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Pani-rosa [81]
3 years ago
6

The intensity of the Sun's light in the vicinity of the Earth is about 1000W/m^2. Imagine a spacecraft with a mirrored square sa

il of dimension 5.5 km. Estimate how much thrust (in newtons) this craft will experience due to collisions with the Sun's photons. [Hint: assume the photons bounce off the sail with no change in the magnitude of their momentum.}
Physics
1 answer:
NeX [460]3 years ago
8 0

Explanation:

Formula to represent thrust is as follows.

             F = \frac{dP}{dt}

                = \frac{2p}{dt}

or,           p = \frac{E}{c}

         \frac{p}{dt} = \frac{W}{c}

                  F = \frac{2IA}{c}

                     = \frac{2 (1000 W/m^{2})(5.5 \times 10^{3} m)^{2}}{3 \times 10^{8} m/s}

                     = 201.67 N

Thus, we can conclude that the thrust is 201.67 N.

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Projectile's horizontal range on level ground is R=v20sin2θ/g. At what launch angle or angles will the projectile land at half o
seraphim [82]

Answer:

\theta = 15^o \: or\: 75^o

Explanation:

As we know that the formula of range is given as

R = \frac{v^2sin2\theta}{g}

now we know that

maximum value of the range of the projectile is given as

R_{max} = \frac{v^2}{g}

now we need to find such angles for which the range is half the maximum value

so we will have

\frac{R}{2} = \frac{v^2}{2g} = \frac{v^2sin(2\theta)}{g}

sin(2\theta) = \frac{1}{2}

2\theta = 30 or 150

\theta = 15^o \: or\: 75^o

7 0
3 years ago
A wooden plaque is in the shape of an ellipse with height 30 centimeters and width 22 centimeters. Find an equation for the elli
Volgvan

Answer:

Explanation:

height of Ellipse =30 cm  

i.e. 2 a=30

Width of Ellipse =22 cm

i.e. 2 b=22

Equation of a vertical Ellipse is

\frac{x^2}{b^2}+\frac{y^2}{a^2}=1

a=15 ,b =11

\frac{x^2}{11^2}+\frac{y^2}{15^2}=1

at y=4 cm

\frac{x^2}{11^2}+\frac{4^2}{15^2}=1  

x=\frac{11}{15}\times \sqrt{15^2-4^2}

x=10.6 cm  

5 0
3 years ago
A current of 9 A flows through an electric device with a resistance of 43 Ω. What must be the applied voltage in this particular
NeTakaya
Answer:
387 volts

Explanation:
Ohm's law is used to relate voltage, current and resistance.
The formula is as follows:V = I * R
where:
V is the applied voltage (measured in volts)
I is the current flowing (measured in amperes)
R is the resistance (measured in ohm)

In the given, we have:
current (I) = 9 amperes
resistance (R) = 43 ohm

Substitute with the givens in the above formula to get the voltage as follows:
V = 9 * 43
V = 387 volts

Hope this helps :)
4 0
3 years ago
Read 2 more answers
6. A 2-kg ball B is traveling around in a circle of radius r1 = 1 m with a speed (vB)1 = 2 m/s. If the attached cord is pulled d
kipiarov [429]

Answer:

Explanation:

Given that,

Mass of ball m = 2kg

Ball traveling a radius of r1= 1m.

Speed of ball is Vb = 2m/s

Attached cord pulled down at a speed of Vr = 0.5m/s

Final speed V = 4m/s

Let find the transverse component of the final speed using

V² = Vr²+ Vθ²

4² = 0.5² + Vθ²

Vθ² = 4²—0.5²

Vθ² = 15.75

Vθ =√15.75

Vθ = 3.97 m/s.

Using the conservation of angular momentum,

(HA)1 = (HA)2

Mb • Vb • r1 = Mb • Vθ • r2

Mb cancels out

Vb • r1 = Vθ • r2

2 × 1 = 3.97 × r2

r2 = 2/3.97

r2 = 0.504m

The distance r2 to the hole for the ball to reach a maximum speed of 4m/s is 0.504m

The required time,

Using equation of motion

V = ∆r/t

Then,

t = ∆r/Vr

t = (r1—r2) / Vr

t = (1—0.504) / 0.5

t = 0.496/0.5

t = 0.992 second

7 0
3 years ago
Suppose the student in (Figure 1) is 68kg, and the board being stood on has a 12kg mass. What is the reading on the left scale?
lesantik [10]

The equilibrium conditions allow to find the results for the balance forces are:

  • F₁ = 225.4 N
  • F₂ = 558.6 N

When the acceleration is zero we have the equilibrium conditions for both linear and rotational motion.

            ∑ F = 0

            ∑ τ = 0

           

Where F are the forces and τ the torques.

The torque  is the product of the force and the perpendicular distance to the point of support,

The free-body diagrams are diagrams of the forces without the details of the bodies, see attached for the free-body diagram of the system.

We write the translational equilibrium condition.

           F₁ - W₁ - W₂ + F₂ = 0

We write the equation for the rotational motion, set our point of origin at scale 1, and the counterclockwise turns are positive.

         F₂ 2 - W₁ 1 - W₂ 1.5 = 0\frac{W_1  \ 1 + W_2 \ 1.5}{2}

Let's calculate F₂

         F₂ = \frac{W_1 \ 1 + W_2 \ 1.5 }{2}  

         F₂ = (m g + M g 1.5)/ 2

         F₂ = \frac{(12 + 68 \ 1.5 ) \  9.8}{2}  

         F₂ = 558.6 N

We substitute in the translational equilibrium equation.

         F₁ = W₁ + W₂ - F₂

         F₁ = (m + M) g - F₂

         F₁ = (12 +68) 9.8 - 558.6

         F₁ = 225.4 N

In conclusion using the equilibrium conditions we can find the forces of the balance are:

  • F₁ = 225.4 N
  • F2 = 558.6 N

Learn more here:  brainly.com/question/12830892

5 0
2 years ago
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