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larisa86 [58]
3 years ago
8

Summarize Newton’s First Law

Physics
1 answer:
Tamiku [17]3 years ago
8 0
Newton's first law is "An object at rest stays at rest and an object in motion stays in motion with the same speed and in the same direction unless acted upon by an unbalanced force."

Basically it means an object won't move unless something moves it, and an object won't stop moving unless something stops it.
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Suppose you designed a spacecraft to work by photon pressure. The sail was a completely absorbing fabric of area 1.0 km2 and you
Alekssandra [29.7K]

Answer:

(a) F = 6.14 *10⁻⁴ N

(b) P = 6.14* 10⁻¹⁰ Pa

(c) t = 27.2 min

Explanation:

Area of sail A = 1.0 km² = 1.0 * 10⁶m²

Wavelength of light  λ = 650 nm = 650 * 10⁻⁹ m

Rate of impact of photons R = 1 mol/s = 6.022 * 10²³ photons/s

(a)

Momentum of each photon is Ρ = h/λ = 6.625 * 10⁻³⁴ / 650 * 10⁻⁹

      = 1.0192 * 10⁻²⁷ kg.m/s

Since the photons are absorbed completely, in every collision the above momentum is transferred to the sail.  

Momentum transferred to the sail per second is product of rate of impact of photons and momentum transferred by each photon.

dp/dt = R * h/ λ

This is the force acting on the sail.

F = R * h/λ = 6.022 * 10²³ * 1.0192 * 10⁻²⁷ = 6.14 *10⁻⁴ N

F = 6.14 *10⁻⁴ N

b)

Pressure exerted by the radiation on the sail = Force acting on the sail / Area of the sail

P = F/A =  6.14 * 10⁻⁴ / 1.0 * 10⁶ =  6.14* 10⁻¹⁰ Pa

P = 6.14* 10⁻¹⁰ Pa

c)  

Acceleration of spacecraft a = F/m = 6.14 * 10⁻⁴ /1.0 = 6.14 * 10⁻⁴m/s²

As the spacecraft starts from rest, initial speed u=0,m/s ,

final speed is u = 1.0 m/s after time t  

v = u+at  

t = 1.0 - 0/ 6.14 * 10⁻⁴ =  1629s = 27.2 min

t = 27.2 min

4 0
3 years ago
If Earth was with no tilt, would we still have seasons at all? If so, how would they be different?
murzikaleks [220]

Answer:

If earth had no tilt, we would have no seasons.

Explanation:

As stated in the answer, if the earth had no tilt we wouldn't have seasons. The earth all around the globe would maintain the same temperature,

And due to the no tilt it would also change our orbit to a bit larger slant, in January when we are at our closest to the sun we WOULD have a mini summer. For the North and South Pole, they would remain cold.

8 0
3 years ago
Doctor prescribed drug which is 500mg for patient change this unit into gram
ololo11 [35]

Answer:

.5 grams

Explanation:

1 gram is equal to 1000 milligrams (mg)

6 0
3 years ago
Two identical tiny spheres of mass m =2g and charge q hang from a non-conducting strings, each of length L = 10cm. At equilibriu
Citrus2011 [14]

Answer:

0.247 μC

Explanation:

As both sphere will be at the same level at wquilibrium, the direction of the electric force will be on the x axis. As you can see in the picture below, the x component of the tension of the string of any of the spheres should be equal to the electric force of repulsion. And its y component will be equal to the weight of one sphere. We can use trigonometry to find the components of the tensions:

F_y:  T_y - W = 0\\T_y = m*g = 0.002 kg *9.81m/s^2 = 0.01962 N

T_y = T_*cos(50)\\T = \frac{T_y}{cos(50)} = 0.0305 N

T_x = T*sin(50) = 0.0234 N

The electric force is given by the expression:

F = k*\frac{q_1*q_2}{r^2}

In equilibrium, the distance between the spheres will be equal to 2 times the length of the string times sin(50):

r = 2*L*sin(50) = 2 * 0.1m * sin(50) 0.1532 m

And k is the coulomb constan equal to 9 *10^9 N*m^2/C^2. q1 y q2 is the charge of each particle, in this case, they are equal.

F_x = T_x - F_e = 0\\T_x = F_e = k*\frac{q^2}{r^2}

q = \sqrt{T_x *\frac{r^2}{k}} = \sqrt{0.0234 N * \frac{(0.1532m)^2}{9*10^9 N*m^2/C^2} } = 2.4704 * 10^-7 C

O 0.247 μC

8 0
3 years ago
Suppose a nonconducting sphere, radius r2, has a spherical cavity of radius r1 centered at the sphere's center. Assuming the cha
leva [86]

Answer:

Explanation:

a ) Between r = 0 and r = r₁

Electric field will be zero . It is so because no charge lies in between r = 0 and r = r₁ .

b ) From r = r₁ to r = r₂

At distance r , charge contained in the sphere of radius r

volume charge density x 4/3 π r³

q = Q x r³ / R³

Applying Gauss's law

4πr² E = q / ε₀

4πr² E = Q x r³ / ε₀R³

E= Q x r / (4πε₀R³)

E ∝ r .

c )

Outside of r = r₂

charge contained in the sphere of radius r = Q

Applying Gauss's law

4πr² E = q / ε₀

4πr² E = Q  / ε₀

E = Q  / 4πε₀r²

E ∝ 1 / r² .

6 0
3 years ago
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