Two manned satellites approach one another at a relative velocity of v=0.190 m/s, intending to dock. The first has a mass of m1=4.00×103 kg and the second a mass of m2=7.50×103 kg. If the two satellites collide elastically rather than dock, what is their final relative velocity?

Answers

Answer 1

Answer:

Their final relative velocity is 0.190 m/s

Explanation:

The relative velocity of the satellites, v = 0.190 m/s

The mass of the first satellite, m₁ = 4.00 × 10³ kg

The mass of the second satellite, m₂ = 7.50 × 10³ kg

Given that the satellites have elastic collision, we have;

[tex]v_2 = \dfrac{2 \cdot m_1}{m_1 + m_2} \cdot u_1 - \dfrac{m_1 - m_2}{m_1 + m_2} \cdot u_2[/tex]

[tex]v_2 = \dfrac{ m_1 - m_2}{m_1 + m_2} \cdot u_1 + \dfrac{2 \cdot m_2}{m_1 + m_2} \cdot u_2[/tex]

Given that the initial velocities are equal in magnitude, we have;

u₁ = u₂ = v/2

u₁ = u₂ = 0.190 m/s/2 = 0.095 m/s

v₁ and v₂ = The final velocities of the satellites

We get;

[tex]v_1 = \dfrac{2 \times 4.0 \times 10^3}{4.0 \times 10^3 + 7.50 \times 10^3} \times 0.095 - \dfrac{4.0 \times 10^3- 7.50\times 10^3}{4.0 \times 10^3+ 7.50\times 10^3} \times 0.095 = 0.095[/tex]

[tex]v_2 = \dfrac{ 4.0 \times 10^3 - 7.50\times 10^3}{4.0 \times 10^3 + 7.50 \times 10^3} \times 0.095 + \dfrac{2 \times 7.50\times 10^3}{4.0 \times 10^3+ 7.50\times 10^3} \times 0.095 = 0.095[/tex]

The final relative velocity of the satellite, [tex]v_f[/tex] = v₁ + v₂

∴ [tex]v_f[/tex] = 0.095 + 0.095 = 0.190

The final relative velocity of the satellite, [tex]v_f[/tex] = 0.190 m/s


Related Questions

15millas a km alguien pliss para ahorita porfa lo sigo

Answers

Answer:

X = 24.135 kilometres

Explanation:

Given the following data;

Distance = 15 miles

To convert the value in miles to kilometers;

Conversion:

1 mile = 1.609 kilometres

15 miles = X kilometres

Cross-multiplying, we have;

X = 1.609 * 15

X = 24.135 kilometres

to all the physicians please help this is for my assignment

Answers

Answer:

Q. 1. Newton's Law of gravitation states that all bodies in the universe exerts a force of attraction on all other bodies in the universe with a proportional force to both the product of the masses of the bodies and inversely proportional to the square of the distance between their centers

Mathematically, we have;

[tex]F = G \times \dfrac{m_1 \times m_2}{R^2}[/tex]

Where;

m₁, and m₂ are the masses of the bodies

R = The distance between their centers

G = The gravitational constant = 6.6743 × 10⁻¹¹ N·m²/kg²

The gravitational constant, G, is the Newton's law of gravitation's constant of proportionality between the force of attraction that exist two bodies and the product of their masses divided by the square of the distance between their centers

Q. 2. Newton's law of gravitation in vector form is presented as follows;

[tex]\underset{F_{12}}{\rightarrow} = -G \times \dfrac{m_1 \times m_2}{R_{21}^2} \cdot \hat R_{12}[/tex]

The above equation gives the gravitational force of attraction of body 1 on body 2, with the negative sign and unit vector indicating that the force of of gravity is towards body 1

The force of gravity of body 2 on 1 is presented as follows;

[tex]\underset{F_{12}}{\rightarrow} = -G \times \dfrac{m_1 \times m_2}{R_{12}^2} \cdot \hat R_{21}[/tex]

