°
c) t = 6.8 s
Explanation:
In the situation of the problem, you can assume that the trajectory of the hawk and the trajectory of the mouse form a rectangle triangle.
One side of the triangle is the horizontal trajectory of the hawk after 2.00s of flight, the other side of the triangle is the distance traveled by the mouse when it is falling down. And the hypotenuse is the trajectory of the hawk when it is trying to recover the mouse.
(a) In order to calculate the diving speed of the hawk, you first calculate the hypotenuse of the triangle.
One side of the triangle is c1 = (18.0m/s)(2.0s) = 36m
The other side of the triangle is c2 = 230m - 3m = 227 m
Then, the hypotenuse is:
[tex]h=\sqrt{(36m)^2+(227m)^2}=229.83m[/tex] (1)
Next, it is necessary to calculate the falling down time of the mouse, this can be done by using the following formula:
[tex]y=y_o+v_ot+\frac{1}{2}gt^2[/tex] (2)
yo: initial height = 230m
vo: initial vertical speed of the mouse = 0m/s
g: gravitational acceleration = -9.8m/s^2
y: final height of the mouse = 3 m
You replace the values of the parameters in (2) and solve for t:
[tex]3=230-4.9t^2\\\\t=\sqrt{\frac{227}{4.9}}=6.8s[/tex]
The hawk traveled during 2.00 second in the horizontal trajectory, hence, the hawk needed 6.8s - 2.0s = 4.8 s to travel the distance equivalent to the hypotenuse to catch the mouse.
You use the value of h and 4.8s to find the diving speed of the hawk:
[tex]v_d=\frac{229.83m}{4.8s}=47.88\frac{m}{s}[/tex]
The diving speed of the Hawk is 47.88m/s
(b) The angle is given by:
[tex]\theta=cos^{-1}(\frac{c_1}{h})=cos^{-1}(\frac{36m}{229.83m})=80.9 \°[/tex]
Then angle between the horizontal and the trajectory of the Hawk when it is descending is 80.9°
(c) The mouse is falling down during 6.8 s
Two guitarists attempt to play the same note of wavelength 6.50 cm at the same time, but one of the instruments is slightly out of tune. Consequently, a 17.0-Hz beat frequency is heard between the two instruments. What were the possible wavelengths of the out-of-tune guitar’s note? Express your answers, separated by commas, in centimeters to three significant figures IN cm.
Answer:
The two value of the wavelength for the out of tune guitar is
[tex]\lambda _2 = (6.48,6.52) \ cm[/tex]
Explanation:
From the question we are told that
The wavelength of the note is [tex]\lambda = 6.50 \ cm = 0.065 \ m[/tex]
The difference in beat frequency is [tex]\Delta f = 17.0 \ Hz[/tex]
Generally the frequency of the note played by the guitar that is in tune is
[tex]f_1 = \frac{v_s}{\lambda}[/tex]
Where [tex]v_s[/tex] is the speed of sound with a constant value [tex]v_s = 343 \ m/s[/tex]
[tex]f_1 = \frac{343}{0.0065}[/tex]
[tex]f_1 = 5276.9 \ Hz[/tex]
The difference in beat is mathematically represented as
[tex]\Delta f = |f_1 - f_2|[/tex]
Where [tex]f_2[/tex] is the frequency of the sound from the out of tune guitar
[tex]f_2 =f_1 \pm \Delta f[/tex]
substituting values
[tex]f_2 =f_1 + \Delta f[/tex]
[tex]f_2 = 5276.9 + 17.0[/tex]
[tex]f_2 = 5293.9 \ Hz[/tex]
The wavelength for this frequency is
[tex]\lambda_2 = \frac{343 }{5293.9}[/tex]
[tex]\lambda_2 = 0.0648 \ m[/tex]
[tex]\lambda_2 = 6.48 \ cm[/tex]
For the second value of the second frequency
[tex]f_2 = f_1 - \Delta f[/tex]
[tex]f_2 = 5276.9 -17[/tex]
[tex]f_2 = 5259.9 Hz[/tex]
The wavelength for this frequency is
[tex]\lambda _2 = \frac{343}{5259.9}[/tex]
[tex]\lambda _2 = 0.0652 \ m[/tex]
[tex]\lambda _2 = 6.52 \ cm[/tex]
This question involves the concepts of beat frequency and wavelength.
The possible wavelengths of the out-of-tune guitar are "6.48 cm" and "6.52 cm".
The beat frequency is given by the following formula:
[tex]f_b=|f_1-f_2|\\\\[/tex]
f₂ = [tex]f_b[/tex] ± f₁
where,
f₂ = frequency of the out-of-tune guitar = ?
