Answer:
While Bohr's atomic model hypothesizes that electrons move in particular energy levels around the nucleus, the electron cloud model suggests that electrons move in an unpredictable pattern but are more likely to be in certain regions than others.
Explanation:
At an airport, two business partners both walk at 1.5 m/sm/s from the gate to the main terminal, one on a moving sidewalk and the other on the floor next to it. The partner on the moving sidewalk gets to the end in 60 ss, and the partner on the floor reaches the end of the sidewalk in 90s.
Required:
What is the speed of the sidewalk in the Earth reference frame?
Answer:
[tex]v=0.8m/s[/tex]
Explanation:
From the question we are told that
Distance [tex]d=1.5m/sm/s[/tex]
Time [tex]t_1=60s[/tex]
Time [tex]t_2=90s[/tex]
Generally the the equation for the distance traveled is mathematically given as
[tex]d=vt[/tex]
[tex]d=1.5*90[/tex]
[tex]d=138m[/tex]
Generally equation for speed of side walk is mathematically given as
[tex]d=(v+u)t[/tex]
[tex]v=\frac{d}{t}-u[/tex]
[tex]v=\frac{138}{60}-1.5[/tex]
[tex]v=0.8m/s[/tex]
1. (6x + 8)(5x - 8)
a. 30x2 + 49x + 20
2. (5x + 6(5x - 5)
b. 24x3 + 8x2 + 6x + 4
3. (6x + 3)(6x - 4)
c. 25x2 + 5x - 30
4. (6x + 5)(5x + 5)
d. 30x2 - 8x - 64
e. 36x2 - 6x - 1
5. (4x + 2) (6x2 - x + 2)
Answer:
form 1 question??????????
A student is provided with a battery-powered toy car that the manufacturer claims will always operate at a constant speed. The student must design an experiment in order to test the validity of the claim. Which of the following measuring tools can the student use to test the validity of the claim?
a. Photogates placed at the beginning, end, and at various locations along the track that the car travels on.
b. A meterstick to measure the distance of the track that the car travels on.
c. A motion detector that is oriented perpendicular to the direction that the car travels.
d. A mass balance to determine the mass of the car
Answer:
a. Photogates placed at the beginning, end, and at various locations along the track that the car travels on.
b. A meterstick to measure the distance of the track that the car travels on.
Explanation:
Physics can be defined as the field or branch of science that typically deals with nature and properties of matter, motion and energy with respect to space, force and time.
In this scenario, a student is provided with a battery-powered toy car that the manufacturer claims will always operate at a constant speed. The student must design an experiment in order to test the validity of the claim.
Therefore, to test the validity of the claim, the student should use the following measuring tools;
a. Photogates placed at the beginning, end, and at various locations along the track that the car travels on. This device is typically used to measure time with respect to the rate of change of the interruption or block of an infra-red beam.
b. A meterstick to measure the distance of the track that the car travels on.
Hence, with these two devices the student can effectively measure or determine the validity of the claim.
One of the disadvantages of experimental research is that __________.
A.
it isn’t easily replicated
B.
it doesn’t often reflect reality
C.
the results aren’t generalizable
D.
conditions are not controllable
Please select the best answer from the choices provided
Answer:
B
Explanation:
Two 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 7.35 cm from the zero volt plate (and 2.65 cm from the other) is 533 V
Answer:
[tex]E=6.8Kv/m[/tex]
Explanation:
From the question we are told that
Distance b/w plate [tex]d=10cm=>0.1m[/tex]
P_1 Potential at 7.35 [tex]V=533v[/tex]
Generally the equation for electric field at a distance is mathematically given as
[tex]E=\frac{v}{d}[/tex]
[tex]E=\frac{533}{7.85*10^-^2}[/tex]
[tex]E=6789.808917[/tex]
[tex]E=6.8*10^3[/tex]
[tex]E=6.8Kv/m[/tex]
How much force is needed to accelerate a 65 kg rider AND her 215 kg motor scooter 8 m/s?? (treat
the masses as like terms)
Answer:
Force = 2240 Newton.
Explanation:
Given the following data;
Mass A= 65kg
Mass B = 215kg
Acceleration = 8m/s²
To find the force;
Force is given by the multiplication of mass and acceleration.
Mathematically, Force is;
[tex] F = ma[/tex]
Where;
F represents force.
m represents the mass of an object.
a represents acceleration.
