What machine would benefit from slowing lights speed

Suppose that the radius of the circular path is r when the speed of the rocket is v and the acceleration of the rocket has magni

SCORPION-xisa [38]

Answer:

A3

Explanation:

4 0

3 years ago

What is the magnification of an astronomical telescope whose objective lens has a focal length of 74 cm and whose eyepiece has a

Novay_Z [31]

Answer:

The magnification of an astronomical telescope is -30.83.

Explanation:

The expression for the magnification of an astronomical telescope is as follows;

$M=-\frac{f_o}{f_e}$

Here, M is the magnification of an astronomical telescope, $f_e$ is the focal length of the eyepiece lens and $f_o$ is the focal length of the objective lens.

It is given in the problem that an astronomical telescope having a focal length of objective lens 74 cm and whose eyepiece has a focal length of 2.4 cm.

Put $f_o=74 cm$ and $f_e=2.4 cm$ in the above expression.

$M=-\frac{74}{2.4}$

M=-30.83

Therefore, the magnification of an astronomical telescope is -30.83.

5 0

3 years ago

If an object is projected upward from ground level with an initial velocity of 144144 ft per sec, then its height in feet after

Liula [17]

Answer:

4.5 s, 324 ft

Explanation:

The object is projected upward with an initial velocity of

$v_0 = 144 ft/s$

The equation that describes its height at time t is

$s(t) = -16t^2 + 144 t$ (1)

where t, the time, is measured in seconds.

In order to find the time it takes for the object to reach the maximum height, we must find an expression for its velocity at time t, which can be found by calculating the derivative of the position, s(t):

$v(t) = s'(t) = -32t +144$ (2)

At the maximum heigth, the vertical velocity is zero:

v(t) = 0

Substituting into the equation above, we find the corresponding time at which the object reaches the maximum height:

$0=-32t+144\\t=\frac{144}{32}=4.5 s$

And by substituting this value into eq.(1), we also find the maximum height:

$s(t) = -16(4.5)^2+144(4.5)=324 ft$

3 0

3 years ago

If I keep F constant in F=ma, what is the relationship between m and a?

Jobisdone [24]

Answer:

If F is a constant, we can take f = 1

f = m×a

ma = 1

therefore we can say that force is hence proportinal to the product of mass and acceleration.

6 0

2 years ago

How would a spinning disk's kinetic energy change if its moment of inertia was five times larger but its angular speed was five

Keith_Richards [23]

Answer:

The kinetic energy of a spinning disk will be reduced to a tenth of its initial kinetic energy if its moment of inertia is made five times larger, but its angular speed is made five times smaller.

Explanation:

Let us first consider the initial characteristics of the angular motion of the disk

moment of inertia = $I$

angular speed = ω

For the second case, we consider the characteristics to now be

moment of inertia = $5I$ (five times larger)

angular speed = ω/5 (five times smaller)

Recall that the kinetic energy of a spinning body is given as

$KE = \frac{1}{2}Iw^{2}$

therefore,

for the first case, the K.E. is given as

$KE = \frac{1}{2}Iw^{2}$

and for the second case, the K.E. is given as

$KE = \frac{1}{2}(5I)(\frac{w}{5} )^{2} = \frac{5}{50}Iw^{2}$

$KE = \frac{1}{10}Iw^{2}$

this is one-tenth the kinetic energy before its spinning characteristics were changed.

This implies that the kinetic energy of the spinning disk will be reduced to a tenth of its initial kinetic energy if its moment of inertia is made five times larger, but its angular speed is made five times smaller.

6 0

3 years ago