Answer:
1800 J
Explanation:
GPE= mgh
60*3*10
=1800 J
5.000 km =
3.125
mi
8.000 fl oz =
mL
Answer:
236.588 mL
Explanation:
The formula for an approximate result is to multiply the volume value by 29.574
[tex]8.000 \times 29.574 = 236.588[/tex]
Is that what you were asking for?
A 27.0 kg box is pulled with a 113.6 N force at an angle of 35.1 degrees along a
surface that has a coefficient of friction of 0.4. What is the net force on the box?
Answer: 2,509.42
Explanation: using the equation net force=force x cos(angle) we can find the net force by multiplying the kg and the Newtons to get 3,067.2 then find the cos of 35.1 then multiple those two together to get 2,509.42
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?
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
if the momentum of a 1,400 kg car is the same as the truck in question 17, what is the velocity of the car?
Answer:
Explanation:
momentum is mass times velocity
p = mv
so take the momentum of the truck in question 17 and divide by the mass of this car
v = p/m = p / 1400
Please help me.............................
Answer:
[tex]a[/tex]
Explanation:
is a corect anser
A 10kg object is 15 meters up a hill. Find its potential energy
Answer:
Explanation:
Relative to an origin at the bottom of the hill,
PE = mgh = 10(9.8)(15) = 1470 J
7. What is the velocity of an object with a distance of 90m south and a time of
5s?
Answer:
Explanation:
v= s/t
V =90m/5s
V = 8m/s
please help 9.2.1 project in science just ned an example
Answer:
Give me what kind of example you need please so I can help you. Put it in the comments.
Explanation:
A wheel in the shape of a flat, heavy, uniform, solid disk is initially at rest at the top of an inclined plane of height 2.00 m when it begins to roll down the incline. If rolling and sliding friction are neglected, what is the linear velocity, in m/s, of the center-of-mass of the wheel when it reached the bottom of the incline?
Answer:
Explanation:
If friction is neglected, the wheel cannot roll and can only slide frictionlessly and will have the same velocity at the bottom of the ramp as if it had been in free fall as it has converted the same amount of potential energy.
mgh = ½mv²
v = √(2gh) = √(2(9.81)(2.00)) = 6.26418... = 6.26 m/s
However if we do not ignore all friction and the wheel rolls without slipping down the slope, the potential energy becomes linear and rotational kinetic energy
mgh = ½mv² + ½Iω²
mgh = ½mv² + ½(½mR²)(v/R)²
2gh = v² + ½v²
2gh = 3v²/2
v = √(4gh/3) =√(4(9.81)(2.00)/3) = 5.11468... = 5.11 m/s
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:
Understanding what motivates anyone is not easy because each individual has different
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
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.
What is the amplitude of this wave ?
Hope you could get an idea from here.
Doubt clarification - use comment section.
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]
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
help please i don’t know
Answer:
Explanation:
Potential energy at the top of the slide
PE = mgh = 49(9.8)(3) = 1,440.6 J
Energy converted to work of friction
W = Fd = 35(10) = 350 J
Converted potential that becomes kinetic energy
1440.6 - 350 = 1090.6 J
KE = ½mv²
v = [tex]\sqrt{2KE/m}[/tex]
v = [tex]\sqrt{2(1090.6)/49}[/tex]
v = 6.671902...
v = 6.7 m/s
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
Answer:
The total amount of energy, kinetic plus potential, remains the same.
Explanation:
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.
When a baseball curves to the right (a curveball) , air is flowing faster over the right side than over the left side. at the same speed all around the baseball, but the ball curves as a result of the way the wind is blowing on the field. faster over the left side than over the right side. faster over the top than underneath.
Answer:
faster over the left side than over the right side.
Explanation:
due to ball rotation, the right side is more closely matched to the speed of the air passing by as the ball progresses. This causes the air to stick more closely to the right side of the ball and that air stays with the ball surface as the spin moves it to the back of the ball and therefore leftward. As every action has an equal and opposite reaction the leftward force moving air causes the ball to experience an equal rightward force.
When the baseball curves to the right (a curveball), then the ball moves faster over the left side than over the right side.
What direction does a curveball move?The ball, which is thrown with a spin, is curve in the direction in which the front of the ball turns.
When a baseball curves to the right (a curveball),
For this condition, the pressure of air should be high on the left side than the pressure on right side.Molecules of the air on right side pushed backward by this spinning ball.The left side with high pressure push the ball towards right side where the pressure is low.Due to higher pressure, the ball move faster on the left side than the right side.Hence, when the baseball curves to the right (a curveball), then the ball moves faster over the left side than over the right side.
Learn more about the curveball movement here;
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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.
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]
Learn more about conservation of linear momentum here:https://brainly.com/question/24424291
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~
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 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.
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:
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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a heat engine is a device that uses to produce useful work
A heat engine is a device that uses to produce useful work.
Definition - a device for producing motive power from heat, such as a gasoline engine or steam engine.
So..
If this is a true or false question.. Your answer is:
TRUE
Answer:
HEAT
Explanation:
A stereo speaker produces a pure "A" tone, with a frequency of 220.0 Hz.
What is the period of the sound wave produced by the speaker?
T=
What is the wavelength water of the same sound wave as it enters some water, where it has a speed of about 1480 m/s?
λwater=
What is the wavelength air of this sound wave as it travels through air with a speed of about 341 m/s?
λair=
(a) The period of the sound wave is 0.005 s.
(b) The wavelength of the wave when the speed of the wave is 1480 m/s is 6.73 m.
(c) The wavelength of the sound wave as it travels through air is 1.55 m.
The given parameters;
Frequency of the wave, F = 220 HzThe period of the sound wave is calculated as follows;
[tex]T = \frac{1}{f} \\\\T = \frac{1}{220} \\\\T = 0.005 \ s[/tex]
The wavelength of the wave when the speed of the wave is 1480 m/s is calculated as follows;
[tex]v = f\lambda \\\\\lambda = \frac{v}{f} \\\\\lambda = \frac{1480}{220} \\\\\lambda = 6.73 \ m[/tex]
The wavelength of the sound wave as it travels through air with a speed of about 341 m/s;
[tex]\lambda = \frac{v}{f} \\\\\lambda = \frac{341}{220} \\\\\lambda = 1.55 \ m[/tex]
Learn more about wave equation here: https://brainly.com/question/4692600
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
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]
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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