Which light bulbs are coated on the inside with a powder?

A woman can see clearly with her right eye only when objects are between 45 cm and 155 cm away. Prescription bifocals should hav

Yuri [45]

Answer:

Explanation:

given,

A woman can see object between 45 cm and 155 cm

glasses is 2 cm from the eye

to able to read a book distance = 25 cm

For the object distant apart

$\dfrac{1}{f}= \dfrac{1}{p} +\dfrac{1}{q}$

$\dfrac{1}{f}= \dfrac{1}{\infty}+\dfrac{1}{-153}$

$\dfrac{1}{f} = - 6.53 \times 10^{-3}$

hence, D = - 0.653 m

For the object at the near point

$\dfrac{1}{f}= \dfrac{1}{p} +\dfrac{1}{q}$

$\dfrac{1}{f}= \dfrac{1}{23} +\dfrac{1}{-43}$

$\dfrac{1}{f}= 0.0202 cm$

hence, D = 2.02 m

8 0

3 years ago

A neon sign that requires a voltage of 11,000 v is plugged into a 120-v wall outlet. what turns ratio (secondary/primary) must a

Arisa [49]

The equation that relates the voltages and the number of turns in a transformer is
$\frac{V_s}{V_p}= \frac{N_s}{N_p}$
where $V_s = 120 V$ is the voltage in the secondary coil, $V_p=11000 V$ is the voltage in the primary coil, and $N_s$ and $N_p$ are the number of turns on the secondary and primary coils.

Using the numbers, we find the ratio between the number of turns:
$\frac{N_s}{N_p}= \frac{V_s}{V_p}= \frac{120 V}{11000 V}=0.0109$

8 0

3 years ago

A rock is thrown upward from a bridge into a river below. The function f(t)=−16t2+44t+88 determines the height of the rock above

babymother [125]

1) 88 ft

2) 4.09 s

3) 1.38 s

4) 118.2 m

Explanation:

1)

For an object thrown upward and subjected to free fall, the height of the object at any time t is given by the suvat equation:

$h(t) = h_0 + ut - \frac{1}{2}gt^2$ (1)

where

$h_0$ is the height at time t = 0

$u$ is the initial vertical velocity

$g=32 ft/s^2$ is the acceleration due to gravity

The function that describes the height of the rock above the surface at a time t in this problem is

$f(t)=-16t^2+44t+88$ (2)

By comparing the terms with same degree of eq(1) and eq(2), we observe that

$h_0 = 88 ft$

which means that the rock is at height h = 88 ft when t = 0: therefore, this means that the height of the bridge above the water is 88 feet.

2)

The rock will hit the water when its height becomes zero, so when

$f(t)=0$

which means when

$0=-16t^2+44t+88$

First of all, we can simplify the equation by dividing each term by 4:

$0=-4t^2+11t+22$

This is a second-order equation, so we solve it using the usual formula and we find:

$t_{1,2}=\frac{-b \pm \sqrt{b^2-4ac}}{2a}=\frac{-11\pm \sqrt{(11)^2-4(-4)(22)}}{2(-4)}=\frac{-11\pm \sqrt{121+352}}{-8}=\frac{-11\pm 21.75}{-8}$

Which gives only one positive solution (we neglect the negative solution since it has no physical meaning):

t = 4.09 s

So, the rock hits the water after 4.09 seconds.

3)

Here we want to find how many seconds after being thrown does the rock reach its maximum height above the water.

For an object in free fall motion, the vertical velocity is given by the expression

$v=u-gt$

where

u is the initial velocity

g is the acceleration due to gravity

t is the time

The object reaches its maximum height when its velocity changes direction, so when the vertical velocity is zero:

$v=0$

which means

$0=u-gt$

Here we have

$u=+44 ft/s$ (initial velocity)

$g=32 ft/s^2$ (acceleration due to gravity)

Solving for t, we find the time at which this occurs:

$t=\frac{u}{g}=\frac{44}{32}=1.38 s$

4)

The maximum height of the rock can be calculated by evaluating f(t) at the time the rock reaches the maximum height, so when

t = 1.38 s

The expression that gives the height of the rock at time t is

$f(t)=-16t^2+44t+88$

Substituting t = 1.38 s, we find:

$f(1.38)=-16(1.38)^2 + 44(1.38)+88=118.2 m$

So, the maximum height reached by the rock during its motion is

$h_{max}=118.2 m$

Which means 118.2 m above the water.

3 0

4 years ago

By what factor must we increase the amplitude of vibration of an object at the end of a spring in order to double its maximum sp

strojnjashka [21]

Answer:

$A'=2A$

Explanation:

According to the law of conservation of energy, the total energy of the system can be expresed as the sum of the potential energy and kinetic energy:

$E=U+K=\frac{kA^2}{2}\\E=\frac{kx^2}{2}+\frac{mv^2}{2}=\frac{kA^2}{2}$

When the spring is in its equilibrium position, that is $x=0$ , the object speed its maximum. So, we have:

$\frac{k(0)^2}{2}+\frac{mv_{max}^2}{2}=\frac{kA^2}{2}\\A^2=\frac{mv_{max}^2}{k}\\A=\sqrt{\frac{mv_{max}^2}{k}}$

In order to double its maximum speed, that is $v'{max}=2v_{max}$ . We have:

$A'=\sqrt{\frac{m(v'_{max})^2}{k}}\\A'=\sqrt{\frac{m(2v_{max})^2}{k}}\\A'=\sqrt{\frac{4mv_{max}^2}{k}}\\A'=2\sqrt{\frac{mv_{max}^2}{k}}\\A'=2A$

6 0

3 years ago

WILL GIVE BRAINLIEST!! Part 1: Other than distance from the sun, which factor affects the temperature of a planet?

Svet_ta [14]

The thickness of the ozone layer (for earth) and the atmosphere.
Mercury is the closest planet to the sun but it isn’t the hottest because it has no atmosphere. Although Venus is the second planet from the sun, it is the hottest because of it’s thick atmosphere and the clouds that trap heat in.

8 0

4 years ago