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saveliy_v [14]
4 years ago
15

Can work done=mass*acceleration*displacement(work=m*a*s)

Physics
1 answer:
Airida [17]4 years ago
5 0

no, work is = force * distance or displacement


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What type of electromagnetic radiation is being used in the picture?
pashok25 [27]

Answer: b i think

Explanation:

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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:

<h2>a) </h2>

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)

<h2>B)</h2>

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}

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3 years ago
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Answer:

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2 years ago
Erica is participating in a road race. The first part of the race is on a 5.2-mile-long straight road oriented at an angle of 25
rjkz [21]

Answer:

8.6 miles

Explanation:

In order to calculate the straight-line distance between the starting point and the ending point, we need to resolve each part of the displacement into two perpendicular directions, let's say horizontal (x) and vertical (y) directions.

In the first part, Erica moves for 5.2 miles 25∘ north of east. Resolving into the two directions:

d_{1x} = 5.2 cos 25^{\circ}=4.7 mi\\d_{1y} = 5.2 sin 25^{\circ} =2.2 mi

In the second part, Erica moves 5.0 miles north, so the components are

d_{2x} = 0\\d_{2y} = 5.0 mi

So the components of the net displacement are

d_x = d_{1x}+d_{2x}=4.7+0 = 4.7 mi\\d_y = d_{1y}+d_{2y} =2.2+5.0 = 7.2 mi

And the magnitude of the net displacement is

d=\sqrt{d_x^2 +d_y^2}=\sqrt{4.7^2+7.2^2}=8.6 mi

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3 years ago
G How much buoyancy force, in N, a person with a mass of 70 kg experiences by just standing in air
sattari [20]

Answer:

686.7N

Explanation:

Given data

Mass= 70kg

We know that the buoyant force experienced by the person is equal to the weight of the person

Hence the weight is

Weight = mass* Acceleration

Weight= 70*9.81

Weight= 686.7N

Therefore the weight is 686.7N

5 0
3 years ago
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