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3 years ago

How to calculate displacement?

Physics

1 answer:

Blizzard [7]3 years ago

4 0

-- Take a straight ruler.

-- Lay it down with the 'zero' mark at the start point.

-- Rotate it around the start point until the end point is also touching the edge of the ruler.

-- From the marks on the ruler, read the straight-line distance from the start point to the end point.

-- Without moving the ruler, observe and write down the DIRECTION from the start point to the end point.

-- The Displacement is the straight-line distance and direction from the start point to the end point.

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Which force is always pulling but never pushes and is directly influenced by mass

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Read 2 more answers

A charge of 8.5 × 10–6 C is in an electric field that has a strength of 3.2 × 105 N/C. What is the electric force acting on the

Sonja [21]

2.72 N

Explanation:

Step 1:

From the basic formula in electrostatics

F = E * q

where F = Force due to charges

E = Electric field strength

q = Charge

Step 2:

From the given question

q= $8.5*10^{-6} C$

E = $3.2 * 10^{5} N/C$

F = $8.5 * 10^{-6} * 3.2 * 10^{5} = 2.72$ N

8 0

3 years ago

A 1.30-m string of weight 0.0125 N is tied to the ceiling at its upper end, and the lower end supports a weight W. Neglect the v

atroni [7]

Answer:

1. t = 0.0819s

2. W = 0.25N

3. n = 36

4. y(x , t)= Acos[172x + 2730t]

Explanation:

1) The given equation is

$y(x, t) = Acos(kx -wt)$

The relationship between velocity and propagation constant is

$v = \frac{\omega}{k}=\frac{2730rad/sec}{172rad/m}\\\\$

v = 15.87m/s

Time taken, $t = \frac{\lambda}{v}$

$= \frac{1.3}{15.87}\\\\=0.0819 sec$

t = 0.0819s

The velocity of transverse wave is given by

$v = \sqrt{\frac{T}{\mu}}$

$v = \sqrt{\frac{W}{\frac{m}{\lambda}}}$

mass of string is calculated thus

mg = 0.0125N

$m = \frac{0.0125N}{9.8N/s}$

m = 0.00128kg

$\omega = \frac{v^2m}{\lambda}$

$\omega = \frac{(15.87^2)(0.00128)}{1.30}$

$\omega =$ 0.25N

The propagation constant k is

$k=\frac{2\pi}{\lambda}$

hence

$\lambda = \frac{2\pi}{k}\\\\\lambda = \frac{2 \times 3.142}{172}$

$\lambda =$ 0.036 m

No of wavelengths, n is

$n = \frac{L}{\lambda}\\\\n = \frac{1.30m}{0.036m}\\$

n = 36

The equation of wave travelling down the string is

$y(x, t)=Acos[kx -wt]\\\\becomes\\\\y(x , t)= Acos[(172 rad.m)x + (2730 rad.s)t]$

$without, unit\\\\y(x , t)= Acos[172x + 2730t]$

7 0

3 years ago

A person stands on the ball of one foot. The normal force due to the ground pushing up on the ball of the foot has magnitude 750

White raven [17]

Answer:

the force exerted on the foot by the tibia would be 2975 N

Explanation:

Given the data in the question;

To maintain equilibrium between the foot and the ball vertically, the addition normal normal force $N^>$ (750 N) and the tension in the Achilles tendon $F^>_{Achilles}$ (2225 N) must be equal to the force exerted on the foot by the tibia;

| $N^>$ | + | $F^>_{Achilles}$ | = | $F^>_{Tibia}$ |

so force exerted on the foot by the tibia will be;

| $F^>_{Tibia}$ | = | $N^>$ | + | $F^>_{Achilles}$ |

so we substitute IN OUR VALUES

| $F^>_{Tibia}$ | = 750 N + 2225 N

| $F^>_{Tibia}$ | = 2975 N

Therefore, the force exerted on the foot by the tibia would be 2975 N

3 0

2 years ago

How to calculate displacement?​

How to calculate displacement?