Answers
Answer:
The types of energy is bond breaking and bond forming in chemical energy.
Explanation:
During Chemical reaction energy is required either for breaking up bonds in case of reactants and building bonds to form products.
The chemical reaction in which energy is released is called exothermic reactions, which is released due to making up the bonds.
The chemical reaction in which energy is absorbed is called endothermic reactions, in which energy is absorbed for breaking up the bonds.
Related Questions
A wagon of dog treats (combined mass 55 kg) is rolling at 2.1 m/s. A dog with mass 21 kg dives into the wagon, colliding with just enough momentum to make both stop. If the collision between the dog and the wagon lasts 0.1 s, what is the magnitude of the average force that will be exerted on the dog by the collision with the wagon
Answers
Answer:
Explanation:
An impulse results in a change of momentum
If the wagon and dog both stop, they must have had equal and opposite momentums
FΔt = mΔv
F = mΔv/Δt = m(v₁ - v₀)/(t₁ - t₀)
v₁ = t₀ = 0
F = m(v₀)/t₁
F = 55(2.1)/0.1 = 1155 N
We could have also figured the dog's initial velocity and used the dog's mass in the equation as well. Result would be identical.
Help!
A man standing in front of a plane mirror finds his image to be at a distance of 6m from himself. The distance of man from the mirror is
Answers
Answer:
The distance of man from the mirror is 3 m
Refer to the attachment
[tex]\Large\textsf{Hope \: It \: Helped}[/tex]
The distance of the man from the mirror is 3m
From the characteristics of image formed in a plane mirror,
Virtual and erectFormed behind the mirrorSame size as the objectLaterally inverted Distance of the image behind the mirror is the same as the distance of the object from the mirror.⇒ From the last point,
Let the distance of the image behind the mirror be xAlso the distance of the man from the mirror is x⇒ From the question,
x+x = 62x = 6x = 6/2x = 3 mHence, The distance of the man from the mirror is 3m
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An object following a straight-line path at constant speed
A.) has no forces acting on it.
B.) has a net force acting on it in the direction of motion.
C.) has zero acceleration.
D.) must be moving in a vacuum.
E.) none of the above
Answers
An object following a straight-line path at constant speed is option C.) has zero acceleration.
Are there any forces acting on a moving item traveling in a straight line at a constant speed?There are no forces operating on a body if it is travelling straight ahead at a steady speed. There are no forces operating on a body if it is travelling straight ahead at a steady speed.
Note that the physics concept of acceleration measures how quickly an object's motion is changing. An object's speed or velocity is what largely defines its motion.
Therefore, An object is considered to be accelerating when its velocity changes over time and as such since acceleration of the object is said to be zero, one can say that the net force acting on it is also zero.
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AP Physical problem the wording is really throwing me off and im totally lost on how to do this. I would love some help please and thank you!
Answers
Explanation:
a) Here is the free-body diagram. Note that I included the components of the weight mg (shown in dotted arrows) for use in the other parts of the problem.
b) The component of the weight parallel to the plane (shown in the diagram as a dotted arrow along the x-axis) is [tex]mg\sin15[/tex] and it is equal to
[tex]mg\sin15 = (25\:\text{kg})(9.8\:\text{m/s}^2)\sin15 = 63.4\:\text{N}[/tex]
c) Applying Newton's 2nd law to the y-axis, we can write
[tex]y:\;\;\;N - mg\cos15 = 0 \Rightarrow N = mg\cos15[/tex]
[tex]N = (25\:\text{kg})(9.8\:\text{m/s}^2)\cos15 = 236.7\:\text{N}[/tex]
d) The component of the weight mg into the plane is the same as the normal force, hence it's also 236.7 N.
e) To solve for the coefficient of friction, we apply Newton's 2nd law to the x-axis:
[tex]x:\;\;\;mg\sin15 - F_f = 0[/tex]
[tex]\Rightarrow F_f = mg\sin15\;\;(2)[/tex]
where [tex]F_f[/tex] is the frictional force defined as [tex]F_f = \mu N[/tex] so we can use Eqn(1) on Eqn (2) to write
[tex]\mu (mg\cos15) = mg\sin15[/tex]
Solving for [tex]\mu,[/tex] we get
[tex]\mu = \dfrac{\sin15}{\cos15} = \tan15 = 0.27[/tex]
Identify the type of chemical reaction:
CaCO3 -->CaO + CO2
Answers
Explanation:
decomposition reaction.....
