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3 years ago

What is any motion in which an object is moving along a curved path

Physics

1 answer:

Usimov [2.4K]3 years ago

4 0

It's called "circular motion"

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How quickly a leaf grows is proportional how big [ie the surface area] the leaf is. If the area of the leaf grows from 2cm2 to 3

marusya05 [52]

Answer: 9 days

Explanation:

Step 1

Let the rate of Leaf growth r be defined as, $\frac{Increase in area}{time taken}$ = $\frac{A1 - A}{t}$

where A is initial area of the leaf, A1 is the final area of the leaf and t is the time taken for the increase in Area.

Express the proportional relationship in equation.

Given that rate of leaf growth, r is proportional to the surface area of the leaf A. we have r ∝ A.

r = kA, where k is the rate constant.

therefore, k = $\frac{r}{A}$

when A = 2 $cm^{2}$ , A1 = 3

so k = $\frac{\frac{3 - 2}{3}}{2}$

= $\frac{1}{3}$ ÷ 2

= 0.33 ÷ 2

k = 0.167

After calculating the rate constant k, we then find the time t when A1 is 5 $cm^{2}$

we have r = k × A1 = $\frac{A1 - A}{t}$

so, 0.167 × 2 = $\frac{5 - 2}{t}$

0.33 = $\frac{3}{t}$ .

t = 3/0.33

Therefore, t = 9 days.

3 0

3 years ago

When a person plucks a guitar string, the number of half wavelengths that fit into the length of the string determines the _____

Vedmedyk [2.9K]

Answer:

Frequency

Explanation:

Each half wavelength has a point of largest amplitude (aka a node). Depending on the wavelength each node oscillates at a certain rate of swings per unit of time. The latter is referred to as frequency and measure in Hertz [Hz].

8 0

3 years ago

Need help with this physics question!

Fynjy0 [20]

Answer:

omg i need help with the same answer lol

Explanation:

i wish i can help but i need help on this hehe

6 0

3 years ago

A radio wave transmits 38.5 W/m2 of power per unit area. A flat surface of area A is perpendicular to the direction of propagati

Margaret [11]

Answer:

$P=2.57\times 10^{-7}\ N/m^2$

Explanation:

Given that,

A radio wave transmits 38.5 W/m² of power per unit area.

A flat surface of area A is perpendicular to the direction of propagation of the wave.

We need to find the radiation pressure on it. It is given by the formula as follows :

$P=\dfrac{2I}{c}$

Where

c is speed of light

Putting all the values, we get :

$P=\dfrac{2\times 38.5}{3\times 10^8}\\\\=2.57\times 10^{-7}\ N/m^2$

So, the radiation pressure is $2.57\times 10^{-7}\ N/m^2$ .

3 0

3 years ago

A 54 kg person stands on a uniform 20 kg, 4.1 m long ladder resting against a frictionless wall.

SVETLANKA909090 [29]

A) Force of the wall on the ladder: 186.3 N

B) Normal force of the ground on the ladder: 725.2 N

C) Minimum value of the coefficient of friction: 0.257

D) Minimum absolute value of the coefficient of friction: 0.332

Explanation:

The free-body diagram of the problem is in attachment (please rotate the picture 90 degrees clockwise). We have the following forces:

$W=mg$ : weight of the ladder, with m = 20 kg (mass) and $g=9.8 m/s^2$ (acceleration of gravity)

$W_M=Mg$ : weight of the person, with M = 54 kg (mass)

$N_1$ : normal reaction exerted by the wall on the ladder

$N_2$ : normal reaction exerted by the floor on the ladder

$F_f = \mu N_2$ : force of friction between the floor and the ladder, with $\mu$ (coefficient of friction)

Also we have:

L = 4.1 m (length of the ladder)

d = 3.0 m (distance of the man from point A)

Taking the equilibrium of moments about point A:

$W\frac{L}{2}sin 21^{\circ}+W_M dsin 21^{\circ} = N_1 Lsin 69^{\circ}$

where

$Wsin 21^{\circ}$ is the component of the weight of the ladder perpendicular to the ladder

$W_M sin 21^{\circ}$ is the component of the weight of the man perpendicular to the ladder

$N_1 sin 69^{\circ}$ is the component of the normal force perpendicular to the ladder

And solving for $N_1$ , we find the force exerted by the wall on the ladder:

$N_1 = \frac{W}{2}\frac{sin 21^{\circ}}{sin 69^{\circ}}+W_M \frac{d}{L}\frac{sin 21^{\circ}}{sin 69^{\circ}}=\frac{mg}{2}\frac{sin 21^{\circ}}{sin 69^{\circ}}+Mg\frac{d}{L}\frac{sin 21^{\circ}}{sin 69^{\circ}}=\frac{(20)(9.8)}{2}\frac{sin 21^{\circ}}{sin 69^{\circ}}+(54)(9.8)\frac{3.0}{4.1}\frac{sin 21^{\circ}}{sin 69^{\circ}}=186.3 N$

Here we want to find the magnitude of the normal force of the ground on the ladder, therefore the magnitude of $N_2$ .

We can do it by writing the equation of equilibrium of the forces along the vertical direction: in fact, since the ladder is in equilibrium the sum of all the forces acting in the vertical direction must be zero.

Therefore, we have:

$\sum F_y = 0\\N_2 - W - W_M =0$

And substituting and solving for N2, we find:

$N_2 = W+W_M = mg+Mg=(20)(9.8)+(54)(9.8)=725.2 N$

Here we have to find the minimum value of the coefficient of friction so that the ladder does not slip.

The ladder does not slip if there is equilibrium in the horizontal direction also: that means, if the sum of the forces acting in the horizontal direction is zero.

Therefore, we can write:

$\sum F_x = 0\\F_f - N_1 = 0$

And re-writing the equation,

$\mu N_2 -N_1 = 0\\\mu = \frac{N_1}{N_2}=\frac{186.3}{725.2}=0.257$

So, the minimum value of the coefficient of friction is 0.257.

Here we want to find the minimum coefficient of friction so the ladder does not slip for any location of the person on the ladder.

From part C), we saw that the coefficient of friction can be written as

$\mu = \frac{N_1}{N_2}$

This ratio is maximum when N1 is maximum. From part A), we see that the expression for N1 was

$N_1 = \frac{W}{2}\frac{sin 21^{\circ}}{sin 69^{\circ}}+W_M \frac{d}{L}\frac{sin 21^{\circ}}{sin 69^{\circ}}$

We see that this quantity is maximum when d is maximum, so when

d = L

Which corresponds to the case in which the man stands at point B, causing the maximum torque about point A. In this case, the value of N1 is:

$N_1 = \frac{W}{2}\frac{sin 21^{\circ}}{sin 69^{\circ}}+W_M \frac{L}{L}\frac{sin 21^{\circ}}{sin 69^{\circ}}=\frac{sin 21^{\circ}}{sin 69^{\circ}}(\frac{W}{2}+W_M)$

And substituting, we get

$N_1=\frac{sin 21^{\circ}}{sin 69^{\circ}}(\frac{(20)(9.8)}{2}+(54)(9.8))=240.8 N$

And therefore, the minimum coefficient of friction in order for the ladder not to slip is

$\mu=\frac{N_1}{N_2}=\frac{240.8}{725.2}=0.332$

Learn more about torques and equilibrium:

brainly.com/question/5352966

#LearnwithBrainly

7 0

3 years ago