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2 years ago

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A force of constant magnitude F and fixed direction acts on an object of mass m that is initially at rest. If the force acts for

a time interval ∆t over a displacement ∆x , what is the magnitude of the resultant change in the linear momentum of the object?

1 answer:

ZanzabumX [31]2 years ago

5 0

Answer:

$\Delta P = F \Delta t$

Explanation:

As we know Newton's II law

$F$ = Rate of change in momentum

so we will have

$F = \frac{\Delta P}{\Delta t}$

now we will have

$\Delta P = F \Delta t$

so here we can say that change in momentum of the object is the product of force and interval of time for which the force is acting on it.

so we will have

$\Delta P = F \Delta t$

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The Lyman series comprises a set of spectral lines. All of these lines involve a hydrogen atom whose electron undergoes a change

mihalych1998 [28]

Answer:

a) 1.2*10^-7 m

b) 1.0*10^-7 m

c) 9.7*10^-8 m

d) ultraviolet region

Explanation:

To find the different wavelengths you use the following formula:

$\frac{1}{\lambda}=R_H(1-\frac{1}{n^2})$

RH: Rydberg constant = 1.097 x 10^7 m^−1.

(a) n=2

$\frac{1}{\lambda}=(1.097*10^7m^{-1})(1-\frac{1}{(2)^2})=8227500m^{-1}\\\\\lambda=1.2*10^{-7}m$

(b)

$\frac{1}{\lambda}=(1.097*10^7m^{-1})(1-\frac{1}{(3)^2})=9751111,1m^{-1}\\\\\lambda=1.0*10^{-7}m$

(c)

$\frac{1}{\lambda}=(1.097*10^7m^{-1})(1-\frac{1}{(4)^2})=10284375m^{-1}\\\\\lambda=9.7*10^{-8}m$

(d) The three lines belong to the ultraviolet region.

8 0

2 years ago

In a power plant, pipes transporting superheated vapor are very common. Superheated vapor flows at a rate of 0.3 kg/s inside a p

grigory [225]

Answer: $h=160.84 W/m^2-K$

Explanation:

Given

mass flow rate=0.3 kg/s

diameter of pipe=5 cm

length of pipe=10 m

Inside temperature=22

Pipe surface =100

Temperature drop=30

specific heat of vapor(c)=2190 J/kg.k

heat supplied $Q=mc\Delta T=0.3\times 2190\times (30)$

Heat due to convection =hA(100-30)

$A=\pi d\cdot L$

$A=\pi 0.05\times 10=1.571 m^2$

$Q_{convection}=h\times 1.571\times (100-22)=122.538 h$

$Q=Q_{convection}$

19,710=122.538 h

$h=160.84 W/m^2-K$

5 0

2 years ago

Suppose you are talking by interplanetary telephone to your friend, who lives on the Moon. He tells you that he has just won a n

Savatey [412]

Answer:

The friend on moon will be richer.

Explanation:

We must calculate the mass of gold won by each person, to tell who is richer. For that purpose we will use the following formula:

W = mg

m = W/g

where,

m = mass of gold

W = weight of gold

g = acceleration due to gravity on that planet

<u>FOR FRIEND ON MOON</u>:

W = 1 N

g = 1.625 m/s²

Therefore,

m = (1 N)/(1.625 m/s²)

m(moon) = 0.6 kg

<u>FOR ME ON EARTH</u>:

W = 1 N

g = 9.8 m/s²

Therefore,

m = (1 N)/(9.8 m/s²)

m(earth) = 0.1 kg

Since, the mass of gold on moon is greater than the mass of moon on earth.

<u>Therefore, the friend on moon will be richer.</u>

7 0

2 years ago

Explain how the forces need to change so the aeroplane can land

Fofino [41]

When an airplane is flying straight and level at a constant speed, the lift it produces balances its weight, and the thrust it produces balances its drag. However, this balance of forces changes as the airplane rises and descends, as it speeds up and slows down, and as it turns.

7 0

2 years ago

A small rock is launched straight upward from the surface of a planet with no atmosphere. The initial speed of the rock is twice

Scorpion4ik [409]

If gravitational effects from other objects are negligible, the speed of the rock at a very great distance from the planet will approach a value of $\sqrt{3} v_{e}$

<u>Explanation:</u>

To express velocity which is too far from the planet and escape velocity by using the energy conservation, we get

Rock’s initial velocity , $v_{i}=2 v_{e}$ . Here the radius is R, so find the escape velocity as follows,

$\frac{1}{2} m v_{e}^{2}-\frac{G M m}{R}=0$

$\frac{1}{2} m v_{e}^{2}=\frac{G M m}{R}$

$v_{e}^{2}=\frac{2 G M}{R}$

$v_{e}=\sqrt{\frac{2 G M}{R}}$

Where, M = Planet’s mass and G = constant.

From given conditions,

Surface potential energy can be expressed as, $U_{i}=-\frac{G M m}{R}$

R tend to infinity when far away from the planet, so $v_{f}=0$

Then, kinetic energy at initial would be,

$k_{i}=\frac{1}{2} m v_{i}^{2}=\frac{1}{2} m\left(2 v_{e}\right)^{2}$

Similarly, kinetic energy at final would be,

$k_{f}=\frac{1}{2} m v_{f}^{2}$

Here, $v_{f}=\text { final velocity }$

Now, adding potential and kinetic energies of initial and final and equating as below, find the final velocity as

$U_{i}+k_{i}=k_{f}+v_{f}$

$\frac{1}{2} m\left(2 v_{e}\right)^{2}-\frac{G M m}{R}=\frac{1}{2} m v_{f}^{2}+0$

$\frac{1}{2} m\left(2 v_{e}\right)^{2}-\frac{G M m}{R}=\frac{1}{2} m v_{f}^{2}$

'm' and $\frac{1}{2}$ as common on both sides, so gets cancelled, we get as

$4\left(v_{e}\right)^{2}-\frac{2 G M}{R}=v_{f}^{2}$

We know, $v_{e}=\sqrt{\frac{2 G M}{R}}$ , it can be wriiten as $\left(v_{e}\right)^{2}=\frac{2 G M}{R}$ , we get

$4\left(v_{e}\right)^{2}-\left(v_{e}\right)^{2}=v_{f}^{2}$

$v_{f}^{2}=3\left(v_{e}\right)^{2}$

Taking squares out, we get,

$v_{f}=\sqrt{3} v_{e}$

4 0

2 years ago