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2 years ago

Will give brainliest

Physics

1 answer:

natali 33 [55]2 years ago

3 0

Answer:

a = 5 m/s^2

Explanation:

The force is constant. The distance does not matter if there is no friction. This is an example of Newton's second law.

F = m * a

F = 10 N

m = 2 kg

a = ?

10 = 2 * a Divide by 2

10/2 = a

a = 5 m/s^2 second from the bottom.

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With what minimum speed must you toss a 200 g ball straight up to just touch the 14-m-high roof of the gymnasium if you release

Brut [27]

When they say use energy, you want to use

Total energy = potential energy + kinetic energy = mgh + 1/2mv²

I assume you mean 200 g ball,

so, we know the total distance traveled is going to be 13 - 1.3 = 11.7 m

If the ball just makes it to the top ( 13 m ) , then the ball will stop moving and the kinetic energy will be 0,

therefore, the potential energy at the top will be the total energy of the system = mgh

from this, we say that mgh = 1/2mv² solve for v

<span> v = sqrt (2gh) = 15.2 m/s </span>

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3 years ago

Match the following items.

Ray Of Light [21]

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2 years ago

Using Mirror equation A, Calculate The Frequency Of The Long Stand And The Shortest Wave Length Of That An Object Is Placed Of A

Leya [2.2K]

The Image distance and Magnification of The Image will be 30 cm and 3.

<h3>What is focal length?</h3>

The focal length of the lens, which is often expressed in millimeters, is the distance between the lens and the image sensor when the subject is in focus.

Given data;

Focal length,f=?

Image distance,v=?

Object distance,u= 10 cm

Magnification,m= 2.85

The focal length is half of the radius;

f=R/2

f=30 Cm/2

f= 15 Cm

The mirror equation is found as;

$\rm \frac{1}{f} =\frac{1}{v} +\frac{1}{u} \\\\\ \frac{1}{15}=\frac{1}{v}+\frac{1}{10} } \\\\\ v= -30 \ cm$

The magnification of the lens is found as;

$\rm m=\frac{30}{10}\\\\ m=3$

Hence, the image distance and magnification of The image will be 30 cm and 3.

To learn more about the focal length refer;

brainly.com/question/16188698

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1 year ago

Which of the following pieces of laboratory equipment is not directly used to make measurements

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I believe the answer is a test tube.

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3 years ago

Read 2 more answers

Se lanza verticalmente una esfera con una rapidez de 30m/se. Determinar la rapidez de la esfera a una altura de 40m (g=10m/s2)

sergeinik [125]

$v^2-{v_0}^2=2a(x-x_0)$

dónde $v$ es la velocidad de la esfera, $v_0$ es suya velocidad inicial, $a=-g$ la aceleración debida a la gravedad, $x$ la posición, y $x_0$ la posición inicial. Tomamos $x_0=0\,\mathrm m$ a referirse a la posición de la esfera en el momento que la esfera fue lanzada.

Entonces

$v^2-\left(30\,\dfrac{\mathrm m}{\mathrm s}\right)^2=2\left(-10\,\dfrac{\mathrm m}{\mathrm s^2}\right)(40\,\mathrm m)$

$\implies v^2=100\,\dfrac{\mathrm m^2}{\mathrm s^2}\implies v=\pm10\,\dfrac{\mathrm m}{\mathrm s}$

Esto nos dice que la esfera alcanza una altura de 40 m en dos momentos - una vez hacia arriba y una vez hacia abajo. Sin embargo, independientemente del signo de la velocidad, sabemos que suya magnitud es 10 m/s, y así tenemos una rapidez de 10 m/s también en ambos momentos.

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3 years ago