Answer:
200 m/s
Explanation:
as momentum is a product of mass and speed, and mass is not changing, four times the speed will result in four times the momentum.
p = mv
4p = m(4v)
At 200m/s velocity of the train, the momentum of the train will be four times its original momentum. This is because the velocity is directly proportional to the momentum of an object.
What is Momentum?Momentum is the tendency of an object to remain in motion. It is the product of the mass of the object and the velocity of the object. It is denoted by "p". Momentum is a vector quantity because it has both the magnitude and direction. The unit of Momentum is g.m/s.
The momentum can be calculated with the formula:
p = m × v
where, p = momentum of the object,
m = mass of the object,
v = velocity of the object.
The momentum is directly proportional to the mass and velocity of the object. Hence, to increase the momentum of the train by four times the mass or velocity of the train needs to be increased by four times.
The increase in the velocity of the train by four times will increase the momentum of the train.
Original Velocity of the train = 50 m/s
Increased Velocity of the train = 50 × 4 = 200 m/s
Increased new Momentum of the train = mass × new velocity
Increased Momentum = 20 metric ton × 200m/s
Increased Momentum = 4000 metric ton. m/s
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A student catches a can of soda dropped from some unknown height by Mr. Fineman. If
the can was dropped from rest, and it is traveling at 8.9 m/s before it arrives in the
student's hand...
Answer:
v² = u² + 2gh
8.9² = 0² + 2(9.8)h
h = 4.0 m
v = at
t = 8.9/9.8
t = 0.91 s
Answer This!!!!!!
I'll give brainliest to whoever gets it right.
And 100 Points!
Answer: Time (days) = 88, the (mass) = 0.2180
Explanation:
Time (days) = 88, the (mass) = 0.2180
A car travels a certain distance from A to B with a speed of 60km/hr and then returns along the same path to the starting point with a speed of 40km/hr. Find the average speed and average velocity.
a) Km/hr
b) m/s
wrong answers will be reported!
Answer:
Explanation:
Speed is total distance traveled over time taken to do so.
If AB is measured in kilometers, time (t) for the whole trip is
t = AB/60 + AB/40
t = 2AB/120 + 3AB/120
t = 5AB/120 hrs
Average speed is distance over time
s = 2AB / (5AB/120)
s = 2(120)/5
s = 48 km/hr
s = 48(1000 m/km / 3600 s/hr) = 13.333333.... 13 m/s
Velocity is displacement over time.
As displacement is zero, velocity is zero
v = 0 km/hr = 0 m/s
Pretty harsh reporting answers just because they are wrong.
Please help me.............................
Answer:
[tex]a[/tex]
Explanation:
is a corect anser
When you put a pot of water on the stove, the stove transfers thermal energy to the water. As the water gains large
amounts of energy, how does it transfer this energy to its surrounding environment?
Answer:
It releases some of the energy into the atmosphere as hot steam.
Explanation:
Which of the following describes the motion of a block while it is in equilibrium? The block:
A. moves at a constant speed
B. slows down gradually to stop
C. speeds up for a bit, then moves at a constant speed
D. Accelerates constantly
The statement that describes the motion of a block while it is in equilibrium is: The block moves at a constant speed.
EQUILIBRIUM:A state of equilibrium in physics refers to a state of rest or the forces exerted on the object is in a balanced state.
In dynamic equilibrium, the acceleration of a body is zero. This means that the body is moving at a uniform speed.
Therefore, the statement that describes the motion of a block while it is in equilibrium is: The block moves at a constant speed.
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A circular disk of radius 0.200 m rotates at a constant angular speed of 2.50 rev/s. What is the centripetal acceleration (in m/s2) of a point on the edge of the disk?
[tex]a_c = 3.14\:\text{m/s}^2[/tex]
Explanation:
First, we need to convert the given angular speed [tex]\omega[/tex] from rev/s to rad/s:
[tex]2.50\:\dfrac{\text{rev}}{\text{s}}×\dfrac{2\pi\:\text{rad}}{1\:\text{rev}} = 15.7\:\text{rad/s}[/tex]
The centripetal acceleration [tex]a_c[/tex] is defined as
[tex]a_c = \dfrac{v^2}{r}[/tex]
Recall that [tex]v = r\omega[/tex] so we can write [tex]a_c[/tex] as
[tex]a_c = \dfrac{(r\omega)^2}{r} = \omega^2r[/tex]
[tex]\;\;\;\;\;=(15.7\:\text{rad/s})^2(0.200\:\text{m}) = 3.14\:\text{m/s}^2[/tex]
what is the pressure exerted by a force of 25 N on an area of 5m square
Answer:
pressure = force / area
then pressure = 25 / 5 = "5" N/m^2
What is Acceleration?