First of all, we would have to find the total mass.
Total mass = Mass A + Mass B
Total mass = 65 + 215
Total mass = 280kg
Substituting into the equation, we have
[tex] Force = 280 * 8 [/tex]
Force = 2240 Newton.
A violin has a string of length
0.320 m, and transmits waves at
622 m/s. At what frequency does
it oscillate?
Answer:
1.9kHz
Explanation:
Given data
wavelength [tex]\lambda= 0.32m[/tex]
velocity [tex]v= 622 m/s[/tex]
We know that
[tex]v= f* \lambda\\\\f= v/ \lambda[/tex]
substitute
[tex]f= 622/ 0.32\\\\f= 1943.75\\\\f= 1.9kHz[/tex]
Hence the frequency is 1.9kHz
Answer:
971.2
Explanation:
It was right on acellus :)
An RC car is carrying a tiny slingshot with a spring constant of 85 N/m at 0.2 m off the ground at 5.6 m/s. The sling shot is pulled back 3.5 cm from a relaxed state and shoots a 25 g steel pellet in the same direction the car is moving. What is the velocity of the steel pellet relative to the ground as it leaves the sling shot
Answer:
The velocity of the steel pellet relative to the ground when it leaves the sling shot is approximately 5.960 meters per second.
Explanation:
Let suppose that RC car-slingshot-steelpellet is a conservative system, that is, that non-conservative forces (i.e. friction, air viscosity) can be neglected. The velocity of the steel pellet can be found by means of the Principle of Energy Conservation and under the consideration that change in gravitational potential energy is negligible and that the RC car travels at constant velocity:
[tex]\frac{1}{2}\cdot (m_{C}+m_{P})\cdot v_{o}^{2} + \frac{1}{2}\cdot k\cdot x^{2} = \frac{1}{2}\cdot m_{C}\cdot v_{o}^{2} + \frac{1}{2}\cdot m_{P}\cdot v^{2}[/tex]
[tex]\frac{1}{2}\cdot m_{P}\cdot v_{o}^{2} + \frac{1}{2}\cdot k\cdot x^{2} = \frac{1}{2}\cdot m_{P}\cdot v^{2}[/tex]
[tex]m_{P}\cdot v_{o}^{2} + k\cdot x^{2} = m_{P}\cdot v^{2}[/tex]
[tex]v^{2} = v_{o}^{2} + \frac{k}{m_{P}}\cdot x^{2}[/tex]
[tex]v = \sqrt{v_{o}^{2}+\frac{k}{m_{P}}\cdot x^{2} }[/tex] (1)
Where:
[tex]v_{o}[/tex] - Initial velocity of the steel pellet, measured in meters per second.
[tex]v[/tex] - Final velocity of the steel pellet, measured in meters per second.
[tex]k[/tex] - Spring constant, measured in newtons per meter.
[tex]m_{P}[/tex] - Mass of the steel pellet, measured in kilograms.
[tex]m_{C}[/tex] - Mass of the RC car, measured in kilograms.
[tex]x[/tex] - Initial deformation of the spring, measured in meters.
If we know that [tex]v_{o} = 5.6\,\frac{m}{s}[/tex], [tex]k = 85\,\frac{N}{m}[/tex], [tex]m_{P} = 0.025\,kg[/tex] and [tex]x = 0.035\,m[/tex], then the velocity of the steel pellet relative to the ground when it leaves the sling shot is:
[tex]v = \sqrt{\left(5.6\,\frac{m}{s} \right)^{2}+\frac{\left(85\,\frac{N}{m} \right)\cdot (0.035\,m)^{2}}{0.025\,kg} }[/tex]
[tex]v \approx 5.960\,\frac{m}{s}[/tex]
The velocity of the steel pellet relative to the ground when it leaves the sling shot is approximately 5.960 meters per second.
A sinusoidal wave is traveling on a string with speed 19.3 cm/s. The displacement of the particles of the string at x = 6.0 cm is found to vary with time according to the equation y = (2.6 cm) sin[1.8 - (5.8 s-1)t]. The linear density of the string is 5.0 g/cm. What are (a) the frequency and (b) the wavelength of the wave? If the wave equation is of the form
Answer:
Explanation:
equation of wave is given by the following equation
y = (2.6 cm) sin[1.8 - (5.8 s-1)t].
