Answer:
the speed of the first spacecraft as viewed from the second spacecraft is 0.95c
Explanation:
Given that;
speed of the first spacecraft from earth v[tex]_a[/tex] = 0.80c
speed of the second spacecraft from earth v[tex]_b[/tex] = -0.60 c
Using the formula for relative motion in relativistic mechanics
u' = ( v[tex]_a[/tex] - v[tex]_b[/tex] ) / ( 1 - (v[tex]_b[/tex]v[tex]_a[/tex] / c²) )
we substitute
u' = ( 0.80c - ( -0.60c) ) / ( 1 - ( ( 0.80c × -0.60c) / c² ) )
u' = ( 0.80c + 0.60c ) / ( 1 - ( -0.48c² / c² ) )
u' = 1.4c / ( 1 - ( -0.48 ) )
u' = 1.4c / ( 1 + 0.48 )
u' = 1.4c / 1.48
u' = 0.9459c ≈ 0.95c { two decimal places }
Therefore, the speed of the first spacecraft as viewed from the second spacecraft is 0.95c
what is the magnitude of an electric field (in 106 n/c) that balances the weight of a plastic sphere of mass 2.1 g that has been charged to 3.0 nc
Answer:
[tex]E=6.86\times 10^6\ N/C[/tex]
Explanation:
Given that,
Mass of the sphere, m = 2.1 g = 0.0021 kg
Charge, q = 3 nC
We need to find the magnitude of the electric field that balanced the weight of sphere. Let it is E. So,
qE = mg
[tex]E=\dfrac{mg}{q}[/tex]
Put all the values,
[tex]E=\dfrac{0.0021\times 9.8}{3\times 10^{-9}}\\\\E=6.86\times 10^6\ N/C[/tex]
So, the magnitude of the elecric field is [tex]6.86\times 10^6\ N/C[/tex].
If a 5.0 kg box is pulled simultaneously by a 10.0 N force and a 5.0 N force, then its acceleration must be Group of answer choices 1.0 m/s2. We cannot tell from the information given. 2.2 m/s2. 3.0 m/s2.
Answer:
We cannot tell from the information given
Explanation:
Given;
mass of the box, m = 5 kg
first force, F₁ = 10 N
second force, F₂ = 5 N
(I) Assuming the two forces are acting horizontally in opposite direction, the resultant force on the box is calculated as;
∑Fx = 10 N - 5 N
= 5 N
Apply Newton's second law of motion;
∑Fx = ma
a = ∑Fx/m
a = 5 / 5
a = 1 m/s² in the direction of the 10 N force.
(II) Also, if the two forces are acting in the same direction, the resultant force is calculated as;
∑Fx = 10 N + 5 N
∑Fx = 15 N
a = 15 / 5
a = 3 m/s²
Therefore, the information given is not enough to determine the acceleration of the box.
which one of the following is not a simple machine
1 ladder
2 wheel Barrow
3 pulley
4 electric pole
Engineers are working on a design for a cylindrical space habitation with a diameter of 7.50 km and length of 29.0 km. The habitation will simulate gravity by rotating along its axis. With what speed (in rad/s) should the habitation rotate so that the acceleration on its inner curved walls equals 8 times Earth's gravity
Answer:
The speed will be "0.144 rad/s".
Explanation:
Given that,
Diameter,
d = 7.50 km
Radius,
R = [tex]\frac{7.5}{2} \ Km[/tex]
Acceleration on inner curve,
= 8 times
Now,
As we know,
⇒ [tex]\omega^2R=8g[/tex]
or,
⇒ [tex]\omega=\sqrt{\frac{8g}{R} }[/tex]
On substituting the values, we get
⇒ [tex]=\sqrt{\frac{8\times 9.8}{\frac{7.5}{2}\times 10^3 } }[/tex]
⇒ [tex]=\sqrt{\frac{78.4}{3750} }[/tex]
⇒ [tex]=\sqrt{0.0209}[/tex]
⇒ [tex]=0.144 \ rad/s[/tex]
Plants get their food by which of the following ways?