Answer:
Depends on what are you looking at. All of the following are valid definitions, it just depends on which way you are analizing the problem.
From a kinematics point of view:
Acceleration is, by definition, is a vector quantity that measures the rate of change of the rate of change in position (add brackets if it helps visualizing the idea). This leads to the following different definitions - which are more like means of calculating it
the second derivative of position with respect to time [tex]a = \ddot x= \frac{d}{dt}({\frac d{dt}} x)[/tex];the first derivative of velocity with respect to time [tex]a = \dot v = \frac d{dt} v[/tex].From a dinamics point of view
Acceleration is the effect of a force applied to a body, and measures the ratio of the force applied to a body of mass m and the mass itself (which is another formulation of Newton second law):
[tex]a = \frac Fm[/tex]
Acceleration is the rate of change of velocity
To define acceleration, We need to know more about motion.
Motion: This can be defined as the change in position of a body from one point to another. When an object accelerates, it undergoes motion.
Definition
Acceleration can be defined as the rate of change of velocity. The S.I unit of acceleration is meter-per-squared seconds. (m/s²)
The formula of acceleration is
a = (v-u)/t................. Equation 1⇒ Where:
a = accelerationu = initial velocityv = final velocityt = timeHence, Acceleration is the rate of change of velocity
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the distance between two charges q a and q b is r and the force between them is F. What is the force between them if the distance between them is doubled?
The force will be reduced to 1/4 of the original
Explanation:
According to Coulomb's law, the force between two charges [tex]q_a\:\text{and}\:q_b,[/tex] separated by a distance r is given by
[tex]F = k\dfrac{q_aq_b}{r^2}[/tex]
where k is the Coulomb constant.
Now let F' be the force between the two charges when their separation distance is doubled. We can write this force as
[tex]F' = k\dfrac{q_aq_b}{(2r)^2} = k\dfrac{q_aq_b}{4r^2}[/tex]
[tex]\;\;\;\;\;= \frac{1}{4}\left(k\dfrac{q_aq_b}{r^2}\right) = \frac{1}{4}F[/tex]
Therefore, the force will be reduced to a quarter of its original value.
Hope you could understand.
If you have any query, feel free to ask.
Is this right for the second one
Answer:
Yes.
Explanation:
There is no movement in magnetic, chemical, electrostatic, or nuclear (potential) energy. The other options for that question can't be right. Mechanical energy is a form of kinetic, so B cannot be true. Thermal energy is also kinetic, which makes C and D incorrect as well.
9.The force of gravity between two asteroids is 10,000 newtons (N).
When the distance between the asteroids is halved, what will be the force of gravity between them?
Answer:
F1 = G m1 m2 / R^2 force of attraction
F2 = G m1 m2 / (R/2)^2
F2 / F1 = 4 the force of gravity will be quadrupled
Which of the following is the equation for acceleration?
Answer:
A
Explanation:
acceleration = velocity / time
Hope that helps
Answer:
D
Explanation:
Average acceleration = final velocity- initial velocity, divided by elapsed time.
ā = V - Vo/t
= change in velocity ÷ change in time
Find the required angular speed, ω, of an ultracentrifuge for the radial acceleration of a point 1.60 cm from the axis to equal 300000 g (where g is the acceleration due to gravity).
Express your answer numerically in revolutions per minute.
Answer:
Explanation:
300000(9.8) = ω²(0.0160)
ω = 13555 rad/s
13555 rad/s (60 s/min/ 2π rad/rev) = 129,445 rev/min
An object, initially traveling at a velocity of 73 m/s, experiences an acceleration of -9.8 m/s^2. How much time will it take it to come to rest?
