g as measured from the earth, a spacecraft is moving at speed .80c toward a second spacecraft moving at speed .60c back toward the first spacecraft. What is the speed of the first spacecraft as viewed from the second spacecraft

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

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.

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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a particle of violet light has exactly the same energy as a particle of red light

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]

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

Answer:

hydrogen

Explanation:

hope it helps bye.

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%

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

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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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].

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

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.

Explanation:

thankyou fo

r the poimts

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].

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

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.

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.

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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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]

Answer:

I think the particles transferred between brandi's body and the door knob causing the shock is the electrons.

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.

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.

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

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.

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

Answer:

C

Explanation:

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