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1 year ago

A train accelerates steadily from 4.m.s.-1 to 20.m.s.-1 in 100 seconds

Physics

1 answer:

kozerog [31]1 year ago

6 0

Answering question assuming that we need to calculate how far it will travel in a given time.

A train accelerates steadily from 4 m/s to 20 m/s in 100 seconds then it will travel 1200 m in a given time.

Given:

Initial velocity u = 4m/s

Final velocity v = 20m/s

Time t = 100s

We need to find out the distance covered in given time.

Distance =?

For finding distance, we need to find acceleration first.

Here we can apply kinematic equations as acceleration is steady as given in question.

So applying the first kinematic equation to find acceleration first.

a = v-u/t

a = 20-4/100

a = 16/100

a = 0.16m/s²

Now applying second kinematic equation

S=ut+ 1/2at²

s = 4 x 100+ 1/2 x 0.16 x 100 x 100

s = 400 + 800 = 1200 m

Thus a train accelerates steadily from 4 m/s to 20 m/s in 100 seconds then it will travel 1200 m in a given time.

To learn more about the Kinematic equation application please click on the link brainly.com/question/26469691

#SPJ9

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a sound wave is an example of a a. transverse wave. b. longitudinal wave. c. standing wave. d. surface wave

Leona [35]

Answer: Longitudinal wave

Explanation:

Longitudinal wave are the oscillations that are parallel to the direction of energy transfer that means the vibrations are in line with the direction where the energy is travelling.

A key feature of sound wave is that they cause sound particles to vibrate. The region where the particles are close together are called compressions and regions where particles are further apart they are called rarefactions.

The other options explanation:

-Transverse waves are where the oscillations are perpendicular to the energy of transfer.

-A standing wave is where the waves are travelling back and forth where there are some fixed points in the system whilst other vibrate with highest amplitude

-Surface waves have both the characteristics of longitudinal and transverse waves

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3 years ago

A 225-kg object and a 525-kg object are separated by 3.80 m. (a) Find the magnitude of the net gravitational force exerted by th

pav-90 [236]

Answer: $F_{net}=3.383\times 10^{-7}\ N$

Explanation:

Given

Mass of first object $m_1=225\ kg$

Mass of second object $m_2=525\ kg$

Distance between them $d=3.8\ m$

$m_3=61\ kg$ object is placed between them

So force exerted by $m_1$ on $m_3$

$F_{13}=\frac{Gm_1m_3}{1.9^2}$

$F_{13}=\frac{6.674\times 10^{-11}(225\times 61)}{1.9^2}$

$F_{13}=2.5374141274×10^{−7}\ N$

Force exerted by $m_2\ on\ m_3$

$F_{23}=\frac{Gm_2m_3}{1.9^2}$

$F_{23}=\frac{6.674\times 10^{-11}(525\times 61)}{1.9^2}$

$F_{23}=5.920632964\times 10^{-7}\ N$

So net force on $m_3$ is

$F_{net}=F_{23}-F_{13}$

$F_{net}=5.920632964\times 10^{-7}-2.5374141274\times 10^{-7}$

$F_{net}=3.383\times 10^{-7}\ N$

i.e. net force is towards $m_2$

(b)For net force to be zero on $m_3$ , suppose

So force exerted by $m_1$ and $m_2$ must be equal

$F_{13}=F_{23}$

$\Rightarrow \frac{Gm_1m_3}{x^2}=\frac{Gm_2m_3}{(3.8-x)^2}$

$\Rightarrow \frac{m_1}{x^2}=\frac{m_2}{(3.8-x)^2}$

$\Rightarrow (\frac{3.8-x}{x})^2=\frac{m_2}{m_1}$

$\Rightarrow \frac{3.8-x}{x}=\sqrt{\frac{525}{225}}$

$\Rightarrow 3.8-x=1.52752x$

$\Rightarrow 3.8=2.52x$

$\Rightarrow x=1.507\ m$

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3 years ago

A 0.614mole sample of ideal gas at 12degree occupies a volume of 4.32.what is the pressure of the gas

Kaylis [27]

Answer:

336.9520 atm

Explanation:

The Gas Equation is as follows;-

Pressure×Volume=Number of Moles × Universal Gas Constant ×Temperature(in Kelvin)

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Temperature 12°C or 12+273.15 ie 285.15°F

Volume-4.32 L

Universal Gas Constant-8.314 J/mol·K

Pressure -?(in atm)

Plugging in all the values in the Gas Equation:-

Pressure= $\frac{0.614 × 8.314× 285.15}{4.32} atm$

Pressure=336.9520 atm

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3 years ago

HELP ASAP! <br>why is rolling friction much smaller than sliding friction? Also give an example.

IceJOKER [234]

Answer:

Rolling friction is much smaller than sliding friction because Rolling friction is considerably less than sliding friction as there is no work done against the body that is rolling by the force of friction. For a body to start rolling a small amount of friction is required at the point where it rests on the other surface, else it would slide instead of roll.

Rolling Friction example: Anything with weels (cars,skateboards) or a ball rooling.

Sliding Friction example: Bicycle brakes,skinning your knee walking,writing.

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3 years ago

If a car's speed triples, how does the momentum and kinetic energy of the

Harrizon [31]

Answer: When the car speed triples, momentum also triples but Kinetic energy increases 9 times or by 9 fold.

Explanation:

The momentum of a car (an object) is

p= mv

where

m is =the mass of the object( in this case car)

v is its= velocity

While the kinetic energy is is given by the formulae

K=1/2mv²

To determine how momentum and kinetic energy of the car changes when the speed of the object triples, We have that the new velocity,

v¹= 3v

So that the momentum change becomes

p¹=mv¹=m (3v)= 3mv

mv=p

therefore p¹= 3p

we can see that the momentum also triples.

And the kinetic energy change becomes

K¹=1/2m(v¹)²= 1/2m (3v)²

= 1/2m9v²= 1/2 x m x 9 x v²=9 x1/2mv²

1/2mv²=K

K¹= Kinetic energy = 9k

but Kinetic energy increases 9 times

7 0

3 years ago

A train accelerates steadily from 4.m.s.-1 to 20.m.s.-1 in 100 seconds​

A train accelerates steadily from 4.m.s.-1 to 20.m.s.-1 in 100 seconds