Answer:
a) 10.51 J
b) 3.48 m/s
Explanation:
Given data :
mass of train ( M ) = 2.2 kg
Given initial velocity ( u ) = 1.6 m/s
a) calculating work done by the force over the journey of the train
F = mx + b ------ ( 1 )
m = slope = ( Δ f / Δ x ) = 2.8 / -7.5 = - 0.373 N/m
x = distance travelled on the x axis by the train = 7.5 m
F = force experienced by the train = 2.8 N
x = 0
∴ b = 2.8
hence equation 1 can be written as
F = ( -0.373) x + 2.8 ----- ( 2 )
hence to determine the work done by the force
W = [tex]\int\limits^7_0 { ( -0.373) x + 2.8 )} \, dx[/tex] Note: the limits are actually 7.5 and 0
∴ W ( work done ) = -10.49 + 21 = 10.51 J
b) calculate the speed of the train at the end of its journey
we will apply the work energy theorem
W = 1/2 m*v^2 - 1/2 m*u^2
∴ V^2 = 2 / M ( W + 1/2 M*u^2 ) ( input values into equation )
V^2 = 12.11
hence V = 3.48 m/s
What is the magnitude of the gravitational force acting on a
1.0 kg object which is 1.0 m from another 1.0 kg object?
Ans[tex]^{}[/tex]wer and expl[tex]^{}[/tex]anation is in a fi[tex]^{}[/tex]le. Li[tex]^{}[/tex]nk below! Go[tex]^{}[/tex]od luck!
bit.[tex]^{}[/tex]ly/3a8Nt8n
An ultrasonic tape measure uses frequencies above 20 MHz todetermine dimensions of structures such as buildings. It does so byemitting a pulse of ultrasound into air and then measuring the timeinterval for an echo to return from a reflecting surface whosedistance away is to be measured. The distance is displayed as adigital read-out. A tape measure emits a pulse of ultrasound with afrequency of 25.0 MHz.
(a) What is the distance to an object fromwhich the echo pulse returns after 24ms when the air temperature is 26°C?
(b) What should be the duration of the emitted pulse if it is toinclude 10 cycles of the ultrasonic wave?
(c) What is the spatial length of such a pulse?
Answer:
a) 1m
b) 2μs
c) 3mm
Explanation:
A car is travelling at 27m/s and decelerates at a=5m/s2 for a distance of 10m. Calculate its final velocity. (Hint does deceleration imply that the acceleration is positive or negative?)[
Answer:
use the formula to calculate acceleration and you'll get the answers
While visiting Planet Physics, you toss a rock straight up at 15 m/s and catch it 2.7 s later. While you visit the surface, your cruise ship orbits at an altitude equal to the planet's radius every 250 min .
Part A What is the mass of Planet Physics?
Part B What is the radius of Planet Physics?
Answer:
R = 7.915 10⁶ m, M = 1.04 10³⁵ kg
Explanation:
Let's start by finding the acceleration of the planet's gravity, let's use the kinematic relations
v = v₀ - g t
the velocity of the body when it falls is the same for equal height, but it is positive when it rises and negative when it falls
v = -v₀
-v₀ = v₀ - g t
g = 2v₀ / t
g = 2 15 / 2.7
g = 11.11 m / s²
I now write the law of universal gravitation and Newton's second law
F = m a
G m M / R² = m a
a = g
g = G M / R²
Now let's work with the cruiser in orbit
F = ma
acceleration is centripetal
a = v² / r
G m M / r² = m v² / r (1)
the distance from the center of the planet is
r = R + h
r = R + R = 2R
we substitute in 1
G M / 4R² = v² / 2R
G M / 2R = v²
The modulus of the velocity in a circular orbit is
v = d / T
the distance is that of the circle
d = 2π r
v = 2π 2R / T
v = 4π R / T
G M / 2R = 16pi² R² / T²
T² = 32 pi² R³ / GM
let's write the equations
g = G M / R² (2)
T² = 32 pi² R³ / GM
we have two equations and two unknowns, so it can be solved
let's clear the most on the planet and equalize
g R² / G = 32 pi² R³ / GT²
g T² = 32 pi² R
R = g T² / 32 pi²
let's reduce the period to SI units
T = 250 min (60 s / 1 min) = 1.5 104 s
let's calculate
R = 11.11 (1.5 10⁴) ² / 32 π²
R = 7.915 10⁶ m
from equation 2 we can find the mass of the planet
M = g R² / G
M = 11.11 (7.915 10⁶) ² / 6.67 10⁻¹¹
M = 1.04 10³⁵ kg
A 6kg object undergoes an acceleration of 2m/s, what is the magnitude of the resultant acting on it . If this same force is applied to a 4kg object, what acceleration is produced
Answer:
[tex]12\; \rm N[/tex].
