What occurs as a ray of light passes from al inilo water?

An upward moving object must be experiencing (or at least usually does experience) an upward force.

gayaneshka [121]

Gravity pulls objects down to the earth

8 0

2 years ago

If the average pitcher is releasing the ball from a height of 1.8 m above the ground, and the pitcher's mound is 0.2 m higher th

mina [271]

The catcher can catch the ball at a height of 0.96 m from the ground.

The distance between the pitcher's mound and the catcher's box is about 60'6", which translates to 18.44 m. An average pitcher can pitch with speeds ranging from 88 mph to 97 mph, which is from 39.3 m/s to 43.4 m/s.

Assume the pitcher pitches a ball horizontally with a speed of 40 m/s. If the catcher catches the ball in a time t, then the ball travels a horizontal distance x of 18.44 m and at the same time falls through a height y.

The horizontal motion of the ball is uniform motion since no force acts on the ball ( assuming no air resistance) and hence the acceleration of the ball along the horizontal direction is zero.

Therefore,

$x=ut$

Calculate the time t by substituting 18.44 m for x and 40 m/s for u.

$t=\frac{x}{u} \\ =\frac{18.44 m}{40 m/s} \\ =0.461s$

The ball is acted upon by the earth's gravitational attraction and hence it accelerates downwards with an acceleration equal to the acceleration due to gravity g.

Since a horizontal projection is assumed, the ball has no component of velocity in the downward direction.

Therefore, for vertical motion, which is an accelerated motion, the distance y, the ball falls in the time t taken by it to reach the catcher's box is given by the equation,

$y=\frac{1}{2} gt^2$

Substitute 9.8 m/s² for g and 0.461 s for t.

$y=\frac{1}{2} gt^2\\ y=\frac{1}{2}(9.8 m/s^2)(0.461s)^2=1.04 m$

The pitcher releases the ball at a height of 1.8 m from a mound which is at a height of 0.2 m. Thus, the ball is released at a height of 2.0 m from the ground. It falls through a distance of 1.04 m in the time it takes to reach the catcher.

Therefore, the height at which the catcher needs to keep his glove so as to catch the ball is given by, $(2.0 m)-(1.04 m)=0.96 m$

The catcher needs to hold his glove at a height of 0,96 m from the ground.

8 0

3 years ago

Consider a long rod of mass, m, and length, l, which is thin enough that its width can be ignored compared to its length. The ro

ehidna [41]

Answer:

Explanation:

The rod will act as pendulum for small oscillation .

Time period of oscillation

$T=2\pi\sqrt{\frac{l}{g} }$

angular frequency ω = 2π / T

= $\omega=\sqrt{\frac{g}{l} }$

b )

ω = 20( given )

velocity = ω r = ω l

Let the maximum angular displacement in terms of degree be θ .

1/2 m v ² = mgl ( 1 - cosθ ) ,

[ l-lcosθ is loss of height . we have applied law of conservation of mechanical energy .]

.5 ( ω l )² = gl( 1 - cos θ )

.5 ω² l = g ( 1 - cosθ )

1 - cosθ = .5 ω² l /g

cosθ = 1 - .5 ω² l /g

θ can be calculated , if value of l is given .

4 0

3 years ago

If a person weighs 600N on earth, how much does he weigh on the moon

Snezhnost [94]

Answer:

His weight would be 100 N

Explanation:

5 0

3 years ago

Say I have a series circuit with 20v and four 65 ohm resistors, what is the current in each resistor?

Komok [63]

Data:

$E = 20 V$
$R_{1} = 65\Omega$
$R_{2} = 65\Omega$
$R_{3} = 65\Omega$
$R_{4} = 65\Omega$

Now that we have all the values we need properly identified, simply calculate the equivalent total resistance of the circuit and the intensity of the total electric current using the Ohm's Law:

 $R_{T} = R_{1} + R_{2} + R_{3} + R_{4}$
$R_{T} = 65 + 65 + 65+ 65$
$R_{T} = 260\Omega$

Like this:

$I_{T} = \frac{E}{ R_{T} }$

$I_{T} = \frac{20}{ 260 }$
$I_{T} = 0,076923076...$

$\boxed{\boxed{I_{T} \approx 0,07A}}$
Answer:
The intensity of the total electric current
 $\boxed{\boxed{I_{T} \approx 0,07A}}$

P.S:. Since the association is in series, the current of 0.07A is the same for all resistors.

4 0

3 years ago