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3 years ago

Is there any difference between reflection and echo???

Physics

1 answer:

Fofino [41]3 years ago

6 0

Answer:

reflection is the act of reflecting or the state of being reflected while echo is

reflection of sound waves from a surface back to the initial listener

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Tammy leaves the office, drives 26 km due

fredd [130]

Answer:

72.98 km

Explanation:

Her displacement is simply the distance from her final position to her initial position.

Now, I've drawn and attached a triangle diagram to depict this her movement.

Point O is her initial starting point.

Point A is the first point she gets to after travelling north while point B is the final point after travelling north east.

From the triangle, the displacement will be the distance OB which is denoted by x and can be solved from cosine rule.

Thus;

x² = 62² + 26² - 2(62 × 26)cos 120

x² = 4520 + 806

x² = 5326

x = √5326

x = 72.98 km

4 0

2 years ago

If an object has a fast velocity, the dots on a ticker tape diagram will be _____.

Gala2k [10]

Answer:

If an object has a fast velocity, the dots on a ticker tape diagram will be far apart.

7 0

2 years ago

Read 2 more answers

a man hikes 6.6 km north along a straight path with an average velocity of 4.2 km/h to the north. he rest at a bench for 15 min.

SSSSS [86.1K]

Answer:

2.6h

Explanation:

I attached the image below of the work hope you can see it. Hope this helps!

6 0

2 years ago

A force of 44 N will stretch a rubber band 88 cm (0.080.08 m). Assuming that Hooke's law applies, how far will aa 11-N forc

Setler79 [48]

Answer:

The rubber band will be stretched 0.02 m.

The work done in stretching is 0.11 J.

Explanation:

Force 1 = 44 N

extension of rubber band = 0.080 m

Force 2 = 11 N

extension = ?

According to Hooke's Law, force applied is proportional to the extension provided elastic limit is not extended.

F = ke

where k = constant of elasticity

e = extension of the material

F = force applied.

For the first case,

44 = 0.080K

K = 44/0.080 = 550 N/m

For the second situation involving the same rubber band

Force = 11 N

e = 550 N/m

11 = 550e

extension e = 11/550 = 0.02 m

The work done to stretch the rubber band this far is equal to the potential energy stored within the rubber due to the stretch. This is in line with energy conservation.

potential energy stored = $\frac{1}{2}ke^{2}$

==> $\frac{1}{2}* 550* 0.02^{2}$ = 0.11 J

3 0

2 years ago

Let the masses of blocks A and B be 4.50 kg and 2.00 kg , respectively, the moment of inertia of the wheel about its axis be 0.4

Free_Kalibri [48]

Answer:

Accelerations of both the sides is 0.6125 $m/s^{2}$ , A moves downwards whereas B moves upwards.

$\alpha$ =6.125 rad/ $s^{2}$

Tension on side A = 4.5 × g= 44.1 m/ $s^{2}$

Tension on side B= 2.0 × g= 19.6 m/ $s^{2}$

Explanation:

As both, the blocks A and B are attached due to the constraint they can only possess a single acceleration a.

Observe the figure attached, let the tension with Block A be $T_{2}$ and the tension attached with Block B be $T_{1}$ .

Tensions will be only be due to the weight of the blocks as no other force is present.

$T_{2}$ = 4.5 × g= 44.1 m/ $s^{2}$

$T_{1}$ = 2.0 × g= 19.6 m/ $s^{2}$

Now, lets make a torque equation about the center of the wheel and find the alpha

$T_{2}$ ×R- $T_{1}$ ×R= MI( Moment of Inertia of Wheel)× Alpha

where, R= Radius of the wheel=0.100m and

Alpha( $\alpha$ )= Angular acceleration of the wheel

MI of the wheel= 0.400 kg/ $m^{2}$

$(44.1-19.6)R=0.400\alpha$

$\alpha = \frac{24.5 * 0.100}{0.400}$

$\alpha$ =6.125 rad/ $s^{2}$

Acceleration = R × $\alpha$

= 0.1 * 6.125

=0.6125 $m/s^{2}$

Accelerations of both the sides is 0.6125 $m/s^{2}$ , A moves downwards whereas B moves upwards.

7 0

3 years ago