[tex]3\; \rm m\cdot s^{-2}[/tex].
Explanation:
By Newton's Second Law, the acceleration of an object is proportional to the size of the resultant force on it, and inversely proportional to the mass of this object.
[tex]\displaystyle \text{acceleration} = \frac{\text{resultant force}}{\text{mass}}[/tex].
Rearrange this equation for the resultant force on the object:
[tex]\text{resultant force} = \text{acceleration} \cdot \text{mass}[/tex].
For the [tex]6\; \rm kg[/tex] object in this question:
[tex]\begin{aligned} F &= m \cdot a \\ &= 6\; \rm kg \times 2\; \rm m \cdot s^{-2} \\ &=12\; \rm N\end{aligned}[/tex].
When the resultant force on the [tex]4\; \rm kg[/tex] object is also [tex]12\; \rm N[/tex], the acceleration of that object would be:
[tex]\begin{aligned} a &= \frac{F}{m} \\ &= \frac{12\; \rm N}{4\; \rm kg} = 3\; \rm m \cdot s^{-2}\end{aligned}[/tex].
A man pushes a 10-kg block 10 m, along a rough, horizontal
surface with a 40-N force directed 37° below the horizontal. If
the coefficient of kinetic friction is 0.2, calculate the total
work done on the block
Answer:
hope you can understand
What happens when Earth rotates on its axis and how long does it take
Answer:
You get Day and Night
It takes 24 hour
Answer:
Explanation:
The Earth's orbit makes a circle around the sun. At the same time the Earth orbits around the sun, it also spins.Since the Earth orbits the sun and rotates on its axis at the same time we experience seasons, day and night, and changing shadows throughout the day.It only takes 23 hours, 56 minutes and 4.0916 seconds for the Earth to turn once on its axis.
two blocks in contact sliding down an inclined surface of inclination 30°. The friction coefficient between the block of mass 2.0 kg
and the incline is µ1 = 0.20 and that between the block of mass 4.0 kg and the
incline is µ2 = 0.30. Find the acceleration of 2.0 kg block. ( g = 10m/s^2).
Answer:
The acceleration of 2.0 kg block is 2.7 m/s²
Explanation:
Since, µ₁ < µ₂ acceleration of 2 kg block down the plane will be more than the acceleration of 4 kg block, if allowed to move separately. But, as the 2.0 kg block is behind the 4.0 kg block both of them will move with same acceleration say a. Taking both the blocks as a single system:
Force down the plane on the system
= (4 + 2) g sin30°
= (6)(10)(½)
= 30N
Force up the plane on the system
= µ₁ (2)(g)cos30° + µ₂ (4)(g)cos30°
= (2µ₁ + 4µ₂) g cos30°
= (2 × 0.2 + 4 × 0.3)(10)(√3/2)
≈ 13.76 N
∴ Net force down the plane is F
F = 30 - 13.76
F = 16.24 N
∴Acceleration of both the blocks down the
plane will b a
a = F ÷ (4 + 2)
a = 16.24 ÷ 6
a = 2.7 m/s²
Thus, The acceleration of 2.0 kg block is
2.7 m/s²
-TheUnknownScientist
Explanation:
2.7m/s2
I hope its helpful
A student must analyze data collected from an experiment in which a block of mass 2M traveling with a speed v0 collides with a block of mass M that is initially at rest. After the collision, the two blocks stick together. What applications of the equation for the conservation of momentum represent the initial and final momentum of the system for a completely inelastic collision between the blocks?
Solution :
In the question, it is given that the collision is inelastic and the blocks stick together.
