How do the properties of waves correspond with observations of waves?

A square piece of tin has 12 inches on a side. An open box is formed by cutting out equal square pieces at the corners and bendi

Citrus2011 [14]

Answer:

Explanation:

Given a square Piece whose side is 12 inches

Now square pieces are cut from each corner to make it a open box

Suppose x is the length of square piece at each corner

then

base square has a length of $12-2x$

Dimension of new box is $(12-2x)\times (12-2x)\times x$

Volume $V=(12-2x)\times (12-2x)\times x$

$V=\left ( 12-2x\right )^2\cdot x$

For maximum volume differentiate with respect to x we get

$\Rightarrow\frac{\mathrm{d} V}{\mathrm{d} x}=2\times \left ( 12-2x\right )\times \left ( -2\right )\cdot x+\left ( 12-2x\right )^2=0$

we get x=6 and 4 but at x=6 volume becomes zero therefore x=4 is valid

$V=\left ( 12-2\cdot 4\right )^2\cdot 4$

$V=4^3$

$V=64\ in.^3$

6 0

3 years ago

What do electric forces between charges depend on

denis23 [38]

Answer:

On the magnitude of the charges, on their separation and on the sign of the charges

Explanation:

The magnitude of the electric force between two charges is given by

$F=k\frac{q_1 q_2}{r^2}$

where

k is the Coulomb's constant

q1, q2 are the magnitudes of the two charges

r is the separation between the charges

From the formula, we see that the magnitude of the force depends on the following factors:

- magnitude of the two charges

- separation between the charges

Moreover, the direction of the force depends on the sign of the two charges. In fact:

- if the two charges have same sign, the force is repulsive

- if the two charges have opposite signs, the force is attractive

7 0

3 years ago

A 1.00 kg object is attached to a horizontal spring. the spring is initially stretched by 0.500 m, and the object is released fr

valina [46]

The spring is initially stretched, and the mass released from rest (v=0). The next time the speed becomes zero again is when the spring is fully compressed, and the mass is on the opposite side of the spring with respect to its equilibrium position, after a time t=0.100 s. This corresponds to half oscillation of the system. Therefore, the period of a full oscillation of the system is
$T=2 t = 2 \cdot 0.100 s = 0.200 s$
Which means that the frequency is
$f= \frac{1}{T}= \frac{1}{0.200 s}=5 Hz$
and the angular frequency is
$\omega=2 \pi f = 2 \pi (5 Hz)=31.4 rad/s$

In a spring-mass system, the maximum velocity of the object is given by
$v_{max} = A \omega$
where A is the amplitude of the oscillation. In our problem, the amplitude of the motion corresponds to the initial displacement of the object (A=0.500 m), therefore the maximum velocity is
$v_{max} = A \omega = (0.500 m)(31.4 rad/s)= 15.7 m/s$

6 0

3 years ago

A man whose mass is 69 kg and a woman whose mass is 52 kg sit at opposite ends of a canoe 5 m long, whose mass is 20 kg. Suppose

dusya [7]

Answer:

the canoe moved 1.2234 m in the water

Explanation:

Given that;

A man whose mass = 69 kg

A woman whose mass = 52 kg

at opposite ends of a canoe 5 m long, whose mass is 20 kg

now let;

x1 = position of the man

x2 = position of canoe

x3 = position of the woman

Now,

Centre of mass = [m1x1 + m2x2 + m3x3] / m1 + m2 + m3

= ( 69×0 ) + ( 52×5) + ( 20× 5/2) / 69 + 52 + 20

= (0 + 260 + 50 ) / ( 141 )

= 310 / 141

= 2.19858 m

Centre of mass is 2.19858 m

Now, New center of mass will be;

52 × 2.5 / ( 69 + 52 + 20 )

= 130 / 141

= 0.9219858 m { away from the man }

To get how far, the canoe moved;

⇒ 2.5 + 0.9219858 - 2.19858

= 1.2234 m

Therefore, the canoe moved 1.2234 m in the water

5 0

3 years ago

A ball traveling at a speed ν0 rolls off a desk and lands at a horizontal distance x0 away from the desk, as shown in the figure

klasskru [66]

Answer:

$3x_0$

Explanation:

The horizontal distance covered by the ball in the falling is only determined by its horizontal motion - in fact, it is given by

$d=v_x t$

where

$v_x$ is the horizontal velocity

t is the time of flight

The time of flight, instead, is only determined by the vertical motion of the ball: however, in this problem the vertical velocity is not changed (it is zero in both cases), so the time of flight remains the same.

In the first situation, the horizontal distance covered is

$d=v_0 t = x_0$

in the second case, the horizontal velocity is increased to

$v_x' = 3v_0$

And so the new distance travelled will be

$d' = v_x' t = 3 v_0 t = 3 x_0$

So, the distance increases linearly with the horizontal velocity.

5 0

3 years ago