Answer:
D
Explanation:
the energy is coming up and down
The answer is:
D) The kinetic energy of the car
In an attempt to reduce the extraordinarily long travel times for voyaging to distant stars, some people have suggested traveling at close to the speed of light. Suppose you wish to visit the red giant star Betelgeuse, which is 430 lyly away, and that you want your 20,000 kgkg rocket to move so fast that you age only 36 years during the round trip.
A. How fast (v) must the rocket travel relative to earth?
B. How much energy is needed to accelerate the rocket to this speed?
C. How many times larger is this energy than the total energy used by the United States in the year 2000, which was roughly 1.0 x 10^20 J?
Answer:
a) [tex]v=0.999124c[/tex]
b) [tex]E=7.566*10^{22}[/tex]
c) [tex]E_a=760 times\ larger[/tex]
Explanation:
From the question we are told that
Distance to Betelgeuse [tex]d_b=430ly[/tex]
Mass of Rocket [tex]M_r=20000[/tex]
Total Time in years traveled [tex]T_d=36years[/tex]
Total energy used by the United States in the year 2000 [tex]E_{2000}=1.0*10^20[/tex]
Generally the equation of speed of rocket v mathematically given by
[tex]v=\frac{2d}{\triangle t}[/tex]
[tex]v=860ly/ \triangle t[/tex]
where
[tex]\triangle t=\frac{\triangle t'}{(\sqrt{1-860/ \triangle t)^2}}[/tex]
[tex]\triangle t=\frac{36}{(\sqrt{1-860/ \triangle t)^2}}[/tex]
[tex]\triangle t=\sqrt{(860)^2+(36)^2}[/tex]
[tex]\triangle t=860.7532[/tex]
Therefore
[tex]v=\frac{860ly}{ 860.7532}[/tex]
[tex]v=0.999124c[/tex]
b)
Generally the equation of the energy E required to attain prior speed mathematically given by
[tex]E=\frac{1}{\sqrt{1-(v/c)^2} }-1(20000kg)(3*10^8m/s)^2[/tex]
[tex]E=7.566*10^{22}[/tex]
c)Generally the equation of the energy [tex]E_a[/tex] required to accelerate the rocket mathematically given by
[tex]E_a=\frac{E}{E_{2000}}[/tex]
[tex]E_a=\frac{7.566*10^{22}}{1.0*10^{20}}[/tex]
[tex]E_a=760 times\ larger[/tex]
PLEASE HELP! URGENT
Explain how the forces need to change so an aeroplane can land
Answer:
it changes by taking the air from below the plane and curving it to the top causing draw wich slows it down then the weight that pulls it down to land.
please put me brainliest
Explanation:
How did the projected storm track change from image A TO B TO C TO D
An electromagnetic wave has a frequency of 1.0 x 10^5. What is the wavelength of the wave?
Answer:
The frequency , speed and wavelength of an electromagnetic wave are related by the formula
Speed = frequency x wavelength
frequency = speed / wavelength
substituting the values
frequency = 3 x #10 ^8# m /s / 1 x #10^15# m
= 3 x #10^-7# /s
The wavelength of the wave will be 3000 m. The relationship between the wave's wavelength, time, and speed is used to solve the problem.
What is wavelength?The distance between two successive troughs or crests is known as the wavelength. The peak of the wave is the highest point, while the trough is the lowest.
The given data in the problem is;
[tex]\lambda[/tex] is the wave's wavelength
c denotes the wave's speed=3×10⁸ m/secv
v is the frequency = 1 ×10⁵ Hz
The relationship between the wave's wavelength, time, and speed is given as
[tex]\rm \lambda = \frac{c}{v} \\\\ \rm \lambda = \frac{3 \times 10^8}{1.0 \times 10^5} \\\\ \lambda=30000 \ m[/tex]
Hence the wavelength of the wave will be 3000 m.
To learn more about the wavelength refer to the link;
brainly.com/question/7143261
The two uniform, slender rods B1and B2, each of mass 2kg, are pinned together at P, and then B1is suspended from a pin at O. (This arrangement is called a double pendulum.) The counterclockwise couple C0, having moment 150????⋅m, is applied to B2beginning at t=0. Find the angular accelerations of B1and B2upon application of the couple, and the force exerted on B2at P.