A. Making it from sunlight
B.Eating other plants
C. Absorbing it from the soil
D.Taking nutrients from the air
Answer:
C
Explanation:
A man who works for a moving company is loading a box onto a moving van. He pushes a 200N box up a 5m long ramp. If he pushes with a force of 60N and the ramp is 1m high, what is the efficiency of the inclined plane
Answer:
η = 0.667 = 66.7%
Explanation:
The efficiency of the man can be given by the following formula:
η = output/input
where,
η = efficiency of man = ?
output = potential energy gain of the box = Wh
input = work done by man = Fd
Therefore,
[tex]\eta = \frac{Wh}{Fd}[/tex]
where,
W = weight of box = 200 N
h = height gained by box = 1 m
F = force exerted by man = 60 N
d = length of ramp = 5 m
Therefore,
[tex]\eta = \frac{(200\ N)(1\ m)}{(60\ N)(5\ m)}[/tex]
η = 0.667 = 66.7%
Which image shows an example of potential energy?
Answer:
D
Explanation:
Potential energy involves the change of an object's position, which in this case a rocket is increasing its vertical displacement from the ground.
When a rocket is increasing its vertical displacement from the ground, it exhibits both potential and kinetic energy. Therefore option D is correct.
At the initial stage, when the rocket is on the ground and not moving, it possesses potential energy. This potential energy is in the form of stored energy due to its elevated position above the ground.
As the rocket launches and gains altitude, it continues to accumulate potential energy because it is moving higher against the force of gravity.
Simultaneously, as the rocket moves upward, it also gains kinetic energy. Kinetic energy is the energy associated with the rocket's motion.
The faster the rocket moves, the greater its kinetic energy becomes. As the rocket ascends, its speed increases, resulting in an increase in kinetic energy.
Therefore, in the context of a rocket increasing its vertical displacement from the ground, both potential energy (due to its height) and kinetic energy (due to its motion) are present.
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Consider an electron confined in a region of nuclear dimensions (about 5 fm). Find its minimumpossible kinetic energy in MeV. Treat this problem as one-dimensional, and use the relativistic relationbetweenEandp. Give your answer to 2 significant figures. (The large value you will find is a strongargument against the presence of electrons inside nuclei, since no known mechanism could contain anelectron with this much energy.)
Answer:
39.40 MeV
Explanation:
Determine the minimum possible Kinetic energy
width of region = 5 fm
From Heisenberg's uncertainty relation below
ΔxΔp ≥ h/2 , where : 2Δx = 5fm , Δpc = hc/2Δx = 39.4 MeV
when we apply this values using the relativistic energy-momentum relation
E^2 = ( mc^2)^2 + ( pc )^2 = 39.4 MeV ( right answer ) because the energy grows quadratically in nonrelativistic approximation,
Also in a nuclear confinement ( E, P >> mc )
while The large value will portray a Non-relativistic limit as calculated below
K = h^2 / 2ma^2 = 1.52 GeV
A certain 20-A circuit breaker trips when the current in it equals 20 A. What is the maximum number of 100-W light bulbs you can connect in parallel in an ideal 120-V dc circuit without tripping this circuit breaker
Answer: 28
Explanation:
Given
Circuit breaker current is [tex]I=20\ A[/tex]
Power of the light bulb is [tex]P=100\ W[/tex]
Voltage of the DC-circuit is [tex]V=120\ V[/tex]
If the resistance are connected in parallel, they must have same voltage i.e. 120 V
So, Resistance is given by
[tex]\Rightarrow R=\dfrac{V^2}{P}\\\\\Rightarrow R=\dfrac{120^2}{100}\\\\\Rightarrow R=144\ \Omega[/tex]
For the 20 A current and 120 V battery, net resistance is
[tex]\Rightarrow R_{net}=\dfrac{120}{20}\\\\\Rightarrow R_{net}=6\ \Omega[/tex]
Suppose there are n resistance in the circuit connected in parallel.
[tex]\Rightarrow \dfrac{144}{n}=R_{net}\\\\\Rightarrow n=\dfrac{144}{6}\\\\\Rightarrow n=28.8\approx 28\ \text{for current to be less than 20A}[/tex]
Thus, there can maximum of 28 bulbs.
In principle, when you fire a rifle, the recoil should push you backward. How big a push will it give
Answer:
The answer is " because it is in the opposite direction of the bullet".
