In a closed system that has 45 J of mechanical energy, the gravitational

A jogger runs 10.0 blocks do east, 5.0 blocks due South, and another two. Zero blocks do east. Assume all blocks are equal size,

Annette [7]

Jogger moves in three displacements

d1 = 10 blocks East

d2 = 5 blocks South

d3 = 2 blocks East

now we can say

total displacement towards East direction will be

$d_x = 10 + 2= 12 blocks$

Total displacement towards South

$d_y = 5 block$

now to find the net displacement we can use vector addition

$d = \sqrt{d_x^2 + d_y^2}$

$d = \sqrt{12^2 + 5^2}$

$d = 13 blocks$

<em>so magnitude of net displacement will be equal to 13 blocks</em>

6 0

2 years ago

A wave is propagating from left to right in a medium. The particles in the medium are also vibrating from left to right. What ki

Anna71 [15]

Based on the direction of propagation compared to direction of vibration, waves are classified into:
1- Transverse waves: The direction of propagation of the wave is perpendicular to the direction of vibration of the medium particles.
2- Longitudinal waves: The direction of propagation of the wave is the same as the direction of vibration of the medium particles.

For the question we have here, since the direction of the wave is the same as the direction of vibration of particles, therefore, this wave is a longitudinal wave

6 0

2 years ago

Some plants disperse their seeds when the fruit splits and contracts, propelling the seeds through the air. The trajectory of th

Anton [14]

Answer:

Option B, 93 cm

Explanation:

An diagram of the seed's motion is attached to this solution.

This is very close to a projectile motion question. And the quantity to be calculated, how far along the grant a seed released would travel is called the Range.

And this would be obtained from the equations of motion,

First of, the height of the plant is related to some quantities of the motion with this relation.

H = u(y) t + 0.5g(t^2)

U(y) = initial vertical component of velocity = 0 m/s, H = height at which motion began, = 20cm = 0.2 m

That means t = √(2H/g)

The horizontal distance covered, R,

R = u(x) t + 0.5g(t^2) = u(x) t (the second part of the equation goes to zero as the vertical component of the acceleration of this motion is 0)

(substituting the t = √(2H/g) derived from above

R = u(x) √(2H/g)

Where u(x) = the initial horizontal component of the bomb's velocity = maximum initial speed, that is, 4.6 m/s, H = vertical height at which the seed was released = 20 cm = 0.2 m, g = acceleration due to gravity = 9.8 m/s2

R = 4.6 √(2×0.2/9.8) = 0.929 m = 0.93 m = 93 cm. Option B.

QED!