Answer:
hello the diagram relating to this question is attached below
a) angular accelerations : B1 = 180 rad/sec, B2 = 1080 rad/sec
b) Force exerted on B2 at P = 39.2 N
Explanation:
Given data:
Co = 150 N-m ,
a) Determine the angular accelerations of B1 and B2 when couple is applied
at point P ; Co = I* ∝B2'
150 = ( (2*0.5^2) / 3 ) * ∝B2
∴ ∝B2' = 900 rad/sec
hence angular acceleration of B2 = ∝B2' + ∝B1 = 900 + 180 = 1080 rad/sec
at point 0 ; Co = Inet * ∝B1
150 = [ (2*0.5^2) / 3 + (2*0.5^2) / 3 + (2*0.5^2) ] * ∝B1
∴ ∝B1 = 180 rad/sec
hence angular acceleration of B1 = 180 rad/sec
b) Determine the force exerted on B2 at P
T2 = mB1g + T1 -------- ( 1 )
where ; T1 = mB2g ( at point p )
= 2 * 9.81 = 19.6 N
back to equation 1
T2 = (2 * 9.8 ) + 19.6 = 39.2 N
The best leaper in the animal kingdom is the puma, which can jump to a height of 3.7m when leaving the ground at an angle of 45 degrees. with what speed must the animal leave the ground to reach that height?
Answer:
v = 12 m/s
Long, boring, and convoluted explanation:
First, let's lay out our information.
- max height = 3.7 m
- 0 = 45°
- gravitational acceleration constant = 9.8 [tex]\frac{m}{s^2}[/tex]
Since the puma leaves the ground at a 45 ° angle, its motion will follow a curved path as seen in many projectile motion problems, where the object is being influenced solely by the force of gravity. And because the puma leaves the ground at an angle, its initial velocity is broken down into its horizontal and vertical components. We were also told, though indirectly, that the max height is 3.7m because the puma can reach up to that height. Gravity is always given to be 9.8 [tex]\frac{m}{s^2}[/tex]
Because we are dealing with maximum height and gravity, we have to use the vertical component of the velocity, vsin ( θ ) , and not the horizontal component, vcos ( θ ) .
Given its max height, the acceleration due to gravity, and the angle, we can now solve for the speed at which the puma leaves the ground using the following equation: vsin ( θ ) = [tex]\sqrt{2hg}[/tex]
Where vsin ( θ ) is the vertical component of the initial velocity and h and g are max height and gravitational acceleration constant respectively.
Plugin, rearrange and solve
v sin ( θ ) = [tex]\sqrt{2hg}[/tex]
v sin ( 45 ∘ ) = √ 2 × 3.7 × 9.8
v ( 0.71 ) = [tex]\sqrt{72.52}[/tex]
v ( 0.71 ) = 8.52
v = 8.52 /0.71
v = 12 m s
A house brick has a volume of 1900 cm³ and a weight in air of 80N.What is its apparent weight in water?The density of water is 1.00 g cm³?
Answer:
61 N
Explanation:
We'll begin by calculating the mass of the brick when placed in water. This can be obtained as follow:
Volume of brick = 1900 cm³
Density of water = 1 g/cm³
Mass of brick in water =…?
Density = mass / volume
1 = mass of brick in water / 1900
Cross multiply
Mass of brick in water = 1 × 1900
Mass of brick in water = 1900 g
Next, we shall convert 1900 g to Kg.
1000 g = 1 Kg
Therefore,
1900 g = 1900 g × 1 Kg / 1000 g
1900 g = 1.9 Kg
Next, we shall determine the weight in water. This can be obtained as follow:
Mass (m) = 1.9 Kg
Acceleration due to gravity (g) = 10 m/s²
Weight (W) =?
W = m × g
W = 1.9 × 10
W = 19 N
Thus, the weight of the brick in water is 19 N.
Finally, we shall determine the apparent weight of the brick in water. This can be obtained as follow:
Weight in air = 80 N
Weight in water = 19 N
Apparent weight =?
Apparent weight = weight in air – weight in water
Apparent weight = 80 – 19
Apparent weight = 61 N
pls help giving brainliest!
• Explain how Ohm's Law is like a river.
o What represents Voltage?
o What represents Current?
o What represents Resistance?
Answer:
Pressure of water = Voltage
Flow rate of water = Current
Obstructions present in the river = Resistance
Explanation:
We can describe Ohm's Law by using the river analogy,
The flow of water from a mountain to lower parts can be considered as change from high potential to a lower potential.The flow of water can also be considered as flow of electrons, that is current in a circuitThere are also obstructions in rivers that can be considered as resistance in an electric circuit.Some fundamental particles such as quarks (and some leptons, such as the electron) have an electric charge and use photons as mediating particles to communicate with other particles with electric charge. Similarly, quarks, identified by their color charge, use a mediating particle to interact with other particles with color charge, that is, other quarks. What is the particle that mediates interactions among quarks
Answer: hello the options to your question is missing below are the missing
options :
the W boson the gluon the Z boson the tau particleanswer : The Gluon ( option 2 )
Explanation:
The particle that mediates the interactions between quarks is known as the Gluon.