Comparing it with standard form of wave
y = A sin ( ωt - kx )
we get
ω = 5.8
2πn = 5.8
n = .92 per second
kx = 1.8
k x 6 = 1.8
k = 0.3
[tex]\frac{2\pi}{\lambda}[/tex] = 0.3
λ = 20.9 cm
If a cyclist travels 30 km in 2 h, What is her average speed?
Answer:
15km/h
Explanation:
→ Speed = Distance ÷ Time
30 ÷ 2 = 15km/h
Fill in the blank with the correct word below (from the reading_):
helps you track your progress once you have made a lifestyle
change.
Self-monitoring
Healthy food
Regular xxercise
Goals
Answer:I think it’s self monitoring sorry if wrong
Explanation:
Answer:
It self monitoring
Explanation:
I took the test
When a drag strip vehicle reaches a velocity of 60 m/s, it begins a negative acceleration by releasing a drag chute and applying its brakes. While reducing its velocity back to zero, its acceleration along a straight line path is a constant -7.5 m/s2 . What displacement does it undergo during this deceleration period
Answer:
240 meters
Explanation:
The distance traveled by the vehicle can be calculated using the following equation:
[tex] v_{f}^{2} = v_{0}^{2} + 2ax [/tex] (1)
Where:
x: is the displacement
[tex]v_{f}[/tex]: is the final speed = 0 (reduces its velocity back to zero)
[tex]v_{0}[/tex]: is the initial speed = 60 m/s
a: is the acceleration = -7.5 m/s²
By solving equation (1) for x we have:
[tex] x = \frac{v_{f}^{2} - v_{0}^{2}}{2a} = \frac{0 - (60 m/s)^{2}}{2*(-7.5 m/s^{2})} = 240 m [/tex]
Therefore, the vehicle undergoes 240 meters of displacement during the deceleration period.
I hope it helps you!
Two spherical objects are separated by a distance that is 1.08 x 10-3 m. The objects are initially electrically neutral and are very small compared to the distance between them. Each object acquires the same negative charge due to the addition of electrons. As a result, each object experiences an electrostatic force that has a magnitude of 3.89 x 10-21 N. How many electrons did it take to produce the charge on one of the objects
Answer:
the charge on the object is 71.043×10^-20 C and the number of electron is 4.44
Explanation:
from coulumbs law, The force that is acting over both charge can be computed as
F=( kq1q2)/r^2..............eqn(1)
Where
F=electrostatic force= 3.89 x 10-21 N
k= column constant= 9 x 10^9 Nm^2/C^2
q1 and q2= magnitude of the charges
r= distance between the charges= 1.08 x 10-3 m.
Since both charges are experiencing the same force, eqn(1) can be written as
F=( kq^2)/r^2.
We can make q subject of the formula
q= √(Fr^2)/k
= √[(3.89 x 10^-21× (1.08 x 10^-3)^2]/8.99 x 10^9
q= 71.043×10^-20 C
Hence, the charge is 71.043×10^-20 C
From quantization law, the number of electron can be computed as
N=q/e
Where
N= number of electron
q= charges
=1.6×10^-19C
N=71.043×10^-20/1.6×10^-19
=4.44
Hence, the charge on the object is 71.043×10^-20 C and the number of electron is 4.44
A particle with charge Q and mass M has instantaneous speed uy when it is at a position where the electric potential is V. At a later time, the particle has moved a distance R away to a position where the electric potential is V2 ) Which of the following equations can be used to find the speed uz of the particle at the new position?
a. 1/2M(μ2^2-μ1^2)=Q (v1-v2)
b. 1/2M(μ2^2-μ1^2)^2=Q(v1-v2)
c. 1/2Mμ2^2=Qv1
d. 1/2Mμ2^2=1/4πx0 (Q^2/R)
Answer:
A
Explanation:
Ke = 1/2 MV^2
A group of 25 particles have the following speeds: two have speed 11 m/s, seven have 16 m/s , four have 19 m/s, three have 26 m/s, six have 31 m/s, one has 37 m/s, and two have 45 m/s.
Requiredd:
a. Determine the average speed.
b. Determine the rms speed.
c. Determine the most probable speed.
Answer:
a) Average speed is 24.04 m/s
b) the rms speed is 25.84 m/s
c) the most probable speed is 16 m/s
Explanation:
Given the data in the question;
a) Determine the average speed.
