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3 years ago

Which statement best describes beaches?

Physics

2 answers:

mylen [45]3 years ago

6 0

A is a good example, i suppose.

Ludmilka [50]3 years ago

6 0

I would say A. Beaches are always changing due to factors such as waves adding and eroding sand

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If the mass of a wheel is increased by a factor of 7 and the radius is increased by a factor of 15, by what factor is the moment

inn [45]

Answer:

The moment of inertia will be increased by a factor of 1575

Explanation:

Let the moment of inertia of a wheel be given as;

$I = \frac{1}{2} mr^2\\\\$

where;

I is the moment of inertia of the wheel

m is mass of the wheel

r is radius of the wheel

when the mass is increased by a factor of 7 and the radius is increased by a factor of 15

$I_1 = \frac{1}{2}mr^2\\ \\I_2 = \frac{1}{2}(7m)(15r)^2\\\\I_2 = \frac{1}{2}(7m)(225r^2)\\\\I_2 = 7\times 225[\frac{1}{2}(mr^2)]\\\\I_2 = 1575[\frac{1}{2}(mr^2)]\\\\Recall, \frac{1}{2}(mr^2) = I_1 \\\\I_2 = 1575 I_1$

Thus, the moment of inertia will be increased by a factor of 1575

3 0

3 years ago

A triangular plate with height 6 ft and a base of 8 ft is submerged vertically in water so that the top is 4 ft below the surfac

solmaris [256]

Answer:

$F=4673.95lbs$

Explanation:

Recall weight density of water 62.5 lb/ft^3 and the height 6ft base 8ft

The width of the triangle

$\frac{W(x)}{7}=\frac{(8-x)}{2}$

The hydrostatic force is the pressure times area of the submerged

$W(x)=\frac{7*(8-x)}{4}$

$F=\int\limits {p} \, dA$

$F=\int\limits^8_2 {pgx*\frac{7}{4}*(8-x)} \, dx$

$F=\int\limits^8_2 {62.5*\frac{7}{4}*x*(8-x)} \, dx$

$F=109.375\int\limits^8_4{(8x-x^2)} \, dx$

$F=109.375[4x^2-\frac{1}{3}x^3]|4,8$

$F=109.375*42.73=4673.95lbs$

7 0

3 years ago

Suppose that volumes of four stars in the Milky Way are 2.7 x 1018 km3, 6.9 x 1012 km3, 2.2 x 1012 km3, and 4.9 x 1021 km3. What

Irina18 [472]

Answer:

The order is Star 3 < Star 2 < Star 1 < Star 4

Explanation:

Lets name the stars with their written order.

Star 1: 2.7 x 10^18 km3

Star 2: 6.9 x 10^12 km3

Star 3: 2.2 x 10^12 km3

Star 4: 4.9 x 10^21 km3

Star with lowest power of 10 has the least volume. Therefore, star 2 and star 3 would be the stars with least volume. Star 2's coefficient is better than Star 3. Thus, the order will be Star 3 < Star 2 < Star 1 < Star 4.

3 0

3 years ago

An 85,000 kg stunt plane performs a loop-the-loop, flying in a 260-m-diameter vertical circle. at the point where the plane is f

konstantin123 [22]

A) When the plane is flying straight down, there are three forces acting on it:
- the centripetal force $F=m \frac{v^2}{r}$ , directed toward the center of the circle (so, horizontally)
- the weight of the plane: $W=mg$ , downward, so vertically
- a third force, given by the propulsion of the plane, which is accelerating it towards the ground (because the problem says that the plane has an acceleration of a=12 m/s^2 towards the ground)

The radius of the circle is $r= \frac{260 m}{2} = 130 m$ , so the centripetal force acting on the plane is
$F_c=m \frac{v^2}{r} = \frac{(85000 kg)(55 m/s)^2}{130 m}=1.98 \cdot 10^6 N$
On the vertical axis, we have two forces: the weight
$W=mg=(85000 kg)(9.81 m/s^2)=8.34 \cdot 10^5 N$
and the other force F given by the propulsion. Since we know that their sum should generate an acceleration equal to $a=12 m/s^2$ , we can find the magnitude of this other force F by using Newton's second law:
$F+mg=ma$
$F=m(a-g)=(85000kg)(12 m/s^2-9.81 m/s^2)=1.86 \cdot 10^5 N$

So, the net force acting on the plane will be the resultant of the centripetal force (acting in the horizontal direction) and the two forces W and F (acting in the vertical direction):
$R= \sqrt{(F_c^2+(W+F)^2}=$
$= \sqrt{(1.98\cdot 10^6N)^2+(8.34 \cdot 10^5N+1.86 \cdot 10^5 N)^2} =2.23 \cdot 10^6 N$

(b) The tangent of the angle with respect to the horizontal is the ratio between the sum of the forces in the vertical direction (taken with negative sign, since they are directed downward) and the forces acting in the horizontal direction, so:
$\tan \theta = \frac{-(W+F)}{F_c}= -0.5$
And so, the angle is
$\theta = \arctan (-0.5)=-26.8 ^{\circ}$

7 0

3 years ago

Explain why the entering liquid pressure has no effect on the opening force of a balanced Port TXV

Ber [7]

Because its is a liquid and not a solid

3 0

3 years ago