Answer:
1.5kg
Explanation:
Given data
mass m1= 2.5kg
mass m2=??
velocity of mass one v1= 1.5m/s
velocity of mass two v2= 4.1m/s
common velocity after impact v= 2.5m/s
Let us apply the formula for the conservation of linear momentum for inelastic collision
The expression is given as
m1v1+ m2v2= v(m1+m2)
substitute
2.5*1.5+ m2*4.1= 2.5(2.5+m2)
3.75+4.1m2= 6.25+2.5m2
collect like terms
3.75-6.25= 2.5m2-4.1m2
-2.5= -1.6m2
divide both sides by -1.6
m2= -2.5/-1.6
m2= 1.5 kg
Hence the second mass is 1.5kg
An observer on Earth sees rocket 1 leave Earth and travel toward Planet X at 0.3c. The observer on Earth also sees that Planet X is stationary. An observer on Planet X sees rocket 2 travel toward Earth at 0.4c. What is the speed of rocket 1 according to an observer on rocket 2?
Answer:
0.625 c
Explanation:
Relative speed of a body may be defined as the speed of one body with respect to some other or the speed of one body in comparison to the speed of second body.
In the context,
The relative speed of body 2 with respect to body 1 can be expressed as :
[tex]$u'=\frac{u-v}{1-\frac{uv}{c^2}}$[/tex]
Speed of rocket 1 with respect to rocket 2 :
[tex]$u' = \frac{0.4 c- (-0.3 c)}{1-\frac{(0.4 c)(-0.3 c)}{c^2}}$[/tex]
[tex]$u' = \frac{0.7 c}{1.12}$[/tex]
[tex]u'=0.625 c[/tex]
Therefore, the speed of rocket 1 according to an observer on rocket 2 is 0.625 c
A parallel plate capacitor creates a uniform electric field of and its plates are separated by . A proton is placed at rest next to the positive plate and then released and moves toward the negative plate. When the proton arrives at the negative plate, what is its speed
Complete Question
A parallel plate capacitor creates a uniform electric field of 5 x 10^4 N/C and its plates are separated by 2 x 10^{-3}'m. A proton is placed at rest next to the positive plate and then released and moves toward the negative plate. When the proton arrives at the negative plate, what is its speed?
Answer:
[tex]V=1.4*10^5m/s[/tex]
Explanation:
From the question we are told that:
Electric field [tex]B=1.5*10N/C[/tex]
Distance [tex]d=2 x 10^{-3}[/tex]
At negative plate
Generally the equation for Velocity is mathematically given by
[tex]V^2=2as[/tex]
Therefore
[tex]V^2=\frac{2*e_0E*d}{m}[/tex]
[tex]V^2=\frac{2*1.6*10^{-19}(5*10^4)*2 * 10^{-3}}{1.67*10^{-28}}[/tex]
[tex]V=\sqrt{19.2*10^9}[/tex]
[tex]V=1.4*10^5m/s[/tex]
a coach is travelling east wards at 12.6 m/s after 12 second its velocity is 9.5 m/s in the same direction. what is the acceleration and direction of its acceleration?
pls do it with the formula
thx mates :)
[tex]\\ \rm\longmapsto a=\dfrac{v-u}{t}[/tex]
[tex]\\ \rm\longmapsto a=\dfrac{12.6-9.5}{12}[/tex]
[tex]\\ \rm\longmapsto a=\dfrac{3.1}{12}[/tex]
[tex]\\ \rm\longmapsto \overrightarrow{a}=0.25m/s^2[/tex]
What is the energy of a photon with a frequency of 3.6 × 1015 Hz? Planck’s constant is 6.63 × 10–34 J•s.
1.8 × 10–49 J
2.4 × 10–19 J
1.8 × 10–18 J
2.4 × 10–18 J
We know
[tex]\boxed{\sf E=hv}[/tex]
[tex]\\ \sf\longmapsto E=6.63\times 10^{-34}J\times 3.6\times 10^{15}s^{-1}[/tex]
[tex]\\ \sf\longmapsto E=23.86\times 10^{-19}J[/tex]
[tex]\\ \sf\longmapsto E=2.38\times 10^{-18}J[/tex]
[tex]\\ \sf\longmapsto E=2.4\times 10^{-18}J[/tex]
Answer:
D!!!!!
