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3 years ago

What is the minimum value an eccentricity can be?

Physics

2 answers:

damaskus [11]3 years ago

7 0

Answer:

A eccentricity of zero would be a circle, because if it's zero you only have one point

Explanation:

Mila [183]3 years ago

4 0

The minimum eccentricity would be zero.

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Which statements about the two fossil images are correct?

tangare [24]

What fossil images, there are no pictures attached.

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4 years ago

Read 2 more answers

Pick the right statement about normal force: (section 5.7)

Montano1993 [528]

Answer:

A. Normal force is always perpendicular to the area of contact between an object and support.

Explanation:

Normal force is defined as the contact force. If there is no contact between the surfaces, they cannot applies a force which is normal on each other. For e.g, the surfaces of a cubical box and the cart cannot applies a force of normal on each other because of no contact.

If, when there is a contact between two surfaces they applies a normal force on each other, and this force is perpendicular to the each other . This normal force is necessary to prevent object to penetrating into other.

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3 years ago

A projectile is fired upward at an angle θ above the horizontal with an initial speed v0. At its maximum height, what are its ve

aliya0001 [1]

Answer:

$\vec{v}_{\rm max} = v\cos(\theta)(\^x)\\|\vec{v}_{\rm max}| = v\cos(\theta)\\\vec{a} = \vec{g} = -9.8\^y$

Explanation:

The equations of kinematics will be used to solve this question:

$y - y_0 = v_{y_0}t + \frac{1}{2}a_yt^2\\v_y^2 = v_{y_0}^2 + 2a_y(y - y_0)\\v_y = v_{y_0} + a_yt$

At its maximum height, the projectile has zero velocity in the y-direction. But its velocity in the x-direction is unaffected.

First, let's apply the above equations to the x-direction.

There is no acceleration in the x-direction. So, its velocity in the x-direction is constant during the motion.

$v_x = v_{x_0} + a_xt = v_{x_0} + 0\\v_x = v_{x_0} = v\cos(\theta)$

Therefore, the velocity vector of the projectile is

$v_{max} = v_x = v\cos(\theta)$

The speed of the projectile is the same.

The acceleration vector is constant during the motion and equal to the gravitational acceleration, which is -9.8 downwards.

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4 years ago

Calculate change in height of a 2kg ball moving a speed of 10m/s up a frictionless ramp until it stops

Nataliya [291]

The change in the height of the object is 5.1 m.

<h3>Conservation of mechanical energy</h3>

The principle of conservation of mechanical energy states that the total energy of an isolated system is always conserved.

The change in the height of the object is calculated by applying the principle of conservation of mechanical energy as follows;

P.E = K.E

$mg \Delta h = \frac{1}{2} mv^2\\\\g \Delta h = \frac{1}{2}v^2\\\\\Delta h = \frac{v^2}{2g} \\\\\Delta h = \frac{(10)^2}{2(9.8)} \\\\\Delta h = 5.1 \ m$

Thus, the change in the height of the object is 5.1 m.

Learn more about conservation of mechanical energy here: brainly.com/question/6852965

6 0

2 years ago

Find the x and y components of each of the following vectors.

Furkat [3]

Answer:

x =

$- 15 \times \frac{ \sqrt{3} }{2} \frac{m}{s}$

y = 7.5m/s

8 0

3 years ago