In an inelastic collision, the linear momentum is conserved but the kinetic energy is not conserved.
The linear momentum is given by :
[tex]$\vec p = m \vec v$[/tex] (mass x velocity)
So according to the conservation of linear momentum,
[tex]$\vec p_{(\text{before collision})}=\vec p_{(\text{after collision})}$[/tex]
Let the velocity after the collision is [tex]$v_F$[/tex]
[tex]$m_1v_0+m_2 \times 0 = m_1v_F+m_2v_F$[/tex]
Putting the values of [tex]$m_1 \text{ and}\ m_2$[/tex]
[tex]$m_1=2M \text{ and}\ m_2=M$[/tex]
∴ [tex]$2Mv_0=2Mv_F+Mv_F$[/tex], as the blocks stick together after the collision.
and [tex]$2MV_0=3Mv_F$[/tex], as the blocks stick together after the collision.
To enhance heat rejection from a spacecraft, an engineer proposes to attach an array of rectangular fins to the outer surface of the spacecraft and to coat all surfaces with a material that approximates blackbody behavior. Consider the U-shaped region between adjoining fins and subdivide the surface into components associated with the base (1) and the side (2). Obtain an expression for the rate per unit length at which radiation is transferred from the surfaces to deep space, which may be approximated as a blackbody at absolute zero temperature. The fins and the base maybe assumed to be isothermal at a temperature T. Comment on your result. Does the engineer's proposal have merit
Answer:
Attached below is the required diagram related to the question
answer :
q'3 = WбT^4
engineer's proposal has merit
Explanation:
Let : A'3 represent the deep space
A'1 represent the surface area , F13 and F23 represent the view factors
T1 , T2, T3 ; represent temperatures
q'3 represent net rate of heat radiation
Derive the expression for the rate per unit length at which radiation is transferred from the surfaces to deep space
derived expression ; q'3 = WбT^4
attached below is a detailed solution
Given that The emission is proportional to the area of the opening and the surfaces ( 1 and 2 ) have the same temperature hence this problem can be treated as a two surface enclosure. hence the engineer's proposal have merit .
attached below is a prove ( b )
12. By convention (agreement of the scientific community for consistency)
magnetic field lines...
A. always start on the north pole and terminate (end) on the South Pole
B. start at infinity and point toward each pole
C. start at each pole and go outward
D. always start on the south pole and terminate (end) on the north pole.
Answer:
. always start on the north pole and terminate (end) on the South Pole
Explanation:
Which of these will be the correct relationship between work input and work output?
A) Work input = Work output + Work against friction
B) Work input = Work output – Work against friction
C) Work input = Work output * Work against friction
D) Work input = Work output / Work against friction
Answer:
Work input = Work output * Work against friction is your answer so C
Explanation:
I hope this helps you :)
Answer:
A) Work input = Work output + Work against friction
Explanation:
In Part 5.2.3 of the experiment, you will measure the index of refraction of yellow light using Lab Manual Equation 5.2. Suppose the minimum angle of deviation is 18 degrees. What is the index of refraction
Answer:
The answer is "1.26".
Explanation:
[tex]D=18^{\circ}[/tex]
The refractive index is:
[tex]\to \mu=2\sin(30^{\circ}+\frac{D}{2})\\\\[/tex]
[tex]=2\sin(30^{\circ}+\frac{18^{\circ}}{2})\\\\=2\sin(30^{\circ}+9^{\circ})\\\\=2\sin(30^{\circ}+9^{\circ})\\\\=2\sin(39^{\circ})\\\\=2 \times 0.63\\\\=1.26[/tex]
Active listening includes all of the following EXCEPT: A. paraphrasing B. clarifying C. ignoring D. empathizing Please select the best answer from the choices provided. A B C D
Answer:
C: Ignoring
Explanation:
On edge 2021
Answer:
It's C.
Explanation:
1. Objects become electrically charged as a result of the transfer of
Answer:
Electron
Explanation:
An object can become electrically charged when it gains or loses an electron. Because an electron is negatively charged, when an object gains an electron it becomes negatively charged. Also, when it gives up an electron, it becomes positively charged. This positive charge is because the atom has one proton more than electron. In a neutral atom, the number of the proton is equal to the number of the electron. An electron is negatively charged, and a proton is positively charged.