Answer:
caco₃--cao+co₂
calcium oxide + carbon die oxide gives us calcium carbonate
this is the reaction of Acidic oxide(cao) and Basic oxide (co₂) to form salt.
A solid non-conducting sphere of radius R carries a charge Q distributed uniformly throughout its volume. At a radius r (r < R) from the center of the sphere the electric field has a value E. If the same charge Q were distributed uniformly throughout a sphere of radius 2R the magnitude of the electric field at a radius r would be equal to:__________
Answers
Answer:
Hence the answer is E inside [tex]= KQr_{1} /R^{3}[/tex].
Explanation:
E inside [tex]= KQr_{1} /R^{3}[/tex]
so if r1 will be the same then
E [tex]\begin{bmatrix}Blank Equation\end{bmatrix}[/tex] proportional to 1/R3
so if R become 2R
E becomes 1/8 of the initial electric field.
Answer:
The electric field is E/8.
Explanation:
The electric field due to a solid sphere of uniform charge density inside it is given by
[tex]E =\frac{\rho r}{3}[/tex]
where, [tex]\rho[/tex] is the volume charge density and r is the distance from the center.
For case I:
[tex]\rho = \frac{Q}{\frac{4}{3}\pi R^3}[/tex]
So, electric field at a distance r is
[tex]E = \frac { 3 Q r}{3\times 4\pi R^3}\\\\E = \frac{Q r}{4\pi R^3}[/tex]
Case II:
[tex]\rho = \frac{Q}{\frac{4}{3}\pi 8R^3}[/tex]
So, the electric field at a distance r is
[tex]E' = \frac { 3 Q r}{3\times 32\pi R^3}\\\\E' = \frac{Q r}{8\times 4\pi R^3}\\\\E' = \frac{E}{8}[/tex]
Use the sentence to answer the question.
Light is affected by gravity.
Which inference can be made based on this fact?
(1 point)
Light behaves differently in space than on Earth.
Light behaves differently in space than on Earth.
Gravity causes light to refract.
Gravity causes light to refract.
Light moves faster in space than on Earth.
Light moves faster in space than on Earth.
Stronger gravity causes an increase in light.
Answers
Answer:
Light behaves differently in space than on Earth.
Explanation:
Because the gravity field is greater near earth than in most of space. Not the areas near stars, black holes, pulsars, and such but in the vast emptyness between the clumpy spots.
(0.50kg)(6.0mls)+(1.00kg)(-12.00mls)
Answers
Answer:
-9.0 × 10-6 m3 kg
Explanation:
I'm not sure if that's what you're looking for nor do I know how to explain it.
During take-off a 8kg model rocket is burning fuel causing its speed to increase
at a rate of 4m/s2 despite experiencing a 90N drag.
What’s is the strength of the thrust?
(Answer unit is in N)( and the answer isn’t 212)
Answers
The strength of the thrust is 122 newtons.
The motion of the rocket is described by the second Newton's law, whose model is shown below:
[tex]\Sigma F = F - D = m\cdot a[/tex] (1)
Where:
[tex]F[/tex] - Thrust, in newtons[tex]D[/tex] - Drag, in newtons[tex]m[/tex] - Mass of the rocket, in kilograms[tex]a[/tex] - Net acceleration of the rocket, in meters per square secondIf we know that [tex]D = 90\,N[/tex], [tex]m = 8\,kg[/tex] and [tex]a = 4\,\frac{m}{s^{2}}[/tex], then the strength of the thrust is:
[tex]F = D + m\cdot a[/tex]
[tex]F = 90\,N + (8\,kg)\cdot \left(4\,\frac{m}{s^{2}} \right)[/tex]
[tex]F = 122\,N[/tex]
The strength of the thrust is 122 newtons.
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Find the ratio of the diameter of aluminium to copper wire, if they have the same
resistance per unit length. Take the resistivity values of aluminium and copper to
be 2.65× 10−8 Ω m and 1.72 × 10−8 Ω m respectively
Answers
Answer:
1.24
Explanation:
The resistivity of copper[tex]\rho_1=2.65\times 10^{-8}\ \Omega-m[/tex]
The resistivity of Aluminum,[tex]\rho_2=1.72\times 10^{-8}\ \Omega-m[/tex]
The wires have same resistance per unit length.