Explanation:
a student standing between two walls shouts once.he hears the first echo after 3 seconds and the next after 5 seconds. calculate the distance between the walls.
Explanation:
It took [tex]t_1 =1.5\:\text{s}[/tex] for the sound to reach the 1st wall and at the same time time, the same sound took [tex]t_2 = 2.5\:\text{s}[/tex] to reach the 2nd wall. Assuming that the sound travels at 343 m/s, then let [tex]x_1[/tex] be the distance of the person to the 1st wall and [tex]x_2[/tex] be the distance to the 2nd wall. So the distance between the walls X is
[tex]X = x_1 + x_2 = v_st_1 + v_st_2 = v_s(t_1 + t_2)[/tex]
[tex]\:\:\:\:\:= (343\:\text{m/s})(4.0\:\text{s}) = 1372\:\text{m}[/tex]
Flag question
Consider the pressure and force acting on the
dam retaining a reservoir of water. Suppose the
dam is 500-m wide
and the water is 80.0-m
deep at the dam, as illustrated below. What is
the average pressure on the dam due to the
water?
Answer:
P = density (p) * g * h
P = 1000 kg/m^3 * 9.8 m/s^2 * 40 m = 392,000 N/m^2
since kg m / s^2 = Newtons
The average pressure is 1/2 (pressure at 0m + pressure 80 m) for liquid of uniform density
What is the connection between speed, friction, and radius of the curve when turning when driving a car.
Answer:
hhhbbbbbbbbbbbbbbnnnnnbbhb
given A=4i-10j and B= 7i+5j find b such that A+bB is a vector pointing along the x-axis (i.e has no y component)
Answer:
-4/7
Explanation:
Given the following
A=4i-10j and B= 7i+5j
A+ bB = 4i-10j + (7i+5j)b
A+ bB = 4i-10j + 7ib+5jb
A+ bB =
The vector along the x-axis is expressed as i + 0j
If the vector A+ bB is pointing in the direction of the x-axis then;
[tex]A+ bB * \frac{i+0j}{|i+0j|} = 0 \\ (4+7b)i-(10-5b)j* \frac{i+0j}{\sqrt{1^2+0^2} } = 0\\(4+7b)i-(10-5b)j *(i+0j) = 0\\4+7b-0 =0\\7b=-4\\b = -4/7[/tex]
Hence the value of b is -4/7
The value of [tex]\beta[/tex] such that [tex]\vec C = \vec A + \beta \cdot \vec B = c\,\hat{i}[/tex] is 2.
According to the statement, we have following system of vectorial equations:
[tex]\vec A = 4\,\hat {i} - 10\,\hat{j}[/tex] (1)
[tex]\vec {B} = 7\,\hat{i} + 5\,\hat{j}[/tex] (2)
[tex]\vec C = \vec A + \beta \cdot \vec B = c\,\hat{i}[/tex] (3)
By applying (1) and (2) in (3):
[tex](4\,\hat{i}-10\,\hat{j}) + \beta\cdot (7\,\hat{i}+5\,\hat{j}) = c\,\hat{i}[/tex]
[tex](4+7\cdot \beta)\,\hat{i} +(-10+5\cdot \beta)\,\hat{j} = c\,\hat{i}[/tex]
And we get two scalar equations after analyzing each component:
[tex]4+7\cdot \beta = c[/tex] (4)
[tex]-10+5\cdot \beta = 0[/tex] (5)
We solve for [tex]\beta[/tex] in (5):
[tex]\beta = 2[/tex]
And for [tex]c[/tex] in (4):
[tex]c = 4+7\cdot (2)[/tex]
[tex]c = 18[/tex]
The value of [tex]\beta[/tex] such that [tex]\vec C = \vec A + \beta \cdot \vec B = c\,\hat{i}[/tex] is 2.
Please see this question related to Sum of Vectors for further details: https://brainly.com/question/11881720
In the diagram, disk 1 has a moment of inertia of 3.4 kg · m2 and is rotating in the counterclockwise direction with an angular velocity of 6.1 rad/s about a frictionless rod passing through its center. A second disk rotating clockwise with an angular velocity of 9.3 rad/s falls from above onto disk 1. The two then rotate as one in the clockwise direction with an angular velocity of 1.8 rad/s. Determine the moment of inertia, in kg · m2, of disk 2.
