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3 years ago

1. What are three appliances in your house that use a motor that turns electrical energy into mechanical energy?

Physics

1 answer:

AleksAgata [21]3 years ago

6 0

A vacuum is an electrical motor and<span> which it converts electrical energy into mechanical energy.

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A go-cart and rider have a mass of 14 kg. If the cart accelerates at 6m/s^2 during a 40 m sprint in 100 seconds, how much power

lara [203]

Answer:

P = 33.6 [N]

Explanation:

To solve this problem we must use Newton's second law, which tells us that the sum of forces on a body is equal to the product of mass by acceleration.

∑F = m*a

where:

F = forces [N]

m = mass = 14 [kg]

a = acceleration = 6 [m/s²]

$F = 14*6\\F = 84 [N]$

In the second part of this problem we must find the work done, where the work in physics is known as the product of force by distance, it is important to make it clear that force must be applied in the direction of movement.

$W = F*d$

where:

W = work [J]

F = force = 84 [N]

d = displaciment = 40 [m]

$W = 84*40\\W = 3360 [J]$

Finally, the power can be calculated by the relationship between the work performed in a given time interval.

$P=W/t\\$

where:

P = power [W]

W = work = 3360 [J]

t = time = 100 [s]

Now replacing:

$P=3360/100\\P=33.6[W]$

The power is given in watts

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3 years ago

Select all that Apple

egoroff_w [7]

B is the correct answer since acid corrode on metals such as carbon, steel, zinc and such.

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4 years ago

Suppose a rocket is traveling through space and moving at a speed close to the speed of light. Three minutes pass on the rocket

Sedbober [7]

Answer:

Explanation:

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3 years ago

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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:

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