[tex]f_b[/tex] = beat frequency = 17 Hz
f₁ = frequency of in-tune guitar = [tex]\frac{speed\ of\ sound\ in\ air}{\lambda_1}=\frac{343\ m/s}{0.065\ m}=5276.9\ Hz[/tex]
Therefore,
f₂ = 5276.9 Hz ± 17 HZ
f₂ = 5293.9 Hz (OR) 5259.9 Hz
Now, calculating the possible wavelengths:
[tex]\lambda_2=\frac{speed\ of\ sound}{f_2}\\\\\lambda_2 = \frac{343\ m/s}{5293.9\ Hz}\ (OR)\ \frac{343\ m/s}{5259.9\ Hz}\\\\[/tex]
λ₂ = 6.48 cm (OR) 6.52 cm
Learn more about beat frequency here:
https://brainly.com/question/10703578?referrer=searchResults
The Sun's energy comes from which nuclear reaction?
A. Nuclear fission
B. Gamma decay
C. Positron emission
D. Nuclear fusion
SUBMIT
Sun's energy comes from the nuclear fusion taking place inside. In nuclear fusion two light nuclei fuses together to form a heavy nuclei with the release of greater amount of energy.
What is nuclear fusion :Nuclear fusion is the process of combining two light nuclei to form a heavy nuclei. In this nuclear process, tremendous energy is released. This is the source of heat and light in stars.
On the other hand, nuclear fission is the process of breaking of a heavy nuclei into two lighter nuclei. Fission also produces massive energy. But in comparison, more energy is produced by nuclear fusion.
Nuclear fission is used in nuclear power generators. The light energy and heat energy comes form the nuclear fusion of hydrogens to form helium nuclei. Hence, option D is correct.
Find more on nuclear fusion:
https://brainly.com/question/12701636
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The power dissipated in each of two resistors is the same. The current across resistor A is triple that across resistor B. If the resistance of resistor B is R, what is the resistance of A?
Answer:
Explanation:
this is the answer to your question
a wall, a 55.6 kg painter is standing on a 3.15 m long homogeneous board that is resting on two saw horses. The board’s mass is 14.5 kg. The saw horse on the right is 1.00 m from the right. How far away can the painter walk from the saw horse on the right until the board begins to tip?
Answer:
0.15 m
Explanation:
First calculating the center of mass from the saw horse
[tex]\frac{3.15}{2} -1=0.575 m[/tex]
from the free body diagram we can write
Taking moment about the saw horse
55.9×9.81×y=14.5×0.575×9.81
y= 0.15 m
So, the painter walk from the saw horse on the right until the board begins to tip is 0.15 m far.
Three masses are located in the x-y plane as follows: a mass of 6 kg is located at (0 m, 0 m), a mass of 4 kg is located at (3 m, 0 m), and a mass of 2 kg is located at (0 m, 3 m). Where is the center of mass of the system?
A. (1 m, 2 m)
B. (2 m, 1 m)
C. (1 m, 1 m)
D. (1 m, 0.5 m)
E. (0.5 m, 1 m)
D. (1m, 0.5m)
Explanation:
The center of mass (or center of gravity) of a system of particles is the point where the weight acts when the individual particles are replaced by a single particle of equivalent mass. For the three masses, the coordinates of the center of mass C(x, y) is given by;
x = (m₁x₁ + m₂x₂ + m₃x₃) / M ----------------(i)
y = (m₁y₁ + m₂y₂ + m₃y₃) / M ----------------(ii)
Where;
M = sum of the masses
m₁ and x₁ = mass and position of first mass in the x direction.
m₂ and x₂ = mass and position of second mass in the x direction.
m₃ and x₃ = mass and position of third mass in the x direction.
y₁ , y₂ and y₃ = positions of the first, second and third masses respectively in the y direction.
From the question;
m₁ = 6kg
m₂ = 4kg
m₃ = 2kg
x₁ = 0m
x₂ = 3m
x₃ = 0m
y₁ = 0m
y₂ = 0m
y₃ = 3m
M = m₁ + m₂ + m₃ = 6 + 4 + 2 = 12kg
Substitute these values into equations (i) and (ii) as follows;
x = ((6x0) + (4x3) + (2x0)) / 12
x = 12 / 12
x = 1 m
y = (6x0) + (4x0) + (2x3)) / 12
y = 6 / 12
y = 0.5m
Therefore, the center of mass of the system is at (1m, 0.5m)
A 5.3 kg block initially at rest is pulled to the right along a frictionless, horizontal surface by a constant horizontal force of magnitude 21 N. Find the block's speed after it has moved through a horizontal distance of 6.4 m.
