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3 years ago

Can someone please help me ?

Physics

1 answer:

mario62 [17]3 years ago

5 0

Explanation:

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10. A man throws a water balloon down off the edge of a building. If he wants the water

stepladder [879]

Answer:

intinal speed should be 10

Explanation:

v = v0 + at

30 = v0 +10.2

then v0= 10

5 0

2 years ago

FIGURE 2 shows a 1.5 kg block is hung by a light string which is wound around a smooth pulley of radius 20 cm. The moment of ine

Sindrei [870]

Answer:

At t = 4.2 s

Angular velocity: 6. 17 rad /s

The number of revolutions: 2.06

Explanation:

First, we consider all the forces acting on the pulley.

There is only one force acting on the pulley, and that is due to the 1.5 kg mass attached to it.

Therefore, the torque on the pulley is

$\tau=Fd=mg\cdot R$

where m is the mass of the block, g is the acceleration due to gravity, and R is the radius of the pulley.

Now we also know that the torque is related to angular acceleration α by

$\tau=I\alpha$

therefore, equating this to the above equation gives

$mg\cdot R=I\alpha$

solving for alpha gives

$\alpha=\frac{mgR}{I}$

Now putting in m = 1.5 kg, g = 9.8 m/s^2, R = 20 cm = 0.20 m, and I = 2 kg m^2 gives

$\alpha=\frac{1.5\cdot9.8\cdot0.20}{2}$ $\boxed{\alpha=1.47s^{-2}}$

Now that we have the value of the angular acceleration in hand, we can use the kinematics equations for the rotational motion to find the angular velocity and the number of revolutions at t = 4.2 s.

The first kinematic equation we use is

$\theta=\theta_0+\omega_0t+\frac{1}{2}\alpha t^2$

since the pulley starts from rest ω0 = 0 and theta = 0; therefore, we have

$\theta=\frac{1}{2}\alpha t^2$

Therefore, ar t = 4.2 s, the above gives

$\theta=\frac{1}{2}(1.47)(4.2)^2$

$\boxed{\theta=12.97}$

So how many revolutions is this?

To find out we just divide by 2 pi:

$\#\text{rev}=\frac{\theta}{2\pi}=\frac{12.97}{2\pi}$ $\boxed{\#\text{rev}=2.06}$

Or about 2 revolutions.

Now to find the angular velocity at t = 4.2 s, we use another rotational kinematics equation:

$\omega^2=w^2_0+2\alpha(\Delta\theta)_{}$

Since the pulley starts from rest, ω0 = 0. The change in angle Δθ we calculated above is 12.97. The value of alpha we already know to be 1.47; therefore, the above becomes:

$\omega^2=0+2(1.47)(12.97)$ $w^2=38.12$ $\boxed{\omega=6.17.}$

Hence, the angular velocity at t = 4.2 w is 6. 17 rad / s

To summerise:

at t = 4.2 s

Angular velocity: 6. 17 rad /s

The number of revolutions: 2.06

3 0

1 year ago

HELP!!! Why is plasma considered the most common state of matter in the universe?

olasank [31]

Clue : There is plasma in sun

4 0

3 years ago

Read 2 more answers

A child standing on a bus remains still when the bus is at rest. When the bus moves forward and then slows down, the child conti

pochemuha

B. Newton's First Law, I'm pretty sure. The first states that an object in motion stays in motion, and an object at rest stays at rest until an outside force is applied, and that seems pretty relevant.

4 0

4 years ago

Read 2 more answers

A train traveled from Station A to Station B at an average speed of 80 kilometers per hour and then from Station B to Station C

Vinil7 [7]

Answer:

75 kmh⁻¹

Explanation:

$v_{ab}$ = Speed of train from station A to station B = 80 kmh⁻¹

$d_{ab}$ = distance traveled from station A to station B

$t_{ab}$ = time of travel between station A to station B

we know that

$Time = \frac{distance}{speed}$

$t_{ab} = \frac{d_{ab}}{v_{ab}} = \frac{d_{ab}}{80}$

$d_{bc}$ = distance traveled from station B to station C

$v_{bc}$ = Speed of train from station B to station C = 60 kmh⁻¹

$t_{bc} = \frac{d_{bc}}{v_{bc}} = \frac{d_{bc}}{60}$

Total distance traveled is given as

$d = d_{ab} + d_{bc}$

Total time of travel is given as

$t = t_{ab} + t_{bc}$

Average speed is given as

$v_{avg} = \frac{d}{t} \\v_{avg} = \frac{d_{ab} + d_{bc}}{t_{ab} + t_{bc}}\\v_{avg} = \frac{d_{ab} + d_{bc}}{(\frac{d_{ab}}{80} ) + (\frac{d_{bc}}{60} ) }$

Given that :

$d_{ab} = 4 d_{bc}$

$v_{avg} = \frac{4 d_{bc} + d_{bc}}{(\frac{4 d_{bc}}{80} ) + (\frac{d_{bc}}{60} ) }\\v_{avg} = \frac{4 + 1}{(\frac{4 }{80} ) + (\frac{1}{60} ) }\\v_{avg} = 75 kmh^{-1}$

$v_{ab}$ = Speed of train from station A to station B = 80 kmh⁻¹

$t_{ab}$ = time of travel between station A to station B

$d_{ab}$ = distance traveled from station A to station B

we know that

$distance = (speed) (time)$

$d_{ab} = v_{ab} t_{ab}\\d_{ab} = 80 t_{ab}$

$d_{bc}$ = distance traveled from station B to station C

$v_{bc}$ = Speed of train from station B to station C = 60 kmh⁻¹

$t_{bc}$ = time of travel for train from station B to station C

we know that

$distance = (speed) (time)$

$d_{bc} = v_{bc} t_{bc}\\d_{bc} = 60 t_{bc}$

Total distance traveled is given as

$d = d_{ab} + d_{bc}\\d = 80 t_{ab} + 60 t_{bc}$

Total time of travel is given as

$t = t_{ab} + t_{bc}$

Average speed is given as

$v_{avg} = \frac{d}{t} \\v_{avg} = \frac{d_{ab} + d_{bc}}{t_{ab} + t_{bc}}\\v_{avg} = \frac{80 t_{ab} + 60 t_{bc}}{t_{ab} + t_{bc}}$

Given that :

$t_{ab} = 3 t_{bc}$

$v_{avg} = \frac{80 t_{ab} + 60 t_{bc}}{t_{ab} + t_{bc}}\\v_{avg} = \frac{80 (3) t_{bc} + 60 t_{bc}}{(3) t_{bc} + t_{bc}}\\v_{avg} = \frac{(300) t_{bc}}{(4) t_{bc}}\\v_{avg} = 75 kmh^{-1}$

4 0

3 years ago