What uses of iron are due to its chemical properties

A fisherman is fishing from a bridge and is using a "42.0-N test line." In other words, the line will sustain a maximum force of

lara31 [8.8K]

Answer:

(a) 42 N

(b)36.7 N

Explanation:

Nomenclature

F= force test line (N)

W : fish weight (N)

Problem development

(a) Calculating of weight of the heaviest fish that can be pulled up vertically, when the line is reeled in at constant speed

We apply Newton's first law of equlibrio because the system moves at constant speed:

∑Fy =0

F-W= 0

42N -W =0

W = 42N

(b) Calculating of weight of the heaviest fish that can be pulled up vertically, when the line is reeled with an acceleration whose magnitude is 1.41 m/s²

We apply Newton's second law because the system moves at constant acceleration:

m= W/g , m= W/9.8 , m:fish mass , W: fish weight g:acceleration due to gravity

∑Fy =m*a

m= W/g , m= W/9.8 , m:fish mass , W: fish weight g:acceleration due to gravity

F-W= ( W/9.8 )*a

42-W= ( W/9.8 )*1.41

42= W+0.1439W

42=1.1439W

W= 42/1.1439

W= 36.7 N

8 0

3 years ago

In a Young’s interference experiment, the two slits are separated by 0.150 mm and the incident light includes two wavelengths: l

dezoksy [38]

Answer:

2.52 × 10⁻² cm

Explanation:

The distance of bright fringe from the center of the screen is given by the formula

$y = \frac{m\lambda D}{d}$

Here, wavelength is λ, Distance of the screen from the slits is D, seperation between the

slits is d.

Separation between the slits, d = 0.15 mm

$= 0.15 * 10^{-3} m$

Distance of the screen from the slits = 1.40 m

We have a wavelength, λ1 = 540 nm

= $540 * 10^{-9} m$

By substituting all these values in the above equation we get

y1 = mλD/d

$y1 = m(540 \times 10^{-9} m)(1.40 m)/(0.15 \times 10^{-3} m)$

$y1 = m(5.04 * 10^{-3} m)$

We have a wavelength, λ2 = 450 nm

= $450 * 10^-9 m$

By substituting all these values in the above equation we get

$y_2 = \frac{m\lambda D}{d}$

$y_2 = m(450 * 10^{-9} m)(1.40 m)/(0.15 * 10^{-3} m)$

$y_2 = m'(4.20 * 10^{-3} m)$

According to the problem, these two distance are coincides with each other.

So,

$y_1 = y_2$

$m(5.04 * 10^{-3} m) = m'(4.20 * 10^{-3} m)$

by testing values, the above equation is satisfied only when, m = 5 and m' = 6

Then from the above we have

y1 = y2 = 0.0252 m

= 2.52 × 10⁻² cm

8 0

3 years ago

If a 12 kg cat is sitting 5 m up in a tree, how much PE does it have?

Helen [10]

Answer:

588 J

Explanation:

PE (potential energy) = (mass) x (gravity) x (height)

mass = 12 kg

gravity = 9.8m/s^2

height = 5 m

PE = (12) x (9.8) x (5) = 588 J (Joules)

5 0

3 years ago

Q4. Consider the skier on a slope shown in the figure below. Her mass including equipment is 55.0 kg.

Shkiper50 [21]

Answer:

Part a)

When there is no friction then acceleration is

$a = 4.14 m/s^2$

Part b)

if there is friction force along the inclined then acceleration is

$a = 3.33 m/s^2$

Explanation:

Part a)

As we know that the skier is on inclined plane

So here if there is no friction then net force along the inclined plane is given as

$F = mg sin\theta$

now acceleration of the skier is given as

$a = \frac{F}{m}$

$a = g sin\theta$

$a = 9.81(sin25)$

$a = 4.14 m/s^2$

Part b)

if there is friction force along the inclined then net force along the inclined plane is given as

$F = mg sin\theta - F_f$

now acceleration of the skier is given as

$a = \frac{F}{m}$

$a = g sin\theta - \frac{F_f}{m}$

$a = 9.81(sin25) - \frac{45}{55}$

$a = 3.33 m/s^2$

5 0

4 years ago

The ________ is the maximum value or strength of the signal over time; typically this value is measured in volts.

Harrizon [31]

Peak value is what we say to the maximum value of the signal..

or rms value..root mean square value

5 0

3 years ago