Answer:
C your answer would be C
Explanation:
It should be right
The loaded car of a roller coaster has mass M = 320 kg. It goes over the highest hill with a speed v of 21.4 m/s. The radius of curvature R of the hill is [01] m. (a) What is the force (N) that the track must exert on the car? (positive is up) (b) What must be the force (N) that the car exerts on a 61 kg passenger?
This question is incomplete, the complete question is;
The loaded car of a roller coaster has mass M = 320 kg. It goes over the highest hill with a speed v of 21.4 m/s. The radius of curvature R of the hill is 15.8 m.
(a) What is the force (N) that the track must exert on the car? (positive is up)
(b) What must be the force (N) that the car exerts on a 61 kg passenger?
Answer:
a) the force (N) that the track must exert on the car is -6139.14 N
b) the force (N) that the car exerts on a 61 kg passenger is -1170.27 N
Explanation:
Given the data in the question;
Let N represent the force that the track must exerted on the car
Net force on the car Fnet = Mg + N
so
M × a = Mg + N
N = Ma - Mg
N = Ma - M(v²/R)
we substitute
N = (320kg × 9.8m/s²) - ( 320 × ((21.4m/s)² / 15.8 m) )
N = 3136 - ( 320 × 28.9848 )
N = 3136 - 9275.136
N = -6139.14 N
Therefore, the force (N) that the track must exert on the car is -6139.14 N
b) What must be the force (N) that the car exerts on a 61 kg passenger?
Let N represent the force that the car exerts on 61kg passengers
so
Net force of passengers Fnet = mg + N
Ma = Mg + N
N = Ma - Mg
N = Ma - M(v²/R)
N = (61kg × 9.8m/s²) - ( 61 × ((21.4m/s)² / 15.8 m) )
N = 597.8 - ( 61 × 28.9848)
N = 597.8 - 1768.0728
N = -1170.27 N
Therefore, the force (N) that the car exerts on a 61 kg passenger is -1170.27 N
The centripetal force of the track on the car moving in the circular path is [tex]1.465 \times 10^6 \ N[/tex].
The force (N) that the car exerts on a 61 kg passenger is 597.8 N.
Centripetal force of the track
The centripetal force of the track on the car moving in the circular path is calculated as follows;
[tex]F_c = \frac{mv^2}{r}\\\\ F_c = \frac{320 \times 21.4^2}{0.1} \\\\F_c = 1.465 \times 10^6 \ N[/tex]
Normal force of the passengerThe force (N) that the car exerts on a 61 kg passenger is equal to the force the passenger exerts on the car based on Newton's third law of motion.
F = mg
F = 61 x 9.8
F = 597.8 N
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Cameron is standing on the edge of a 60 m high cliff. He horizontally throws a football
with an initial velocity of 10 m/s. How long does it take for the football to hit the
ground?
Answer:34.6 m/s
Explanation: It is asking how long meaning the answer is in time
your "A" never changes, while your "Y" changes depending on strength of gravity. is it mass or weight?
Answer:
A - mass. B - Weight
Explanation:
This is because weight varies with the strength of gravity. Mass is just the amount of matter in an object
If you have a 0.125 kg lead piece at
20.0°C, how much heat must you
add to melt it? (Remember, you
must warm it to its melting point
first.)
Material
Lead
Melt Pt (°C)
327
L (1/kg)
2.32.104
Boil Pt (°C) Lv (1/kg)
1750 8.59.105
c (1/(kg*c)
128
(Unit = J)
Answer:
7,812 J
Explanation:
Using the relation:
Q = mcΔθ
Q = quantity of heat
C = specific heat capacity of lead
Δθ = temperature change (T2 - T1)
M = mass of substance
Q = mass * specific heat * Δθ
Q = 0.125kg * 128 * (327 – 20)
Q = 0.125 * 128 * 307
Q = 4912 J
For melting:
Q = mass * Hf
0.125 * (2.32 * 10^4)
= 2,900 J
Total = 4,912 J + 2,900 J = 7,812 J
In the Styrofoam ball investigation, it is likely that the charges on the ball and rod are
?
A. the same
B. opposite
C. constantly changing.
Answer:
answer = opposite
they are not changing being in a place as being they-self
the same would not be true because its different then the Styrofoam and it could only be different and its make the most sense aswell.