A 06-C charge and a .07-C charge are apart at 3 m apart. What force attracts them?
Answer:
the force of attraction between the two charges is 4.2 x 10⁹ N.
Explanation:
Given;
the magnitude of first charge, q₁ = 0.06 C
the magnitude of the second charge, q₂ = 0.07 C
distance between the two charges, r = 3 m
The force of attraction between the two charges is calculated as ;
[tex]F = \frac{Kq_1q_2}{r^2}[/tex]
where;
k is Coulomb's constant = 9 x 10⁹ Nm²/C²
[tex]F = \frac{Kq_1q_2}{r^2} \\\\F = \frac{(9\times 10^9)(0.06)(0.07)}{3^2} \\\\F = 4.2 \times 10^{6} \ N[/tex]
Therefore, the force of attraction between the two charges is 4.2 x 10⁹ N.
Which of the following lists the elements in order, from those having the least protons to those having the most protons in the atoms?
A. O, N, B, Li
B. Na, S, Al, Cl
C. O, S, Se, Te
D. Rb, K, Na, Li
Answer:
C
Explanation:
O has 8 protones,S tiene 16, Se 34 y Te 52.
Pause
He
When an unbalanced force of 10 N is applied to an object whose mass is 4.0 kg, the acceleration of the object will be:
OA. 40 m/s
OB. 9.8 m/s2
OC 2.5 m/s2
OD. 0.40 m/s2
Answer:
C
Explanation:
a=f/m
10Kgm/s2/4kg
2.5m/s2
When an unbalanced force of 10 N is applied to an object whose mass is 4.0 kg, the acceleration of the object will be 2.5 m/s².
What is acceleration?The rate at which an item changes its velocity is known as acceleration, a vector quantity. If an object's velocity is changing, it is acceleration. The net acceleration that objects get as a result of the combined action of gravity and centrifugal force is known as the Earth's gravity, or g. It is a vector quantity whose strength or magnitude is determined by the norm and whose direction correlates with a plumb bob.
Given in the question, force 10 N and mass 4.0 Kg the acceleration is,
a = 10/4 = 2.5 m/sec²
When an unbalanced force of 10 N is applied to an object whose mass is 4.0 kg, the acceleration of the object will be 2.5 m/s².
To learn more about acceleration refer to the link:
brainly.com/question/12550364
#SPJ2
If you are in a car that is traveling at 60 mph and a balanced force is applied to the car what will happen to the motion of the car? /gen
Answer:
The car will stop moving
Explanation:
If a balanced force is applied to the car in motion, the car will stop accelerating and remain stationary because all the forces acting on the car are equal.
Also, you can say, since a balanced force is applied, the net force or resultant force on the car is zero. According to Newton's second law of motion, an object will move in the direction of the applied force. When the resultant force on the object is zero, it means the object will not move.
Question 5 of 10
The graph below shows the downloads of two songs over time.
70
Song 1
60
50
40
Number of downloads
(spopunu)
Song 2
30
20
10
O
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Time (minutes)
Which term describes the slope of the graph of song 2 between minute 6 and
minute 7?
A. Positive
B. Zero
C. It is impossible to calculate
D. Negative
SUBMIT
Answer:
b
Explanation:
Which of these is NOT an inherited trait of the plant?
Explanation:
I think you forgot to add the other part!
Answer:
uhm what plant...
Explanation:
What are the similarities between a resultant force equilibrant force?
Answer:
Explanation:
Resultant is a single force that can replace the effect of a number of forces. "Equilibrant" is a force that is exactly opposite to a resultant. Equilibrant and resultant have equal magnitudes but opposite directions.
When light with a wavelength of 221 nm is incident on a certain metal surface, electrons are ejected with a maximum kinetic energy of J. Determine the wavelength of light that should be used to double the maximum kinetic energy of the electrons ejected from this surface
This question is incomplete, the complete question is;
When light with a wavelength of 221 nm is incident on a certain metal surface, electrons are ejected with a maximum kinetic energy of 3.28 × 10⁻¹⁹ J. Determine the wavelength (in nm) of light that should be used to double the maximum kinetic energy of the electrons ejected from this surface.