Explanation:
Given:
[tex]m_1= 70 \ kg\\\\m_2 = 0.01\ kg\\\\v_2 = 500\ \frac{m}{s}\\\\v_1 = ?[/tex]
Using formula:
It recoils the velocity of the rifle [tex]= -0.07142 \ \ \frac{m}{s}[/tex]
Its negative sign displays the opposite direction of the rifle.
A balloon contains 0.04m3 of air at a pressure of 1.20 x 105Pa. Calculate the pressure required to reduce its volume to 0.025m3 at constant temperature.
Answer:
you can simply answer p1v1=p2v2 p2 = p1v1÷ p2 then inter what given in the formula you get it pl=1.20x 105pa vl= 0.40m3 v2=0.025m3
After the big bang started the universe 13.78 billion years ago, the universe began to rapidly expand. This rapid expansion caused the universe to cool
down into a network of gaseous clouds composed mainly of which element?
O Oxygen
Hydrogen
O Neon
O Carbon
Answer:
hydrogen
Explanation:
hope it helps bye.
In an experiment, a student brings up the rotational speed of a piece of laboratory apparatus to 24 rpm. She then allows the apparatus to slow down uniformly on its own, and counts 236 revolutions before the apparatus comes to a stop. The moment of inertia of the apparatus is known to be 0.076 kg m2. What is the magnitude of the torque on the apparatus
Answer:
T = 6.43 x 10⁻⁵ N.m
Explanation:
First, we will calculate the deceleration of the apparatus by using the third equation of motion:
[tex]2\alpha \theta = \omega_f^2-\omega_i^2[/tex]
where,
α = angular decelration = ?
θ = angular displacement = (236 rev)(2π rad/rev) = 1482.83 rad
ωi = initial angular speed = (24 rpm)(2π rad/1 rev)(1 min/ 60 s) = 2.51 rad/s
ωf = final angular speed = 0 rad/s
Therefore,
[tex]2\alpha(1482.83\ rad) = (0\ rad/s)^2-(2.51\ rad/s)^2\\\\\alpha = -\frac{(2.51\ rad/s)^2}{2965.66\ rad} \\\\\alpha = - 8.46\ x\ 10^{-4}\ rad/s^2[/tex]
negative sign shows deceleration
Now, for torque:
T = Iα
where,
T = Torque = ?
I = moment of inertia = 0.076 kg.m²
Therefore,
T = (0.076 kg.m²)(8.46 x 10⁻⁴ N.m)
T = 6.43 x 10⁻⁵ N.m
3. A skater moves across the ice a distance of 12 m before a constant frictional force of 15 N cause him to stop. His initial speed is 2.2 m/s . Calculate the skater's mass.
Explanation:
thankyou fo
r the poimts
Suppose you are pushing a 3 kg box with a force of 25 N (directed parallel to the ground) over a distance of 15 m. Afterward, the box (initially at rest) is traveling at a speed of 6 m/s. How much work (in Joules) did friction do in this process
Answer: 321 J
Explanation:
Given
Mass of the box [tex]m=3\ kg[/tex]
Force applied is [tex]F=25\ N[/tex]
Displacement of the box is [tex]s=15\ m[/tex]
Velocity acquired by the box is [tex]v=6\ m/s[/tex]
acceleration associated with it is [tex]a=\dfrac{F}{m}[/tex]
[tex]\Rightarrow a=\dfrac{25}{3}\ m/s^2[/tex]
Work done by force is [tex]W=F\cdot s[/tex]
[tex]W=25\times 15\\W=375\ J[/tex]
change in kinetic energy is [tex]\Delta K[/tex]
[tex]\Rightarrow \Delta K=\dfrac{1}{2}m(v^2-0)\\\\\Rightarrow \Delta K=\dfrac{1}{2}\times 3\times 6^2\\\\\Rightarrow \Delta K=\dfrac{1}{2}\times 3\times 36\\\\\Rightarrow \Delta K=54\ J[/tex]
According to work-energy theorem, work done by all the forces is equal to the change in the kinetic energy
[tex]\Rightarrow W+W_f=\Delta K\quad [W_f=\text{Work done by friction}]\\\\\Rightarrow 375+W_f=54\\\Rightarrow W_f=-321\ J[/tex]
Therefore, the magnitude of work done by friction is [tex]321\ J[/tex]
If two dogs are pulling a bone with force-20Newtons in opposite direction, then the resultant force is
Answer:
Newtons third law of motion: Balanced forces
Every action has a corresponding and opposing response, according to Newton's third law of motion. As a result, forces always work in pairs. Once more, tug-of-war is a prime illustration.
