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3 years ago

When performing the spike the ball should be hit with the ?

Physics

1 answer:

VMariaS [17]3 years ago

5 0

Answer:

When you are performing spike it's most effective to strike the ball from the right or left side at a sharp downward angle. Whether you are spiking the ball from the right or left front position, position yourself behind the 10-foot line (attack line), which is the line that is about four steps away from the net.

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Wind power holds great promise for the united states because of the _____________, and experts believe wind energy could meet as

SIZIF [17.4K]

Wind<span> power holds great promise for the united states because of the great wind power (capacity), and experts believe wind energy could meet as much as 20 percent of the nation’s energy needs . Texas is the state with most wind capacity. On the second place is Iowa and oh the third place Oklahoma. The wind farms produce clean energy that does not harm the environment.</span>

6 0

3 years ago

A merry-go-round with a rotational inertia of 600 kg m2 and a radius of 3.0 m is initially at rest. A 20 kg boy approaches the m

nekit [7.7K]

Answer:

The velocity of the merry-go-round after the boy hops on the merry-go-round is 1.5 m/s

Explanation:

The rotational inertia of the merry-go-round = 600 kg·m²

The radius of the merry-go-round = 3.0 m

The mass of the boy = 20 kg

The speed with which the boy approaches the merry-go-round = 5.0 m/s

$F_T \cdot r = I \cdot \alpha = m \cdot r^2 \cdot \alpha$

Where;

$F_T$ = The tangential force

I = The rotational inertia

m = The mass

α = The angular acceleration

r = The radius of the merry-go-round

For the merry go round, we have;

$I_m \cdot \alpha_m = I_m \cdot \dfrac{v_m}{r \cdot t}$

$I_m$ = The rotational inertia of the merry-go-round

$\alpha _m$ = The angular acceleration of the merry-go-round

$v _m$ = The linear velocity of the merry-go-round

t = The time of motion

For the boy, we have;

$I_b \cdot \alpha_b = m_b \cdot r^2 \cdot \dfrac{v_b}{r \cdot t}$

Where;

$I_b$ = The rotational inertia of the boy

$\alpha _b$ = The angular acceleration of the boy

$v _b$ = The linear velocity of the boy

t = The time of motion

When the boy jumps on the merry-go-round, we have;

$I_m \cdot \dfrac{v_m}{r \cdot t} = m_b \cdot r^2 \cdot \dfrac{v_b}{r \cdot t}$

Which gives;

$v_m = \dfrac{m_b \cdot r^2 \cdot \dfrac{v_b}{r \cdot t} \cdot r \cdot t}{I_m} = \dfrac{m_b \cdot r^2 \cdot v_b}{I_m}$

From which we have;

$v_m = \dfrac{20 \times 3^2 \times 5}{600} = 1.5$

The velocity of the merry-go-round, $v_m$ , after the boy hops on the merry-go-round = 1.5 m/s.

5 0

2 years ago

A train has an acceleration of magnitude 0.90 m/s2 while stopping. A pendulum with a 0.55-kg bob is attached to a ceiling of one

anygoal [31]

The angle of the pendulum with the vertical is $5.2^{\circ}$

Explanation:

As the train decelerates, the bob of the pendulum will feel a force given by

$F=ma$

where

m = 0.55 kg is the mass of the bob

$a=0.9 m/s^2$ is the magnitude of the acceleration

In the horizontal direction.

The pendulum will be inclined at an angle $\theta$ from the vertical, so it will be in equilibrium, and therefore the horizontal component of the tension in the string must be equal to the net force F of the previous equation:

$T sin \theta = ma$ (1)

where T is the tension in the string.

We also know that the bob is in equilibrium along the vertical direction: so the vertical component of the tension must be equal to the weight of the bob,

$T cos \theta = mg$ (2)