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3 years ago

What an image of a distant object is brought into focus in front of a persons retina the fact is called

1 answer:

7 0

When the image of a distant object is brought into focus of front of a person's retina, the defect is called: nearsightedness.

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At a given time of day, the ratio of the height of an object to the length of its shadow is the same for all objects. If a 4-f

katrin2010 [14]

Answer:35.2 ft

Explanation:

Given

height of stick =4 ft

shadow length =2.8 ft

Angle of elevation of sun is

$tan\theta =\frac{4}{2.8}$

let the height of tree be h

as $\theta$ will remain same thus

$tan\theta =\frac{h}{24.64}$

$\frac{4}{2.8}=\frac{h}{24.64}$

h=35.2 ft

8 0

3 years ago

A 95kg fullback (football player for those not into sports) moving south with a speed of 5.0 m/s has a perfectly inelastic colli

Lunna [17]

Answer:

a. $v=3.11mls$ , $29.4^{0}$

b. $K.E =-697.8J$

Explanation:

To calculate the values in the question, a deep understanding of perfect inelastic collision is important.

When two bodies undergo inelastic collision, two important parameters must be well understood i.e

Momentum: the momentum is always conserved in perfectly inelastic collision. i.e the total momentum after collision is the sum of the individual momentum before collision

Kinetic energy: Kinetic energy is not conserved due to dissipative force.

a.To calculate the velocity, we first find the total momentum before collision

Momentum of player 1 $p_{1} =mv=95kg*5m/s\\p_{1} =475kgm/s\\$

Momentum of player 2 $p_{2} =mv=90kg*3m/s\\p_{1} =270kgm/s\\$

Hence the total momentum $p_{12}=p_{1}+p_{2}\\$

Note, since the direction of movement before collision is due south and due north respectively we have to represent the velocity using the rectangular coordinate

Hence $p_{12}=(m_{1}+m_{2})v=p_{1}i+p_{2}j\\$

$(95+90)v=475i+270j\\$

$v=2.57i+1.45j\\$

solving for the resultant velocity, we have

$v=\sqrt{2.75^{2} +1.45^{2}}\\ v=3.11mls$

To calculate the direction of movement, we have

$\alpha =tan^{-1}=\frac{v_{j} }{v_{i}}\\ \alpha =tan^{-1}=\frac{1.45}{2.57}\\\alpha =29.4^{0}$

b. to calculate the decrease in total kinetic energy, before collision, the total kinetic was

$K.E_{initial} =\frac{1}{2}m_{1}v_{1}^{2}+\frac{1}{2}m_{2}v_{2}^{2}.\\K.E_{initial} =((1/2)*95*5^{2})+((1/2)*90*3^{2})\\K.E_{initial} =1187.5+405\\K.E_{initial} =1592.5J\\$

And the final kinetic energy after collision is

$K.E_{final} =\frac{1}{2}(m_{1}+m_{2} )v^{2}\\ K.E_{final} =\frac{1}{2}(95+90)* 3.11^{2}\\ K.E_{final} =894.7J$

The decrease in Kinetic energy is

$K.E =K.E_{final}- K.E_{initial}=894.7-1592.5$

$K.E =-697.8J$

The negative sign indicate a decrease in Kinetic energy

4 0

3 years ago

If Frequency F, velocity v, and density D are considered fundamental units, the dimensional formula for momentum will be :

gizmo_the_mogwai [7]

Let's see

Momentum be P

$\\ \rm\Rrightarrow P=[Frequency]^a[velocity]^b[Density]^c$

$\\ \rm\Rrightarrow [P]=[F]^a[v]^b[D]^c$

$\\ \rm\Rrightarrow [M^1L^1T^{-1}]=[T^{-1}]^a[L^1T^{-1}]^b[M^1L^{-3}]^c$

$\\ \rm\Rrightarrow MLT^{-1}=T^{-a}L^bT^{-b}M^cL^{-3c}$

$\\ \rm\Rrightarrow MLT^{-1}=T^{-a-b}L^{b-3c}M^c$

On comaparing

b-3c=1
b-3=1
b=1+3
b=4

and

-a-b=-1
-a-4=-1
-a=-1+4=3
a=-3

So the unit is

DV⁴/F³

5 0

2 years ago

Read 2 more answers

Which of the following EM waves do not kill cells?<br> Radio waves<br> UV light<br> Xrays<br> Gamma

Novosadov [1.4K]

Answer:

The answer is below!!

Explanation:

Gamma-rays

Gamma-rays have the smallest wavelengths and the most energy of any other wave in the electromagnetic spectrum. These waves are generated by radioactive atoms and in nuclear explosions. Gamma-rays can kill living cells, a fact which medicine uses to its advantage, using gamma-rays to kill cancerous cells.

Hope I Helped!!!

7 0

2 years ago

What name is given to the rate of flow of<br> electric charge?

RSB [31]

Answer:

<h2>Electric charge</h2>

Explanation:

The rate of the flow of electric charge is known as electric current. <u>By convention, the direction of electric current is always the direction of net flow of positive charge.</u>

3 0

2 years ago