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3 years ago

Can a body having zero velocity move with uniform speed? Give an example.

1 answer:

3 0

Sure. Body can move with uniform speed, and having zero velocity, when velocity becomes zero due to change in direction over time t.

For Example. - An Object is moving with uniform speed in a circular path, then after one complete revolution, it's velocity is zero, but speed still remains uniform

Hope this helps!

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Two balls, each with a mass of 0.5 kg, collide on a pool table. Is the law of conservation of momentum satisfied in the collisio

Dmitriy789 [7]

During the collision between two balls on the pool table there is no external force along the line of collision between them

Since there is no external force on it so here we can say

$F = 0 = \frac{\Delta P}{\Delta t}$

here we have

$\Delta P = 0$

so we can say

$P_i = P_f$

since there is no external force so we can say during the collision the momentum of two balls will remain conserved

7 0

3 years ago

Read 2 more answers

The Rigidbody component adds collision to a GameObject

kifflom [539]

Rigidbodies are components that allow a GameObject<u> to react to real-time physics. </u>

Explanation:

Rigidbodies are components that allow a GameObject to react to real-time physics. This includes reactions to forces and gravity, mass, drag and momentum. You can attach a Rigidbody to your GameObject by simply clicking on Add Component and typing in Rigidbody2D in the search field.
A rigidbody is a property, which, when added to any object, allows it to interact with a lot of fundamental physics behaviour, like forces and acceleration. You use rigidbodies on anything that you want to have mass in your game.
You can indeed have a collider with no rigidbody. If there's no rigidbody then Unity assumes the object is static, non-moving.
If you had a game with only two objects in it, and both move kinematically, in theory you would only need a rigidbody on one of them, even though they both move.

7 0

3 years ago

A 6 kilogram block in outer space is moving at -100 m/s (to the left). It suddenly experiences three forces as shown below.

Alika [10]

Newton's second law and the kinematic relations allow to find the results for the questions about forces and the movement of the block are:

B) the force applied to maintain the system is equilibrium is: F = 0.39 N with an angle of tea = 180º

C) The maximum force is: F = 24 N

D) The time to stop the block is: t = 25 s

Newton's second law establishes a relationship between the net force, the mass, and the acceleration of the body. In the special case that the acceleration is zero it is called the equilibrium condition.

B) They indicate a diagram of forces on the block, let's look for the components of the force that the block maintains with zero acceleration, in the attached we have a free-body diagram including the force applied to keep the system in equilibrium.

x-axis

-10 + 12 sin 60 + Fₓ = 0

Fₓ = 10- 12 sin 60 = -0.39 N

y-axis

12 cos 60 - 6 + F_y = 0

F_y = 6 - 12 cos 60 = 0 N

We can give the result of the force in two ways:

Form of coordinates F = -0.39 i ^ N

Form of module and angle.

Let's use Pythagoras' theorem to find the modulus.

$F = \sqrt{F_x^2 + F_y^2 } \\F = \sqrt{0.39^2 +0^2}$

F = 0.39N

We use trigonometry for the angle.

$tan \theta = \frac{F_y}{F_x}$

tan θ= 0º

The component of the force is negative therefore this angle is in the second quadrant, to measure the angle from the positive side of the x axis in a counterclockwise direction.

θ = 180 + θ'

θ = 180 + 0

θ = 180º

C) if the three forces can be moved and the maximum force occurs when they are all linear.

10+ 6 + 6 + F = 0

F = -24 N

D) if we maintain this force and eliminate the other three, the block stops, let's look for its acceleration.

$a = \frac{F}{m}$

a = $\frac{24}{6}$

a = 4 m / s²

The acceleration is in the opposite direction of the initial velocity of the block v₀ = -100 m / s

If we use kinematic relations.

v = v₀ - a t

Final velocity when stopped is zero

t = $\frac{0-v_o}{a}$

t = 100/4

t = 25 s

In conclusion using Newton's second law and the kinematics relations we can find the results for the questions about the forces and the motion of the block are:

B) the force applied to maintain the system is equilibrium is: F = 0.39 N with an angle of tea = 180º

C) The maximum force is: F = 24 N

D) The time to stop the block is: t = 25 s

Learn more about Newton's second law here: brainly.com/question/25545050

3 0

3 years ago

What kind of energy does a rubber band have when it is stretched?

RideAnS [48]

Elastic potential energy. When you stretch a rubber band it has the "potential" to do work, to fly in a given direction. In doing so it changes it's elastic potential energy to kinetic energy.

5 0

3 years ago

Read 2 more answers

Assume that a satellite orbits mars 150km above its surface. Given that the mass of mars is 6.485 X 10^23kg, and the radius of m

Kisachek [45]

<span>3598 seconds The orbital period of a satellite is u=GM p = sqrt((4*pi/u)*a^3) Where p = period u = standard gravitational parameter which is GM (gravitational constant multiplied by planet mass). This is a much better figure to use than GM because we know u to a higher level of precision than we know either G or M. After all, we can calculate it from observations of satellites. To illustrate the difference, we know GM for Mars to within 7 significant figures. However, we only know G to within 4 digits. a = semi-major axis of orbit. Since we haven't been given u, but instead have been given the much more inferior value of M, let's calculate u from the gravitational constant and M. So u = 6.674x10^-11 m^3/(kg s^2) * 6.485x10^23 kg = 4.3281x10^13 m^3/s^2 The semi-major axis of the orbit is the altitude of the satellite plus the radius of the planet. So 150000 m + 3.396x10^6 m = 3.546x10^6 m Substitute the known values into the equation for the period. So p = sqrt((4 * pi / u) * a^3) p = sqrt((4 * 3.14159 / 4.3281x10^13 m^3/s^2) * (3.546x10^6 m)^3) p = sqrt((12.56636 / 4.3281x10^13 m^3/s^2) * 4.458782x10^19 m^3) p = sqrt(2.9034357x10^-13 s^2/m^3 * 4.458782x10^19 m^3) p = sqrt(1.2945785x10^7 s^2) p = 3598.025212 s Rounding to 4 significant figures, gives us 3598 seconds.</span>

8 0

3 years ago