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3 days ago

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Simone is walking her dog on a leash. The dog is pulling with a force of 32 N to the right and Simone is pulling backward with a

force of 16 N. What is the net force on them?

2 answers:

ValentinkaMS [1.1K]3 days ago

6 0

Conclusion:

The total net force acting on the objects is 16 N, directed towards the right.

Clarification:

It is stated that,

The force exerted by the dog, $F_1 = 32\ N$ (to the right)

The force exerted by Simone, $F_2 = -16\ N$ (backward)

Here, assume the backward direction is negative and the right direction is positive.

The net force will move in the direction where the larger force is present. The net force can be calculated as:

$F=F_1+F_2$

$F=32+(-16)$

F = 16 N

Thus, the net force amounts to 16 N, acting towards the right.

Sav [1K]3 days ago

5 0

Conclusion:

16 N (to the right)

Clarification:

The force pulling to the right, F1 = 32 N

The force pulling to the left, F2 = 16 N

The net force is calculated as F1 - F2

F = 32 - 16 = 16 N (to the right)

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When a mass of 25 g is attached to a certain spring, it makes 20 complete vibrations in 4.0 s. what is the spring constant of th

kicyunya [1011]

Response: The spring constant is 25 N/m.

Details:

The body’s mass is 25 g, which converts to 0.025 kg (since 1 kg = 1000 g).

The total oscillations are 20 in 4 seconds.

Oscillations per second = $\frac{20}{4}=5$

Spring's frequency of vibration is = $5 s^{-1}=5 Hz$

The spring constant 'k' can be derived from the relationship involving frequency, mass, and spring constant.

$Frequency=\frac{1}{2\pi}\times \sqrt{\frac{k}{m}}$

$5 s^{-1}=\frac{1}{2\times 3.14}\times \sqrt{\frac{k}{0.025 kg}}$

$k=24.649 N/m\approx 25 N/m$

The spring constant is 25 N/m.

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7 days ago

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Astronomers determine that a certain square region in interstellar space has an area of approximately 2.4 \times 10^72.4×10 7

Sav [1095]

Answer:

1.5 × 10³⁶ light-years

Explanation:

A particular square area in interstellar space measures roughly 2.4 × 10⁷² (light-years)². To find the area of a square, the following formula is utilized:

A = l²

where,

A represents the area of the square

l denotes the length of one side of the square

Thus, l = √A = √2.4 × 10⁷² (light-years)² = 1.5 × 10³⁶ light-years

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6 days ago

A block of mass m, initially held at rest on a frictionless ramp a vertical distance H above the floor, slides down the ramp and

kicyunya [1011]

Answer:

$\displaystyle W=-m.g.H$

Explanation:

Transformation of Energy

Also known as energy conversion, this refers to the process in which energy shifts from one type to another. In this context, three energy forms are involved. When the object is stationary at the ramp's peak, it possesses gravitational potential energy, calculated as

$U=m.g.H$

As the object descends the frictionless ramp, it converts all its potential energy into kinetic energy, represented as

$\displaystyle K=\frac{m.v^2}{2}$

Thus,

$K=m.g.H$

Ultimately, when the object encounters a rough surface, all energy converts to thermal energy. The work performed by the friction force corresponds to the alteration in kinetic energy, as all velocity is lost in this process:

$\displaystyle W=\Delta E=K_f-K=0-K=-\frac{m.v^2}{2}$

Given the kinetic energy equals the initial potential energy:

$\boxed{\displaystyle W=-m.g.H}$

The negative sign indicates that the work acted against the direction of movement, meaning the force and displacement are 180° apart.

This outcome is independent of the distance D needed to halt the block or the kinetic friction coefficient.