What force in opposite direction follow newton law?The third law of motion by Newton states that equal, but diametrically opposed forces always act in pairs. There is an equal but opposite reaction to every action, to put it another way.
The forces are balanced if the pullers are exerting equal force but going in the opposite direction on either side of the rope. There is hence no motion.
Although equal and opposite in nature, action and reaction forces cannot be balanced since they act on separate things and do not cancel one another out.
Therefore, This means that when you push against a wall, the wall pushes back against you with an equal amount of force.
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HELP ME PLEASE !!!!!!!!!!!!
Answer:
D
Explanation:
Because the y axis is meter. If it is straight line at time and meter graph then it velocity and speed is 0
POR FA ME PODRIAN AYUDAR. URGENTE!!!!
¿Cuál es la energía cinética de un 0,01kg bala viajando a una velocidad de 700Sra?
Answer:
K.E = 2450 Joules
Explanation:
Given the following data;
Mass = 0.01 kg
Velocity = 700 m/s
To find the kinetic energy;
La energía cinética (K.E) se puede definir como una energía que posee un objeto o cuerpo debido a su movimiento.
Matemáticamente, la energía cinética viene dada por la fórmula;
[tex] K.E = \frac{1}{2}mv^{2}[/tex]
Sustituyendo en la fórmula, tenemos;
K.E = ½ * 0.01 * 700²
K.E = 0.005 * 490000
K.E = 2450 Joules
Which statements are true of noble gases?
Check all that apply.
A. They are metalloids.
B. Their valence shells are full of electrons.
C. They are not very reactive.
D. All of the noble gases have at least two electron shielding layers.
Please helppppppp!!!!
Answer:
I think the particles transferred between brandi's body and the door knob causing the shock is the electrons.
MCQs. How does the heat from the Sun reach the Earth.(options)a: conduction b: convertion c: radiation d: combustion.
the organelle involved in cell secretion is
Answer:
Golgi apparatus (bodies).
Explanation:
A cell can be defined as the fundamental or basic functional, structural and smallest unit of life for all living organisms. Some living organisms are unicellular while others are multicellular in nature.
Generally, cells have the ability to independently replicate themselves. In a cell, the "workers" that perform various functions or tasks for the survival of the living organism are referred to as organelles. Some examples of cell organelles found in all living organisms such as trees, birds, and bacteria include; nucleus, cytoplasm, cell membrane, golgi apparatus, mitochondria, lysosomes, ribosomes, chromosomes, endoplasmic reticulum, vesicles, etc.
Golgi apparatus is also referred to as Golgi bodies and it functions as a packaging unit in living organisms, especially eukaryotic cells because it prepares protein and lipid molecules for export by chemically tagging them.
Hence, the Golgi apparatus (bodies) is an organelle involved in cell secretion through the transportation (export) of proteins and lipids out of a cell.
In Case 1, a mass M hangs from a vertical spring having spring constant k and is at rest in its equilibrium position. In Case 2 the mass has been lifted a distance D vertically upward. If we define the potential energy in Case 1 to be zero, what is the potential energy of Case 2
Answer: hello your question is incomplete attached below is the complete question
answer : 1/2 KD^2 ( option A )
Explanation:
P.E ( potential energy ) = mgd
In case 1 P.E = 0 i.e. mgd = 0
Given that in case 2 the Mass M had moved through the Distance D by the compression of the spring
The potential energy of the M in case 2
= P.E of M at rest + P.E of the spring
= 0 + 1/2 KD^2
The electric field of a negative infinite line of charge: Group of answer choices Points perpendicularly away from the line of charge and decreases in strength at larger distances from the line charge Points parallel to the line of charge and decreases in strength at larger distances from the line charge Points parallel to the line of charge and increases in strength at larger distances from the line charge Points perpendicularly away from the line of charge and increases in strength at larger distances from the line charge Points perpendicularly toward the line of charge and increases in strength at larger distances from the line charge Points perpendicularly toward the line of charge and decreases in strength at larger distances from the line charge
Answer:
Points perpendicularly toward the line of charge and decreases in strength at larger distances from the line charge
Explanation:
The electric field for a uniform line of charge is given by E = λ/2πε₀r where λ = charge density and r = distance from line of charge.
