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PolarNik
3 months ago
14

A 5.0-kilogram box is sliding across a level floor. The box is acted upon by a force of 27 newtons east and a frictional force o

f 17 newtons west. What is the magnitude of the acceleration of the box?

Physics
2 answers:
Sav [3.1K]3 months ago
8 0

The acceleration of the box has a magnitude of 2.0 m/s²

\texttt{ }

Expanded explanation

Newton's second law of motion indicates that the total force acting on an object is directly proportional to the mass and acceleration of that object.

\large {\boxed {F = ma }

F = Force ( Newton )

m = Mass of the Object ( kg )

a = Acceleration ( m )

Now, let’s approach this problem!

\texttt{ }

Provided:

mass of box = m = 5.0 kg

force applied = F = 27 N

frictional force = f = 17 N

Query:

find the box's acceleration = a =?

Solution:

We will apply Newton's Law of Motion to resolve this:

\Sigma F = ma

F - f = ma

27 - 17 = 5.0a

10 = 5.0a

a = 10 \div 5.0

a= 2.0 \texttt{ m/s}^2

\texttt{ }

Final answer:

The box's acceleration has a magnitude of 2.0 m/s²

\texttt{ }

Further learning

  • Effects of Gravity:
  • Impact of Earth’s Gravity on Objects:
  • Acceleration Due To Gravity:
  • Newton's Laws of Motion:
  • Illustration of Newton's Laws:

\texttt{ }

Answer Clarifications

Grade Level: High School

Subject Focus: Physics

Chapter: Dynamics

Yuliya22 [3.3K]3 months ago
6 0

Answer:

The box experiences an acceleration with a magnitude of 2 m/s².

Explanation:

Details provided:

Mass of the box: m=5.0 kg

Force acting towards the east: F=27 N

Frictional force acting towards the west: f=17 N

Let’s denote the acceleration as a m/s².

Thus, the net force acting on the box in the east direction is stated as:

F_{net}=F-f=27-17=10\textrm{ N}

According to Newton's second law of motion,

F_{net}=ma\\10=5.0\times a\\a=\frac{10}{5.0}=2\textrm{ }m/s^2

Thus, the magnitude of the box's acceleration is determined to be 2 m/s².

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A baseball is hit high into the upper bleachers of left field. Over its entire flight the work done by gravity and the work done
serg [3582]

Answer:

Explanation:

As the baseball ascends, gravitational forces as well as air resistance act downward, whereas the displacement is moving upward which results in an angle of 180° between the force and displacement. Therefore, the work done by both the gravitational force and air resistance is negative, confirming option (d) as accurate.

6 0
3 months ago
The intensity level of a "Super-Silent" power lawn mower at a distance of 1.0 m is 100 dB. You wake up one morning to find that
Maru [3345]

Answer:

The correct option is 80 dB.

Explanation:

The transformation of sound intensity level into sound intensity utilizes the formula

[D] = 10 log (I/I₀)

Where I₀ = 10⁻¹² W/m²

[D] results in 100 dB

100 = 10 log (I/I₀)

Log (I/I₀) converts to 10

(I/I₀) = 10¹⁰

I is determined as I₀ × 10¹⁰ = 10⁻¹² × 10¹⁰ = 10⁻² W/m²

Sound intensity inversely relates to the square of the distance from the source.

I ∝ (1/d²)

I can be expressed as k/d²

When d = 1 m, the intensity is 10⁻² W/m²

Thus, 0.01 = k/1

Providing that k = 0.01 W

For d = 20 m, we can calculate I

I = 0.01/20² = 2.5 × 10⁻⁵ W/m²

With four neighbors mowing their lawns concurrently,

I = 4 × 2.5 × 10⁻⁵ = 10⁻⁴ W/m⁻²

The sound intensity level in decibels is represented as

[D] = 10 log (I/I₀)

[D] = 10 log (10⁻⁴/10⁻¹²)

[D] = 10 log (10⁸)

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4 0
3 months ago
A delivery man starts at the post office, drives 40 km north, then 20 km west, then 60 km northeast, and finally 50 km north to
Yuliya22 [3333]



assuming north-south is along the Y-axis and east-west along the X-axis

X = total X-displacement

from the graph, total displacement in the X-direction is computed as

X = 0 - 20 + 60 Cos45 + 0

X = 42.42 - 20

X = 22.42 m


Y = total Y-displacement

from the graph, total displacement in the Y-direction is computed as

Y = 40 + 0 + 60 Sin45 + 50

Y = 90 + 42.42

Y = 132.42 m

To calculate the magnitude of the net displacement vector, we apply the Pythagorean theorem, yielding

magnitude: Sqrt(X² + Y²) = Sqrt(22.42² + 132.42²) = 134.31 m

Direction: tan⁻¹(Y/X) = tan⁻¹(132.42/22.42) = 80.4 deg north of east


4 0
3 months ago
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