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4 years ago

Through which type of media can mechanical waves move ?

Physics

1 answer:

natta225 [31]4 years ago

5 0

Answer: Solids, liquids, and gasses.

Explanation:

well a mechanical wave is a disturbance in matter that carries energy from one place to another, and a medium (or media because you put the question in pluralised form) is matter through which a wave travels. So, Solids, liquids, and gasses.

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A student launches a small rocket which starts from rest at ground level. At a height of h = 1.25 km the rocket reaches a speed

Novay_Z [31]

The acceleration of the rocket will be "56.2 m/s²".

According to the question,

The initial speed during launch,

u = 0 m/s

The speed at fuel running out point,

$v_f$ = 375 m/s

Height,

h = 1.25 km

= 1250 m

As we know,

→ $vf^2 = u^2+2ah$

or,

→ $a = \frac{v_f^2-u^2}{2h}$

By putting the values, we get

→ $=\frac{(375)^2-(0)^2}{2\times 1250}$

→ $=\frac{140625}{2500}$

→ $= 56.2 \ m/s^2$

Thus the above solution is correct.

Learn more:

brainly.com/question/11038449

4 0

2 years ago

A 1.0-kg ball of putty is released from rest and falls vertically 1.5 m until it strikes a hard floor, where it comes to rest in

topjm [15]

<span>We first calculate the velocity of the ball when it hits the ground; this is equal to the square root of the quantity (2*g*d) where g is the acceleration of gravity (9.8 m/s^2) and d is the distance fallen, 1.5m.

So, we get a velocity of sqrt(2*9.8*1.5) = 5.42 m/s.

We can calculate the impulse force applied to the putty ball by using Newton's second law, which states that the applied force is equal to the product of mass and acceleration, where acceleration can be further decomposed as the change in velocity divided by the change in time. Thus, inputting the known values, we have:

F = ma = m(dv/dt) = 1.0*5.42/0.045 = 120.4 newtons.</span>

8 0

3 years ago

A small block with mass 0.0350kg slides in a vertical circle of radius 0.525m on the inside of a circular track. During one of t

kirill115 [55]

Answer:

w = -0.475N

Explanation:

$K.E_{a} + P.E_{a} + W_{fr} = K.E_{b} + P.E_{b}\\K.E = 0.5mv^{2} \\Normal force at point B, N_{B} = 0.665N\\Normal force at point A, N_{A} = 3.85N\\$

To get Va and Vb

$F = mv_{A} ^{2} /R................(1)\\F = N_{A} - mg.........................(2)\\mv_{A} ^{2} /R = N_{A} - mg\\v_{A} ^{2} = R (N_{A}/m - g)\\v_{A} = \sqrt{ R (N_{A}/m - g)}$

R = 0.525 m

m = 0.0350 kg

g = 9.8 m/s²

$v_{A}= \sqrt{0.525(3.85 /0.0350 - 9.8)} \\v_{A} = 7.25 m/s$

K.Ea = 0.5 * 0.035 * 7.25²

K.Ea = 0.92 J

Since point A is at the bottom of the path, h = 0 m

P.Ea = 0 m

For Vb

$F = mv_{B} ^{2} /R................(1)\\F = N_{B} - mg.........................(2)\\mv_{B} ^{2} /R = N_{B} - mg\\v_{B} ^{2} = R (N_{B}/m - g)\\v_{B} = \sqrt{ R (N_{B}/m - g)}$