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3 years ago

6

A light microscope allows for more magnification than electron microscope because it uses a beam of visible light. true false

1 answer:

Igoryamba3 years ago

4 0

False because light microscopes have low resolve and magnification.

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In a crash or collision, why is it advantageous for an occupant to extend the time during which the collision is taking place?

Veseljchak [2.6K]

Answer:

Explanation:

During a car collision momentum of vehicle ceases within a fraction of seconds so Force due to the impulse is huge.

Impulse is defined as the product of average force and time. If we can increase the period of collision for the same impulse then the average force imparted will be less.

If we can increase the time period then damage due to collision will be less.

3 0

2 years ago

A ray diagram is shown.

Rufina [12.5K]

Answer:

X

Explanation:

8 0

3 years ago

Read 2 more answers

The propeller of an airplane is at rest when the pilot starts the engine; and its angular acceleration is a constant value. Two

Maurinko [17]

Answer:0.318 revolutions

Explanation:

Given

Initially Propeller is at rest i.e. $\omega _0=0 rad/s$

after $t=10 s$

$\omega =10 rad/s$

using $\omega =\omega _0+\alpha t$

$10=0+\alpha \cdot 10$

$\alpha =1 rad/s^2$

Revolutions turned in 2 s

$\theta =\omega _0t+\frac{\alpha t^2}{2}$

$\theta =0+\frac{1\times 2^2}{2}$

$\theta =2 rad$

To get revolution $\frac{\theta }{2\pi }$

= $\frac{2}{2\pi}=0.318\ revolutions$

3 0

3 years ago

Read 2 more answers

four equal charges of +3microcolumb are placed at the four corners of a square that is 40cm on a side. Find the force on any one

Reil [10]

Answer:

i can see your question

pls fast

6 0

3 years ago

You are standing on a sheet of ice that covers the football stadium parking lot in Buffalo; there is negligible friction between

Anni [7]

Answer:

0.074m/s

Explanation:

We need the formula for conservation of momentum in a collision, this equation is given by,

$m_1u_1+m_2u_2 = m_1v_1+m_2v_2$

Where,

$m_1$ = mass of ball

$m_2$ = mass of the person

$u_1$ = Velocity of ball before collision

$u_2$ = Velocity of the person before collision

$v_1$ = velocity of ball afer collision

$v_2$ = velocity of the person after collision

We know that after the collision, as the person as the ball have both the same velocity, then,

$v_1 = v_2$

$m_1u_1 + m_2u_2 = (m_1+m_2)v_2$

Re-arrenge to find $v_2$ ,

$v_2 = \frac{m_1u_1+m_2u_2}{m_1+m_2}$

Our values are,

$m_1$ = 0.425kg

$u_1$ = 12m/s

$m_2$ = 68.5kg

$u_2$ = 0m/s

Substituting,

$v_2 = \frac{(0.425)(12)+(68.5)(0)}{0.425+68.5}$

$v_2 = 0.074m/s$

<em />

<em>The speed of the person would be 0.074m/s after the collision between him/her and the ball</em>

7 0

3 years ago