Both ball and rod has opposite charges on their bodies because of presence of different charges on it.
What are the charges on ball and rod?In the Styrofoam ball investigation, it is likely that the charges on the ball and rod are opposite because Styrofoam ball is negatively charged due to the presence of electrons while on the other hand, the rod is positively charged because of lining of positive charges on the rod.
So we can conclude that both ball and rod has opposite charges on their bodies because of presence of different charges on it.
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A 0.25 kg beach ball rolling at a speed of 7 m/s collides with a heavy exercise ball at rest. The beach ball bounces straight back with a speed of 4 m/s. That is the change in momentum of the beach ball? What is the impulse exerted on the beach ball? What is the impulse exerted on the exercise ball?
The impulse exerted on the beach ball is 2.75 kgm/s.
The impulse exerted on the exercise ball is - 2.75 kgm/s.
What is impulse?
This is the force applied to an object that acts over a period of time.
The impulse exerted on the beach ball is the change in the momentum of the ball and it is calculated as follows;
J = ΔP
J = m(v - u)
J = 0.25(7 - (-4))
J = 0.25(7 + 4)
J = 2.75 kgm/s
The impulse exerted on the exercise ball is equal in magnitude but opposite in direction to the beach ball.
Thus, the impulse exerted on the exercise ball is - 2.75 kgm/s.
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TWO forces, one of 12N and another or 24N
act on body in such a way that they make an angle of 90° with each other. Find the resaltant of two forces.
Answer:
26.833 N
Explanation:
The computation of the resaltant of two forces is shown below:
Given that
Force A = 12N
Force B = 24N
Based on the above information
Resultant R is
[tex]=\sqrt{A^2 + B^2 + 2AB \times cos \theta}\\\\=\sqrt{144 + 576 + 2\times 24\times 12\times cos90^{\circ}}\\\\=\sqrt{144+576+576\times 0}\\\\=\sqrt{720}[/tex]
=26.833 N
Allison and Heather are going to conduct an experiment to see whether salt affects the growth of plants. They assemble five groups of identical plants and give the plants in each group water with a different salt concentration. What is the outcome variable (dependent variable) for their experiment?
A. Salt concentration in plant tissue
B. Salt concentration in plant water
C. Amount of water absorbed by plants
D. Average mass of plants in each group
Answer:b
Explanation:
Guess
What is the Radiation left over from the big bang called?
Answer:
The Big Bang theory predicts that the early universe was a very hot place and that as it expands, the gas within it cools. Thus the universe should be filled with radiation that is literally the remnant heat left over from the Big Bang, called the “cosmic microwave background", or CMB.
Explanation:
The Big Bang theory suggest that the universe in early stage was at very hot place and which can be expanded, the gas within it cools. It is in an infinite universe and it has no edge.
What is big bang theory ?The Big Bang theory is a cosmological model which explain the existence of the observable universe from the earliest periods to the large-scale evolution.
The model describes the mechanism behind the universe expansion from an initial state of high density and temperature, it is very important concept as a lot of research is going on in this field to find out exactly how the universe began billions of years ago.
The universe began to cool down in order to allow the formation of particles become atoms after its initial phase of expansion, Primordial elements such as Hydrogen, Helium, and Lithium are condensed through gravity are formed early stars and galaxies.
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if a current of 5A flows for 2minutes, find the quantity of electricity transfered
because there the sperm and eggs are combining together to produce
so thats why they look alike
Answer:
yes. that is how a baby is conceived.
If the mass of the object doubles then the acceleration is when the force is kept the same
Answer:
Halved
Explanation:
F=ma
Let case 1 (original) be:
[tex]F_{1}=m_{1} a_{1} \\[/tex]
Case 2 (new) be:
[tex]F_{2}=m_{2} a_{2}[/tex]
Mass is double:
[tex]m_{2}= 2m_{1}[/tex]
Force kept the same:
[tex]F_{1} =F_{2}[/tex]
Combine the equation and gives:
[tex]\frac{F_{1} }{F_{2}} =\frac{m_{1} a_{1} }{m_{2}a_{2} }\\\frac{F_{1} }{F_{1}} =\frac{m_{1} a_{1} }{2m_{1}a_{2} }\\1=\frac{a_{1} }{2a_{2} }\\a_{2}=\frac{1}{2} a_{1}[/tex]
Acceleration is halved
The sides of a right triangle that has any given vector for the hypotenuse are called _____
A. Scalar
B. Component
C. Resultant
D. Vector
Answer:
They are the resultant vector.
