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Valentin [98]
3 years ago
6

Please please help me please

Physics
2 answers:
Julli [10]3 years ago
8 0

Answer:

Could I get brainest answer

I think it is B and e

bazaltina [42]3 years ago
5 0

Answer:

The answer is D hope it helps

Explanation:

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After the ball leaves the woman's hand, the only relevant force acting on the ball is the gravity, which produces an acceleration of a=g=9.81 m/s^2 on the baseball, directed downward. 
Gravity acts on the ball for the whole duration of the motion, so the acceleration of the ball is g=9.81 m/s^2 at every point of the motion, and therefore also when the baseball is at its highest point.
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A positive charge Q is placed at the origin. Four other positive charges are placed on a circle in the x-y plane centered on Q a
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6 0
3 years ago
a concave lens creates a virtual image at -47.0 cm and a magnification of +1.75. what is the focal length?
Lilit [14]

The focal length of given concave lens will be -26.85 cm

The height of an image to the height of an object is the ratio that is used to determine a lens' magnification. Additionally, it is provided in terms of object and image distance. It is equivalent to the object distance to image distance ratio.

Given  concave lens creates a virtual image at -47.0 cm and a magnification of +1.75.

We have to find focal length

The focal length can be found out by following way:

Magnification = m = +1.75

m = hi/h

hi = -47 cm

1.75 = -47/h

h = -26.85 cm

So the focal length of given concave lens will be -26.85 cm

Learn more about magnification factor here:

brainly.com/question/6947486

#SPJ10

8 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) .

<h2>(a):</h2>

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>

<h2>(b):</h2>

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
4 years ago
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