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3 years ago

Map.

Physics

1 answer:

Karo-lina-s [1.5K]3 years ago

7 0

Answer:

The object has decreasing acceleration and decreasing velocity.

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Which of the following classifications of star temperature is coolest?

Oliga [24]

The hottest would be the O type and the coolest is M

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3 years ago

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The following represents a mass attached to a spring oscillating in simple harmonic motion. X(t) = 4.0 cos(3.0t +0.10) units of

kolbaska11 [484]

Answer:

a) A = 4.0 m , b) w = 3.0 rad / s , c) f = 0.477 Hz , d) T = 20.94 s

Explanation:

The equation that describes the oscillatory motion is

x = A cos (wt + fi)

In the exercise we are told that the expression is

x = 4.0 cos (3.0 t + 0.10)

let's answer the different questions

a) the amplitude is

A = 4.0 m

b) the frequency or angular velocity

w = 3.0 rad / s

c) angular velocity and frequency are related

w = 2π f

f = w / 2π

f = 3 / 2π

f = 0.477 Hz

d) the period

frequency and period are related

T = 1 / f

T = 1 / 0.477

T = 20.94 s

e) the phase constant

Ф = 0.10 rad

f) velocity is defined by

v = dx / dt

v = - A w sin (wt + Ф)

speed is maximum when sine is + -1

v = A w

v = 4 3

v = 12 m / s

g) the angular velocity is

w² = k / m

k = m w²

k = 1.2 3²

k = 10.8 N / m

h) the total energy of the oscillator is

Em = ½ k A²

Em = ½ 10.8 4²

Em = 43.2 J

i) the potential energy is

Ke = ½ k x²

for t = 0 x = 4 cos (0 + 0.1)

x = 3.98 m

j) kinetic energy

K = ½ m v²

for t = 00.1 ²

v = A w sin 0.10

v = 4 3 sin 0.10

v = 1.98 m / s

3 0

3 years ago

Consider the motion of a 4.00-kg particle that moves with potential energy given by U(x) = + a) Suppose the particle is moving w

gtnhenbr [62]

Correct question:

Consider the motion of a 4.00-kg particle that moves with potential energy given by

$U(x) = \frac{(2.0 Jm)}{x}+ \frac{(4.0 Jm^2)}{x^2}$

a) Suppose the particle is moving with a speed of 3.00 m/s when it is located at x = 1.00 m. What is the speed of the object when it is located at x = 5.00 m?

b) What is the magnitude of the force on the 4.00-kg particle when it is located at x = 5.00 m?

Answer:

a) 3.33 m/s

b) 0.016 N

Explanation:

a) given:

V = 3.00 m/s

x1 = 1.00 m

x = 5.00

$u(x) = \frac{-2}{x} + \frac{4}{x^2}$

At x = 1.00 m

$u(1) = \frac{-2}{1} + \frac{4}{1^2}$

= 4J

Kinetic energy = (1/2)mv²

$= \frac{1}{2} * 4(3)^2$

= 18J

Total energy will be =

4J + 18J = 22J

At x = 5

$u(5) = \frac{-2}{5} + \frac{4}{5^2}$

$= \frac{4-10}{25} = \frac{-6}{25} J$

= -0.24J

Kinetic energy =

$\frac{1}{2} * 4Vf^2$

= 2Vf²

Total energy =

2Vf² - 0.024

Using conservation of energy,

Initial total energy = final total energy

22 = 2Vf² - 0.24

Vf² = (22+0.24) / 2

$Vf = \sqrt{frac{22.4}{2}$

= 3.33 m/s

b) magnitude of force when x = 5.0m

$u(x) = \frac{-2}{x} + \frac{4}{x^2}$

$\frac{-du(x)}{dx} = \frac{-d}{dx} [\frac{-2}{x}+ \frac{4}{x^2}$

$= \frac{2}{x^2} - \frac{8}{x^3}$

At x = 5.0 m

$\frac{2}{5^2} - \frac{8}{5^3}$

$F = \frac{2}{25} - \frac{8}{125}$

= 0.016N

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4 years ago

What is the value of x?<br><br> 6x−2(x 4)=12

Ymorist [56]

Answer: The value of x is -6.

Explanation:

To calculate the value of 'x', we need to solve each function happening inthe equation.

The equation provided to us is $6x-2(4x)=12$

To solve this, we will multiply 4x with 2 and then subtract the like terms and finally, we evaluate the value of 'x'.

$6x-8x=12\\\\-2x=12\\\\x=\frac{12}{-2}\\\\x=-6$

Hence, the value of x will be -6.

5 0

4 years ago

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Why are road accidents at high speeds very much worse than road accidents at low speeds?

Triss [41]

Answer:

The momentum makes it worse.

Explanation:

The momentum of vehicles running at faster speeds is very high and causes a lot of damage to the vehicles.

4 0

3 years ago

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