To determine the average speed, we simply divide total some of speed by number of particles;
Average speed = [(2×11 m/s)+(7×16 m/s)+(4×19 m/s)+(3×26 m/s)+(6×31 m/s)+(1×37 m/s)+(2×45 m/s)] / 25
= 601 / 25
= 24.04 m/s
Therefore, Average speed is 24.04 m/s
b) Determine the rms speed
we know that (rms speed)² = sum of square speed / total number of particles
so
(rms speed)² = [(2×11²)+(7×16²)+(4×19²)+(3×26²)+(6×31²)+(1×37²)+(2×45²)] / 25
(rms speed)² = 16691 / 25
(rms speed)² = 667.64
(rms speed) = √ 667.64
(rms speed) = 25.84 m/s
Therefore, the rms speed is 25.84 m/s
c) Determine the most probable speed.
Most particles (7) have velocity 16 m/s
i.e 7 is the maximum number of particle for a particular speed ,
Therefore, the most probable speed is 16 m/s
A small sphere of reference-grade iron with a specific heat of 447 J/kg K and a mass of 0.515 kg is suddenly immersed in a water-ice mixture. Fine thermocouple wires suspend the sphere, and the temperature is observed to change from 15 to 14C in 6.35 s. The experiment is repeated with a metallic sphere of the same diameter, but of unknown composition with a mass of 1.263 kg. If the same observed temperature change occurs in 4.59 s, what is the specific heat of the unknown material
Answer:
The specific heat of the unknown material is 131.750 joules per kilogram-degree Celsius.
Explanation:
Let suppose that sphere is cooled down at steady state, then we can estimate the rate of heat transfer ([tex]\dot Q[/tex]), measured in watts, that is, joules per second, by the following formula:
[tex]\dot Q = m\cdot c\cdot \frac{T_{f}-T_{o}}{\Delta t}[/tex] (1)
Where:
[tex]m[/tex] - Mass of the sphere, measured in kilograms.
[tex]c[/tex] - Specific heat of the material, measured in joules per kilogram-degree Celsius.
[tex]T_{o}[/tex], [tex]T_{f}[/tex] - Initial and final temperatures of the sphere, measured in degrees Celsius.
[tex]\Delta t[/tex] - Time, measured in seconds.
In addition, we assume that both spheres experiment the same heat transfer rate, then we have the following identity:
[tex]\frac{m_{I}\cdot c_{I}}{\Delta t_{I}} = \frac{m_{X}\cdot c_{X}}{\Delta t_{X}}[/tex] (2)
Where:
[tex]m_{I}[/tex], [tex]m_{X}[/tex] - Masses of the iron and unknown spheres, measured in kilograms.
[tex]\Delta t_{I}[/tex], [tex]\Delta t_{X}[/tex] - Times of the iron and unknown spheres, measured in seconds.
[tex]c_{I}[/tex], [tex]c_{X}[/tex] - Specific heats of the iron and unknown materials, measured in joules per kilogram-degree Celsius.
[tex]c_{X} = \left(\frac{\Delta t_{X}}{\Delta t_{I}}\right)\cdot \left(\frac{m_{I}}{m_{X}} \right) \cdot c_{I}[/tex]
If we know that [tex]\Delta t_{I} = 6.35\,s[/tex], [tex]\Delta t_{X} = 4.59\,s[/tex], [tex]m_{I} = 0.515\,kg[/tex], [tex]m_{X} = 1.263\,kg[/tex] and [tex]c_{I} = 447\,\frac{J}{kg\cdot ^{\circ}C}[/tex], then the specific heat of the unknown material is:
[tex]c_{X} = \left(\frac{4.59\,s}{6.35\,s} \right)\cdot \left(\frac{0.515\,kg}{1.263\,kg} \right)\cdot \left(447\,\frac{J}{kg\cdot ^{\circ}C} \right)[/tex]
[tex]c_{X} = 131.750\,\frac{J}{kg\cdot ^{\circ}C}[/tex]
Then, the specific heat of the unknown material is 131.750 joules per kilogram-degree Celsius.
Consider a swimmer that swims a complete round-trip lap of a 50 m long pool in 100 seconds. The swimmer's... average speed is 0 m/s and average velocity is 0 m/s. average speed is 0.5 m/s and average velocity is 0.5 m/s. average speed is 1 m/s and average velocity is 0 m/s. average speed is 0 m/s and average velocity is 1 m/s.What is the swimmers average speed and average velocity?