Gluon is known as an elementary particle that acts as the exchange particle between quarks. It's function is similar to the activity involving the exchange of photons in the electromagnetic force involving two charged particles
A graph of a wave is shown if the frequency f is 8.0 Hz what is the waves speed
Answer:
C
[tex]6 = \frac{3}{2} \lambda \\ wavelength : \: \lambda = 4 \: m \\ from \: wave \: equation \\ v = f\lambda \\ v = 8 \times 4 \\ v = 24 \: {ms}^{ - 1} [/tex]
The speed of the waves will be = [tex]24\dfrac{m}{s}[/tex]
Given data is frequency f = 8.0 Hz
What will be the speed of the wave?As we know that the speed of the wave is given by the formula below
[tex]V=f\times \lambda[/tex]
Since
[tex]6=2 \lambda[/tex]
[tex]\lambda =3m[/tex]
now
[tex]V=8\times3[/tex]
[tex]V=24m[/tex]
The speed of the waves will be = [tex]24\dfrac{m}{s}[/tex]
To know more about waves speed follow
https://brainly.com/question/13761336
A person is dragging a packing crate of mass 74.9 kg across a rough horizontal floor where the coefficient of kinetic friction is 0.35. He exerts a force F at and angle 43.0 degrees above the horizontal. What is the Force F such that the crate moves at a constant speed
Answer:
351.28 N
Explanation:
Let F be the force on the object and f be the frictional force. The component of the force acting in the horizontal direction causing the object to move is FcosФ where Ф is the angle between F and the horizontal = 43.0°. The frictional force on the packing crate f = μN where μ = coefficient of kinetic friction = 0.35 and N = normal force = W = weight of the packing crate = mg where m = mass of crate = 74.9 kg and g = acceleration due to gravity = 9.8 m/s². So, f = μN = μW = μmg
So, the net force on the packing crate is
FcosФ - f = ma
FcosФ - μmg = ma
Since the crate moves at constant speed, its acceleration a = 0
So, FcosФ - μmg = ma
FcosФ - μmg = m(0)
FcosФ - μmg = 0
FcosФ = μmg
F = μmg/cosФ
Substituting the values of the variables into the equation, we have
F = μmg/cosФ
F = 0.35 × 74.9 kg × 9.8 m/s²/cos43.0°
F = 256.907 kg-m/s²/0.73135
F = 351.28 kg-m/s²
F = 351.28 N
Discuss how planck's hypothesis explain the observed blackbody spectrum.
Answer:
Planck's radiation law, a mathematical relationship formulated in 1900 by German physicist Max Planck to explain the spectral-energy distribution of radiation emitted by a blackbody (a hypothetical body that completely absorbs all radiant energy falling upon it, reaches some equilibrium temperature, and then reemits
Explanation:
An astronomy class is so excited by the discovery of planets around other stars that they decide to do a library exhibit on the subject so that everyone in the school can learn about it. In this exhibit they want to pay tribute to both the astronomers of today who have done the work AND some of the scientists of the past whose work was essential to making the discoveries possible (and directly related to the techniques involved). Which of the following scientists of the past should definitely be included in the exhibit?
George Herbig Ejnar Hertzsprung Ptolemy
Gerard Kuiper Christian Doppler
Answer:
Christian Doppler
Explanation:
The Scientist with the most significant contribution to the discovery of planets around other stars is Christian Doppler and his work that made this discovery possible is the Principle of DOPPLER EFFECT
Christian Doppler was an Austrian scientist and physicist whose principle Doppler effect explained how observed frequency of light and sound waves are affected by a relative motion of both the source and detector
Suppose that you'd like to find out if a distant star is moving relative to the earth. The star is much too far away to detect any change in its brightness as it moves toward or away from the earth. Instead we can use the Doppler effect to determine its relative speed. For this problem we are going to look at the spectral lines from hydrogen, specifically the one with a wavelength of 656.46 nm.
Required:
What is V(star) , the speed at which the star is moving relative to the earth?
Answer:
The speed will be "18km/s". A further explanation is given below.