The resistance of a wire is given by :
[tex]R=\rho \dfrac{l}{A}\\\\R=\rho \dfrac{l}{\pi (\dfrac{d}{2})^2}\\\\\dfrac{R}{l}=\rho \dfrac{1}{\pi (\dfrac{d}{2})^2}[/tex]
According to given condition,
[tex]\rho_1 \dfrac{1}{\pi (\dfrac{d_1}{2})^2}=\rho_2 \dfrac{1}{\pi (\dfrac{d_2}{2})^2}\\\\\rho_1 \dfrac{1}{{d_1}^2}=\rho_2 \dfrac{1}{{d_2}^2}\\\\(\dfrac{d_2}{d_1})^2=\dfrac{\rho_1}{\rho_2}\\\\\dfrac{d_2}{d_1}=\sqrt{\dfrac{\rho_1}{\rho_2}}\\\\\dfrac{d_2}{d_1}=\sqrt{\dfrac{2.65\times 10^{-8}}{1.72\times 10^{-8}}}\\\\=1.24[/tex]
So, the required ratio of the diameter of Aluminum to Copper wire is 1.24.
A 0.75 kg model car is moving west at a speed of 9.0 m/s when it collides head-on with a 2.00 kg model truck that is traveling east at a speed of 10.0 m/s. After the collision, the 0.75 kg model car is now moving east at 11 m/s. What is the speed and direction of the model truck after the collision?
Answers
Answer:
2.5 m/s east
Explanation:
Let east be the positive direction for velocity.
The change in momentum of the 0.75 kg model car is ...
m1·v2 -m1·v1 = (0.75 kg)(11 m/s) -(0.75 kg)(-9 m/s)
= (0.75 kg)(20 m/s) = 15 kg·m/s
The change in momentum of the 2.0 kg model car is the opposite of this, so the total change in momentum is zero.
m2·v2 -m2·v1 = (2 kg)(v2 m/s) -(2 kg)(10 m/s) = 2(v2 -10) kg·m/s
The required relation is ...
15 kg·m/s = -2(v2 -10) kg·m/s
-7.5 = v2 -10 . . . . divide by -2
2.5 = v2 . . . . . . . add 10
The velocity of the model truck after the collision is 2.5 m/s east.
Help me in my hw,A train starts from rest.Its velocity becomes 90km/hr after 1 min,Calculate the acceleration of train and distance covered by the train.Answer it ASAP
Answers
Answer:
I am serious about that
Explanation:
.......
Check if correct or not:
Directions: Using what you learned about energy describe the energy transfer or transformations for each of the items below.
1. Clapping Your Hands:
Kinetic- sound
2. Dropping Your Pencil:
3. The Toaster:
Electric-Thermal/Heat
4. A Cat Lying in a Sunny Window:
Light-Thermal/heat
5. Lifting a Book Over Your Head:
kinetic-potential
6. The Radio:
Electric-sound
Tell me if correct or not
Answers
Answer:
Looks good to me
Explanation:
#2 should probably be turning potential energy to kinetic.
Compare the time it
takes the light to travel from your
teacher to your eye with the time
it takes sound to travel the same
distance.
Answers
Answer:
Light takes less time than sound.
Explanation:
Let's say, the teacher and the student are at a distance "d" from each other.
The medium around them would be air.
And,
The speed of light in air is approx. 3× 10⁸ m/s
while, the speed of sound in air is approx. 330 m/s
We have a formula that establishes the relation between speed, distance and time.
[tex] \boxed{ \mathsf{speed = \frac{distance}{time} }}[/tex]
Our hunt for time — Speed in both the scenarios is known to us whereas the distance is same.
Sound
[tex] \mathsf{330 = \frac{d}{time_{s}} }[/tex]
[tex] \underline{\mathsf{time _{s} = \frac{d}{330} }}[/tex]
Light
[tex] \mathsf{3 \times {10}^{8} = \frac{d}{time _{l} } }[/tex]
[tex] \underline{ \mathsf{ time _{l} = \frac{d}{3 \times {10}^{8}} }}[/tex]
The best way of comparison is finding their ratio.
[tex] \implies \mathsf{\frac{ time_{s}}{time_{l} } = \frac{ \frac{d}{330} }{ \frac{d}{3 \times {10}^{8} } } }[/tex]
simplifying the fraction
[tex] \implies \mathsf{\frac{ time_{s}}{time_{l} } = \frac{d \times (3 \times {10}^{8} )}{330 \times d}}[/tex]
d gets canceled and we're left with the following expression
[tex] \implies \mathsf{\frac{ time_{s}}{time_{l} } = \frac{ (3 \times10 \times {10}^{7} )}{330}}[/tex]
30, being a common factor in the numerator as well as denominator, gets canceled out. and in its place remains 1/ 11
(why?
