1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
vaieri [72.5K]
2 years ago
6

I need help with the problem below

Physics
1 answer:
sdas [7]2 years ago
4 0

Answer:

Explanation:

a) F = ma

a = F/m

a = 9(800) / 1 x 10⁹ = 7.2 x 10⁻⁶ m/s

b) t = v/a

t = 200 / 7.2 x 10⁻⁶

t = 2.8 x 10⁷ s       about 10½ months

c) v² = u² + 2as

s = (v² - u²) / 2a

s = (200² - 0²) / (2( 7.2 x 10⁻⁶))

s = 2.8 x 10⁹ m    nearly 7 times around the earth

And all this assumes NO FRICTION.

You might be interested in
Colossal Cave was formed years ago by underground running water. Today, it is the home to many animals, like bats. These interac
jeyben [28]

Answer:

d. interaction atmosphere and biosphere interaction

Explanation:

hydrosphere, lithosphere, and biosphere interaction.

6 0
3 years ago
an elevator suspended by a vertical cable is moving downward but slowing down. the tension in the cable must be:
vladimir1956 [14]

Answer:

F = M a is the vector equation involved

F = T - M g      are the forces acting on the elevator   (scalar equation)

T - M g = M a

T = M (a + g)    remember this a scalar

If a is slowing down then it must have a positive acceleration upwards

Therefor the tension in the cable must be greater than zero

When the tension increases to M g, a has increased to zero

For a to be zero, no acceleration, T = M g

6 0
3 years ago
A person kicks a ball, giving it an initial velocity of 20.0 m/s up a wooden ramp. When the ball reaches the top, it becomes air
Alex Ar [27]

Answer:

(a) Height is 4.47 m

(b) Height is 4.37 m

Solution:

As per the question:

Initial velocity of teh ball, v_{o} = 20.0 m/s

Angle made by the ramp, \theta = 22.0^{\circ}

Distance traveled by the ball on the ramp, d = 5.00 m

Now,

(a) At any point on the projectile before attaining maximum height, the velocity can be given by the eqn-3 of motion:

v^{2} = v_{o}^{2} - 2gH

where

H = dsin22^{\circ} = 5sin22^{\circ}

g = 9.8 m/s^{2}

v^{2} = 20^{2} - 2\times 9.8\times 5sin22^{\circ}

v = \sqrt{400 - 19.6\times 5sin22^{\circ}} = 19.06 m/s

Now, maximum height attained is given by:

h = \frac{(vsin\theta)^{2}}{2g}

h = \frac{(19sin(22^{\circ}))^{2}}{2\times 9.8} = 2.60 m

Height from the ground = 5sin22^{circ} + 2.86 = 1.87 + 2.60 = 4.47m

(b) now, considering the coefficient of friction bhetween ramp and the ball, \mu = 0.150:

velocity can be given by the eqn-3 of motion:

v^{2} = v_{o}^{2} - 2gH - \mu gd

v^{2} = 20^{2} - 2\times 9.8\times 5sin22^{\circ} - 0.150\times 9.8\times 5

v = \sqrt{400 - 19.6\times 5sin22^{\circ} - 0.150\times 9.8\times 5} = 18.7 m/s

Now, maximum height attained is given by:

h = \frac{(vsin\theta)^{2}}{2g}

h = \frac{(18.7sin(22^{\circ}))^{2}}{2\times 9.8} = 2.50 m

Height from the ground = 5sin22^{circ} + 2.86 = 1.87 + 2.50 = 4.37 m

6 0
3 years ago
An ideal photo-diode of unit quantum efficiency, at room temperature, is illuminated with 8 mW of radiation at 0.65 µm wavelengt
nadya68 [22]

Answer:

I = 4.189 mA    V = 0.338 V

Explanation:

In order to do this, we need to apply the following expression:

I = Is[exp^(qV/kT) - 1]   (1)

However, as the junction of the diode is illuminated, the above expression changes to:

I = Iopt + Is[exp^(qV/kT) - 1]   (2)

Now, as the shunt resistance becomes infinite while the current becomes zero, we can say that the leakage current is small, and so:

I ≅ Iopt

Therefore:

I ≅ I₀Aλq / hc  (3)

Where:

I₀A = Area of diode (radiation)

λ: wavelength

q: electron charge (1.6x10⁻¹⁹ C)

h: Planck constant (6.62x10⁻³⁴ m² kg/s)

c: speed of light (3x10⁸ m/s)

Replacing all these values, we can get the current:

I = (8x10⁻³) * (0.65x10⁻⁶) * (1.6x10⁻¹⁹) / (6.62x10⁻³⁴) * (3x10⁸)

I = 4.189x10⁻³ A or 4.189 mA

Now that we have the current, we just need to replace this value into the expression (2) and solve for the voltage:

I = Is[exp^(qV/kT) - 1]

k: boltzman constant (1.38x10⁻²³ J/K)

4.189x10⁻³ = 9x10⁻⁹ [exp(1.6x10⁻¹⁹ V / 1.38x10⁻²³ * 300) - 1]

4.189x10⁻³ / 9x10⁻⁹ = [exp(38.65V) - 1]

465,444.44 + 1  = exp(38.65V)

ln(465,445.44) = 38.65V

13.0508 = 38.65V

V = 0.338 V

6 0
4 years ago
How much energy (in kW-h) does a 900 Watt stove use in a week if it is used for 1.5 hours each day?
ioda

Answer:

9.45 kWh

Explanation:

Energy = Power × time

E = 900 W × (1.5 h/day × 7 day)

E = 9450 Wh

E = 9.45 kWh

3 0
3 years ago
Other questions:
  • Kyle lays a mirror flat on the floor and aims a laser at the mirror. The laser beam reflects off the mirror and strikes an adjac
    6·1 answer
  • Which of the following terms relate to the meaning of "physical"?
    11·2 answers
  • In the simple Bohr model of the eighth excited state of the hydrogen atom, the electron travels in a circular orbit around a fix
    5·1 answer
  • Lila is using a Bunsen burner. She has all of her chemicals on her workstation. Which would be the best lab practice?
    10·2 answers
  • An object that is not accelerating or decelerating has zero net force acting on it. Choose 1 answer: Choose 1 answer: A True B F
    6·2 answers
  • 1. Complete the table with the appropriate equivalences
    10·1 answer
  • Which of these correctly describes whether a girl holding a ball in the same position is doing work on the ball?
    14·1 answer
  • A balloon-powered car rolls across the floor at a speed of 0.711 m/s. How long does it take to cover 8.25 m?
    8·1 answer
  • Why is the sky blue? The earth is surrounded by an atmosphere.
    12·1 answer
  • Animal-vehicle collisions can happen any time of the year, but_______ is the peak season for deer- car collisions.
    7·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!