The gravitational force of attraction of body 2 on body 1 is therefore, equal in magnitude and opposite in direction of the gravitational force of body 1 on body 2 (towards body 2)

[tex]-\underset{F_{12}}{\rightarrow} = G \times \dfrac{m_1 \times m_2}{R_{21}^2} \cdot \hat R_{12} = G \times \dfrac{m_1 \times m_2}{R_{21}^2} \cdot -(\hat R_{21}) = -G \times \dfrac{m_1 \times m_2}{R_{21}^2} \cdot \hat R_{21}[/tex]

[tex]-\underset{F_{12}}{\rightarrow} = -G \times \dfrac{m_1 \times m_2}{R_{21}^2} \cdot \hat R_{21} = \underset{F_{21}}{\rightarrow}[/tex]

[tex]-\underset{F_{12}}{\rightarrow} = \underset{F_{21}}{\rightarrow}[/tex]

Explanation:

Reference frame definitely changes when also changes

Answers

Reference frame definitely changes when the body is changing. That is the reason that in order to describe the position of a point that moves relative to a body that is moving relative to the Earth, it is usually convenient to use a reference frame attached to the moving body.

What unit is used in MKS system and FPS system​

Answers

The "second" is the base unit of time in both systems.

1. A bicycle initially moving with a velocity
5.0 m s-1 accelerates for 5 s at a rate of 2 m s? Wh
will be its final velocity ?

Answers

Answer:

[tex]\boxed {\boxed {\sf 15 \ m/s \ or \ 15 \ m*s^{-1}}}[/tex]

Explanation:

We are asked to find the final velocity. We are given the acceleration, time, and initial velocity, so we can use the following kinematics formula.

[tex]v_f= v_i+ at[/tex]

In this formula, [tex]v_f[/tex] is the final velocity, [tex]v_i[/tex] is the initial velocity, [tex]a[/tex] is the acceleration, and [tex]t[/tex] is the time.

The bicycle has an initial velocity of 5.0 m *s⁻¹ or m/s, acceleration of 2 m/s², and a time of 5 seconds.

[tex]\bullet \ v_i = 5.0 \ m/s \\\bullet \ a= 2\ m/s^2\\\bullet \ t= 5 \ s[/tex]

Substitute the values into the formula.

[tex]v_f=5.0 \ m/s + ( 2\ m/s^2 * 5 \ s)[/tex]

Solve inside the parentheses.

[tex]\frac {2 \ m}{s^2}* 5 \ s = \frac{ 2 \ m}{s} * 5 = \frac{ 10 \ m}{s} = 10 \ m/s[/tex]

[tex]v_f= 5.0 \ m/s + (10 \ m/s)[/tex]

Add.

[tex]v_f= 15 \ m/s[/tex]

The units can also be written as:

[tex]v_f= 15 \ m*s^{-1}[/tex]

The bicycle's final velocity is 15 meters per second.

What is the connection of H ions at a ph=2?

Answers

Answer:

Explanation:

High concentrations of hydrogen ions yield a low pH (acidic substances), whereas low levels of hydrogen ions result in a high pH (basic substances). The overall concentration of hydrogen ions is inversely related to its pH and can be measured on the pH scale

The mass of objects is 4kg and it has a density of 5gcm^-3. what is the volume ​

Answers

Answer:

4kg×5gm^3=60

Explanation:

the object if heavy

can uh help in in this question step by step​

Answers

Initial velocity=u=72km/h

Convert to m/s

[tex]\\ \sf \longmapsto 72\times \dfrac{5}{18}=5(4)=20m/s[/tex]

Final velocity=v=0m/sTime=2s=t

[tex]\\ \sf \longmapsto Acceleration=\dfrac{v-u}{t}[/tex]

[tex]\\ \sf \longmapsto Acceleration=\dfrac{0-20}{2}[/tex]