=> 30÷330 = 1÷11)
[tex] \implies \mathsf{\frac{ time_{s}}{time_{l} } = \frac{ 1\times {10}^{7} }{11}}[/tex]
taking timeₛ to the numerator on the other side.
[tex] \implies \mathsf{time_{s} = \frac{ 1\times {10}^{7} }{11}\times time_{l}}[/tex]
Therefore, we get timeₛ is approx. 10⁶ times the timeₗ.
That's a big difference, no wonder light's way much faster than sound.
As lesser the time taken to cover a distance, faster is the wave.
The sound takes about 874,000 times MORE time than the light takes.
Your friend's Frisbee has become stuck 19 m above the ground in a tree. You want to dislodge the Frisbee by throwing a rock at it. The Frisbee is stuck pretty tight, so you figure the rock needs to be traveling at least 4.1 m/s when it hits the Frisbee.
If you release the rock 1.8 m above the ground, with what minimum speed must you throw it?
Answers
Answer:
18.36 m/s
Explanation:
We can solve this using conservation of energy. The energy in the system will be conserved since there are no outside forces acting upon it so the potential energy and kinetic energy will be equal. Giving us this formula to start:
1/2mv^2=mgh
m=mass
g=gravity
h=height
v=velocity
We can start by figuring out the total height the rock travels which we can do by subtracting the height of the frisbee by the height the rock started at.
19m-1.8m=17.2m
Now we can plug in our variables to solve for velocity.
First we negate mass since its on both sides and cancels out leaving us with.
1/2v^2=gh
Plug in.
1/2v^2=(9.8)(17.2)
1/2v^2=168.56
v^2=337.12
v=18.36m/s
If you are good at activities that require agility, what are you able to do well?
1.maintain your heart rate in the upper ranges
2.quickly change the direction of your movement
3.hold a difficult position for a long period of time
4.strategize about the best way to win a game
Answers
Answer:
2.quickly change the direction of your movement
Explanation:
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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why is it so important that you take care of your nervous system?
Answers
Answer:
The nervous system handles the stress response, which, if overworked, can eventually lead to diseases ranging from high blood pressure to diabetes.
Explanation:
hope I helped
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=
Answers
(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]
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Which of the following is most likely to be a secondary source
Answers
Answer:
analyze, assess or interpret an historical event, era, or phenomenon,.
Explanation:
Secondary sources are works that analyze, assess or interpret an historical event, era, or phenomenon, generally utilizing primary sources to do so. Secondary sources often offer a review or a critique. Secondary sources can include books, journal articles, speeches, reviews, research reports, and more.
During what time period does the balloon drift in a counterclockwise direction?
Answers
please help 9.2.1 project in science just ned an example
Answers
Answer:
Give me what kind of example you need please so I can help you. Put it in the comments.
Explanation:
The diagram below shows a 5.00-kilogram block
at rest on a horizontal, frictionless table.
5.00-kg
block
Table
Which of the following is the correct name and strength of the force holding the block up?
Answers
The name and strength of the force holding the block up is 50 N upward - Normal force.
The given parameters:
Mass of the block, m = 5 kgThe weight of the block acting downwards due to gravity is calculated as follows;
W = mg
where;
g is acceleration due to gravity = 10 m/s²W = 5 x 10
W = 50 N (downwards)
Since the block is at rest, an a force equal to the weight of the block must be acting upwards. This force is known as normal reaction.
Fₙ = 50 N (upwards)
Thus, the name and strength of the force holding the block up is 50 N upward - Normal force.
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The block will remain on the table because the normal force balances with the weight of the block. The correct answer is 50 N upward normal force
From the diagram shown a 5.00-kilogram block at rest on a horizontal, frictionless table. The weight of the block will act downward which will be
Weight W = mg
let g = 10 m/[tex]s^{2}[/tex]
W = 5 x 10
W = 50 N
The block will also produce an equal but in opposite direction of a normal force which is equal to the weight of the block. That is,
Normal force N = 50 N
The block will remain on the table because the normal force balances with the weight of the block.
Therefore, the correct name and strength of the force holding the block up is 50 N upward normal force.