7.4 s
Explanation:
Given:
[tex]v_0 = 73\:\text{m/s}[/tex]
[tex]v = 0[/tex]
[tex]a = -9.8\:\text{m/s}^2[/tex]
[tex]t = ?[/tex]
To solve the time it takes for the object to come to a stop, we are going to use the equation below:
[tex]v = v_0 + at \Rightarrow t = \dfrac{v - v_0}{a}[/tex]
Using the given values above, we get
[tex]t = \dfrac{0 - 73\:\text{m/s}}{-9.8\:\text{m/s}^2}[/tex]
[tex]\;\;\;\;= 7.4\:\text{s}[/tex]
A body is moving along a circular path with variable speed, it has both radial and tangential acceleration.
Select one:
True
False
Answer:
True; ar = v^2 / R Radial acceleration because it moves in a circular path
at = α R = tangential acceleration because its speed changes
a = at + ar total acceleration equals sum of radial and tangential
A mars surface exploration vehicle drops a rock off a 1.00 I'm high vertical Cliff. The sound of the rock landing at the base of the cliff is recorded by instruments on the vehicle 27.1 seconds later. Calculate the acceleration due to gravity on Mars given that the speed of sound on Mars is 320 m/s
The acceleration due to gravity on Mars is 11.81 m/s².
The given parameters:
Height of the cliff, h = 1 mTime of motion of the sound wave, t = 27.1 sSpeed of sound in mass, v = 320 sThe equation of motion to determine the acceleration due to gravity on the moon is calculated as follows;
[tex]s = vt + \frac{1}{2} gt^2[/tex]
where;
s is the distance traveledt is the time of motionSince the time measured is two way time, the new equation for the total distance traveled is calculated as;
[tex]v = \frac{2d}{t} \\\\2d = vt\\\\d = \frac{vt}{2} \\\\d = \frac{320 \times 27.1}{2} \\\\d = 4,336 \ m[/tex]
The acceleration due to gravity is calculated as follows;
[tex]s = vt + \frac{1}{2} gt^2\\\\4,336 = 0 \ + \ \frac{1}{2} \times g \times (27.1)^2\\\\4,336 = 367.21g\\\\g = \frac{4,336}{367.21} \\\\g = 11.8 1 \ m/s^2[/tex]
Thus, the acceleration due to gravity on Mars is 11.81 m/s².
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This is an image of a satellite traveling around Earth. Explain what are the two forces that are keeping the satelite around Earth without flying off or hitting the ground.
Answer:
One force will be gravity & inertia.
Explanation:
Bioth are combine to keep Earth in orbit around the sun, and the moon in orbit around Earth
When laser light of wavelength 632.8 nm passes through a diffraction grating, the first bright spots occur at ± 17.8 ∘ from the central maximum. What is the line density
Answer:
Depends on the weight i would be 7.120
A 30-cm-tall, 4.0-cm-diameter plastic tube has a sealed bottom. 250 g of lead pellets are poured into the bottom of the tube, whose mass is 30 g, then the tube is lowered into a liquid. The tube floats with 5.0 cm extending above the surface. What is the density of the liquid
The density of the liquid will be equal to [tex]\rho=0.892 \ \dfrac{g}{cm^3}[/tex]
What is density?The density of an object is defined as the ratio of the mass of an object to the volume of the object.
Volume of tube = 2^2 * pi * 30 = 377 cm^3
Volume of tube submerged = 25* 377 / 30 = 314 cm^3
Buoyancy = weight of liquid displaced
Volume of liquid displaced = 314 cm^3
Mass of tube and lead = 250 + 30 = 280 g
Now from the mass density by definition
[tex]\rho = \dfrac{m}{v}[/tex]
[tex]m=\rho \times v[/tex]
Mass of liquid displaced = Mass being supported
[tex]314 \times \rho = 280 g[/tex]
[tex]\rho= \dfrac{280}{ 314 } = .892 \frac{g}{cm^3}[/tex]
Thus the density of the liquid will be equal to [tex]\rho=0.892 \ \dfrac{g}{cm^3}[/tex]
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The density of the liquid is 1.67 g/[tex]cm^3[/tex].
The volume of the tube is
30 * 4 * 3.14 * 0.25 = 94.2 [tex]cm^3[/tex].
The mass of the lead pellets and the plastic tube is
30 + 250 = 280 g.
The volume of the lead pellets is
250 / 11.34 = 22 [tex]cm^3[/tex].
The volume of the liquid that the tube displaces is
94.2 - 22 = 72 [tex]cm^3[/tex].
The density of the liquid is
280 / 72 = 1.67 g/ [tex]cm^3[/tex].
Therefore, the density of the liquid is 1.67 g/ [tex]cm^3[/tex].
