Answers
Answer:
Id say beat it with a hammer and light it on fire. Then if it doesn't work, sacrifice it.
Explanation:
Related Questions
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?
Answers
(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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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
Answers
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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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
Answers
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.
Please help me.............................
Answers
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?
Answers
Answer:
It releases some of the energy into the atmosphere as hot steam.
Explanation:
what is the pressure exerted by a force of 25 N on an area of 5m square
Answers
Answer:
pressure = force / area
then pressure = 25 / 5 = "5" N/m^2
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?
Answers
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]
What is Acceleration?
Answers
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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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.
Answers
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
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?
Answers
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.
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
Answers
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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Is this right for the second one
Answers
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.
A body is moving along a circular path with variable speed, it has both radial and tangential acceleration.
Select one:
True
False
Answers
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
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
Answers
Answer:
Depends on the weight i would be 7.120
a 0.015 kg bullet traveling at 500 m/s strikes a 1.0 kg block of wood that is balanced on a table edge 0.92 m above the ground as shown to the right. the bullet buries itself in the block. calculate the horizontal distance, dx where the block hits the floor.
Answers
The horizontal distance where the block hits the floor is 3.2 m.
The given parameters:
mass of the bullet, m₁ = 0.015 kgspeed of the bullet, u₁ = 500 m/smass of block wood, m₂ = 1.0 kgheight of the table, h = 0.92 mThe final velocity of the bullet-block system after the collision is calculated by applying the principle of conservation of linear momentum;
[tex]m_1u_1 + m_2u_2 = v(m_1 + m_2)\\\\0.015(500) + 1(0) =v(0.015 + 1)\\\\7.5 = 1.015v\\\\v = \frac{7.5}{1.015} \\\\v = 7.39 \ m/s[/tex]
The time taken for the bullet-block system to fall to the floor after collision is calculated as follows;
[tex]h = v_0_yt + \frac{1}{2} gt^2\\\\h = 0 + \frac{1}{2} gt^2\\\\h = \frac{1}{2} gt^2\\\\t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2\times 0.92}{9.8} }\\\\t = 0.43 \ s[/tex]
The horizontal distance where the block hits the floor is calculated as follows;
[tex]X = v_x t\\\\X = 7.39 \times 0.43\\\\X = 3.2 \ m[/tex]
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Mechanical energy conservation states that
The total amount of energy will eventually be destroyed.
Potential energy will be conserved, but kinetic energy will be destroyed.
The total amount of energy, kinetic plus potential, remains the same.
Kinetic energy will be conserved, but potential energy will be destroyed.
Pls hurry I’ll give 50 points
Answers
Answer:
The total amount of energy, kinetic plus potential, remains the same.
Explanation:
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
Answers
Answer:
oK so here's what you should do is add .09 and 0.6
Explanation:
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.
Answers
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
Hey there!
having a question which is from ( JEE Advanced 2014 )
[tex]\huge\color{black}\boxed{\colorbox{pink}{Question♡}}[/tex]
A rocket is moving in a gravity free space with a constant acceleration of 2 −2 along +x direction (see figure). The length of a chamber inside the rocket is 4 m. A ball is thrown from the left end of the chamber in +x direction with a speed of 0.3 −1 relative to the rocket. At the same time, another ball is thrown in –x direction with a speed of 0.2 −1 from its right end relative to the rocket. The time in seconds when the two balls hit each other is?
[tex] \\ \\ [/tex]
proper explanation needed~
thankyou~
Answers
The time taken for the two balls to hit each other is 8 s.
The given parameters:
Acceleration of the rocket, a = 2 m/s²Length of the chamber, s = 4 mSpeed of the first ball, = V1 = 0.3 m/sSpeed of the second ball, V2 = 0.2 m/sThe time taken for the two balls to hit each other is calculated by applying relative velocity formula as shown below;
[tex](V_1 - (-V_2) )t = s\\\\(V_1 + V_2) t = s\\\\(0.3 + 0.2) t = 4\\\\0.5t = 4\\\\t = \frac{4}{0.5} \\\\t = 8 \ s[/tex]
Thus, the time taken for the two balls to hit each other is 8 s.
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a stone is thrown down off a bridge with a velocity of 22 m/s. what is its velocity after 1.5 seconds has passed?
Answers
Answer:
Velocity of the stone after 1.5 seconds has passed = 37 m/s
Explanation:
Initial velocity (u) = 22 m/s
Time (t) = 1.5 sec
Acceleration due to gravity (g) = 10 m/s²
By using kinematics equation:
v = u + gt
v = 22 + 10 × 1.5
v = 22 + 15
v = 37 m/s
Final velocity (v) = 37 m/s
Which of the following is the equation for acceleration?
Answers
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
I need help. please look at the image below and let me know I need this by 7:20 am pst.
Answers
Answer:
3(1.5) = 4.5 V
Explanation:
What is the amplitude of this wave ?
Answers
Hope you could get an idea from here.
Doubt clarification - use comment section.
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?
Answers
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
Pls answer ASAP pls bc I’m tryna get my grade up please
Answers
Answer:
The right answer for this question is 85%.
(I had the same question.)
Answer This!!!!!!
I'll give brainliest to whoever gets it right.
And 100 Points!
Answers
Answer: Time (days) = 88, the (mass) = 0.2180
Explanation:
Time (days) = 88, the (mass) = 0.2180
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
Answers
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
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?
Answers
[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]
37. The progressive loss of material from a surface by the mechanical action of a fluid on a surface is called
Answers
Answer:
The progressive loss of material from a surface by the mechanical action of a fluid on a surface is called erosion.
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
Answers
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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