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2 years ago

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Physics

1 answer:

Greeley [361]2 years ago

6 0

Answer:

Gravitational force is proportional to the masses of interacting objects

Explanation:

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Jill does twice as much work as Jack does and in half the time. Jill's power output is Group of answer choices one-fourth as muc

Musya8 [376]

Answer:

Second Choice.

Explanation:

Jack's Power = W/t

Jill's Power = 2W/(0.5)*t

2/0.5 = 4

Jill's Power = 4*W/t

Jill's Power is 4 times greater than Jack's

Second Choice

3 0

2 years ago

The total distance a group of movers move a refrigerator is 22 m. If they complete this task in 15 minutes, their efforts would

lianna [129]

Answer:

Their efforts would be expressed in units of Joules per second

Explanation:

The unit of their efforts can be derived from the formula of power which is given by the product of mass, acceleration and distance (the product is energy with unit joules) divided by time taken to complete the task (unit is seconds)

Therefore, the unit of their efforts would be joules per second

4 0

3 years ago

On the way to school, Jed traveled 100m north, 300m east, 100 north, 100m east, and 100m north. A.) Find the total distance trav

Oduvanchick [21]

Answer:

Total distance = 700 m

Displacement = 500 m

Explanation:

Notice that Jed travelled a total of 3 x 100 m = 300 m in the North direction, and 300 m + 100 m = 400 m in the East direction. Therefore the total distance he travelled is: 300 + 400 = 700 m.

But the actual displacement is given by the Pythagorean theorem as the hypotenuse of a right angle triangle of legs 300 m and 400 m:

displacement = $\sqrt{300^2+400^2} =\sqrt{250000} =500\,m$

5 0

3 years ago

Which of the following objects is accelerating?

Schach [20]

Answer: A flower pot falling

Explanation:

3 0

3 years ago

Read 2 more answers

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