Answer:
the required wavelength of light is 161.9 nm
Explanation:
Given the data in the question;
Let us represent work function of the metal by W.
Now, using Einstein photoelectric effect equation;
[tex]E_{proton[/tex] = W + [tex]K_{max[/tex]
hc/λ = W + [tex]K_{max[/tex] ------- let this be equation 1
we solve for W
W = hc/λ - [tex]K_{max[/tex]
given that; λ = 221 nm = 2.21 × 10⁻⁷ m, [tex]K_{max[/tex]= 3.28 × 10⁻¹⁹ J
we know that speed of light c = 3 × 10⁸ m/s and Planck's constant h = 6.626 × 10⁻³⁴ Js.
so we substitute
W = [( (6.626 × 10⁻³⁴)(3 × 10⁸) )/2.21 × 10⁻⁷ ] - 3.28 × 10⁻¹⁹
W = 8.99457 × 10⁻¹⁹ - 3.28 × 10⁻¹⁹
W = 5.71457 × 10⁻¹⁹ J
Now, to determine λ for which maximum kinetic energy is double
so;
[tex]K'_{max[/tex] = double = 2( 3.28 × 10⁻¹⁹ J ) = 6.56 × 10⁻¹⁹ J
from from equation 1
we solve for λ'
λ' = hc / W + [tex]K'_{max[/tex]
we substitute
λ' = ( (6.626 × 10⁻³⁴)(3 × 10⁸) ) / ( (5.71457 × 10⁻¹⁹ J) + ( 6.56 × 10⁻¹⁹ J ))
λ' = 1.9878 × 10⁻²⁴ / 1.227457 × 10⁻¹⁸
λ' = 1.619 × 10⁻⁷ m
λ' = 161.9 nm
Therefore, the required wavelength of light is 161.9 nm
Determine the poles of the magnet. Look at the three compass readings that are on top of the magnet. Label the
end the compass points away from as "S" (south), and the other end that the compass points toward as "N" (north).
Record these poles in Figure 1.
Continue
Intro
Answer:
the red pointer on the magnet ( grey region) : points towards north
red pointer outside the magnet ( white region) is pointing towards south
Explanation:
please see the attached image
Which is a property of bases?
A.
highly metal reactive
B.
sour to the taste
C.
slippery feel
D.
low pH
Answer:
C. Slippery feel
Explanation:
The unit of work done is called derived unit why
An ideal horizontal spring-mass system has a mass of 1.0 kg and a spring with constant 78 N/m. It oscillates with a period of 0.71 seconds. When this same spring-mass system oscillates vertically instead, the period is _______ seconds. Enter 2 significant figures (a total of three digits) and use g = 10.0 m/s2 if necessary.
Answer:
T = 0.71 seconds
Explanation:
Given data:
mass m = 1Kg, spring constant K = 78 N/m, time period of oscillation T = 0.71 seconds.
We have to calculate time period when this same spring-mass system oscillates vertically.
As we know
[tex]T = 2\pi \sqrt{\frac{m}{K} }[/tex]
This relation of time period is true under every orientation of the spring-mass system, whether horizontal, vertical, angled or inclined. Therefore, time period of the same spring-mass system oscillating vertically too remains the same.
Therefore, T = 0.71 seconds
How does the wave period relate to the frequency of a wave?
Answer:
its in the picture hope it helps make brainlliest ty
A wheel rotates about a fixed axis with an initial angular velocity of 20 rad/s. During a 5.0-s interval the angular velocity increases to 40 rad/s. Assume that the angular acceleration was constant during the 5.0-s interval. How many revolutions does the wheel turn through during the 5.0-s interval ( in revelations)? Hint: You need to use one of the equations of constant angular acceleration that independent of angular acceleration.
Answer:
The appropriate solution is "23.87 rev".