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11 days ago

A man makes a 27.0 km trip in 16 minutes. (a.) How far was the trip in miles? (b.) If the speed limit was 55 miles per hour, wa

Maru [1053]

Answer:

(a) 16.777 miles

(b) Yes, he exceeded the speed limit

Explanation:

(a)

We need to perform the necessary calculations to convert kilometers to miles:

$27km*\frac{1000m}{1km} *\frac{1mi}{1609.34m} =16.77706389mi$

Thus, the distance of the trip in miles is:

$d=16.77706389mi$

(b)

Next, we will compute the man's speed during the journey:

$v=\frac{d}{t}$

Before that, we must convert minutes to hours:

$16min*\frac{1h}{60min} =2.666666667h$

The resulting speed is:

$v=\frac{16.77706389mi}{2.666666667h} =62.91398959\frac{mi}{h}$

Consequently:

$62.91398959\frac{mi}{h}>55\frac{mi}{h}$

Thus, it can be concluded that the driver was speeding

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15 hours ago

Derive an algebraic equation for the vertical force that the bench exerts on the book at the lowest point of the circular path i

Keith_Richards [1021]

a)

i) 120 s

ii) 1.57 m/s

b)

i) Refer to the attached diagram

ii) Up

c) $N=mg+m\frac{v_b^2}{R}$

d) Greater than

Explanation:

The problem does not provide full details: consult the attachments for the complete text.

a)

The revolution period of the book equals the total duration needed for the book to make one full revolution.

By examining the graph, we can approximate the revolution period by calculating the time difference between two successive points of the book's motion that share the same shape.

We could use the time difference between two adjacent crests to estimate the period. The first crest is observed at t = 90 s, and the following crest appears at t = 210 s.

This results in the revolution period being

T = 210 - 90 = 120 s

ii)

The tangential speed of the book is computed as the ratio of the distance traveled over one revolution (i.e., the circumference of the wheel) to the revolution period.

Mathematically:

$v_b=\frac{2\pi R}{T}$

where

R represents the wheel radius

T = 120 s indicates the period

Based on the graph, the book reaches a maximum at x = +30 m and a minimum at x = -30 m, giving the diameter of the wheel as

d = +30 - (-30) = 60 m

This means the radius calculates to

R = d/2 = 30 m

So, the final speed is

$v_b=\frac{2\pi (30)}{120}=1.57 m/s$

b)

i) Please consult the attached free-body diagram for the book when at its lowest point.

Two forces act on the book at the lowest position:

- The weight of the book, represented as

$W=mg$

where m denotes the book's mass and g stands for gravitational acceleration. This force functions downward.

- The normal force the bench exerts on the book is represented by N. This force acts upward.

ii)

While at its lowest position, the book maintains a horizontal motion at constant speed.

Nevertheless, the book is undergoing acceleration. Acceleration is defined as the rate of velocity change, which is vectorial, having both speed and direction. While the speed remains unchanged, the direction changes (upward), indicating the book has upward net acceleration.

According to Newton's second law, the net vertical force acting on the book corresponds with the vertical acceleration:

$F=ma$

where F = net force, m = mass, a = acceleration. Thus, if a is non-zero, the upward net force must exist in line with the direction of the acceleration.

c)

As discussed in part b), there are two forces influencing the book at the lowest point:

- The weight, $W=mg$ , directed downward

- The normal force from the bench, N, directed upward

Given that the book is in uniform circular motion, the net force must match the centripetal force $m\frac{v_b^2}{R}$ , leading us to the equation:

$N-mg=m\frac{v_b^2}{R}$

where

$v_b$ represents the speed of the book

R stands for the radius of the circular path.

We derive an expression for the normal force:

$N=mg+m\frac{v_b^2}{R}$

d)

As per the discussions in parts c) and d):

- The normal force acting on the book at its lowest point becomes

$N=mg+m\frac{v_b^2}{R}$

- The weight (gravitational force) of the book is

$W=mg$

Upon comparing these two equations, we conclude:

$N>W$

Thus, it is evident that the normal force exerted by the bench exceeds the weight of the book.

4 0

13 days ago