[tex]\\ \sf \longmapsto Acceleration=\dfrac{-20}{2}[/tex]

[tex]\\ \sf \longmapsto Acceleration=a=-10m/s^2[/tex]

Distance be s

Using second equation of kinematics

[tex]\\ \sf \longmapsto s=ut+\dfrac{1}{2}at^2[/tex]

[tex]\\ \sf \longmapsto s=20(2)+\dfrac{1}{2}(-10)(2)^2[/tex]

[tex]\\ \sf \longmapsto s=40+(-20)[/tex]

[tex]\\ \sf \longmapsto s=40-20[/tex]

[tex]\\ \sf \longmapsto s=20m[/tex]

Now

Mass=m=5000kg

Using newtons second law

[tex]\\ \sf \longmapsto Force=ma[/tex]

[tex]\\ \sf \longmapsto Force=5000(-10)[/tex]

[tex]\\ \sf \longmapsto Force=-50000N[/tex]

Force is in opposite direction so its negative

[tex]\\ \sf \longmapsto Force=50kN[/tex]

Answer . The acceleration of the truck is 10m/[tex]s^{2}[/tex], and the distance covered is 40 m. Have attached the picture for solution.

Hope that helps.

the moon revolves around the earth in a nearly circular orbit kept by gravitational force exerted by the earth work done will be

Answers

Answer:

Zero because the applied force is perpendicular to the motion of the object.

No work is done on an object moving is a circular path about a central attractive force.

Any work done in such a case would result in a change in the orbit.

A 250–g piece of gold is at 19 °C. 5.192 kJ of energy is added to it by heat. The specific heat of gold is 129 J/(kg·°C). Calculate its final temperature.




We heat a 25–g sample of metal from 10 °C to 100 °C. 1.082 kJ of energy is added to it by heat. Calculate
the specific heat of the metal.

Answers

Answer:

A. DT is given by Q= MCs DT

m = mass of the substances

Cs= is it's specific heat capacity

Ck= Q

Mk ×DTk

=250 × 9 × 5

129

=Dt = 180.1085271

answer is 180degree C.

Explanation:

B. = 25×10 ×100

1.082

=2500

1.082

= 23105.360 g/kj.

The final temperature is 180 degree. and the specific heat of the metal is 23105.360 g/kj.

How to calculate the specific heat?

Q = m . C . ΔT

Q = heat; m = mass; C is the specific heat and

ΔT = Final T° - Initial T°

Q = C lat . m

Q = Heat

m = mass

C lar = Latent heat of fusion

A) DT is given by Q= M Cs DT

where, m = mass of the substances

Cs= is it's specific heat capacity

Ck= Q

Mk × DTk

=250 × 9 × 5

129 =Dt = 180.1085271

Thus, the final temperature is 180 degree.

B) We heat a 25–g sample of metal from 10 °C to 100 °C. 1.082 kJ of energy is added to it by heat = 25×10 ×100

=2500

1.082

Q = 23105.360 g/kj

Hence, the specific heat of the metal is 23105.360 g/kj.

Learn more about heat here;

https://brainly.com/question/12909121

#SPJ2

an object is sliding down in clean plane the velocity change at a constant rate from 10 cm to 15 CM in 2 second what is it acceleration ?​

Answers

Initial velocity=10m/s=u

Final velocity=v=15m/s

Time=t=2s

[tex]\boxed{\sf Acceleration=\dfrac{v-u}{t}}[/tex]

[tex]\\ \sf\longmapsto Acceleration=\dfrac{15-10}{2}[/tex]

[tex]\\ \sf\longmapsto Acceleration=\dfrac{5}{2}[/tex]

[tex]\\ \sf\longmapsto Acceleration=2.5m/s^2[/tex]

Answer: a = 2.5 cm/s²

Explanation:

Acceleration = (Final velocity - Initial Velocity)/time taken

a = v-u/t

Initial velocity = 10 cm/s

Final velocity = 15 cm/s

Time = 2 seconds

a = (15-10)/2

a = 5/2

a = 2.5 cm/s²

Therefore the acceleration is 2.5 cm/s²

please click thanks and mark brainliest if you like :)

A comet of mass 2 × 10^8 kg is pulled toward the star. If the comet's initial velocity is very small, and the comet starts moving toward the star from 700,000,000 km away, how fast is it going right before it hits the surface of the star? (Assume that it does not lose any mass by melting as it approaches the star.)