Please help 25 points!
Three waves with frequencies of 1 Hertz (Hz), 3 Hz, and 9Hz travel at the same speed. Which of the following statements is correct?
A. The 1 Hz wave contains the most energy.
B. The crests of all three waves are of equal height.
C. The wavelength of the 9Hz wave is three times that of the 3 Hz wave.
D. The 1 Hz wave has the longest wavelength.
Answer:
B
Explanation:
The crest of all three waves are of equal height
Two identical conducting spheres are placed with their centers 0.30 m apart. One is given a charge of 12 X10^-9 C and the other is given a charge of -18 X 10^-9 C. a. Find the electric force exerted on one sphere by the other. b. The sphere are connected by a conducting wire. After equilibrium has occurred, find the electric force between the two spheres.
Answer:
Explanation:
Force between two charged conducting sphere
= k x Q₁ x Q₂ / r² , k is a constant Q₁ and Q₂ are charges and r is distance between them .
= 9 x 10⁹ x 12 x 10⁻⁹ x 18 x 10⁻⁹ / .30²
= 21600 x 10⁻⁹
= 2.16 x 10⁻⁵ N .
b )
After the spheres are joined together , there is redistribution of charge and remaining charge will be equally shared by them .
Charge on each sphere = (12 - 18 ) x 10⁻⁹ / 2
= - 3 x 10⁻⁹ C .
Force = 9 x 10⁹ x 3 x 10⁻⁹ x 3 x 10⁻⁹ / .30²
= 900 x 10⁻⁹ N .
A soccer ball was kicked over the edge of a wall and traveled 35 m horizontally at a speed of 5.6m/s. Calculate the vertical height of the wall.
Answer:
Are you sure it was soccer ball? Or meine hearts
Explanation:
Albert Bandura emphasized the idea of __________, which is the belief one has in one’s own ability to succeed. A. operant conditioning B. determinism C. self-efficacy D. self-worth
Answer:
C
Explanation:
Albert Bandura emphasized the idea of Self efficacy which is the belief one has in one’s own ability to succeed.
What is Self efficacy?
A person's self-efficacy relates to their confidence in their ability to carry out the behaviors required to achieve particular performance goals (Bandura, 1977, 1986, 1997).
The belief in one's capacity to exercise control over one's own motivation, behavior, and social environment is known as self-efficacy. The goals for which people strive, the amount of effort put out to obtain goals, and the possibility of achieving particular levels of behavioral performance are all influenced by these cognitive self-evaluations.
Self-efficacy beliefs, unlike conventional psychological notions, are anticipated to change according to the operating domain and the environment in which an action occurs.
Therefore, Albert Bandura emphasized the idea of Self efficacy which is the belief one has in one’s own ability to succeed.
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A 450.0 kg roller coaster is traveling in a circle with radius 15.0m. Its speed at point A is 28.0m/s and its speed at point B is 14.0 m/s. At point A the cart is already moving with circular motion. a) Draw free bodydiagramsfor the cartatpointsAand B(two separate free body diagrams). b) Calculate the acceleration of the cartat pointsAandB(magnitude and direction). c) Calculate the magnitude of the normal force exerted by the trackson the cartat point A. d) Calculate the magnitude of the normal force exerted by the tracks on the cart at point B.
Answer:
b) a = 52.26 m / s², a ’= 13.06 m / s², c) N = 2.79 10⁴ N, d) N = 1.89 10³ N
Explanation:
a) In the attached we can see the free body diagrams for the two positions, position A in the lower part of the circle and position B in the upper part of the circle
b) Let's start at point A
Let's use that the acceleration is centripetal
a = v² / r
let's calculate
a = 28² / 15.0
a = 52.26 m / s²
as they relate it is centripetal it is directed towards the center of the circle, therefore for this point it is directed vertically upwards
Point B
a ’= 142/15
a ’= 13.06 m / s²
in this case the acceleration is vertical downwards
c) The values of the normal force
point A
let's use Newton's second law
∑ F = m a
N- W = m a
N = mg + ma
N = m (g + a)
N = 450.0 (9.8 + 52.25)
N = 2.79 10⁴ N
d) Point B
-N -W = m (-a)
N = ma -m g
N = m (a-g)
N = 450.0 (14.0 - 9.8)
N = 1.89 10³ N
Which statement best compares potential and kinetic energy?