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3. A car travelling at 12 m/s into a stationary truck of about 10 times the cars mass. a. If the collision was completely inelastic, what velocity would the two travel at if the stuck together? b. If the collision was completely elastic, what would be the velocities of the car and truck after the collision? c. In order to exert a force of only 3500N on the truck during the collision, how much time would the collision have to take?
(a) The final velocity of the two vehicles if the collision was inelastic is 1.1 m/s.
(b) For the elastic collision, the final velocity of the car is 9.81 m/s backwards and the final velocity of the truck is 2.19 m/s forward.
(c) The time taken to exert the given force is 0.00625 m (s).
The given parameters;
Initial velocity of the car, u₁ = 12 m/sInitial velocity of the truck, u₂ = 0Mass of the car, = mMass of the truck, = 10m(a) The final velocity of the two vehicles if the collision was inelastic is calculated as follows;
[tex]m_1 u_1 + m_2u_2 = v(m_1+ m_2)\\\\12m + 10m(0) = v(m + 10m)\\\\12m = v(11m)\\\\v = \frac{12m}{11m} \\\\v = 1.1 \ m/s[/tex]
(b) The final velocity of the two vehicles if the collision was elastic is calculated as follows;
[tex]m_1 u_1 + m_2u_2 = m_1v_1 + m_2v_2\\\\\12m \ + \ 10m(0) = mv_1 + 10mv_2\\\\12m = m(v_1 + 10v_2)\\\\12 = v_1 + 10 v_2\ \ - --(1)[/tex]
Apply one-directional velocity equation:
[tex]u_1 +v_1 = u_2 + v_2\\\\12 + v_1 = 0 + v_2\\\\12+ v_1 = v_2 \ \ --- (2)[/tex]
Substitute the value of [tex]v_2[/tex] into equation (1);
[tex]12 = v_1 + 10(12 + v_1)\\\\12= v_1 + 120 + 10v_1\\\\12- 120 = 11v_1\\\\-108 = 11v_1\\\\v_1 = \frac{-108}{11} \\\\v_1 = -9.81 \ m/s\\\\[/tex]
Solve for [tex]v_2[/tex];
[tex]v_2 = 12 + v_1\\\\v_2 = 12 - 9.81\\\\v_2 = 2.19 \ m/s[/tex]
Thus, for the elastic collision, the final velocity of the car is 9.81 m/s backwards and the final velocity of the truck is 2.19 m/s forward.
(c)
The change in the momentum of the truck is calculated as;
[tex]\Delta P = m_2(v_2 - u_2)\\\\\Delta P = 10m(2.19)\\\\\Delta P = 21.9m[/tex]
The time taken to exert the given force is calculated as follows;
[tex]Ft = \Delta P\\\\t = \frac{\Delta P}{F} \\\\t = \frac{21.9 \ m}{3500} \\\\t = 0.00625 \ m (seconds)[/tex]
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Pls answer ASAP pls bc I’m tryna get my grade up please
Answer:
The right answer for this question is 85%.
(I had the same question.)
A solid, uniform sphere with a mass of 2.5 kg rolls without slipping down an incline plane starting from rest at a vertical height of 19 m. If the sphere has a radius of 0.60 m, what is the angular speed of the sphere at the bottom of the incline plane
Answer:
1/2 m v^2 + 1/2 I ω^2 = m g h conservation of energy
I = 2/5 m R^2 inertia of solid sphere
1/2 m v^2 + 1/5 m ω^2 R^2 = m g h
1/2 v^2 + 1/5 v^2 = g h
v^2 = 10 g h / 7 = 1.43 * 9.80 * 19 m^2/s^2 = 266 m^2/s^2
v = 16.3 m/s
v = R ω
ω = 16.3 / .6 = 27.2 / sec
I need help. please look at the image below and let me know I need this by 7:20 am pst.
Answer:
3(1.5) = 4.5 V
Explanation:
Light and Reflection
Diagram Skills
E
STI
500
Mirrot
Flat Mirrors
1. The point of a 20.0 cm
D
pencil is placed 25.0 cm
from a flat mirror. Its
eraser is 15.0 cm from
the mirror. Three of the
light rays from the
pencil's point hit the
mirror with incident
angles of 0°, 20°, and
50° at points A, B, and C as shown.
a. Use a protractor to draw the reflected rays from points A, B, and C.
b. Where do reflected rays or their extensions intersect?
