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mylen [45]
3 years ago
8

An object experiences an acceleration of -6.8m/s^2. As a result, it accelerates from 54m/s to a complete stop. how much distance

did it travel during that acceleration
Physics
1 answer:
Kipish [7]3 years ago
6 0

Answer:

210 m

Explanation:

Given:

a = -6.8 m/s²

v₀ = 54 m/s

v = 0 m/s

Find: Δx

v² = v₀² + 2aΔx

(0 m/s)² = (54 m/s)² + 2 (-6.8 m/s²) Δx

Δx ≈ 210 m

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A student, who weighs 720N, is standing on a bathroom scale and riding an elevator that is moving downwards with a speed that is
jasenka [17]

Answer:

1) The mass of the student is approximately 73.39 kg

2) The net force on the student is approximately 947.523 N

3) The value the scale will read is approximately 96.59 kg

Explanation:

The given parameters are;

The weight of the student = 720 N

The speed at which the elevator is decreasing = 3.1 m/s²

1) The weight of the student = The mass of the student × The acceleration due to gravity

The acceleration due to gravity is a constant = 9.81 m/s²

Substituting the known values gives;

720 N = The mass of the student × 9.81 m/s²

∴ The mass of the student = 720 N/(9.81 m/s²) ≈ 73.39 kg

2) The forces acting on the student are;

i) The force of gravity which is the weight of the student acting downwards

ii) The inertia force of the slowing elevator acting downwards in the same direction as the weight of the student

The net force, F_{net} = The weight of the student + The inertia force of the slowing elevator

∴ The net force, F_{net} = 720 N + 73.39 kg × 3.1 m/s² ≈ 947.523 N

3) The scale will read the mass of the student as follows;

Mass reading of student on the scale = Force on scale/9.81

∴ Mass reading of student on the scale = 947.523/9.81 ≈ 96.59 kg

The value the scale will read = 96.59 kg.

3 0
3 years ago
A 4.3-kg block slides down an inclined plane that makes an angle of 30° with the horizontal. Starting from rest, the block slide
-BARSIC- [3]

Answer:

0.56

Explanation:

Let the coefficient of friction is μ.

m = 4.3 kg, θ = 30 degree, initial velocity, u = 0, s = 2.7 m, t = 5.8 s

By the free body diagram,

Normal reaction, N = mg Cosθ = 4.3 x 9.8 x Cos 30 = 36.49 N

Friction force, f = μ N = 36.49 μ

Net force acting on the block,

Fnet = mg Sinθ - f = 4.3 x 9.8 x Sin 30 - 36.49 μ

Fnet = 21.07 - 36.49μ

Net acceleartion, a = Fnet / m

a = (21.07 - 36.49μ) / 4.3

use second equation of motion

s = ut + 1/2 a t^2

2.7 = 0 + 1/2 x (21.07 - 36.49μ) x 5.8 x 5.8 / 4.3

By solving we get

μ = 0.56

8 0
3 years ago
True or False<br><br> An object will slow down and stop as long as no forces act upon it
My name is Ann [436]

Answer:

false

Explanation:

Understand Newton’s first law of motion. Experience suggests that an object at rest will remain at rest if left alone, and that an object in motion tends to slow down and stop unless some effort is made to keep it moving. An object will go on forever if there were no forces to act upon it. In space, a rocket will go forever at the speed it projects.

5 0
3 years ago
Read 2 more answers
Jenny wants to determine if a new brand name washing detergent is better than her old detergent. She washes ten dirty t-shirts w
zmey [24]

The independent variable would be the shirts you are using the dependent variable would be the detergent.

Hope this helps!

6 0
3 years ago
An owl has a mass of 4.00 kg. It dives to catch a mouse, losing 800.00 J of its GPE. What was the starting height of the owl, in
vesna_86 [32]

Answer:

height =20m

Explanation:

gpe=mgh

800=4×10×x

40x=800

x=20

3 0
3 years ago
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