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2 years ago

Where is the distance on a position time graph

Physics

1 answer:

S_A_V [24]2 years ago

4 0

Answer:

The distance represents the difference of the first position and last position of the body.

Explanation:

For example, if y axis represents the position axis, and the first position is 3, second 9 we can see that the distance is a (positive) projection of one position into another. 9-3=6

Hope this helps.

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The power radiated by the sun is 3.90 1026 W. The earth orbits the sun in a nearly circular orbit of radius 1.50 1011 m. The ear

WITCHER [35]

Answer:

3234.2 W

Explanation:

Since intensity I = Power/Area. The intensity of the light from the sun, I = power radiated by sun/area of sphere of radius, r = 1.5 × 10¹¹ m.

So, I = 3.9 10²⁶W/4π(1.5 × 10¹¹ m)² = 2.069 × 10³ W/m².

Now, the power radiated on the patch of area 0.570 m² at the equator is

P = Icos27/A = 2.069 × 10³ W/m² cos27/0.570 m² = 1843.49/0.570 = 3234.2 W

6 0

2 years ago

horizontal force pushes block up a 20.0 incline with an initial speed 12.0 m//s. a) how high up the plane does slide before comi

Annette [7]

Answer:

Explanation:

We shall apply conservation of mechanical energy .

initial kinetic energy = 1/2 m v²

= .5 x m x 12 x 12

= 72 m

This energy will be spent to store potential energy . if h be the height attained

potential energy = mgh , h is vertical height attined by block

= mg l sin20 where l is length up the inclined plane

for conservation of mechanical energy

initial kinetic energy = potential energy

72 m = mg l sin20

l = 72 / g sin20

= 21.5 m

deceleration on inclined plane = g sin20

= 3.35 m /s²

v = u - at

t = v - u / a

= (12 - 0) / 3.35

= 3.58 s

it will take the same time to come back . total time taken to reach original point = 2 x 3.58

= 7.16 s

7 0

2 years ago

What is it called when sediment is dropped and comes to rest?

larisa86 [58]

I think d would be the answer.

3 0

3 years ago

Read 2 more answers

A sculpture is suspended in equilibrium by two cables, one from a wall and the other

aleksandrvk [35]

Answer:

$T_1=6655.295917 \approx 6655.3N$

Explanation:

From the question we are told that

Angle of cable 2 $\theta=37.0\textdegree$

Weight of sculpture $W=5000 N$

Generally the Tension from cable 2 $T_2$ is mathematically given by

$T_2sin37\textdegree=5000N$

$T_2=5000N/sin37\textdegree$

$T_2=8308.2N$

Generally the Tension from Cable 1 $T_1$ is mathematically given by

$T_1=T_2 cos37\textdegree$

$T_1=8308.2* cos 37\textdegree$

$T_1=6655.295917 \approx 6655.3N$

7 0

3 years ago

Using a scale diagram, calculate the resultant force acting on a sailing boat when an easterly wind provides 2, point, 50, k, N,

EastWind [94]

Answer:

F = 3.6 kN, direction is 9.6º to the North - East

Explanation:

The force is a vector, so one method to find the solution is to work with the components of the vector as scalars and then construct the resulting vector.

Let's use trigonometry to find the component of the forces, let's use a reference frame where the x-axis coincides with the East and the y-axis coincides with the North.

Wind

X axis

F₁ = 2.50 kN

Tide

cos 30 = F₂ₓ / F₂

sin 30 = F_{2y} / F₂

F₂ₓ = F₂ cos 30

F_{2y} = F₂ sin 30

F₂ₓ = 1.20cos 30 = 1.039 kN

F_{2y} = 1.20 sin 30 = 0.600 kN

the resultant force is

X axis

Fₓ = F₁ₓ + F₂ₓ

Fₓ = 2.50 +1.039

Fₓ = 3,539 kN

F_y = F_{2y}

F_y = 0.600

to find the vector we use the Pythagorean theorem

F = $\sqrt{F_x^2 +F_y^2}$

F = $\sqrt{ 3.539^2 + 0.600^2 }$

F = 3,589 kN

the address is

tan θ = F_y / Fₓ

θ = tan⁻¹ $\frac{F_y}{F_x}$

θ = tan⁻¹ $\frac{0.6}{3.539}$ 0.6 / 3.539

θ = 9.6º

the resultant force to two significant figures is

F = 3.6 kN

the direction is 9.6º to the North - East

7 0

2 years ago