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3 years ago

GIVE DEFINITION Coastal Plain Peninsula Plateau

Physics

1 answer:

ozzi3 years ago

8 0

Answer:

Coastal plain: a flat low-lying land adjacent to the sea

Peninsula: A piece of land surrounded by water or projecting out into a body of water ( example: Florida)

Plateau: An area of relatively level high ground its usually flat also

hope this helps...

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What are dangers of synthetic drugs

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These are some potential dangers of synthetic drug abuse

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3 years ago

A mass is attached to an ideal spring. At time t = 0 the spring is at its natural length and the mass is given an initial veloci

JulijaS [17]

Answer:

t = T/4

Explanation:

The power delivered to the mass by the spring is work done by the spring per second.

$P = \frac{dW}{dt}$

The work done by the spring is equal to the elastic potential energy stored in the spring.

$U = \frac{1}{2}kx^2$

The maximum energy stored in the spring is at the amplitude of the oscillation.

$U_{max} =\frac{1}{2}kA^2$

So the first time the mass reaches to its amplitude can be found by the following equation of motion:

$x = A\cos(\omega t + \phi)\\\phi = \pi/2 ~because ~at ~t= 0, ~ x = 0\\0 = A\cos(0 + \pi/2)\\x = A\cos(\omega t + \pi/2)$

When the mass reaches the amplitude:

$A = A\cos(\omega t + \pi/2)\\1 = \cos(\omega t + \pi/2)\\\omega t + \pi/2 = \pi$

because cos(π) = 1.

$\omega t = \pi/2$

Using ω = 2π/T,

$\omega t = \pi/2\\\frac{2\pi}{T}t = \pi/2\\t = \frac{T}{4}$

4 0

4 years ago

A 3.0-kg block starts at rest at the top of a 37° incline, which is 5.0 m long. Its speed when it reaches the bottom is 2.0 m/s.

Mama L [17]

Answer: $f_{r}$ = 16.49N

Explanation: The object is placed on an inclined plane at an angle of 37° thus making it weight have two component,

$W_{x}$ = horizontal component of the weight = mgsinФ

$W_{y}$ = vertical component of weight = mgcosФ

Due to the way the object is positioned, the horizontal component of force will accelerate the object thus acting as an applied force.

by using newton's law of motion, we have that

mgsinФ - $f_{r}$ = ma

where m = mass of object=5kg

a = acceleration= unknown

Ф = angle of inclination = 37°

g = acceleration due to gravity = 9.8 $m/s^{2}$

$f_{r}$ = frictional force = unknown

we need to first get the acceleration before the frictional force which is gotten by using the equation below

$v^{2} = u^{2} + 2aS$

where v = final velocity = 2m/s

u = initial velocity = 0m/s (because the object started from rest)

a= unknown

S= distance covered = length of plane = 5m

$2^{2} = 0^{2} + 2*a*5\\\\4= 10 *a\\\\a = \frac{4}{10} \\a = 0.4m/s^{2}$

we slot in a into the equation below to get frictional force

mgsinФ - $f_{r}$ = ma

3 * 9.8 * sin 37 - $f_{r}$ = 3* 0.4

17.9633 - $f_{r}$ = 1.2

$f_{r}$ = 17.9633 - 1.2

$f_{r}$ = 16.49N

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It transforms it to mechanical

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