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3 years ago

Are the correct? PLEASE HELP ASAP

Physics

2 answers:

Trava [24]3 years ago

7 0

Energy is the ablity to dow work.

Power is the time rate at which work is done.

independent variable often along x axis, dependent along y axis.

You can set the independent value, from which the dependent follows.

Archy [21]3 years ago

6 0

Energy is the ablity to dow work.

Power is the time rate at which work is done.

independent variable often along x axis, dependent along y axis.

You can set the independent value, from which the dependent follows.

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Two objects were lifted by a machine.One object had a mass of 2 kilograms and was lifted at a speed of 2 m/sec.The other had a m

NikAS [45]

Answer:

While lifting two object the machine needs the different momentum for different mass object.

Explanation:

Momentum is the quantity of motion contained in an object. Usually it is measured by the product of mass and velocity.
Momentum of first mass = 2 kg × 2 m/sec = 4 kg m/sec
Momentum of second mass = 4 kg × 3 m/sec = 12 kg m/sec
So the machine requires higher mass in motion for second object ( i.e. momentum) than the first one while lifting.

7 0

3 years ago

1200 meters is less than 1 kilometer

SashulF [63]

<span>1200 meters is less than 1 kilometer
</span>is false

4 0

3 years ago

Read 2 more answers

8. An unpowered flywheel is slowed by a constant frictional torque. At time t = 0 it has an angular velocity of 200 rad/s. Ten s

allsm [11]

Answer:

a) $\omega = 50\,\frac{rad}{s}$ , b) $\omega = 0\,\frac{rad}{s}$

Explanation:

The magnitude of torque is a form of moment, that is, a product of force and lever arm (distance), and force is the product of mass and acceleration for rotating systems with constant mass. That is:

$\tau = F \cdot r$

$\tau = m\cdot a \cdot r$

$\tau = m \cdot \alpha \cdot r^{2}$

Where $\alpha$ is the angular acceleration, which is constant as torque is constant. Angular deceleration experimented by the unpowered flywheel is:

$\alpha = \frac{170\,\frac{rad}{s} - 200\,\frac{rad}{s} }{10\,s}$

$\alpha = -3\,\frac{rad}{s^{2}}$

Now, angular velocities of the unpowered flywheel at 50 seconds and 100 seconds are, respectively:

a) t = 50 s.

$\omega = 200\,\frac{rad}{s} - \left(3\,\frac{rad}{s^{2}} \right) \cdot (50\,s)$

$\omega = 50\,\frac{rad}{s}$

b) t = 100 s.

Given that friction is of reactive nature. Frictional torque works on the unpowered flywheel until angular velocity is reduced to zero, whose instant is:

$t = \frac{0\,\frac{rad}{s}-200\,\frac{rad}{s} }{\left(-3\,\frac{rad}{s^{2}} \right)}$