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3 years ago

Home Run

1 answer:

8 0

The force acting on the ball are unbalanced. Reactionary momentum force (that originated as a result of the swing of the bat) is the most powerful.

Yes friction is acting on the ball. In course of journey it would slow the ball down and make it trace a parabolic path rather than straight path as intended by hitter.

Explanation:

As the hitter hits the ball, momentum of the bat due to swing (mass of the bat*velocity provided by the batsman swinging action of bat) gets transferred on the ball on its impact with the bat.

Since ball’s mass is quite small as compared to the bat, the velocity of the ball increases by the same factor by which the ball’s mass is lower than the bat’s mass. This velocity causes forward motion of the ball (of course in the direction of bat’s motion, here the batsman intends to send the ball straight away hence the ball would move straight).

Various forces on ball is-

Reactionary momentum force -bat’s force (most powerful force)
The frictional force of the air (opposing the motion of the ball through the air)
Gravity force (pulling the ball down to the Earth)

As a combined effect of these force when all the force remains unbalanced, the ball moves away in the straight path under the impact of bats momentum which was most powerful of all.

Frictional force and Gravity force continue acting on the ball. While frictional forces decrease the ball velocity through the air, gravity force pulls it down thus deflecting its direction. Under the combined impact of declining bats momentum, friction force and gravity force, the ball traces a parabolic path (in accordance with the first law of motion from Newton)

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A tugboat tows a ship at a constant velocity. The tow harness consists of a single tow cable attached to the tugboat at point A

Y_Kistochka [10]

Answer:

The tensions in $T_{BC}$ is approximately 4,934.2 lb and the tension in $T_{BD}$ is approximately 6,035.7 lb

Explanation:

The given information are;

The angle formed by the two rope segments are;

The angle, Φ, formed by rope segment BC with the line AB extended to the center (midpoint) of the ship = 26.0°

The angle, θ, formed by rope segment BD with the line AB extended to the center (midpoint) of the ship = 21.0°

Therefore, we have;

The tension in rope segment BC = $T_{BC}$

The tension in rope segment BD = $T_{BD}$

The tension in rope segment AB = $T_{AB}$ = Pulling force of tugboat = 1200 lb

By resolution of forces acting along the line A_F gives;

$T_{BC}$ × cos(26.0°) + $T_{BD}$ × cos(21.0°) = $T_{AB}$ = 1200 lb

$T_{BC}$ × cos(26.0°) + $T_{BD}$ × cos(21.0°) = 1200 lb............(1)

Similarly, we have for equilibrium, the sum of the forces acting perpendicular to tow cable = 0, therefore, we have;

$T_{BC}$ × sin(26.0°) + $T_{BD}$ × sin(21.0°) = 0...........................(2)

Which gives;

$T_{BC}$ × sin(26.0°) = - $T_{BD}$ × sin(21.0°)

$T_{BC}$ = - $T_{BD}$ × sin(21.0°)/(sin(26.0°)) ≈ - $T_{BD}$ × 0.8175

Substituting the value of, $T_{BC}$ , in equation (1), gives;

- $T_{BD}$ × 0.8175 × cos(26.0°) + $T_{BD}$ × cos(21.0°) = 1200 lb

- $T_{BD}$ × 0.7348 + $T_{BD}$ ×0.9336 = 1200 lb

$T_{BD}$ ×0.1988 = 1200 lb

$T_{BD}$ ≈ 1200 lb/0.1988 = 6,035.6938 lb

$T_{BD}$ ≈ 6,035.6938 lb

$T_{BC}$ ≈ - $T_{BD}$ × 0.8175 = 6,035.6938 × 0.8175 = -4934.1733 lb

$T_{BC}$ ≈ -4934.1733 lb

From which we have;

The tensions in $T_{BC}$ ≈ -4934.2 lb and $T_{BD}$ ≈ 6,035.7 lb.

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3 years ago

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dalvyx [7]

Answer:

Beta radiation

Explanation:

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Answer:

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