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2 years ago

What is the relationship between wave amplitude and wave energy?

Physics

1 answer:

Nataliya [291]2 years ago

3 0

Answer:

waves

Explanation:

they are both waves

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A woman and her dog are out for a morning run to the river, which is located 4.0 KM away. The woman runs at 2.5 M/S in a straigh

VLD [36.1K]

River shore is located at distance

$d = 4 km$

speed of the woman is given as

$v_1 = 2.5 m/s$

now the time taken by the woman to cover the distance is

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

$t = \frac{4000}{2.5} = 1600 s$

for the same time interval the dog will run to and fro with speed 4.5 m/s

so the total distance moved by the dog is given by

$d = v* t$

$d = 4.5 * 1600$

$d = 7200 m$

<em>so the total distance that dog will move is 7200 m or 7.2 km</em>

8 0

3 years ago

Two identical 9.10-g metal spheres (small enough to be treated as particles) are hung from separate 300-mm strings attached to t

Musya8 [376]

Answer:

$n = 1.266\times 10^{12}$

Explanation:

Given data:

mass of sphere is 10 g

Angle between string and vertical axis is $\theta = 13 degree$

thickness of string 300 mm = 0.3 m

$sin\theta =\frac{2}{0.3 m}$

r =0.3 sin 13 = 0.067 m

$Fe = \frac{ kq_1 q-2}{d^2}$

$Fe = \frac{kq^2}{(2r)^2} = mg tan\theta$

$q^2 = mg tan\theta \frac{(2r)^2}{k}$

$= 0.0091 \times 9.8 tan13 \times \frac{(2\times 0.067)^2}{9\times 10^9}$

$q^2 = 4.10\times 10^{-14}$

$q = 2.026 \times 10^{-7} C$

q = ne

$n = \frac{1.6\times 10^{-19}}{2.02\times 10^{-7}}$

$n = 1.266\times 10^{12}$

3 0

3 years ago

Tammy leaves the office, drives 34 km due north, then turns onto a second highway and continues in a direction of 35◦ north of e

Natasha2012 [34]

Answer:

The total displacement is 102 km $51^o$ north of east.

Explanation:

We can treat this problem as a trigonometric one, so we need to calculate the total displacement on the north and east.

$d_n=34km+79km*sin(35^o)=79km$

and

$d_e=79*cos(35^o)=65km$

The total displacement is given by:

$D=\sqrt{(79km)^2+(65km)^2}=102km$

with an agle of:

$\alpha =arctg(\frac{79km}{65km})=51^o$

4 0

3 years ago

Question 4 of 20

Mnenie [13.5K]

Answer:

Explanation:

F = G m1 m2 / r^2 now double r

F = G m1m1/ (2r)^2

F = 1/4 G m1m2/r^2 <===== this is 1/4 of the original

5 0

2 years ago

A box is pulled to the right with a force of 65 N at an angle of 58 degrees to the horizontal. The surface is frictionless. The

Citrus2011 [14]

The free-body diagram is missing, but I assume the only forces acting on the box are the force F pushing the box, the weight of the object and the normal reaction of the surface.

Since the weight and the normal reaction acts in the vertical (y) direction, the only force acting on the box in the horizontal (x) direction is the horizontal component of the force F, which is given by
$F_x = F \cos 58^{\circ} = (65 N)(\cos 58^{\circ} )=34.4 N$
And so this is the net force in the x-direction.

5 0

3 years ago

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