What is the minimum energy required to lift an object weighing 200 Newtons at a height of 20 meters?

Harrizon [31]

Answer:

Hope this helped :

Explanation:

When light passes from a medium with one index of refraction (m1) to another medium with a lower index of refraction (m2), it bends or refracts away from an imaginary line perpendicular to the surface (normal line). As the angle of the beam through m1 becomes greater with respect to the normal line, the refracted light through m2 bends further away from the line.

At one particular angle (critical angle), the refracted light will not go into m2, but instead will travel along the surface between the two media (sine [critical angle] = n2/n1 where n1 and n2 are the indices of refraction [n1 is greater than n2]). If the beam through m1 is greater than the critical angle, then the refracted beam will be reflected entirely back into m1 (total internal reflection), even though m2 may be transparent!

4 0

3 years ago

Read 2 more answers

A 12,500 N alien UFO is hovering about the surface of Earth. At time , its position can be given as () = ((0.24 m/s^3)^3 + 25 m)

stepladder [879]

a) $F=(3675i-4543k)N$

b) 5843 N

Explanation:

a)

The position of the UFO at time t is given by the vector:

$r(t)=(0.24t^3+25)i+(4.2t)j+(-0.43t^3+0.8t^2)k$

Therefore it has 3 components:

$r_x=0.24t^3+25\\r_y=4.2t\\r_z=-0.43t^3+0.8t^2$

We start by finding the velocity of the UFO, which is given by the derivative of the position:

$v_x=r'_x=\frac{d}{dt}(0.24t^3+25)=3\cdot 0.24t^2=0.72t^2\\v_y=r'_y=\frac{d}{dt}(4.2t)=4.2\\v_x=r'_z=\frac{d}{dt}(-0.43t^3+0.8t^2)=-1.29t^2+1.6t$

And then, by differentiating again, we find the acceleration:

$a_x=v'_x=\frac{d}{dt}(0.72t^2)=1.44t\\a_y=v'_y=\frac{d}{dt}(4.2)=0\\a_z=v'_z=\frac{d}{dt}(-1.29t^2+1.6t)=-2.58t+1.6$

The weight of the UFO is W = 12,500 N, so its mass is:

$m=\frac{W}{g}=\frac{12500}{9.8}=1276 kg$

Therefore, the components of the force on the UFO are given by Newton's second law:

$F=ma$

So, Substituting t = 2 s, we find:

$F_x=ma_x=(1276)(1.44t)=(1276)(1.44)(2)=3675 N\\F_y=ma_y=0\\F_z=ma_z=(1276)(-2.58t+1.6)=(1276)(-2.58(2)+1.6)=-4543 N$

So the net force on the UFO at t = 2 s is

$F=(3675i-4543k)N$

b)

The magnitude of a 3-dimensional vector is given by

$|v|=\sqrt{v_x^2+v_y^2+v_z^2}$

where

$v_x,v_y,v_z$ are the three components of the vector

In this problem, the three components of the net force are:

$F_x=3675 N\\F_y=0\\F_z=-4543 N$

Therefore, substituting into the equation, we find the magnitude of the net force:

$|F|=\sqrt{3675^2+0^2+(-4543)^2}=5843 N$

7 0

2 years ago

Q.Solve the following circuit find total resistance RT. Also find value of voltage across resister RC.

vagabundo [1.1K]

Answer:

R_total = 14.57 Ω , V_C = 1.176 V

Explanation:

To solve this circuit we are going to find the equivalent resistance of each branch, let's remember

* Serial resistance

$R_{eq}$ = ∑ $R_{i}$

* For resistance in parallel

1 / R_{eq} = ∑ 1/R_{i}

We solve the two branches of the wheatstone bridge

Series resistors

Branch B

R_B = Rb + R4

R_B = 2 + 18

R_B = 20 Ω

Branch C

R_C5 = Rc + R5

R_C5 = 3 + 12

R_C5 = 15 Ω

Resistance in parallel R_B and R_C5

1 / R_BC = 1 / R_B + 1 / R_C5

1 / R_BC = 1/20 + 1/15 = 0.116666

R_BC = 8.57 Ω

Now we have a single branch, we solve the series resistance

R_total = R_A + R_BC

R_total = 6 + 8.57

R_total = 14.57 Ω

b) they ask us for the voltage in the resistance R_C

Let's remember that the voltage in a series circuit is the sum of the voltages

10 = V_a + V_BC

10 = i R_a + i R_BC = i (R_a + R_BC)

i = 10 / (R_a + R_BC)

i = 10 / (14.57)

i = 0.6863 A

The current in the series circuit is constant

V_BC = i R_BC

V_BC = 0.6863 8.57

V_BC = 5.8819 V

This voltage is divided in the bridge, for the two branches in parallel it is the same, but the resistance is different in each branch.

Branch C

V_BC = i R_C5

i = V_BC / R_C5

i = 5.8819 / 15

i = 0.39213 A

In this branch we have two resistors in series, let's remember that the current in a series circuit is constant

V_C = i R_C

V_C = 0.39213 3

V_C = 1.176 V

3 0

3 years ago

What is the potential energy of a 5kg rock that is 7m high on a hill

Goryan [66]

Explanation:

Gravitational potential energy = mgh = (5)(9.81)(7) = 343.35J.

4 0

2 years ago

A vector has an x component of 25.0 units and a y

sammy [17]

Here is the answer to your question

8 0

3 years ago