Answer:
I = 3.6 kg•m²
Explanation:
Conservation of angular momentum
Let's assume CW is the positive direction
3.4(-6.1) + I(9.3) = 3.4(1.8) + I(1.8)
I(9.3 - 1.8) = 3.4(1.8 + 6.1)
I(7.5) = 3.4(7.9)
I = 3.4(7.9)/(7.5) = 3.5813333333...
The moment of inertia of the second disk will be [tex]I=3.58\ kg-m^2[/tex]
What is moment of inertia?The moment of inertia is defined as the product of mass of section and the square of the distance between the reference axis and the centroid of the section.
here it is given that
MOI of disk one [tex]I_1=3.4\ kg-m^2[/tex]
Angular velocity [tex]w_1=6.1\ \frac{rad}{s}[/tex]
Angular velocity of disk two [tex]w=1.8\ \frac{rad}{s}[/tex]
MOI of the disk two [tex]I=?[/tex]
The final angular velocity [tex]w_f= 1.8\ \frac{rad}{sec}[/tex]
Now from the conservation of the momentum the angular momentum before collision will be equal to the angular momentum after collision.
[tex]I_1w_1+I_2w_2=(I_1+I_2)w_f[/tex]
Now put the values in the formula
[tex](3.4\times 6.10)+(I_2\times 9.3)=(3.4+I_2)\times 1.8[/tex]
[tex]I_2=3.58\ kg-m^2[/tex]
Thus the moment of inertia of the second disk will be [tex]I=3.58\ kg-m^2[/tex]
To know more about moment of inertia follow
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1 of 3 : please help got an extra day for a test and i don’t get this (must show work) points and brainliest!
Answer:
y = 1/2at^2
we could also write it as-
y = (at^2)/2
2y = at^2
2y/a = t^2
√2y/a = t
hope it helps
A gymnast weighs 450 N. She stands on a balance beam of uniform construction which weighs 250 N. The balance beam is 3.0 m long and is supported at each end. If the support force at the right end is four times the force at the left end, how far from the right end is the gymnast
Answer:
x = 9.32 cm
Explanation:
For this exercise we have an applied torque and the bar is in equilibrium, which is why we use the endowment equilibrium equation
Suppose the counterclockwise turn is positive, let's set our reference frame at the left end of the bar
- W l / 2 - W_{child} x + N₂ l = 0
x = [tex]\frac{-W l/2 + n_2 l}{W_{child}}[/tex] 1)
now let's use the expression for translational equilibrium
N₁ - W - W_(child) + N₂ = 0
indicate that N₂ = 4 N₁
we substitute
N₁ - W - W_child + 4 N₁ = 0
5 N₁ -W - W_{child} = 0
N₁ = ( W + W_{child}) / 5
we calculate
N₁ = (450 + 250) / 5
N₁ = 140 N
we calculate with equation 1
x = -250 1.50 + 4 140 3) / 140
x = 9.32 cm
If the mass of an object is 15 kg and the velocity is -4 m/s, what is the momentum?
momentum p= m x v = 15 x -4 = -60 N.s
a method of reducing friction
Answer:
Lubrication
Explanation:
People oil/lubricate bicycle chains because the chain turns around the cogs and rub together so this help with friction.
Hope this helps :)
Answer:
The method of reducing friction are :
i) In moving parts of machine friction can be reduced by using a ball bearing between the moving surfaces
ii) The bodies of aeroplane ,ship ,boat etc are made streamlined to reduce friction.
iii) Friction can be reduced by polishing rough surfaces. For example : carrom boards are highly polished to reduce friction.
I hope this help you:)
a person lifts 60kg on the surface of the earth, how much mass can he lift on the surface of the moon if he applies same magnitude of force
Explanation:
Hey there!
According to the question;
A person can lift mass of 60 kg on earth.
mass(m1) = 60kg
acceleration due to gravity on earth (a) = 9.8m/s²
Now;
force (f) = m.a
= 60*9.8
= 588 N
Since, there is application of same magnitude of force on moon,
mass(m) =?
acceleration due to gravity on moon (a) = 1.67m/s²
Now;
force (f) = m.a
588 = m*1.67
m = 352.09 kg
Therefore, the person who can lift the mass of 60 kg on earth can lift mass of 352 kg on moon.