Answer:
average speed is 1 m/s and average velocity is 0 m/s.
Explanation:
Given that :
Length of round trip = 50 m
Time taken = 100 seconds
The average speed :
Total distance / total time taken
Length of complete round trip :
(50 + 50) m, total. Distance = 100 m
100 / 100 = 1m/s
The average velocity :
Total Displacement / total time taken
Total Displacement of round trip = end point - start point = 0
0 / 100 = 0
Average speed is 1 m/s and average velocity is 0 m/s.
The average speed is defined as the ratio of distance to time. Speed is a scalar quantity hence it does not take direction into account while velocity is a vector quantity hence it takes direction into account.
The speed is obtained from;
Speed = Distance/time = 2(50 m)/100 s = 1 m/s.
The velocity is 0 m/s since it is complete round-trip lap.
Learn more about speed: https://brainly.com/question/7359669
What is the period of an objects motion?
HELP PLEASE!!!
Running at 3.0 m/s, Burce, the 50.0 kg quarterback, collides with Max, the 100.0 kg tackle, who is traveling at 6.0 m/s in the other direction. Upon collision, Max continues to travel forward at 2.0 m/s.If the collision between the players lasted for 0.04 s, determine the impact force on either during the collision
Answer:
10kN
Explanation:
Given data
m1= 50kg
u1= 3m/s
m2= 100kg
u2= 6m/s
v1= 2m/s
time= 0.04s
let us find the final velocity of Bruce v1
from the conservation of linear momentum
m1u1+m2u2=m1v1+m2v2
substitute
50*3+100*6= 50*v1+100*2
150+600=50v1+200
750-200=50v1
550= 50v1
divide both sides by 50
v1= 550/50
v1=11 m/s
From
F= mΔv/t
for Bruce
F=50*(11-3)/0.04
F=50*8/0.04
F=400/0.04
F=10000
F=10kN
for Max
F=100*(6-2)/0.04
F=100*4/0.04
F=400/0.04
F=10000
F=10kN
A car enters a 105-m radius flat curve on a rainy day when the coefficient of static friction between its tires and the road is 0.4. What is
the maximum speed which the car can travel around the curve without sliding
Static friction (magnitude Fs) keeps the car on the road, and is the only force acting on it parallel to the road. By Newton's second law,
Fs = m a = W a / g
(a = centripetal acceleration, m = mass, g = acceleration due to gravity)
We have
a = v ² / R
(v = tangential speed, R = radius of the curve)
so that
Fs = W v ² / (g R)
Solving for v gives
v = √(Fs g R / W)
Perpendicular to the road, the car is in equilibrium, so Newton's second law gives
N - W = 0
(N = normal force, W = weight)
so that
N = W
We're given a coefficient of static friction µ = 0.4, so
Fs = µ N = 0.4 W
Substitute this into the equation for v. The factors of W cancel, so we get
v = √((0.4 W) g R / W) = √(0.4 g R) = √(0.4 (9.80 m/s²) (105 m)) ≈ 20.3 m/s
How could a change in straight line motion due to unbalanced forces be predicted from an understanding of inertia?
Answer:
If the force goes in the direction of movement, the speed must increase and if the net force goes in the opposite direction, the speed must decrease.
Explanation:
The principle of inertia or Newton's first law states that every body remains static or with constant velocity if there is no net force acting on it.
Based on this principle, if we have a net force, the velocity of the body changes by having an unbalanced force acting.
If the force goes in the direction of movement, the speed must increase and if the net force goes in the opposite direction, the speed must decrease.
The electric field between two parallel plates is uniform, with magnitude 628 N/C. A proton is held stationary at the positive plate, and an electron is held stationary at the negative plate. The plate separation is 4.22 cm. At the same moment, both particles are released.
A. Calculate the distance (in cm) from the positive plate at which the two pass each other.
B. Repeat part (a) for a sodlum lon (Nat) and a chlorlde lon (CI).
Answer:
Answer is explained in the explanation section below.
Explanation:
Solution:
Data Given:
Electric Field between two parallel plates = 628 N/C
Separation = 4.22 cm
a) In this part, we are asked to calculate the distance from positive plate at which the electron and proton pass each other.