A brass ring of diameter 10.00 cm at 19.0°C is heated and slipped over an aluminum rod with a diameter of 10.01 cm at 19.0°C. Assuming the average coefficients of linear expansion are constant. What if the aluminum rod were 10.02 cm in diameter?
Answer:
the final temperature is [tex]\mathbf{T_f = -377.2^0 C}[/tex]
Explanation:
The change in length of a bar can be expressed with the relation;
[tex]\Delta L = L_f - L_i[/tex] ---- (1)
Also ; the relative or fractional increase in length is proportional to the change in temperature.
Mathematically;
ΔL/L_i ∝ kΔT
where;
k is replaced with ∝ (the proportionality constant )
[tex]\dfrac{ \Delta L}{L_i}=\alpha \Delta T[/tex] ---- (2)
From (1) ;
[tex]L_f = \Delta L + L_i[/tex] --- (3)
from (2)
[tex]{ \Delta L}=\alpha \Delta T*{L_i}[/tex] ---- (4)
replacing (4) into (3);we have;
[tex]L_f =(\alpha \Delta T*{L_i} ) + L_i[/tex]
On re-arrangement; we have
[tex]L_f = L_i + \alpha L_i (\Delta T )[/tex]
from the given question; we can say that :
[tex](L_f)_{brass}}} = (L_f)_{Al}[/tex]
So;
[tex]L_{brass} + \alpha _{brass} L_{brass}(\Delta T) = L_{Al} + \alpha _{Al} L_{Al}(\Delta T)[/tex]
Making the change in temperature the subject of the formula; we have:
[tex]\Delta T = \dfrac{L_{Al}-L_{brass}}{\alpha _ {brass} L_{brass}-\alpha _{Al}L_{Al}}[/tex]
where;
[tex]L_{Al}[/tex] = 10.02 cm
[tex]L_{brass}[/tex] = 10.00 cm
[tex]\alpha _{brass}[/tex] = 19 × 10⁻⁶ °C ⁻¹
[tex]\alpha_{Al}[/tex] = 24 × 10⁻⁶ °C ⁻¹
[tex]\Delta T = \dfrac{10.02-10.00}{19*10^{-6} \ \ {^0}C^{-1} *10.00 -24*10^{-6} \ \ {^0}C^{-1} *10.02}[/tex]
[tex]\Delta T[/tex] = −396.1965135 ° C
[tex]\Delta T[/tex] ≅ −396.20 °C
Given that the initial temperature [tex]T_i = 19^0 C[/tex]
Then ;
[tex]\Delta T = T_f - T_i[/tex]
[tex]T_f = \Delta T + T_I[/tex]
Thus;
[tex]T_f =(-396.20 + 19.0)^0 C[/tex]
[tex]\mathbf{T_f = -377.2^0 C}[/tex]
Thus; the final temperature is [tex]\mathbf{T_f = -377.2^0 C}[/tex]
In 1949, an automobile manufacturing company introduced a sports car (the "Model A") which could accelerate from 0 to speed v in a time interval of Δt. In order to boost sales, a year later they introduced a more powerful engine (the "Model B") which could accelerate the car from 0 to speed 2.92v in the same time interval. Introducing the new engine did not change the mass of the car. Compare the power of the two cars, if we assume all the energy coming from the engine appears as kinetic energy of the car.
Answer: [tex]\frac{P_B}{P_A}[/tex] = 8.5264
Explanation: Power is the rate of energy transferred per unit of time: P = [tex]\frac{E}{t}[/tex]
The energy from the engine is converted into kinetic energy, which is calculated as: [tex]KE = \frac{1}{2}.m.v^{2}[/tex]
To compare the power of the two cars, first find the Kinetic Energy each one has:
K.E. for Model A
[tex]KE_A = \frac{1}{2}.m.v^{2}[/tex]
K.E. for model B
[tex]KE_B = \frac{1}{2}.m.(2.92v)^{2}[/tex]
[tex]KE_B = \frac{1}{2}.m.8.5264v^{2}[/tex]
Now, determine Power for each model:
Power for model A
[tex]P_{A}[/tex] = [tex]\frac{m.v^{2} }{2.t}[/tex]
Power for model B
[tex]P_B = \frac{m.8.5264.v^{2} }{2.t}[/tex]
Comparing power of model B to power of model A:
[tex]\frac{P_B}{P_A} = \frac{m.8.5264.v^{2} }{2.t}.\frac{2.t}{m.v^{2} }[/tex]
[tex]\frac{P_B}{P_A} =[/tex] 8.5264
Comparing power for each model, power for model B is 8.5264 better than model A.
