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2 years ago

Basketball player Darrell Griffith is on record as

Physics

1 answer:

Gre4nikov [31]2 years ago

5 0

Explanation:

We use the equation

h = $\frac{gt^2}{2}$ , where

h is the height traveled,

g is the acceleration due to gravity and

t is the time taken to reach height h.

We can now calculate t to be

$\sqrt{\frac{2*1.2 m}{9.81 m/s^2} }$

= 0.495 s

Let v be the initial velocity of the player.

The player deaccelarates from v m/s to 0 m/s in 0.495 s at the rate of 9.81 m/s^2.

v = 9.81 m/s^2 x 0.495 s = 4.85 m/s

The player takes 0.3 s to increase his velocity from 0 m/s to 4.85 m/s. So his average accelaration is

4.85 m/s / 0.3 s = 16.2 m/s^2

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An object is placed in front of a convex lens of a length 10cm. What is the nature of the image formed if the object distance is

Lady_Fox [76]

${\mathfrak{\underline{\purple{\:\:\: Given:-\:\:\:}}}} \\ \\$

$\:\:\:\:\bullet\:\:\:\sf{Focal\:length=10\:cm}$

$\:\:\:\:\bullet\:\:\:\sf{Object \ distance = -15\:cm}$

$\\$

${\mathfrak{\underline{\purple{\:\:\:To \:Find:-\:\:\:}}}} \\ \\$

$\:\:\:\:\bullet\:\:\:\sf{Nature \: of \:the\:image}$

$\\$

${\mathfrak{\underline{\purple{\:\:\: Solution:-\:\:\:}}}} \\ \\$

<h3>☯ <u>By using formula of Lens</u> </h3>

$\\$

$\dashrightarrow\:\: {\boxed{\sf{\dfrac{1}{u} + \dfrac{1}{v} = \dfrac{1}{f}}}}$

$\\$

$\dashrightarrow\:\: \sf{\dfrac{1}{v}-\dfrac{1}{-15}=\dfrac{1}{10}}$

$\\$

$\dashrightarrow\:\: \sf{\dfrac{1}{v}+\dfrac{1}{15}=\dfrac{1}{10}}$

$\\$

$\dashrightarrow\:\: \sf{\dfrac{1}{v} = \dfrac{1}{10} - \dfrac{1}{15}}$

$\\$

$\dashrightarrow\:\: \sf{\dfrac{1}{v} = \dfrac{1}{30}}$

$\\$

$\dashrightarrow\:\: \sf{ v = 30 \ cm}$

$\\$

<h3>☯ <u>Now, Finding the magnification </u></h3>

$\\$

$\dashrightarrow\:\: \sf{ m = \dfrac{-30}{-15}}$

$\\$

$\dashrightarrow\:\: \sf{m = -2}$

$\\$

<h3>☯ <u>Hence</u>, $\\$ </h3>

$\:\:\:\:\star\:\:\:\sf{Image \ distance = 30 \ cm}$

$\:\:\:\:\star\:\:\:\sf{Nature = Real \ \& \ inverted}$

3 0

2 years ago

If a sample emits 2000 counts per second when the detector is 1 meter from the sample, how many counts per second would be obser

Alona [7]

Answer:

<h2><em>6000 counts per second</em></h2>

Explanation:

If a sample emits 2000 counts per second when the detector is 1 meter from the sample, then;

2000 counts per second = 1 meter ... 1

In order to know the number of counts per second that would be observed when the detector is 3 meters from the sample, we will have;

x count per second = 3 meter ... 2

Solving the two expressions simultaneously for x we will have;

2000 counts per second = 1 meter

x counts per second = 3 meter

Cross multiply to get x

2000 * 3 = 1* x

6000 = x

<em>This shows that 6000 counts per second would be observed when the detector is 3 meters from the sample</em>

5 0

2 years ago

Newton's law of gravity was inconsistent with Einstein's special relativity because

Lubov Fominskaja [6]

Answer:

Mass and thus force depends on the reference frame chosen

Explanation:

This can be explained as Newton's law of gravity provides action which are instantaneous at a distance and involves the evaluation of all the quantities at present time or at the instant they occur.

If the body undergoes a change in its mass distribution there will be an immediate change in its gravitational force without any lag.

Now, if we talk about special relativity, it would be absurd to say that an information can travel faster than light. The effect is in synchronization with the cause in one reference frame where the effect occurs after the cause for some observer in some other reference frame.

In order to observe Newton's law of gravity all the observer's in different reference frames must observe the same phenomena which could only be possible if time were absolute and in special relativity, time is not absolute.

Therefore, Newton's law of gravity was inconsistent with the Einstein's Special Relativity.

3 0

2 years ago

Fnet = fa + ff or fnet = fa - ff

skelet666 [1.2K]

Answer:first of all what is your question and i can give and example which is Use them when you have 2 forces named Fa & FF or Fg & Ff acting in opposite directions on an object and you need to know the resultant of your 2 forces.

Explanation:

i searched it up

5 0

2 years ago

What is the IMA of the following pulley system?<br><br>34567

Lynna [10]

Answer:

IMA of given system = $\frac{F_{r} }{F_{e} }$

Explanation:

The "Ideal Mechanical advantage" (IMA) of given pulley is $\frac{F_{r} }{F_{e} }$ .

Ideal Mechanical advantage of a system is defined by the ratio of achieved or output force to the implied force. In the pulley system above, output force is the resistant force denoted by $F_{r}$ . The input force is analogous or equivalent to the effort applied i.e. $F_{e}$ .

Hence by dividing these two forces we calculate the IMA of the above mentioned pulley system which is $\frac{F_{r} }{F_{e} }$ .
Its mathematical reference would be:

IMA = $\frac{F_{r} }{F_{e} }$

6 0

2 years ago