O Objects always have more potentiał energy than kinetic energy.
O Kinetic energy increases and potential energy decreases when the velocity of an object increases
O Only potential energy decreases when an object's height increases.
O Objects always have more kinetic energy than potential energy.
Answer:
Kinetic energy increases and potential energy decrease when velocity of an object increase.
Question 3 (5 points)
Yissel was going to be late to Mr. Scharff's science class. Just as the bell was about to ring. Vissel ran the last little bit of the hallway at 2.47
meters/second for 8 seconds to beat the bell. How far away was Yissel from Mr. Scharff's classroom when she started to run?
A girl and a boy are riding on a merry-go-round that is turning at a constant rate. The girl is near the outer edge, and the boy is closer to the center. Who has greater angular displacement?
a) boy
b) girl
c) both have the same angular displacement
Answer:
c) both have the same angular displacement
Explanation:
In this scenario, girl and a boy are riding on a merry-go-round that is turning at a constant rate. The girl is near the outer edge, and the boy is closer to the center. Therefore, both have the same angular displacement.
Common transparent tape becomes charged when pulled from a dispenser. If one piece is placed above another, the repulsive force can be great enough to support the top piece's weight. Assuming equal point charges (only an approximation), calculate the magnitude of the charge if electrostatic force is great enough to support the weight of a 12.0 mg piece of tape held 0.55 cm above another. (The magnitude of this charge is consistent with what is typical of static electricity.)
Answer:
q = 2 10⁻⁸ C
Explanation:
For this exercise we use the translational equilibrium equation
F_e -A =
F_e = W
the electric force is given by Coulomb's law
F_e = [tex]k \frac{q_1q_2}{r^2}[/tex]
in this case they indicate that the loads on the tapes are equal
F_e = k q² / r²
we substitute
k q² / r² = m g
q = [tex]\sqrt{ \frac{mg r^2}{k} }[/tex]
calculate
q = [tex]\sqrt { \frac{ 12 \ 10^{-3} \ 9.8 (0.55 \ 10^{-2})^2 }{9 \ 10^9} }[/tex]
q = [tex]\sqrt{ 3.9526 \ 10^{-16}[/tex]
q = 1,999 10⁻⁸ C
q = 2 10⁻⁸ C
Force = mass X acceleration. Acceleration due to gravity is 9.8 m/s/s regardless of height or mass. This means a single object will hit the ground with the same force regardless of the height it is dropped from. If this is true, I normally drop a pumpkin from different heights during this unit. why does dropping the pumpkin stay together when dropped from one meter but break apart when dropped from a height of 5 meters?
Given that Carbon-14 has a half-life of 5700 years, determine how long it would take for
this reduction to occur.
Answer:It will take about 3000 years
Explanation:
A fan has four identical, symmetrically placed blades. The blades are rotating clockwise at twenty revolutions per second.
A) What is the smallest time interval between stroboscope flashes that will make the fan blades appear motionless?
B) What is the highest frequency (in flashes per second) at which a stroboscope will make the
fan blades appear to stand still? Show your calculation.
C) The same questions as (a) and (b), but someone has put a yellow dot on one blade, and now you want the yellow dot to appear to be standing still. Explain, and show your calculation.
D) Now the stroboscope is set for nineteen flashes per second, and the yellow dot appears to be slowly rotating. Which direction does it appear to rotate, clockwise or counterclockwise? Explain, and show your calculation.
E) The same as (d), but the stroboscope is set for twenty-one flashes per second. Explain, and show your calculation.