Answers

Answer:

The speed of the comet at the surface of the star is approximately 1,208,694.7 m/s

Explanation:

Question parameter obtained online; The mass of the star, M = 5 × 10³¹ kg

Explanation;

The given mass of the comet, m = 2 × 10⁸ kg

The initial velocity of the comet, v → 0

The distance of the comet from the star, d = 700,000,000 km

The gravitational potential at d = G·M·m/d

The kinetic energy of the comet, K.E. = m·v²/2

The kinetic energy of the comet at d = m·(0)²/2 = 0

The gravitational potential at the surface of the star, R = G·M·m/R

The kinetic energy of the comet at the surface of the star, R = m·(v)²/2 = 0

Where;

M = The mass of the star = 5 × 10³¹ kg

[tex]M_{Sun}[/tex] = The mass of the Sun = 1.989 × 10³⁰ kg

M/[tex]M_{Sun}[/tex] = 5 × 10³¹/(1.989 × 10³⁰) ≈ 25

G = The universal gravitational constant = 6.67430 × 10⁻¹¹ N·m²/kg²

R = The radius of the star

Therefore, we have;

m·(0)²/2 - G·M·m/d = m·v²/2 - G·M·m/R

∴ v = √((G·M·m/R - G·M·m/d)×2/m) = √(2·G·M(1/R - 1/d))

Therefore; v = (2 × 6.67430 × 10⁻¹¹ × 5 × 10³¹ × (1/R - 1/700,000,000,000))

v = 81696389149.1×√(1/R - 1/700,000,000,000).

The speed of the comet at the surface of the star, v = 81696389149.1×√(1/R - 1/700,000,000,000)

The mass radius relationship is given as follows;

[tex]\dfrac{R}{R_{Sun}} = 1.30 \times \left(\dfrac{M}{M_{Sun}} \right)^{\dfrac{1}{2} }[/tex]

[tex]R = R_{Sun} \times 1.30 \times \left(\dfrac{M}{M_{Sun}} \right)^{\dfrac{1}{2} }[/tex]

The radius of the Sun = 696,340,000 M

∴ R ≈ 696,340,000 × 1.3 × √(25.14) = 4538865694.76

R = 4538865694.76 m

v = 81696389149.1×√(1/4538865694.76 - 1/700,000,000,000) ≈ 1208694.7  m/s

10. Match the following varibles to their relationship in Newton's 2nd Law. Questions 1. Force and Acceleration 2. Mass and Acceleration 3. Speed and Distance Answer Choices A. Direct Relationship B. Inverse Relationship C. Not in Newton's 2nd Law​

Answers

Explanation:

based on the above information

1.A

2.B

3. C

A CROW BAR WITH LENGTH 200 CM IS USED TO LIFT A LOAD OF 600N . IF THE DISTANCE BETWEEN FULCRUM AND LOAD IS 0.75. CALCULATE ; a, effort b, MA c, VR

Answers

Answer:

a. Effort = 960 Newton

b. Mechanical advantage (M.A) = 0.625

c. Velocity ratio (V.R) = 1.67

Explanation:

Given the following data;

Load = 600 NLength of crowbar = 200 cmLength of load arm = 0.75 m

Conversion:

100 cm = 1 m

X cm = 0.75 m

Cross-multiplying, we have;

X = 0.75 * 100 = 75 cm

First of all, we would find the effort arm;