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What is the maximum speed at which a car can round a curve of 25-m radius on a level road if the coefficient of static friction between the tires and road is 0.80?
Answers
I assume the curve is flat and not banked. A car making a turn on the curve has 3 forces acting on it:
• its weight, mg, pulling it downward
• the normal force from contact with the road, n, pushing upward
• static friction, f = µn, directed toward the center of the curve (where µ is the coefficient of static friction)
By Newton's second law, the net forces on the car in either the vertical or horizontal directions are
∑ F (vertical) = n - mg = 0
∑ F (horizontal) = f = ma
where a is the car's centripetal acceleration, given by
a = v ²/r
and where v is the maximum speed you want to find and r = 25 m.
From the first equation, we have n = mg, and so f = µmg. Then in the second equation, we have
µmg = mv ²/r ==> v ² = µgr ==> v = √(µgr )
So the maximum speed at which the car can make the turn without sliding off the road is
v = √(0.80 (9.80 m/s²) (25 m)) = 14 m/s
Which technological device makes an energy conversion in the same way that a human ear makes an energy conversion?
a.) a loudspeaker
b.) a headphone
c.) a light bulb
d.) a microphone
I think it's c because of the concept of mechanical energy to electrical energy but I'm not sure
Answers
Answer:
I THINK C
Explanation:
BECAUSE A Light Emitting Diode (LED) glows even when a weak electric current passes through it.
Two divers, G and H, are at depths 20 m and 40 m respectively
below the water surface in lake. The pressure on G is P, while
the pressure on H is P2 if the atmospheric pressure is equivalent
to 10 m of water, then the value of P2/P1 is.
A. 1.67.
B. 2.00.
C. 0.50.
D. 0.60.
Answers
Answer:
B
Explanation:
P1/P1 = 40/20
=2
A cyclist rides in a circle with speed 8.1 m/s. What is his centripetal
acceleration if the circle has a radius of 27 m?
Answers
Explanation:
We know that the tangent velocity is 8.1 m/s. We also know that the tangent velocity can be written in the following way:
Vt = ωr with ω being the angular velocity.
We now calculate ω:
ω = Vt/r = 8.1 m/s / 27m = 0.3 rad/s
Now that we have ω we can calculate the centripetal aceleration:
a = ω^2 * r = ( 0.3 )^2 * 27 = 2.43 m/s^2
8 N to the left , and 4 N to the right. Find the net force. Is this balanced?
Answers
Explanation:
12N by first law of newton is net force after colloision
The equation below can be used to calculate a change in gravitational potential energy. What units must be used for h? Give the full name, not the abbreviation.
e=m x g x h
Answers
Answer:
h = change in vertical position (height)
has units of distance.
Explanation:
The equation below can be used to calculate a change in gravitational potential energy then the units used for the height would be meters.
What is mechanical energy?
The sum of all the energy in motion (total kinetic energy) and all the energy that is stored in the system (total potential energy) is known as mechanical energy.
As given in the problem, the equation below can be used to calculate a change in gravitational potential energy, the units used for the height of the object would be in meters, which is the SI unit of the length
The gravitational potential energy = mass×acceleration ×height
Thus, the unit of height used in the gravitational potential energy formula would be meter.
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The spaceship Enterprise 1 is moving directly away from earth at a velocity that an earth-based observer measures to be 0.66c. A sister ship, Enterprise 2, is ahead of Enterprise 1 and is also moving directly away from earth along the same line. The velocity of Enterprise 2 relative to Enterprise 1 is 0.34c. What is the velocity of Enterprise 2, as measured by the earth-based observer
Answers
Answer:
The answer is "0.82 c".
Explanation:
Given:
Spacecraft speed 1 is [tex]u = + 0.66 \ c[/tex]
Space velocity 2 relative to spacecraft 1 is [tex]v = + 0.34\ c[/tex]
The spacecraft velocity 2 measured by the Earth observation
[tex]\to u' = \frac{u +v}{1 + ( \frac{uv}{c^2})}[/tex]
[tex]= \frac{0.66 \ c +0.34\ c}{ 1+ (\frac{0.66\ c \times 0.33\ c }{c^2})}\\\\ = \frac{1 \ c }{ 1+ (\frac{0.2178\ c^2 }{c^2})}\\\\ = \frac{1 \ c }{ 1+ (0.2178 )}\\\\ = \frac{1 \ c }{ 1.2178 }\\\\=0.82\ c[/tex]
