A 5 kilograms bowling ball is dropped out a window. It hits the ground, and bounces upward. The velocity change of the ball is noted to be 15 m/s downward and 12 m/s upward. What is the contact time for the ball if the force applied on the ball from the ground is equal to 10 N?
Answer:
13.5
Explanation:
Mass: 5kg
Initial Velocity: -15
Final Velocity: 12
Force: 10
We can use the equation: Vf = Vi + at
We need to find acceleration, and we can use the equation, F=ma,
We have mass and the force so it would look like this, 10=5a, and 5 times 2 would equal 10, so acceleration would be 2.
Now we have all the variables to find time.
Back to Vf = Vi + at, plug the numbers in, 12 = -15 + 2(t)
Plugging them in into desmos gives 13.5 for time.
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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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
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.
Learn more about equilibrium at: https://brainly.com/question/13524990
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:
yayy here you are f, r, e, e, p, o, i, n, t, s
Answer:
Albert Einstein Albert Einstein was a German-born theoretical physicist, widely acknowledged to be one of the greatest physicists of all time. Einstein is best known for developing the theory of relativity, but he also made important contributions to the development of the theory of quantum mechanics.Explanation:
Thank you so much buddy !!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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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
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.
Please help me.............................
Answer:
[tex]a[/tex]
Explanation:
is a corect anser
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 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.
A volleyball that has an initial momentum of
−
1.0
kg
⋅
m
s
−1.0kg⋅
s
m
minus, 1, point, 0, start text, k, g, end text, dot, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction changes direction after a hand hits it with a force of
150
N
150N150, start text, N, end text for
0.01
seconds
0.01seconds0, point, 01, start text, s, e, c, o, n, d, s, end text.
A hand hits a volleyball. An arrow points to the right, in the directio
Answer:
10-
Step-by-Step Explanation:
Define the term dimension
Answer:
Q1. A measurable extent of a particular kind, such as length, breadth, depth, or height.
Q2. A dimensional constant is a physical quantity that has dimensions and has a fixed value. Some of the examples of the dimensional constant are Planck's constant, gravitational constant, and so on.
Q3. Physical quantities which posses dimensions and have variable are called dimensional variables. Examples are length, velocity, and acceleration etc.
Q4. Dimensionless variables are the quantities which doesn't have any dimensions the the value is a variable. Eg: angle = arc/ radius. Dimensions = L/L. = 1. So angle does not have any dimensions and the value can vary.
Q5. Principle of Homogeneity states that dimensions of each of the terms of a dimensional equation on both sides should be the same. This principle is helpful because it helps us convert the units from one form to another.
Q6. Dimensional analysis has been around a long time, Newton called it the "Great principle of Similitude", but the modern form can be traced back to James Clerk Maxwell. It was Maxwell who distinguished mass [A/], length [£], and time [7"] as the independent dimensions from which others could be derived.
Q7. Mass, length, time, temperature, electric current, amount of light, and amount of matter.
Q8. Dimensional analysis is used to convert the value of a physical quantity from one system of units to another system of units. Dimensional analysis is used to represent the nature of physical quantity. The expressions of dimensions can be manipulated as algebraic quantities.
Hope that helps. x
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:
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.
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.)
What initial speed v is required if the blocks m1 =2.5 kg and m2=1.5 kg are to travel a distance d =7.0cm before coming to rest? Assume the coefficient of kinetic friction between m1 and the tabletop is ųk=0.21
Answer:
OPTRIMUM PRIDE URGH URGH URGH
Explanation:
AHHAAHAHAHAHA
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
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
Four men are pushing on a broken car. One man pushes on the car with 345 N, another with 203 N, another
with 291 N and 101 N. The friction between the car and road pushes in the opposite direction with a force of 940 N.
1.) Make a claim: What is the net force on the car?
2.) Cite evidence: What evidence supports your claim?
3.) Make a claim: Will the car move?
4.) Provide reasoning: Why do you say that?
Answer:
1. 3 N
2. The total force being put on the car by the men is 943 while the friction is going in the opposite direction with a force of 940.
3. Yes
4. The net force is not equal to zero
Explanation:
1. 345 + 203 + 291 + 101 = 943 + -940 = 3
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
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
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 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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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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Can someone help me with this problem?
Answer:
no
Explanation:
u flipped it i cant see
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
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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#SPJ3
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