Mirror
B
c. What is the distance between the pencil's head and its image?
d. Would a person's eye located at point D perceive one of the reflected rays
drew? Will the person be able to see the image? Explain.
e. What if the eye is located at point E?
f. Draw incident rays from the eraser of the pencil to point A and to poin
The law of reflection allows to find the results for the questions about ray reflection in a plane mirror are:
a) Attachment we see a diagram of the incident and reflected rays, incident and reflected angles are equal.
b) The extension of the reflected rays is what forms the image.
c) The image's distance is 20 cm behind the flat mirror.
d) The point D (normal for an angle of 50º) cannot perceive the rays coming from point A, B, C
e) the Rays at points A, B, C cannot perceive in the point E.
f) attachment we see the rays that come out from the pencil eraser.
g) The image is behind the mirror at 15 cm.
The geometric interaction describes the interaction of light rays with surfaces, looking for where the rays are directed, it is described by two phenomenological laws:
Refraction. Establishes a relationship between incident rays and those transmitted by material means. Reflection. It establishes that the angle of incidence and reflection of the rays is the same.[tex]\theta_i = \theta_{r}[/tex]
From these two general laws, geometric optics establishes a relationship for the formation of the image, called the constructor's equation.
[tex]\frac{1}{f} = \frac{1}{p} + \frac{1}{q}[/tex]
Where f is the focal length, p and q are the distance to the object and the image, respectively.
In this exercise, the medium is a mirror, which is why it must comply with the law of reflection.
a) In the attachment we see a diagram of the incident and reflected rays for the three points.
According to the law of reflection, the incident and reflected angles are equal.
b) From the diagram we can see that the extension of the reflected rays is what forms the image, which is called virtual and is located behind the mirror.
c) In the diagram we see two rays to form the image, we see that the distance to the object is equal to the distance to the image.
From the constructor's equation a plane mirror has an infinite radius.
p = -q
Therefore the image's distance is 20 cm behind the flat mirror. Therefore the distance to the object and the image are the same, the negative sign indicates that the image is behind the mirror.
d) A person located at point D (normal for an angle of 50º) cannot perceive the rays coming from point A, B, C since their angle of reflection is not equal to the incident angle.
To perceive a ray it must have an angle of incidence of 25º.
e) Point E is located very far from the pencil, so the incident angle increases as does the reflected angle.
the Rays at points A, B, C cannot perceive.
f) In the attachment we see the rays that come out from the pencil eraser, they indicate that the distance to the plane mirror is 15.0 cm,
g) The image is behind the mirror at 15 cm.
In conclusion using the law of reflection we can find the results for the questions are:
a) Attachment we see a diagram of the incident and reflected rays, incident and reflected angles are equal.
b) The extension of the reflected rays is what forms the image.
c) The image's distance is 20 cm behind the flat mirror.
d) The point D (normal for an angle of 50º) cannot perceive the rays coming from point A, B, C
e) the Rays at points A, B, C cannot perceive in the point E.
f) attachment we see the rays that come out from the pencil eraser.
g) The image is behind the mirror at 15 cm.
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How to fix this? I need to save both
Answer:
Id say beat it with a hammer and light it on fire. Then if it doesn't work, sacrifice it.
Explanation:
37. The progressive loss of material from a surface by the mechanical action of a fluid on a surface is called
Answer:
The progressive loss of material from a surface by the mechanical action of a fluid on a surface is called erosion.
I need a short answer ?
Answer:
Explanation:
7a) t = d/v = 100/45cos14.5 = 2.29533...= 2.30 s
7b) h = ½(9.81)(2.29533/2)² = 6.46056... = 6.45 m
or
h = (45sin14.5)² / (2(9.81)) = 6.47 m
which rounds to the same 6.5 m when limiting to the two significant digits of the initial velocity.
Cody hits up food king and uses a scale to weigh the mass of an apple. if the spring potential energy in the scale is .09 j and is spring is stretched 0.6 meters, calculate the spring constant
Answer:
oK so here's what you should do is add .09 and 0.6
Explanation:
A pendulum with a length of 2 m has a period of 2.8 s. What is the period of a pendulum with a length of 8 m
Answer:
P = 2 pi (L / g)^1/2
P2 / P1 = (8 / 2)^1/2 = 2
The period would be twice as long or 5.6 sec.