How much greater is the light-collecting area of a 6-meter telescope than a 3-meter telescope?.

Sav [38]

Answer:

In physics, equations of motion are equations that describe the behavior of a physical system in terms of its motion as a function of time.[1] More specifically, the equations of motion describe the behaviour of a physical system as a set of mathematical functions in terms of dynamic variables. These variables are usually spatial coordinates and time, but may include momentum components. The most general choice are generalized coordinates which can be any convenient variables characteristic of the physical system.[2] The functions are defined in a Euclidean space in classical mechanics, but are replaced by curved spaces in relativity. If the dynamics of a system is known, the equations are the solutions for the differential equations describing the motion of the dynamics.

6 0

2 years ago

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A mass weighing 24 pounds, attached to the end of a spring, stretches it 4 inches. Initially, the mass is released from rest fro

svetoff [14.1K]

Answer:

The equation of motion is $x(t)=-$ $\frac{1}{3} cos4\sqrt{6t}$

Explanation:

Lets calculate

The weight attached to the spring is 24 pounds

Acceleration due to gravity is $32ft/s^2$

Assume x , is spring stretched length is ,4 inches

Converting the length inches into feet $x=\frac{4}{12} =\frac{1}{3}feet$

The weight (W=mg) is balanced by restoring force ks at equilibrium position

mg=kx

$W=kx$ ⇒ $k=\frac{W}{x}$

The spring constant , $k=\frac{24}{1/3}$

= 72

If the mass is displaced from its equilibrium position by an amount x, then the differential equation is

$m\frac{d^2x}{dt} +kx=0$

$\frac{3}{4} \frac{d^2x}{dt} +72x=0$

$\frac{d^2x}{dt} +96x=0$

Auxiliary equation is, $m^2+96=0$

$m=\sqrt{-96}$

= $\frac{+}{} i4\sqrt{6}$

Thus , the solution is $x(t)=c_1cos4\sqrt{6t}+c_2sin4\sqrt{6t}$

$x'(t)=-4\sqrt{6c_1} sin4\sqrt{6t}+c_2$ $4\sqrt{6}$ $cos4\sqrt{6t}$

The mass is released from the rest x'(0) = 0

$=-4\sqrt{6c_1} sin4\sqrt{6(0)}+c_2$ $4\sqrt{6}$ $cos4\sqrt{6(0)}$ =0

$c_2$ $4\sqrt{6} =0$

$c_2=0$

Therefore , $x(t)=c_1$ $cos 4\sqrt{6t}$

Since , the mass is released from the rest from 4 inches

$x(0)= -4$ inches

$c_1 cos 4\sqrt{6(0)} =-\frac{4}{12}$ feet

$c_1=-\frac{1}{3}$ feet

Therefore , the equation of motion is $-\frac{1}{3} cos4\sqrt{6t}$

7 0

2 years ago

⦁ A car going 50 m/s is brought to rest in a distance of 20.0 m as it strikes a pile of dirt. How large an average force is exer

gtnhenbr [62]

Answer:

the average force exerted by seatbelts on the passenger is 5625 N.

Explanation:

Given;

initial velocity of the car, u = 50 m/s

distance traveled by the car, s = 20 m

final velocity of the after coming to rest, v = 0

mass of the passenger, m = 90 kg

Determine the acceleration of the car as it hit the pile of dirt;

v² = u² + 2as

0 = 50² + (2 x 20)a

0 = 2500 + 40a

40a = -2500

a = -2500/40

a = -62.5 m/s²

The deceleration of the car is 62.5 m/s²

The force exerted on the passenger by the backward action of the car is calculated as follows;

F = ma

F = 90 x 62.5

F = 5625 N

Therefore, the average force exerted by seatbelts on the passenger is 5625 N.

8 0

2 years ago

Come si esprime il numero 0,00123 in notazione scientifica?

Inessa [10]

Answer:

0,00123 = 1,2*10^{-3}

Explanation:

To write down correctly the number 0,00123 in scientific notation, you take into account that the scientific notation demands that there in only one number after the comma ( , ). Furthermore, it is necessary that you move the comma to the right of the first number different of zero, in this case the number 1. To do this you move the comma three positions.

Then, you have to multiply the expresion 1.23 by 10 with an exponential -3 (because of the movement of the comma in three positions). That is:

0,00123 = 1,23*10^{-3}

But it is mandatory that nly one number can stay after the comma, so, you approximate the number three. In this case, the number is lower than 5, hence, you approximate 3 to 0.

Finally, you have:

0,00123 = 1,2*10^{-3}

3 0

2 years ago

A vector starts at the point (0.0) and ends at (2,-7) what is the magnitude of the displacement

Leto [7]

Answer:

|x| = √53

Explanation:

We are told that the vector starts at the point (0.0) and ends at (2,-7) .

Thus, magnitude of displacement is;

|x| = √(((-7) - 0)² + (2 - 0)²)

|x| = √(49 + 4)

|x| = √53

5 0

2 years ago