Solution:
First of all:
Force on proton due to the Electric field between the plates is:
[tex]F_{p}[/tex] = [tex]q_{p}[/tex]E
and, we know that, F = ma
So,
[tex]m_{p}[/tex]a = [tex]q_{p}[/tex]E
a = [tex]\frac{q_{p}.E }{m_{p} }[/tex] Equation 1
So,
The distance covered by the electron is:
S = ut + 1/2[tex]at^{2}[/tex]
Here, u = 0.
S = 1/2[tex]at^{2}[/tex]
Put equation 1 into the above equation:
S = 1/2 x ([tex]\frac{q_{p}.E }{m_{p} }[/tex] )[tex]t^{2}[/tex] Equation 2
So,
Similarly, the distance covered by electron will be:
(D-S) = 1/2 x ([tex]\frac{q_{e}.E }{m_{e} }[/tex] )[tex]t^{2}[/tex] Equation 3
We know that the charge of electron is equal to the charge of proton so,
[tex]q_{p}[/tex] = [tex]q_{e}[/tex] = q
By dividing the equation 2 by equation 3, we get:
[tex]\frac{S}{D-S}[/tex] = [tex]\frac{m_{e} }{m_{p} }[/tex]
Solve the above equation for S,
S[tex]m_{p}[/tex] = [tex]m_{e}[/tex]D - [tex]m_{e}[/tex]S
So,
S = [tex]\frac{m_{e}.D }{(m_{e} + m_{p}) }[/tex]
Plugging in the values,
As we know the mass of electron is 9.1 x [tex]10^{-31}[/tex] and the mass of proton is 1.67 x [tex]10^{-27}[/tex]
S = [tex]\frac{9.1 . 10^{-31} . 4.22 }{(9.1 . 10^{-31} + 1.67 . 10^{-27} }[/tex]
S = 0.002298 cm (Distance from the positive plate at which the two pass each other)
b) In this part, we to calculate distance for Sodium ion and chloride ion as above.
So,
we already have the equation, we need to put the values in it.
So,
S = [tex]\frac{m_{Cl}.D }{(m_{Cl} + m_{Na}) }[/tex]
As we know the mass of chlorine is 35.5 and of sodium is 23
S = [tex]\frac{35.5 . 4.22}{(35.5 + 23)}[/tex]
S = 2.56 cm
The tub of a washer goes into its spin-dry cycle, starting from rest and reaching an angular speed of 4.0 rev/s in 9.0 s. At this point, the person doing the laundry opens the lid, and a safety switch turns off the washer. The tub slows to rest in 15.0 s. Through how many revolutions does the tub turn during this 24 s interval
Answer:
48 rev
Explanation:
a) we can calculate the distance covered by the tube using the formula:
θ = (ω + ωo)t/2
where ω is the final angular speed, θ is the distance covered, t is the time taken, ωo is the initial angular speed.
Firstly, we calculate the distance covered from 0 to 9 s then from 9s to 24 s.
within 9s, the tub runs from rest (0) to 4 rev/s, hence:
t = 9s, wo = 0, w = 4 rev/s = (4 * 2π) rad/s = 8π rad/s. Hence:
θ = (ω + ωo)t/2 = (0 + 8π)9 / 2 = 36π rad
θ = 36π rad = (36π)/2π rev = 18 rev
Also, within 15 s, the tub runs from 4 rev/s to rest, hence:
t = 15 s, wo = 4 rev/s = 8π rad/s, w = 0 rad/s. Hence:
θ = (ω + ωo)t/2 = (8π + 0)15 / 2 = 60π rad
θ = 60π rad = (60π)/2π rev = 30 rev
Therefore the total revolutions by the tube during 24 s interval = 30 rev + 18 rev = 48 rev
25 points!
A 6 kg object accelerates from 5 m•s to 25 m•s in 30 seconds. What was the net force acting on the
object? Give your answer in Newtons to one significant figure and without a unit.
(Show Work)
Answer:
6N
Explanation:
Given parameters:
Mass of object = 6kg
Initial velocity = 5m/s
Final velocity = 25m/s
Time = 30s
Unknown:
Net force acting on the object = ?
Solution:
From Newton's second law of motion:
Force = mass x acceleration
Acceleration is the rate of change of velocity with time
Acceleration = [tex]\frac{Final velocity - Initial velocity }{time}[/tex]
Force = mass x [tex]\frac{Final velocity - Initial velocity }{time}[/tex]
So;
Force = 6 x [tex]\frac{25 - 5}{30}[/tex] = 6N
It has been argued that power plants should make use of off-peak hours (such as late at night) to generate mechanical energy and store it until it is needed during peak load times, such as the middle of the day. One suggestion has been to store the energy in large flywheels spinning on nearly frictionless ball-bearings. Consider a flywheel made of iron, with a density of 7800 kg/m^3 , in the shape of a uniform disk with a thickness of 11.3 cm.
Required:
a. What would the diameter of such a disk need to be if it is to store an amount of kinetic energy of 14.1 MJ when spinning at an angular velocity of 93.0 rpm about an axis perpendicular to the disk at its center?
b. What would be the centripetal acceleration of a point on its rim when spinning at this rate?
Answer:
Explanation:
kinetic energy = 14.1 MJ = 14.1 x 10⁶ J
Let radius of flywheel be r .
volume of flywheel = π r² x t where t is thickness
= 3.14 x r² x .113 m³
= .04 r² m³
mass = volume x density
= .04 r² x 7800 = 312.73 r²kg
moment of inertia I = 1 / 2 mass x radius²
= .5 x 312.73 r² x r²
= 156.37 r⁴ kg m²
angular velocity ω = 2π x 93/60
= 9.734 rad /s
kinetic energy = 1/2 Iω² where ω is angular velocity
= .5 x 156.37 r⁴ x 9.734²
= 7408.08 r⁴
Given
7408.08 r⁴ = 14.1 x 10⁶
r⁴ = .19 x 10⁴
r = .66 x 10
= 6.60 m .
Diameter = 13.2 m
b )
centripetal acceleration of a point on its rim = ω² r
= 9.734² x 6.6
= 625.35 m /s²
Two pieces of amber are hung from threads. Piece A is charged by rubbing piece A with fur. Piece B is charged by rubbing piece B with silk. Nylon is used to rub and charge a third piece of amber. Piece A and B are both repelled by the third piece of amber. This means:____.
Answer:
ieces A and B must also have the same type of charges
Explanation:
In electrostatics, charges of the same sign repel and charges of different signs attract.
If we apply this to our case, we have that part A and C repel each other, therefore they have the same type of charge.
Also part A and C repel each other, therefore they have the same type of charge.
If we use the transitive property of mathematics, pieces A and B must also have the same type of charges
A certain white dwarf star was once an average star like our Sun. But now it is in the last stage of its evolution and is the size of our Moon but has the mass of our Sun.
Estimate gravity on the surface on this star.
I know the solution of this question it is as the picture shows but I only need to add *10^3 to the lower part of the division lower part to get the correct answer. But I don't know why I should add it can anyone explain?
Answer:
4.384 * 10^13
Explanation:
Given the expression :
[(6.67 * 10^-11) * (1.99 * 10^30)] ÷ [(1.74*10^3)*(1.74*10^3)]
Applying the laws of indices
[(6.67 * 1.99) *10^(-11 + 30)] ÷ [(1.74 * 1.74) * 10^3+3]
13.2733 * 10^19 ÷ 3.0276 * 10^6
(13.2733 / 3.0276) * 10^(19 - 6)
4.3840996 * 10^13
= 4.384 * 10^13
Answer:
[tex]g=4.38*10^{7} m/s^2[/tex]
Explanation:
To solve this, we need to know the mass of the sun, and the radius of the moon
[tex]M_{s} = 1.989*10^{30}kg \\R_{m} = 1737400m[/tex]
Now we can plug our values into our equation:
[tex]g=G*\frac{M_{E} }{r^{2} }[/tex]
This gives us:
[tex]g=6.67*10^{-11}*\frac{1.989*10^{30}}{1737400^{2}}[/tex]
This equals:
[tex]g=4.38*10^{7} m/s^2[/tex]
Like charges will exert a force of
a. positive
b. negative
c. attraction
d. repulsion
e. neutral
Answer:
D- Repulsion
Explanation:
A positively charged object will exert a repulsive force upon a second positively charged object.
Which of the following is a mixture?
a air
biron
Chydrogen
d nickel
Answer:
it will option option A hope it helps
3. What is the SI unit of force? What is this unit equivalent to in terms of fundamental units?
4. Why is force a vector quantity?
Answer:
force = mass * acceleration
therefore the SI unit is kg*m/s2 or newton's
it's a vector quantity because it has both direction(acceleration) and size (mass)