Hope it helps!
An object moving with a constant
acceleration changes its velocity from
10ms' to 20 ms' in five seconds. What is the
distance travelled in five seconds
Answer:
Acceleration:
[tex]{ \tt{a = \frac{v - u}{t} }} \\ { \tt{a = \frac{20 - 10}{5} }} \\ { \tt{a = 2 \: m {s}^{ - 2} }}[/tex]
From third equation:
[tex]{ \bf{ {v}^{2} = {u}^{2} + 2as}} \\ { \tt{s = \frac{ {20}^{2} - {10}^{2} }{2 \times 2} }} \\ = { \tt{s = 75 \: m}}[/tex]
Answer:
Formula = m/s
Explanation:
The answer is 10 m / 5 seconds = 2 meters distance
The answer is 20 m / 5 seconds = 4 meters distance
Which of the following is not true about Triton, the large moon of Neptune? It is more reflective than Earth's Moon. It is larger than Earth's Moon. It is in a retrograde orbit. It has a thin atmosphere. It has nitrogen geysers.
Answer:
Triton is the largest of Neptune's 13 moons. It is unusual because it is the only large moon in our solar system that orbits in the opposite direction of its planet's rotation―a retrograde orbit. ... Like our own moon, Triton is locked in synchronous rotation with Neptune―one side faces the planet at all times.
A tuning fork with a frequency of 335 Hz and a tuning fork of unknown frequency produce beats with a frequency of 5.3 when struck at the same time. A small piece of putty is placed on the tuning fork with the known frequency and it's frequency is lowered slightly. When struck at the same time, the two forks now produce a beat frequency of 8 Hz. 1)What is frequency of tuning fork which originally had a frequency of 335 Hz after the putty has been placed on it
Answer:
Explanation:
Unknown fork frequency is either
335 + 5.3 = 340.3 Hz
or
335 - 5.3 = 329.7 Hz
After we modify the known fork, the unknown fork frequency equation becomes either
(335 - x) + 8 = 340.3
(335 - x) = 332.3
x = 2.7 Hz
or
(335 - x) + 8 = 329.7
(335 - x) = 321.7
x = 13.3 Hz
IF the unknown fork frequency was 340.3 Hz,
THEN the 335 Hz fork was detuned to 335 - 2.7 = 332.3 Hz
IF the unknown fork frequency was 329.7 Hz,
THEN the 335 Hz fork was detuned to 335 - 13.3 = 321.7 Hz
what is meant by specific latent heat of vaporization of water is -2.26mjkg^-1 or -2.26mj/kg?
Answer:
The specific latent heat of a substance is the amount of energy required to change the state of one kilo of the substance without change in it temperature.The latent heat of vaporization or evaporation is the heat given to some mass to convert if from the liquid to the vapor phase.
A 2.0 kg puck is at rest on a level table. It is pushed straight north with a constant force of 5N for 1.50 s and then let go. How far does the puck move from rest in 2.25 s?
Answer:
d = 6.32 m
Explanation:
Given that,
The mass of a puck, m = 2 kg
It is pushed straight north with a constant force of 5N for 1.50 s and then let go.
We need to find the distance covered by the puck when move from rest in 2.25 s.
We know that,
F = ma
[tex]a=\dfrac{F}{m}\\\\a=\dfrac{5}{2}\\\\a=2.5\ m/s^2[/tex]
Let d is the distance moved in 2.25 s. Using second equation of motion,
[tex]d=ut+\dfrac{1}{2}at^2\\\\d=0+\dfrac{1}{2}\times 2.5\times (2.25)^2\\\\d=6.32\ m[/tex]
So, it will move 6.32 m from rest in 2.25 seconds.
A physical pendulum in the form of a planar object moves in simple harmonic motion with a frequency of 0.680 Hz. The pendulum has a mass of 2.00 kg, and the pivot is located 0.340 m from the center of mass. Determine the moment of inertia of the pendulum about the pivot point.