Effort arm = length of crow bar - length of load arm

Effort arm = 200 - 75

Effort arm = 125 cm

Next, we would determine the mechanical advantage (M.A) of the crow bar;

[tex] M.A = \frac {Effort \; arm}{Load \; arm} [/tex]

Substituting the values into the formula, we have;

[tex] M.A = \frac {125}{200} [/tex]

M.A = 0.625

To find the effort of the crow bar;

[tex] M.A = \frac {Load}{Effort} [/tex]

Making "effort" the subject of formula, we have;

[tex] Effort = \frac {Load}{M.A} [/tex]

[tex] Effort = \frac {600}{0.625} [/tex]

Effort = 960 Newton

Lastly, we would determine the velocity ratio (V.R);

[tex] V.R = \frac {length \; of \; effort \; arm}{length \; of \; load \; arm} [/tex]

[tex] V.R = \frac {125}{75} [/tex]

V.R = 1.67

round off 20.96 to 3 significant figures. a.20.9 b.20 c.21.0 d.21​

Answers

Answer:

option c. 21.0

Explanation:

It was given that to find 3 significant figures. So the answer is 21.0

12 x cos 50 = ?

Does anyone have the answer ? I forgot my my calculator.

Answers

12 x cos 50 = 7.713451316...

the answer is 7.713451316

what ia measurement in science?
= The process of comparing an unknown quantities with an standard known quantities is called measurement.​

Answers

Yes it is the measurement in science

pls help me asap with this​

Answers

Answer:

a) cos30=adj/hyp

cos30= horizontal force/10

horizontal force= 8.66 N

rest of a is in the picture.

b) i believe you can continue.

Many people believe that if the human race continues to use energy as we are now, without change, we'll witness a significant worldwide environmental impact in this century. Research this topic and discuss this possibility. Include concrete examples of specific environmental consequences of global warming.

Answers

Answer:

It is correct to say that if the human race continues to use energy as it is now, without change, we will witness negative environmental impacts around the world in this century.

As a concrete example, we can cite the means of transport that use fossil fuels, such as cars and buses, which release polluting gases into the atmospheric layer and cause the greenhouse effect, contributing to global warming.

To solve these problems, it is necessary to raise the awareness of individuals, so that there is more and more interest and search for environmentally responsible solutions, such as the large-scale production of electric cars, which do not pollute the environment.

Sort the processes based on the type of energy transfer they involve. condensation freezing deposition sublimation evaporation melting thermal energy added thermal energy removed

Answers

Answer:

condensation - thermal energy removed

freezing -thermal energy removed

deposition - thermal energy removed

sublimation - thermal energy added

evaporation - thermal energy added

melting - thermal energy added

Explanation:

Thermal energy is heat energy. Processes in which heat is added involve the addition of thermal energy while processes in which heat energy is removed involves removal of thermal energy.

Condensation involves a change from gas to liquid, freezing involves a change from liquid to solid while deposition involves the settling of mobile particles at a place. All these processes involve a decrease in energy of particles.

On the other hand, sublimation is a direct change from solid to gas, melting involves a change from solid to liquid while evaporation involves a change from liquid to gas. All these processes occur when energy is added to the particles in a system.

Answer:

condensation - thermal energy removed

freezing -thermal energy removed

deposition - thermal energy removed

sublimation - thermal energy added

evaporation - thermal energy added

melting - thermal energy added

5. a. Answer the following questions. What is density? Write a formula by showing the relation among density mass and volume.​

Answers

Answer:

Density is how compact something is. The relationship is M/V=D (Mass divided by Volume equals Density).

Explanation:

WHAT IS DENSITY:

Density is the degree of compactness of a substance.

EXAMPLE:

"a reduction in bone density"

FORMULA OF DENSITY:

The formula for density is d = M/V, where d is density, M is mass, and V is volume.

Answer the following questions. 3 A student runs 2 m/s. What does this mean?