Answer:
A) t = 1.249 10⁻² s, B) f = 80 Hz, C) f = 20 Hz,
D) slowly advancing an angle of approximately Δθ = 0.05 rad each flash
E) In each flash it seems to go backward an angle of Δθ = -0.05 rad
Explanation:
A) To make it appear that the blades are immobile, it implies that every time the light turns on, a blade should be in the same position, therefore, as we have 4 blades, they must rotate an angle of 2π/4,
θ = π / 2
θ = 1.57 rad
taking the angle let's use the endowment kinematics relations
θ = w₀ t + ½ α t²
in general the fans rotate at constant speed α= 0
θ = w₀ t
t = θ / w₀
let's reduce the magnitudes to the SI system
w₀ = 20 rev / s (2π rad / 1rev) = 125.66 rad / s
let's calculate
t = 1.57 / 125.66
t = 1.249 10⁻² s
B) the fastest speed for the blades to rotate is when one blade of a complete turn , we use the relationship between the fecuance and the period
f = 1 / T
f = 1 / 1.25 10⁻²
f = 80 Hz
C) we have two possibilities:
* a yellow dot is placed on each sheet
In this case the angular velocity of the blade is the same at all points, therefore the results obtained should not change
* a yellow dot is placed on a single sheet.
Here for the point to remain fixed the angle of rotation must be
θ= 2π rad
the time is
t = 2π / 125.66
t = 5 10⁻² s
the maximum frequency is
f = 1/5 10⁻²
f = 20 Hz
D) The copy strobe rotates at f = 19 Hz, the time between each flash is
t = 1/19
t = 5.26 10⁻² s
this time is higher, so the angle turned is large
θ = w t
θ = 125.66 5.26 10⁻²
θ = 6.61 rad
the relationship between this angle and the angle of a circle is
θ = 1,052
We can see that it is this time the blade rotates 1 complete turns, for this the position of the blade changes us, for the other 0.052 rad the blade rotates a little more than the circumference therefore it seems that it is slowly advancing an angle of approximately
Δθ = 0.05 rad each flash
E) in this case changes the flash speed
t = 1/21
t = 4.76 10⁻² s
the angle rotated is
θ = 125.66 4.76 10⁻²
θ = 5.984 rad
θ / 2π = 0.95
in that case, the blade did not complete the turn, therefore in each flash it seems to go backward an angle
Δθ = -0.05 rad
A rocket, with a mass of 5100 kg, has an engine that provides a net upward force of 8.0 x 10^5 N. It starts from rest and reaches a maximum speed of 900 m/s. How long does it take to reach that maximum velocity?
Answer:
5.7375 seconds
Explanation:
The computation of the time required to reach that maximum velocity is shown below:
Given that
Mass = m = 5100 kg
Net upward force F = 8 × times 10^5 N
Initial speed = V_i = 0
Maximum speed = V = 900 m.s
Based on the above information
Impluse J = m(V - V_i)
= 5100 (900 - 0)
= 459 × 10^4 kg m.s
As we know that
J = FT
So
T = J ÷ F
= (459 × 10^4) ÷ (8 × 10^5)
= 5.7375 seconds
A student is conducting an experiment to compare the resistivity of two unknown materials by using two wires, each made of one of the materials and each connected in a circuit. The student measures the potential difference across and current in the wires. What must be the same to be able to compare the resistivities using just the potential difference and current measurements?
Answer:
is there a. b. c or d?
Explanation:
4 Two friction disks A and B are to be brought into contact withoutslipping when the angular velocity of disk A is 240 rpm counterclockwise. Disk A starts from rest at time t = 0 and is given a constantangular acceleration with a magnitude α. Disk B starts from rest attime t = 2 s and is given a constant clockwise angular acceleration,also with a magnitude α. Determine (a) the required angular acceleration magnitude α, (b) the time at which the contact occurs
This question is incomplete, the missing image is uploaded along this answer below;
Answer:
a) the required angular acceleration magnitude α is π rad/s² or 3.14 rad/s²
b) the time at which the contact occur is 8 seconds
Explanation:
Given the data in the question;
first we convert the given angular velocity to rad/s
angular velocity = 240 rpm = ((240/60) × 2π ) = 8π rad/s
so
ωA = 8π rad/s
next we determine angular acceleration at point A; so
ωA = at
8π rad/s = at -------let this be equation
thus, angular acceleration of disk A is ωA and rotates in counter clockwise direction.
