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3 years ago

SECTION BTHEORY QUESTIONS(1a) Define the following(a) Work:

Physics

1 answer:

Vesna [10]3 years ago

6 0

Work is the amount of energy transferred

Explanation:

In physics, work is a measure of the energy transfer occurring in a process. Typically, we talk about work when energy is converted from one form into another.

For instance, work is done when a force is applied on an object. The work done on the object is given by:

$W=Fd cos \theta$

where

F is the magnitude of the force

d is the displacement

$\theta$ is the angle between the direction of the force and of the displacement

We notice the following:

No work is done when the force is perpendicular to the displacement ( $cos 90^{\circ}=0$ )
The work is maximum when the force is parallel to the displacement

Whenever work is done, there is also an energy transfer taking place. For instance, in the previous example, when the force is applied to the object, the object will accelerate (assume there is no friction), and will gain kinetic energy: therefore, there is a transfer of energy to the object.

Learn more about work:

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A .5 kg air puck moves to the right at 3 m/s, colliding with a 1.5kg air puck that is moving to the left at 1.5 m/s.

arlik [135]

Answer:

part (a) v = 1.7 m/s towards right direction

part (b) Not an elastic collision

part (c) F = -228.6 N towards left.

Explanation:

Given,

Mass of the first puck = $m_1\ =\ 5\ kg$
Mass of the second puck = $m_2\ =\ 3\ kg$
initial velocity of the first puck = $u_1\ =\ 3\ m/s.$
Initial velocity of the second puck = $u_2\ =\ -1.5\ m/s.$

Part (a)

Pucks are stick together after the collision, therefore the final velocities of the pucks are same as v.

From the conservation of linear momentum,

$m_1u_1\ +\ m_2u_2\ =\ (m_1\ +\ m_2)v\\\Rightarrow v\ =\ \dfrac{m_1u_1\ +\ m_2u_2}{m_1\ +\ m_2}\\\Rightarrow v\ =\ \dfrac{5\times 3\ -\ 1.5\times 1.5}{5\ +\ 1.5}\\\Rightarrow v\ =\ 1.7\ m/s.$

Direction of the velocity is towards right due to positive velocity.

part (b)

Given,

Final velocity of the second puck = $v_2\ =\ 2.31\ m/s.$

Let $v_1$ be the final velocity of first puck after the collision.

From the conservation of linear momentum,

$m_1u_1\ +\ m_2u_2\ +\ m_1v_1\ +\ m_2v_2\\\Rightarrow v_1\ =\ \dfrac{m_1u_1\ +\ m_2u_2\ -\ m_2v_2}{m_1}\\\Rightarrow v_1\ =\ \dfrac{5\times 3\ -\ 1.5\times 1.5\ -\ 1.5\times 2.31}{5}\\\Rightarrow v_1\ =\ 1.857\ m/s.$

For elastic collision, the coefficient of restitution should be 1.

From the equation of the restitution,

$v_1\ -\ v_2\ =\ e(u_2\ -\ u_1)\\\Rightarrow e\ =\ \dfrac{v_1\ -\ v_2}{u_2\ -\ u_1}\\\Rightarrow e\ =\ \dfrac{1.857\ -\ 2.31}{-1.5\ -\ 3}\\\Rightarrow e\ =\ 0.1\\$

Therefore the collision is not elastic collision.

part (c)

Given,

Time of impact = t = $25\times 10^{-3}\ sec$

we know that the impulse on an object due to a force is equal to the change in momentum of the object due to the collision,

$\therefore I\ =\ \ m_1v_1\ -\ m_1u_1\\\Rightarrow F\times t\ =\ m_1(v_1\ -\ u_1)\\\Rightarrow F\ =\ \dfrac{m_1(v_1\ -\ u_1)}{t}\\\Rightarrow F\ =\ \dfrac{5\times (1.857\ -\ 3)}{25\times 10^{-3}}\\\Rightarrow F\ =\ -228.6\ N$

Negative sign indicates that the force is towards in the left side of the movement of the first puck.

3 0

3 years ago

igor_vitrenko [27]

Answer:

$a=40\ m/s^2$

Explanation:

Given that,

Initial speed of a shuttlecock, u = 30 m/s

Final speed of the shuttlecock, v = 10 m/s

Time, t = 0.5 s

We need to find its average acceleration. The acceleration of an object is equal to the change in speed divided by time taken. It is given by :

$a=\dfrac{v-u}{t}\\\\a=\dfrac{10-30}{0.5}\\\\a=-40\ m/s^2$

So, the average acceleration of badminton shuttlecock is $40\ m/s^2$ .

3 0

3 years ago

Ezekiel is standing inside a spaceship ( S′ frame) that moves to the right at a speed of v=0.5c , where c is the speed of light.

vova2212 [387]

Answer:

c. Both signals are simultaneous.

Explanation:

The speed of light is the same in all frame of reference.

Since an observer inside the ship receives both signals at the same time, then the signals from both flashlight are simultaneously.

3 0

3 years ago

A 20 cm tall object is placed in front of a concave mirror with a radius of 31 cm. The distance of the object to the mirror is 9

Aloiza [94]

Answer:

The focal length of the concave mirror is -15.5 cm

Explanation:

Given that,

Height of the object, h = 20 cm

Radius of curvature of the mirror, R = -31 cm (direction is opposite)

Object distance, u = -94 cm

We need to find the focal length of the mirror. The relation between the focal length and the radius of curvature of the mirror is as follows :

R = 2f

f is the focal length

$f=\dfrac{R}{2}$

$f=\dfrac{-31}{2}$

f = -15.5 cm

So, the focal length of the concave mirror is -15.5 cm. Hence, this is the required solution.

4 0

3 years ago

How does the toaster serve as an example of Convection energy transfer?

blsea [12.9K]

Check bing for the answer

3 0

3 years ago

SECTION BTHEORY QUESTIONS(1a) Define the following(a) Work:​

SECTION BTHEORY QUESTIONS(1a) Define the following(a) Work: