The speed of a 2.0-kg object changes from 30 m/s to 40 m/s during a 5.0-second time interval.
During this same time interval, the velocity of the object changes its direction by 90°. What is the
magnitude of the average total force acting on the object during this time interval?
a. 30 N
b. 20 N
c. 15 N
d. 40 N
e. 10 N
Which is the correct answer?
Answer:
F = 2 * 30 / 5 = 12 N to stop forward motion
F = 2 * 40 / 5 = 16 N to accelerate to 90 degrees
(12^2 + 16^2)^1.2 = 20 N average force applied
The magnitude of the average total force acting on the object during this time interval is 20 N.
The given parameters:
Mass of the object, m = 2.0 kgInitial velocity, u = 30 m/sFinal velocity, v = 40 m/sTime of motion, t = 5.0 sThe magnitude of the average total force acting on the object during this time interval is calculated as follows;
[tex]F = \frac{mv }{t} \\\\F_1 = \frac{2(40)}{5} \\\\F_1 = 16\ N\\\\F_2= \frac{2(30)}{5} \\\\F_2 = 12 \ N\\\\F = \sqrt{F_1^2 + F_2^2} \\\\F = \sqrt{16^2 + 12^2} \\\\F = 20 \ N[/tex]
Learn more about resultant force here: https://brainly.com/question/25239010
Define Potential Energy
Begin by defining potential energy in your own words within one concise eight word sentence
Answer:
potential energy is a type of energy an object has because of it's position
Help!!
A table is pushed across the floor for a distance of 32 m with a force of 320 N in 150 seconds. How much power was used?
A.70.2W
B.68.3W
C.56.7W
D.49.8W
Compute the work done on the table:
W = Fd = (320 N) (32 m) = 10,240 J
Divide this by the given time duration to get the power output:
P = W/∆t = (10,240 J) / (150 s) ≈ 63.3 W
A rocket at fired straight up from rest with a net upward acceleration of 20 m/s2 starting from the ground. After 4.0 s, the thrusters fail and the rocket continues to coast upward with insignificant air resistance. (a) What is the maximum height reached by the rocket
Answer:
The maximum height reached by the rocket is 486.53 m
Explanation:
Given;
initial velocity of the rocket, u = 0
acceleration of the rocket, a= 20 m/s²
duration of the rocket first motion, t = 4 s
The distance traveled by the rocket before its thrust failed
h₁ = ut + ¹/₂at²
h₁ = 0 + ¹/₂ x 20 x 4²
h₁ = 160 m
The second distance moved by the rocket is calculated as follows;
The velocity of the rocket before its thrust failed;
v = u + at
v = 0 + 20 x 4
v = 80 m/s
This becomes the initial velocity for the second stage
At maximum height, the final velocity = 0
[tex]v_f^0 = v_i^2 - 2gh_2\\\\0 = (80)^2 - (2 \times 9.8)h_2\\\\0 = 6400 - 19.6h_2\\\\19.6h_2 = 6400\\\\h_2 = \frac{6400}{19.6} \\\\h_2 = 326.53 \ m[/tex]
The maximum height reached by the rocket = h₁ + h₂
= 160 + 326.53
= 486.53 m
If the loading is 0.4, the coinsurance rate is 0.2, the number of units of medical care is 100, and the number of units of medical care is 1. What is the premium of this insurance?
Answer:
72 is the premimum of the insurance.
Explanation:
Below is the given values:
The loading = 0.4
Coinsurance rate = 0.2
Number of units = 100
Total number of units = 100 * 0.4 = 40
Remaining units = 60 * 0.2 = 12
Add the 60 and 12 values = 60 + 12 = 72
Thus, 72 is the premimum of the insurance.
A swimmer heading directly across a river that is 200 m wide reaches the opposite bank in 6 min 40 s. During this swim, she is swept downstream 480 m. How fast can she swim in still water
Answer:
The speed of the swimmer in stil water is 0.5 m/s
Explanation:
Given;
total time taken to swim across = 6 mins 40 s = (6 x 60s) + 40 s = 400 s
width of the river, = 200 m
Please find the image attached for explanation.
I need help with physics question.
(D)
Explanation:
Assuming that the charge q is moving perpendicular to the magnetic field B, the magnitude of the force experienced by the charge is
F = qvB = (2.9×10^-17 C)(4.0×10^5 m/s)(1.7T)
= 2.0×10^-11 N
Kaseem is taking his bicycle for a ride. His bicycle is a system, and its main purpose is to provide transportation. What is the main input into this system? What is the desired output of this system?
Please help I need this done within 30 mins
It may be thinner and more dense? I’m not too experienced in the study of Earth’s crust. However, I know enough to remember that the earths crust is thin.
what is the frequency of a wave related to
Answer:
Frequency is the number of complete oscillations or cycles or revolutions made in one second.
What is the relationship between organ systems and organs? organs are made from one type of organ system organ systems are made from one type of organ organs are made from different types of organ systems organ systems are made from different types of organs
A lens with a focal length of 15 cm is placed 45 cm in front of a lens with a focal length of 5.0 cm .
Required:
How far from the second lens is the final image of an object infinitely far from the first lens?
Answer:
the required distance is 6 cm
Explanation:
Given the data in the question;
f₁ = 15 cm
f₂ = 5.0 cm
d = 45 cm
Now, for first lens object distance s = ∝
1/f = 1/s + 1/s' ⇒ 1/5 = 1/∝ + 1/s'
Now, image distance of first lens s' = 15cm
object distance of second lens s₂ will be;
s₂ = 45 - 15 = 30 cm
so
1/f₂ = 1/s₂ + 1/s'₂
1/5 = 1/30 + 1/s'₂
1/s'₂ = 1/5 - 1/30
1/s'₂ = 1 / 6
s'₂ = 6 cm
Hence, the required distance is 6 cm
The distance of the final image from the first lens will be is 6 cm.
What is mirror equation?The mirror equation expresses the quantitative connection between object distance (do), image distance (di), and focal length (fl).
The given data in the problem is;
f₁ is the focal length of lens 1= 15 cm
f₂ s the focal length of lens 2= 5.0 cm
d is the distance between the lenses = 45 cm
From the mirror equation;
[tex]\frac{1}{f} = \frac{1}{s} +\frac{1}{s'} \\\\ \frac{1}{5} = \frac{1}{\alpha} +\frac{1}{s'} \\\\[/tex]
If f₁ is the focal length of lens 1 is 15 cm then;
[tex]s'=15 cm[/tex]
f₂ s the focal length of lens 2= 5.0 cm
s₂ = 45 - 15 = 30 cm
From the mirror equation;
[tex]\frac{1}{f_2} = \frac{1}{s_1} +\frac{1}{s_2'} \\\\ \frac{1}{5} = \frac{1}{30} +\frac{1}{s_2'} \\\\ \frac{1}{s_2'}= \frac{1}{5} -\frac{1}{30} \\\\ \frac{1}{s_2'}= \frac{1}{6} \\\\ \rm s_2'= 6 cm[/tex]
Hence the distance of the final image from the first lens will be is 6 cm.
To learn more about the mirror equation refer to the link;
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E=kq/r^2 chứng minh điện thế V=kq/r từ mối liên hệ giữa điện trường E và điện thế V
Answer:
hindi ko maintindihan teh
7. An electric train moving at 20km/hrs
. Accelerates to a speed of 30km/hrs. in
20 sec, find the distance travelled in meters during the period of
acceleration
Answer
NB:
- speed, U is measure in m/s
- acceleration, a is measured in m/s²
-time t in seconds , s
Therefore conversation must be made
Speed U = 20km/hrs
=20km÷1hr
But 20km= 20×1000=20000m
1hr= 1×60min×60sec=3600s
U=20000÷3600=5.56m/s
a=30km/hrs
=30km÷1hr
But 30km=30×1000=30000
1hr=3600s
a=30000÷3600=8.33m/s²
From the equation of motion
S=Ut + ½ at².
Where s= distance
S = 5.56m/s × 20s + ½(8.33m/s²)(20s)²
S = 1777.3m
How far did you travel in 10 hours if you drove at a constant speed of 5km/hr? *
Answer:
you drove 50km
Explanation:
10×5 hope this helps
Answer:
50 Km
Explanation:
This is how far you have got on your journey if traveling like this.
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Hope this Helps
assuming a filament in a 120W light bulb acts like a prefect blackbody, what is the temperature of the hottest portion of the filament if it has a surface area of 6.4×10^_5m^2. The stefan- boltzmann constant is 5.67×10^-8W/(m2.k2) A. 12OOk B. 2400K C. 2100K
Answer:
T = 2398 K
Explanation:
To calculate the emission of the light bulb we use the law is Stefan
P = σ A e T⁴
as they indicate that the filament is a black body, the emissivity is equal to 1 (e = 1)
T = [tex]\sqrt[4]{\frac{P}{ \sigma A} }[/tex]
let's calculate
T =[tex]\sqrt[4]{\frac{120}{5.67 \ 10^{-8} \ 6.4 \ 10^{-5}} }[/tex]
T = [tex]\sqrt[4]{33.06878 \ 10^{12} }[/tex]
T = 2,398 10³ K
T = 2398 K
Problem
A charged particle is moving in the presence of uniform magnetic field. The mass of the particle
is m = 10−6 kg its charge is Q = 10−5 C and the magnetic field vector is B~ = (1T, 0, 0). At the
beginning the velocity vector of the particle is ~v0 = (12 m/s, 0, 5 m/s).
a.) How large will the x component of the velocity of the particle be in t = 2 s?
b.) Where will the particle be in t = 3.14 s?
c.) How large will the magnitude of the velocity be in t = 2.5 s?
Answer:
Answer is a I checked the work
Two identical loudspeakers 2.0 m apart are emitting sound waves into a room where the speed of sound is 340 m/sec. John is standing 5.0m in front of one of the speakers, perpendicular to the line joining the speakers, and hears a maximum in the intensity of the sound. What is the lowest possible frequency of sound for which this is possible?
Answer: The lowest possible frequency of sound for which this is possible is 212.5 Hz.
Explanation:
It is known that formula for path difference is as follows.
[tex]\Delta L = (n + \frac{1}{2}) \times \frac{\lambda}{2}[/tex] ... (1)
where, n = 0, 1, 2, and so on
As John is standing perpendicular to the line joining the speakers. So, the value of [tex]L_{1}[/tex] is calculated as follows.
[tex]L_{1} = \sqrt{(2)^{2} + (5)^{2}}\\= 5.4 m[/tex]
Hence, path difference is as follows.
[tex]\Delta L = (5.4 - 5) m = 0.4 m[/tex]
For lowest frequency, the value of n = 0.
[tex]\Delta L = (0 + \frac{1}{2}) \times \frac{\lambda}{2} = \frac{\lambda}{4}[/tex]
[tex]\lambda = 4 \Delta L[/tex]
where,
[tex]\lambda[/tex] = wavelength
The relation between wavelength, speed and frequency is as follows.
[tex]\lambda = \frac{\nu}{f}\\4 \Delta L = \frac{\nu}{f}\\[/tex]
where,
[tex]\nu[/tex] = speed
f = frequency
Substitute the values into above formula as follows.
[tex]f = \frac{\nu}{4 \Delta L}\\f = \frac{340}{4 \times 0.4 m}\\= 212.5 Hz[/tex]
Thus, we can conclude that the lowest possible frequency of sound for which this is possible is 212.5 Hz.
Which conclusion can be made based on the information in the table?
Wave speed and wavelengths can vary inversely to produce the same frequency.
O Frequency and wave speed can vary directly to produce the same wavelength.
O Wavelengths and frequency can vary inversely to produce the same wave speed.
O Frequency and wavelengths can vary directly to produce the same wave speed.
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Previous Activity
Answer:
The correct option is (b).
Explanation:
The relation between the wavelength and frequency is given by :
[tex]\lambda=\dfrac{v}{f}[/tex]
Where
v is the wave speed
f is the frequency of a wave
It is clear from the above equation that the wavelengths and frequency can vary inversely to produce the same wave speed.
A girl and her bicycle have a total mass of 40.0 kg. At the top of the hill her speed is 5.0 m/s, and her speed doubles as she rides down the hill. The hill is 10.0 m high and 100 m long. How much kinetic energy and potential energy is lost to friction
Answer:
The kinetic energy and potential energy lost to friction is 2,420 J.
Explanation:
Given;
total mass, m = 40 kg
initial velocity of the girl, Vi = 5 m/s
hight of the hill, h = 10 m
length of the hill, L = 100 m
initial kinetic energy of the girl at the top hill:
[tex]K.E_{i} = \frac{1}{2} mv_i^2 = \frac{1}{2} \times 40 \times (5)^2\\\\K.E_{i} = 500 \ J[/tex]
initial potential energy of the girl at the top hill:
[tex]P.E_{i} = mgh_i = 40 \times 9.8 \times 10\\\\P.E_{i}= 3920 \ J[/tex]
Total energy at the top of the hill:
E = 500 J + 3920 J
E = 4,420 J
At the bottom of the hill:
final velocity = double of the initial velocity = 2 x 5 m/s = 10 m/s
hight of the hill = 0
final kinetic energy of the girl at the bottom of the hill:
[tex]K.E_{f} = \frac{1}{2} mv_f^2 \\\\K.E_f = \frac{1}{2} \times 40 \times (10)^2 = 200 0 \ J[/tex]
final potential energy of the girl at the bottom of the hill:
[tex]P.E_f = mgh_f = 40 \times 9.8 \times 0 = 0[/tex]
Based on the principle of conservation of energy;
the sum of the energy at the top hill = sum of the energy at the bottom hill
The energy at the bottom hill is less due to energy lost to friction.
[tex]E_{friction} \ + E_{bottom}= E_{top}\\\\E_{friction} = E_{top} - E_{bottom}\\\\E_{friction} = 4,420 \ J - 2,000 \ J\\\\E_{friction} = 2,420 \ J[/tex]
Therefore, the kinetic energy and potential energy lost to friction is 2,420 J.
Hai điện tích điểm Q1 = 8 C, Q2 = –6
C đặt tại hai điểm A, B cách nhau 0,1
m trong không khí. Tính cường độ điện
trường do hai điện tích này gây ra tại
điểm M, biết MA = 0,2 m
Answer:
English please
Explanation:
I don't understand the question
which team won the champions league in 2020 2021
Answer:
Chelsea F.C
Explanation:
Chelsea F.C
Soccer
A ball is thrown horizontally at a speed of 24 meters per second from the top of a cliff. If the ball hits the ground 6.0 seconds later, approximately how high is the cliff?
Answer:
144 meters
Explanation:
it takes 6 seconds to hit the ground right and the ball lays off 24 m per second .
so by the time the ball hits the ground 6 seconds passed. so that means the cliff is 6.0×24=144
Are you aware of human rights violation happening in the community and explain
Answer:
Individuals who commit serious violations of international human rights or humanitarian law, including crimes against humanity and war crimes, may be prosecuted by their own country or by other countries exercising what is known as “universal jurisdiction.”
A disk of charge is placed in the x-y plane, centered at the origin. The electric field along the axis of a positive disk of charge... points towards the disk along the z-axis. points away from the disk along the z-axis. always points in the positive z-direction. none of these choices
Answer:
Points away from the disk along the z-axis.
Explanation:
Along the axis of the disk, which is the z - axis, the total vertical electric field components of the charged disk sum up while the horizontal components cancel out. Thus, leaving only vertical components of electric field along the axis of the disk.
Since the disk is positively charged and electric field lines point away from a positive charge, the electric field along the axis of a positive disk of charge points away from the disk along the z-axis.
Calculate the rms speed of helium atoms near the surface of the Sun at a temperature of about 5100 K. Express your answer to two significant figures and include the appropriate units.
Answer:
[tex]V_{rms}=5.6*10^3m/s[/tex]
Explanation:
From the question we are told that:
Temperature [tex]T=5100K[/tex]
Generally the equation for RMS Speed is mathematically given by
[tex]V_{rms}=\sqrt{\frac{3kT}{m}}[/tex]
Where
[tex]K=Boltzman's constant[/tex]
[tex]K=1.38*10^{-23}[/tex]
And
[tex]M=molecular mass[/tex]
[tex]M=4*1.67*10^{-27}[/tex]
[tex]V_{rms}=\sqrt{\frac{3(1.38*10^{-23})5100}{4*1.67*10^{-27}}}[/tex]
[tex]V_{rms}=5.6*10^3m/s[/tex]
A car is stopped for a traffic signal. When the light turns green, the car accelerates, increasing its speed from zero to 9.41 m/s in 4.24 s. What is the magnitude of the linear impulse experienced by a 67.0 kg passenger in the car during this time
Answer:
the impulse experienced by the passenger is 630.47 kg
Explanation:
Given;
initial velocity of the car, u = 0
final velocity of the car, v = 9.41 m/s
time of motion of the car, t = 4.24 s
mass of the passenger in the car, m = 67 kg
The impulse experienced by the passenger is calculated as;
J = ΔP = mv - mu = m(v - u)
= 67(9.41 - 0)
= 67 x 9.41
= 630.47 kg
Therefore, the impulse experienced by the passenger is 630.47 kg
A locomotive pulls 11 identical freight cars. The force between the locomotive and the first car is 150.0 kN, and the acceleration of the train is 2 m/s2. There is no friction to consider. 1) Find the force between the tenth and eleventh cars. (Express your answer to two significant figures.)
Answer:
The force between the 10 th car and the 11 th car is 13636.4 N.
Explanation:
Force, F = 150 kN
acceleration, a = 2 m/s^2
Let the mass of each car is m. \Total numbers of cars = 11
F = n m a
150000 = 11 x m x 2
m = 6818.18 kg
The force between the 10 th and 11 th car is
T = ma = 6818.18 x 2 = 13636.4 N
g Calculate the final speed of a solid cylinder that rolls down a 5.00-m-high incline. The cylinder starts from rest, has a mass of 0.750 kg, and has a radius of 4.00 cm.
Answer:
[tex]V=8.08m/s[/tex]
Explanation:
From the question we are told that:
Height[tex]h=5.00m[/tex]
Mass [tex]m=0.750kg[/tex]
Radius [tex]r=4.00cm=>0.04m[/tex]
Generally the equation for Total energy is mathematically given by
[tex]mgh=\frac{1}{2}mv^2+\frac{1}{2}Iw^2[/tex]
Therefore
[tex]V=\sqrt{\frac{4gh}{3}}[/tex]
[tex]V=\sqrt{\frac{4*9.8*5}{3}}[/tex]
[tex]V=8.08m/s[/tex]
You swing a bat and hit a heavy box with a force of 1273 N. The force the box exerts on the bat is Group of answer choices less than 1273 N if the box moves. exactly 1273 N whether or not the box moves. None of the above choices are correct. exactly 1273 N only if the box does not move. greater than 1273 N if the bat bounces back. greater than 1273 N if the box moves.
Answer:
exactly 1273 N whether or not the box moves.
Explanation:
In the case when the bat is swing and it is hitted to a heavy box having a force of 1273 N so here the force of the box that exert on the box should be accurately 1273 N even if the box is moved or not. As the third law of the newton should be equivalent & the opposite reaction
Therefore as per the given situation, the above represent the answer
