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2 years ago

How can wasted energy be made useful?

Physics

1 answer:

murzikaleks [220]2 years ago

7 0

Answer:

by using it in a such place or thing which needs it or which can work with it

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Question 30

stellarik [79]

Answer: $0.69\°$

Explanation:

The angular diameter $\delta$ of a spherical object is given by the following formula:

$\delta=2 sin^{-1}(\frac{d}{2D})$

Where:

$d=16 m$ is the actual diameter

$D=1338 m$ is the distance to the spherical object

Hence:

$\delta=2 sin^{-1}(\frac{16 m}{2(1338 m)})$

$\delta=0.685\° \approx 0.69\°$ This is the angular diameter

3 0

3 years ago

Consider the points below. P(1, 0, 1), Q(−2, 1, 4), R(6, 2, 7) (a) Find a nonzero vector orthogonal to the plane through the poi

kozerog [31]

Answer:

a) (0, -33, 12)

b) area of the triangle : 17.55 units of area

Explanation:

We know that the cross product of linearly independent vectors $\vec{A}$ and $\vec{B}$ gives us a nonzero, orthogonal to both, vector. So, if we can find two linearly independent vectors on the plane through the points P, Q, and R, we can use the cross product to obtain the answer to point a.

Luckily for us, we know that vectors $\vec{A} = \vec{P}-\vec{Q}$ and $\vec{B} = \vec{R} - \vec{Q}$ are living in the plane through the points P, Q, and R, and are linearly independent.

We know that they are linearly independent, cause to have one, and only one, plane through points P Q and R, this points must be linearly independent (as the dimension of a plane subspace is 3).

If they weren't linearly independent, we will obtain vector zero as the result of the cross product.

So, for our problem:

$\vec{A} = \vec{P} - \vec{Q} \\\\\vec{A} = (1,0,1) - (-2,1,4)\\\\\vec{A} = (1 +2,0-1,1-4)\\\\\vec{A} = (3,-1,-3)$

$\vec{B} = \vec{R} - \vec{Q} \\\\\vec{B} = (6,2,7) - (-2,1,4)\\\\\vec{B} = (6 +2,2-1,7-4)\\\\\vec{B} = (8,1,3)$

$\vec{A} \times \vec{B} = (A_y B_z - B_y A_z) \ \hat{i} - ( A_x B_z-B_xA_z) \ \hat{j} + (A_x B_y - B_x A_y ) \ \hat{k}$

$\vec{A} \times \vec{B} = ( (-1) * 3 - 1 * (-3) ) \ \hat{i} - ( 3 * 3 - 8 * (-3)) \ \hat{j} + (3 * 1 - 8 * (-1) ) \ \hat{k}$

$\vec{A} \times \vec{B} = ( - 3 + 3 ) \ \hat{i} - ( 9 + 24 ) \ \hat{j} + (3 + 8 ) \ \hat{k}$

$\vec{A} \times \vec{B} = 0 \ \hat{i} - 33 \ \hat{j} + 12 \ \hat{k}$

$\vec{A} \times \vec{B} =(0, -33, 12)$

We know that $\vec{A}$ and $\vec{B}$ are two sides of the triangle, and we also know that we can use the magnitude of the cross product to find the area of the triangle:

$|\vec{A} \times \vec{B} | = 2 * area_{triangle}$

so:

$\sqrt{(-33)^2 + (12)^2} = 2 * area_{triangle}$

$\sqrt{1233} = 2 * area_{triangle}$

$35.114= 2 * area_{triangle}$

$17.55 \ units \ of \ area = area_{triangle}$

5 0

3 years ago

I need help plz and thank you this is do

love history [14]

Answer:

Oh I am sorry this is my first time on brainly i dont how to exit and sorry but dont know the answer

Explanation:

4 0

3 years ago

A weightlifter is attempting a biceps curl with a 200 n barbell. the moment arm of the barbell about the elbow joint is 40 cm. t

CaHeK987 [17]

Weight of the barbell W = 200 Ndistance of the joint is r = 40 cm = 0.4 mtorque created by the weight at the joint is τ = F*r = 200 N*0.4 m = 80 N.mat equilibrium condition , Στ = force*distance - 80 N.m = 0 F'*0.4 - 80 N.m = 0 F'*0.4 = 80 force F' = 200 N

4 0

3 years ago

Q1. Helmut rides 5km in 2h on his bike. His speed was *

baherus [9]

Answer:

Speed of bike = 2.5 km/h

Distance travel = 1,000 km (Approx.)

Explanation:

Given:

Distance cover by Helmut = 5 km

Time taken = 2 hour

Find:

Speed of bike

Computation:

Speed = Distance / Time

Speed of bike = 5 / 2

Speed of bike = 2.5 km/h

Given:

Speed of plane = 250 km/h

time taken = 3 hr 58 min = 3.967 hr

Find:

Distance travel

Computation:

Distance = Speed x time

Distance travel = 250 x 3.967

Distance travel = 991.669

Distance travel = 1,000 km (Approx.)

8 0

3 years ago