Next we determine the velocity of point C;
Vc = rA × ωA
where Vc is velocity at point C, rA is radius of A ( 150/1000)m, { from the diagram }
so we substitute
Vc = 0.15m × 8π
Vc = 1.2π m/s
for angular velocity at point B;
Vc = rB × ωB
where rB is the radius of B ( 200/1000)m
we substitute
1.2π = 0.2 × ωB
ωB = 1.2π / 0.2
ωB = 6π rad/s
Thus, the wheel B rotates with an angular velocity of 6π rad/s in clock wise direction.
Now,
a) Determine the required angular acceleration magnitude α
we find the the angular acceleration of disk B after 2 seconds, using the expression;
ωB = at
where angular acceleration is a and t is time ( t - 2)
we substitute
ωB = at
6π = a( t - 2) -------- let this be equation 2
now, lets substract equation 1 form equation 2
(6π = a( t - 2)) - (8π = at)
(6π = at - 2a) - ( 8π = at)
-2π = 0 + -2a
2π = 2a
a = 2π/2
a = π rad/s² or 3.14 rad/s²
Therefore, the required angular acceleration magnitude α is π rad/s² or 3.14 rad/s²
b) determine the time at which the contact occurs;
from equation 1
8π = at
we substitute in the value of a
8π = π × t
t = 8π / π
t = 8 seconds
Therefore, the time at which the contact occur is 8 seconds
3. Sodium-24 has a half-life of 15 hours. If a sample of sodium-24 has an
original activity of 800 Bq, what will
its activity be after:
i) 15 hours?
ii) 30 hours?
iii) 45 hours?
iv) 60 hours?
Answer:
See explanation
Explanation:
From the formula;
0.693/t1/2 = 2.303/t log (Ao/A)
t1/2 = half life of Sodium-24
Ao = initial activity of Sodium-24
A= activity of Sodium-24 at time = t
So,
0.693/15 = 2.303/15 log (800/A)
0.0462 = 0.1535 log (800/A)
0.0462/0.1535 = log (800/A)
0.3 = log (800/A)
Antilog(0.3) = (800/A)
1.995 = (800/A)
A = 800/1.995
A = 401 Bq
ii) 0.693/15 = 2.303/30 log (800/A)
0.0462 = 0.0768 log (800/A)
0.0462/0.0768 = log (800/A)
0.6 = log (800/A)
Antilog (0.6) = (800/A)
3.98 = (800/A)
A = 800/3.98
A = 201 Bq
iii)
0.693/15 = 2.303/45 log (800/A)
0.0462 = 0.0512 log (800/A)
0.0462/0.0512 = log (800/A)
0.9 = log (800/A)
Antilog (0.9) = (800/A)
7.94 = (800/A)
A = 800/7.94
A= 100.8 Bq
iv)
0.693/15 = 2.303/60 log (800/A)
0.0462 = 0.038 log (800/A)
0.0462/0.038 = log (800/A)
1.216 = log (800/A)
Antilog(1.216) = (800/A)
16.44 = (800/A)
A = 800/16.44
A = 48.66 Bq
.You have always been impressed by the speed of the elevators in your apartment building. You wonder about the maximum acceleration for these elevators during normal operation, so you decide to measure it by using your bathroom scale. While the elevator is at rest on the ground floor,you get in, put down your scale, and stand on it. The scale reads 50 kg. You continue standing on the scale when the elevator goes up, carefully watching the reading. During the trip to the 10th floor, the greatest scale reading was
Answer:
5.51 m/s^2
Explanation:
Initial scale reading = 50 kg
assume the greatest scale reading = 78.09 kg
Determine the maximum acceleration for these elevators
At rest the weight is = 50 kg
Weight ( F ) = mg = 50 * 9.81 = 490.5 N
At the 10th floor weight = 78.09 kg
Weight at 10th floor ( F ) = 78.09 * 9.81 = 766.11 N
F = change in weight
Change in weight( F ) = ma = 766.11 - 490.5 (we will take the mass as the starting mass as that mass is calculated when the body is at rest)
50 * a = 275.61
Hence the maximum acceleration ( a ) = 275.61 / 50 = 5.51 m/s^2