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3 years ago

Franny drew a diagram to compare images produced by concave and convex lenses.

Physics

2 answers:

Stella [2.4K]3 years ago

7 0

Answer:

B. virtual

Explanation:

Answer on edge 2020

yanalaym [24]3 years ago

4 0

Answer:

virtual

Explanation:

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A puck moves 2.35 m/s in a -22 degree direction. A hockey stick pushes it for 0.215 s, changing its velocity to 6.42 m/s in a 50

Whitepunk [10]

Answer:

Displacement in Y direction is 0.434 m

Explanation:

initial velocity of the puck is 2.35 m/s at -22 degree

so here it is given as

$v_i = 2.35 cos22 \hat i - 2.35 sin22 \hat j$

$v_i = 2.18 \hat i - 0.88 \hat j$

final velocity is given as 6.42 m/s at 50 degree

so we have

$v_f = 6.42 cos50 \hat i + 6.42 sin50\hat j$

$v_f = 4.13 \hat i + 4.92\hat j$

now displacement in Y direction is given as

$\Delta y = \frac{v_f + v_i}{2}t$

$\Delta y = \frac{-0.88 + 4.92}{2} (0.215)$

$\Delta y = 0.434 m$

5 0

3 years ago

sketch a graph of position vs. time for a person starting 0.4 m away and standing still. Include an explanation of your predicti

Korvikt [17]

Answer:

Straight line parallel to time axis.

Explanation:

The slope of the position time graph gives the velocity.

As the man is still, that means the velocity is zero. So, the slope of the graph is zero. It is a straight line parallel to time axis.

5 0

3 years ago

Some coastal regions of the world have cooler summers and warmer winters than inland regions at the same latitude. What accounts

maks197457 [2]

Water holds in heat very well. Keep the temperature more steady and average. The areas around the water will also have a less variant change in temperature as a result. This property of water is known as high specific heat.

8 0

3 years ago

NEEED HELP!!!

Amanda [17]

Answer:

The correct choice is A. Yes, the more exercise the better.

5 0

2 years ago

Given a particle that has the velocity v(t) = 3 cos(mt) = 3 cos (0.5t) meters, a. Find the acceleration at 3 seconds. b. Find th

DiKsa [7]

Answer:

a. $\rm -1.49\ m/s^2.$

b. $\rm 50.49\ m.$

Explanation:

<u>Given:</u>

Velocity of the particle, v(t) = 3 cos(mt) = 3 cos (0.5t) .

The acceleration of the particle at a time is defined as the rate of change of velocity of the particle at that time.

$\rm a = \dfrac{dv}{dt}\\=\dfrac{d}{dt}(3\cos(0.5\ t ))\\=3(-0.5\sin(0.5\ t.))\\=-1.5\sin(0.5\ t).$

At time t = 3 seconds,

$\rm a=-1.5\sin(0.5\times 3)=-1.49\ m/s^2.$

<u>Note</u>:<em> The arguments of the sine is calculated in unit of radian and not in degree.</em>

The velocity of the particle at some is defined as the rate of change of the position of the particle.

$\rm v = \dfrac{dr}{dt}.\\\therefore dr = vdt\Rightarrow \int dr=\int v\ dt.$

For the time interval of 2 seconds,

$\rm \int\limits^2_0 dr=\int\limits^2_0 v\ dt\\r(t=2)-r(t=0)=\int\limits^2_0 3\cos(0.5\ t)\ dt$

The term of the left is the displacement of the particle in time interval of 2 seconds, therefore,

$\Delta r=3\ \left (\dfrac{\sin(0.5\ t)}{0.05} \right )\limits^2_0\\=3\ \left (\dfrac{\sin(0.5\times 2)-sin(0.5\times 0)}{0.05} \right )\\=3\ \left (\dfrac{\sin(1.0)}{0.05} \right )\\=50.49\ m.$

It is the displacement of the particle in 2 seconds.

7 0

3 years ago