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3 years ago

Shelby the Skater's rocket provides a forward force of

Physics

1 answer:

DochEvi [55]3 years ago

4 0

Answer:

$a=3.17\ m/s^2$

Explanation:

Given that,

Force force acting on the Skater's rocket is 220 N

Resistive force acting on it, F' = 30 N

Mass of Shelby, m = 60 kg

We need to find the acceleration of Shelby. Net force acting on Shelby is given by :

Net force = 220-30

= 190 N

The formula of net force is :

F = ma

a is acceleration of Shelby

$a=\dfrac{F}{m}\\\\a=\dfrac{190}{60}\\\\a=3.17\ m/s^2$

So, the acceleration of Shelby is $3.17\ m/s^2$

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What is potential difference

lisabon 2012 [21]

Answer:

VOLTAGE

Explanation:

Potential difference is the difference in the amount of energy that charge carriers have between two points in a circuit. ... A potential difference of one Volt is equal to one Joule of energy being used by one Coulomb of charge when it flows between two points in a circuit

5 0

3 years ago

Read 2 more answers

Eric drops a 2.20 kg water balloon that falls a distance of 45.08 m off the top of a

Marianna [84]

Answer:

972 J

Explanation:

At the bottom, all the gravitational potential energy was converted into kinetic energy. If you calculate the GPE, its value will be the same that the KE at the bottom. The GPE can be calculated this way:

GPE = mass×gravity×heigth

GPE = 2.2×9.8×45.08 ≈ 972

4 0

3 years ago

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

Which option is an example of a physical property?

Aleksandr-060686 [28]

state of matter is tha ans

3 0

3 years ago

Which is not true of the weak nuclear force?

slavikrds [6]

Answer:

The weak nuclear force is one of the fundamental forces:

Is a force that is responsible of some radiative phenomenoms, like the beta decay.

The intensity is way smaller than the one of the strong interaction, but this is because it acts in a way smaller range. And this can produce atractive and repulsive interactions.

So options D and B can be discarded.

And this force can be repulsive or attractive depending on the phenomena, so the options that are wrong are A and B, because the interaction is both things, not only one.

7 0

3 years ago