A 2.5 m long diving board weighs 120 N. It has two supports, one at the end and another at a distance of 1.0 m from that end. A 100 N diver stands at the other end of the board. What is the magnitude and direction of the force exerted by the support at the end of the diving board
Answer:
46.67 N Upwards (with a clockwise moment)
Explanation:
length of board = 2.5 m
weight of board = 120 N
the board has two supports, say support A and support B
support A is at one end,
support B is at 100 m from the other end.
weight of diver = 100 N
diver stands on the other end of the board.
Magnitude of support A at the end of the board
To get the magnitude and force exerted by the support at the end of the board (support A, we take moment of the forces about support B
Moment of a force is the product of force and perpendicular distance of the force about a center.
The weight of the board acts at the center of the board (1.25 m from each end of the board). That is 2.5 m from the support B.
moment of board's weight about support B is 120 x 0.25 = 30 N-m
The moment due to the weight of the board acts anticlockwise.
Weight of the diver acts at the opposite side of the board, and it acts 1 m from support B.
Moment of diver about support B is 100 x 1 = 100 N-m
Th moment due to the diver acts clockwise.
The moment due to the reaction at support A acts at a distance 1.5 m from support B
If the reaction force on support A is Fa, then the reaction about support B is Fa x 1.5 = 1.5Fa.
The moment due to support A acts clockwise.
According to moment laws, the total clockwise movement must be equal to the total anticlockwise movement.
Total clockwise movements = 100 N-m + 1.5Fa
Total anticlockwise moment = 30 N-m
according to moment laws,
100 + 1.5Fa = 30
1.5 Fa = 30 - 100 = -70
Fa = -70/1.5 = -46.67 N
The magnitude of the force exerted at support A is equal but opposite to the reaction at support A and is equal to 46.67 N
13. Under what condition (if any) does a moving body experience no energy even though there
is a net force acting on it?
(2 marks)
Answer:
When the Net Forces are equal to 0
Explanation:
Momentum of a body can be defined as product of mass and velocity. It is in the same direction as in velocity. When the momentum of a body doesn't change, it is said to be conserved. If the momentum of a body is constant, the the net forces acting on a body becomes zero. When net forces acting on a body is zero, it means that no kinetic energy is being lost or gained, hence the kinetic energy is also conserved. If no energy is being gained or lost, it means that the body will experience no energy.
Consider two identical springs. At the start of an experiment, Spring A is already stretched out 3 cm, while Spring B remains at the zero position. Both springs are then stretched an additional three centimeters. What conclusion can you draw about the force required to stretch these springs during the experiment
Answer:
Explanation:
In this interesting exercise we have that spring A is 3 cm longer, due to previous experiments if these experiments did not reach the non-linear elongation point, the cosecant Km of the spring must remain the same, therefore when we lengthen the two springs these the longitudinal are lengthened.
As a consequence of the above according to Hockey law, the prediction of lengthening is the same, therefore the outside is the same in two two systems
F = K Δx
An electron is initially moving at 1.4 x 107 m/s. It moves 3.5 m in the direction of a uniform electric field of magnitude 120 N/C. What is the kinetic energy of the electron at the end of the motion
Answer:
K.E = 15.57 x 10⁻¹⁷ J
Explanation:
First, we find the acceleration of the electron by using the formula of electric field:
E = F/q
F = Eq
but, from Newton's 2nd Law:
F = ma
Comparing both equations, we get:
ma = Eq
a = Eq/m
where,
E = electric field intensity = 120 N/C
q = charge of electron = 1.6 x 10⁻¹⁹ C
m = Mass of electron = 9.1 x 10⁻³¹ kg
Therefore,
a = (120 N/C)(1.6 x 10⁻¹⁹ C)/(9.1 x 10⁻³¹ kg)
a = 2.11 x 10¹³ m/s²
Now, we need to find the final velocity of the electron. Using 3rd equation of motion:
2as = Vf² - Vi²
where,
Vf = Final Velocity = ?