If λ is negative, E is negative so it points in the negative direction towards the line of charge.
Also, since for negative charges, electric field lines end up in them, the electric field for an infinitely long negative line of charge points towards the charge perpendicular to it.
Also as r increases, E decreases since E ∝ 1/r
So, the electric field decreases at larger distances from the line of charge.
So, the electric field of a negative infinite line of charge Points perpendicularly toward the line of charge and decreases in strength at larger distances from the line charge.
tell me about Orion nebula
Keesha is looking at a beetle with a magnifying glass. She wants to see an upright, enlarged image at a distance of 25 cm. The focal length of the magnifying glass is +5.0 cm. Assume that Keesha's eye is close to the magnifying glass.
(a) What should be the distance between the magnifying glass and the beetle?
(b) What is the angular magnification?
Answer:
a) p = 4.167 cm, b) m = + 6
Explanation:
a) For this exercise we must use the equation of the constructor
[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 case the distance to the image q = 25 cm and the focal length is f = 5.0 cm
Since the object and its image are on the same side of the lens, the distance to the image by the sign convention must be negative.
[tex]\frac{1}{p } = \frac{1}{f} - \frac{1}{q}[/tex]
[tex]\frac{1}{p} = \frac{1}{5} - \frac{1}{-25}[/tex]
[tex]\frac{1}{ p}[/tex] = 024
p = 4.167 cm
b) angular magnification
m = h ’/ h = - q / p
m = - (-25) /4.167
m = + 6
the positive sign indicates that the image is straight and enlarged
A body initially at rest is accelerated at a constant rate for 5.0 seconds in the positive x direction. If the final speed of the body is 20.0 m/s, what was the body's acceleration?
Answer:
[tex]a=4\ m/s^2[/tex]
Explanation:
Given that,
Initial speed of a body, u = 0
Final speed of the body, v = 20 m/s
Time, t = 5 s
We need to find the acceleration of the body. We know that the acceleration of an object is equal to the rate of change of velocity divided by time taken. So,
[tex]a=\dfrac{v-u}{t}\\\\a=\dfrac{20-0}{5}\\\\a=4\ m/s^2[/tex]
So, the body's acceleration is equal to [tex]4\ m/s^2[/tex].
What is the speed acquired by a freely falling object 4 seconds after being dropped from a rest position? Use units of meter per second (m/s) and assume acceleration from gravity is 10 m/s2.
speed = 40 m/s
Explanation:
Since the object is dropped, V0y = 0.
Vy = V0y - gt
= -(10 m/s^2)(4 s)
= -40 m/s
This means that its velocity is 40 m/s downwards. Its speed is simply 40 m/s.
The speed acquired by a freely falling object 4 seconds after being dropped from a rest position would be 40 meters/seconds.
What are the three equations of motion?There are three equations of motion given by Newton
The first equation is given as follows
v = u + at
the second equation is given as follows
S = ut + 1/2×a×t²
the third equation is given as follows
v² - u² = 2×a×s
Keep in mind that these calculations only apply to uniform acceleration.
As given in the problem, we have to find the speed acquired by a freely falling object 4 seconds after being dropped from a rest position,
By using the first equation of motion,
v = u + at
initial velocity(u) = 0 m/s
acceleration(a) = 10 m/s²
v = 0 + 10×4
v = 40 meters/seconds
Thus, the speed acquired by a freely falling object 4 seconds after being dropped from a rest position would be 40 meters/seconds.
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Which statement is true?
a particle of violet light has less energy than a particle of red light
a particle of violet light has more energy than a particle of red light
a particle of violet light has exactly the same energy as a particle of red light
particles of light do not have any energy, regardless of what color the light is
a particle of violet light has exactly the same energy as a particle of red light
An atom has 20 protons and 22 neutrons and 18 electrons. The charge of this atom is: ________
Answer:
the number of electrons should equal to the the number of protons in a neutral atom
if there is a inequality between the numbers it means the atom has a + or - charge
The charge of this atom=+(20-18)=+2