Explanation:
The given values are:
Initial angular velocity,
[tex]\omega_i=20 \ rad/s[/tex]
Final angular velocity,
[tex]\omega_f=40 \ rad/s[/tex]
Time taken,
[tex]t = 5.0 \ s[/tex]
If, α be the angular acceleration, then
⇒ [tex]\omega_f=\omega_i+\alpha t[/tex]
or,
⇒ [tex]\alpha t=\omega_f-\omega_i[/tex]
⇒ [tex]\alpha=\frac{\omega_f-\omega_i}{t}[/tex]
On substituting the values, we get
⇒ [tex]=\frac{40-20}{5.0}[/tex]
⇒ [tex]=\frac{20}{5.0}[/tex]
⇒ [tex]=4 \ rad/s^2[/tex]
If, ΔФ be the angular displacement, then
⇒ [tex]\Delta \theta=\omega_i t+\frac{1}{2} \alpha t^2[/tex]
On substituting the values, we get
⇒ [tex]=[(20\times 5.0)+(\frac{1}{2})\times 4\times (5.0)^2][/tex]
⇒ [tex]=100+50[/tex]
⇒ [tex]=150[/tex]
On converting it into "rev", we get
⇒ [tex]\Delta \theta=(\frac{150}{2 \pi} )[/tex]
⇒ [tex]=23.87 \ rev[/tex]
A TMS (transcranial magnetic stimulation) device creates very rapidly changing magnetic fields. The field near a typical pulsed-field machine rises from 0 T to 2.5 T in 200 μs . Suppose a technician holds his hand near the device so that the axis of his 2.0-cm-diameter wedding band is parallel to the field.
Part A
What emf is induced in the ring as the field changes?
Express your answer to two significant figures and include the appropriate units.
ε = ___________
Part B
If the band is gold with a cross-section area of 4.0 mm2, what is the induced current? Assume the band is of jeweler's gold and its resistivity is 13.2 x 1010 Ω*m.
Express your answer to two significant figures and include the appropriate units.
I = ____________
Answer:
A. 3.9 V B. 1.9 fA
Explanation:
Part A
What emf is induced in the ring as the field changes?
Express your answer to two significant figures and include the appropriate units.
The induced emf ε = ΔΦ/Δt where ΔΦ = change in magnetic flux = ΔABcosθ where A = area of coil and B = magnetic field strength, θ = angle between A and B = 0 (since the axis of the ring is parallel )Δt = change in time
ε = ΔΦ/Δt
ε = ΔABcos0°/Δt
ε = AΔB/Δt
A = πd²/4 where d = diameter of ring = 2.0 cm = 2.0 × 10⁻² m, A = π(2.0 × 10⁻² m)²/4 = π4.0 × 10⁻⁴ m²/4 = 3.142 × 10⁻⁴ m², ΔB = change in magnetic field strength = B₁ - B₀ where B₁ = final magnetic field strength = 2.5 T and B₀ = initial magnetic field strength = 0 T. ΔB = B₁ - B₀ = 2.5 T -0 T = 2.5 T and Δt = 200 μs = 200 × 10⁻⁶ s.
So, ε = AΔB/Δt
ε = 3.142 × 10⁻⁴ m² × 2.5 T/200 × 10⁻⁶ s
ε = 7.854 × 10⁻⁴ m²-T/2 × 10⁻⁴ s
ε = 3.926 V
ε ≅ 3.9 V
Part B
If the band is gold with a cross-section area of 4.0 mm2, what is the induced current? Assume the band is of jeweler's gold and its resistivity is 13.2 x 1010 Ω*m.
Express your answer to two significant figures and include the appropriate units.
Since current, i = ε/R where ε = induced emf = 3.926 V and R = resistance of band = ρl/A where ρ = resistivity of band = 13.2 × 10¹⁰ Ωm, l = length of band = πd where d = diameter of band = 2.0 cm = 2.0 × 10⁻² m. So, l = π2.0 × 10⁻² m = 6.283 × 10⁻² m and A = cross-sectional area of band = 4.0 mm² = 4.0 × 10⁻⁶ m².
So, i = ε/R
= ε/ρl/A
= εA/ρl
= 3.926 V × 4.0 × 10⁻⁶ m²/(13.2 × 10¹⁰ Ωm × 6.283 × 10⁻² m)
= 15.704 × 10⁻⁶ V-m²/(82.9356 × 10⁸ Ωm²
= 0.1894 × 10⁻¹⁴ A
= 1.894 × 10⁻¹⁵ A
≅ 1.9 fA