Vi = Initial Velocity = 1.4 x 10⁷ m/s
s = distance = 3.5 m
Therefore,
(2)(2.11 x 10¹³ m/s²)(3.5 m) = Vf² - (1.4 x 10⁷)²
Vf = √(1.477 x 10¹⁴ m²/s² + 1.96 x 10¹⁴ m²/s²)
Vf = 1.85 x 10⁷ m/s
Now, we find the kinetic energy of electron at the end of the motion:
K.E = (0.5)(m)(Vf)²
K.E = (0.5)(9.1 x 10⁻³¹ kg)(1.85 x 10⁷ m/s)²
K.E = 15.57 x 10⁻¹⁷ J
The barricade at the end of a subway line has a large spring designed to compress 2.00 m when stopping a 1.10 ✕ 105 kg train moving at 0.350 m/s. (a) What is the force constant (in N/m) of the spring? N/m (b) What speed (in m/s) would the train be going if it only compressed the spring 0.600 m? m/s (c) What force (in N) does the spring exert when compressed 0.600 m? 2020 N (in the direction opposite to the train's motion)
Answer:
(a) k = 1684.38 N/m = 1.684 KN/m
(b) Vi = 0.105 m/s
(c) F = 1010.62 N = 1.01 KN
Explanation:
(a)
First, we find the deceleration of the car. For that purpose we use 3rd equation of motion:
2as = Vf² - Vi²
a = (Vf² - Vi²)/2s
where,
a = deceleration = ?
Vf = final velocity = 0 m/s (since, train finally stops)
Vi = Initial Velocity = 0.35 m/s
s = distance covered by train before stopping = 2 m
Therefore,
a = [(0 m/s)² - (0.35 m/s)²]/(2)(2 m)
a = 0.0306 m/s²
Now, we calculate the force applied on spring by train:
F = ma
F = (1.1 x 10⁵ kg)(0.0306 m/s²)
F = 3368.75 N
Now, for force constant, we use Hooke's Law:
F = kΔx
where,
k = Force Constant = ?
Δx = Compression = 2 m
Therefore.
3368.75 N = k(2 m)
k = (3368.75 N)/(2 m)
k = 1684.38 N/m = 1.684 KN/m
(c)
Applying Hooke's Law with:
Δx = 0.6 m
F = (1684.38 N/m)(0.6 m)
F = 1010.62 N = 1.01 KN
(b)
Now, the acceleration required for this force is:
F = ma
1010.62 N = (1.1 kg)a
a = 1010.62 N/1.1 x 10⁵ kg
a = 0.0092 m/s²
Now, we find initial velocity of train by using 3rd equation of motion:
2as = Vf² - Vi²
a = (Vf² - Vi²)/2s
where,
a = deceleration = -0.0092 m/s² (negative sign due to deceleration)
Vf = final velocity = 0 m/s (since, train finally stops)
Vi = Initial Velocity = ?
s = distance covered by train before stopping = 0.6 m
Therefore,
-0.0092 m/s² = [(0 m/s)² - Vi²]/(2)(0.6 m)
Vi = √(0.0092 m/s²)(1.2 m)
Vi = 0.105 m/s
A 60.0-kg boy is surfing and catches a wave which gives him an initial speed of 1.60 m/s. He then drops through a height of 1.57 m, and ends with a speed of 8.50 m/s. How much nonconservative work was done on the boy
Answer:
Work = 1167.54 J
Explanation:
The amount of non-conservative work here can be given by the difference in kinetic energy and the potential energy. From Law of conservation of energy, we can write that:
Gain in K.E = Loss in P.E + Work
(0.5)(m)(Vf² - Vi²) - mgh = Work
where,
m = mass of boy = 60 kg
Vf = Final Speed = 8.5 m/s
Vi = Initial Speed = 1.6 m/s
g = 9.8 m/s²
h = height drop = 1.57 m
Therefore,
(0.5)(60 kg)[(8.5 m/s)² - (1.6 m/s)²] - (60 kg)(9.8 m/s²)(1.57 m) = Work
Work = 2090.7 J - 923.16 J
Work = 1167.54 J
A fox locates rodents under the snow by the slight sounds they make. The fox then leaps straight into the air and burrows its nose into the snow to catch its meal. If a fox jumps up to a height of 85 cm , calculate the speed at which the fox leaves the snow and the amount of time the fox is in the air. Ignore air resistance.
Answer:
v = 4.08m/s₂
Explanation:
During the deceleration of an ascending elevator, the normal force on the feet of a passenger is _____ her weight. During the deceleration of a descending elevator, the normal force on the feet of a passenger is _____ her weight.
Answer: Smaller than ; larger than
Explanation:
When the elevator is moving in the upward direction, then the force acting on it is negative in nature because of
N= mg +ma, (g is gravity and a is acceleration)
here ma is negative so the N= mg-ma
Hence, it feels smaller than its original weight.
When the elevator is moving downward , then the force acting will be positive in nature
N= mg+ma,
here ma will be positive so it feels larger the original weight of passenger.