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lana [24]
3 years ago
14

What are some ways houses along the coastlines can protect themselves from storm surges?

Physics
1 answer:
BARSIC [14]3 years ago
8 0
Build walls around the coast
You might be interested in
You are riding on a school bus and suddenly get thrown forward. What did the bus just do?
sergij07 [2.7K]
The bus is going forward and suddenly stops
5 0
2 years ago
A large solar panel on a spacecraft in Earth orbit produces 1.0 kW of power when the panel is turned toward the sun. What power
Mandarinka [93]

Answer:

e*P_s = 11 W

Explanation:

Given:

- e*P = 1.0 KW

- r_s = 9.5*r_e

- e is the efficiency of the panels

Find:

What power would the solar cell produce if the spacecraft were in orbit around Saturn

Solution:

- We use the relation between the intensity I and distance of light:

                                  I_1 / I_2 = ( r_2 / r_1 ) ^2

- The intensity of sun light at Saturn's orbit can be expressed as:

                                  I_s = I_e * ( r_e / r_s ) ^2

                                  I_s = ( 1.0 KW / e*a) * ( 1 / 9.5 )^2

                                  I_s = 11 W / e*a

- We know that P = I*a, hence we have:

                                  P_s = I_s*a

                                  P_s = 11 W / e

Hence,                       e*P_s = 11 W

3 0
2 years ago
A physics student of mass 43.0 kg is standing at the edge of the flat roof of a building, 12.0 m above the sidewalk. An unfriend
Dmitry_Shevchenko [17]

Answer:

The speed of the student just before she lands, v₂ is approximately 8.225 m/s

Explanation:

The given parameters are;

The mass of the physic student, m = 43.0 kg

The height at which the student is standing, h = 12.0 m

The radius of the wheel, r = 0.300 m

The moment of inertia of the wheel, I = 9.60 kg·m²

The initial potential energy of the female student, P.E.₁ = m·g·h₁

Where;

m = 43.0 kg

g = The acceleration due to gravity ≈ 9.81 m/s²

h = 12.0 h

∴ P.E.₁ = 43 kg × 9.81 m/s² × 12.0 m = 5061.96 J

The kinetic rotational energy of the wheel and kinetic energy of the student supporting herself from the rope she grabs and steps off the roof, K₁, is given as follows;

K_1 = \dfrac{1}{2} \cdot m \cdot v_{1}^2+\dfrac{1}{2} \cdot I \cdot \omega_{1}^2

The initial kinetic energy, 1/2·m·v₁² and the initial kinetic rotational energy, 1/2·m·ω₁² are 0

∴ K₁ = 0 + 0 = 0

The final potential energy of the student when lands. P.E.₂ = m·g·h₂ = 0

Where;

h₂ = 0 m

The final kinetic energy, K₂, of the wheel and student is give as follows;

K_2 = \dfrac{1}{2} \cdot m \cdot v_{2}^2+\dfrac{1}{2} \cdot I \cdot \omega_{2}^2

Where;

v₂ = The speed of the student just before she lands

ω₂ = The angular velocity of the wheel just before she lands

By the conservation of energy, we have;

P.E.₁ + K₁ = P.E.₂ + K₂

∴ m·g·h₁ + \dfrac{1}{2} \cdot m \cdot v_{1}^2+\dfrac{1}{2} \cdot I \cdot \omega_{1}^2 = m·g·h₂ + \dfrac{1}{2} \cdot m \cdot v_{2}^2+\dfrac{1}{2} \cdot I \cdot \omega_{2}^2

Where;

ω₂ = v₂/r

∴ 5061.96 J + 0 = 0 + \dfrac{1}{2} \times 43.0 \, kg \times v_{2}^2+\dfrac{1}{2} \times 9.60 \, kg\cdot m^2 \cdot \left (\dfrac{v_2}{0.300 \, m} }\right ) ^2

5,061.96 J = 21.5 kg × v₂² + 53.\overline 3 kg × v₂² = 21.5 kg × v₂² + 160/3 kg × v₂²

v₂² = 5,061.96 J/(21.5 kg + 160/3 kg) ≈ 67.643118 m²/s²

v₂ ≈ √(67.643118 m²/s²) ≈ 8.22454363 m/s

The speed of the student just before she lands, v₂ ≈ 8.225 m/s.

5 0
2 years ago
In an electricity experiment, a 1.10 g plastic ball is suspended on a 56.0 cm long string and given an electric charge. A charge
Shalnov [3]

Answer:

Tension, T = 0.0115 N                      

Explanation:

Given that,

Mass of the plastic ball, m = 1.1 g

Length of the string, l = 56 cm

A charged rod brought near the ball exerts a horizontal electrical force F on it, causing the ball to swing out to a 21.0 degree angle and remain there. According to attached figure :

T\cos\theta=mg

T is tension in the string

T=\dfrac{mg}{\cos\theta}\\\\T=\dfrac{1.1\times 10^{-3}\times 9.8}{\cos(21)}\\\\T=0.0115\ N

So, the tension in the string is 0.0115 N.

8 0
3 years ago
0.A 20-g bullet moving at 1 000 m/s is fired through a one-kg block of wood emerging at a speed of 100 m/s. What is the kinetic
Fiesta28 [93]

Answer:

KE = 0.162 KJ

Explanation:

given,

mass of bullet (m)= 20 g = 0.02 Kg

speed of the bullet (u)= 1000 m/s

mass of block(M) = 1 Kg

velocity of bullet after collision (v)= 100 m/s

kinetic energy = ?

using conservation of momentum

m u = m v + M V

0.02 x 1000 = 0.02 x 100 + 1 x V

20 = 2 + V

V = 18 m/s

now,

Kinetic energy of the block

KE = \dfrac{1}{2}mv^2

KE = \dfrac{1}{2}\times 1 \times 18^2

KE = 162 J

KE = 0.162 KJ

4 0
3 years ago
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