Explanation:
According to the question, the values are:
Wavelength,
[tex]\lambda = 656.46 \ nm[/tex]
[tex]\Delta \lambda = 0.04[/tex]
[tex]c=3\times 10^8[/tex]
As we know,
⇒ [tex]\frac{\Delta \lambda}{\lambda} =\frac{v}{c}[/tex]
On substituting the values, we get
⇒ [tex]\frac{656.46}{0.04} =\frac{v}{3\times 10^8}[/tex]
⇒ [tex]v=\frac{656.46}{0.04} (3\times 10^8)[/tex]
⇒ [tex]=16411.5\times 3\times 10^8[/tex]
⇒ [tex]=18280 \ m/s[/tex]
or,
⇒ [tex]=18 \ km/s[/tex]
The speed at which the star is moving relative to the earth is 18 km/s
Calculation of the speed:Since the wavelength is 656.46 nm.
And, Δλ=0.04nm
Here Wavelength should represent the distance between the same points in the cycles of a signal propagated in space or it should be along a wire.
Now the speed should be
[tex]= 0.04 \div 656.46 \times 3 \times 10^8[/tex]
= 1820m/s
= 18 km.s
Therefore, The speed at which the star is moving relative to the earth is 18 km/s
learn more about speed here: https://brainly.com/question/24692367
Your spaceship lands on a moon of a small planet that orbits a distant star. As you initially circled the moon, you measured its diameter to be 5480 km. After land you observe that a simple pendulum that had a frequency of 3.50 Hz on Earth now has a frequency of 1.82 Hz. What is the mass of the moon
Answer:
[tex]m=3*10^2^3kg[/tex]
Explanation:
From the question we are told that:
Moon diameter [tex]D_m=5480Km \approx 2740000m[/tex]
Earth's frequency [tex]F_e=3.50Hz[/tex]
Planet's frequency [tex]F_p=1.80Hz[/tex]
Generally the equation for Frequency and Gravity relationship is mathematically given by
[tex]\frac{F_p}{F_e}=\sqrt{ \frac{g_p}{g_e} }[/tex]
Therefore gravity of Moon is given as
[tex]g_p= g_e*(\frac{f_p}{f_e} )^2[/tex]
[tex]g_p= 9.8*(\frac{1.82}{3.50} )^2[/tex]
[tex]g_p=2.64992m/s^2[/tex]
Generally the equation for mass of spherical body is mathematically given by
[tex]m=\frac{g_pD_m^2}{G}[/tex]
where G= gravitational constant
[tex]m=\frac{2.64992m/s^2*(2740000m)^2}{6.67*10^{-11}}[/tex]
[tex]m=2.98268956*10^2^3kg[/tex]
[tex]m=3*10^2^3kg[/tex]
The light waves transfer their oscillations and energy to other object in what way?
The frequency of the light waves physically causes the object to move
The energy of the light wave transfers to the electrons of the material which causes them to gain that energy
The wavelength of the light changes the velocity of the atoms in the material
The amplitude of the light will affect how much of it shines on the object
Answer: b the energy of light...
Explanation:
Rudy plans to conduct an experiment using three rosebushes of the same variety and size. His hypothesis is that the plant
receiving the most sunlight at the end of three months will show the most growth. In his plan below, he has listed how much
water, fertilizer, and sunlight each plant will receive. Why is his experiment flawed?
Answer: B) He has too many independent variables.
Explanation: Nothing in the experiment is a constant or being measured.
Tendons are, essentially, elastic cords stretched between two fixed ends; as such, they can support standing waves. These resonances can be undesirable. The Achilles tendon connects the heel with a muscle in the calf. A woman has a 20-cm long tendon with a cross-section area of 130 mm^2. The density of tendon tissue is 1100 kg/m^3.
Required:
For a reasonable tension of 600 , what will be the fundamental resonant frequency of her Achilles tendon?
Answer:
161.938 Hz
Explanation:
the computation of the fundamental resonant frequency is shown below
p = 1100 kg/m^3
A = 130 mm^2
= 130 ×10^-6 m^2
T = 600 N
L = 20 cm
= 0.2 m
Now the linear density of tendon is
= 1100 kg/m^3 × 130 ×10^-6
= 0.143 kg/m
Now the wave of the string is
= √600 ÷ √0.143
= 64.775 m/s
Now finally the fundamental resonant frequency is
= 64.775 ÷ (2 × 0.2)
=161.938 Hz
PLS HELP, WILL GIVE BRAINLIEST!!!
A piece of clay with a mass of 350 g is dropped to the floor from a height of 1.5 m (A). Calculate the KE and PE of the clay at
a. (A = 1.5 m).
b. (B = 70 cm), and
c. (C = 0 m, just before hitting the ground). d. Make an energy graph at each point
Answer:
b bro it's b bro
what part of the rocket reaches space
Answer:
7
Explanation:
There is really no explanation needed
A capacitor is constructed of two large, identical, parallel metal plates separated by a small distance d
a. True
b. False
When measuring speed of sound in air, speed of light = 4f(length of column of air + 0.3 * diameter of tube).
