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2 years ago

How does a push or pull affect motion?

Physics

1 answer:

STatiana [176]2 years ago

8 0

Answer:

The motion of an object acted on by a force depends partly on the strength of the push or pull. The stronger the push or pull, the faster the object will move.

Explanation:

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Which of these waves cannot travel in a vacuum?

gayaneshka [121]

Light. The black hole is a vacuum, and not even light can escape it's suction

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3 years ago

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An ideal monatomic gas at temperature T is held in a container. If the gas is compressed isothermally, that is at constant tempe

OlgaM077 [116]

Answer:

a) 0 J

b) W = nRTln(Vf/Vi)

c) ΔQ = nRTln(Vf/Vi)

d) ΔQ = W

Explanation:

a) To find the change in the internal energy you use the 1st law of thermodynamics:

$\Delta U=\Delta Q-W$

Q: heat transfer

W: work done by the gas

The gas is compressed isothermally, then, there is no change in the internal energy and you have

ΔU = 0 J

b) The work is done by the gas, not over the gas.

The work is given by the following formula:

$\\W=nRTln(\frac{V_f}{V_i})$

n: moles

R: ideal gas constant

T: constant temperature

Vf: final volume

Vi: initial volume

Vf < Vi, then W < 0 and the work is done on the gas

c) The gas has been compressed. Thus, its temperature increases and heat has been transferred to the gas.

The amount of heat is equal to the work done W

$\Delta U = \Delta Q-W\\\\0=\Delta Q-W\\\\\Delta Q=W=nRTln(\frac{V_f}{V_i})$

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3 years ago

A projectile is launched at an angle of 30 and lands 20 s later at the same height as it was launched. (a) What is the initial s

Pavlova-9 [17]

Answer:

(a) 196 m/s

(b) 490 m

(c) 3394.82 m

(d) 2572.5 m

Explanation:

First of all, let us know one thing. When an object is thrown in the air, it experiences two forces acting in two different directions, one in the horizontal direction called air resistance and the second in the vertically downward direction due to its weight. In most of the cases, while solving numerical problems, air resistance is neglected unless stated in the numerical problem. This means we can assume zero acceleration along the horizontal direction.

Now, while solving our numerical problem, we will discuss motion along two axes according to our convenience in the course of solving this problem.

<u>Given:</u>

Time of flight = t = 20 s
Angle of the initial velocity of projectile with the horizontal = $\theta = 30^\circ$

<u>Assume:</u>

Initial velocity of the projectile = u
R = Range of the projectile during the time of flight
H = maximum height of the projectile
D = displacement of the projectile from the initial position at t = 15 s

Let us assume that the position from where the projectile was projected lies at origin.

Initial horizontal velocity of the projectile = $u\cos \theta$
Initial horizontal velocity of the projectile = $u\sin \theta$

Part (a):

During the time of flight the displacement of the projectile along the vertical is zero as it comes to the same vertical height from where it was projected.

$\therefore u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow u\sin \theta t=\dfrac{1}{2}(g)t^2\\\Rightarrow u=\dfrac{gt^2}{2\sin \theta t}\\\Rightarrow u=\dfrac{9.8\times 20^2}{2\sin 30^\circ \times 20}\\\Rightarrow u=196\ m/s$

Hence, the initial speed of the projectile is 196 m/s.

Part (b):

For a projectile, the time take by it to reach its maximum height is equal to return from the maximum height to its initial height is the same.

So, time taken to reach its maximum height will be equal to 10 s.

And during the upward motion of this time interval, the distance travel along the vertical will give us maximum height.

$\therefore H = u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow H = 196\times \sin 30^\circ \times 10 + \dfrac{1}{2}\times(-9.8)\times 10^2\\ \Rightarrow H =490\ m$

Hence, the maximum altitude is 490 m.

Part (c):

Range is the horizontal displacement of the projectile from the initial position. As acceleration is zero along the horizontal, the projectile is in uniform motion along the horizontal direction.

So, the range is given by:

$R = u\cos \theta t\\\Rightarrow R = 196\times \cos 30^\circ \times 20\\\Rightarrow R =3394.82\ m$

Hence, the range of the projectile is 3394.82 m.

Part (d):

In order to calculate the displacement of the projectile from its initial position, we first will have to find out the height of the projectile and its range during 15 s.

$\therefore h = u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow h = 196\times \sin 30^\circ \times 15 + \dfrac{1}{2}\times(-9.8)\times 15^2\\ \Rightarrow h =367.5\ m\\r = u\cos \theta t\\\Rightarrow r = 196\times \cos 30^\circ \times 15\\\Rightarrow r =2546.11\ m\\\therefore D = \sqrt{r^2+h^2}\\\Rightarrow D = \sqrt{2546.11^2+367.5^2}\\\Rightarrow D =2572.5\ m$

Hence, the displacement from the point of launch to the position on its trajectory at 15 s is 2572.5 m.

6 0

2 years ago

the frequency of a beam of uv light is 1.0 ×10 ^15hz what is the energy in one quantum of this light express it in ev?

kap26 [50]

Answer:

4.14 eV

Explanation:

f = 1.0 ×10^15 Hz

h= 6.63×10^-34 J s ( this is called PLANCK 'S CONSTANT)

ENEGY = E = ?

E = hf ( THIS IS FORMULA FOR ENERGY OF ONE QUANTA OR ONE PHOTON )

E= 6.63×10^-34×1.0 ×10^15

E = 6.63×10^-19 J

As 1eV = 1.6×10^-19 J so changing energy in eV from joules we will divide energy by 1.6×10^-19

hence E in eV = 6.63×10^-19/(1.6×10^-19)

E = 4.14 eV

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3 years ago

How is a conducter different from an insulater

Maru [420]

Answer:

Conductors have magnetic fields; insulators do not have magnetic fields. Conductors do not have magnetic fields; insulators do have magnetic fields. ... In a conductor, electric current cannot flow freely; in an insulator, it can flow freely.

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3 years ago

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