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castortr0y [4]
3 years ago
14

State and explain newton second law of motion also describe the concept of force, represent it quantiatively and derive the unit

of force
this is a long question only answer if you know how to solve it you will be REWARDED with points​
Physics
1 answer:
Molodets [167]3 years ago
8 0

Explanation:

Newton's second law of motion states F=ma which means force is equal to mass multiplied by acceleration which in simple terms means If you give mass force it will accelerate the concept of force in physics is any interaction that when unopposed will change the motion of an object.

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For a projectile launched from the vertically from the surface the earth with initial velocity v0, the velocity of the projectil
UNO [17]

Answer:

(dr/dt) = √(v₀² + k²/r - k²/R)

Explanation:

v² = v₀² + k²[(1/r) - (1/R)]

v = velocity of Body launched from the surface of the earth

v = initial velocity of body

k = a constant

r = distance from the centre of the earth

R = radius of the earth

v² = v₀² + k²/r - k²/R

v = √(v₀² + k²/r - k²/R)

But v = dr/dt

(dr/dt) = √(v₀² + k²/r - k²/R)

(dr/√(v₀² + k²/r - k²/R)) = dt

To go one step beyond and integrate the differential equation

∫ (dr/√(v₀² + k²/r - k²/R)) = ∫ dt

Integrating the left hand side from 0 to r and the right hand side from 0 to t

Note, v₀, k and R are all constants

(- 4r²/k²)[√(v₀² + k²/r - k²/R)] = t

√(v₀² + k²/r - k²/R) = (- k² t/4r²)

(v₀² + k²/r - k²/R) = (- k² t/4r²)²

(v₀² + k²/r - k²/R) = (k⁴t²/16r⁴)

(k⁴t²/16r⁴) + (k²/r) = [v₀² - (k²/R)]

k² [(k²t²/16r⁴) + (1/r)] = [v₀² - (k²/R)]

[(k²t²/16r⁴) + (1/r)] = [(v₀²/k²) - (1/R)]

3 0
3 years ago
A cat with mass 4.0 kg jumps down to the floor from a bookcase 2.0 m high. What is the cat's change in
eduard

Answer: -7J

Explanation:

To determine the change in potential energy, use the equation ΔPE=mgΔh

Δ

PE=mg

Δ

h

, where m is the mass, g= 9.8 m/s2

g= 9.8 m/s

2

, and Δh is the change in height. Hence, we have that the potential energy PE=4.0 kg×9.8 m/s2×-2.0 m=-78 J

PE=4.0 kg

×

9.8 m/s

2

×

-2.0 m=-78 J

. To verify that this is correct, note that since the cat changes the potential energy to kinetic energy by jumping, the potential energy decreases. Hence, the potential energy should be negative.

3 0
3 years ago
Read 2 more answers
In the visible spectra of stars, absorption lines of hydrogen are produced when atoms are excited from n = 2 to higher levels (t
grigory [225]

Answer:

3. relatively high temperature, about 10,000 K, so that significant numbers of electrons are excited from the ground state, n = 1, to the first excited state, n = 2, but not too many of them have been ejected completely from the atoms

Explanation:

If hydrogen absorption lines are very strong in the visible spectrum of a particular star that means the population of electron in n = 2 is very high so on being exited they absorb radiation in Balmer series and give rise to absorption spectrum. The average temperature required to excite electron in hydrogen atom from n=1 to n = 2   is 10000K .

4 0
3 years ago
A ray of laser light travels through air and enters an unknown material. The laser enters the material at an angle of 36 degrees
V125BC [204]

Answer:1.27

Explanation:

Given

incident angle i=36^{\circ}

refracted angle r=27.5^{\circ}

Suppose n_2 is the refractive index of material then using Snell's law we  can write

n_1\sin i=n_2\sin r

where n_1=refractive index of air

1\times \sin (36)=n_2\times \sin (27.5)

n_2=\dfrac{0.5877}{0.4617}

n_2=1.27

3 0
3 years ago
A 3.53-g lead bullet traveling at 428 m/s strikes a target, converting its kinetic energy into thermal energy. Its initial tempe
Taya2010 [7]

Complete question:

A 3.53-g lead bullet traveling at 428 m/s strikes a target, converting its kinetic energy into thermal energy. Its initial temperature is 40.0°C. The specific heat is 128 J/(kg · °C), latent heat of fusion is 24.5 kJ/kg, and the melting point of lead is 327°C.

(a) Find the available kinetic energy of the bullet. J

(b) Find the heat required to melt the bullet. J

Answer:

Part (a) the available kinetic energy of the bullet is 323.32 J

Part (b) the heat required to melt the bullet is 216.17 J

Explanation:

Given;

mass of the bullet = 3.53 g = 0.00353 kg

velocity of the bullet = 428 m/s

initial temperature of the bullet = 40.0°C

final temperature of the bullet =  327°C

specific heat capacity, c= 128 J/(kg · °C)

latent heat of fusion, Hf  = 24.5 kJ/kg

Part (a) the available kinetic energy of the bullet. J

KE = ¹/₂ × mv²

KE = ¹/₂ × 0.00353 × 428²

     = 323.32 J

Part (b) the heat required to melt the bullet. J

This is the thermal energy required to increase the temperature of the bullet and the heat energy required to melt the bullet.

Quantity of heat required to raise the temperature of the bullet:

Q = mcΔT

   = 0.00353 × 128 × (327-40)

   = 0.00353 × 128 × 287

   = 129.68 J

Quantity of heat required to melt the bullet:

Q = mH_f

Q = 0.00353 × 24500

   = 86.49 J

TOTAL energy required to melt the bullet = 129.68 J + 86.49 J

                                                                      = 216.17 J

3 0
3 years ago
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