Answer:
Therefore, the moment of inertia is:
[tex]I=0.37 \: kgm^{2} [/tex]
Explanation:
The period of an oscillation equation of a solid pendulum is given by:
[tex]T=2\pi \sqrt{\frac{I}{Mgd}}[/tex] (1)
Where:
I is the moment of inertiaM is the mass of the pendulumd is the distance from the center of mass to the pivotg is the gravityLet's solve the equation (1) for I
[tex]T=2\pi \sqrt{\frac{I}{Mgd}}[/tex]
[tex]I=Mgd(\frac{T}{2\pi})^{2}[/tex]
Before find I, we need to remember that
[tex]T = \frac{1}{f}=\frac{1}{0.680}=1.47\: s[/tex]
Now, the moment of inertia will be:
[tex]I=2*9.81*0.340(\frac{1.47}{2\pi})^{2}[/tex]
Therefore, the moment of inertia is:
[tex]I=0.37 \: kgm^{2} [/tex]
I hope it helps you!
Four identical balls are thrown from the top of a cliff, each with the same speed. The
first is thrown straight up, the second is thrown at 30° above the horizontal, the third
at 30° below the horizontal, and the fourth straight down. How do the speeds and
kinetic energies of the balls compare as they strike the ground? Ignore the effects of
air resistance. Explain fully using the concepts from this unit.
The comparison of the speeds and kinetic energy of the identical balls are as follows
The speed and the kinetic energy of the first and fourth ball are equal, while the speed and kinetic energy of the second and third balls are equal
The reason for the above comparison results areas follows;
Known parameters;
First ball is thrown straight up
Second ball is thrown 30° above the horizontal
Third ball it thrown 30° below the horizontal
The fourth ball is thrown straight down
Unknown:
Comparison of the speed and kinetic energy of the four balls
Method:
The kinetic energy, K.E. = (1/2) × m × v²
The velocity of the ball, v = u × sin(θ)
Where;
u = The initial velocity of the ball
θ = The reference angle
For the first ball thrown straight up, we have;
θ = 90°
∴ [tex]v_y[/tex] = u
The final velocity of the ball as it strikes the ground is v₂ = u² + 2gh
Where;
h = The height of the cliff
∴ Kinetic energy of first ball, K.E.₁ = (1/2) × m × (u₁² + 2gh)²
For the second ball thrown 30° to the horizontal, we have;
K.E. = (1/2) × m × ((u×sin30)² + 2·g·h)² = K.E. = (1/2) × m × ((0.5·u)² + 2·g·h)²
Kinetic energy K.E.₂ = (1/2) × m × ((0.5·u₂)² + 2·g·h)²
For the third ball thrown at 30° below the horizontal, we have;K.E. = (1/2) × m × ((u×sin30)² + 2·g·h)² = K.E. = (1/2) × m × ((0.5·u)² + 2·g·h)²
Kinetic energy K.E.₃ = (1/2) × m × ((0.5·u₃)² + 2·g·h)²
For the fourth ball thrown straight down, we have;Kinetic energy K.E.₄ = (1/2) × m × (u₄² + 2gh)²
Therefore, as the ball strike the ground, the speed and the kinetic energy of the first and fourth ball are equal, while the speed and kinetic energy of the second and third balls are equal
u₁ = u₄, K.E₁ = K.E.₄, u₂ = u₃, K.E₂ = K.E.₃
Learn more about object kinetic energy of objects in free fall here;
https://brainly.com/question/14872097
The area around a charged object that can exert a force on other charged objects is an electric ___
A 55kg bungee jumper has fallen far enough that her bungee cord is beginning to stretch and resist her downward motion . Find the ( magnitude and direction ) exerted on her by the bungee cord at an instant when her downward acceleration has a magnitude of 7.1m/s2
Answer:
148.5 N
Explanation:
Given that,
The mass of a bungee jumper, m = 55 kg
The downward acceleration, a = 7.1 m/s²
We need to find the net force acting on the jumper. As it is moving in downward direction, net force is :
T = m(g-a)
Put all the values,
T = 55(9.8 - 7.1)
= 148.5 N
So, the force exerted on the bungee cord is 148.5 N.
Answer:
The downward force is 148.5 N.
Explanation:
mass, m = 55 kg
downwards acceleration, a = 7.1 m/s^2
Let the force is F.
