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hjlf
3 years ago
6

To describe velocity, you need to know ____. (1 point)

Physics
2 answers:
kati45 [8]3 years ago
8 0
The correct answer that would best complete the given statement above would be the first option. <span>To describe velocity, you need to know speed and direction. When we say velocity, this is the speed of something in a given direction. To solve for velocity, its formula is similar to speed which is distance over time, and just include the direction. Hope this answer helps.</span>
son4ous [18]3 years ago
3 0
<span>To describe velocity, you need to know speed and direction of motion.</span>
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Answer:

746 watts

Explanation:

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3 years ago
How much work is done when you push a crate horizontally with 130
Harrizon [31]
Work= Force in the direction of displacement*displacement.

You know the force in the direction of displacement (horizontally) and the displacement. So,

W=130*11=1430 

Therefore, the work done is 1,430 Joules


3 0
3 years ago
A package is dropped from a helicopter moving upward at 15 m/s
daser333 [38]

The distance the package above the ground when it was released, s ≈ 530 meters

<h3 /><h3>What are kinematic equations?</h3>

The kinematic equation of motion gives the interrelationships of the variables of motion.The correct option for the distance the package above the ground when it was released, is the third option;

It is given that:

The velocity of the helicopter from which the package was dropped = 15 m/s

The time it takes the package to strike the ground = 12 seconds

The required parameter:

The height of the package from the ground when it was dropped

The kinematic equation of motion relating distance, s, time, t, acceleration due to gravity, g, initial velocity, u, and final velocity, v, is applied as follows;

The package continues the upward motion for some time, t₁, given as follows;

Upward motion of the package

v = u - g·t₁

v = 0 at highest point reached by the package;

Therefore;

0 = 15 m/s - 9.81 m/s²  × t₁

t₁ = 15 m/s/(9.81 m/s²) ≈ 1.5295022 seconds

The time the package takes to return to the initial starting point, t₂ = t₁

The time the package falls after returning to the point it was dropped, t₃, is given as follows;

t₃ = t - (t₂ + t₁) = t - 2 × t₁

∴ t₃ = 12 s - 2 × 1.5295022 s ≈ 8.940996 s

From the symmetry of the motion of a projectile, the velocity of the package when returns to its staring point where it was dropped = u (Downwards) = 15 m/s

The distance the package falls, s, which is the distance the package above the ground when it was released, is given as follows;

s = u·t + (1/2)·g·t²

s = 15× 8.940996  + (1/2) × 9.81 × 8.940996² = 526.22755346 ≈ 530

The distance the package falls, s ≈ 530 m = The height of the

The distance the package above the ground when it was released, s ≈ 530 meters

Learn more about the kinematic equations of motion here:

brainly.com/question/16995301

#SPJ4

7 0
2 years ago
1)what is the momentum of a 40,000,000kg tanker traveling at 5 m/s?
Fittoniya [83]

Answer:

1)4*10^7 * 5= 2*10^8

2)50/10=5

5 0
3 years ago
An incident ray that passes through the vertex of a convex lens:
WINSTONCH [101]
The answer is refracts parallel to the axis of the lens
7 0
3 years ago
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