The final velocity of the cannon ball when it strikes the earth is 43.36 m/s.
The given parameters:
Height of the hill, h = 50 mMass of the cannon, m = 5 kgVelocity of the ball, v = 30 m/sThe final velocity of the cannon ball when it strikes the earth is calculated by applying the principle of conservation of energy as follows;
[tex]P.E_i + K.E_i = P.E_f + K.E_f\\\\mgh_i + \frac{1}{2} mv_i^2 = mgh_f + \frac{1}{2} mv_f^2\\\\gh_i + \frac{1}{2} v_i^2 = g(0) + \frac{1}{2} v_f^2\\\\2gh_i + v_i^2 = v_f^2\\\\v_f = \sqrt{2gh_i + v_i^2 } \\\\v_f = \sqrt{(2 \times 9.8 \times 50) \ \ + \ \ 30^2} \\\\v_f = 43.36 \ m/s[/tex]
Thus, the final velocity of the cannon ball when it strikes the earth is 43.36 m/s.
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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
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]
Four identical balls are thrown from the top of a cliff, each with the same speed. The
first is thrown straight up, the second is thrown at 30° above the horizontal, the third
at 30° below the horizontal, and the fourth straight down. How do the speeds and
kinetic energies of the balls compare as they strike the ground? Ignore the effects of
air resistance. Explain fully using the concepts from this unit.
The comparison of the speeds and kinetic energy of the identical balls are as follows
The speed and the kinetic energy of the first and fourth ball are equal, while the speed and kinetic energy of the second and third balls are equal
The reason for the above comparison results areas follows;
Known parameters;
First ball is thrown straight up
Second ball is thrown 30° above the horizontal
Third ball it thrown 30° below the horizontal
The fourth ball is thrown straight down
Unknown:
Comparison of the speed and kinetic energy of the four balls
Method:
The kinetic energy, K.E. = (1/2) × m × v²
The velocity of the ball, v = u × sin(θ)
Where;
u = The initial velocity of the ball
θ = The reference angle
For the first ball thrown straight up, we have;
θ = 90°
∴ [tex]v_y[/tex] = u
The final velocity of the ball as it strikes the ground is v₂ = u² + 2gh
Where;
h = The height of the cliff
∴ Kinetic energy of first ball, K.E.₁ = (1/2) × m × (u₁² + 2gh)²
For the second ball thrown 30° to the horizontal, we have;
K.E. = (1/2) × m × ((u×sin30)² + 2·g·h)² = K.E. = (1/2) × m × ((0.5·u)² + 2·g·h)²
Kinetic energy K.E.₂ = (1/2) × m × ((0.5·u₂)² + 2·g·h)²
For the third ball thrown at 30° below the horizontal, we have;K.E. = (1/2) × m × ((u×sin30)² + 2·g·h)² = K.E. = (1/2) × m × ((0.5·u)² + 2·g·h)²
Kinetic energy K.E.₃ = (1/2) × m × ((0.5·u₃)² + 2·g·h)²
For the fourth ball thrown straight down, we have;Kinetic energy K.E.₄ = (1/2) × m × (u₄² + 2gh)²
Therefore, as the ball strike the ground, the speed and the kinetic energy of the first and fourth ball are equal, while the speed and kinetic energy of the second and third balls are equal
u₁ = u₄, K.E₁ = K.E.₄, u₂ = u₃, K.E₂ = K.E.₃
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A rectangular loop of wire with sides 0.129 and 0.402 m lies in a plane perpendicular to a constant magnetic field (see part a of the drawing). The magnetic field has a magnitude of 0.888 T and is directed parallel to the normal of the loop's surface. In a time of 0.172 s, one-half of the loop is then folded back onto the other half, as indicated in part b of the drawing. Determine the magnitude of the average emf induced in the loop.
Answer:
[tex]0.2677\ \text{V/m}[/tex]
Explanation:
A = Area of loop = [tex]0.129\times0.402[/tex]
B = Magnetic field = [tex]0.888\ \text{T}[/tex]
t = Time taken = [tex]0.172\ \text{s}[/tex]
Electric field is given by
[tex]E=B\dfrac{dA}{dt}\\\Rightarrow E=0.888\times\dfrac{0.129\times 0.402}{0.172}\\\Rightarrow E=0.2677\ \text{V/m}[/tex]
The emf induced is [tex]0.2677\ \text{V/m}[/tex].
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
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.
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
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.
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)
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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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
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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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?
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.
8 N to the left , and 4 N to the right. Find the net force. Is this balanced?
Explanation:
12N by first law of newton is net force after colloision
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?
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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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:__________
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]
Choose one. 5 points
Use the equation from week 3:
frequency =
wavespeed
wavelength
and the wavelength you found in #3 to calculate the frequency of this photon (remember the speed of
light is 3E8 m/s);
7.6E14 Hz
6.0E14 Hz
4,6E14 Hz
The frequency is 4,6E14 Hz.
What is the frequency?
Frequency is the fee at which modern changes direction in step with 2nd. it's far measured in hertz (Hz), a worldwide unit of degree wherein 1 hertz is identical to 1 cycle in line with 2d. Hertz (Hz) = One hertz is the same as 1 cycle in step with the second. Cycle = One entire wave of alternating present-day voltage.
Frequency describes the number of waves that pass a hard and fast place in a given quantity of time. So if the time it takes for a wave to skip is half of 2d, the frequency is 2 per 2nd. If it takes 1/one hundred of an hour, the frequency is a hundred in step with hour.
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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
Answer:
I THINK C
Explanation:
BECAUSE A Light Emitting Diode (LED) glows even when a weak electric current passes through it.
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]
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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.
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.
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.
Answer:
B
Explanation:
P1/P1 = 40/20
=2
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
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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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
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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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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Which of the following is most likely to be a secondary source
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.
Identify the type of chemical reaction:
CaCO3 -->CaO + CO2
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.
15 . A scientist who studies the whole environment as a working unit .
Botanist
Chemist
Ecologist
Entomologist
Answer:
Ecologist.
Your answer is Ecologist.
(Ecologist) is a scientist who studies the whole environment as a working unit.
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?
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
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?
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
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.
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.
Question 3 of 10
What has the same value no matter where it is located in the universe?
A. Volume
B. Weight
C. Mass
D. Density
Reset Selection
Answer:
C. Mass
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
Answer:
Looks good to me
Explanation:
#2 should probably be turning potential energy to kinetic.
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
Answer:
2.quickly change the direction of your movement
Explanation:
A hot-air balloon plus cargo has a mass of 308 kg and a volume of 2910 m3 on a day when the outside air density is 1.22 kg/m3. The balloon is floating at a constant height of 9.14 m above the ground.
Required:
What is the density of the hot air in the balloon?
9514 1404 393
Answer:
1.114 kg/m³
Explanation:
The total mass of the air in the balloon and the balloon + cargo will be the mass of the displaced air. If d is the density of the air in the balloon, then we have ...
2910d +308 = 2910×1.22
Solving for d, we find ...
2910d = 2919(1.22) -308
d = 1.22 -308/2910
d ≈ 1.114 . . . kg/m³
The density of the hot air is about 1.114 kg/m³.
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
Answer:
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Explanation:
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