According to the newton's second law
m g - F = m a
F = m (g - a)
F = 55 (9.8 - 7.1)
F = 148.5 N
If car A passes car B, then car A must be
A. accelerating at a greater rate than car B.
B. moving faster than car B, but not necessarily accelerating
C. accelerating
D. moving faster than car B and accelerating more than car B
Answer:
B. moving faster than car B, but not necessarily accelerating
Explanation:
Velocity is the speed of something. So car A's velocity is greater than car B but does not mean car A is accelerating.
Why are hydraulic brakes used?
Answer:
Hydraulic brake systems are used as the main braking system on almost all passenger vehicles and light trucks. Hydraulic brakes use brake fluid to transmit force when the brakes are applied.
Explanation:
Our system is a block attached to a horizontal spring on a frictionless table. The spring has a spring constant of 4.0 N/m and a rest length of 1.0 m, and the block has a mass of 0.25 kg.
Compute the PE when the spring is compressed by 0.50 m.
Answer
E - 1/2 K x^2 potential energy of compressed spring
E = 1/2 * 4 N / m * (.5 m)^2 = 2 * .5^2 N-m = .5 N-m
During normal beating, the heart creates a maximum 4.10-mV potential across 0.350 m of a person's chest, creating a 1.00-Hz electromagnetic wave. (a) What is the maximum electric field strength created? V/m (b) What is the corresponding maximum magnetic field strength in the electromagnetic wave? T (c) What is the wavelength of the electromagnetic wave?
Explanation:
Given that,
Maximum potential, V = 4. mV
Distance, d = 0.350 m
Frequency of the wave, f = 100 Hz
(a) The maximum electric field strength created is given by:
[tex]E=\dfrac{V}{d}\\\\E=\dfrac{4.1\times 10^{-3}}{0.350 }\\\\E=0.0117\ V/m[/tex]
(b) The corresponding maximum magnetic field strength in the electromagnetic wave is given by :
[tex]B=\dfrac{E}{c}\\\\B=\dfrac{0.0117}{3\times 10^8}\\\\B=3.9\times 10^{-11}\ T[/tex]
(c) The wavelength of the electromagnetic wave can be calculated as :
[tex]\lambda=\dfrac{c}{f}\\\\\lambda=\dfrac{3\times 10^8}{100}\\\\=3\times 10^6\ m[/tex]
So, the wavelength of the electromagnetic wave is [tex]3\times 10^6\ m[/tex].
A magnifying glass produces a maximum angular magnification of 5.4 when used by a young person with a near point of 20 cm. What is the maximum angular magnification obtained by an older person with a near point of 65 cm
How is the sun used to make food?
Answer:
Plants use a process called photosynthesis to make food. During photosynthesis, plants trap light energy with their leaves. Plants use the energy of the sun to change water and carbon dioxide into a sugar called glucose. Glucose is used by plants for energy and to make other substances like cellulose and starch.
Thank you
Answer:
Plants use a process called photosynthesis to make food. During photosynthesis, plants trap light energy with their leaves. Plants use the energy of the sun to change water and carbon dioxide into a sugar called glucose.
A mysterious constant force of 10 N acts horizontally on everything. The direction of the force is found to be always pointed toward a wall in a big hall. Find the potential energy of a particle due to this force when it is at a distance x from the wall, assuming the potential energy at the wall to be zero.
Answer:
it will be 10x
Explanation:
workdone(potential energy before it hits the wall)= horizontal force × distance
=10× x = 10x joules
A mysterious constant force of 10 N acts horizontally on everything. The direction of the force is found to be always pointed toward a wall in a big hall.The potential energy of a particle due to this force is 10x.
What is force?A force is an effect that can alter an object's motion according to physics. An object with mass can change its velocity, or accelerate, as a result of a force. An obvious way to describe force is as a push or a pull. A force is a vector quantity since it has both magnitude and direction.
Given in the question a mysterious constant force of 10 N acts horizontally on everything. The direction of the force is found to be always pointed toward a wall in a big hall the potential energy,
Work done (potential energy before it hits the wall)
= horizontal force × distance
=10× x = 10x joules
The potential energy of a particle due to this force is 10x.
To learn more about force refer to the link:
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