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Gnoma [55]
3 years ago
9

1. What are the different types of waves? Give examples of each type.

Physics
1 answer:
lyudmila [28]3 years ago
4 0

Answer:

See explanation

Explanation:

1) There are two main types of waves;

Mechanical waves and electromagnetic waves.

Mechanical waves usually require a medium for propagation, e.g sound, waves on a spiral spring, etc

Electromagnetic waves do not require a medium for propagation e.g microwaves, x-rays etc

2) The phenomenon is known as refraction. Refraction occurs at the boundary between two media because of the change in the speed of a wave as it moves from one medium to another.

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4 0
3 years ago
Two identical small metal spheres with q1 > 0 and |q1| > |q2| attract each other with a force of magnitude 81 mN when sepa
Gekata [30.6K]

Answer:

Explanation:

Check the attachment for solution

8 0
3 years ago
our 3.80-kg physics book is placed next to you on the horizontal seat of your car. The coefficient of static friction between th
gtnhenbr [62]

Answer:

Explanation:

Maximum force of friction possible = μmg

= .65 x 3.8 x 9.8

= 24.2 N

u = 72 x 1000 / 60 x 60

= 20 m /s

v² = u² - 2as

a = 20 x 20 / (2 x 30)

= 6.67 m / s²

force acting on it

= 3.8 x 6.67

= 25.346 N

Friction force possible is less .

So friction will not be able to prevent its slippage

It will slip off .

4 0
3 years ago
Your starship, the Aimless Wanderer,lands on the mysterious planet Mongo. As chief scientist-engineer,you make the following mea
melisa1 [442]

Answer:

m = 1.26*10²⁵ kg.

Explanation:

Assuming that the mass of the stone is much smaller than the mass of the planet, we can get the mass, applying the Universal Law of  Gravitation to both masses, as follows:

Fg = G* ms* mp / rp²

Now, if we apply Newton's 2nd Law to the mass of the stone, we can get the gravitational acceleration, as follows:

Fg = ms*a = ms*g ⇒ g = G*mp / rp²

First of all, we need to get the value of g.

Assuming that this acceleration is constant, we can appy the kinematic equations to this situation.

We know that the stone is thrown upward with an initial velocity vo = 15 m/s.

At the highest point in the trajectory, just before of changing direction, the stone comes momentarily to a stop.

At this point, applying the definition of acceleration, we can write:

vf = vo -g*t ⇒ 0 = vo -gt ⇒ g = vo/t (1)

We have the total time since the stone was thrown upwards, not the one used for the upward trajectory.

It can be showed, using the expression for the displacement (which is the same in both directions) that the time used for going up, it's the same used to go down, so the time that we need to put in (1). is just the half of the total time.

So, replacing in (1) we get the value of g, as follows:

g = 15 m/s / 4.5 s = 3.33 m/s²

Now, we can replace this value in the equation that gives us g based in the Universal Law of Gravitation, as follows:

g=G*mp / rp² (2)

Before solving for mp, however, we need to get the value of the radius of the planet.

Assuming that it's a perfect sphere, we can get this value from the value of the circumference at the planet's equator:

rp = 2*π*rp / 2*π ⇒ rp = 1.0*10⁵ km / 2*π = 15,915 km.

With this value for  rp, we can solve (2) for mp, as follows:

mp= g*rp² / G = 3.33 m/s² * (15,915 km)² / 6,67*10⁻¹¹ N.m²/kg²

mp = 1.26*10²⁵ kg.

8 0
3 years ago
An object with a mass M = 250 g is at rest on a plane that makes an angle θ = 30 o above the horizontal. The coefficient of kine
liubo4ka [24]

Answer:

v = 79.2 m/s

Solution:

As per the question:

Mass of the object, m = 250 g = 0.250 kg

Angle, \theta = 30^{\circ}

Coefficient of kinetic friction, \mu_{k} = 0.100

Mass attached to the string, m = 0.200 kg

Distance, d = 30 cm = 0.03 m

Now,

The tension in the string is given by:

Mgsin\theta + \mu_{k}Mgcos\theta + Ma = T        (1)

Also

T = m(g + a)

Thus eqn (1) can be written as:

Mgsin\theta + \mu_{k}Mgcos\theta + Ma = m(g - a)

Mgsin\theta + \mu_{k}Mgcos\theta + Ma = mg + ma

mg - Mgsin\theta - \mu_{k}Mgcos\theta = (M - m)a

a = \frac{0.2\times 9.8 - 0.250\times 9.8\times sin30^{\circ} - 0.1\times 0.250\times 9.8\times cos30^{\circ}}{0.250 - 0.200}

a = 10.45\ m/s^{2}

Now, the speed is given by the third eqn of motion with initial velocity being zero:

v^{2} = u^{2} + 2ad

where

u = initial velocity = 0

Thus

v = \sqrt{2ad}

v = \sqrt{2\times 10.45\times 0.03} = 0.792\ m/s

3 0
3 years ago
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