1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Zepler [3.9K]
3 years ago
9

A particle moving along a straight line is subjected to a deceleration ???? = (−2???? 3 ) m/???? 2 , where v is in m/s. If it ha

s a velocity v = 10 m/s and a position s = 15 m when t = 0, determine its velocity and position when t = 25 s.
Physics
1 answer:
dlinn [17]3 years ago
3 0

Answer:

(a). The velocity is 0.099 m/s.

(b). The position is 19.75 m.

Explanation:

Given that,

The deceleration is

a=(-2v^3)\ m/s^2

We need to calculate the velocity at t = 25 s

The acceleration is the first derivative of velocity of the particle.

a=\dfrac{dv}{dt}

\dfrac{dv}{dt}=-2v^3

\dfrac{dv}{-v^3}=2dt

On integrating

int{\dfrac{dv}{-v^3}}=\int{2dt}

\dfrac{1}{2v^2}=2t+C

v^2=\dfrac{1}{4t+2C}....(I)

At t = 0, v = 10 m/s

10^2=\dfrac{1}{4\times0+2C}

C=\dfrac{1}{200}

Put the value of C in equation (I)

v^2=\dfrac{1}{4\times25+2\times\dfrac{1}{200}}

v=\sqrt{\dfrac{1}{4\times25+2\times\dfrac{1}{200}}}

v=0.099\ m/s

The velocity is 0.099 m/s.

(b). We need to calculate the position at t = 25 sec

The velocity is the first derivative of position of the particle.

\dfrac{ds}{dt}=v

On integrating

\int{ds}=\int(\sqrt{\dfrac{200}{800t+1}})dt

s=\dfrac{\sqrt{200}\times2\sqrt{800t+1}}{800}+C'

At t = 0, s = 15 m

15=\dfrac{200}{800}+C'

C'=15-\dfrac{200}{800}

C'=14.75

Put the value in the equation

s=\dfrac{\sqrt{200}\times2\sqrt{800\times25+1}}{800}+14.75

s=19.75\ m

The position is 19.75 m.

Hence, (a). The velocity is 0.099 m/s.

(b). The position is 19.75 m.

You might be interested in
A planet’s gravity is related to which of the following? (Points : 5)
katovenus [111]

-- <span>The gravitational force that you feel when you stand on the surface
of a planet depends on the planet's mass and size.  It has </span><span><span>nothing
to do with the planet's orbit.  (</span>Of course,"size" is also related to the
planet's mass, density, and surface area.)

-- One possible cause of deforestation is the removal of trees without
adequate replanting.

-- According to Hubble’s Law, the farther away a galaxy is, the faster
it is moving away from us

-- Electromagnetic energy can be defined as energy that moves at
the speed of light.  If you conduct experiments to determine whether
the electromagnetic energy is moving in the form of particles or waves,
you find that it behaves as both.</span>


4 0
3 years ago
A man-made satellite of mass 6105 kg is in orbit around the earth, making one revolution in 430 minutes. What is the magnitude o
blondinia [14]

Answer:

A gravitational force of 6841.905 newtons is exerted on the satellite by the Earth.

Explanation:

At first we assume that Earth is represented by an uniform sphere, such that the man-made satellite rotates in a circular orbit around the planet. Hence, the following condition must be satisfied:

\left(\frac{4\pi^{2}}{T^{2}} \right)\cdot r = \frac{G\cdot M}{r^{2}} (1)

Where:

T - Period of rotation of the satellite, measured in seconds.

r - Distance of the satellite with respect to the center of the planet, measured in meters.

G - Gravitational constant, measured in newton-square meters per square kilogram.

M - Mass of the Earth, measured in kilograms.

Now we clear the distance of the satellite with respect to the center of the planet:

r^{3} = \frac{G\cdot M\cdot T^{2}}{4\pi^{2}}

r = \sqrt[3]{\frac{G\cdot M\cdot T^{2}}{4\pi^{2}} } (2)

If we know that G = 6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}}, M = 6.0\times 10^{24}\,kg and T = 25800\,s, then the distance of the satellite is:

r = \sqrt[3]{\frac{\left(6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (6.0\times 10^{24}\,kg)\cdot (25800\,s)^{2}}{4\pi^{2}} }

r \approx 18.897\times 10^{6}\,m

The gravitational force exerted on the satellite by the Earth is determined by the Newton's Law of Gravitation:

F = \frac{G\cdot m\cdot M}{r^{2}} (3)

Where:

m - Mass of the satellite, measured in kilograms.

F - Force exerted on the satellite by the Earth, measured in newtons.

If we know that G = 6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}}, M = 6.0\times 10^{24}\,kg, m = 6105\,kg and r \approx 18.897\times 10^{6}\,m, then the gravitational force is:

F = \frac{\left(6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (6105\,kg)\cdot (6\times 10^{24}\,kg)}{(18.897\times 10^{6}\,m)^{2}}

F = 6841.905\,N

A gravitational force of 6841.905 newtons is exerted on the satellite by the Earth.

4 0
3 years ago
4. How does nearness to water affect the climate zone of an area?
Afina-wow [57]

Answer:

Choice A.

Nearness to a body of water causes an increase in humidity, due to the higher rate of evaporation.

6 0
2 years ago
Read 2 more answers
-g A small block is attached to an ideal spring and is moving in SHM on a horizontal frictionless surface. The amplitude of the
Nina [5.8K]

Answer:

a= 92. 13 m/s²

Explanation:

Given that

Amplitude ,A= 0.165 m

The maximum speed ,V(max) = 3.9 m/s

We know that maximum velocity in the SHM  given as

V(max)  = ω A

ω=Angular speed

A=Amplitude

\omega =\dfrac{3.9}{0.165}\ rad/s

ω=23.63 rad/s

The maximum acceleration given as

a = ω² A

a= (23.63)² x 0.165 m/s²

a= 92. 13 m/s²

Therefore the maximum magnitude of the acceleration will be 92. 13 m/s².

5 0
3 years ago
To move a heavy object, like a refrigerator what could be used to help decrease the frictional force?
Olenka [21]

Answer:

A. Remove everything in the refrigerator to lighten the load.

B. Put a lubricant between the surface of the object and the floor

C. Use round objects, like pencils , to decrease the friction and push the refrigerator over the pencils more easily

Explanation:

Force of friction is a resistance force which acts between two surfaces which are in relative motion. Friction is both boon and bane. Due to friction, we are able to sit, walk etc but also, due to friction there is dissipation of energy. Friction can be reduced by applying lubricants, reducing contact area, reducing the load.

F = μN where N is the normal force which depends on the mass.

Thus, by reducing the load, force of friction can be reduced. Round objects like wheels can also be used. By this the contact area reduces.

8 0
3 years ago
Other questions:
  • Which sentence correctly describes a friction force? A. It acts in the same direction as the motion of an object. B. It acts in
    14·1 answer
  • Magma can partially crystallize at depth and then rise to shallow depths where the remaining magma solidifies. the early-formed
    11·1 answer
  • Three equal point charges, each with charge 1.15 μCμC , are placed at the vertices of an equilateral triangle whose sides are of
    10·1 answer
  • What happens to the strength of the magnetic field as you come closer to the current carrying wire?
    9·1 answer
  • Describe what causes the flow of electric current. (4 points)
    5·1 answer
  • How do i do this? i don’t understand
    9·1 answer
  • (FLUID MECHANICS)
    11·1 answer
  • What happens when a gas changes state to become a liquid
    14·1 answer
  • BERE
    7·2 answers
  • Glucose solution is administered to a patient in a hospital. The density of the solution is 1.308 kg/l. If the blood pressure in
    10·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!