Answers

Answer:

2ms-¹ means that the body under consideration moves 2m in a second, and may be it will continue to move 2m in every 1 second, if there's no external unbalanced force acting on that body (those forces do include frictional forces). mark its brainlist plz. Kaneppeleqw and 6 more users found this answer helpful. Thanks 3.

Answer:

that the student has travels 2 meters every 1 second that passes

sl unit of upthrust and SI unit of pressure​

Answers

Answer:

The SI unit of upthrust is Newton(N).

The SI unit of preesure is Pascal(P).

Thank You

can anyone help me to explain theory of relativity??? ​

Answers

Answer:

The theory of relativity usually encompasses two interrelated theories by Albert Einstein: special relativity and general relativity, proposed and published in 1905 and 1915, respectively. Special relativity applies to all physical phenomena in the absence of gravity. General relativity explains the law of gravitation and its relation to other forces of nature.It applies to the cosmological and astrophysical realm, including astronomy.

The theory transformed theoretical physics and astronomy during the 20th century, superseding a 200-year-old theory of mechanics created primarily by Isaac Newton. It introduced concepts including spacetime as a unified entity of space and time, relativity of simultaneity, kinematic and gravitational time dilation, and length contraction. In the field of physics, relativity improved the science of elementary particles and their fundamental interactions, along with ushering in the nuclear age. With relativity, cosmology and astrophysics predicted extraordinary astronomical phenomena such as neutron stars, black holes, and gravitational waves

Draw a wave that has a wavelength of 3 cm and an amplitude of 1 cm. Label the wavelength, the amplitude, the rest position, and the crest and trough of your wave.

Answers

Answer:

Please find attached, the required wave drawn with MS Excel

Explanation:

Functions that represent waves is given as follows

A general form of the wave equation is A·sin(B·x) + D

Where;

B = 2·π/T

T = The period of the wave = 1/f

D = The vertical shift of the wave = 0

A = The amplitude of the wave = 1 for sine wave

v = The wave velocity

λ = The wavelength of the wave

f = The frequency of the wave

v = f·λ

At constant v, λ ∝ 1/f  

∴ λ ∝ T

Where T = 3, we have;

B = 2·π/T

∴ B = 2·π/3

Therefore, we have the wave with an amplitude of 1 cm, and wavelength, 3 cm, given as follows

y = sin((2·π/3)·x)

Plotting the above wave with MS Excel, we can get the attached wave

Why are road accidents at high speeds very much worse than road accidents at low speeds?

Answers

Answer:

The momentum makes it worse.

Explanation:

The momentum of vehicles running at faster speeds is very high and causes a lot of damage to the vehicles.

Vehicles with high speeds have high momenta (plural of momentum). When they come to rest in very small time they experience high forces
we know that rate of change of momentum is force, lower the time higher is force. So, slow moving vehicles experience less force. It is important to note that if any one vehicle involving has high speed it can do more damage both the vehicles involving.

Using your Periodic Table, which of the elements below is most likely to be a solid at room temperature?
A.) potassium, B.) Hydrogen, C.) Neon, D.) Chlorine

Answers

The answer is definitely Potassium

giving me the points are enough

Answers

Answer:

the product of mass and velocity

....in my syllabus

this is physics practical

Answers

Answer:

well done buddy

Explanation:

Distance travelled by a free falling object in the first second is: a) 4.9m b) 9.8m c) 19.6m d) 10m​

Answers

Time=1s=tAcceleration due to gravity=g=9.8m/s^2Distance=s

In free fall

[tex]\boxed{\sf s=-\dfrac{1}{2}gt^2}[/tex]

[tex]\\ \sf\longmapsto s=-\dfrac{1}{2}\times 9.8(1)^2[/tex]

[tex]\\ \sf\longmapsto s=-4.9(1)[/tex]

[tex]\\ \sf\longmapsto s=-4.9m[/tex]

Take it positive

[tex]\\ \sf\longmapsto s=4.9m[/tex]

Option a is correct
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