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3 years ago

True or false? A tide with the least distance between low and high tides is called a neap tide.

Physics

2 answers:

lord [1]3 years ago

7 0

the only logical answer is true

aksik [14]3 years ago

6 0

The answer is True. Neap tides are when there is the least distance between low and high tides.

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On that system is<br> In an isolated system, the net __<br> zero.

DedPeter [7]

Answer:

the net work is 0

Explanation:

in an isolated system, there is no energy entering or exiting the system, and since work is dependent on the change in energy (W = ΔE) , with no change in energy, there is no work done. the net work done on or by the system must be 0

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3 years ago

Please help me

gtnhenbr [62]

Answer:

Catch these points g thanks

Explanation:

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3 years ago

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A ball of mass m is thrown into the air in a 45° direction of the horizon, after 3 seconds the ball is seen in a direction 30° f

Rzqust [24]

Answer:

Velocity (magnitude) is 98.37 m/s

Explanation:

We use the vertical component of the initial velocity, which is:

$v_{0y}=v_0*sin(45)=\frac{\sqrt{2} }{2}v_0$

Using kinematics expression of vertical velocity (in y direction) for an accelerated motion (constant acceleration, which is gravity):

$v_{y}=v_{0y}+a*t=\frac{\sqrt{2} }{2}v_0-9.8t$

Now we need to find $v_y$ as a function of $v_0$ . We use the horizontal velocity, which is always the same as follow:

$v_x=v_0cos(45\º)=\frac{\sqrt{2} }{2}v_0=v_{t=3}*cos(30\º) \\$

We know the angle at 3 seconds:

$v_y(t=3)=v_{t=3}*sin(30\º)\\v_{t=3}=\frac{v_y}{sin(30\º)}$

Substitute $v_{t=3}$ in $v_x$ and then solve for $v_y$

$\frac{\sqrt{2} }{2}v_0=\frac{v_y*cos(30\º) }{sin(30\º)} \\v_y=\frac{\sqrt{6} }{6}v_0$

With this expression we go back to the kinematic equation and solve it for initial speed

$\frac{\sqrt{6} }{6} v_0 =\frac{\sqrt{2} }{2}v_0-29.4\\v_0(\frac{\sqrt{6}-3\sqrt{2}}{6} )=-29.4\\v_0=98.37 m/s$

3 0

4 years ago

A ball is dropped from a cliff. determine how far the ball fell after 7.5 seconds

grin007 [14]

Answer:

The ball fell 275.625 meters after 7.5 seconds

Explanation:

If an object is left on free air (no friction), it describes an accelerated motion in the vertical direction, powered exclusively by the acceleration of gravity. The formulas needed to compute the different magnitudes involved are

$V_f=gt$

$\displaystyle y=\frac{gt^2}{2}$

Where $V_f$ is the final speed of the object in free fall, assumed positive downwards, t is the time elapsed since the release and y is the vertical distance traveled by the object