Why is multiplied by 4?
Where does the 4 come from?
Answer:
the length of the column in air
Sheryl always expects special treatment wherever she goes because, after all, she is better than everyone else. She could care less how others feel and routinely expects people to worship her. These attributes are typical of ______.
Explanation:
it attributes to Being pride of herself
What is the velocity of an object that travels 200 meters in 20 seconds?
Answer:
10m/s
Explanation:
velocity=displacement/time
=200/20
=10m/s
On the graph of voltage versus current, how do the resistors compare?
- Is the kinetic energy conserved in the collision between
the ball and the pendulum?
Answer:
No, Momentum is conserved.
Explanation:
The ballistic pendulum is a device in which a projectile such as a bullet is fired into a suspended heavy wooden stationary block. Some kinetic energy gets transformed into heat and sound, and some is used to deform the block. However, momentum is conserved.
Please help 23 also 29 the answer choices are reflection or refraction your fraction absorption
Answer:
23. Option B. Hertz
29. Refraction.
Explanation:
23. Determination of the unit of measurement of frequency.
Frequency is simply defined as the number of oscillation made in one second. Mathematically, frequency can be represented as:
Frequency = 1/period
f = 1/T
Period is measured in seconds.
Thus, the unit of frequency becomes
f = 1/T
f = 1/s = s¯¹ (Hertz)
Therefore, the unit of frequency is Hertz.
29. When a wave enters a new medium, the speed of the wave is uttered. This leads to the bending of the wave. When this occurs, we can say refraction has taken place.
How does density affect refraction
Answer:
please give me brainlist and follow
Explanation:
The angle and wavelength at which the light enters a substance and the density of that substance determine how much the light is refracted. ... When light passes from a more dense to a less dense substance, (for example passing from water into air), the light is refracted (or bent) away from the normal.
Current I flows along the positive z-direction in the inner conductor of a long coaxial cable and returns through the outer conductor. The inner conductor has radius a, and the inner and outer radii of the outer conductor are b and c, respectively. (a) Determine the magnetic field in each of the following regions: 0 ≤ r ≤ a, a ≤ r ≤ b, b ≤ r ≤ c, and r ≥ c. (b) Plot the magnitude of H as a function of r over the range from r = 0 to r = 10 cm, given that I = 10 A, a = 2 cm, b = 4 cm, and c = 5 c
Answer:
A) determine magnetic fields
For 0 ≤ r ≤ a
Magnetic field = ∅ [tex]\frac{rI}{2\pi a^2}[/tex]
For a ≤ r ≤ b
Magnetic field = ∅ [tex]\frac{I}{2\pi r}[/tex]
For b ≤ r ≤ c
Magnetic field in the region = ∅ [tex]\frac{I}{2\pi r} [ c^2 - r^2 / c^2-b^2 ][/tex]
For r ≥ c
magnetic filed in the region = 0
B ) attached below
Explanation:
A) Determine the magnetic field in the following regions
i) For 0 ≤ r ≤ a
Magnetic field = ∅ [tex]\frac{rI}{2\pi a^2}[/tex]
attached below is the detailed solution
ii) For a ≤ r ≤ b
Magnetic field = ∅ [tex]\frac{I}{2\pi r}[/tex]
attached below is the detailed solution
iii) For b ≤ r ≤ c
Magnetic field in the region = ∅ [tex]\frac{I}{2\pi r} [ c^2 - r^2 / c^2-b^2 ][/tex]
attached below is the detailed solution
iv) For r ≥ c
magnetic filed in the region = 0 and this is because the net current enclosed in the region = 0
Aluminum has a shear strength of 210 megapascals. When you bend aluminum foil around an edge (i.e., the edge of the box) and pull, you are effectively applying a shear force along the bent edge of the foil. If a roll of household aluminum foil is 30.0 centimeters wide and its thickness is approximately 15.0 micrometers, how much shear force is needed to pull off a sheet
Answer:
The answer is "[tex]945\ N[/tex]"
Explanation:
Aluminum has 210 megapascals of tensile resistance. They choose a shear force only at bent foil edge to bend aluminum foil over an edge (that is the edge of its box) to pull them. When a roll of aluminium domestic foil is 30 cm in width and about 15.0 micrometers.
[tex]\to 0.000015 \times 0.300 = 0.0000045\\\\\to 210000000 = \frac{F}{0.0000045}\\\\\to F=210000000 \times 0.0000045\\\\\to F = 945\ N[/tex]